continuum mechanics via problems and exercises part ii answers and solutions ed by m e eglit d h...

CONTINUUM MECHANICS VIR PROBLEMS RND EXERCISES

PartII: Answers and Solutions

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CONTINUUM MECHANICS VIA PROBLEMS RND EXERCISES

P a r t i Answers And Solutions

Editors

Margarita E. Eglit Moscow State university

Russia Dewey H. Hodges

Georgia Institute of Technology U.SA

Co-Authors Margarita E. Egllt Andrei G. Kullkovsky Alexander N. Golublatnlkov Alexander G. Petrov Jacov A. Kamenjarzh irlna S. Shlklna Vladimir P. Karllkov Elena I. Sveshnlkova

Translation from Russian: A. N. natlushkln Figures: E. N. Paschenko

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Cover illustration: From 'Torus Doubling in Four Weakly Coupled Oscillators" International Journal of Bifurcation and Chaos, Vol. 5, No. 1, February 1995, p. 237.

CONTINUUM MECHANICS VIA PROBLEMS & EXERCISES Partll: Answers and Solutions

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Contents

Preface vii

Nomenclature ix

II ANSWERS AND SOLUTIONS 1

1 Basic Concepts used for Description of Motion and Deformation of a Continuum 3 1 Lagrangian and Eulerian Description of Motion 3 2 Tensors and their Cartesian Components 7 3 Curvilinear Coordinate Systems 8 4 Deformation. Deformation Rate. Vorticity 14 5 Principles of Symmetry and Tensor Functions 30

2 General Laws and Equations of Continuum Mechanics 35 7 Mass Conservation Law. Continuity Equation 35 8 Stress Tensor 37 9 Differential Equations of Motion and Equilibrium 40 10 Using the Conservation Laws in Integral Form for Calculation of Forces

and Moments Acting on Bodies Moving in a Fluid (Control Surface Method) 45

11 Angular Momentum Equations 50

3 Thermodynamics of Continua 55 13 The First Law of Thermodynamics.

Energy Equation. Perfect Gas 55 14 The Second Law of Thermodynamics. Entropy. Gibbs Identity. . . . 62 15 Restrictions Imposed by The Laws of Thermodynamics on the Form

of Constitutive Relationships 65 16 Thermodynamics of media with internal angular momentum 69

v

vi CONTENTS

4 Discontinuity Surfaces in Continuum Media 75 17 Conditions across Discontinuity Surfaces 75 18 Discontinuity Surfaces in the Lagrangian Description 83

5 Fluid Mechanics 87 20 Statics of Fluid 87 21 Dynamics of an Incompressible Ideal Fluid 90 22 Dynamics of an Incompressible Viscous Fluid 102 23 Waves on a Surface of Heavy Incompressible Liquid 113 24 Mechanics of a Compressible Fluid 121

6 Elasticity Theory 165 26 Linearly Elastic Solid 165 27 Nonlinear Elasticity 188 28 Couple Elasticity and Averaging in Media with Microstructure . . . . 190

7 Inelastic Solids 195 29 Plastic Flow Theory 195 30 Rate-Dependent Effects in Solids 203

8 Basic Notions of Relativistic Kinematics and Dynamics. General Properties of Electromagnetic Field 207 31 Lorentz Transformations. Minkowski Space 207 32 Concepts of Relativistic Kinematics and Dynamics 211 33 Maxwell Equations 213

9 Models of Media Interacting with Electromagnetic Field 219 34 Magnetohydrodynamics 219 35 Electrohydrodynamics 228

10 Dimensional Analysis and Modelling 231 37 Examples of Application of Dimensional Analysis 231

Index 251

Preface

This volume contains hints, solutions and answers to Problems presented in Part 1 of the book.

As a rule only the main points of solutions are given. Details (sometimes very time consuming) are left for independent work of the reader. That is why even those problems for which solutions are supplied are sometimes very challenging to solve, and professors can use them for tests and exams.

In any case, even if a reader solves a problem himself, it is very useful to study the solution presented herein after that. Very often one can find not only an alternative way of solution but also additional elements of the theory and comments which enable the reader to achieve a better understanding of the subject.

To find definitions of different notions one may use the Index to this volume and, in addition, the Index to Part 1.

I wish good luck and great success to all readers of the book.

Margarita E. Eglit, Moscow, January 15, 1996

vn

This page is intentionally left blank

Nomenclature

The list of the notations from the Part I of the book is repeated here to the benefits of the students.

Coordinates are usually numbered by the upper indices and denoted by xl (i = 1,2,3). Symbols x, y and z are often used to emphasize the Cartesian coordinates. Lagrangian coordinates are usually denoted by £'.

The symbol Vj stands for the covariant derivative over coordinate x'. In Cartesian coordinates

a - speed of sound

c - speed of light; speed of a characteristic; heat capacity per unit mass; concentration

Cp - specific heat capacity at constant pressure

cv - specific heat capacity at constant volume

dA - the amount of work for time interval dt

dq - the amount of heat per unit mass that enters a system during time interval dt

dq" - uncompensated heat

-jr- - rate of entropy production per unit mass

da - element of area

dV - element of volume

e.i - covariant basis vectors

e' - contravariant basis vectors

IX

X Nomenclature

d, ki - basis vectors of Lagrangian coordinate system in the initial and current states, respectively

dj - components of the strain rate tensor

g - acceleration due to gravity

g - metric tensor

9ij, 9li - components of the metric tensor

i - specific enthalpy, imaginary unit

k - wave number

k - internal angular momentum per unit mass

n - normal to the boundary

p - pressure

Pij - components of Cauchy stress tensor

p„ - stress vector

q - heat flux vector

ql - component of heat flux vector in direction of x'

s - entropy per unit mass

t - time

u - internal energy per unit mass; component of a velocity vector in the direction of Cartesian coordinate x

v - velocity; component of a velocity vector in the direction of Cartesian coordinate y

v - velocity vector

w - component of a velocity vector in the direction of Cartesian coordinate z

w - displacement vector

xx - coordinates

x , y, z - Cartesian coordinates

Nomenclature xi

B - magnetic induction

D - speed of a discontinuity surface

E - Young's modulus

E - electric field strength

T - free energy per unit mass

F - body force per unit mass

Qij - components of couple stress tensor

R - gas constant

S - entropy

T - temperature

U - internal energy; potential of body force

V - volume

W - complex potential

Fr - Proude number

M - Mach number

Pe - Peclet number

Re - Reynolds number

St - Strouhal number

a - coefficient of thermal expansion;

7 - adiabatic exponent, the ratio Cp/cv

6ij,6j - Kronecker symbols

tij - components of a strain tensor

£\jk - components of the Levi-Civita tensor

A - one of the viscosity coefficients; one of the elastic coefficients

H - viscosity coefficient; one of the elastic coefficients

v - kinematic viscosity coefficient

Vj - covariant derivatives

p - density

7r,J - components of the Piola-Kirchhoff stress tensor

a - Poisson coefficient, electrical conductivity coefficient

r - tangential stress

T , J - viscous stresses

tp - potential of velocity

ip - stream function

f* - Lagrangian coordinates

u - vorticity vector; angular velocity

T - velocity circulation

r \ - Christoffel symbols

xii Nomenclature

Chapter 1

Basic Concepts used for Description of Motion and Deformation of a Continuum

1 Lagrangian and Eulerian Description of Motion 1.1 Choose a Cartesian coordinate system 11,12,2:3 in space. Choose coordi

nates (xi, X2.X3) of a point as the Lagrangian coordinates (fi, £2> £3) of the particle situated at this point at the instant t = 0. a) Let the axis x\ be directed along the velocity vector (having constant direction). The motion is the translation of the body in the axis Xi direction by the distance vt. So the motion law is represented by the functions Xi = vt + £i,X2 = 62. £3 = £3. b) Let the X3-axis be directed along the (immovable) rotation axis. The motion is rotation around it by the angle ut. For this rotation, the transformation of the vector of the initial position of a particle into the vector of its position at the instant t is represented by premultiplication an orthogonal matrix, so that

( xi \ / cos ut — sin ut 0 \ / £1 \ x2 = sin ut cos ut 0 £2 •

*3 / V 0 0 1 ) \ 6 / Then, the functions representing the motion law have the form Xi = £1 cos ut — £2 sin ut, x<i = £1 sin ut + ^2 c o s wi, X3 = £3. Another solution: (The correspondence F <-» (01, 02,03) between vectors of Euclidean space and triples of numbers need not be set in the form ax = Xi, a2 = x2, a3 = x3 where (xi,X2,x3) are the components of r in a Cartesian coordinate system. For example, cylindrical coordinates Xi = /?, X2 = <p, X3 = 2 may be used, where R is the distance from the endpoint of the vector r to the axis x3, <p is the angle of the plane passing through r and the axis

3

4 SOLUTIONS. BASIC CONCEPTS

x3 with the plane XiOx3, 2 = x3.) Obviously, a rotation around the axis 23 does not change the cylindrical coordinates R and 2 of the position of a particle and changes the coordinate <p by the quantity wt if the angular velocity is constant. So the motion law is represented by the functions R = RQ, <p = ut + <po, 2 = 20 in cylindrical coordinates; (Ro, tpo, z0) are the Lagrangian coordinates of the particle. Thus, Cartesian coordinate systems are not always the most convenient.

1.2 The velocities of all particles of a rigid body are the same when its motion is translatory,

fi(£, 0 = ui(«) , t*(f, 0 = «a(0 , t*(£. 0 = "s(0 ■ The functions representing the motion law are found using the velocity definition vi(Z,t) = dxi(£,t)/dt,i = 1,2,3

t t t xi =f Ui(T)dr + fi , x2 = J U2(T)CIT + £2 , x3 = J U3(T)(IT + £3 .

0 0 0

1.3 The velocity field: Vi = a£2, v2 = b£i, v3 = 0; the acceleration field: ai = a2 = a3 = 0. The point (x0i, 102,2:03) is occupied, at the instant t = t0, by the particle with the Lagrangian coordinates

_ XQI - at0X02 _ X0 2 - bt0X0l _ 6 ~ 1 - abtl ' ?2 ~ 1 - abtl ' & ~ X°3 ■

1.4 a) The velocity field: Vi = ^ I / T , V2 = 2£2/T, V3 = 2£3t/r2; the acceleration field: ai = 0, a2 = 0, 03 = 2 ^ / T 2 . b) The particle is situated at the point (2a, 76/3,5c).

1.5 The values of the functions £a = ga(xi,x2,x3,t), a = 1,2,3, are Lagrangian coordinates and indicate what particle is situated at the point (x!,X2,x3) at the instant t; d£a/dt = 0, a = 1,2,3.

1.6 The velocity and acceleration fields have the following form in the Eulerian description (a dot denotes differentiation with respect to time), a) The velocity field: Vi = ax\/a, v2 = 6x2/6, t>3 = cx3/c; the acceleration field: ai = dx\/a, a2 = 6x2/6, a3 = cx3/c. b) The velocity field: V\ = 6x2, 2 = ^3 = 0; the acceleration field: a! = 6x2, a2 = a3 = 0. c) The velocity field: Vi = B^x* where B = AA_1; the acceleration field: a, = C^x* where C = AA~l.

1.7 The motion law has the form

Xl=Zl(l+t/T), X2 = &(1 + (t/T)2) , X3 = &(1 + (t/T)3)

where the coordinates of the point occupied by an individual particle at the instant t = 0 are chosen as the Lagrangian coordinates of this particle.

1. Lagrangian and Eulerian Description of Motion 5

1.8 If coordinates of the point occupied by an individual particle at the instant t = 0 are chosen as the Lagrangian coordinates of this particle, the motion law has the form a) Xi = r£i/r0, x2 = r£2/r0, x3 = £3 where

/ t x i / 2

r0 = y/& + & r=U + ^jQ(T)dr\ ;

b) X{ = R£i/Ro where 1/3

c) xi = £\e~At, x<i = £2eBt, 2:3 = £3- The streamlines coincide with particle paths in the cases (a), (b) and (c), although, in the cases (a) and (b), the motions are not steady. In the cases (a) and (c), the streamlines lie in a plane 13 = const, the picture of the streamlines are identical in all such planes and described respectively by the equations

/ \ ~BIA

, s X2 Xi X2 X^ (a) ■ -r = 7- - (c) : T = 7-

In the case (b), the streamlines are all possible straight lines passing through the origin of coordinates.

1.9 If coordinates of the point occupied by an individual particle at the instant t = 0 are chosen as the Lagrangian coordinates of this particle, the motion law has the form

xi = & exp I - J A{r)dT I , x2 = & exp I J B(r)dT 1 Z3 = £3

The streamlines at the instant t = to are the same as for the steady motion with A = A(t0) = const, B = B(t0) = const

1.10 a) No, they cannot: the curve, along which an individual particle moves, is known for every particle; however, the velocity of this motion may have an arbitrary magnitude, b) No, it cannot: the velocity direction is known (it is tangent to the streamline); however, the velocity magnitude may be arbitrary.

1.11 Hints, a) It is translatory motion with speed u along the X3-axis combined with rotation having angular velocity u around this axis, c) The particle paths are determined from the motion law

V V x\ = — cos ut + %! , x2 = — sin uJt + £2 , ^3 = £3

by means of eliminating the time. Answer, a) The streamlines and particle paths coinciding with them are the spirals on the cylinders x\ + x2 = c2 (c = const) with the pitch 2-nu/uj. b) The streamlines and particle paths coinciding with them are the ellipses x\/A + x\/B = c2, 13 = £3 (c = const), c) The particle paths are the circles (xi ~ £i)2 + (x2 — Zi)2 = (y/u) i x3 — £3; the streamlines are, at the instant to, the straight lines

Xi - ci = -(12 - c2) tan wt0) x3 = c3 (ci, c2, c3 = const) .

1.12 a) Yes, they can; b) yes, they can.

1.13 Yes, it can.

•■» §-*(^-i)--*-1.15 0.

1.16 a) It is necessary to test whether the components of the velocity in the Eulerian description Vi(x,t) do not depend upon the time, where x = (xi,i2,£3). The components of the velocity in the Lagrangian description are

^ = OiMCi + Ut, £2 ,6) U

where d\fi is the derivative of the function / ( with respect to the first argument. To change over to the Eulerian description, express, from the relationships

Xi = £ ( 6 + £/<,&, £3), * = 1,2,3,

the arguments of the function /< in terms of x

£i+Ut = gi(x), £2 = ff2(x), 6 = ff3(x).

Then, in Eulerian description, the components of the velocity are

Vi(x,t) = difi{gi(x),g2{x),g3(x)) U

and, obviously, do not depend upon time, b) Since the motion is steady, the streamlines coincide with the particle paths. The path of a particle (foi, £02, £03) is the locus of points having, in all possible instants t, the coordinates

xi = /i(£0l + E^,f02,f03)>

i.e., the curve x* = fi(r, £02, £03) where r is a parameter.

1.17 | ( l / p ) - ^(4nx2v) = 0.

2. Tensors and their Cartesian Components 7

2 Tensors and their Cartesian Components 2.1 a) tu + <22 +13 3 . b) The first two expressions are equal and, at i = 1,2,3,

represent the sums pnux + Pi2«2 + Pi3«3i P2i"i + P22W2 + P23U3, P3i«i +P32W2 + P33U3, respectively. The third and fourth expressions are equal and, at j = 1,2,3, represent the sums pnui +P21U2+P31U3, P12U1+P22U2 +P32«3. Pi3«i +P23W2 +P33«3> respectively which are not, in general, equal to the preceding ones, c) The first six expressions are equal, and the sum represented by the seventh expression differs from them.

2.2 a) 3,3,3; b) n, n, n.

2 3 dA_dA dA(x, t) dt dt ' dxi

2.4 a)Hint: The sets Ty and T'H for the orthonormal bases e* and e'j are bound by the tensor transformation rule. b)TyUjU,- = UjUiUy, in general, TijUiVj and UjUiVj are not equal.

2.6 Hint: The sums obey the tensor transformation rule.

2.7 Hint: The products Byjtf£mn obey the tensor transformation rule. Their sums BijkiEki obey this rule too. To show this, express By*; and ew in terms of the components in another orthonormal basis, e.g. £« = A^pA^e'^, and use the formula AkrAkp = Srp valid for the matrix ||Ay|| of an orthogonal transformation.

2.8 A contraction does not depend upon what letters denote summation indices, hence SyOy = SjjO;i. It remains to note that Sy = Sji, ay = — a,*, and, consequently, sydy = —s^i.

2.9 tij = (Uj + tji)/2 + (tij — tji)/2; the representation is unique.

2.10 Hint: Sy(ui + Vi){uj + Vj) = SyiiiUj + SijUiVj + SijViUj + SijViVj.

2.11 Hint: See Problem 2.9.

2.12 a) 0; b) 0.

2.13 Hint: If ei is an orthonormal basis, the basis e'i = ci, e'2 = e3, e'3 = e2 is orthonormal as well, hence tfu = 0. Express £'12 in terms of £y and deduce that <i3 = 0. Similarly, show that iy = 0 for the other i,j, i ^ j , i.e., the matrix ||ty|| is diagonal. Taking this into account, find such orthonormal basis e% that the condition <*i2 = 0 leads to the equality tu = t22. Similarly, show that tu = t33. Answer. In any orthonormal basis the matrix \\Uj\\ of the components of a tensor t is proportional to the identity matrix: £y = c£y, where c does not depend upon the basis.

2.14 a) The eigenvalues are Xx = —2, A2 = 1, A3 = 3. The corresponding principal axes are directed along the vectors

y/2el+e2, - v / 3e i + e 2 , e3 .

b) The eigenvalues are Xi = —2, A2 = A3 = 2. The corresponding principal axes are directed along the vector \ /3ei + e2 and along any pair of orthogonal vectors lying on the plane of vectors — \/3 d + e2 and e3. Note that the triple of principal axes is not unique in the case of equal eigenvalues.

2.15 a) Hint: J\, J2, J3 are obtained as a result of tensor contraction, hence, they are scalars. Their independence upon the basis can also be proved with the help of the same argumentation that is used in solution of Problem 2.7. b) Hint: 72 and J3 are expressed in terms of J\, J2,Js by the following formulae

h = \{J2y - J2) , h = \{A - 3JiJ2 + 2J3) .

2.16 Yes, they are. They are invariants, since they are determined as the roots of a cubic equation with invariant coefficients.

2.17 If Ai, A2, A3 are the eigenvalues of the tensor, then

11 = Ai + A2 + A3 , 72 = AiA2 + A1A3 + A2A3 , 13 = AiA2A3 ,

J2 = Aj + A2 + A3 , j 3 = Aj + A2 + A3 .

3 Curvilinear Coordinate Systems 3.1 Hint: a) Present the dual basis in the form ek = Xkjej and find the coeffi

cient Xki satisfying the conditions ekei = 8k. b) exe? = gxkek-gilei = g'kg*lek-ei = ffVffw = 9ij-

3.2 a) The basis dual to an orthonormal one coincides with it. b) Yes, it is. c) v1 = -(3a - b - c), v2 = -(36 - c - a), v3 = -(3c - a - b),

e1 = - (3e! - e2 - e3), e2 = -(3e2 - e3 - d ) , e3 = -(3e3 - d - e2).

3.4 Hint: The transformation law of the quantities g,j is found with expressing the basis e'k in terms of ek- To determine the transformation law of the quantities g1', it is sufficient to verify that the matrix with the components

dx" dx'j mn dxm dxn

is reciprocal to the matrix ||<?'ifc||.

3. Curvilinear Coordinate Systems 9

3.5 Hint: For example, the third set of relationships (a) follows from the first one: sm

n = gmisin = gmism = snm.

3.6 a) In general, no, they are not. b) Yes, they are.

3.7 a) e.\ = cosipe'i + smipe'2, e-2 = —rsmipe'i + rcostpe'2, e3 = e'3, so ei = e'i, e2 = 5e'2, e3 = e'3 at Mi;

e.\ = - r -e ' i + ^e>2, e2 = —5e\ + 5v3e'2 , e3 = e'3 at M2. b) Components of the metric tensor are gn = 1, 022 = r2, 033 = 1, gn = gi3 = 923 = 0 0 " = 1, fl22 = ^ , 533 = 1, 512 = fl13 = g23 = 0, flV = ^ The last equality is valid for any coordinate system, c) e1 = ei, e2 = e2/r2, e3 = e3.

, 1 . 1 3.8 e 1 = cos f e.\ sin if e2, e 2 = sin w ei H— cos u? e2.

r r 3.9 The nonzero components: jfn = acos2<f + bs'm2(p,

P'12 = P'21 = (a — &) cos </> sin <p, p'22 = asin2</? + 6cos2y>.

3.10

ei = sin 0 cos A e\ + sin # sin A e'2 + cos 0 e'3 e2 = r cos 9 cos A e'i + r cos 0 sin A e'2 — r sin 9 e'3 e3 = — rsinflsin A e'i + rsinflcos A e'2

Components of the metric tensor are

flu = 1. A22 = r2, g33 = r2 sin20, gu = fli3 = A23 = 0

*» = 1. fl22 = ^ , * 3 3 = ^ ' A12 = fl13 = fl23 = 0

3.11 Hint: The covariant transformation rule can be used (this method yields the solution for any vector field). However, for the special conditions of this problem, a simpler way is to note that v = r/\r\, where r is a vector with components (x' ,x' ,x' ) in the basis e'i, e'2, e'3, and to use this fact. Answer. V\ = 1, i>2 = 0, v3 = 0.

*1 r2 + a*

r2 + a*

3.12 ds2 = (r2 + a2 sin V ) — - + dip2 + dz2 1 T*^ -1- n^ '

3.13 ds2 = (r2 + a2 cos 29) -f— +d92)+ (r2 + a2) sin 29d<p2

1 rZ _l_ nZ I

3.14 a) ei / |e i | ; b) the cosine of the angle equals eY * e2/ |e1 | |e2 | ; c) eVIe1)

3.15

/ h\ (da db \ db da 61 = [l + R) U e * + Tse")• e2 = - d s e * + Tse*

where R = R(s) is the radius of curvature of the given curve;

, ff22=l, 5

12 = 0; ^ = a j .

3.16 a) [x1] - L, [x2] = [x3] = 1; b) [e,] = 1, [e2] = [e3] = L, je1] = 1, [e2] = [e3] = L"1; c) \gn] = 1, [g22] = \933} = L\ [,"] = 1, [g"} = [g33] = L~2; d) [v1] = LT~\ [v2] = [v3] = T"1, [«i] = 1, [«j] = [wj] = I^T"1. Here L, T denote the dimensions of length and time respectively; the dimension of a quantity is denoted with brackets [•], e.g., [v] is the dimension of velocity v, \v] = L/T; [/] = 1 means that / is dimensionless.

3.17 b) u • v = Upjfp! + up2uP2 + "PS^PS- c) For example,

tPl2 = «12|e,||e2| = t^le'lle2! = t}2\ei\\e2\ = t ^ e 1 ^ .

3.18 a) e r = ei = cosipe'x + s\n<pe'2, e^ = e2/r = -simpe'i + cosipe'2, ez - e3 = e'3. b) vP1 = 0, vP2 = wr, vP3 = 0, aP1 = -wr2 , aP2 = <l>r, aP3 = 0.

3.19 Hint: If the physical basis e r , ev, e, (see Problem 3.18(a)) is the local basis of a coordinate system y*, the basis c of the cylindrical coordinate system would be bound with it by the relationships

dr r

dy^e

d<p dyl

% 2

Mdy2

+ -dte*

^dy3

+ W

dip

dy3

Taking into account the relation of the bases e* and e r , ev , ez (Problem 3.18(a)), show that the first two of these relationships are inconsistent.

3.20 Hint: a) Verify that 6^ = b(eitej) and use this fact. Answer, b) ||<7r,||.

3.21 a) Hint: Verify that a'. = e' • ae} and use this fact, b) The elements of the matrix ||a'J| of the operator equal a', = 6J — n'rij.

<"=H

3.22 The tensor t may be only spherical (Problem 2.13). However, the equality ii2 = 0, unlike Problem 2.13, is not only valid in Cartesian coordinates. In particular, let us consider the coordinate system x1 = x'1 + x'2, x2 = 2x;1 + x'2, x3 = x'3. Using the tensor transformation rule, we find that t\\ = £22 = 0 from the condition t!y± = 0, and £33 = in = 0. Consequently, t = 0.

3.23 Hint: Consider the coordinate systems x' and x" = 2x' and show that the sums a,J + bij and a"J + 6^ are not bound by the tensor transformation rule unless a = 0 or b = 0.

3.24 Hint: Find the components of the metric tensor in the coordinate system x' and, then, the contravariant components W. The contravariant components of the sum are

a 1 1+6 1 1 = 2 , a12 + 6I2 = a 2 1+6 2 1 = - l , a22 + b22 = 1

and the other components equal zero.

3.25 Hint: Express the both sides of the considered equality in terms of Qijki, using the components of the metric tensor.

3.26 a) Hint: Components of B, e in different coordinate system are bound by the tensor transformation rule. Application of this rule yields the same property for the sums under consideration.

3.27 Hint: See the hint to Problem 3.23. Yes, they do, although they do not if Cartesian coordinate systems only are considered (cf. Problem 2.15).

3.28 Ji = h = t% Ji = V-t'-i, J3 = tyWi.

3.29 a) In a coordinate system with basis ej, the eigenvalues and eigenvectors v = v'ei are determined by any of the following three systems UjiP = XgijV3, tl*Vj = Xg'-'Vj, t\\Ap = \vx. b) Hint: Use the criterion for existence of a nonzero solution of the last system in the formula of preceding item. l\ = <!;, I2 = §(£\fj - ^ ^ i ) , h = de t | | ^ | | .

3.30 Hint: Verify that, when changing over to another coordinate system x ' 1 , ! ' 2 ^ ' 3 , the set of numbers t'kim = e'k • (e'( x e'm) is expressed in terms of ey* with the use of the tensor transformation rule.

3.31 Hint: c) Let the vectors of the basis e' be decomposed in a right Cartesian basis e" with the coefficients Ak

t (et = Akie'k). Then d • (ei x e3) = det.4 where A = ||i4*;||. Express det A in terms of g = det \\gij\\ taking into account that AAT = llfloll-


3.32 Hint: b) Show that e123 = e1 • (e2 x e3). Express e1 • (e2 x e3) in terms of det ||p ,J|| = l/y/g (see the hint to Problem 3.31).

3.35 b) Using the formulae from Problems 3.34 and 3.33, we find that

a x (6 x c)

= t^al[tvqrVd')e} = ( - « $ + ^ b M a . e , = (-a,6V + a , ^ ) ^ = b(o • c) - c(a • b)

3.37 a) No, it does not. b) det ||aV|| = i,,k a\a32ak

3 = -tijk ipqra.lpa>qakr\ The

elements of the matrix ||6* || reciprocal to ||a|,|| are

3.38 Hint: a) Use the formula for elements of a reciprocal matrix (Problem 3.37).

3.39 Hint: Use the relationships bk{a?k = 6j determining the elements of the

reciprocal matrix.

3.40 No, they do not. They do not obey the tensor transformation rule.

3.41 Hint: The Christoffel symbols equal zero in a Cartesian coordinate system, but at least one of them is nonzero in a curvilinear one.

3.42 No, it is not.

3.43 a) V ^ = 0; b) V,eljfc = 0.

3.44 In both cylindrical and polar coordinate systems the Christoffel symbols are r22 = —r, r2

2 = T2,, = 1/r and the other rjfc = 0 (here the indices run the values 1,2,3 and 1,2 in the cases (a) and (b) respectively).

3.45 For a sperical coordinate system the Christoffel symbols are r2 2 = - r , Ti, = - r s in 2 0 , T2

2 = T2, = 1/r, r23 = -sinflcosfl, T3

3 = P3, = 1/r, T33 = T3

2 = cot 6, the other r)k = 0.

3.46

rM / , h\~l h dR „. / h\~l 1 r» - " i1 + 7*) t TeTfa- r l 2 = ( 1 + HJ R r?1 = "(1 + ^ ) ' It' ^ = r2

2 = r22 = o

3.47 Hint: Use the formula for the differential of the determinant of a matrix (Problem 3.38(b)) and the relationship

P = i u (99u ,dgik_ ^9ik\ = j- udgu ik 29 \dxk dx* dx' J 29 dxk'

3.48 Hint: a) It is sufficient to verify the equality in a Cartesian coordinate system, b) Use the result of the preceding problem.

3.51 Hint: Use the symmetry of Christoffel symbols r j t = T^.

3.53 In the desired coordinate system, kn = 1, k12 = k'3 = 0; therefore the covariant transformation rule results in the relationships dxl/dxn = kx. In connection with this fact, the following system of differential equation is considered

dxi

Let xl = /*(s; a, b, c) be its solution with the initial data x^O) = a, x2(0) = b, x3(0) = c where (a, b, c) is any point of a vicinity of a nonsingular point of this system — that point where k =fi 0. Without loss of generality, it can be assumed that the coordinates of this point are x1 = x2 = x3 = 0, and the vector fc(0, x2,x3) directs outward to the surface x1 = 0 (in the considered vicinity). Then the solutions x' = / ' (s ;0 , b, c) with the initial data on the surface x1 = 0 can be used for introduction of the new coordinate system

x1 = F'(x'1,x'2,x'3), F<(x'1,x'2,x'3) := /'(x'1;0,x'2,x'3)

which has the required property.

3.54 b) It is sufficient to show that the formula is valid in a Cartesian coordinate system x' with basis e . Consider also the curvilinear coordinate system x" with basis e[ in which v = e\. The condition divv = 0 due to the formula from Problem 3.48b is reduced to the equality dy/g'/dx'1 = 0 where g' = det Wg'^W- Hence, ■y/ff7 depends only on x'2, x13: y/g1 = fix12, x13). It is convenient to express e\ in terms of e12 and e'3 by the formula from Problem 3.Id

e12 x e'3

e, = V = en ■ {en x e'3) V

where V* = l/y/g1 (see solution to Problem 3.32). Thus, we have

v = e'^fie^xe'3

dx12 t dx'3 k dx*2 dx'3 , k

= *-dx7*XMe =*-dxJdx^*Xe

dx^dx^_ lJk dxi dxk

This results in the following formula for the component v* in the Cartesian coordinate system xk

i dxa dx'3 ijk V dxi dxk

We define two scalar fields a, P that take values a{p) = x ' 2^) , /?(p) = x,3(p) at any point p, xl being the above-chosen (fixed) coordinate system. The previous formula takes the form

To obtain the final formula vi = A* L Jt

dxi dxk ' it remains to be shown that two functions ip, x of two variables can be chosen such that the scalar fields a = ip(a,P), b = x(a,P) satisfy the condition

da 80 iik da db ijk

The condition is equivalent to the equality

ufa ff\ — ^-e«* =(d±dX_d±^X\da_dP_ ijk *y )P' dxi dxk \da dp dp da) dxi dxk

which is obviously valid if

/ fl> dtp dx dip dx

This equahty holds, e.g., for the functions

4>(a,P) = J<p(a,p)da, X(<x,P) = P-

4 Deformation. Deformation Rate. Vorticity. Deformation. Strain tensors.

4.1 These material line elements displace parallel to themselves. The relative elongation of an element directed along the axis Xi equals a, that of an element perpendicular to the axis x\ equals zero. If a > 0, extension and, if —1 < a < 0, compression of these elements take place.

4. Deformation. Deformation Rate. Vorticity. 15

4.2 Hint: The components of the strain tensors can be calculated according to the formulae relating them to the components of the displacement field. Answer: The displacement field has the form

w = a£iei in the Lagrangian description

w = xiei in the Eulerian description. 1 + a

Only the components e n and £n of the Green and Almansi strain tensors do not vanish; they are

1„ . ,2 - lr- 1

* = ! [ ( ! + a ) 8 - l ] e i e i > e = I [ l - - ^ _ J e 1 e 1

4.3 a) Hint: Use the formula

ds2 - dsl = 2 eaP d£a d£0

where ds and ds0 are the lengths of the material line element before and after deformation. Answer: The relative elongation equals

\ / l + 2ea0dad0 - 1

where da are the components of the unit vector the direction of which coincides with that of d£. b) The relative elongation of each element equals Jl + a + a2/2 — 1.

4.4 a) Hint: The angle made by the material line elements can be found by calculating of the scalar product of the vectors to which their positions correspond after deformation

d ^ - f ^ e , , dx<2> = §i^2)e, • Take into account that

dxi dn o

and use the answer to Problem 4.3a to find the values of |dx(1)| and |dx'2'|. Answer: The angle <p made by the elements is determined by the value of

(2°ea0 + 6a0)da»df COS (p ■

v /i+2iKAd(1)41V1+2^42)^2)

where d^ and dp are the components of the unit vectors the direction of which coincides with those of d£^ and d^ respectively, b) The angle made by the elements is

2a + a2

if = arccos 2 + 2a + a2 '

4.5 The relative change in volume equals a.

4.6 Answer: a) The segments initially parallel to the axis x3 are translated without elongation remaining parallel to themselves; the segments initially parallel to the axis x2 are translated without elongation and rotated by an angle of 7r/2 around the axis x3; the segments initially parallel to the axis i i are translated, rotated in the same way, and extended as much as (1 + 6) times, b) Only the components en and £22 of the Green and Almansi tensors do not vanish, they are

2ll = ^[(l + * ) 2 - l ] , *22 = i f l - :

e = | [ ( l + 6 ) 2 - l ] c , C l , e=\ 1

(1 + 6)2

(1 + 6)2 e2e2

c) When |6| -C 1, one obtains e « 6 e 1 e 1, e w 6e2e2. Consequently, for this case (because of the presence of the finite rotation), the Green and Almansi tensors do not coincide even for small strains.

4.7 Answer: The location of the material line element emanating from a particle £ and characterized by a vector d£ in the undeformed state is characterized by the vector dx with the components

dxi = d£i + ak cos(fc£i) d£i , dx2 = d£2 , dx3 = d£3 .

Thus, a small vicinity of the particle £ experiences a uniaxial extension in the direction of the axis i i . The relative elongation of the material line element emanating from the particle £ and parallel to the axis Xi is I = ak cos(fc£i). The Green strain tensor 1 8 o 1

e = [ak cos(fc£i) + -a2k2 cos2(fc£i)]eiei and vanishes in the particles for which £1 = 7r/2fc + rmr/k where m = 0, ± 1 , . . . or £1 = £(± arccos(-2/aA;) + 27rn where |2/orfc| < 1, n = 0, ± 1 , . . . .

4.8 The components of the Green strain tensor eap = (a+a?/2)6a0 are obtained from the displacement field u = a£aea. Using the answer to Problem 4.3a, one finds that the relative elongation of any material line element equals a. If a > 1, extension takes place, and, if a < 1, compression.


4.9 Answer. The Green strain tensor is

o a . a2

£ = o ( e i e 2 + e 2 e i ) + ~o e2e2 '

its eigenvalues are o O2 0? O4 ° AI'2 = T ± V T + 1 6 ' A3 = 0 '

and the corresponding principal axes are

2 ° ei + - Ai,2e2 , e 3 .

a

The Almansi strain tensor is

a . a2

£ = 9 ( e i e 2 + e 2 e i ) ~~ T e 2 C 2 '

its eigenvalues are a2 / ^ a 7 ,

Aw = - T ± V T + l 6 ' A3 = 0 ' and the corresponding principal axes are

2 . ei + - Ai,2e2 , e 3 .

a

4.10 The displacement field is u(£, £) = af2ei in the Lagrangian description and w(x,t) = ax2e\ in the Eulerian description. Calculation of the components of the strain tensors leads again to the formula of the answer to Problem 4.9. The small strain tensor is e" ' = f (e ie 2 + e 2 d ) .

4 .11 Hint: From the formula of the answer to Problem 4.3a, one can obtain the relative elongation of any material line element (when the Green strain tensor is known). Conversely, one can obtain all the material line elements experiencing relative elongations of a given value. The Green strain tensor for a simple shear is determined by solution to Problem 4.9. a) The relative elongations of all the material line elements parallel to the axis X\ or 13 in the initial state equal zero: l\ = £3 = 0, and the relative elongations of those parallel to the axis x2 in the initial state equal l2 = y ' l + a'2 — 1. b) The material line elements, whose relative elongations equal zero, are material line elements, the direction of which characterized by the vector d£ - d£aea with df2 = 0 or with d£i + | d£2 = 0 at the initial instant. In other words, these are the material line elements lying on a plane £2 = const or on a plane f! + I f2 = const at the instant t = 0. Or, these are the material line elements lying on a plane x2 = const or on a plane X\ — | x2 = const at the instant t.


4.12 0.

4.13 Hint: The relative elongations are determined by the formula of the answer to Problem 4.3a. Use the following property of principal axes and eigenvalues: the maximal and minimal values of the quadratic form e ap da da on unit vectors d are reached when d is directed along one of the principal axes; the values equal the corresponding eigenvalues. Answer: The maximal relative elongation 1 ^ = 0.04 is experienced by a material line element with the direction d^^ = (ei + e2)/\/2, the minimal relative elongation Zmin = —0.02, an element with the direction d^a = (—e.\ + e2)/\/2- The relative change in volume equals 0.03.

4.14 The Green and Almansi strain tensor have the matrices of components in the coordinate system x*

/ 0 6/2 0 \ || !Q/3|| = 6/2 62/2 6/2

\ 0 6/2 62/2 )

( 0 6/2 -62/2 \ 6/2 -62/2 6(l + 62)/2

\ -62/2 6(1 + 62)/2 -62(1 + 62)/2 / eyll =

4.15 The displacement field is

w(x, t) = b(x2 - bx3)ei + 6x3e2 ,

the small strain tensor is

e ( 1 ) = 9~(eie2 + e 2 e i + e 2 e 3 + e z e ^ ■

4.16 Let us examine the element the location of which in the deformed state is characterized by the vector dx^K Its location in the undeformed state is characterized by the vector

d£w = ^— dx\'ea = 3— 6'kdsea = ^— eads . OXk OXk OXi

The scalar product

on the other hand, equals ^ ( 0 . ^U) = (! + /.)(! + 1.) c o s ^ ds2

and from this the proved formula follows.

4 .17 An eigenvalue of a Green strain tensor A a is its component e a Q in the orthonormal basis of the principal axes. Prom the equality

ha = ll(l + la)2-l} (la is the relative elongation of a material line element directed along the corresponding principal axis at t = 0), the relation 1 + 2 £aa > 0 follows. Here, it is taken into account that la > — 1 according to the definition of relative elongation. The inequality for the eigenvalues of an Almansi strain tensor is determined similarly with the help of the equality

e«=l-[-(l + la)2 + l}

proved in solution to Problem 4.16.

4.18 Let i j be the spatial Cartesian coordinates, fa be the corresponding La-grangian coordinates, and F = | |F f a | | be the distortion matrix. Then the components of the Green and Almansi strain tensors have the form

T-a0 = ^(FhaFkf) - <5a/j) , £ij = ^(<5jj - H^H^j)

where the matrix \\H^j\\ is the inverse of matrix \\Fia\\. Let A be an eigenvalue of the Green tensor, r]a be the components of the corresponding eigenvector, and A and tft be an eigenvalue and the components of the corresponding eigenvector of the Almansi strain tensor. This is equivalent to satisfaction of the relationships

1 ° 1

-^(FkaFkp - 6a0)j]P =\r)a , -(6ij - H^Hy^yj = Xy{

or, alternatively, of the relationships

FkcFkpV0 = (1 + 2 A h , , H^H^Vj = (1 - 2\)yi . (a4.1) If a material line element is directed along the vector rj before deformation, it is directed along the vector F • rj with the components Fiarja in the deformed state. Let r)i and A satisfy the first equality in (a4.1). To prove the statements (a) and (b), it remains to be verified that yi = FiaT]i and the number A for which

1 - 2A = —^—s-1 + 2A

satisfies the second equality in (a4.1), i.e., that the relationship

H-yiH^jFjaTIc = ^FipT]0 1 + 2 A


is valid. Since the matrix | | i /7 j | | is inverse to the matrix ||.FjQ||, this relationship is equivalent to the equality

1 + 2A or, alternatively, to the equality

1 + 2 A

which is valid according to the problem statement. The statement (b) is proved similarly.

4.19 Consider a symmetric matrix U determined by its principal axes (let them coincide with the principal axes of the Green strain tensor corresponding to the distortion F: eap = \{FiaFip — Sap)) and eigenvalues (let them be equal to k{ =

I o o o

V 1 + 2 Ai where A« axe the eigenvalues of the Green tensor; the equality 1 + 2 At > 0 is established in solution to Problem 4.17). The matrix U is not degenerate, so the matrix R is uniquely determined by the relationship F = RU. Let us show that it is orthogonal. It is sufficient for this to indicate a triple of mutually orthogonal unit vectors which remain to be of unit length and mutually orthogonal after the linear transformation determined by the matrix R. It is obvious that this transformation leaves the directions of the principal axes of the tensor e mutually orthogonal, since distortion has this property (see Problem 4.18). The transformation determined by the matrix U also leaves them mutually orthogonal due to its definition. At last, let us verify that the transformation determined by the matrix R preserves the length of each vector of these directions. In fact, let e\, e2, e3 be the orthonormal basis formed by unit vectors of these directions. The matrix of the components of the tensor e in

o ° this basis is diagonal and has the components £aa =Aa- Consider, e.g., a material line element with the direction e\. Its relative elongation l\, at the considered distortion F, is determined by the corresponding component of the strain tensor e

In other words, it is elongated 1 + l\ = y l + 2 en = \j 1 + 2 Ai times. On the other

hand, it is elongated the same number of times k\ = y 1 + 2 Ai if its transformation is determined by the matrix U. Thus, the transformation determined by the matrix R does not change its length, Q.E.D.

4.20 Hint: Use the statement proved in solution to Problem 4.18a. Answer: The following three vectors remain mutually orthogonal after deformation: one of

them is directed along the Xi-axis in the initial state, and the other two vectors are directed along any orthogonal directions in the plane x2,x3. The maximal relative elongation is reached by the material line elements directed along the Xi-axis in the initial state.

4.21 Hint: Compare the eigenvalues of the strain tensors that do not change during the additional rotation or translation.

4.22 Hint: Express the distortion matrix in the form F = RU where R is an orthogonal matrix, and U is a symmetric positive-definite matrix. For this purpose, one can find the matrix U — its principal axes and eigenvalues are expressed in terms of the principal axes and eigenvalues of the Green strain tensor (see Problem 4.18). The matrix R is then sought in the form R = FU'1. The particle £ is translated to the point with the coordinates Xi = £i + a^2, ^2 — 6>> £3 = &• The transformation of its small vicinity is formed by extensions along three mutually perpendicular directions followed by a rotation. The matrices of the three-axial extension and the rotation in the coordinate system xi have the form

U = u> 1 o \ - J O 0 0 1 /

, R = ( R' 1 0

- - J 0 ^ 0 0 1

The rotation takes place around the x3-axis by the angle <j> = — arctan | . The coefficient of elongation in the X3-direction equals 1, i.e., the length of a material line element directed along it does not change. The two other principal directions of the matrix U are

a + \/4 + a2 a - \/4 + a2

ei H -z e2 , ex + e2 ,

and the coefficients of elongation in their directions are

a2 + y/4a2 + a4 \1 / 2 / a2 - yAla2 + a4 V / 2

4.23 a) f1 = xi /( l + a(t)), f2 = x2, f3 = x3. b) The coordinate lines of the concomitant system are straight lines parallel to the axes Xi, x2, x3; its basis is e.\ = [1 + a(t)} ei, e2 = e2, e3 = e3; the components of the metric tensor in it are

<?n = [1+ a{t)}2 , £22 = 333 = 1 , 9ij = 0 at i jt j .


c) The components of the Green strain tensor (see Problem 4.2) are e n = ^[(1 + a)2 — 1], with the other components equal zero. As usual, the covariant components of the Almansi strain tensor in a concomitant coordinate system coincide with them: £ap =£Q/3- The mixed and contravariant components of the Almansi strain tensor in the concomitant coordinate system are

I1 I £ l ~ 2

1

1 -

2(1 +a) 2

1 (l + a)2J )

[ (l + a)M the other components equal zero.

4.24 The Lagrange coordinates 77 are bound with Lagrangian coordinates £ by the relationships rj = x(£, t,), i.e.,

i?1 = [! + «(*.)]&. f2 = 6 . ^ = 6 -Using the Lagrangian coordinates rj, the formulae describing the motion is written in the form

1 + a(t) 1 + a(U)

The concomitant coordinate system r), for the instant t = 0, sets correspondence between the point with spatial coordinates 1 = (i i , 2:2,13) and the triple of numbers V1, V2, V3

V = l + a(t.) 1 + a(0) Zl , V = x2 V =*3

Thus, the coordinate lines of the concomitant coordinate system 77 at the instant t = 0 . , ,. „ , . . . . . 1 + a(0) are the straight lines parallel to the axes x\, £2, £3; its basis is ei = ;—r-ei,

1 + a(t.) e2 = e2, e3 — 63; the components of the metric tensor in it are gap = ea ■ e@ where

ffn 1 + o(0)

§22 = 933 = 1 , 9aff = 0 at a ^ 0

4.25

l + a(*.),

The coordinate line f' of the concomitant coordinate system is the straight line z2 = 0, 13 = 0, the coordinate line £2 is the straight line x\ = a(t)x2, £3 = 0 (it varies in time), and the coordinate line £3 is the straight line Xi = 0, x2 = 0. The basis vectors of the concomitant coordinate system are e.\ — ei, e2 = a(t) e\ +e2, 63 = e3; the components of the metric tensor in it are gu = g22 = 1, 912 = o.(t), 322 = 1 + a2(t), g13 = $23 = 0.

4.26 a) A cross section £3 = const remains in its plane; the circumference o o

(£i)2 + (&)2 =R2 (R is the initial radius of the rod) is transformed into the circumference concentric to it of the radius R =RJl + (a£3)2; a radial segment rotates by the angle /? = arctan(af3); a longitudinal segment is transformed into a segment of a straight line lying on the hyperboloid into which the surface of the rod is transformed, e.g., a segment for the points of which £2 = 0 is transformed into a segment of the

o o

straight line X\ =R, x^ = a RX3. b) The displacement field is

w(x, t) = (xi sin2 0 — x2 sin /? cos /?)ei -I- (xt sin /? cos /? + x2 sin2 P)&2 ,

where P = arctan(ax3). c) If |Q| <§; 1, the components of the small strain tensor are e13 — e31 — O 2 ' 2 3 _ 3 2 _ O J '

the other components equal zero. The material line element of the elements emanating from a point x that experiences the maximal relative elongation (of the value I yx? + X\) is the element directed along the vector with the components (—sin ip,cos <p, 1). d) In the cylindrical coordinates r, if, z (see Problem 3.7), the motion is described by the formulae

r = rQ\J\ + (a(t) ZQ)2 , ip = ip0 + arctan(a(<) ZQ) , z = ZQ

where ro, ifio, Zg are the coordinates of the initial position of a particle.

4.27 a) Hint: The Green strain tensor is determined by the general formula

o o o a o B o \ (dxl dx* o e =ea0 e ep , ea0 = - ^ — — 9ij- gaP

where, for the considered case, the cylindrical coordinates x1 = r, x2 = (p, x3 = z and the Lagrangian coordinates £' = ro, £2 = <po, f3 = zo are used. The components of the metric tensor in this coordinate system are

S11 = 1 , 522 = r2 , 333 = 1 , gij = 0 at i ^ j .

The quantities dx%/d^a are obtained from the given formulae describing the motion. It must be taken into account that, in the general formula, the components g^ are calculated for the point x(£, t), and gap and eQ for the point £. Answer: The Green strain tensor for the particle (ro, tpo, ^o) at the instant t is

e (ro, <Po, ZQ, t) = - i + fW-i el(r0)el(r0) +

\[{ro + f(r0,t))2-rl]e2(r0)e2(r0)

where e1, e2 are the vectors of the basis dual to the basis eu e2, e^ of the cylindrical coordinate system, b) Hint: The covariant components of the Almansi strain tensor in the concomitant coordinate system coincide with the components of the Green tensor in the basis e^ro), e2(r0), e3(ro) found in the item (a). The basis of the concomitant coordinate system ea = (dxx/d£a)ei remains to be found:

e\(r0)ip0,zo,t) = i + §-M ei{r0 + f(r0,t)) ,

e2{r0, <fo, zo, t) = e2(r0 + /(r0 , t)) , e3{ro, <Po, zo, t) = e3(r0 + /(r0 , t))

and, then, the basis dual to it:

e\r0, ip0, zo, t) = gf — - el(r0 + f(r0, t)) ,

e2(r0, ipo, z0, t) = e2(r0 + /(r0 , t)) , e3(r0, <p0, zo, t) = e3(r0 + /(r0 , t)) .

The Almansi strain tensor in the particle (ro, <pa, ^o) at the instant t is

e{ro,fo, zo,t) =

l ( l + & ( r o . 0 ) 2 - l 2 l + lt(ro,0

e1(r0 + f{r0,t))e1{ro + f{ro,t)) +

\ [(ro + f(ro, t))2 - r2} e2(r0 + f(r0, t)) e2(r0 + f(r0, t))

where e1, e2 are the vectors of the basis dual to the basis ei, e2, e3 of the cylindrical coordinate system, c) Hint: The relative elongations in the considered case can be found by physical reasoning. For example, the material particles forming a circumference of radius ro before deformation are situated along the circumference of radius r0 + f(ro, t) at the instant (; so the relative elongation of a material line element tangent to the coordinate line (p equals / (r0 ) t)/r0. On the other hand, to calculate the relative elongations according to the general method is also interesting. However, it is impossible to use immediately the known formula

Ioo = ^[(l + U 2 - l ] , since it is valid only for the components of the Green strain tensor in Cartesian coordinate systems. To apply this formula, it is necessary to find the physical components of the Green strain tensor first, i.e., its components in the orthonormal basis

e1 e2 e3

and to apply this formula to them. The values of \ei\ are known, since |e' |2 = 1/gu-Answer: The relative elongations of material line elements emanating from a point (r0, <fio, z0) directed along the coordinate lines r0, tfio, Zo equal respectively

or0 r0

4.28 The Almansi strain tensor and the corresponding linearized strain tensor are expressed in terms of the displacement field (given in the Eulerian description)

e(x) = ~2^eiei + e 2 C 2 ) ' e<1)W = ° •

The covariant components of the Green strain tensor in the basis e, coincide with the covariant components of the Almansi strain tensor in the concomitant coordinate system and, consequently, are not zero; also are of order 1.

4.29 The components of the linearized strain tensor e C are expressed in terms of the derivatives of the displacement field

u{S,t) = {Rvr{t)$r-Za)ea.

They are

and, in general, are of order 1. On the other hand, the Green and Almansi strain tensors equal zero for any orthonormal matrix ||fly||. This can be verified by straightforward calculation which, however, is not necessary: the considered mapping determined by the orthogonal matrix does not change the distances between particles, and, consequently, e = e = 0.

Strain ra te tensor, vorticity, divergence of velocity.

4.30 Setting Xi be a Cartesian coordinate system, and ej its basis, the velocity a(t) a(t) field v and strain rate field e have the form: a) v = —r^ xiei, e = — -r- eiei. b) v = a(t) a(t)

a(t) x2elt e = —-(e 1 e 2 + e2e1). c) v = b(t)(x2 - b(t)x3)ei + b(t)x3e2,

e = —[eiei + e2ex + e2e3 + e3e2 - &(*)(eie3 + e3ex)).


4.31 a) e u = A, e22 = B, eff = \(2A - B), e2d2> = | ( 2 B - A),

1 9 e ^ = — -(A + B); the other components ey, ey equal zero, b) en = at, e u ' = -at,

3 ^ e22 = e33 = —3a*i *n e other components ey, ey equal zero, c) ei3 = e^ = -/3t;

the other components ey, ey equal zero.

4.32 en = 2/t, and the other components ey equal zero. There is no change in volume (since divu = 0).

k 4.33 It is convenient to write the strain rate tensor e = — (eie2 + e2ei) in the

basis e\ = —7z{e\ + e2) , e'2 = —M-ex + e2) , e'3 = e3

so that the considered material line elements are directed along e\, e'2. It is written in the form e = ^e^ej — e2e'2), and its component e'12 equals zero. Consequently, the rate of change in angle between the considered elements equals zero.

4.34 a) The particle paths are the circumferences r = const, z = const. The speed is |v| = \k\/r, and the physical components of the velocity are i»ph = k/r, uph = '"ph = 0- b) erv = —k/r, and the other components of the strain rate tensor equal zero, c) The vorticity equals zero, d) The principal axes of the strain rate tensor are directed along the vectors e^ ± \ ev and ez. They rotate with time in an individual particle, e) The motion of a material line element (a segment, but not a solid body !) is composed of a translation, a rotation around a lateral axis and a longitudinal extension (six degrees of freedom). The angular velocity of rotation of material line elements directed along the principal axes of the strain rate tensor equals zero at this instant.

4.35 Hint: Those material line elements rotate whose direction is not parallel to the plane x2, x3 and to the axis i i .

4.36 Those material line elements which are directed along the vectors e\ ± e2 rotate with the angular velocity u> = —-d(t)e3 . The angular velocity of the material line elements directed along the axes Xi and x3 equals zero, and that of the element directed along the axis x2 equals 2u>. Thus, the angular velocities of material line elements are different and not necessarily equal to the vorticity vector; see also Problem 4.34e.

4.37 n.


4.38 Let us use a Cartesian coordinate system X{ and denote wy = -(dvj/dxi — dvi/dij). In Cartesian coordinate systems, uik = u>ij, where u>k are the components of the vorticity vector, w^ are called the components of the vorticity tensor, and the values of the indices k, i, j form a cyclic permutation of 1,2,3. The following identity is valid

dtJkj _ deii _ °eik

dxi dxk dij where e^ are the components of the strain rate tensor (this identity is verified by substitution of the expressions for wjy, e^, e^ in terms of the derivatives of the velocity components vj. So, if ei3 do not depend upon the coordinates, then w^ and, consequently, ujk and u do not depend upon the coordinates.

4.39 Let v(ii, t) be the velocity at an arbitrary point of the medium with the coordinates xt at the instant t and v(xt + dxit t) be the velocity of a particle in an infinitesimal vicinity of the considered point at the same instant. According to the Cauchy-Helmholtz formula,

v(xi + dxi, t) = v(xi, t) + etj dxjei +u x dr , dr = eidii

If eij = 0, the vector u> may depend only upon t (Problem 4.38). In the case e^ = 0, the Cauchy-Helmholtz formula can be rewritten in the form

dv = u> x dr

for fixed t. Integrating this equality along any line connecting a certain point O with an arbitrary point A of the continuum and taking into account that u = const at ( = const results in

v(A) - v(0) = u; x (rA - r0) ■ Denoting v(0) = vo, TA—TO = T, V(A) = v, one obtains the Euler formula

» = « o + w x r .

The vector of instantaneous angular velocity is the vorticity vector.

4.40 a) Hint: Make use of symmetry relations Ty = T^. b) Hint: To prove the first formula, make use of the definitions of vorticity and the tensor w^e'e7. The second formula arises from the first one with account of Levi-Civita tensor property (see Problem 3.33).

4.41 Hint: Using the Lagrangian description show that the following formula is valid for acceleration components in a concomitant coordinate system

. _ dx' dv* _ dva d \v\2


(here, Vi and va are the velocity components in a spatial Cartesian coordinate system and concomitant coordinate system, respectively). The previous formula implies the following expression for the components of curl a in a concomitant coordinate system

(curl a) = e1 p — —- . v ' dt d£a

Then the condition curl a = 0 is reduced to the relationship

d_ dv^ _ d_ dv^ _ _ dt dia dt d& ~ '

Due to the preceding problem (a) this results in dojap/dt = 0. Each of the components a;7 equals u>ap/y/g or —Cjap/y/g due to the preceding problem (b) and the formulae for eali~i (see Problem 3.32), hence, d(Cjiy/g)/dt = 0).

4.42 For an instant in time t0, consider a vorticity line passing through the particle (£,£»,£!)• Let s be a parameter and £a(s) be the parametric description of the hne in concomitant coordinate system, in particular, £a(0) = f°. Then the definition of the vorticity line results in the following equations for the functions £Q(s)

dt;1 dt* dt3

vfoto) "2((i,to) LJ3(^,t0) equivalent to the equations

dt1 dt,2 d£3

^ ( £ . * o ) v ^ * o ) *2{£,to)Jg{Z,to) &(t,to)y/g(Z,to) '

According to (b) of the preceding problem, the functions ua\/g actually do not depend upon to- Hence, the solution £a(s) of this system of differential equations satisfying the condition £a(0) = £" does not depend upon time. That means that the vorticity line under consideration (£i,£?i£«) passes through the same particles £a(s) at any moment.

Compatibility conditions 4.43 Hint: Verify by a straightforward calculation that curlgradi/? = 0.

4.44 The potential of the velocity field of simple extension along the axis i i

where / is an arbitrary function of time.


4.45 a) Hint: Take into account that the components of Levi-Civita tensor in a Cartesian coordinate system equal 1, —1, or 0. b) Hint: The expressions on the left-hand sides of each of the equalities are components of a tensor. It is sufficient to verify that there exists a coordinate system in which all of them vanish.

4.46 Hint: The statement that e is the field of strain rate for a velocity field means that there exists a vector field v satisfying the relationships (in a Cartesian coordinate system)

1 (dVi dvi 2 \dxj dii

The condition of solvability with respect to v for such relationships is known as the compatibility conditions for components of small strain tensor. Answer: The necessary and, in the case of a simply connected domain, also sufficient condition is satisfaction of the equalities

CkipCii"d^dx~r 4.47 a) No, it is not if even one of the constants A, B, C is not equal zero.

b) Yes, it is.

4.48 Hint: Consider for example the components f\j and possibility to find a function V\ for which the relationship dv\/dxj = fij holds. It is possible if and only if

dxk

Answer: If and only if the given components fij in curvilinear coordinate system satisfy the conditions e'^V/t/y = 0, then there exists a vector field v for which VjV> = fij.

4.49 Hint: Introduce a tensor with the components u>ji = ejimwm. Then the question is if there exists a vector field v for which the relationships

dvi dvj _ dvi dvj _ dxj dii X] ' dxj dxi ; t

hold, that is, dvi/dij = e^ + Wji. Then the statement from the preceding problem results in the necessary and sufficient conditions in the following form

eikj dx~ ^6ij + "^ = ° ' Answer: If and only if the given components e^ and LJ1 in a curvilinear coordinate

system xk satisfy the conditions e'^Vtey = V^w', then there exists a velocity field v for which e and u are the strain rate and vorticity fields.

5 Principles of Symmetry and Tensor Functions. 5.1 A u = e*», A3 = 1.

5.2 R}j = n'rij + (<5J — nlrij) cos cp — e'jfcnfc sin <p, where e* -fc are the components of Levi-Civita tensor.

5.3 There exists at least one real eigenvalue A for a three-by-three matrix. Let n be the eigenvector (normalized to length unity) corresponding to it. Then, contracting twice the orthogonality condition for the matrix R^Rjgij = gu with the vector n, one obtains A2 = 1. The value A = — 1 corresponds to reflection in the plane perpendicular to the vector n .

5.4 The components of any orthogonal matrix are the set of components of three orthonormal vectors the choice of the orientation of which is restricted by the representation

Rxj = ±n'rij + (6* - n'n.j) cos <p - e'jkn* sin <p .

5.5 Two eigenvalues are imaginary with different signs, and one eigenvalue is zero.

oo i oo f _ i \ n + l oo 5.6 e s = £ - i s" , ln(g + S) = £ ^ - ^ S " , ( g _ s)"1 = £ S" where g

^ 0 n ! n=l " n O is the metric tensor.

5.7 Contract the power series of F(S) with each of the eigenvectors of the matrix S. The eigenvectors of S and F(S) coincide. Their eigenvalues are related by the same power series.

5.8 The equality does not depend upon the basis and is obviously valid if the matrix is reduced to triangular form (optionally, with transformations described by complex-valued matrices).

5.9 Reduce S to diagonal form (optionally, with transformations described by complex-valued matrices).

5.10 See Problem 5.9.

5.11 See Problem 5.9.

5.12 If A2 = A3 * Ax, F(S) = | — ^ 5 F(At) + ? — ^ F(A2). If A, = A2 = Aj — A 2 A2 — Aj

A3, F(S) = gF(Ax).

5. Principles of Symmetry and Tensor Functions. 31

5.13 Orthogonal matrices with determinant equal to 1.

5.14 a) The rotations around the i3-axis (Problem 5.1), reflection in the plane (x2, x3), and all their possible products, b) The transformations of Problem 5.1 supplemented with their products with the reflection in the plane (x^x2). c) The union of products of transformations a) and b). d) The transformations of Problem 5.1. e) The transformations of Problem 5.1 supplemented with their products with a rotation of magnitude -K radians around the x'-axis.

5.15 Answer: All possible powers and products of the reflection in the coordinate planes of the coordinate system whose axes coincide with the principal axes of the tensor S. The symmetry group of Problem 5.14c for the cases of two coinciding eigenvalues of the tensor S, the complete orthogonal group for the case of three coinciding eigenvalues.

5.16 a) The conditions for invariance of the second rank tensor T = T^eje, have the form b'kb'iT1'1 = T*J where 6'* are the components of an arbitrary orthogonal matrix. It is sufficient to consider invariance relative to several particular transformations. Due to invariance relative to the reflections in the coordinate planes, all non-diagonal elements of the matrix (T*J) equal zero. Next, invariance relative to the rotations of magnitude TT/2 radians around the coordinate axes leads to the required expression T = kg where A: is an arbitrary scalar. That expression is invariant relative to the complete orthogonal group, b) Three combinations are possible: g*igkl, g^g'1, gugh^- Linear independence of the corresponding fourth rank tensors is proved by contradiction. Let there exist a linear dependence of the form agijgkl + Pg^g*1 + igilgki = 0 at a2 + 01 + -y2 > 0. Then, by means of contractions with gij, gik, gu, one obtains the system of equations 3a + /? + 7 = 0, a + 3/? + 7 = 0, a + 0 + 37 = 0. It follows from here that a = 0 = 7 = 0, i.e., the supposition about linear dependence of the tensors is not valid.

5.17 Following the method presented in solution to Problem 5.16, one deduces that, for all the cases, the invariant symmetric second rank tensor has the form fcig + ^2^363; the invariant nonzero antisymmetric tensor equal to k3e • e$ is allowed in the cases b) and d); here, ka (a = 1,2,3) are arbitrary scalars.

5.18 Applying the method presented in solution to Problem 5.16, one obtains kig + foS + k3S • S where ka (a = 1,2,3) are arbitrary scalars.

5.19 Determine the transformation rule for the components of the inverse matrix, using the result of solution to Problem 3.36b.

5.20 gVg", p V ' , s V j > 9ijBkl, gilcB^1, guB^k, gklB^, ^Bik, gikBil, B^Bkl

where B = 6363. Linear independence of the tensors is proved by contradiction, using contractions with g and B as in Problem 5.16.


5.21 a) a = fb where / is an arbitrary scalar function of \b\. b) c = f\a + ^ 6 where f\, / 2 are arbitrary scalar functions of \a\, \b\, a • b.

\b\ 5.22 $ = / T/(T) dr + const.

o

5.23 Use the dependence $(Ji , J2i J3) in a coordinate system whose axes coincide with the principal axes of the tensor S.

case, 5.24 Remove the parentheses in the polynomial of Lagrange. In the general

v^ A2A3F(A1) _ (A2 + A3)F(A,) ^ (A! - A2)(A, - A3) ' £ (A, - A2)(A1 - A3) '

z r F(Al) ^ ( ^ - A ^ A x - A , )

where the symbol l ^c / denotes the sum of terms over all possible cyclic permutation of indices 1, 2, 3.

5.25 The tensor of inertia has the form

I = J{\r\2gii-xix')pdVeiej v

where r = x'e* is the radius-vector relative to the center of mass, p is the density, V is the volume of the body. a) Due to symmetry, the tensor of inertia for a homogeneous sphere is spherical: I = fcg (Problem 5.16). Carrying out the calculations in a spherical coordinate system, one obtains

1 2 2 the expression for the scalar k = -1\ = - MR where M, R are the mass

and radius of the solid sphere. b) In = - M(b2 + c2), I22 = 7 M(a2 + c2), 5 5

/33 = - M{a2 -I- 62), Uj — 0 if i ^ j where M, a, 6, c are the mass and semiaxes 5

of the ellipsoid.

5.26 A regular tetrahedron can be inscribed in a cube so that its edges would be diagonals of the cube. Consequently, the symmetry group of a regular tetrahedron is a subgroup of the symmetry group of a cube. In turn, the symmetry group of a regular octahedron, which can be inscribed in a cube so that its vertices would be the centers of the faces of the cube, coincides with the symmetry group of a cube. Therefore, it is sufficient to prove that the indicated tensor of inertia I (Problem 5.25), as a second rank tensor invariant relative to the symmetry group of the body, is spherical in the case of regular tetrahedron. Due to invariance relative to rotations of magnitude IT

5. Principles of Symmetry and Tensor Functions. 33

radians around the three mutually perpendicular axes passing through the midpoints of opposite edges of the tetrahedron, all the non-diagonal (in the coordinate system with those axes) elements of the matrix (/y) equal zero. Next, invariance relative to rotations by angles 27r/3, 47r/3 around one of the axes passing through the vertices of the tetrahedron and perpendicular to its faces leads to the final result.

5.27 In the case of Problem 5.14d, all the components of any invariant tensor field in the cylindrical coordinate system (r,<p,z) are arbitrary functions of r, z. In the other cases, certain relationships of evenness relative to z arise, and certain components of a vector v and tensor T become zero, a) v* = 0, Trv = T^ = Tzv = Tvz = 0, the other components, including the scalar, are arbitrary functions of r, z; b) vz, Trz, T"', Tvz, Tzv are odd functions of z, while the others are even functions of z; dependence upon r is arbitrary; c) the union of the restrictions a) and b); e) vz, vv, TTZ, Tzr, TTV, T1^ are odd functions of z, while the others are even functions of z; dependence upon r is arbitrary.

5.28 In a spherical coordinate system (r,0,<p), an invariant scalar, vector and second rank tensor have the form $(?"), v = vT(r)eT and T = Trr(r)erer +

Chapter 2

General Laws and Equations of Continuum Mechanics

7 Mass Conservation Law. Continuity Equation.

7.2 — / pdV = — I pvn da where V is the space volume, E is its boundary, v v E

is the velocity of the medium.

7.3 b) If i , y, z are Cartesian coordinates, then the continuity equation has the form

dp dp dp dp di+Vxd-X

+V^+Vzdz (dvx dvy dvz\

7.4 For the proof one can use the fact that, if V is a material volume, then

V £ V

7.5 For an incompressible medium V'Vjy> = A<p = 0; for compressible one |f + V^pViip) = 0; here VV = 9ijVj<P-

7.7 In an arbitrary curvilinear coordinate system, the formula

,. 1 dvaV9 , ., , d l V V = ^ ~ ^ ' ^ d e t ^ )

is valid (see Problem 3.48). In an orthogonal system, <7y = 0 at i =fi j , det(</y) = 9n922933- The physical components u* of the vector v are defined by the formulae

35

36 SOLUTIONS. GENERAL LAWS AND EQUATIONS

v = vk ejt = ufce/t/|ejt|, i.e., ux — v'y/gTi (no summation over i), see Problems 3.17, 3.18. In an orthogonal coordinate system, the continuity equation is

r dp d/yg22ff33 u1 dpy/gngssv? dpy/gng22u3

y/9dt + dx* + dx* + 8x3 ~ ° ' a) In a cylindrical coordinate system r, <p, z, using the notations u1 = ur, u2 = uv, v? = uz, one obtains

dp dpriir dpuv dprux

dt dr d<p dz

b) In a spherical coordinate system r, 6 (polar angle), A (longitude),

r'smB— + s i n 0 ^ + r — + r -^— = 0 . at or do o\

7.8 dp 1 dpur" =

dt r" dr where v = 0,1,2, respectively, for motions with plane, cylindrical and spherical symmetry, u = uT.

7.9 Use the formulae relating the volumes of a small material particle in the initial and final states (see Section 4).

7.10 These formulae can be derived from the relationship poy9 = p\fg where O O O O , o i

g = det(<7,;), g — det(<7y), using the equalities g^ =9ij -I- 2£y = <?ik(<5* + 2 £*) and 9 ij = fjij — 2E{J = gik(6j — 2e*) (see Section 4). In the linear approximation, the continuity equation can be written in the form p = po(l — / I ( E ) ) = Po(l — A(£))-

7.11 dC | 3(C + h)vs __ Q

dt dx 7.12

1 dS dv 1 dp „ , d d d " " 7 + ^~ + ~^7 = °> w h e r e T* = ^7 + v°^~ S0 dt ox p0 dt dt at ox

7.13

dt jpAdV = j (^ + diw(pAv)\ dV = Jp(j£+ vaVaA\ dV V V

Here the continuity equation is used.

8. Stress Tensor 37

7.14 Calculate divw.

7.15 a) vy = Ax/r2; b) v = gra.d<p, since curlv = 0; A<p = divw = 0; c) the equation of the particle paths is dx/vz = dy/vy, i.e., xdx + ydy = 0.

7.16 In the considered case, the velocity potential (f is determined as the solution of the Neumann problem A< = 0, dtp/dn\r = T(xl,t), the equation and boundary condition for ip as well as the relation between the velocity and <p are linear, the equation Aip = 0 is obtained only from the continuity equation.

8 Stress Tensor 8.1 Let the x-axis be directed along the bar velocity, and the z-axis be

vertical. Then pnz = - 1 kgf/0.02 m2 = -490.5 N/m2. Coulomb's law states that the friction force is proportional to the normal force. So Ip^l = 0.3|pn2|, p n l = 147.15 N/m2;pny = 0.

8.2 a) Write the equality expressing the momentum conservation law for a small volume of the medium in the form of a tetrahedron, three faces of which lies in the coordinate planes of a Cartesian coordinate system, and the orientation of the fourth one is determined by the normal n. Divide all terms by the area of the tetrahedron base and find the limit equality as the tetrahedron shrinks to a point, b) It is necessary for / p(a — F) dV to be small of 3rd order with respect to the linear size of the volume.

v It is sufficient for p(a — F) to be finite, c) p12 is the projection of the stress vector acting on the surface element normal to the x2-axis along the I'-axis.

8.3 Consider a small volume of the medium having, at the initial instant, the form of a tetrahedron oriented with respect to an initial Lagrangian coordinate system as described in the solution to the Problem 8.2.

8.4 a) Use the definition (8.2) and the relationships

A A - dSk A ° ^ pdarij = pdank— = p0daQ nk —

where n,, hk are the components of n in Eulerian and Lagrangian coordinate systems in the deformed state. To obtain the second part of these relationships consider a particle in the shape of a small cylinder having da0 as the base and arbitrary vector dro = d£l e i as the element at the initial state. At the deformed state it has the base da and the element dr = d£' e*. Write the equality expressing the fact that the mass of the particle is the same in both states, b) In a Cartesian coordinate


system, ir1"1 is the projection onto the f fc-axis of the force acting on the surface element perpendicular to the f'-axis at the initial instant divided by the area of this element before deformation.

8.5 a) Use the formulapn = p^^Cj and the fact that n ' = e ' / |e ' | for coordinate surface x' = const, b) Physical components p*{' of p" are P^'Jgjj/g" (no summation over i, j). c) They are the stress vectors acting on the coordinate surfaces. Their components are jPx.

8.6 Use the equation (6.8). For details see Section 11.

8.7 p\ « 213 Pa, p2 » -56 Pa, p3n « 32 Pa, \pn\ « 223 Pa, p n n « 56 Pa,

pnT ~ 216 Pa, cosfl = p„„/|pn | « 0.25, 6 « 75°.

8.8 P n = - ^ e 2 + | e 3 .

8.9 p„„, = P^rijTiu, p„ in = p^nijUi = p%uTij; as p y = pji then p n m = p n , n .

8.10 a) a sphere; b) two parallel planes x1 = ±c; c) a hyperbolic cylinder

8.11 a) pn = pyn Je' = pyX^e'/M = 2grad7"/|r| where r = x-'e,, .F = pyx'x^, .F = c is the equation of the stress surface, b) Any stress surface is a surface of the second order for which three orthogonal axes are known to exist such that r || n at the point where the axes cross the surface (r is the radius-vector of the point, n is the normal to the surface at this point), c) Only on a sphere, r \\n for all its points.

8.12 a) In the principal coordinate system of the stress tensor, e.g., on the surface element perpendicular to the axis x1, p n i = pn, pn 2 = P21 = 0, p„3 = P31 = 0 (according to the formula of Cauchy (8.1)). b) p n n = pynJ'n';in the principal coordinate system pnn = pi(ni)2 +P2(n2)2 + p3(n3)2, with (ni)2 + (n2)2 -I- (n3)2 = 1; an extremum of pnn corresponding to n , satisfies the conditions

a b n n - A ( n 2 + n2

+ n 2 - l ) ] = 0 |

OTli

(A is a Lagrange multiplier). If pi > p2 > p3 then these conditions show that p„„ reaches an extremal value if n is parallel to one of the principal axes. Then (p„n)majt = Pii (Pnn)min = ?3- c) The principal components are pi = —8, p2 = 4, p3 = 16. The direction cosines of the principal axes are (0, l / \ /2, — \/y/2), (l/v^3, —1/>/3, —1/-/3), (-2/VS, -l/v/6, -l/v/6).

8. Stress Tensor 39

8.13 The deviator of a tensor p is defined by the formula

(<*) * k-

Plj = Va ~ gP k9ij where <7y are the components of the metric tensor equal to Sij in Cartesian coordinates. For the given stress tensor, the spherical component and the deviator are

/ 4 0 0 \ / 8 4 4 \ 0 4 0 , 4 - 4 8

^ 0 0 4 / \ 4 8 - 4 /

8.14 The principal components pi = P2 = 1, ?3 = —2. The vectors (1/\Z2, 1/V2, 0), (l/v/6, - l / v /6 , 2/v/6), (1/v^, - l / \ / 3 , -1/V3) can be taken as director vectors of the principal axes; the transformation to the principal coordinate system y{ has the form y1 = (x1 + x2)/\/2, y2 = (x1 - x2 + 2x3)/v/6, J/3 = (x1 - x2 - x3)/v/3.

8.15 Due to the symmetry, it is sufficient to consider 0 < 8 < n. The principal components are p1|2 = C cos | (1 ± sin |)/27rr, p3 = 0; for 0 < 8 < rr the largest of p* ispi; a = (38 + 7r)/4.

8.16 a) The most convenient coordinate system for the calculations is the Cartesian system for which the considered surface elements he in the coordinate planes. In this coordinate system, |p n l | 2 + |p„2|2 + | p ^ | 2 = P^Pij- The value of this sum does not depend upon the coordinate system and upon the directions of n4. b) Ii = P1+P2+P3, h = P l P 2 +P2P3 +P3P1, ^3 =PlP2P3-

8.17 pnn = fn2, p2T = [fntf - tfn})2.

8.19 Extremal tangential stresses equal (p< — Pj)/2; the surface elements acted on by them go through one of the principal axis of the stress tensor and make the angles 45° and 135° with the other principal axes. The surface pnT max — k = 0 is a hexahedral prism. The distances equal fc\/2 and 4k/-</6, respectively.

8.20 Let us choose a Cartesian coordinate system with axes x, y, z directed along the edges of the parallelepiped with initial lengths o, b, c respectively. Then a) on the endwalls, 7Ti3 = 0, 7:23 = 0, 7r33 = P/ab where P = \P\ for extension, P = —\P\ for compression. On the faces perpendicular to x-axis, p n = 0, so nu = 0, 7T2i = 0, 7T3i = 0. On the faces perpendicular to j/-axis, p n = 0, so n^ = 0, 7r22 = 0, 7r32 = 0. b) All the components of the Piola-Kirchhoff tensor are equal to zero except 7T33 = P/ab. c) To find the Cauchy stress tensor, it is necessary to know the areas of the faces where the forces act in the deformed state, d) For this problem, the Piola-Kirchhoff tensor does not depend upon the shape of the cross section of the cylinder.

8.21 The components of the Piola-Kirchhoff tensor would be ir2i — o, 7ri3 = ""31 = ""12 = "22 = ^23 = 7T32 = ^33 = nU = 0-

8.22 Choose a Cartesian coordinate system such that rr^-axis coincides with the axis of the bar. In accordance with the conditions of the problem, the components of the stress tensor are p11 = P/S, and the other p u = 0. The stress vector on a surface element with the normal n(ni,n2,n3) is pn = p11niei = (P/S)niei. The values of the normal and tangential components are, respectively, pnn = pun\, p£T = (P11)2"? (1 - " ? ) . a n d t n e i r maxima are (pnn)max = p11 = P/S, (rii = ±1); |p„T|m(0( = p»/2 = P/2S, (m = ±4).

8.23 Use the formula (of Cauchy) pn = p^rijet for the proof.

9 Differential Equations of Motion and Equilibrium

9.2 a) A concomitant Lagrangian coordinate system (the coordinates £', the basis vectors e<, the components of the metric tensor £y) moves together with the medium. The rate of change of a basis vector in a particle is

de,(e,t) =d_dL=dv_ ^.,*. _ (dvk , . ^ dt dt dp d?

so

> „ » t » Idvk . /At \ -

dv(p, t) (dvk , i r , „k . „ dt \ dt

b) The equations of motion have the form

p ( ^ + t > i V , i ) f c ) = p F f c + Vipfci .

c) The expressions of ViVk and Vjpfcl contain the Christoffel symbols

'" 29 [dp dp dp) " o o

In their turn, <?y =9 y + 2ey where 9 y are the components of the metric tensor in the Lagrangian system in the initial state, and £y are the components of the strain tensor. Consequently, the equations of motion in a Lagrangian coordinate system contain terms with derivatives of £y with respect to p.

9. Differential Equations of Motion and Equilibrium 41

dv d Vi o

~dt=~dT ei =

w22/, Fz = 0.

d2W{

dt2

9.3 The momentum conservation law for a material volume can be written in the form

— j p0v dV0 = j poF dV0 + j tTno d°o V0 V0 E0

where Vo is the volume occupied by the considered medium in the initial state, Eo is its boundary; v = v(£', t), F = F(£\ t), and pn(f , t) are the velocity, the density of body forces, and the stress vector at the instant ^respectively; ir,^ do^ = pndcr (according to the definition of TT^), and podVo = pdV (according to the mass conservation law). Using the facts that 7r^, = -ir^noj eit the volume Vo is arbitrary, and all the functions are the necessary number of times differentiable, one can obtain from here the differential equation

dv o p0— = p0F + VjTT,3ei.

If the initial state does not vary with time (e^ = const), then

e*

9.4 Fx = u2x, Fy

9.5 Fx = g,Fy = 0, Ft = 0.

9.6 a) F = 0, b) pF1 = br - a/r, F2 = F3 = 0, c) pF1 = br + cr sin20 - 2a/r, pF2 = -bcot6 + c sin<? cos9, pFz = 0.

9.7 a) From the equation (9.1) with v = 0, p'-7 = — pg'i it follows the equation of equilibrium of a fluid in the form pF — gradp = 0. b)pg — gradp = 0, where g is the gravitational acceleration. c) Since curl gradp = 0, curlp.F = 0 at equilibrium. Expanding the scalar product of the last equality and F, one obtains the condition F ■ curl F = 0.

9.8 The equation for the deflection of a membrane w can be obtained from the equality to zero of the sum of projections onto the z-axis of all forces acting on a small rectangle element of the membrane ABCD (see Figure a9.1) with lengths of the sides Ax and Ay. The tension T acts along the external normal to the contour of ABCD in the plane tangent to the membrane surface. The sum of projections of the forces caused by tension T acting on sides AD and BC equals, up to infinitesimals of higher orders,


w . MOT n Side view

Figure a9.1

Here it is taken into account that Ax, Ay are small, as well as the deflection of the membrane is small as compared with the size of the membrane, i.e., dw/dx is small. Considering similarly the sum of projections of the tension forces acting on sides AB and DC as well as distributed normal load q, one can obtain

/ d f„dw\ d f„dw\ \ A A

or, since T = const, Aw

dx

-q/T.

9.10 It follows from the equations of equilibrium that the expressions pn dy — P12 dx and P21 dy — P22 dx are total differentials of some functions F and $. The condition P12 = P21 yields the fact that F dy — $ dx is also a total differential of some function U. Thus, F = dU/dy, $ = dU/dx,

P11 = 9F dy

&U dy2' P22 = - dx

&U dx2' P12

aF dx dxdy

9.11 b) On the lateral surface n3 = 0, pn = pljrijei = (p31ni + p32n2)e3; ni, ^2 can be regarded as the components of the normal to the contour of the rod cross section. Introducing the unit vector r tangent to this contour (T • n = 0) and expressing P31 and P32 in terms of stress function T, one can transform the condition p n = 0 to the form

^ T 2 + ^ r 1 = ( g r a d ^ . r ) = — = 0 ,

i.e., T = const on the contour of the cross section, c) Rx = JJdF/dydydx = 0, s

Ry = ffdF/dxdydx = 0 (S is the area of the cross section), because T = const s

on the contour of the section (use the Stokes formula); Mx = JJdF/dxzdydx = 0, s

9. Differential Equations of Motion and Equilibrium 43

Mv = - fi df/dyzdydx = 0, Mz = - JJfedF/dx - ydJr/dy)dydx; the moment of s s

the forces acting at the section (with the normal =(0,0,1)) has only the component

directed along the axis of the rod, so it is a torsional moment.

9.12 The equations of motion for an ideal fluid (Euler equations) have the form

pa = pF — gradp

9.13 Use the momentum conservation law and the fact that in an ideal fluid P„ = -pn.

9.14 The expression for the acceleration is transformed as follows

a = -£ + wfcVfc«4c' = -£ + v^VkJ + vk{VkVi - V^Je* =

-£ + Vil^Jei + 2[utxv]iei , u> = -cur lv .

9.15 Project the Euler equation onto a streamline or vortex line. Use i) the expression for acceleration obtained in the problem 9.14, ii) the fact that the projection of grad $ on a certain direction is the derivative of $ along this direction, iii) the fact that, for a steady motion, the quantities p and p on any given line C are functions only of s, a parameter on C: p = p(s), p = p(s), so, if dp/ds ^ 0, p on C can be regarded as a function of p (in general, differently determined for different lines £).

9.16 a) From the Euler equations, one can obtain grad(d<p/dt+v2/2+V — U) = 0. Hence, d<p/dt + v2/2 + V - U = F(t).

9.17 Use the expression rjfc in terms of the components of the metric tensor and the fact that g^ = 0 at i ^ k in an orthogonal system.

9.18 Introducing the physical components of velocity u1 = «,., u2 = u^,, u3 = uz, (v = v'e{ = u'ej/|ej|), as well as those of the density of body force and pressure gradient, and using formulae for the components of the acceleration from the Problem 9.17, one obtains the Euler equations in a cylindrical coordinate system r, tp, z

dUr dUr UvO%. dUr U2, I dp

at or r dip dz r par


In a spherical coordinate system r, 6 (polar angle), A (longitude), the Euler equations have the form

dur dur u» dur u\ duT u"g + u\ I dp dt dr r 88 rsind dX r pdr

dug dug UgdUg U\ dug UTUg COt 9 2 1 dp dt or r 06 rsmO oX r r pr dd

dux , dux ugdux , ux dux uûx cot 8 1 dp

d£ ar r w rsinfl aA r r prsmffdX

9.19 Use the formula e^ = (VtWi + V ^ ) .

9.20 Use the formula £& = ^(V tu/ + V/ru*). 9.21 Expand the scalar product of the equation of motion and the relationship

dr = vdt and integrate the obtained equality over a volume V. The density of work of internal surface forces equals a) —p,J VjVj dt/p, b) 0.

9.22 a) For a spherical coordinate system, vr = v(r,t), vg = vx = 0. The equations of continuity and motion have the form

d(vr2) dv dv 1 dp dr dt dr p dr

The boundary conditions v = drc/dt, p = 0 at r = rc (rc is the radius of the cavity), oo

p —► 0 at r —► oo. The initial condition Eun = / 4nr2pv2dr/2 —► E0 at t —> 0, rc —► 0.

b) v = ( f ) 1 / 5 ^ 1 / 5 / * - 2 , p = c2p(l - (rc/r)3)/2rr2, rc = (2.5c)2/5t2/5, c = Ê0/2np, c) pm a x = 3pc 2 /8^r c

3 = 3pc*/5/8Â(5c/2)6'5; r j w = ^4>c. d) v = v(t,r,E0,p), p = p(t,r,Eo,p); using the II-theorem of dimensionality theory one can write v = (E0/(pt3)y/5 / ( 0 , P = (£o/(p*3))2/5MO- w h e r e € = (£o/p)-1 /5rA2 /5- If Poo f 0, then v = (E0/(pt3))1/5 *({,»?), where 77 = rp^/Eft, i.e., the problem is not self-similar. e) If Pa, ^ 0, oscillations take place: at first the radius of cavity increases up to rcmax = f/SEo/Anpoo, then the cavity begins to collapse, etc. f) No, they are not, because, for such a motion to begin, at the initial instant the infinite quantity of energy must be released on a unit area of the plane charge or on unit length of the charge distributed over a straight line. To verify this, substitute the expressions for

00 v obtained from the continuity equation into E = 2 f pv2dr/2 (for an explosion on a

00 plane) or into E = / 2irrpv2dr/2 (for an explosion on a straight line).

9. Control Surface Method 45

10 Using the Conservation Laws in Integral Form for Calculation of Forces and Moments Acting on Bodies Moving in a Fluid (Control Surface Method)

10.1 a) Write the corresponding laws for an individual volume of a medium and use the formula (7.10) for differentiation of an integral over a moving volume, b) The energy conservation law for a steady motion has the form

J p(j + u)vnda = J P(F v + ^ ] dV + J (pn-v -q'n) da .

10.2 Use the Gauss formula.

10.3 a) Let us denote the surfaces of the bodies A, B by E^, EB and the inner surface of the tube T by E7-. By definition,

R=- J Pnda,M=- j (rxpn + Qn)da ,

W = - J (pn-v-q'n)da E ^ + E B + E T

where n is the normal to the surface outward with respect to the fluid. Take the volume occupied by the fluid between the sections Ei, E2 as a control volume; then the control surface is Ei + E2 + E^ + EB + ET. Using the laws of conservation of momentum, angular momentum and energy (see Problem 10.1) as well as the condition for the fluid not to flow through E^, Eg, ET, i.e. the condition vn = 0 on E.4, Eg, ET , one obtains

R= j {Pn~ Pvvn) da , M = j {r xpn + Qn- p(r xv + k)vn) da , E1+E2 E1+E2

W= J {pn ■ v - q'n - p(u + v2/2)vn) da . E,+E2

b) On account of the gravity force, the terms / pgdV = mg and / pr x gdV — v v

I / pr dV 1 x g = mr0 x g = r0 x mg arise in the formulae for R and for M

respectively; here, m is the mass of the fluid in the volume V, g is the acceleration


of gravity, and r0 is the radius-vector of the center of gravity for the fluid in the volume V. c) Let us mark the parameters of the stream in the sections Ei, E2 by the indices 1,2. The fluid flows into the control volume through the surface Ei and flows out through E2; therefore v\ = — wi»ii, (pn)i = —pifii = p\V\/v\, v2 = v2n2, and (Pn)2 = — P2TI2 = —P2V2/V2, where n\, n2 are the normals to Ei, E2 directed outward with respect to the control volume. Let Si, S2 be the areas of Ei, E2, r*, r j be the radius-vectors of the geometrical centers of the sections Ei, E2. Then P1V1S1 = p2v2S2 due the mass conservation law, and

R = (P\ + PlV^SiVi/vi - (p2 + P2VI)S2V2/V2 , M = {pi+ piw?)SirJ x vx/vi - (p2 + P2vl)S2r\ x v2/v2 , W = (il-i*2)G, G = pivA = P2V2S2

where i* denotes the expression v2/2 + u + p/p, u is the internal energy per unit mass; i* is called the total enthalpy per unit mass (see Section 24). d) Yes, they are valid, since continuity of the parameters of the stream at the points of the volume V is not necessary for using the formulae (10.1) - (10.3).

10.4 a) Let us take as a control volume the part of the jet cut by its cross sections Eo and E, disposed far upstream and downstream from the place where the jet encounters the wall. The control volume is shown with dashed fines on the figure in the text of the problem. The boundary of this volume is E = E0 + E, -I- Ej -I- £„, where Ej and E„, are the corresponding parts of the surfaces of the jet and wall respectively.

The force acting on the front part of the wall is Rj — I pnw da in the presence of

the jet and RQ= ponw da in the absence of the jet, therefore the desired force

is R = Rj — RQ = (p — Po)nw da = (p- Po)nw da\ here, E^o is the whole front L u O 2-HD

surface of the wall, nw is the normal to it outward with respect to the fluid; it is possible to replace integration over £„*> by integration over £„, because the pressure in the jet far from the place, where the jet encounters the wall, becomes uniform over a section and equal to the pressure po on its external boundary everywhere including the wall. The momentum conservation law can be written in the form

/ pvvn da = — I pn da = — I (p — Po)n da

(see Problem 10.2). Projecting this relationship onto the direction of nw and taking into account that vn = vo on Eo, vn = 0 on Ej and Ew, p = p0 on E0, £, , E ; , v-nw = 0 on £, , v ■ nw = v0 sin a on E0, one obtains the answer: R = poS0v% sin a. b) For


the plane jet, let us consider the control volume V the boundary of which includes besides Eo, £,, T,j, T.w (see item (a)) also two longitudinal sections (parallel to the plane of the flow) the distance between which is unity. Since, according to the problem statement, all the parameters of the stream on them are identical, and the normals to them external with respect to V are opposite, the total contribution of integrals over these sections in all the relationships used equals zero. Therefore the formula for R obtained in item a) can be applied also to a plane jet with S0 = Z • 1. Consequently, R = PQIVQ sin a. To obtain the answer to the other question of the problem, let us introduce the Cartesian coordinate system with x-axis directed along the normal to the wall (along n„,) and y-axis directed along the wall in such a direction that the projection v0 onto y-axis is non-negative (on the figure: upward along the wall). Let us denote with v\, v2, l\, l2 the velocities and thicknesses of the jets spreading "upward" and "downward" over the wall (far from the point O), viy = vi, v2y = v2. Using the mass and the momentum conservation laws projected onto t/-axis, one obtains respectively lv0 = l\Vi + l2v2, Ivl cos a + l\v2 = l2v2. The Bernoulli integral (10.4) written for the streamlines at the surface of the jet (where p = p0 = const) yields v\ = vo and v2 = v0. Therefore li = 1(1 + cos a)/2, l2 = 1(1 — cos a) /2. The point of application of the total force R is determined by the following condition: if R is applied to the point C, its moment relative to any point equals the sum of moments of all the forces the sum of which equals R, i.e.,

vc x R= I r x (p — po)n dcr

(there may be no such point in the general case). Let us consider the moments of the forces relative to the point O. Using the angular momentum conservation law, one obtains

- ( , -i2

rc x R= / r x pvvn do = r x pviVi dx+ / r x pv2v2 dx . E 0 0

Since the vectors r c , R, r, v\, v2 lie in the x-y plane, the last equality projected onto the axes x and y is an identity, and that projected onto 2-axis yields —ycR = p(v\l\ - vjll)/2, i.e., yc = -(1/2) cot a. c)« 9m/s.

10.5 Examine the control volume shown with the dashed lines on Figure 10.3. Use the Bernoulli integral for the calculations, a) R = pv2S(l — cos a); b) R = 2pv2S.

10.6 a = 2S0(pi - Po)/m.

10.7 5 0 /5 = 0,5.


10.8 a) The component of force exerted on a body by the fluid that is parallel to the velocity of the body relative to the fluid at infinity is called a drag force (or, simply, a drag), and the component of this force perpendicular to the velocity is called a lift force (or, simply, a lift). The drag equals zero in this case (the d'Alembert paradox), and lift can be nonzero, b) l)The total drag of all bodies equals zero, 2) the drag of the body with the cavity equals zero.

10.9 a) The difference between the force acting on the body from the side of the moving fluid and the force acting in the absence of motion equals R = ^PV2SSQ/(S — S0)2 and is directed along the velocity v\. b) As S —► oo and for finite S0, v2 —» v\, Pi —* Pii ft —* 0. c) The magnitude of the force R remains the same as in item a), but the direction of this force becomes opposite to the direction of the velocity vi.

10.10 Examine the control volume shown with the thin line on Figure 10.7 (with unit width in the direction perpendicular to the plane of the flow). Sections AD and BC are parallel to period vector I, and AB and DC are identical curves shifted by one period. The mass and energy conservation laws yield plvin = plv2n = G or v\n = 2n = ^n where n ± I; R = (pi — p2)in + G(t>i — v2). Since the flow is periodic, T = (v2i—vu)l where T is the circulation around circuit ABCD; so Ri = G(vu—v2i) = —pTvn. Using the Bernoulli integral, one obtains R„ = l(pi - p2) = pT(yu + v2i)/2. Consequently, R is orthogonal to the vector Vi + v2, \R\ = pT\v\ + v2\/2, i.e., R = p{v\ + v2)/2 x r. b) v2i —> vu, v2 —» vi = » M at / —> oo and T = const, so R = pVoo x r.

10.11 The velocity of incompressible (and, for many cases, compressible) fluids, when the motion is caused by a body moving in the fluid, is known to tend to zero at a large distance from the body in all directions exept the backward direction along which a stream is formed referred to as concurrent. The supposition about the absence of viscous stresses and uniformity of pressure far from the body allows one to neglect the broadening of the stream cross section far from the body. As the control volume choose a rectangular parallelepiped with all the faces disposed sufficiently far from the body and with the lateral faces parallel to the velocity of the body, a) Since this control volume moves together with the body, the motion in a system moving with the body is steady, and the control surface method yields R= pv2 da — pu2 da

St Si where R is the drag, S\ and S2 are the front and back faces of the control volume (perpendicular to the velocity of the body), and u is the velocity of the concurrent stream relative to the body far from the body. Using the mass conservation law I pvda = puda = / dm one obtains R = / (v — u) dm = / pv(v — u) da. b) Since

Sl 52 52 52 52 the body moves inside the control volume, the motion in a system at rest relative to the control volume is unsteady. Thus, the rate of change in momentum equal to


/ pvvc da must be taken into account in the calculation of R. The control surface

method yields R = pvvc da where vc is the velocity magnitude in the concurrent s2

stream relative to the surrounding fluid. Since u = v — vc the two formulae obtained for R coincide.

10.12 Reasoning as in item (a) of the preceding problem, one concludes that the force acting on the rocket equals the difference of the momentum fluxes through S\ and 52- For a case that the momentum and mass fluxes through Si can be neglected,

one obtains R= — I puu da where u is the velocity of the fluid relative to the rocket

in the jet flowing from the rocket. If u is constant along the section, R — mu.

10.13 R = -((p ' - Po)S + pv2S)n; R = -pv2Sn, where S is the area of the outlet nozzle cross section.

10.14 a) R is the diagonal of the parallelogram constructed with the vectors (P2 + P2V%)S2v2/v2 and (pi+piv2)SiVi/vi. b) If the lines parallel to v\ and v2 passing through the geometrical centers of the input and output sections of the tube intersect at point O, this point can be regarded as the point of application for the resultant of the forces exerted on the tube by the fluid, c) 6.75 kgf.

10.15 Let us mark with the indices 1 and 2 parameters in the inlet and outlet sections of the tube. Since Si = S2, it follows from the mass conservation law that vi = v2. Then it follows from the Bernoulli integral that pi = p2. Applying the formula for R obtained in solution of Problem 10.3, one deduces that the force R acting on the tube makes the angle 45° with Vi and v2, and its magnitude equals V2Si(pi + pv2i)/2.

10.16 Apply the momentum conservation law to the fluid contained between sections A and A' shown on Figure 10.10. Answer: p'2 — p2 = p(vi — v2)2/2.

10.17 To calculate the force R, let us choose the system moving together with the body. There is a steady flow around the unmoving body, and one can apply the formulae from the solution of Problem 10.3, considering the sections Ei and E2 of the stream far ahead and behind of the body. It should be assumed in these formulae that Si = S2 = S, and Vi = v2 = —v; then, R = (p2 -pi)Sv/v + mg where m = pV, V is the volume of the fluid contained between the sections Ei and E2, and g is the acceleration of gravity. Use the Bernuolli integral 10.4. Taking into account that U = —gz (z is the vertical coordinate, and the z-axis is directed upward), one obtains


P2~Pi= P9(zi — z2). If the body moves upward, then Ei is the upper section, E2 is the lower one,

Rz = pgS{zx - z2) - pgV = pg(S{zx - z2) -V) = pg(Vx - V2) .

Thus, an upward-directed force acts on the body and the bubble, magnitude of which equals the weight of the displaced fluid (force of Archimedes). If the body moves downward, the answer is the same.

10.18 Choose the control surface composed of the surfaces of the blades, ax-ially symmetrical housing, part of the spinner, and circular cones S\ and 52- For the moment of the pressure forces acting on the blades of the rotor, one obtains Mz = / r^uabspu,, da - j r2ut2abs/wn da where vtiabs, itoaba are the absolute trans-

verse velocities of the medium (the velocity is the sum of axial, radial, and transverse components).

11 Angular Momentum Equations. 11.1 Consider the angular momentum conservation law for a small portion of

the medium in the form of tetrahedron (see Problem 8.2). This tetrahedron has a basis perpendicular to n and lateral sides parallel to coordinate planes. Divide all terms of this equation by the value of the basis area and then take the limit of the equation as the altitude of the tetrahedron is tending to zero.

11.2 Reduce this law to the relation containing only volume integrals. Suppose all functions under the integral sign to be continuous. Use the fact that the equality should be valid for any volume.

11.3 Euler formula v = u> x r describes the velocity of the points distribution in a rigid body with fixed point O, where u> is instantaneous angular velocity, r is the radius-vector of the point of the body with origin in point O. Calculate the angular momentum K of this body about the point O.

K = I r x vpdV = I r x (u> x r)pdV = f(u>\r\2- r(r -u))pdV . V V V

r = xx&i in a Cartesian coordinate system, so

K = uqej, J(\r\2&*> - x"x")pdV = I™ujqep

11. Angular Momentum Equations. 51

where I™ = /(|r|26w - xpxq)pdV. The tensor I = / M e p e , is called the tensor of v central moments of inertia. In this calculation the coordinate system can be either

fixed (Eulerian) or rotating with the body (Lagrangian). Let ep be the basis vectors in a Lagrangian coordinate system and P 9 , ujq be the components of I, w in this system. Obviously /p, do not depend upon the time t because the Lagrangian coordinates of the particles of the medium are constant. The vectors ep depend upon t and also the vector ep can be regarded as the radius-vector of a particle of the medium. Hence dep/dt = (JJ x ep (using the Eulerian formula). So

dK d t A - (dCj„ . . dep — = -(/^ep) = / - ^ e p + W , - ^

= ( / " ^ + / * < w " " p ) e, = M

where M is the total moment acting on the body.

dv' /p ' J + p , i 1 \ 11.4 a) p — = pF + Vj; I + - e'^VitQ'* where eyfc are the compo

nents of Levi-Civita tensor. b) The internal angular momentum equation gives

E ( P ' J - P J i ) ( e , x ej) = V]Qi*ei = 0.

11.5 According to the conditions of the problem the total angular momentum does not change and it is equal to zero. Hence, its projection on the symmetry axis of the bar is Iui + rrvy~1M = 0 or 7 = —mM/Iu. It is emphasized that the coefficient 7 is usually negative.

dfij _ dfij 11.6 Jaumann derivative is a tensor DjT = ——e^e, where —— is the total

at dt derivative and e is the basis which rotates with angular velocity u> = -cur lu . If the continuum moves as a rigid body e* is the Lagrangian basis of the continuum. The components of the inertia moment tensor I of a rigid body are constant in the Lagrangian coordinate system, hence £>jl = 0.

dn 11.7 a) A material straight line, b) 1? = n x — .

11.8 Since I = /(g — nn) where / is the eigenvalue of I, k = I • ft = I ft.

11.9 M = aH + @n(H ■ n), a,0 = const, M x H = /3(H • n)n x H.


11.10 Qij = kigijs/knk + k2v{nj + k3vjn{ = fci$y vfcn*

ka = const (a = 1,2,3).

11.11 Qij = {kig^ + k7nin^Vknk+k2Vi^ + k3Vini + (k4giJ + k1onin^nkTilVknl

+(fc5nJ V'fi; + fegri'V^n, + fartVffi* + fcsn*V(nj)n'. When V.TC is changed to Vn, the terms with k4, kio, ks, kg vanish, since n'Vjnj = 0.

11.12 The tensor Q is expressed in terms of the quadratic potential U = ai(Vin')2 + (a2VW + a3Vjn')Vin., + a4ninjVinkVjnk.

11.13 - (py - Pji) = eijk{agkl + bnknl)(n, - w,).

11. L.14 a.) I (nx — n) = /?(n • i f )n x i f - 2(apjt/ + bnkm)(Q.' - w')e*.

b) 0 = 0. c) Pij ~ Pji - 2a{niD]Tij - njDjrii).

11.15 pF = 0(H-n)HiVnt.

11.16 a) Ty = 6i5i;+62ey+63e,fceJ': + 64ejjte_,/nA!n'+65(eifcnfcnj+ej/tn':ni)+66njnJ where the coefficients ba (a = 1 , . . . , 6) are arbitrary functions of the three eigenvalues of the tensor e and the two scalars of e and n.

b) For the linear case, b3 = b4 = 0, 62 = const, 65 = const, 61 = Ciefcfc + C2eklnkni,

be = C3ekk + c4eklnkni, where Ci, c2, c3, c4 are constant.

11.17 If the orientation vector n is homogeneous and stationary we have dn/dt = 0 and, hence fi = 0. Using the internal angular momentum equation M x H + A(u> — n(w • n)) = 0 and the formula for M, we can deduce the equation

P(H ■ n)n x H + X[u - n(w • n)] = 0 (all . l)

a) If H || u) we obtain A(|u;|2 — u>2) = 0 by scalar multiplication of equation (all . l) by u> (\n\ — 1). Hence u> || n , so n \\ H and equation (all . l) is satisfied completely.

b) If i f J- u; we obtain A(if • n)(u> • n) = 0 considering the scalar product of equation (all . l) and if. 1) If i f • n = 0 then A(u> — n(w • n)) = 0 and hence n || u;. 2) If w • n = 0 then we have (5(H -n)^ x i f ) + Aw = 0. Let us look for a solution for n in the form n = cif + d(if x w). Then 0cd\H|2(if x u>) x i f + Au; = 0. Using the formula (if x w) x i f = \H\2u - (u> ■ H)H = \H|2u> we obtain the equation

11. Angular Momentum Equations. 53

(5cd\H\A + A = 0. The equation \n\2 = 1 now has the form c2\H\2 + d2\H|2|u;|2 = 1. Hence

^ AW \H\2 p2\H\* '

The solutions of this equation exist only if inequality \H\4 > 4A2|u>|2//?2 is satisfied. In this case there are two different positions of vector n (regardless of the direction) corresponding to the solution

For this flow components of the strain rate tensor e and vorticity vector w are e^ = e2i = e, u3 = —e; all other components are equal to zero. In cases a) and b) item 1) vector n is perpendicular to the flow plane and anisotropy of the medium does not affect the stress value; we have pu/fan) = M- In the case b) item 2), when vector n lies in the flow plane, the effective viscosity coefficient is equal to fj, + A/4.

11.18 The equations of internal angular momentum and magnetization lead to the system of equations for f2 and M

MxH = X{n-u), nxH = (M- XH)/T.

In expansion of the solution in a Taylor series with respect to the small dimensionless parameter |u>|r, it is necessary to introduce the dimensionless value |/2|/|w| = C(l) . Then, in the zeroth approximation with respect to |u»|r —> 0,

M = xH , fl = u .

In a first approximation

M = XH + TUXH, H = u + (T/A)(W x H) x H ,

p12/2e12 = / i + ( r / 4 ) | H ± | 2

where H± is the component of H perpendicular to u.

4A2|u>[:

Chapter 3

Thermodynamics of Continua

13 The First Law of Thermodynamics. Energy Equation. Perfect Gas.

13.1 Consider the energy conservation law for a small volume of a medium shaped like a tetrahedron similar to that used in deriving the Cauchy formula for stresses (see Problem 8.1).

13.3 a) 0; b) pdivvdt/p = -pdp/p2 = pdV, V = \/p; c) -T^e{jdt/p, —2p.exjeij dt/p; d) (—p + Adivv)dp/p2 - 2/xey'ey- dt/p.

o

13.4 Use the fact that e^ = diij/dt in a concomitant Lagrangian system (if <7y do not depend upon t).

13.5 The energy equation is

The equation of internal energy is

, fdT kdT\ dq ,_ , du

13.6 As u = u(p,T) the equation of internal energy takes the form du = (du/dT) dT + {du/dp) dp = (p/p2) dp + dq. When V = const, dp = 0, dq = (du/dT) dT; therefore cv = (dq/dT)v=collst = du/dT .

13.7 For an elastic medium u = u(ey, T); du = (du/dT) dT + (du/d£ij) cfcy = (pv/p) dei:i + dq; when <fey = 0, cc = (dq/dT)Cij=COIlst = du(T, e{j)/dT = cT/TQ « c.

55

56 SOLUTIONS. THERMODYNAMICS OF CONTINUA

13.8 du = cvdT = pdp/p2 + dq; when p = const, dp = —(p/T)dT, dq = CpdT = (cv + R)dT; when p = const, dq = cvdT\ consequently, Cp = cv + R. This relationship is called Mayer's formula.

13.9 du = pdp/p2 + dq + Tmkekmdt/p. When p = const, dq = cdT = cvdT -Tmkekm/p. So c < cv if r^efan > 0.

13.10 Write the equation of internal energy with u — i— p/p.

13.11 Prom the equation of internal energy at dq = 0 and from the relationships u = u(p, T),p = p(p, T), one obtains for an ideal fluid

It follows from this equality that T = const at p = const. Therefore T = T(p). For a viscous fluid, the equation of internal energy yields

£*--(7-£)*-7«* Consequently, variation of T is not caused only by variation of p.

13.12 a) T/T0 = (p/po)7"1, P/Po = {p/po)y where 7 = c„/cv = R/cv +1 is the adiabatic exponent, p0, T0, po axe values of p, T, p in a certain state of the particle. The relationship

p = Cp1, C = const (al3.1)

is referred to as the Poisson adiabat. b) No, they are not, unless Tmkemk can be neglected.

13.13 a) c(T - T0) = - Q ( 3 A + 2p,)I1(e)T0/p, T0 is the temperature at e{j = 0, p,j = 0. b) During isothermal processes jfy = \J\{e)gij + 2/i£yj during adiabatic processes pij = A ^ J ^ e ) ^ + 2p,£jj, where X^ = A + a2(3A + 2p)2To/pc;

13.14 1) Since the strains do not vary during the process, dq — cedT = du — (du/dT) dT, cc = du/dT = cT/T0 « c. 2) Since, during the process, p does not vary, i.e., p'-J are constant (do not vary in time), Hooke's law yields that de^ = ag^dT. The equation of internal energy yields

Thus, Cp = CT/TQ + Za2 (3A + 2p)T0/p « c + 3a2 (3A + 2p)T0/p. This answer does not depend upon the form of the strain state.

13. The First Law of Thermodynamics. Perfect Gas. 57

13.15 T = T0 exp{M2/2cMA).

13.16 With the use of equation of the Poisson adiabat (al3.1) one obtains a = \frpfp = y/fRT-

13.17 For the process under consideration, d£|tin = 0 and by the kinetic energy theorem dA^ = —dA®. Denote the total mass of the gas by M ; p, p, T are assumed to be the same for all the particles. (I) The energy equation for the part (I) of the

^ P ,V rv

P4

I T=T,

in

\fh_

\ n

PF

Figure al3.1

process (Figure al3.1)

AU} = McvATi = 0 = Q[e) + A\e)

where AC/i is the total change in gas internal energy, Q\ is the total amount of heat received by gas and xf' is the work done by external forces in the process (I). With the use of the formula for dA® from the solution to Problem 13.3b one obtains

P2 l(e) iW

/ RTi Af> = -A\" = M I '-^-dp = -MFTi ]n{pi/pa) < 0

Q\ ,(«) _ Af] > 0, A I gas = _ 4 « A)" > 0

The gas receives the amount of heat Qj and perform the work A\ gas- (II) <5|j = 0, p = Cpy, 7 = Cp/cv = 1 + R/cv, C = const,

AUU

P3

'„ = - 4 ' = M\%dp=^- (Prl - PV) = Mcv(T2 - T,) ; J fr 7 - 1

_ A® An gas = A™ = McATt - T2) .

The gas performs work ,4n gas. (Ill) is similar to (I); (IV) is similar to (II), and P1/P2 = AJ/AS- As a result of the whole cycle, AU = 0, Q(e) = MR(Ti-T2) \a(px/p2)\ Agas = Q^- The efficiency equals to Agaa/Qi = 1 — T2/T\. It neither depend on M nor the degree of expansion during an isothermal process.

13.18 a) dqW _ divq = K ^

dt b)

^ = idiv(«gxadT) = «AT+ ^ ( g r a d T ) 2

at p p pal

13.19 The energy equation is a)

4 KY)+£&(*> 4 V< (-p«' + «3°^-J + P(F -v) + p

dt

b)

4 ( if) + ^ (cT+^) - Vi((-p + Adivt,y+

The equation of internal energy is a)

b)

c)

(dT kdT\ 1 / % J . . , . 2/x H / c A m dg, =v -57 + u N r r = -(Adivw - p) divu + — e y e y + - A T + —

\ at ax* / p p p 1 dt

aQmaBs

The equation of internal energy for the considered media is reduced to the classical heat-conduction equation if vk = 0, K/pc — const, dqmaas/dt = 0.

13.20 For an adiabatic steady motion, the energy equation has the form

d (v2 \ dU d v2 \ du _ . fc,

Using the continuity equation, one obtains

Vk(vkp) = Vfc (pvk^j = Vk{pvk)P- + pvkVk Q = pvkVk (A .

So, pvkd(v2/2 + i — U)/dxk = 0, i = u + p/p. If dr is directed along a streamline, vk = Xdxk; then dxkd{v2/2+i-U)/dxk = d{v2/2+i-U) = 0; i.e., v2/2+i-U = const along a streamline.

13.21 The equation of internal energy in this case has the form pcdT/dt = KAT. Obviously, T = T(z) where z-axis is perpendicular to the plates, so for a stationary heat conduction process in the resting layer dPT/dz2 = 0. The boundary conditions are T = 7\ at z = 0, T = T2 at z = h. The solution is T = Ti+fo-Tjz/h.

13.22 Let the z-axis of a Cartesian coordinate system be directed along the velocity of the moving plate, and the 2-axis be perpendicular to the plates. Obviously, all parameters are independent of y, vy = vz = 0, p,vx,T do not depend upon x, and the accelerations of the particles equal zero. Prom the Navier-Stokes equations (the equations of motion for a linearly viscous fluid, see Problem 9.19), one obtains vx = v0z/h, and exz = v0/2h. The equation of internal energy yields d?T/dz2 = —P-VQ/KH2. Let us denote a = pvl/nh2, and AT = T2 — T\. Integrating the equation of internal energy and using the boundary conditions T = T\ at z = 0, T = T2 at 2 = h, one obtains T = 7\ - a(z2 - zh)/2 + zAT/h. So, dT/dz = 0 at 2 = h/2 + AT/ah = zM. If 2M < h, then {dT/dz)z=h < 0, i.e., the heat flux at the hot plate is directed from the fluid to the plate (or equals zero at zM = h). Consequently, when zM < h, the hot plate is not cooled by the fluid whose temperature at the distance h from the plate equals 7i < T^. This effect is caused by viscosity. The condition zM < h has the form pv2 > 2/cAT. The maximal temperature in the stream at zM < h is Tmax = (Ti+T2)/2+ah2/8+(AT)2/2ah. The numerical values of Tmax are a) 30.02°C, b) 66°C. In the stream of an inviscid fluid, the heat conduction process is the same as in the resting layer (see Problem 13.21)

13.23 The boundary conditions in this problem are T = T\ at 2 = 0, qn = qz = 0 at 2 = h, i.e., dT/dz\z=h = 0 (see Figure al3.2) Using the solution to Problem 13.22, one obtains T = Ti + az(h - z/2) where a = pvl/nh2. The temperature at the heat insulated plate equals T2 = T^+ah2^ = TX+HV^/2K. a) T2 = 30.005°C; b) T2 = 39°C; c) T2 = 22430°C. d) No, it is not. e) When there is heat equilibrium between the thermometer and the fluid the heat flux from the fluid to the thermometer is absent, so the thermometer indicates temperature T2 instead of T t.

13.24 The equation describing a nonstationary heat conduction process in a homogeneous medium at rest has the form dT/dt = x&T where x = n/pc. The


q.=0

77777777777777777777777777T T,

Figure al3.2

T,

Figure al3.3

initial and boundary conditions for this problem are T = T0 at t = 0, z > 0; T = 7\ at 2 = 0, t > 0. Consequently, T can depend only upon the parameters t, z, x, T0, T\. According to the II-theorem of dimensionality theory (see Section 36), T = T\ f(€,T0/Ti), £ = z/y/tx- Substituting this expression for T into the equation of internal energy, one obtains the equation for / : / " + f/ ' /2 = 0. The solution of this

«/a _f2 equation is / = c / e 4> d£i + C\. Using the boundary conditions and the fact that o

c In

/ e - t f d£i = \/7r/2, one obtains T - Ti = (T0 - Ti) • 4 ; / e -4?d&. The function o o ■4j Je_ £ i cJ i is usually denoted by erf(i), so the solution can be written in the form

T —Ti = (T0 — Ti)erf(z/2-v/tx). A solution in which the required functions depend on z and t through only their combination of the form z/ta is referred to as self-similar. The solution of this problem is self-similar.

13.25 Heating a layer is described by the equation dT/dt = ^AT where x = n/pc is the thermometric conductivity (see Problem 13.24). Consequently, the time


for which the temperature at the upper boundary reaches T2 depends only upon the width of the layer h as well as upon \, T\, To, Tz- t\ = t\{h, x, Ti,T0, T2). Application of dimensionality theory (see Section 36), shows that ti = h2f(Ti/T0, T2/T0)/x- Thus, when x, T\, To, T2 are constant, the time t\ is proportional to h2. So to fry one frying-pan of thick cutlets takes twice as much time than to fry consequently two frying-pans of cutlets that are half as thick.

13.26 The density of internal energy is u = cvT, the internal energy of air in the room of volume V is U = cvT ■ pV = cvpV/R. Since the pressure does not vary, U also does not vary.

13.27 The motion equation of the bullet is md?x/dt2 = (p — pa)S where x is the distance along the barrel, S is the cross-sectional area of the barrel, pa is the pressure ahead of the bullet (assumed to be equal to that of atmosphere), p is the pressure behind of the bullet, m is the mass of the bullet. The expansion of the gas behind of the bullet is adiabatic, so p = P0P11 Po, with p = Mo/Sx where M0 is the mass of the gunpowder turning into the gas. Let us denote the length of the barrel with /. The speed of the bullet reaches the maximal value at the end of the barrel if P(0 = Pa, i-e-. I = {PolPaY^Vo/S where V0 = M0/po- The initial pressure p0 can be estimated as follows: the initial internal energy per unit mass of the gas arising at the expense of burnout of the gunpowder is cvT0, it equals the combustion heat Q; Po = poBTo = p0RQ/cv = poQ{l - 1)- The calculation yields I « 110 cm.

13.28 The mass of air « 144 kg.

13.29 A » 400 J.

13.30 The final temperature T is equal to 24.5°C. It is obtained from the e q u a t i o n Cwater • Tnw$xer(T — T0 water) = Cstone ' m s tone (^0 stone ~ ■* )•

13.31 900 K. See the solution to Problem 13.17.

13.32 To estimate the quantity of required work, let us assume that the volume initially has the shape of a sphere. Then V0 = AT:RI/3 = 47rr3n/3 where RQ is the radius of the initial volume, r is the radius of a droplet, n is the number of droplets. Consequently, E - E0 = 3V0(l/r - 1/flb) « 3V0/r = 6 x 106 cm2; A = a(E - E0)) w 43.8 J; T - T0 = A/mc w 0.01 °C, Ah = A/mg « 4.46 m.

14 The Second Law of Thermodynamics. Entropy. Gibbs Identity.

14.1 Use the formula dS = dQ{e)/T and solution to Problem 13.17. AS, = MR In P1/P2 > 0; AS,, = 0; AS,,, = MR\np3/pA < 0; AS,V = 0; AS = AS, + AS,, + AS,,, + ASIV = 0

14.2 s = cv InT/p 7 - 1 + const = cv Inp/p1 + const = cv lnu /p 7 - 1 + const = Cj,lnT/p'7-1^7 + const where 7 = Cp/cv = R/cv + 1 is the adiabatic exponent.

14.3 No, they do not.

14.4 s = (3A + 2fM)aIi(£ij)/po + c(T — T0)/T0 + s0, so is the value of entropy at T = T0, e y = 0.

PoF = 2 J i + M M - "(3A + 2/i)(T - T0) J i -

- poSoU - To) + const 2Tf 0 A T1

Po" = -^-^? + M -a(3A + 2fj.)(s - s0)J\+ I c

T0po{s - s0)2 . . - + T0po(s - s0) + const 2c

where A^ = A + (3A + 2fj.)2a2T0/(p0c)

14.5

u = cvT + const = Aip~'~les/cv + const = -.——r- + const = (7 - l)P

A2p 1 e'l"* + const

T = cvT(Bx - ln(r/p7 _ 1)) + const = B2p1~le'lev{cv - s) + const =

--(B3 - hi(p/p7)) + const = Btf^e'^Bs - s) + const 7 - i-P

{ = 0^ + const = C1p7~V / c v + const = —— - + const = 7 - l p

C2p 1 e',Cr + const T

* = CpT(D! - 7 In ——) + const = D2p7_1e s / cv(cv - s) + const =

7 P 7 - l p

Here, J4J, S<, CJ, D{ are constant

-(£>3 - ln(p/p7)) + const = D4p 1 e*/c»(£>5 - s) + const

14. The Second Law of Thermodynamics 63

14.6 Use the energy equation and the second law of thermodynamics.

14.7 For an incompressible fluid, s = s(T). This follows from the Gibbs identity and from the fact that u = u(T). So s = const when T = const. For an adiabatic motion, ds = d\s > 0.

14.8 During isothermal compression of a perfect gas, u = const, then, from the Gibbs identity, one obtains Tds = —p dp/p2 < 0 when dp > 0.

14.9 For an incompressible fluid u = u(T), du/dT = c > 0. The equation of internal energy for adiabatic processes yields du = cdT = 0 for an ideal fluid and du = cdT = Tkmeicm dt/p > 0 for a viscous fluid.

14.10 The entropy of each particle and of the whole layer does not vary, because the state of each particle of a resting medium does not vary during a stationary process. Since the heat conduction process is irreversible, for unit time per unit mass, the quantity of entropy d^s/dt = -(qkVkT)/'(pT2) = K.(dT/dz)2/{pT2) = K^Tf/ipT^h2) where AT = T2 - 7\ is produced. The entropy production is maximal at the cold boundary and minimal at the hot one. The total entropy production in the layer per unit area of the plates is

diS } dis _, K ( A T ) 2

— = / p— dz = — — dt J H dt hT{T2

The density of entropy input is

** ^ l „ g * ^ K(AT)2

dt p kT pT^h2

i.e., the entropy departing each particle is more than the entropy entering it. The quantity of entropy entering the whole layer from each unit area of the hot plate for unit time is (—qn/T)? = /tAT//iT2; conversely, the quantity of entropy transmitted to unit area of the cold plate from the layer of the medium is /cAT//i7\. As a result, the quantity of entropy K A T ( 1 / T I — l/T2)/h exactly equal to the produced entropy diS/dt departs into the surrounding medium.

14.11 The entropy of each particle and of the whole layer does not vary

diS 1 k 1 k pfvoV, K (dT\2

Tt=pTT ekm-pl*qVkT--p{-h) +pT2-\Tz)

If zM < h entropy is transmitted from the fluid both to the cold plate and to the hot one. At T2 = T\ the quantity of entropy departing the fluid through unit area of every plate for unit time is /ca/i/27\, i.e., the numerical values for the area 1 cm2 and the time 1 s are a) « 17 erg/K, b) « 7 • 103 erg/K.

14.12 At melting, entropy increases by the quantity Q/T « 150cal/K; at freezing, it decreases by the same value.

14.13 The entropy increases by the quantity Q/T « 700 cal/K.

14.14 The expression for the density of entropy of a linearly thermoelastic medium has the form (see Problem 14.4)

g q J i f o ) , c (T-Tp) , M 3 A + 2M) A^ s = 7- . H — 1- So, t, = — , a (l-2a)p T0 "' X + p, ' 2(A + /x)

where T0 is the temperature at which the strains equal zero in absence of stresses. If a long rod is extended by forces applied to its ends, the stresses and strains in it can be assumed to be independent of the coordinates, with pn = T/T, and the other components of the stress tensor equal to zero [T is the value of the extending force, E is the area of a cross section, the x'-axis is directed along the axis of the rod), a) During isothermal extension at T = T0 = const, according to Hooke's law, p n = EEU, e22 = £33 = -aeu; if en = 0.001, T = pu ■ £ = 21.6 x 103 N, Jifoj) = 5 x 10~4. The change in entropy of the rod is AS = mAs = lT,aE/{\ - 2a) Ji(£y) « 0.24 J/K. b) During adiabatic extension, the entropy remains constant, the temperature decreases: T - T0 = -EaToJi(eij)/(cp(l — 2a)). The change in temperature causes temperature strains, so, during simple adiabatic extension (see Problem 13.13), Pn = £ad£ii! £22 = £33 = —ffadEii) "adiabatic" Young's modulus E^ and Pois-son's coefficient CTad are bound with "isothermal" E and a by the relationships £ad = £(1 + A,), o^ = a{\ + A2), where A, = 6/{E/2{l - 2a) + 5(1 + a)), A2 = Ai(l + a)/a, 6 = A - A^ = E2a2T0/(l - 2a)2pc. For this problem, 6 w 21.85 x 108 N/m2, £«, = E • 1.01, T^ w 21.8 x 103 N, T - T0 « -0.2°K.

14.15 Since there is no energy input from the outside to the gas during the considered process, the total energy of the gas does not vary. This means that the final temperature of the gas equals its initial temperature. The final density of the gas is half as large as the initial density, so the pressure equals 0.5po- The change in entropy of the gas is AS = M ■ cv \n(Tp0

1~l/(T0p1'-1)) = MR In2 = Ro In2 « 1.38 cal/K.

14.16 The solution satisfying the boundary condition v = v0 on the piston and the mass, momentum and energy conservation laws on the discontinuities I and II has the form

PMI = j + 7(7 + 1) 0 ± 7«o Po 4ag ao ^ 16a2,

1+(7+l)%2

14. Restrictions on the Form of Constitutive Relationships 65

piston

Dtt

v = 0 Po Po

II V = V„

Pn Pn

I D,

Pi p. v = 0

Po Po

/> /> / / / / ■7-7-7-7-7-7-7-7-7-7" / / ? / } } J } } ? / 7 / } } } } > } 7

PO

Figure al4.1

(7 - 1) + (7 + l)p,,„/p0

(7 + 1) + (7 - 1)PI,II/PO

1 + (7 + l)p,.„/po)

where a%

0,.„ = ± ^ h . 7Po/Po, 7 = Cp/cv, the "+" sign is for region I, "-" is for region II,

Z?in are the speeds of discontinuities I or II relative to the walls of the tube in the direction of motion of the piston. For this problem, pi/po = 1.490, pi/po = 1.327, Di = 401.454 m/s, pu/po = 0.658, pu/p0 = 0.743, Du = -283.201 m/s. The change in entropy of unit mass of the gas caused by passing the jump I is ASi = cv In (p,Po/PoP7) = 0.00044 cal/g • K; for the jump II it is A5„ = —0.00050 cal/g • K < 0. Consequently, the given solution for region II contradicts the second law of thermodynamics. For region II there exists continuous solution (see Problem 24.30)

15 Restrictions Imposed by The Laws of Thermodynamics on the Form of Constitutive Relationships.

15.1 Use the Gibbs identity and the condition under which ds is the total differential of s(p,T).

15.2 Using the first and second laws of thermodynamics for an elastic medium, one obtains du = pij de^/p + Tds (the Gibbs identity for elastic media). Writing out the condition under which ds is the total differential of the function s(£ij, T), one obtains the desired formula.

15.3 The Gibbs identity can be transformed as follows: du = pdp/p2 + Tds = T(Rdp/p + ds) = Td(R\np + s) = Tdtp, tp = s + Rkip. Hence, u = const if

<p = const, i.e., u = u(<p), T = du/d<p = T(ip), so (p = <p{T), i.e., u = u(T) (this deduction is valid if it is supposed beforehand that s, u may depend upon T, p, but do not depend on the derivatives of T, p with respect to time). When p = const, du = dq and du = (du/dT)dT, dq = cvdT, consequently cv = du/dT = cv(T), u=fcv(T)dT.

15.4 du = f(p)Tdp/p2 + Tds = Td<p where tp = J f{p) dp/p2 + s; the other arguments are similar to those in the solution to the Problem 15.3.

15.5 Use the formula of Problem 15.1.

15.6 Using the Gibbs identity, one obtains

du = T(Rdp/{p(l - bp)) + ds) - adp

i.e., d(u + ap) = Tdip where (p = s + R Inp — R ln(l — bp). So u + ap = f((p), T = d(u + ap)/dip = T(tp), y = <p(T), u + ap = fi(T). When p = const, du = f[ (T) dT = cv dT; consequently, f\ = / cv dT. Besides, when p = const, du = T dtp, i.e., dip = cv dT/T, <p = / c v dT/T. Hence, s = /c v dT/T + R ln((l - bp)/p) + const.

15.7 Use the Gibbs identity and the definitions of T, i, \t (see Section 12)

15.8 Use the formulae of Problem 15.7.

15.9 b) To obtain the formulae of Murnaghan, it is necessary to write the equations of thermodynamics in the form including Eulerian components of the corresponding tensors. Differentiating with respect to time the equalities relating components of the strain tensor in a Lagrangian £* and Eulerian x' coordinate systems

„ _ dxk dxl

one obtains the following formula for the derivatives of the components of the strain tensor in an Eulerian system with respect to time in terms of components of the strain rate tensor eap

^ f = eaM^ ~ W? ~ tftf) + "«*(e?«f + $$) ■ Substituting these expressions into the equality

du . du . paP J± _ , — ds + -— deij = — eaP dt + Tds

as OEij p

and using the fact that the quantities eap, ds are independent and the fact that {du/dcij) deij must not contain terms with u>a0, since the internal energy does not vary at a rotation of a medium as a rigid body, one obtains the desired formulae.

15. Restrictions on the Form of Constitutive Relationships 67

15.10 Aijkl = Aiilk, Aijkl = A>w due to symmetry of p^ and eki. Use the formulae of the Problem 15.9 (taking into account that p « po) to show that dpij/deki = dpkl/deih that is Aijkl = AkliK

15.11 Use the fact that the rate of entropy production due to viscosity can be written in the form

diS A + 2u , , , 2u 7 , d. d 1 , dis

It = i r ^ + 7J2(<)'e" s e ° " ^Jl(e'^ *" - ° • 15.12 a)

dT -~dp V*i;mdi + s dT - dq" + H ■■* dt < 0 Pz P PT

b)

c)

dt - — V*i/mdi - Tds - dg** + ?—£-dt < 0

dv Tmk akS7kT <M-JL Vkvmdt + sdT- dq" + ^—^dt < 0

P P PT

15.13 The dissipation inequality has the form

du p\ j (du ^\ j_ , du j , du fc ~km

— -JL\dp+l — -T)ds + - dekm + j-jdq* ekmdt < 0 dp p2) \ds ) dekm dqk p

Since dekm and dqk are contained only in the underlined terms, and the terms depend upon them linearly, these terms can be both more and less than zero if du/dekm =fc 0, du/dqk ^ 0. This contradicts the dissipation inequality. Hence, u = u{p,s). For quasi-static processes, Tkmekm dt = 0, and p and s can vary independently. So du/dp = p/p2, du/ds = T; moreover, the last equalities are always valid, since u, p, T do not depend explicitly upon ekm-

15.14 The first and second laws of thermodynamics yield (p = dp/dt)

du , du , du ,. du ... p . p.'p Ysds + d-p

dp+d-pdp+d-pdp=7dp+7

One can see that du/dp = 0 , since dp =P dt is a part only of the singly-underlined term and can vary independently of the values of the quantities ds/dt, dp/dt, d?p/dt2. Next, dp = p<dt is a part of the two doubly-underlined terms, so du/dp = fip/p2, i.e., u = w2/2p2 + f(p,s).

15.15 The deductions are obtained (similar to Problem 15.14) from the relationship representing the first and second laws of thermodynamics

15.16 Let u = u(sij, Vfc£y, s), Tds = dq, then

du , du __ du , p,j , , . . _ , ^— cfey + —-— dV tey + —ds = —deij+ dq" +Tds . OEij dVkEij OS p

Next, dVfeEy = Vfcdey, so

du __ _ / du \ , __, ( du , dVk£ij = -Vfc p ^ = — efey + Vfc p ^ = — cfei. "aV t £y " ^ " V dV*£y7 —J '" VraVfc£y

Consequently, it is possible to assume

pij du 1 „ / du \ J . . 1 / du J

p 0£y p V dVkEijJ p V 9"fc£y

15.17 In the considered case under condition the dissipation inequality has the form

-^-dp+—dT + —— dejbn r d p -#P dT dekm p2

r i jey d 0. But, since ey can have any value either greater or less than zero, and T1J = const, r'-'ey can be either greater or less than zero; this fact contradicts the dissipation inequality. For example, let Couette flow be considered (see Problem 13.22) for this medium for which vx = ky, vy = 0, vz = 0, exy = eyx = fc/2, the other ey = 0, T = const. Then the dissipation inequality takes the form (r12 + pJc)fc > 0. It is not fulfilled at 0 > k > - r 12/p if r 1 2 > 0 .

15.18 21.

15.19 6.

16. Thermodynamics of media with internal angular momentum. 69

15.20 Supposing that the dependencies between thermodynamic forces and fluxes are linear, one obtains

q = - L l l g r a d T / T 2 - L 1 2 g r a d ( ^ ^ ) ,

J = -L2l gradT/T2 - L22grad ( ^ ^ ) . (al5.1)

By Onsager's principle Z*i2 = L2\. Next, one has

grad(/i*/r) = - / i f cgradT/r2 + ( g r a d e r ,

grad/ifc = - s f c g r a d T + f - ^ | grade,

So the relationships (al5.1) can be rewritten in the form

q = - ( L „ + ALi2)gradT/T2 - Ll2Bgrade , J = -{L21 + AL22)graAT/T2 - L22J3grade ,

A = fi2-nl+T(s2-si), B = -f^^~-- (a15-2)

The coefficient a = BL\2 determines Duffor's effect, and the coefficient 0 = (L21 + ,4L22)/T2, Soret's effect. Since Ln = L2U a = B{/3T2 - AL22). The formulae (al5.2) make it obvious that the heat conductivity coefficient in the mixture equals (Ln + AL\2)/T2, and the diffusion coefficient equals BL22.

16 Thermodynamics of media with internal angular momentum.

16.1 a) Using the formula for the velocity distribution v = vc + u x r where vc is the velocity of the center of mass, r is the radius-vector relative to the center of mass, one obtains from the kinetic energy theorem in differential form

d(v2/2) . „ {i P^^-L=pF>vt + viVjP'> ,

taking into account that ey = 0 and integrating over the volume of the body V,

\ S ( M ^ + W " * ) = Jp*'-vdV + Jpn-vda + Jpijujij dV

where M, I, u> are the mass, the inertia tensor relative to the center of mass and the vorticity vector, Uy = CytW*. b) Due to the equation of angular momentum VjQ,J = e,J*Pj/t, the work of internal surface forces connected with the antisymmetric part of the stress tensor is

Jfluij dV = J Vj(Qiju>i) dV = J Qn ■ udo = 0 V V £

where it is taken into account that VjW, = 0.

™2

c) For a rotator, £kin rot = — .

16.2 a) i p j t ( v 2 + m2) = pF-v + ph-f2 + V , ^ + Q«n0

- p % - Q^ilt -pijeijk(nk - u>k).

b) -py£y*(fi* - tJk) = 0 if pij = p>{ or Slk = wfc. c) According to the theory presented in the introduction to Sections 12 and 16, the change in the sum of the kinetic energy of a medium and kinetic energy of internal rotation equals the sum of the amounts of work of external body forces and couples, external surface force and couples, internal surface forces and couples, and internal surface forces on variation of orientation of the medium.

16.3 p ^ = p % + Q ^ f l i + p % t ( n * - UJ") - V + p d<lmaaa d t r ^ , ^ . , . . , , , , . y c v . . „ , ^ • ^ d t ■

16.4 Analyze infinitesimal orthogonal transformations of the coordinate system x" = xl + tx*kXja.k, \a\ —» 0. The function u, as a scalar, does not depend upon o if its arguments are subjected to these transformations. Differentiating the condition of invariance u(n", (V,n*)', gxi) = u(nl, V,n*, g**) with respect to a' and, next, supposing that a = 0, one obtains

( du du _ du _ \ ,,-,

where eu ' are the components of Levi-Civita tensor.

16.5 a ) l ) jti = -f — gV, Q»=0; dp

2) pij = - p . _ tVnk - p2-z-gij, Qij = - p - _ te'fc'nt. Here, it is taken into ac-o\/jnK op aVjnr count that

-(V<n ;) = Vi(ejklnknl) - V ^ V ^ .

b) rii = cos cxat, ri2 = sin cxat, n3 = 0 (a = 2,3).

16.6 The quantity pj£v and Q'^ are determined by the formulae of the preceding du problem, T = —, as

16.7 a) -V,T = «,/«, pV^ = A^efcfc + 2/xe" + \ eiik(Qk - uk),

Qltrev = m9tiekk + ne*J + ce'jk(Qk - Uk). Due to Onsager's principle, m = n = c = 0. Then

a = p ^ + ~pY(A(ekfc)2 + 2 / i e° e* J + b{n ~ w|2) • b) - V ^ = - + a ( a - LJ{),

J (py + p")ta-v = W t + 2 ^ , p^ey* = ^ + 6(fi* - wk);

a=p^ + pY {X{e"k)2 + 2/ie"ey + f ?i(n< ~ W<) + b]f2 ~ W|2) •

c) - (p,J +P7 ') is determined by the linear formula of Problem 11.16 where c2 = c5.

P L V ^ * = ^ + MOfc " "*) + nkni(?f + WiV - w%

-ViT =^-+ cuiQi - wj) + mriji— + a2(nJ - w»)).

16.8 a) a = - i ^ |q + KO(/J - w)|2 + - L (6 - ^ L ) | f l - o f

1 2 1 H—— ((A + -p-)(ek

k)2 + 2/ie':'efi) where efy = ey - - <?yefcfc are the components of pT 3 3

/ca2

the deviator of the tensor e. The desired inequalities are K > 0, b — — > 0,

16.9 a) Due to the equality Qrevn« = 0 ^ d differential conditions for the function u to be scalar,

b) ai+a.2 = 0, bi+b2 = 0,

pLv^y* = (4 - n'nk) ( ^ + 6(0, - «,)) •

16.10 It follows from the evolution equation for u that KT"(X2) + e2(4/i + A(l — n3)) = 0- Then, if x2 = ±— corresponds to the boundaries of the stream,

T = T 0 - ^ ( 4 M + A ( l - n 2 ) ) ( * 2 - ^

e2h Qn = ~2^~ (4/" + A ( 1 ~ " ^

where, for the cases a) and bi), n\ = 1, for the case b2), n3 = 0. Since (2 = 0,

16.11 a) Substitute the quantities

, du ds dg'

/ ^ = ( J l f x J f ) f c - p % f c l

M / d M „ , ,\ H = — + T — « x M , x V dt )

p« = - ^ + 2 / ^ + ^ ( ^ - 0 , * ) ,

into the equation of heat flow, and use the condition for incompressibihty V<t;' = 0. b) The equation of heat flow in the first approximation with respect to \U\T —► 0

yields nT"(y) + e2(4/i + T\HJ_|2) = 0. Then, if y = ±— correspond to the boundaries of the stream,

T = T 0 - | l ( 4 / i + r | H 1 | 2 ) ( 2 /2 - ^

Since dM ~dT 0,

<7i =

dq" ~dT

93 = 0 , «72 = ^ ( 4 M + T | H 1 | 2 ) ( J / 2

= o.

Chapter 4

Discontinuity Surfaces in Continuum Media

17 Conditions across Discontinuity Surfaces 17.1 Let D denote the velocity of the surface E. For a point situated on the

surface dx'/dt = D* and f{xx, t) = 0, so |£ + A J!£ = 0. A unit normal vector to E is n = grad f/\ grad / | , so Dn = D ■ n = D{ J£ / | grad f\ = - f /|grad / | .

17.2 Let us mark the values of all quantities on different sides of E by subscripts 1 and 2 and calculate the variation in the time of the function u(x, t) at a point Xi on the plane E. The point moves with the velocity dx^/dt — D = — §£/§£ (see Problem 17.1), so

( du\ (du\ (^u\ dxd _ (9u\ n (^u\

(j).-(S).-(i).-We wrote these formulae assuming that x^ belongs to side 1 or to the side 2 of the plane E respectively. But at any instant Ui = u2 at the point xj as w(x,t) is continuous on E, so {du/dt)\ = (du/dt)2 or

(S)-(SHt)-(£),-This relation is a kinematic condition for u(x, t) across E. Using the notation [ip] = ipi — ip2 one can rewrite the kinematic condition in the form

dt = -D du dx

75

76 SOLUTIONS. DISCONTINUITY SURFACES

17.3 a) Let us mark the values of all quantities on different sides of E by subscripts 1 and 2 and write the equations at points tending from sides 1 and 2 to a point lying on E. We will have, in the limit

~dt du{

~dt

+ aj. duj dx + b, 0 ,

-I- a*(^)2 + 6' = °

The values of ay and fej for these two systems coincide as they depend only on x, t and Uj which are continuous on E, so

= 0 . dui dt + 0.ij dx

These relations are dynamic conditions for Ui(x, t) across E. b) Using kinematic conditions (see Problem 17.2) one can rewrite the dynamic equations in the form

(ay - D6ij) dui dx

= 0 .

These relations can be regarded as equations for [duj/dx], and the determinant of the system must vanish for at least one of [duj/dx] being non-zero. So

det(ay - D6ij) = 0

and D is an eigenvalue of the matrix ay.

17.4 See solutions to Problems 17.2, 17.3. a) Kinematic conditions:

= 0 . dp dt

+ D dp dx

= 0 , dp dt

+ D dp dx

= 0 , dv dt

+ D dv dx

Dynamic conditions:

dp dt

'dv dt dp at

+ v

+ v

+ v

dp dx dv dx dp dx

+ P

1 + -P

+ 1P

dv' dx dp dx dv' dx

= o ,

= 0 ,

= 0 .

b) Possible speeds of weak discontinuity planes are

Dx = v , D2 = v + sjip/p , D3 = v- sfrPlP

17. Conditions across Discontinuity Surfaces 77

17.5 b) (vnl - D)pi = {vn2 - D)f>2 = m, or [p{vn - £>)] = 0; m(«! - v2) = p„i - pn 2 , or [p(u„ - £>)v - pn] = 0; m ( u ? / 2 + U! - u | / 2 - U2) = P„l • «1 - Pn2 • V2 ~ Qn\ + 9n2. or [p(vn - D)(v2/2 + u) - p n • v + <£] = 0. Here, £> is the speed of the discontinuity surface (along the normal n to it).

17.6 On a tangential discontinuity, vni — D = vn2 - D = 0, i.e., v„i = w„2- The conditions across such a discontinuity are

P„i = P„2. Pm K i - Wra) - 9 i + 9 2 = °

On a contact discontinuity, vTi = VT2; hence

Pnl = P n 2 . 9 n l = 9n2

17.7 It is the condition vn = Dnovv- grad/ = D ■ grad/ = -df/dt at / = 0 (see Problem 17.1). So

M-where ufc are the velocity components of the medium at the points of the surface / = 0. If / = 0 is the surface of a tangential discontinuity (there are media in the vicinities adjacent to the both sides of the surface), this condition must be satisfied for the velocities of the both media.

17.8 Write the required condition in the coordinate system whose coordinate lines x1 and x2 lie on the discontinuity surface x3 = const = C. Obviously, the derivatives dw/dxa (a = 1,2) are continuous, since the vector w(x1,x2,C) is continuous on the surface. If the basis vectors ek of the coordinate system are continuous in the vicinities adjacent to the sides of the surface, one can conclude that VaWk (£ = 1,2,3) are continuous as well. Consequently, the components ea0 (a, j3 = 1,2) are continuous. Continuity of w leads to continuity of the derivative ^ + Dk^ where Dk are the components of the velocity of the point of the discontinuity surface, so

dt dx3 = 0 (al7.1)

Taking into account the identity ^ + ^ f c f p r = w and continuity oidw/dxa (a = 1,2), one obtains [v] + [{D3 - v3)j$] + [va]%p; = 0. This is the required relation. If x* are the coordinates in the initial Lagrangian coordinate system, then this relation has the form [v] + D3 [ 0 ] = 0 immediately following from (al7.1).


17.9 Let us mark the values of parameters of the media in the vicinities adjacent to the different sides of the interface by indices 1 and 2. The kinematic conditions for all the cases have the form |£ + v\t2 • grad / = 0 (see Problem 17.7). The dynamic conditions are (see Problem 17.6) a) pi = p2, q^ = g*2 b) —pin* = — p2fi' + T

2nj,

i,j = 1,2,3; qlx = q'n2 If the second fluid is linearly viscous and isotropic (obeys Navier's law (12.32)),

then —pin* = —p2nt + Xdivv^n* + 2p.e2rij; Ini = 9n2 c) -pin* + T^rij = -prf + r^n,, T\3nj(vTli - vT2i) - q^ + q^2 = 0 e) Pi7"; = P2

knk, P\nj(vni - vT2i) - q^ + q*n2 = 0 If the media are linearly elastic and isotropic (obey Hooke's law (12.35)) and thermal stresses are neglected, then the first group of these relations takes the form Xiie^in* + 2n^v.j = A2(efc

fc)2ni + 2/x2e2:S

17.10 The jump conditions for an ideal fluid are

(vm - D)pi = (vn2 - D)p2 = j

j{vin - «2n) = P2 - Pi; «1T = «2T j{v\/2 + wj - v\j2 - u2) = p2vn2 - p\vnl + q^2 - q^

17.11 a) The equation of the Hugoniot adiabat is

P2 = (7 + 1)P2 ~ (7 ~ l)Pi 7 = f z Pi (7 + l)Pi - (7 - 1)P2 ' cv

b ) S 2 - S i = c v l n ( ^ l ^ i f e - t e ) 7 ) . c) For the two cases: 1) if there is no jump, pi = p2, p\ = P2; 2) if 7 = 1.

17.12 The relations representing conservation of the y-component of the momentum flux and continuity of the y-displacement of the medium on the surface have the form

pD[v] - [T] = 0, [v]-Dfr] = 0, [v]=v2-v1 (al7.2)

where p = const is the density of the medium, D is the speed of the discontinuity relative to the medium, v = dwy/dt is the y-component of the velocity of the medium, and [ ] denotes the jump of the corresponding quantity. These equalities lead to the formula p0D2 = [T(7)]/[7] from which it is obvious that the inequality [T(7)] /[7] > 0 must be satisfied. Thus, the speed of discontinuity is determined by the slope of the secant to the curve T(^), and the velocity of an infinitesimal discontinuity (of "small disturbance") by the slope of the tangent (see Figure al7.1). The relations between the speed of a discontinuity and the speeds of the small disturbances mentioned in the problem are the conditions of evolutionarity (see Section 24). These relations have a geometric interpretation and determine mutual disposition of the tangents and secant


Yi - *Y

Figure al7.1 Evolutionary discontinuities.

to the curve at the points 7 = 71 and 7 = 72- Since the stress T according to the conditions of the problem does not depend upon the entropy s, the internal energy U can be separated into the mechanical and heat parts (see Problem 15.9)

U = U{n) + 0(s) , U{i) = -J r ( 7) d1 , = T > 0 . (al7.3)

To determine whether the entropy increases or decreases, examine the condition for

Figure al7.2

continuity of the energy flux on the discontinuity surface

PD[U) + ^pD[v2} + H = 0 .

Supposing, for simplicity, that Vi = 0 and eliminating [v] and [r] with the use of the first and second of equations (al7.2) one obtains from here

[PU] = i ( T l + T2)[7] , or [p(J] = ^ ^ b ] - \pU) .

It follows from the second of equations (al7.3) that [pU] is equal to the area (with the corresponding sign) contained between the secant and the curve T(7) on the r-7 diagram (see Figure al7.2). If this area is positive then [s] > 0 since the signs of [U] and [s] are the same (see the last condition (al7.3)). Figures al7.2a) and al7.2b) correspond to jumps 71 —» 72 with increase and decrease of entropy respectively. The jump shown on the Figure al7.2b) can not exist as it is inconsistent with the second law of thermodynamics.

17.13 The motion in the vicinities adjacent to the both sides of the discontinuity satisfies the relationships

v pV = c , — = m (al7.4)

where V = 1/p, v is the gas velocity relative to the discontinuity, c and m are constant. From the relationship for the momentum fluxes across the discontinuity, one obtains

p-pA = -m\V-VA)+J

trr Figure al7.3

\p]+m2[V] = J (al7.5)

(where [ ] denotes the jump of the corresponding quantity across the discontinuity). On a p-V diagram, if m and the initial state (point A) are given, the curve representing the first formula in (al7.4) crosses the straight line (al7.5) whose slope is at two points B\ and B2 (or nowhere), see Figure al7.3. For the curve (a!7.4),

-m

dp dV

_\_dp V2dp

a'

where a = Jdp/dp is the speed of sound. Taking into account the equality m = v/V, one obtains

> al < al If the input of momentum J varies continuously, and m = const, then, when J varies monotonically, there is no transition through the speed of sound in the flow.

17.14 a) Let us consider the flow with the velocity field r

v = —y at — h <y < h Mo

T Ji i \ v = —y + Th[ at y > h

Mi \Mo Mi / v = —y — Th\ 1 at y < — h

M2 \Mo M 2 / where r = rxy = const is the value of viscous stress on area elements y = const. The difference of the speeds is At; = v(h) — v(—h) = 2r/i//i0- Assuming that r and Av simultaneously do not tend to infinity, one obtains, for h —> 0, the following boundary conditions at y = 0

An —» 0 at fi0/h —► oo aAv -» 2r at u0/h -» a (al7.6)

r —» 0 at fi0/h —► 0

b) Let the equation of the planes be y = ±H, v(H) = —v(—H) = v0 and there is a layer of low-viscosity liquid at — h < y < h, h —* 0. Using the Navier-Stokes equations (see Problem 9.19), the continuity of rxy and the boundary conditions (al7.6) one obtains

at/io/fc-xjo Aw = v(h) - v{-h) = 0, T = w ( 1 / f f l / w )

a t r t , /A-»a Aw = 2r/a, T = H ( 1 / f t * ) w ) t 3 / . at fi0/h —» 0 Aw = 2w0, r = 0

17.15 Write the relationships across the interface between rain and water in the frame of reference that is connected to the interface

pvn = p0vno (= j ) ; pvl+p = povl0 + p a t m ;

VT = vTo ;

j (c(T -T0) + ^ - ^ \ = p a t mwn - pvn . (al7.7)

Here, the quantities with the index "0" relate to rain, those with no index, to water. The density of internal energy is taken equal to cT. The pressure in the medium modelling rain is assumed to be atmospheric. The work of atmospheric pressure forces appearing in (al7.7) equals patmwn, but not Patm nOi since the work is done on droplets, and the total volume of droplets falling per unit area of the interface per unit time is vn. From the relationships written out, one obtains the values of velocity, pressure and temperature in water under the interface: vn = vnoe, vT = v-^, V = Patm + j 2 ( l - e)/po, T = T0 + wj[0(l - e)2/2c, where e = p0/p.

17.16 Let us suppose that the interface between rain and water is composed of two planes passing through the edge of the roof, the velocity of flow behind of every plane is constant everywhere, its direction is given (the stream is parallel to the roof). Examine one of the pitches of the roof. Choose a Cartesian coordinate system so that the z-axis is directed along the velocity of rain v0, the axis x is perpendicular to v0 and to the edge of the roof. Denote by a and 0 the angles of the pitch and the interface with the direction of v0. From the conditions on the interface (see the Problem 17.15), one obtains vn = vn0e, vT = vTo, i.e., vxcot/3 -vz = -evz0, uxtan/? + vz = vz0 where e = pa/p. Hence, taking into account that vz/vx = cot a, one obtains the equation determining tan/?

t an 2 / 9 - t an /3co ta ( l - e ) /£ r + l /e = 0 ,

i.e, 2 tan /? = cot a (± - l) ± i/cot2 a ( i - l) - \. If cot a > 2y/e/(\ - e), there are two possible angles of the interface with v0 and, respectively, two values of velocity and pressure. The choice of one solution or the other depends on the conditions determining the subsequent flow of water (e.g., the conditions on the lower edge of the pitch) and cannot be done within the given setting of the problem. Thus, solutions of the type under consideration do not exist when cot a < 2y/e/{\ — e).

17.17 Prom the momentum conservation law for the whole volume of the barrel projected onto the horizontal plane, one deduces that the speed of the cart equals the speed of the jet flowing out from the hole (relative to the cart). This speed can be found using the Bernoulli integral along the streamline originating under the interface between water and rain in the barrel and getting into the jet flowing out from the hole. Answer: the speed of the cart equals J2gh + v*e where e = p0/p, p is the density of water.

17.18 When the function <p{p) is smooth, one obtains (the x-axis is directed along the traffic from left to right) ff+^gj^ = 0, or, f?+c(p)gf = 0 where c(p) = <p'(p) is a decreasing function of p. Applying the method of characteristics (see Problems 24.26, 24.28 and solutions to them), one finds the solution of this equation p = p0 at x = x0 + c(p0)t. The characteristics do not intersect if dp/dx < 0. The relationship following from conservation of the number of cars across a discontinuity has the form (Pi — P2)D = tp(p\) — f(pi) where p\ and p2 are the values of p from the left and from the right of the discontinuity, D is its velocity. Assuming that a discontinuity exists because the values of p approaching it from different sides along the characteristics differ, one must take (since the characteristics must approach the discontinuity from both sides) that c(pi) > D > c(p2) ("conditions of evolutionarity", see Section 24). Hence, P2 > Pi- The solution describing a stop of cars on traffic signal change has one discontinuity from the left of which the traffic is unperturbed, and from the right,

18. Discontinuity Surfaces in the Lagrangian Description. 83

v = 0, p = Pmax- Motion starting at a traffic signal change at t = 0 is described by a "centered expansion wave" for which p is obtained from the relationships x = c(p)t at 0 < p < Pmax, p = Pmax at I < c{pmax)t; U = <f{p)/p.

18 Discontinuity Surfaces in the Lagrangian Description.

18.1 Examine a smooth function that is the difference of smooth continuations of the left and the right parts of the function 

dX\df_ dy'j dy>

dip dy- * * } ^ = 0 dyldy'

18.2 11£ = /ij£ where f(y') = 0 is an equation of the discontinuity surface.

18.3 [g^gCj] = u§bfa: where f(yf) = 0 is an equation of the discontinuity surface.

18.4 There are the cofactors of components of the matrix \dyl/dxk\ between the brackets. They, when contracted with df/dx*, contain derivatives only along the discontinuity surface and, consequently, are continuous.

18.5 Differentiating with respect to time and with respect the two Lagrangian coordinates, one obtains [v] = ^ , [ea] = ^ \ hence, §2- [v] = £ [ea] .

u"=const " "

18.6 Examine the mapping £' = tpl(xl,t), £4 = t representing the motion of the medium (f' are Lagrangian coordinates, while x' are Eulerian). The equation of the discontinuity surface f(xl,t) = 0 allows the determination of its speed Dn = - § £ / | V / | and normal n = V / / | V / | . Take, in Problem 18.4, fc = 4.

18.7 Let t = /i(x') be the equation of the surface in Eulerian coordinates z*(€M). Use the relationship /(£«) = fc(i'(€»,/(€«)))•

18.8 Use continuity of df/df.

18.9 [A ,V^] = 0, [« + Z^§£] =0,[^ + U + i (p"0 r - ¥)»L] = o.

18.11 According to the answer to Problem 18.2, fe] = -[v']^. Using the formulae of Problem 18.10, one obtains

,- , r i,r 1 [op op + op dfP) x ■

18.12 From the relationship [p(vn — Dn)] = 0, one obtains

[P] = P-M

where p\, v„i are the values ahead of the discontinuity.

d2x' dfr

18.13 Let xl be Cartesian coordinates, then fT, = ^, ^ -—. Using the M dpdp dxx

answer to Problem 18.3, where t = f(P) is the equation of the discontinuity surface, one obtains

rf i - r„'if^ 2121 1 " J ^'dx'dpdp' 18.14 Using the hint to Problem 18.13, one obtains

[a*] (dxi df dxi df [6pql 2 [dp dp + dp dp

[Upq] 2 [dp dp dp dp 18.15 Using the hint to Problem 18.11 and the formula of Problem 18.7, one

obtains [ Ty = 7*'n f cn/MM _ Tt'ntf] _ Tj"nk[yi]

{D„-vnl)2 Dn-vni Dn-vnl' 18.16 Let some quantity of the medium fill volumes V\ and V% adjacent to the

part of the surface £ bounded by a contour L at a given instant. Then

/ pdV + jads\ \ViUV2 E /

j | / PdV+ ladS\ = 0 .

According to the theory of strong discontinuities, the first integral, at Vi,V2 —► 0, yields J[p{Dn — vn)]f dS where the normal is directed to the side marked by the index

E "1" . The second term equals

l7ddiâ)dS+IaVmdl = l{iGdi{â)+îvaa))dS

18. Discontinuity Surfaces in the Lagrangian Description. 85

where G = det(Ga/j), m is the external normal vector to L perpendicular to n , eaVQ is the covariant derivative on the surface. Using the fact that E is arbitrary, one obtains

M = 7G i\{^a) + V Q ( ™ Q ) = ~[p{Dn -Vn)]* ■

Chapter 5

Fluid Mechanics

20 Statics of Fluid 20.1 Using the identities curl (ca) = c curio + (grade) x a, curl grade = 0,

one obtains, from the equations of equilibrium, curlF = —gradp x F/p. Hence, a) F ■ curlF = 0, b) gradp x gradt/ = 0, i.e., gradp||grad£/ and p = p(U).

20.2 At the depth h measured from the line AB, p = pa + pgh, so T> =

f(p — pa)n dS = npghcS, where hc = I / h dS I /S and n is the normal to E external

to the liquid. According to the moment of resultant theorem, V -ho = f(p — pa)h dS,

i.e., ho = 11 Shc where / = Jh2dS.

20.3 V = pga3/2, H = a/3 where a is the side of the square.

20.4 G>VZ= ixpgH2{R - {H cot Q ) / 3 ) cot a.

20.5 H = (H2 + HiH, + ^22)/3(i?i + H2).

20.6 The forces acting on the slab at the instant when it breaks away are shown on Figure a20.1. R is the resultant of pressure forces per unit width of the slab, R = pgH2/2, P is the weight of the slab, F is the friction force, F = kN, and N = P. The slab could not be shifted if kP > R. The slab does not tip if momo P > momo R- Taking into account that h > H, one obtains the three conditions a > I, afi > p/{2kpc), a02 > p/{3pc), a = h/H, 0 = a/H.

20.7 Use the fact that the equilibrium of the liquid surrounding the body is not disturbed if the volume of the rigid body is replaced by the volume of resting Uquid with the density and pressure distributions satisfying the conditions of equilibrium.

87

SOLUTIONS. FLUID MECHANICS

Figure a20.1

20.8 a) pT = 3p/4, b) R/P = 1/3.

20.9 A free surface, if in equilibrium, should be orthogonal to the vector of body force density F. For relative equilibrium, F = g — a where a is the acceleration of the tank, a) a = arctan(a/p), b) a = arctan(acos 9/(g — a sin #)), a = g sin 8(1 — k cot 6), (see Figure a20.2). The free surface is horizontal if k = tan 6.

Figure a20.2

20.10 a) From the equations of relative equilibrium in the cylindrical coordinate system connected with the vessel (the origin is at the center of the bottom, the z-axis is directed vertically upwards), one obtains p = —pgz + pfl2r2/2 + const. Next, one uses the fact that p = pa on the free surface z = z\ (r) and that the volume of the liquid is known. If H, = Q2a2/4g < H, one obtains Z\ = H — H, + 2r2H,/a2, p = pa + pg(H — H. + 2r2H,/a2 — z). If H, > H, the free surface intersects the bottom of the vessel at r = r\ where r\ = a2(l — JH/H,)\ Z\ = 2H,(r2 — r\)/a2, p = pa + pg[2Hm(r2 — r\)/a2 — z] when r > rj , Z\ = 0, p = pa when r < TV

b) V = (p„ + pgH)ira2. c) The force acting on a body with the volume V submersed into a rotating liquid and resting relative to this liquid is

?A = -Jp(g + n2r)dV.

20. Statics of Fluid 89

20.11 The possible cases of disposition of the free surface z = z2(r) and the interface z = zi(r) are depicted on Figure a20.3. 1) If H, = Q2a2/4g < Hi, zt = Hi- H. + 2r2H,/a2, i = 1,2. 2) If H2 > H. > Hu z2 = H2 - H, + 2r2H,/a2,

Figure a20.3

z\ = 2tf,(r2 - r2)/a2 when r > n ; zx = 0 when r < rx. 3) If H, > H2, z{ = 2/f»(r2 - r?)/a2 when r > rit zt = 0 when r < u where r,2 = a2(l - JHi/H+), Hi = Mi/piira2, H2 = {Mi + M2)/p2Tta2.

20.12 The surfaces p = const coincide with the surfaces U = const, U = —gz + fi2(z2 + 2/2)/2 (the x-axis is directed along the axis of rotation, the z-axis, vertically upwards).

20.13 The potential energy II of the floating log equals —pgVzo + p,gV,z, = pgV(z, — z0) where p,, V„ z„ p, V, ZQ are respectively the densities, volumes and the coordinates of the centers of gravity of the log and of the water displaced by the log (the z-axis is directed vertically upwards). The minimum of II corresponds to the position of the stable equilibrium, and the minimum is reached when the log floats horizontally (the length of the log is supposed to be much more than its diameter, and pi is implied to be not equal to p).

20.14 Integrating the equations of equilibrium, one obtains the barometric formula p(z) = p0 exp I — g / dz/RT 1.

a) p = p0 exp (-gz/RT0), p/p0 = p/p0; P = Po/2 at 2 w 5.65 km.

b) p = p0(l - kz)9lRkT° where A; = AT/100To; p/p0 = (p/pB)n where n = 1/(1 -RAT/lQOg) is the polytropic exponent; n = 7 = 1.4 when AT « 0.98°C.

20.15 Examine the balance of the forces acting on a particle of the fluid when it is displaced vertically from the position of equilibrium.

90 SOLUTIONS. FLUID MECHANICS

20.16 Examine the balance of the forces acting on a particle of the fluid when it is displaced vertically from the position of equilibrium.

20.17 Since F = gradU, one obtains div QgradpJ = — AnGp from the equations of equilibrium. In the case of spherical symmetry, this yields d((r2dp/dr)/p) =

r -4irGr2pdr; M = 4ir f pr2dr. a) p = 2nGp2(a? - r2)/3, M = 47rpa3/3 where a is

o the radius of the star, b) Substituting p = Cp6^5 into the equation of equilibrium, one obtains the equation for p(r). It has the particular solution p = 6(A2 + r 2 ) - 5 / 2

where b = X5^2(9C/2nG)5^, and A is constant. This solution corresponds to a star of infinite radius, but of finite mass M = 4nb/3\2, where A is obtained from the last relationship; p = 27\3C5'2/{{\2 + r2)V2^G)3.

21 Dynamics of an Incompressible Ideal Fluid General propert ies of potential flows of an incompressible fluid

21.1 For each instant t, ip is the solution of the following external Neumann problem: A(p = 0 everywhere outside of the body, (d(p/dn)\av = U-n, (grad^)oo = 0 (dV is the surface of the body, U is the velocity of dV), and, consequently, depends upon the shape of the body and the normal component of the velocity at the points of its surface. The last statement is valid also for v = grad <p, but, in a general case, is not valid for dv/dt and, consequently, for p.

21.2 It is known that, if large pressure p' —» oo whose impulse is finite and equals pt — lim / p'dt acts on a liquid during a small time interval r —+ 0 (i.e.,

T-.0 Jo an impact take place), then v = v0 — (gradp{)/p where v0, v are the velocity fields before and after the impact. If v0 = 0, p = const, a potential velocity field with the potential ip = -pt/p arises after the impact, a) A 0; (dip/dz) = £/i at z = 0, r < a; a; grady> = 0 at z = oo; (r = y/x2 + y2, ip = tp(r, z)). b) m(Uo — U\) = — ptpdS where S is the area of the disk.

s

21.3 Let E be the kinetic energy of a potential flow with v = grad ip, and E' be the kinetic energy of another flow with velocity v'. Then E — E' = pf(—{v — v')2/2 +

{v - v') ■ v) dV. Since f{v -v')-vdV = f V , M ^ - ^)) dV = J <p(vn - v'n)dS = 0, V V dV

E < E' if v' ^ v.

21. Dynamics of an Incompressible Ideal Fluid 91

21.4 Let <fii and y>2 be two solutions of one of the three boundary-value problems. Then it follows from the identity f(vi — v^)2 dV = / (i^i —

v dv f2) {difii/dn - dipi/dn) dS = 0, valid for a simply-connected domain, that vi = u2 where V\ = grady>i, «2 = grad<^2-

21.5 An additional inertia force arises in the reference frame connected with the vessel. As a result, the initial state with dependence of p only on the vertical coordinate does not satisfy the equation of relative equilibrium. The arising relative motion cannot be potential, since, according to the theorem of uniqueness, ip = 0 would be the solution of the Neumann problem. If the fluid is homogeneous, the Kelvin and Lagrange theorems of vorticity conservation are valid, i.e., a rotational motion (and, consequently, any motion) can not arise.

21.6 a) Inside the domain, the potential would satisfy the inequality Aip > 0 at a point of minimum and the inequality A<p < 0 at a point of maximum, and this would contradict the equation Atp = 0; Vi = dip/dx' also satisfy the Laplace equation. b) It follows from the equations of motion with divu = 0 taken into account that Ap = -pViidv'/dt) = -pV^viVjV*) = -p(V;VV)(VjVV) < 0, so p cannot reach a minimal value at an internal point of the domain (since Ap > 0 at a point of minimum).

21.7 Use the identity - ip grady?) and the Gauss-Ostrogradskiy formula.

21.8 Use the Green formula (see Problem 21.7) for the domain obtained from V by cutting off the sphere with center at the point r, or for that obtained from V by cutting off the sphere with center at the point f, and pass to the limit, making the radius of the sphere tend to zero.

21.9 a) When supposing that ^(oo) = 0, the decomposition (with zero additive constant) follows from the last identity of Problem 21.8 if one substitutes the Taylor series

1 1 / - « l

l*"o - r\ \r\ |r | into it. From the condition v^00) = 0, it follows that (grad </?)<» = 0, the inverse, in

general, is not valid, b) Since —-- —- g'k = 0 and —: I x' —-r - Slip 1 gjk = 0, the ox* ox* ox3 \ ox* J

' = /§;«*• f-l{*g-*>y<> surface integrals

dv dv

do not vary as dV continuously deforms taking the shape of the sphere So of the radius r with center at the coordinate origin. Substitute the expansion of y into the integrals over the sphere, then only the term j (the source) contributes to the integral / :

So So

Similarly, only the term <pi = C ; d(l/r)/dxj = —C^rij/r2 (the dipole) contributes to the integral / ' . On the sphere So, dipa/dn = 2CiTij/r3; consequently,

So So

c) / is the flux of the fluid through the boundary of the body. It equals the change in volume of the body, i.e., / = dV/dt, C = —\/{A-K)dV/dt. It follows from the relationships

± [xidV= fx^dS=±(r'Vy dt J J dn dtK ' v dv that

d O' r r = ±(T-vy + z-, c = f. dt p 4TT



Planar potential flows 21.12 a) The existence ofip(x, y) follows from the equation of continuity, b) The

equation A ^ = 0 follows from the condition w = 0.

21.13 The scalar product of the vectors v = dip/dyex — dip/dxey and grad^ equals zero. So the lines ip = const are the streamlines.

21.14 Q = il>{x2,y2)-rl>{xl,yl).

21.15 r = (p(x2,y2) - <p(xi,yi).

21.16 a) ip = (Q/2TT) In r, V = (Q/27r)e; b) tp = (r/2ir)e, ip = -(r/27r) In r; c) <p = rn cos TIE, ip = rn sin ne where r, e are the polar coordinates. Q = fv-ndl =

c § dtp is the flow rate through the contour £ enclosing the coordinate origin where c

the source is positioned; r = §v ■ dl = / d<p is the circulation over this contour; n c c

determines the angle a = ir/n inside of which the flow takes place.

21.17 a) The flow is the superposition of a translatory stream and the flow from a dipole positioned at the point 2 = 0. The function dW/dz has a pole of second order at the point z = 0 and zeros at the points z = ±a. The equation of the streamlines is ip = u(r — a?/r) sin e = const, z = re"1. The equation of the contour r — a. b) The flow is the superposition of a translatory stream and the flows created by a source at the point z = —a and a sink at the point z = a. The function dW/dz has poles at the points z = ±a and zeros at the points 2i,2 = ±\ /a 2 + Qa/nU. The equation of the streamlines is

tp = Uy - — arctan -r——- = const. 2TT \ xJ + yJ - a1 J

The contour about which the flow occurs (ip = 0) is an oval passing through z\ and •22-

21.18 The stream function is (r, e are polar coordinates)

tp = (U sin £ - V cos e) [ r ] -\ In - . \ r ) 2n a

The streamline ip = 0 is a circumference of radius a. The velocity at infinity has the components U, V; F is the circulation over the contour enclosing the coordinate origin, and T is an arbitrary constant.

21.19 W(z(Q) = W(Q is an analytic function; dW/dCU = (dW/dz)^ ■ k = k(U — iV); ImV^OIc = Im W(2)|2=aei9 = 0. Thus, to obtain the complex potential W{Q of the flow around the contour C, it is sufficient to determine the conformal mapping of the exterior of the contour C onto the exterior of the circle. This is the essence of the method of conformal mappings.

21.20 The function z{Q = (C + \/C2 ~ b2)/b represents the conformal mapping of the exterior of the segment [—b, b] of the plane ( — x + iy onto the exterior of the circle of unit radius on the plane z, and that which corresponds to the point £B = b (the trailing edge of the plate) is the point ZB = 1. Using the formulae of Problems 21.18, 21.19, one obtains

W(0 = l-cosa [z(0 + -L} - ±usina (2(C) - ^ ) + ^ l n z ( C ) .

The value of T is obtained from Joukovski-Chaplygin postulate:

dW

<=<a

dW dz

dz

<=<B

remains bounded only if (dW/dz)z=ZB = 0. Hence, T = -2nbu sin a.

21.21 To satisfy the boundary condition, a solution, according to the method of mirror reflections, is sought in the form W{z) = Wo(z) + W,(z), where WoM = Q[ln (z — zA)]/2n, and W,(z) is the complex potential of additional sources or sinks at points the positions of which must be found (these points lie out of the region occupied by the fluid), a) The additional source of the same intensity is located at the point zA of mirror reflection relative to the line y — 0 of the point zA\ W = Q( In (z — zA) + In (z — ~ZA~))/2IT, 0 < arg z < 7r; b) the additional sources of the same intensity are located at the points ~zA, —zA, —~zA of mirror reflection relative to the lines y = 0, x = 0 of the point zA\ W = Q In [{z2 - z2

A)(z2 -~ZA*)]/2TT, 0 <argz < 7r/2; c) 2nW = Q In ((z - zA)(z - zB)/(z - Zo)), zB = a2zA/\zA\2, \z - ZQ\ > a. Verify that xj) = Im W = const at all the points of the boundary of the circle (see Figure a21.1). One obtains V(C) = Q{lCAD + ICBD - lCOD)/2ir = Q(lCAD + IOCB)/2TT = g/2 , since AOCB and AOAC are similar, OB ■ OA = OC2, LOCB = LOAC, LCAD + LO AC = v.

Figure a21.1

21.22 Use the Euler formulae sin z = z J [ (1 — (z/nk)2). fc=i

a)W = £ l n ( s i n ^ Y ; 2TT V a /

b) W = In (sin H ) ,

\ II / Q i ( ■ n(z ~a) • *(z + a)\

, , , , , r / . TT(2 - a) . TT(Z -I- a) d) W = -— In sin ——- sin 2iri \ 26 26

Axisymmetric potential motions 21.23 a) In the cylindrical coordinate system x1 = z, x2 = r, x3 = e, the equar

n r . . . , , , 1 (drvz drvT\ tion div v = 0 for an axisymmetric motion takes on the form - — 1—-— = 0,

r \ oz or J from this equation, v, = - - r - , vr = —. The components curl(^e/r)

r or r oz can be calculated according to the formulae (curl a ) ' = —— I -—— — —4-) where

' Jg \dxi dxkJ g = det|0y| = r2, i,j, k forms a cyclic permutation of 1,2,3, and a* are the co-variant components of a vector a. When a = ^e / r , a.\ = 0, 02 = 0, 03 = i/>, so i>2ez + n re r = curl (rpe/r) = « if ve = 0. b) u • grad V> = 0.

1 0i/> 1 W 21.24 See Problems 3.7, 3.10, 3.49. a) vx = — - -^~, v2 = ——— - ^ where h2h3 ox1 h\h3 ox1

hi = y/gii are the Lame parameters,

„ 1 ( d ( h fy\ d ( h dj>\\ w e £ ' M 2 \dxl \hiha dx1) dx2 [h2h3 dx2)) '

b) hi = h2 = 1, h3 = r,

U)

ldr{> I M

2c = - - \^± + ^ - I ^ - H i 0z2 0r2 r drj '

c) /ij = 1, h2 = R, h3 = R sin 0,

1 dy> 1 ay; UR ~ fl2 sin 0 00 ' Ufl ~ flsin0 d~R'

R sin 0 0 f l 2 /£» 00 l^sin 0 00

21.25 a) For axisymmetric flows, v = v(x1,x2), therefore the potential of the velocity may depend upon the angular coordinate x3 only as a hnear function: (f = ip{xl,x2) -r Cx3. For single-valued potentials, C = 0 and v3 = V3f = 0. b) A</5 = 0, 01(1/1) = 0 (the expression cj(ip) see in solution to Problem 21.24).

21.26 2Tr{rP{z2,r2) - i>{zuri)).

21.28 a) <p and ip are superposition of the solutions (for n = 1) considered in Problem 21.27:

f „3 d 1 a3 d 1 \ . / „ a3 \ V = V~{-*^R--ld-zR)=V°°C0Se[R+W*) '

1 n3d2R a3 d2R\ ^ ( t f - a ' J s i n 2 * fl n3d2R a3 d2R\ 2R

where R and 6 are spherical coordinates, b) <p — v^a3 (cos 9)/(2R2), ip = -VooQ.3 (sin2 6)/{2R) where Vgo = - { / , U is the speed of the sphere moving in the direction 0 = 0.

21.29 Let us seek a solution of the form (see Problems 21.27, 21.28) <p -AR cos 6 + B cos 8/R2 where A and B are constants determined by the boundary condition: A = URl/{R\ - R\), B = \UR\R\l{Rl - R\)

„ x d / 1 o \ sin2 6 / 1 a \

v /x2+2 /2 + ( 2 -6 )2 , R, = v/x2 + y2 + ( z - b . ) 2 , 6. = a2/6. b ) | = ± + - -

I / ^' a I ^/(z'-z^ + x2 + y^ 1p Z - b a(z - 6.) / ? - flo r> / - B 7 7

7; = —5- + \ p where flo = <Jx2 + y2 + zl. Q H OK, a

Rotational flows of an ideal incompressible fluid 21.31 Apply the operation curl to the Euler equations written in the form of

Gromeka-Lamb, and use the identity divu> = 0.

21.32 These are verified by calculation.

21.33 a) Since (u> • V)v = u)z dv/dz, the desired equation is obtained from the Helmholtz equation. It also can be obtained immediately from the equation of motion, b) From a) it follows that w is constant along the streamline characterized by a certain value of the stream function ip. Hence, u> — u)(ip).

21.34 a)2u;= - A ^ = -2A(a~2 + ft-2); b) wTtl = 0, vrel = c\u\{\pe)-wexr = grad(^( l /6 2 - l / a 2 )xy /2) .


21.35

J_ 4-7T /

v-y R3 d(

2n L --L f r2'Vy~ 4TT 7

S-xHr- r x

W ^ 2nr2'

where r2 = x2 + y2, the axis z coincides with the line C.

21.36 Let x,y be the coordinates of the point where the given vortex is positioned. Then reflect relative to the coordinate axes as shown on Figure a21.2.

* y

X =

1/x + 1/y = const

Figure a21.2

1 y AIT \y x2 + y'

y = l

4n \X2 + y2 x

21.37 a) Let n be the normal to S and T ± n. The flux of curlu through an element with the area 6 dl and the normal nxr (see Figure a21.3) is 2w • (n x T)S dl. According to the Stokes theorem it equals the circulation V = v (z = 6/2) • rdl —

f)!7

/^ n T

dl 7~

Sy

Figure a21.3

v (z = —6/2) ■ rdl + 0(6). The coordinate origin is chosen lying on 5, z-axis is parallel to n. Taking the limit as 6 —> 0, one obtains (2fi x n —[«])• r = 0, where [v] = v(z = +0) — v(z = —0). Besides, (2fi x n — [«])• n = 0 due to continuity of the

mass flux across 5. Consequently, 2/7 x n = [«], and only the tangential component of the velocity is discontinuous, b) Let us consider the flow of an incompressible fluid for which

!

Ui = const at z > 6 ,

v(z) at \z\<6 ; £ *?> ' Vl* = ^ ' . . . ' ' , ' dvz dz = 0 v2 — const at z < — 6 ' The coordinate origin is chosen lying on S, z-axis is parallel to n . Inside the layer \z\ < 8, 2u> = —(dvy/dz) e\ + (dvx/dz) e2, outside that layer u = 0. In the limit as 6 —» 0, there is a jump of the tangential component of the velocity at z = 0. The plane z = 0 is a vortex sheet on which

+6 l {2 = lim / udz = - ([vx]e2 - [vy]ei) . O-+0 J £

-6

One can see that 2[t2 x e3] = [vxei + vye2] .

21.38 a) Let us choose a coordinate system for which the coordinate lines x1, x2 lie on a meridian plane, and i 3 = e is the azimuth angle. The decomposition of v is v = Vie1 + v2e2 + v^e3. During solution of Problem 21.23, it is shown that vie1 + v2e2 = rot {tpe/r) where e = re3 is the unit vector of the coordinate line e. Introducing the notation w = w3 p n y s = v3/r, one obtains v = curl (tpe/r) + we, div v = 0; c) Use the Helmholtz equation for vorticity.

21.39 a) v • gradrjj = 0. b) Consider a material contour C which is the intersection of a stream surface ip = const with a plane z = const at an instant t. Since C is a circumference of radius r, the circulation of the velocity around C is Tc = 2itrw. When the motion is steady, the contour remains as a circumference belonging to one and the same surface ip = const. According to the Kelvin theorem, Tc = const, i.e., rw = const over ip = const.

21.40 a) v2/2+p/p and rw are constant along a streamline (see Problem 21.39) and, because of the symmetry, are identical for all the streamlines lying on the surface xp = const, b) Use the formulae of Problem 21.38 and the equation of motion projected onto the z-axis in the form 2(urwE — vewT) = dH/dz.

21.41 The stream function satisfies the equation (see Problem 21.38)

d2ij) # V _ l f t y _ _ 2 dz2 dr2 r dr

The boundary condition is ip = const at z2 + r2 = a2. Let us seek the solution in the form ip = Ar2(a2 — z2 — r2). The equation is satisfied if A = c/10.

The coordinate origin is chosen lying on S, z-axis is parallel to n . Inside the layer outside that layer u = 0. In the limit as

there is a jump of the tangential component of the velocity at z = 0. The

21.42 As it follows from the equation, both for ideal and viscous fluids, the formula for the velocity of the motion arising as a result of an impact (see Problem 21.2)

T

is: pv = — gradpt where pt = lim / p' dt. Hence, 2u> = curlu = —gradp x v/p. b) In o

the zeroth approximation with respect to 6, p = const, u> = 0, v = u. In the first approximation, w = —gradp x u/2p.

Integrals of the equations of motion for an ideal incompressible fluid

21.43 r4 = Az where A = 2gS2/iv2v2, S is the area of the opening, v is the speed of lowering of the level, z is the vertical coordinate, and r2 = x2 + y2.

21.44 a) v = 2vx sin 0,p = ?«,+/*& (1-4 sin2 0)/2, i.e. Cp = 2(p-p00)/p*4 = 1 - 4sin20, Cp is called the pressure coefficient; R = 0; b) v = 1.5 00 sin 6, p = Poo + pvlo (1 - 2.25 sin2 0)/2, Cp = 1 - 2.25 sin2 6, R = 0 (see Figure a21.4).

Figure a21.4

21.45 Cavitation arises where v = vmex, andpa+p(u2X) — v2

nax)/2 = 0, i.e., p = 0; v = Umax on the boundary of the body about which the flow occurs, when 8 = n/2 a) p|e=w/2 = 0 if Voo = ,J&pj5p « 12.6 m/s; b) p|9=T/2 = 0 if ««, = ^2pj3p w 8.1 m/s (see Problem 21.44).

21.46 Use the equations of motion in the form of Gromeka-Lamb and the identity 2u> x v = 2uez x v = 2ugraAip.

21.47 From the solution of the Neumann problem for the velocity potential, one finds <p = —a2d/R, v(R, t) = d(f/dR = a2a/R2 where a(t) is the radius of the cavity, and R is the spherical coordinate. The Cauchy-Lagrange integral yields p = Poo — pdf/dt — pv2/2; since p = 0 at R = a, one obtains the equation for


a(t) (of Reyleigh-Lamb) ad + 1.5a2 = — Poo/p which has the energy integral d2a3 = (2poo(ao — a3))/(3p). The time of collapse is

W~!jdw=r0M^° 21.48 ip = cy — ca exp(ky) cos kx, W = cz — cai exp(—ikz) , z = i + iy; the

equation of the streamline ip = 0 is y = a exp(ky) cos kx turns into y = a cos kx with accuracy of order ak when ak are small. If ip = 0 is the free surface of the wave, then (giaAip)2/2 + gy = const or c2/2(l — 2ak cos kx + 0(a2 k2)) +ga cos kx = const along it, i.e., c = ±Jg/k; c is the velocity of propagation of the wave relative to the fluid resting at y = —oo; its amplitude equals a, 2n/k is its length, and k is the wave number.

21.49 v « 1.51/0 in the vicinity of the critical point on the surface of the bubble (see Problem 21.44), and p/p = gR(l - cos 9) - v2/2 + const « (gR - 9U2/4) 62/2 + const. Prom the condition p — const on the boundary of the bubble, one obtains U = 2s/g~R~/3.

21.50 Let us write the condition of pressure equality on the interface between the two liquids in the coordinate system connected with the leading point: pi(v2/2 + gx cos 7) = p2(vl/2+gx cos7) where 7 is the angle between the i-axis and the vertical. Substituting into this the expressions for the speeds in the vicinity of the point O (the flows in the angles a and tr — a) V\ = c\Xl-1'~a^a, and V2 = C2Xa^^~a\ one obtains plc\x2^-°^la - piclx20'^-0* + 2{pi - p2)gxcos-y = 0. When a = TT/3 the terms have, respectively, the orders x4, x, x. The last two terms cancel each other if ci is appropriately chosen, and the first one is unessential as a higher order infinitesimal quantity.

21.51 For a streamline on the free surface of the wave in the vicinity of the corner point in a coordinate system connected with the wave, v2/2 — gx cos(a/2) = const. Since v ~ cx^~a^a, this relationship is valid when a = 27r/3.

Force and moment acting on a body in a stream of an ideal incompressible fluid

21.52 b) Use the identities div (v2ei/2 — Viv) = 0, and div (v2(r x ej)/2 — (r x v)vi) = 0, where e* are the basis vectors of a Cartesian coordinate system.

21.54 a) IF, IM —♦ 0 as S' tends toward infinity, b) Since the problem is linear, the potential tp and the force F can be written in the form

4- rr IT d f A V- dU* f = 2^fkUk, Fi = — J p<pnids= -l^Vik-jf

since ipk is independent of t.

21.55 Use the fact that, for a steady flow, dQ/dt = dK/dt = 0.

21.56 a) dX + idY = — p{nx + iny)dl = pidz, since dz = cos(l,x)dl + i sm(l,x)dl = —sin(n,x)dl + i cos(n,x)dl = i(nx + iny)dl, where I and n are the unit

/ 1 dW dW vectors of the tangent and normal to the contour respectively; p = p\h — 2 dz dz

(Bernoulli integral); b) the contour around which the flow occurs is a streamline; c)X + i V = - I p i / f ^dz = \pi§(^)idz.

21.57 a) F = —/* jjf, b) F = pr x u where a is the radius of the cylinder, p. = pna2 is the apparent mass, r is the vector directed along the axis of the cylinder, and \r\ = r 0 (see Problem 21.18). The formulae of Problem 21.54 are not valid here, since <p is not single-valued.

21.58 Using solutions to Problems 21.20, 21.56c, one obtains

dW

X =

■ U 1 COS (

X -

—2irpbu'

y — i

iY --

' sin2

s"Wc+t,) ■ c

■W(?)'«-a , Y = 2-npbu2 s

= x + iy ,

sin a cos a

One can verify that (X + iY)(cos a + i sin a) = 0, i.e., the force is perpendicular to the velocity vector u, and the magnitude of the force equals 27rpbw2|sin Q| = pu\F\, as it also follows from the formula of Joukowski.

21.59 (p = -a3u ■ R/{2R3), see Problem 21.28b, Q = -pu, F = -pdu/dt where a is the radius, p is the apparent mass, and p = 2npa3/3.

21.60 In a coordinate system moving with the velocity Uoo (and with the acceleration Uoo), the body force of inertia — Uao acts on the fluid in addition, and the sphere moves with the acceleration —Uoo. The presence of the body force of inertia in the fluid results in the action of "Archimedes' force" FA = pVUao on the body, and,

since the sphere moves with acceleration, the fluid exerts on it a force R = pUgo. The total force acting on the unmoving sphere is F = (p + pV)dUoo/dt where p and V are respectively the apparent mass and the volume of the sphere.

21.61 Let us write the velocity of the flow in the form v = v3tmrce + v', 4irvSOUrce = Q(r' - r)/\r' - r\3. Prom the formulae of Problem 21.53, one obtains F = IF = p f [\n — vvn) dS. Let us take S' — the sphere Se of a small radius e

S' V ' with the center in the singular point r. For e —> 0, one finds

F1 = lim -P-r / ( - t / V n , + vn + ifin'nAdS = st

l i m - ^ T (V* dS = pqv'Ur) . £-.0 47re^ J

21.62 At the beginning of its rising to the surface, the gravity force Mg, Archimedes' force FA = —M\g and the drag force R = —fxdu/dt act on the bubble of mass M (here, p. is the apparent mass, Mi = 2p is the mass of the fluid in the volume of the bubble). The equation of motion for the bubble has the form (M + p) du/dt = (Af - Mx)g. Hence, du/dt = g(M - M{)/{M + p.) « -2g if the mass of the bubble is neglected.

21.63 In a coordinate system moving with the stream, the body force of inertia —w arises under the action of which the acceleration of the spherical bubble becomes equal to 2«; (see Problem 21.62). In an unmoving coordinate system, the acceleration of the bubble equals 3w.

22 Dynamics of an Incompressible Viscous Fluid General propert ies of flows of incompressible viscous fluids

22.1 See Section 12.

22.2 Let us denote the potential of the gravity force by U (g = gradW). Using the boundary condition V\QV = 0, one obtains

d ^ m L = / P 9 - v d V d t = /"div(pUv)dVdt= JpUvndSdt = 0 , v v ov

dA^f — / pljrijVidSdt = 0, i.e., dEkia = -Vdt < 0, E^m decreases in time, and a dv

fluid set into motion experiences deceleration.

22. Dynamics of an Incompressible Viscous Fluid 103

22.3 Inertia! body forces arise in a coordinate system connected with the body. One of these forces is the Coriolis force, and the other has potential U\ = fi2r2/2 (r is the distance from the axis of rotation). The work of all body forces equals zero (see Problem 22.2). Consequently, dEfil = -Vdt < 0, so that the fluid tends to come into rotational motion with angular velocity fi.

22 A For the kinetic energy of the motion of a homogeneous viscous fluid relative to a rigid body, dEkin/dt = — V < 0. Hence, if the fluid rests at t = 0, then it rests at t > 0. Motion in a homogeneous ideal fluid, at rest at an initial instant, can be only potential, and, since the solution of the Neumann problem for ip is unique, there is no relative motion. For a nonhomogeneous fluid, dEkin = (Mma88 -Vdt = -fUv'VipdVdt -Vdt (F = gradW). Both for a vis-

v cous and an ideal fluid, dAm,^ ^ 0, so the kinetic energy may increase in time for special motions (see Problems 21.5, 21.42).

22.5 a) divdv/dt = Vi(^V,V) = ( V ^ H V ^ ) = eê^ - 2|w|2; c) since

— I -.—r 1 = Km (ds is the element of the particle path), rii — = ds \\v\J dt

ni\v\ — I |w| ■;—r I = |v|2A"n • m if n • v = 0; d) substitute dv/dt = gradw2/2 + ds \ \v\J

curlu x v into the identity of part b).

22.6 The motion is steady in a coordinate system connected with the body, dEkiD/dt = 0, and dA{^ = Vdt>0. On the other hand, dA^lf = f p^nM dtdS =

av / p^rijUidSdt = —FÛidt where Uj are the components of the velocity of the rigid

av body, n is the inward directed normal to the body, and F* are the components of the force acting on the body. Consequently, F ■ u < 0.

s 22.7 Integrating the equation for kinetic energy, one obtains E\ua = f Fds —

o t JVdt where S = u2/2a is the distance travelled by the sphere at the time t, and o u is its speed. Since V > 0, using the Kelvin principle (Problem 21.3), one obtains s JFds> EyiB > Eli, = M^S = 2npR3aS/3. o

22.8 p\ = - pn ' + /xnJ(V,t/ + V'v,) = -pn* + p,nj(Vjvi - V'u,) + 2/m^V'^ = —pnl — /i(n x curlv)' + 2p.n?VlVj. Using the operator Yk = 6't — n'nk (see Problem 3.21b) (n is the unit normal to dS), transform the last term of the sum in the formula for pl

n so that it would be determined only by the tangential derivatives of the velocity field 7'kVfcfj. Since 0 = VjU1 = I'îVj + nJVnUj (VnVj = nitVkVj is

the derivative of the field v along the normal), nJVvj = n?iltS7kVj + n'nJV„i>j = ■nJ'y'kV'vj — n'7jfcVfcVj. For the velocity of the rigid body, VfcUj = e^k^l'- Because of the condition of adhesion on the surface of the body, the velocities of the fluid and the body as well as their derivatives along the surface coincide, so nP^Vj = nj-yikejikSl' - n{ ■ 0 = -{Q x n ) \

22.9 Using the formulae of the preceding problem for pln, for the tangential stress

on a stream surface (v • n = 0), one obtains 7'kP* = /i(curlu x n) ' -I- 2nniYkVkvj = /x(curl v x n)l+2fib^Vj where 7*fcV':nJ = — 6y', 6^ is the tensor of the second quadratic

d ( v form for the stream surface. Taking into account that — I -—- 1 = Km, Kn • m =

as \\v\J bijV'vj/lvl2, one obtains p n • v/\v\ = /J,(CUT\V X n)v/\v\ + 2^K\v\[n ■ m) .

Unsteady flows 22.11 a) If the x-axis is parallel to the velocity of the plate, then vy =

vz = 0, vx = u(y,t), du/dt = ud2u/dy2. The boundary conditions are u(0, t) = Re[uo exp(—iilt)], u|y_oo —* 0. The derived equation for u is called the heat conduction equation and has particular solutions of the form u = C exp(iky — iQt), k = ±k,, k. = (1 + i)/<5, 6 = J2v/Cl. With the boundary conditions taken into account, u = Re (uoe''fc'y_n'') = Uoe~y/6 cos(y/S — fit). This solution represents a rapidly damped transverse wave; u « 0 outside of the boundary layer with the thickness 8 (in an ideal fluid u = 0 everywhere), b) The tangential stress T on the plate is T = Txy\y=o = —Uoy/Upjl cos (fit + 7r/4). C) Over a vibration period the average value of the energy dissipation per unit area of the plate is

o o

22.12 Let us seek the solution in the form of a superposition of waves corresponding to k = fc, and k = -k, (see the solution of Problem 22.11), i.e., u(y,t) = Re[{deik-V + C2e~ik-y) e'int\, A;, = (1 + i)^Q./2u. Taking into account the boundary conditions UA = u(0,t) = Re(u0 e~,nt), UB = u(h,T) = 0, one obtains

UoRe 'sinh[fc./i(l -y/h)] _int

sinh (k,h)

TA = Txy\y=0 = -tiu0He[k, coth (k.h)} e~lilt

TB = -Txy\y=h = /iUoR-e k. -int

sinh (kmh) max TB —► 0 when f2 ~3> v/h2; max T^/max TA —» 1 when fi <C v/h2.

22.13 a) Considering the problem from the viewpoint of dimensionality theory and taking into account that it is linear, one obtains vx — u = Uof(£), f = y\fvi (self-similar problem). The differential equation and boundary condition for /(£) are W + 2 /" = 0, /(0) = 1, / -» 0 when £ - oo.

u = U0(l-j=Je^dr]=U0 1-erf 2y/ui.

= U0F y/ui

~ x 2 r _ 2 The function erf(i) = —= e v dr), called the error integral, is tabulated, b) Ap-

v o proximate the function U(t) with the step function and use the solution a).

u(y,t) = U(0)F(j=)+J ■ dU(r) dr

y > ( * - T ) .

d r .

The last integral is called Duhamel's integral.

22.14 This problem is similar to Problem 22.13a in essence. The difference consists in the boundary conditions /(+oo) = 1, / ( - co ) = — 1. a) The solution is vx = u = [/erf (y/{2\/irt)\ b) Since erf(2) = 0.99, the thickness of the layer is 6 ~ 2>\/ui. At any point of the stream except infinity, \u\ decreases as t grows, and

0 when t —> oo; c) ux = uiy = 0, UJZ = U v2 = = e x p ( - f - ) . For all t > 0, U

is maximum at y = 0 (at the place of initial discontinuity).

22.15 For vx = u(y, t), one obtains

du d2u dt V dy2 (a22.1)

u(y, t) = 0 at y < 0, t = 0; fidu/dy = T0 at y = 0, t > 0; u —> 0 when y —> -oo. Let us denote r = fidu/dy, then <9r/3t = vd2r/dy2, r = T0 at y = 0, i > 0, T —> 0 at 2/ —» —oo, £ > 0, T = 0 at y < 0, t = 0. According to the solution of Problem 22.13a,

-y/2-M \ e~v dr} . Obtaining from here T = T0 (-A7

/■ r , T0 . / — y \ [ut _j;i

On the free surface, u(0,t) = — J —. At any point 2 / ^ 0 , u(y,t) t —> oo.

u(0,t) when

22.16 a) vx = - £ « - * / « , b) £kin = ^ , Qx = ^ /* s Q 0 w h e r e V« 4v26a 2 V o

a = 1 + 46i/t. c) £:kin — 0, Qx = Q0-22.17 It is obvious that u^ = t>(R, t), vv = vg = 0. It follows from the continuity

equation with the kinematic conditions on the boundaries that v = A(t)/R2, A(t) = RiR2 = R2R2, R%- R\ = 3M/(4irp). The equation of kinetic energy yields

| pRlRl ( l - | l ) = 2R\Rl (pl - P2) - 8 ^ 1 A? ( l - | | ) •

22.18 Prom the equations of Problem 22.17 when (R?-Ri)/Ri = AR/Ri < 1, P l = p2 = 0, one obtains AR = M/4irpR2, R + \2v R/R2 = 0, R - Uu/R = is(Re-12)/Ro where Re = RoRo/v; if Re > 12, R -* 00 when t — 00; if 0 < Re < 12, J W = R' = RQ/{1 - Re/12); if Re < 0, i?min = R'\ if Re = 0, R = R*.

22.19 As a result of the explosion, an expanding spherical cavity of radius Rc{t) is formed; vR = A(t)/R2, v^ = ve = 0, Re = vR\R=Rc, P\R=RC = 0, A(t) = RcRc- From the equation of Problem 22.17 when Ri = Re, R2 —► 00, p\ = p?. = 0, one obtains d{RcRc)ldt + &vRcRc = 0. The initial conditions are Rc(0) = 0, 2-npRcRc\t=0 = E0. Hence, Re = ^Ea/2ixpRz

c - 8v/Rc. If t -> 00, then Re -» 0, i?c —♦ EQ/\2&-KPU2. The Cauchy-Lagrange integral is valid for this motion, so p =

Steady flows 22.20 For a laminar flow, vx = v(z), vv = vz = 0 (the x-axis is parallel to

the bottom, and the z-axis, perpendicular); ucPv/dz2 = —g sin a, (dv/dz)z=h = 0, v\z=o = 0; v = g(2hz - z2) (sin a)/(2v), Vm^ = v(h) = gh2 sin a/2v, vm = 2umax/3. a) umax ~ 1-25 cm/s, uav « 0.83 cm/s. b) umax = 12.5 km/s, uav = 8.3 km/s. The absurdity of the last result is connected with the supposition that the flow is laminar which is invalid for this case.

22.21 vx = vx(y), vy = vz = 0 (the t/-axis is perpendicular to the plates, the i-axis is directed along U); a) vx = Uy/H, TA = Txy\y=o = fiU/H = —TB\ b) dp/dx = -i0 = const, vx = (i0/2p.)y(H-y), TA = ioH/2 = rB\ c) vx = Uy/ H+(i0/2fj,)y(H -y), TA = n{U/H + to/f/2), TB = p.(-U/H + i0H/2).

22.22 In cylindrical coordinates with the z-axis directed along the tube, vT = vv = 0, a) vz = (i0a2/Afj.)(l - r2/a2), vml%x = 2vm = 2Q/ira2 = i0a2/4p.; b) vz = (i0/4/i)(62 -r2 + (b2 - a2) ln(r/6)/ ln(6/o)). c) vz = (i0/2/x)(l - x2/a2 -y2/b2)a2b2/(a2 + b2),Q = {iri0/4n)a3b3/(a2 + b2) = v„irab.

22.23 vmax = (i0a2//x)/(aj), Q = (i0a* / p.) (p(oii) where a is the characteristic size of a cross section of the tube, and Qj are dimensionless parameters determining its shape.

22.24 In cylindrical coordinates with z-axis directed along the axis of the cylinders, vr = vz = 0, f£hys = v(r), and p = p(r). From the Navier-Stokes equations:

dp v2 cPv 1 dv v dr r drl r dr T*

and boundary conditions: v(Ri) = QiRi ,v(R2) = fi2ii2 one obtains v = Ar + B/r where A = (02a2 - ^ / ( a 2 - 1), B = R2^ - ft2)a2/(a2 - 1), a = R2/Rx; M, = —4irp,B = —M2, and Mi and M2 are the moments acting on the internal and external cylinders (per unit length). When i?2 —» oo, 0 2 = 0, the flow is potential, and v = R2£li/r; if Rl = 0, fii = 0, then v = il2r.

22.25 The formulae for v of Problem 22.24 are valid here, vz

( * 2 - f i 2 ) l n f / I n ^ - r ' + tf2 ..... .... ... ... . _ t0 = —dp/dz. For the case when (i?2 — 4(i I Ri Ri 1 Ri)/Rx = e -C 1, let us introduce a new variable y with the formula r =

i e2R2

Ri{\ + ey), then vz « -j/(l — y) as in a plane-parallel Poiseuille stream, and v = Ri[fli + (fi2 — fli)y] as in a plane-parallel Couette stream.

22.26 When the motion is undisturbed, each particle moves along a circular path with radius r = const. It experiences the action of the pressure gradient and the centrifugal inertia force equilibrating the pressure gradient Fin = p?/pr3, where v(r) is the azimuthal velocity of the fluid and p.(r) = pvr = pr2ip is the moment of momentum, conserved even when the motion is disturbed. Let a small particle be displaced from a circular path of radius r0 to a neighboring one of radius r > r0. In the

^ = , F.n = fe) dr pr3 pr3 displaced position, — = —^p, Fin = —. The necessary and sufficient condition

the stability is — > Fin, i.e., dp2/dr > 0 which is equivalent to the inequality or

<pd(ipr2)/dr > 0. Using the solution of Problem 22.24, the stability condition is transformed into the form ( O ^ - f i i f l 2 ) ^ > 0. If fiif22 < 0, then tp changes its sign, and the flow is unstable. When fit > 0 and fi2 > 0 the flow is stable if fi2i^ > QiR2. When fi2 = 0, the flow is unstable; when £l\ = 0, it is stable.

Stream function and vorticity 22.30 ip = ip0 exp(-i/At).

22.31 Since the flow is cylindrically symmetric, vr = vz = 0, v?hys = v(r, t), ur = 0)^ = 0, wz = u{r,t). It follows from the equation of Problem 22.27 that duj/dt = (v/r) d(r du/dr)/dr. The initial condition is T(r, 0) = r 0 = const where

r 2ir

r(r, t) = 2 ur' dip dr'. In view of linear dependence of w upon T and dimen-o o

sionality considerations, one obtains u = (r0/vt) /(£) where £ = r2/ut. The equation for / is / ( 0 + £ / ' ( 0 + 4[/'(£) + £/"(£)] = 0. Integrating, one finds u = {To/Sirut) exp(-r2/4i/i), T = r0[l -exp(-r2 /4i / t)] , t> = (r0/27rr)[l -exp(- r 2 /4 i4) ] .

22.32 The condition for the value of a scalar function /(x1 , x2) to be a maximum (a minimum) at an internal point M, where at least one of the derivatives ViVj/lw / 0, is as follows: V</|M = 0, | |VJV,/ | |M is negative (positive) definite. Since / = w does not depend upon transformation of the coordinates x1, x2, let us carry out the proof in a Cartesian system of coordinates x, y in the plane of the solution. Let there exist a point M at which V w = 0. Due to the equation of Problem 22.28, ViVa;|M = 0, i.e., M is neither at a maximum nor a minimum.

22.33 See solution of Problem 22.32. Since / = ui/r does not depend upon transformation of the coordinates x1, x2, let us carry out the proof in a cylindrical coordinate system z, r (in which the equation of the axis of the flow is r = 0). Let Vi(uj/r) = 0 at a certain point M. Due to the equation of Problem 22.29, ViV'(w/r)|M = 0, i.e., M is neither a point of maximum nor of minimum.

22.34 Write the equation of Problem 22.28 for the stream function ip

dAip dip dAip di> dAip / 0* n d4 d4 \ , n . „„ „, + ^ ^ r - ^ ^ - " k l + 2 Z i ^ + T j K = 0 ' (*22.2) dt dy dx dx dy \dxA dx2dy2 dy*

Express ip in the form ip(x, y, t) = ipo(y)+'<Pi{x, y, t) where ipo is the stream function of the undisturbed flow, ip\/tpo <C 1. Taking into account that ipo satisfies the equation (a22.2) and Atp0 = dU/dy, one obtains

dArPy dArPl 30, d2U a ) -dT + u { y ) - d x ~ - ^ ^ -

(d* n d* a4\ , „ "{w+2dxW + W)i,1 = 0

- dip b) 0 = - ? - = Oatj / = ± d . dy

Low Reynolds number flows. Stokes approximation 22.35 The following transformations (conducted in a Cartesian coordinate sys

tem for simplicity) are obvious:

J(e[]-el])el]dV = JelJ^p^dV =

-f{£;^'--^-i£-M»:-n))}jv, = Y J(V'i ~ Vi)(-P6<3 + 2^eij)ni dS = °

dV

Using the identity

I(v')-I(v) = 2pJ(e'ije[J-etjei})dV = v

= 2tij{2(e'iJ - etJ)etJ) + (e'xj - ey)(ey - etj))dV v

and the proved equality, one obtains

/(»') - I(v) = 2 M | ( e ; , - e0)(e;3 - ey) dV > 0 . v

22.36 Let us assume that any of the boundary-value problems has two solutions (vitp, ey) and (w(,p',ey). Then, using the transformations similar to those of Problem 22.35, one obtains

2fiJ(e'ij-e,J)(e'ij-elj)dV = v

J« - *)[-& - p)6tJ + 2/i(e'y - eyJln, dS . av

For each of the boundary-value problems, the integrals over dV are equal to zero, and, consequently, ey = ey in the region of the flow.

22.37 Using the transformations similar to those of Problem 22.35, one can prove that

2fM f eye'" dV=- f pywW dS = - J py w V dS . v av av

no SOLUTIONS. FLUID MECHANICS

On the boundary of the rigid body, v = U + fi x r, and hence, 2/x/ eê'^ dV = v

U'F + Q'M = UF' + QM'.

22.38 F = pv2l2C{Re), Re = lv/v (see Problem 22.10). Since the Stokes equations and the boundary conditions are linear, F should be linear with respect to v. Hence, C(Re) = A/Re.

22.39 Use the equation of kinetic energy (Problem 22.1) and the fact that F-vOQ = V (Problem 22.6). Since V > Pstoke. (Problem 22.35), F > FStoke8.

22.40 Use the fact that the problem for the Stokes equation is linear, that F*, Uj are the components of vectors and the theorem of reciprocity (Problem 22.37).

22.41 For a variation of the functional I{v') at v' = t>, using the transformation similar to those of Problem 22.35, one obtains 81(v') = / (4/êynJ — 2pn,i — 2fi)6v% dS.

av From the boundary condition, it follows that 61 = 0. Since the functional I(v) is convex, its extremum is a minimum.

22.42 The proof is similar to that of Problem 22.41.

22.43 Let us use the statement of Problem 22.42: I(v') > I(v) where v' and v are, respectively, the solutions of the Navier-Stokes and Stokes equations. It follows from the equation of kinetic energy that Vdt = dAe, i.e., I(v') = —V, V < V, V = 2fJ.feijei>dV.

v

22A4 vr = 3Ur(l - z2/h2)/4h, vz = U(z3/h3 - 3z/h)/2, p = ^ [z2/h2 - 1 -2/i

(r2 — R2)/2h2]+p0 where z = ±h is the equation for the planes, p0 = p\r=R, R = const. 22.45 Outside of the disk and on its external side, there is no flow in practice,

and the pressure can be assumed to be constant and equal to po- Using the solution of Problem 22.44, one obtains pzz\z=h = —p + 2/idvz/dz « 3[iU(r2 - R2)/Ah3 — p0. The drag is found by integration: F — 37r/xf/.R4/8/i3.

22.46 For a thin layer of large extent, the inequalities d/dz » d/dr, vT » vz are obvious. The formula for p in this approximation, p = —3Û(r2 — R2)/4h3 + p0, exhibits a relative error of h2/R2 <tC 1 as compared to Problem 22.45. The drag force is the same as in Problem 22.45.

22.47 b) Using the formulae 2df/8z = df/dx + idf/dy, 2df/dz = df/dx -

idf/dy, f = f(z,~z), the Stokes equation can be written in the form 4t/x 2 = dp

— -—. One can easily verify that this equation is satisfied. az


22.48 One can easily verify the validity of this equation by substitution.

22.49 The unknown constants are obtained from the boundary conditions v$ = 0 at R = a, Vzloo = U, p ^ = p0.

Ua3 cos 8 3aU 3/xa(/ cos 6 H= AW~' G=W~U> P = Po 2¥~ '

dH / dG \ . rr / 3a a3 \ Vfi=dR + {RdR-G)COSe = UcOSe{1-2R + 2R3)

ldH „ . a TJ . Q / 3a a3 \

. , . 3at/ sin 9 _ /•. , ,2 „ , „ r r 2 (curl«)v, = — ^ — , V = n 1 |curlt>faV = 6wfiaU,

v F = 6TTHO.U .

22.50 The statement being proved follows from the fact that Ap = 0.

High Reynolds flows. Laminar boundary layer

22.51 6 ~ y/vL/U

22.52 ff"/2 + /'" = 0, /(0) = /'(0) = 0, /'(oo) = 1.

22.53 The Prandtl equation can be written in the form

d2u d ... . . . . . dU d . , r r ., d , ... .,

- " ^ = « ^ - « ) + - « ) t o + f a K y - « ) l + ^ ^ - « ) ] -

Integrating, one obtains the desired relationships.

22.55 6i = (TT/2 - l)/a, 9 = (1 - VOA*-22.56 It follows from the integral equation of momentum that uUa =

(l~l) T(~U*)' H e D C e ' lla = ylWA-^\fi*JV ~ 3.053y/vx/U, H (du/dy)v=0 « 0.328 y/p»U3/x.

22.57 b) m/'2 - (m + l ) / / " / 2 = m + / '". c) The flow around the wedge with the angle 2nm/(m + 1). This class of flows is obtained by Focker and Sken (1930). d) The boundary conditions are /(0) = /'(0) = 0, /'(oo) = 1.

High Reynolds flows. Laminar boundary layer

22.58 a) On the free surface pnT = 0. Using the relationships of Problem 22.9, one obtains 2u> = (curlv)z = —du'/dy = 2K(U + u') at y = 0 (K is the curvature of

v the streamlines the free surface), b) Since u' = — 2 / wdy, ui\y>s = u'\y>6 = 0, then

o v! ~ w<5 ~ KU6 ~ t//>/Re < t/.

22.59 The components of the velocity in the boundary layer are u = U(x) + u', } d(U +«') , m ,. . u u, .

v — - I ^ dy. Taking into account that JJ ~ —?—-, one assumes that o

u = U(x), v = —ydU/dx and, using the Helmholtz equation for vorticity (see Problem 22.27, obtains the desired equation.

y

22.60 Using the relationship u' = - 2 Judy and integrating the equation for o

u) of Problem 22.59 over y, one obtains the desired equation for u'. Substituting u = U(x)+u' into the Prandtl equation and taking account of the estimation u'/U ~ 1/TRi, one obtains the desired equation for u'.

22.61 a) According to the formula of Problem 22.5c, V = 2fif{v2/a)dS = s

127Tfj,U2a, since v = | U sin 0, dS = 2ira2 sin 6d6. b) The calculation of the velocity exhibits the error ~ Re-1''2, c) Taking into account that F ■ U — V (Problem 22.6), one obtains F = 12wfiUa(l + 0(Re'1/2)).

Turbulent motion 22.62 a) Yes, it is; b) yes, it is; c) 1) no, it is not; 2) yes, it is.

22.63 Use the formula dfjdt = df/dt + V>/VJ following from the definition of df/dt where / = f(x',t), the condition of incompressibility and the properties of averaging.

22.65 vT ~ 1(T2 m2/s = 104 • water.

22.66 divv = 0, dv/dt = F - gradp/p + AhAv + Avd2v/dz2 where A is the two-dimensional Laplacian along horizontal coordinates, and z is the vertical Cartesian coordinate.

22.67 Integrating the Reynolds equation of motion dr^/dy + pud2v/dy2 = 0 and using the hypothesis of Prandtl, one obtains K2y2(dv/dy)2 + v(dv/dy) = r0/p. Neglecting viscous stresses in the region y > 6, one obtains v = — In y + c where v, = JTQ/P is called the dynamical velocity, c = const.

23. Waves on a Surface of Heavy Incompressible Liquid 113

22.68 v = vT = ^ In y+c at y > 6 (see Problem 22.67). For y < 6, one obtains from the equations of laminar motion v dv/dy = To/p that v = iti^ = v2y/u. The condition Re|y={ = 1 yields 8 = v/vt, v(6) = v,. Setting the equality t>iam(<5) = VT{S), one obtains i) = v, + (vm/n) \n(v,y/v).

22.69 dv/dy = f(y,p,To). According to the IT-theorem (see Section 36), (dv/dy)/JTo/py2 = c = const. Hence, v = cJr0/p In y + c\.

22.70 c ^ = -^d iv (q + qT), qT = pcT'v', q is average heat flux, c is heat capacity.

22.71 We have d(qy + qy)/dy = 0 where qy = —kdT/dy, k is the coefficient of molecular heat conductivity, qy = kTdT/dy, kT = avT = an2y2\dv/dy\. For the turbulent stream of Problem 22.67, dv/dy = v,/ny. Neglecting molecular heat conduction and integrating, one obtains T = (qo/anv.) \ny + const, v. = JTQ/p.

23 Waves on a Surface of Heavy Incompressible Liquid

23.1 The potential ip(x,y,z,y) and shape of the free surface (,(x,y,z,y) need to be found. The equation for the potential is Ap = 0. The boundary conditions are (d<p/dn)z=_h = 0 on the bottom, (dip/dt + (gradip)2/2)z=^ + g£ = 0 on the free surface (this dynamical condition follows from p|2=( = Po and the Cauchy-Lagrange integral) and also d(/dt = (dp/dz)z=i (kinematic condition). The initial conditions are ({x,y,0) = (0(x,y), <p(x,y,(0,0) = Po{x,y), at t = 0 (the functions Co(z,2/), <fo(x,y) are given).

23.2 Linearization of the conditions on the free surface yields [dp/dt)z=0+gQ = 0, dQ/dt = (d<p/dz)z=0. So £ = — (d<p/dt)z=0/g, and the problem for p in the domain with a known boundary is as follows: Atp = 0; (dp/dn)z=_h = 0; (dip/dz)z=o = -{d2<p/dt2)z=0/g; vU=o,t=o = Po(x,y); {dp/dt)z=0,t=o = -gCo(x,y); Po, Co are given. Q)(x, y) = (dQ/dt)t=o can be given instead of (p0.

23.3 Seek ip in the form = 0 yields (p = (Aiekz + A2e~kz)el''kx~ut\ From the boundary condition at z = — h, one obtains A\ekh = A2e~kh = 2C, and, if C and k are real, then t) (the solution <p = Reip which differs by the shift of the phase (kx — ujt) by 7r/2 is obtained in a similar way). Using the boundary condition at z = 0 one obtains the dispersion relation u)2 = gk tanh kh. The free surface is C = a cos(fcx — ut), a = Cu cosh kh/g. For a periodic motion, the ampUtude o

and the wave length A = 2ir/k or frequency u must be given, being equivalent to specification of two initial conditions. The points £ = const move with the speed c = u/k (phase speed), a) w2 = gk, b) CJ2 = ghk2.

23.4 a) W(Z) is an analytic function. The stream function is ip = I m W = c(z - aekz cos kx), ip = 0 at z = aekz cos kx « a cos kx (1 + ak cos kx + . . . ) . From the Bernoulli integral on the free surface c2/2 — c2ak cos kx+ga cos kx+c*0(a2k2) = const, one obtains c2 = g/fc. b) In the coordinate system z, x', where the liquid rests at 2 —> -co , z = a cos [k(x' + ct)], x' = x — ct, a is the amplitude, c is the speed of the wave.

23.5 a) vx = U + u', vx — v'. From the equations of Problem 22.60 at U{x) = c, ak -C 1, one obtains the following boundary-value problem for u': cdu'/dx = vd2u'/dz2, du'/dz\z=0 = 2cak2 cos kx, u'|z__00 = 0. Solving this problem, one finds u' = 2cakRe~1^2 exp (fcwRe/2 2) cos[fc(x + zJRe/2) — TT/4],

I/|Z__OO = f %£dz = 2(cak/Re) cos fcx, Re = c/ku » 1. b) Outside of the bound-—oo

ary layer, v = grad(<p + if'). The viscous correction has potential iff. On the external boundary of the boundary layer, dip'/dz\z=0 = «'|z-.-<x» y'U—oo = 0- Hence, if' = 2(ca/Re) efcz cos kx. With the solution of Problem 23.4 taken into account, one obtains f = c[x — aekz sin fcx] + if'.

C(*.t)

23.6 dA/dt = —p dx / gvzdz = —p-r I dx / pzdz

gravity force per unit time.

C(x,t)

Pff/2 / C 2 d x - ( x 2 - x i ) / i 2 = -dEp0t/dt, where di4/dt is the work of

23.7 Use results from the solution of Problems 23.4, 23.5. a) For an ideal liquid, in the coordinate system where the velocity of the liquid equals zero at z —* —oo,

2w/k o cos kx vx-ivz = d(-caiex-p(-ikZ))/dZ,Ekin = % J dx f v2dz = § pa2c2(l+0(ak)),

0 —oo J5p0t « ^7ra2 ss J^kin- b) For a viscous liquid, v = grad(^ + v' where if = Re W{Z), \v'\ ~ cRe-1''2 inside of the boundary layer with the thickness 6 ~ ARe-1'2, and \v'\ ~ cRe-1 outside of it; v'n = 0 on the surface of the wave. Thus, E^n = Eo + Ei + E2, E0 = 0.5pf{gradip)2dV = 0.5pna2c2, £ , = p/div{ipv')dV = p f ifv'ndS = 0,

V V dV

E2 = 0.5pf(v')2dV ~ Re~3/2. c) £ p o t + £ k i n = const for an ideal liquid, d(Epot +

i?kin) = — V dt for a viscous one.

23.8 Write the velocity in the form v = grady + v'. For the dissipated energy, V = V0 + 2?i. According to the formula of Problem 22.5c, D0 = 2(j,f(gradip)2Kds

c where £ is the profile of one period of the wave determined by the equation z = £(x) = a cos A:x, K ds = (," dx. According to the Bernoulli integral, (grad ip)2 —

const — 2o£ on the profile of the wave, so T>0 = —4/ip / CC"^1 = 47r/xâ2fc. Hence, o

2ir/fc 0 2>i«2p / dx J [(d2tp/dxdz)(dv'Jdz) + (dv'Jdz)2/4] dz = D 0 O ( l / v ^ ) .

0 -oo

23.9 a) d(£p0t + ■E'kin) = —2?<&- According to the answers to Problems 23.7, 23.8, £ p o t + £ k i n = prrcV, £> = 4n>c2(afc)2, so da = -2vk2adt, a = a0 exp (—2t/fc2t). b) At t = 2ir/kc: a/ao = exp (-47r/Re), Re = c/kv.

23.10 Seek the solution as the sum of two waves with identical \k\, ui and a running towards to each other (see Problem 23.3) <p = C cosh [k(z + h)] (e^*1-"'' + et(kx+wt)^ p _ j^ e ( p = 2C cosh [k(z + h)] cos wt cos kx, £ = a sin cut cos fcx, a = (2Cw/g) cosh kh (standing wave). The condition at x = 0 is then satisfied, the condition at x = L yields sin kL = 0, k = irn/L or A = Ln/2 (n = 1,2,...). The values k = irn/L are called the eigen-wave numbers; the frequencies corresponding to them (by the dispersion relation) are eigen-frequencies of the basin, and the obtained solutions are called the eigen-oscillations of the basin.

23.11 It follows from the dispersion relation (see Problem 23.3) that for each value of k there exist two waves for which w = LJ\<2 = ±y/gk tanh kh. Since the problem is linear, the solution can be written in the form C(x,t) =

oo oo / fi(k)eikx-iu^tdk+ f f2{k) e*1-^^1 dk. The unknown functions f^k), f2{k)

-oo -oo oo oo

axe obtained from the initial conditions Co = / 9i(k)e dk, Co = / 92{k) elkx dk: -oo -oo

/i(*0 = [m{k) - U29i(k)]/(vi - u;2), .M*) = [î0i(*O - m(k)]/(vi - u2) where 00 00 .

9i(k) = £ / Ux)e-ikxdx, g2(k) = ± / Ux)e~ik'dx. — CO - O O

23.12 C = C1+C2 = 2a cos[(/fci+Aifc/2)x-(w1-l-Aw/2)i] cos[(xAfc-<Au;)/2] = A(x - Ut) cos [kox - v{ko)t] where &o = (î + k2)/2, A = 2a cos [(x - Ut)Ak/2] is the amplitude, and U = Aw/Afc; at Afc —» 0, £/ = dw/dfclfc^ — the group speed.

23.13 a) U = c+kdc/dk, b) c = y/(g tanh fc/i)/fc, t/ = c(l + 2A:/i/sinh 2kh)/2; l)c = U = ,/jfli;2)c = Jgjk, U = c/2.

23.14 Since AA; is small, assume that u(k) ~ cj(ko) + u'(ko)(k - ko). Then oo oo / f{k) e^*-^ dk = e^*01-"0') / f{k) e^-*0 '* dk where x' = x - Ut, U =

- O O —OO

cj'(ko). Denote k-k0 = XAk, then / /(jfe) e'<*-*°>x' dk = Ak / /(A)e iAl 'Afc dX = -oo 0

AkA(x'Ak). The obtained formula makes it obvious that the characteristic length along the x-axis for the function A is inversely proportional to Ak; i.e., A is a slowly varying function of x' if Ak is small. Since AA; is a given constant, C = ReA{x - Ut) e

i(koX-ut\ and A = AAk. The term neglected in the expansion of the frequency w yields a small correction in the phase if d2u)/dk2\k=k0 {Ak)2t <tC 2TT.

23.15 a) The compatibility condition dk/dt + duj/dx = 0 follows from the definitions of k and w. Differentiating the dispersion relation — w + Q(k,x,t) = 0 with respect to x and using the compatibility condition, one obtains the equation dk/dt + (dSl/dk) ■ (dk/dx)+d£l/dx = 0 which is equivalent to the system of equations dk/dt = —dfl/dx, dx/dt = dVt/dk. b) If the undisturbed flow is uniform, then dVt/dx = 0, u> = J7(fc, t), k = k(x, t) and the waves with a given value of k = k0(= const) move with the group speed U(k0) = dCl/dk\k0. A fixed value of phase 9(x, t) = 60 propagates with the speed dx/dt = —9t/6x = u/k.

23.16 According to the solution of the preceding problem, the wave corresponding to a fixed value of k propagates independently of waves with other values of k, and the energy of the waves over an interval [fci,fc2] remains constant. The density of this energy is inversely proportional to the length of the segment [xi, x2] on which the waves with k € [^1, 2] where fci — fo = AA; is small are situated. Hence (see Figure a23.1), Ax « (dx/dk\t=0 + tdU/dk)Ak where U = duj/dk is the

ut i

Ax

X, X2 X

Figure a23.1

group speed. Furthermore the energy density is proportional to the square of the wave amplitude. Hence A(k,t) = A(k,0)/Ja(k)+b(k)t where a(fc) = dx/dk\t=0, b(k) = d2w/dk2.

23.17 a) The Euler equation projected onto the vertical (z-axis), dvz/dt being neglected, yields dp/dz = —pg, consequently, p = Po + pg{C~ z) (hydrostatic pressure distribution over depth). Then the other two equations of motion can be written in the form (23.2). The equation (23.1) is obtained by integrating the equation

divw = 0 over depth using the boundary conditions on z = — h and z = (,. b) d(,/dt + hdvi/dxi = 0; dvi/dt + gdQ/dxi = 0 . These equations lead to the wave equations for «i and C d\/dt2 = ghd2C,/dx2, d2vjdt2 = ghd2Vi/dxl so the velocity of propagation of the waves is c = ±i/gh.

23.18 a) It is obvious that vx = vx(x, t), ( = £(x,t). Eliminating the components of the velocity from the equations of Problem 23.17b (with Fx added to the right part of the equation of motion ), one obtains the nonuniform wave equation for £: d2C/dt2 - c2d2C/dx2 = AKh cos(K.x - at), c2 = gh. Its solution has the form £ = /i (x — ct) + f2(x + ct) + CF- The functions /i and f2 are obtained from the initial conditions. Find the particular solution in the form £F = a c°s {K.x — at) with unknown a. Substitution into the equation yields a = AKh/{c2K2 — a2) b) Resonance takes place at h = a2jgK?.

23.19 The analogy follows from the fact that the system of equations for the "shallow water" approximation (Problem 23.17), with the changes of variable h + £ = p, g(h + C)2/2 = p, coincides with the system describing a motion of a gas (see Section 24).

23.20 The system of equations for the invariants J_, J+ is the characteristic form of the equation for shallow water at h = const (see Problem 24.26).

23.21 Let J_ = M = const. Then vx = M+2a where a = ^g(h + (); J+ = M+ 4a, c+ = M + 3a. Since J+ = const along the lines dx/dt = c+ (see Problem 23.20), a = const along these lines which are straight lines x — c+t = const. Therefore, a = fi(x - c+t), C = /2(z - c+t), vx = f3{x - c+t). The functions fu f2, f3 can be found if, e.g., £(x) and M are given at an initial instant, b) Since C = .M^ -c+t) , each value of C is translated without variation in space with speed c+. Since c+ depends on Ci the shape of the wave changes: if c+ > 0, then the parts of the wave where d(,/dx > 0 become flatter, and the parts where dC,/dx < 0 become steeper and the wave breaks, c) dC,/dt + (v0 + yfgR) dC,/dx + (1.5 \fgh~C,/h) dQ/dx = 0, v0 = vx\(=0-

23.22 Applying the mass and momentum conservation laws for the small volume shaded on Figure a23.2 and taking account of the hydrostatic pressure distribution over depth, one obtains the conditions on a hydraulic jump:

hi (vnl - D) = h2 {vn2 - D) =m } m(vnl-vn2)=g(h2-h2)/2 \ (a23.1) vTl = VT2 )

where v„, vT are the projections of the velocity onto the normal and tangent to the discontinuity surface, D = Dn. b) Let vT = 0, vn\ = v\, vn2 = v2. Eliminating v2 from the conditions on a hydraulic jump (a23.1), one obtains D = vi ± Jgh2 (hi + h2)/2hi.

Figure a23.2

The choice of the sign is determined in item c). c) If n is directed to the side 1, then at D > vi the particles pass from the side 1 to the side 2. Compute the difference of the mechanical (kinetic and potential) energy flowing into the discontinuity and that flowing out of it: AE = pg{ti2 — /ii)3/4/ii/i2. Since dissipation of mechanical energy takes place when passing the discontinuity, the inequality AE > 0 must be satisfied at D > V\, i.e., hi < /i2- This means that the depth of the stream increases after passing the discontinuity, and the discontinuity moves relative to the liquid in the direction of less depth.

23.23 a) dS/dt + dSu/dx = 0, du/dt + udu/dx = -(l/S)dV/dx where V = / g(h — z) dS, the x-axis is directed along the axis of the channel, the z-axis is

s directed vertically upwards, and h is the depth of the stream, b) The analogy becomes obvious if the notation p:= S,p:=V is used. The function p(p) is determined by the function V{S). In case 1), V = gBh2/2 = gS2^; in case 2), V = gh3 (cot 0)/3 = gSl5/3 Vcot 6 (see Figure a23.3). c) c = u± JdV/dS. d) Riemann waves can exist

1) 2)

Figure a23.3

in a channel if the geometry of the channel does not depend on i ; J± = u ± / ( a /5 ) dS where a = JdP/dS, c± = u ± a. The condition for breaking of a wave propagating

along the i-axis in positive direction can be written in the form d2V/dQ2 ■ dS/dx > 0 where Q = 1/5.

23.24 Let vx be the horizontal component of velocity in the coordinate system C

connected with the wave. Then / vx dz = — ch (the mass conservation law), a) Qx =

+00 C +oo +oo

p J dxj(vx + c)dz = pc J (dx; Qz = 0; Epot = (p/2) J g? dx; b) E]dQ = —oo -h -oo —oo

+oo C o +°° (p/2) J dx J(vx + c)2 dz » ^ J C2 dx, vx = -cfc/(A + C) « - c + <</fc at C « A.

- O O —/l - O O

It follows from the Bernoulli integral, v\j1 + 3^ ~ c 2 ^ — (c2//! — <?)£ + . . . = const, that c = ±y/g~K, so E^n « £p0t- To determine the profile of the wave, the more precise Korteweg-de Vries approximation is required (see Problem 23.28).

23.25 a) w2 = ghk2, &(/dt2 = ghd\/dx2; b) u = ±ky/g~h~, d^/dt ± y/g~h~d(/dx = 0.

23.26 a) w2 = p/iJfc2(l - k2h2/3), b) w = ±>/5K*(l - fc2/i2/6).

23.27 Supplement the equation obtained in Problem 23.21c with the terms that take dispersion into account (Problem 23.26b).

23.28 Seek the solution in the form £ = h W(£), f = x — Ut, U = const. Derive the equation for W(£): h2W'"/6+1.5WW'-(U/co-l)W' = 0,co = y/gE. Integrating two times, one obtains tfW^/S + W3 - 2(£//o - \)W2 + AGW + F = 0 where G, F are the constants of integration. It follows from the conditions at x = ±00 that G = F = 0, and the equation takes the form

h3(dW/d£)2/3 = W2(ct - W) (a23.2)

where a = 2(U/CQ— 1). The exact solution of this equation is W = asech2(\/3af/2/i). It is obvious even without integration that W has a maximum at the point where W = a, C = ha = C- If the amplitude of the wave (* is given, one obtains that a = (*/h and C = C* sech2[^/3C*/4/i3(a;-f/0]- T h e SP***1 o f t h i s wave is U = co(l+C*/2/i) > c<>.

23.29 Take the interface between the layers at rest as the plane z = 0. In the approximation of long waves of small amplitude, integrating the equations of continuity and motion with respect to z, one obtains for the upper layer (with index 1) Pi = Po + Piff(G - z + /»i). dvjdt = -gdb/dx, ht dvi/dx + dh/dt - d^/dt = 0; for the lower layer (with index 2) j>2 = Po + Pifl(& - C2 + ^ I ) + fcgib - z),

dv2/dt= -(gpi/p2)dh/dx-g(l- pl/p2)d(2/dx, h2dv2/dx + dC,2/dt = 0; fr > px. Seek the solution of this linear system of equations with constant coefficients in the form va = Vae^kx-Ut\ C<* = Za^kx-'Jt\ a = 1 for the surface wave, a = 2 for internal one. A non-trivial solution exists if the dispersion relation uj2/k2 = c\ 2 = 0.5 [g{hi + h2) ± \Jg2{h\ + h2)2 - 4g2hih2(l - pi/p2)], determining the two possible values of propagation speed of the waves, is satisfied. The ratio of amplitudes of the internal and surface waves is Z2jZ\ = 1 — ghi/c2. If (p2 — Pi)/ P\ <?C 1, one obtains c2 = g(hi + h2), Z2jZ\ = h2/(hx + h2) (as in a homogeneous liquid); c\ = ghih2(l - pi/p2)/{hi + h2), Z2/Zl = ~{p\/p2 + hi/h2)/(l - pi/p2) (these two waves propagate in the opposite directions with small speeds and large amplitudes on the internal boundary).

23.30 Let z = 0 be the location of the undisturbed interface, U = V\—v2 being the relative velocity of the layers. Direct the x-axis along U, then Vi = Uex, and v2 = 0 in the coordinate system moving with the velocity v2. For the disturbances of the velocity potentials <pa(x, z, t) and discontinuity surface £(x, t), one obtains the linear problem: A<pa = 0, d<pi/dz\z=00 = 0, d(p2/dz\z=-.oo = 0, d(,/dt + UdQ/dx = d(pi/dz\z=0, d(,/dt = d<p2/dz\z=0, px (dtfi/dt+Ud<pi/dx)\z=0+Pig^ = P2d<p2/dt\z=0+ p2gQ- The dispersion relation for this problem has the form w = (pxkU/(pi + p2)) ± \j{{pv ~ Pi)gk/{pi + p2)) - P\p2k2U2/{pi + p2)2. Even if p2 > pu I m u > 0 for A; > 9{P\ ~ PO/PiPiU2 (instability of Kelvin-Helmholtz).

23.31 a) dvx/dt = -{l/p)dp/dx ~ 2Slyvz + 2£lzvy, dvy/dt = -(l/p)dp/dy -2Q2vx+2Clxvz, dvjdt = -(1/p) dp/dz-g-2ttxvy+2Slyvx. Forthe Earth2[f2xw]z < g, since fl w 7 • 10"5 s_1. b) For almost horizontal motions, the terms with vz (as compared to the terms with vx, vy) can be neglected in the first two equations of motion. The approximate system is dvx/dt = —(l/p)dp/dx + fvy, dvy/dt = -(1/p) dp/dy - fvx, dvz/dt = —g — (1/p) dp/dz where / = 2f22 = 2S7 sin tp is the Coriolis parameter, and <p is the geographical latitude of the place.

23.32 In the suppositions of shallow water theory for waves of small amplitude, one obtains from the equations of Problem 23.31 dC,/dt + h(dvx/dx + dvy/dy) = 0, dvx/dt-fvy+gdC/dx = 0, dvy/dt+fvx+gd£/dy = 0. a) Seek the particular solution of this system in which vy = 0, then dC,/dt + hdvx/dx = 0, dvx/dt + gdQ/dx = 0, fvx + g dC,/dy = 0. The first two equations are reduced to wave equations; their solution is C = Pi{y) Fi(x - ct) + P2{y) F2{x + ct), vx = c{Px Fi - P2 F2)/h where c2 = gh, Fi and F2 are arbitrary functions, and Pi and P2 are determined by the equation {dP\/dy + fcP\/gh)F\ + (dP2/dy — fcP2/gh)F2 = 0 obtained from the last equation of the system. Taking into account that the functions Fi and F2 are arbitrary, one obtains dP\/dy = —fPi/c and dP2/dy = / P2/c. Consequently £ = Ae-fy/cFi(x - ct) + Befv/cF2(x + ct). In the Northern hemisphere, / > 0, and,

24. Mechanics of a Compressible Fluid 121

it follows from the condition for decay of C as \y\ —> oo that B = 0 for y > 0, A = 0 for y < 0. The obtained solution is called the wave of Kelvin, and c = y/gh is its speed; f/c — (2fi sin ip)/\fgH is the coefficient of amplitude attenuation with increase of the distance from the boundary to the left side in the Northern hemisphere relative to the direction of propagation of this wave, b) Let vy ^ 0. Since the problem is linear, £, vx and vy can be represented in the superposition of waves of the form exp(ikxx + ikyy — iwt) with unknown amplitude. The dispersion relation is w2 — gh{kx + ky) — / 2 = 0. For given real w and kx, there are two roots of this equation ky = kya(uj, kx), a = 1,2 (corresponding to two waves), and it follows from the condition for decay of the solution as \y\ —» oo that those roots should be purely imaginary. Hence, vy = £a=i Ca exp(ikxx — lmkyay — iut) where Im ky\ > 0, Imfcy2 < 0, Ca are constants, vv —► 0 at \y\ —* oo, vy = 0 at y = 0. Hence, Ci = C2 = 0 and vy = 0.

23.33 For vz = 0, dvx/dx + dvy/dy = 0, d(dvy/dx - dvx/dy)/dt + 0vy = 0, P = d(2Q sin <p)/dy (the second equation is obtained as a result of eliminating pressure from the equations of Problem 23.31). Seek the solution of the form v = u0e'(fcl~u't), where vQ is a constant vector. Then vx = u0 = const, vy = i;0c'' fe"u!), kw—iiQ^+P = 0 (the dispersion relation), and the phase speed is c = u/k = Uo—0/k2. In particular, UQ = 0 and c = -p/k2. In the Northern hemisphere 0 > 0, and this wave propagates to the west; \c\ is small since fi is small. For waves observed in nature A ~ 300 — 400 km, and, consequently, c ~ 3 - 5 m/s.

24 Mechanics of a Compressible Fluid Equations describing a motion and state of a compressible fluid

24.1 a) The system of equations includes the continuity equation

-£+pdivv = Q (a24.1) dt

the equations of motion

^ = F - - g r a d p + - V^d (a24.2) dt p p

the energy equation

It {\V2 + U) = {F ' V) ~ ~p diV{PV) + ~p Vj{TiJVi) + ft (a243)

the entropy equation

and the constitutive relations (equations of state)

u = u(p,T) , (a24.5) P = P(P,T), (a24.6)

rij = rij(ekl,T) , ew = 0.5(VtW| + V,vfc) . (a24.7)

Here, F is the density of body forces, rx', and ew are the components of the tensors of viscous stresses and strain rates, respectively, and dq/dt is the heat input per unit mass and per unit time. If F and dq/dt are given, the system is closed. Instead of the energy equation (a24.3), the equivalent equation of internal energy

du dq 1 ,-,- V dp „

¥ = I + rey + (a248) is often used, and, instead of the entropy equation (a24.4), the combination of this equation and equation (a24.8) — the Gibbs identity

^ds du p dp , Tdt = Tt-?i < a 2 4-9>

is used. The equation (a24.9) can be integrated to obtain s = s(p, T) at given u(p, T), p(p,T). b) For this case (see Chapter 3),

rij = Adivvgij + 2p.eij, F = g, ^ = - A T , u = cvT + const, p = RpT dt p

where A, p. are the coefficients of viscosity, K is the coefficient of heat conductivity, cv = const, R = const, and g is the gravitational acceleration. The system of equations has the form

dp — + pdivt; = 0 at dv 1 , v , ... , . — = g gradp+ — grad(divu) + i/Av dt p 3 dT p dp 1 „ K . _ dt p* dt p p

In I I + const = cv In ( — ) + const

where v = p/p is the kinematic coefficient of viscosity, 7 = Cp/cv is the adiabatic exponent, and Cp = R + cv.

24.2 a) Using the continuity equation and the equation of state p = RpT, write the equation of internal energy in the form

dT , dT ^ ,. 1 u K A„, cv — = -cvv' — - RTdwv + -T'^H + - AT

P P dt dxi

The work of viscous stresses can be neglected if at least one of the following inequalities is satisfied:

1 ■r'ea <C cvv

tdT_

-T,]eij < \RTd\vv\ , (a24.10)

- r u e y < P

dT Zv dt

The heat input due to heat conduction can be neglected if at least one of the following inequalities is satisfied:

(a24.11)

Let us introduce the length scale I and the time scale r of the motion (the distance and the time for which parameters of the stream change by a value of the order of these parameters themselves) and denote the typical values of velocity, pressure, density, temperature by v, p, p, T. Then dT/dx* ~ T/l, dT/dt ~ T/T, etc. The inequalities (a24.10) take the form

- A T P

- A T P

- A T P

<

«

•C

{dT CvVdx~* )

flTdivv| ,

dT Cv-dt

vv2 cvTv vv2 KTv I1 T

The first two inequahties are satisfied (or not satisfied) simultaneously, since cv ~ R, and can be written in the form

Re » M2 . (a24.12)

Here, Re = vl/u is the Reynolds number, M = v/a is the Mach number, a2 = (dp/dp), = fRT, a is the speed of sound. (The expression for a2 for a perfect gas is

obtained in the following way. The equation of state p = RpT can be rewritten, using the expression for the entropy of the perfect gas (see Problem 24.1b), in the form

Po \PoJ \ cv J

Here, p0, po, s0 are the values of the parameters in a certain state. Differentiating this formula with respect to p, at s = const, one obtains the formula for a2.) The last inequality (a24.10) is not satisfied for steady motions (r = oo) and is satisfied for unsteady ones if

/ R e ,. Re

where St = VT/1 is the Strouhal number. The first two inequalities (a24.11) are satisfied (or not satisfied) simultaneously and yield

P e » l where Pe = Cplpv/k is the Peclet number. The last inequality (a24.11) is satisfied if

St < Pe .

b) The numerical estimates for such motion are as follows: outside the boundary layer, Re ~ 7 • 107, M « 0.3, Pe « 5 • 107. Thus, in this case, the equation of internal energy can be used in the form du/dt — (p/p2)dp/dt or ds/dt — 0.

24.3 The force of gravity is included only by the projection of the motion equation onto the vertical:

dvz dvz dvz dvz dvz 1 dp dt dt dx v dy dx p dz

(the z-axis is directed vertically upward). If the length scale of the motion / and the velocity scale v have an identical order in all the directions, then

dvz dvx ■>>* "' •"2

~~dt ~ ~~dt where St = VT/1 is the Strouhal number. From the equation of motion projected onto a horizontal, one obtains

Ap dvx v2 ( 1 \ pi dt I V St/

The quantity g can be neglected if

„ < T (! + - ) , ,e. (Fr)» - - > T T T 7 S

where Fr = v/y/gl is the Froude number. If, e.g., the motion is steady (r = oo, 1/St = 0), v ~ 100 m/s, / ~ 10 m, then gl/v2 ~ 10 -3.

V2 V V2 ( 1 \

T + ; = T(1 + SI)


24.4 In this case, it is convenient to take p, s as the parameters of state instead of p, T, since s = const in each particle. In particular, the equation of state is written in the form p = p(p, s) or s = s(p, p). The system of equations is

dp dv 1 dslp.p) -£+pdivv = 0, — = —gradp , - ^ ^ = 0 at at p dt

24.5 a) See Problem 14.2.

1 p u = - + const, s = cv ln(p/p7) 4- const 7 - 1 p

i = u + pjp = CpT + const = (- const, a2 = 7p/p 7 - 1 P

b) The system of equations is

dp dv 1 dip -+pdlvv = 0, ^ = --gradP, - { -

This system is sometimes written in the other form, regarding the pressure p and entropy s as the function sought for and taking into account that dp = a2dp due to the condition ds/dt = 0. Thus,

dp 2 dv 1 ds — + a p d i v v = 0, — = —gradp, — = 0 dt dt p dt

a2 = yp/p, p = Cplhe3/c*, C = p o P o ^ e " ' 0 ^ = const .

24.6 For adiabatic motion of an ideal fluid ds/dt = 0, i.e., s = So(f\£2,£3) where £' are Lagrangian coordinates. Since s = s(p,p) for any homogeneous compressible fluid, then p = p(p, so(£')) f°r an adiabatic motion. Therefore the motion is not barotropic in the general case. If the entropy is identical in all particles, i.e., So does not depend upon £', then the motion is barotropic.

24.7 The circulation of a velocity over a line L from A to B is T&B = fvidxx. L

Let L be a line passing through the same particles of the fluid. Calculate the variation of TAB with time. Introduce a Lagrangian coordinate f on the line L. One may write for points of the line L

dx* Vi = «t(€. t) , dx1 = — aX on L

and

/dx* vi~Qf<% > &*> & = const ,

dvAB ifdvi{^,t)dxi

Jf, VB a v ~ir = j—dr-dzds+JVidtdzdt =

Here, a< = dvi(£,t)/dt are the acceleration components and the equality dx%(£, t)/dt = v* is taken into account. If L is a closed contour, then

dit=hidxi-Under the condition of the problem, a = grad (V — U) where V = / dp/p, U is the potential of the body forces, therefore Oj dx* = d{V — U), and, if U is a single-valued function of the coordinates, then

dt

24.8 Use the Kelvin theorem (Problem 24.7).

24.9 The system consists of the continuity equation

— + (grad ip • grad p) + pAip = 0

and the integral of the equations of motion — the Cauchy-Lagrange integral (see Problem 9.16)

^ + i | g r a d ¥ ) | 2 + 7 ' ( p ) - t / = 0

where V(p) = Jdp/p is the pressure function, and U is the potential of the body forces.

24.10 a) It is the system presented in the solution of Problem 24.9 in which U = —gz (if the 2-axis is directed vertically upward), and

T>(,) = * ^ p 7 - i * . Po 7 - 1 7 - 1

b) Since dp = p dV, then the continuity equation can be written in the form

1 dV dV ^ + A * = ° OT ^ - + ( 7 - 1 ) ^ = 0.

Expressing V and dV/dt in terms of the derivatives of <p with the use of the Cauchy-Lagrange integral, one obtains the equation including only ip.

Motion with small disturbances. Propagat ion of small pressure disturbances. Speed of sound

24.11 The closed system of equations for a barotropic motion of an ideal compressible fluid has the form

dp n dv 1 . . — + pdivu = 0 , — = — g r a d p , p = p(p) . dt dt p

The system has the solution VQ = 0, po = p(po)- As a result of linearization, one obtains

dff j . , „ dv' 1 _, , — + p0divv = 0 , — = gradp ot at po p' = a2p',a2 = l^f\ = const (a24.13)

\dp I P=PO

Differentiating the first equation with respect to t, computing div of the second equation, and subtracting the obtained equations, one eliminates v and obtains the equations for p' and p'

where A is the Laplacian operator. The vector fields v\ and v'2 (v' = v\ + v'2) satisfy the equalities divvj = A<p, divuj = 0, curlu'j = 0. Computing curl the second equation (a24.13) leads to the equation curl (dv'2/dt) — 0. Computing grad the first equation and d/dt the second one (with the formula grad div v — curl (curl v) = At> taken into account) allows one to eliminate p from the system (a24.13) and obtain

The potential ip of the velocity v\ can be taken to satisfy the wave equation

dt2

b) Ap' = 0.

d2<p o A r - a2 A<p = 0

24.12 a) Substituting ip = (p0e,^'tr~u^ into the wave equation, one obtains the relation u2/k2 = a2 between ui, k = |fc| and the coefficient of the equation a. This relation is called the dispersion relation. The speed of travel of the surface / = const along the normal to it n = g rad/ / |grad/ | is determined by the formula D = —df/dt/\g[adf\ (see Problem 17.1). For a monochromatic wave, grad/ = fc, the phase speed equals ui/k = ±a. It does not depend on fc (and, as a consequence, on the wavelength A = 2i:/k). Thus, every monochromatic wave of small amplitude under the considered conditions propagates over a medium at rest with velocity a. For each fc there can exist two waves propagating in opposite directions. The normal to the surface / = const is parallel to the vector fc; i.e., the wave vector indicates the direction of travel of the surface of constant phase along the normal to it. That direction is often called the direction of propagation of the wave, b) Regarding the function <p = ipoR/ee*^ "^ as the velocity potential, one obtains v = grad if = —ipokIme'^ult\ i.e., the velocity vector of the particles is parallel to the wave vector fc.

24.13 The linearized system of equations for small disturbances in the form of plane waves is

dp' dv' „ dv' 1 dp' . , , 9 Idp\

The general solution of the wave equation

ay 2av dt2 dx2

(see Problem 24.11) has the form p'(x, t) = /i(x — at) + /2(x + at) where / i , f2 are arbitrary twice differentiable functions. Substituting this expression for p' into the system of equations, one obtains

P' = tfUdx - at) + f2{x + at)]

v' = — [/!(i - at) - f2(x + at) + C) , C = const ap0

This motion has a potential, ap0(f = Fi(x — at) — F2(x + at) + Cx. Here, Fa = I fa{£) d£j c* = 1,2. Solutions of the form f(x — at) and f(x + at) are called running (or progressive) waves. They carry disturbances along x-axis with the velocity a or —a without variation of the shape, b) Use the solution presented in item a). Obtain / i and f2 from the initial conditions. For t = 0, v'(x, 0) = (/i(x) — /2(z) + C)/poa = 0, i.e., f2(x) = /i(x) + C. Further, p'{x,0) = /i(x) + f2{x) = 2/^x) + C = p'0(x),

i.e., /,(*) = (p'0(x) - C)/2, /2(x) = (p'0(x) + C)/2. For t = 2l/a, p'(x,t) = h{x -21) + f2(x + 21) = 0.5(PQ(Z - 2/) + p'0(x + 21)), the first term being nonzero only at —/ < x — 21 < I, i.e., at / < x < 3/, and the second term being nonzero only at - 3 / < x < —1. The graphs p', v' for t = 0, t = 21/a, t = 3Z/a are presented on the Figure a24.1. c) Since v = dw/dt (w is the displacement), for a monochromatic wave

/

—s\ 1

- / ■ \

p

\

p'

/~\—-

p'

A \ - ^

t=0

t = 2l/a ,»_

" x \y

t = 3l/a -m-

Figure a24.1

, « '

i V'

r\-~~

v'

/^\-~~

(w = A Re e''*1-""' = A cos(kx - ut)), v = Au> sin(kx - ut), v'mgx = ALJ = A- 2ni/ = 0.785 m/s. In a wave running in one direction, as it follows from the solution of the item a), p'/po = v'/a. The wavelength is A = 2ir/k = 2ira/u) = a/v. Therefore, for propagation in air (p'/po)max = 2.3-10-3, A = 0.68 m, in water (p'/po)max = 5.6-10~4, A = 2.8 m.

24.14 In a perfect gas p = pRT, aT = y/KT; for adiabatic oscillations p = Cp7, as = y/^WT. At the temperature T = 288 °K ,aT ss 287 m/s, a, w 340 m/s for air. Thus, what should be called the speed of sound is the speed of propagation of small disturbances for adiabatic processes. The reason is that the Peclet number in this process is of order 104; therefore Pe ~S> 1 (see Problem 24.2).

24.15 If if = <p(r, t), then, in a spherical coordinate system,

A _ dV | 2 ^ = 1 g2(ryQ

dr2 r dr r dr2

Therefore, the wave equation can be written in the form

d2(r<fi) _ o d2{ry) _ dt2 dr2 '

the general solution of which is

f(r,t) /i(r-at) | h{r + at)

Here, / i , f2 are arbitrary twice differentiate functions. Consequently, the expressions for p' and v' are

p = - A ) _ = p 0 a ^ - _ _ j ,

yl = dip = f[(r-at) _ fi{r-at) f2{r + at) _ f2(r + at) dr r r2 r r2

Here, /{, f2 are the derivatives of the functions fi, f2. The solution is a sum of running waves, one of which propagates with speed a away from the center, and the other toward the center. Unlike plane waves, the intensity of spherical waves varies during propagation: the amplitude of divergent waves decreases, and that of convergent waves increases.

24.16 Linearization leads to the equations

dt dx dx dt dx po dx

Seeking for the solution in the form p = p, e^*"-"*), v = vt e't**-"*) yields the dispersion relation (the relation between w and k) w = (U ± a)k. The phase velocity of the waves is U + a or U — a.

24.17 For a stationary observer, \u\ = ak (see Problem 24.12). An observer ahead of the source (the source approaches the unmoving observer) perceives the wave propagating forward, i.e., whose velocity is Ui/k\ — a. In a system moving with the source, the medium has the velocity —U, therefore the phase velocity of the wave is —U + o, and the frequency is u, = (-U + a)ki (see Problem 24.16). Consequently, u>i = o;,/(l — U/a) > w„. An observer behind the source perceives the wave propagating backward; for that wave, w2 = — ak2, w, = —(U + a)k2, u2 = w,/(l + U/a), \UJ2\ < \u,\. Thus, the sound of an approaching source is perceived by a stationary observer as higher, and the sound of a receding source as lower, than in the case of a stationary source. This is the Doppler effect.

24.18 The disturbances propagate along all directions over particles of the medium with speed o (see Problem 24.15) and, simultaneously, are brought away by the stream along the x-axis with speed U. Therefore, the boundary of the perturbed region has the shape shown on Figure a24.2 with the solid thick line for t = t2 and with the dashed line for t = t\. As t —► oo, the whole stream is perturbed if U < a; if U = a, only the region downstream from the plane x = 0 is perturbed; and, if U > a, only the region inside the Mach cone is perturbed, where the Mach cone is a cone directed downstream with apex at the point O and with apex half-angle a such that sin a = a/U = 1/M.

U

U<a U=a U>a

Figure a24.2

24.19 The linearized system for one-dimensional barotropic small disturbances p'(x, t), f/(x, t), t / ( i , t) in a compressible viscous fluid has the form

dff dv' n di/ I dp' # V , 2 , . (dp\

Here, po = const, p0 = const, po = p{po) specify the undisturbed state of fluid, i/i = (X + 2p,)/po, A, p. are the viscosity coefficients. Eliminating p', pf, one obtains the equation for v'(x, t)

~W Ul dtdx2 a dx2~° For the solution to this equation seek a monochromatic wave v' = A e*(fcz_urt) where k is a real number. Substitution of this into the equation for v' yields the dispersion relation (the relation between u and k)

uP + ik2vxu> - o2fc2 = 0 .

Its solution is u = 0.5 (-ik2V! ± k y/4a2 - k2v(\ .

Let k < 2ajv\. Let us denote Rea; = ±w0- A solution of the equation for v'(x, t) has the form

v' = Ae'(kx±wot) • e-05*2"1'

and includes the factor e~05k "'* decreasing with time (compare with the similar decay of transverse waves in an incompressible viscous fluid, see Problem 22.11).

24.20 Let us direct the x-axis to the side of region (2). There are incident Pio = /o(' — x/ai) and reflected j / n = fi(t + x/ai) waves in the region (1), so that the

pressure disturbance is expressed by the sum p$x,t) = f0(t - x/a{) + fi(t + x/a$. And there is only the transmitted wave in the region (2): p'2{x, t) = ip(t — x/a2). For the velocities of particles, respectively,

vi = [f0{t - x/a{) - fi{t + x/ax)] , v2 = ipit ~ x/a2) pi<*i P2CL2

(see Problem 24.13). On the boundary x = 0, the conditions for a contact discontinuity (Section 17) p\ — p2, vi = v2 should be satisfied. These conditions yield the equations that allow /1 and <p to be obtained from /o given:

h « = /o(0 ^ 7 , v(t) = /o(0 r r ? where ^ = rr • 1 + C, l + c, a2p2

Thus, p'n/Pio = ( l - C ) / ( l + 0 ) P2/P10 = 2 / ( 1 + 0 at the interface, a) For propagation of the wave from air to water, a\p\ <S a2p2, ( -C 1, consequently, p'n w p'10, p2 w 2p'10. b) If medium (1) is water and medium (2) is air, then ( > 1, and, consequently, P'n ~ —P'io> Pi ~ 0 (sound disturbances barely propagate from water to air).

24.21 Let the axes y and x be directed , respectively, along the interface and along the normal to it, so that f = x cos 9 + y sin 9. Thus, <p0 = A exp[i(xfccos# + yksin9 — uit)] in the incident wave, and the wave vector has the form k = e\k cos 0 + e2k sin 9. Similarly, write the disturbance in the reflected and refracted waves in the form <pi = A\ exp[i(—xk\ cos#i -I- yki sin#i — ui\t)] and <p2 = i42exp[i(a;A:2Cos#2 + yk2 sin 92 — u2t)\ respectively. The fact that the interface exists at x = 0 cannot affect the dependence of all functions on y and t; i.e., the exited waves should have the same frequency and y-component of the wave vector as those of the incident wave {ijj\ =LJ2 = u, fci sin#i = fc2sin#2

= £sin#). The speeds of the incident and reflected waves are identical (equal to the speed of sound in the medium (1)) a.i = w/k = u}/k\. Consequently, k = k\ and 9 = 6\. In the second medium, a2 = u/k2 = a,ik/k2. Hence, sin 92 = {0.2/0.1) sin 9. If a2/ai 3> 1, then there is no transmitted wave for angles 9 that are not too small (total internal reflection). This property of wave propagation near interfaces between layers with different acoustic properties is the basis for the waveguide effect — sound disturbances do not emerge from a layer in which the speed of sound is less than that in the surrounding regions and, consequently, do not scatter their energy and decay in less degree. The continuity conditions for the pressure and velocity component normal to the interface allow to find the amplitudes of the reflected and refracted waves (A\ and A2) to be found.

24.22 Let the perfect gas forming the first phase have a density pi, a mass m-y and occupy a volume Vi. Similarly, p2, m2, V2 are the density, mass and volume of the incompressible fluid. The total mass m = m\ + m.2 occupies the volume V = Vi + V2, so that the density of the mixture is p = m/V = (piVi + p2V2)/V.

The mass concentration of the first phase is a = mi/m, that of the second phase is m2/m = 1 — a. The density of the mixture is obviously p = p(pi,p2,a). Write this relation in the explicit form: Vi/V = ap/pt (as it follows from the definition of a) , similarly, V2/V = (V — V\)/V = 1 — apj p\. Substituting into the definition of p, one obtains

p = P1P2 ( a 2 4 H )

(1 -a)pi +ap2 Since there is no heat exchange, the entropy of each phase does not vary: Si = const, s2 = const. The entropy per unit mass of the mixture is

m1sl+ m2s2 . . , . s = = asi + (1 — a)s2 = s(a) .

m By the definition of speed of sound, a2 = {dp/dp)3=consi. Hence, a2 = (dp/dp)a=C0QS^. By the statement of the problem, the pressures in both phases are identical and can be represented by the pressure in the perfect gas: p = px = Apj. Consequently,

where Oi is the speed of sound in the compressible phase. The value of (dp\/dp)a can be obtained from the expression (a24.14), taking into account that fo = const. As a result, one obtains

(1 - a)pi + ap2 a[a) = ai -= P2Va

The found expression has a minimum at a, = Pi/(p2 — pi)- And also, if p2 > 2pi, then a(a,) < oi, and, consequently, there is a range of a values (a2 < a < 1) for which the speed of sound in the mixture is less than that in the compressible phase (in the incompressible phase a2 = 00). a) a = ai ■ 0.08 ~ 27 m/s. b) a = ax ■ 0.11 ~ 37 m/s.

Motion with small disturbances. Steady flow around a thin body

24.23 a) Let us direct the i-axis along the direction of the stream velocity v. Let vx = v0 + v'x, vy = v'y, vz = v'z, p = p0 +p\ p = Po + p1 (po, Po are the pressure and density in the unperturbed stream). All the quantities with primes as well as their derivatives with respect to the coordinates are assumed to be small (the derivatives with respect to time vanish, since the flow is steady). With accuracy up to terms of the first order, the continuity equation is

dd , . v0 - r - + po div v = 0 .

ox

The Cauchy-Lagrange integral, in the given case, has the form .,2

-I- V{p) = C , C = const , V = I P

?-+V(p) = C, C = const, V=f—. 2 J p

Since v —> v0, p —> po as x —» —oo, C = v%/2 + P(po). With accuracy up to small quantities of the first order,

v2 = v v = vl + 2v0v'x, V(p)=V(p0)+(-^) \ V ) p=vo

(p-po) = P(po) + - , Po

therefore the Cauchy-Lagrange integral yields

Pov0v'x + p' - 0 , or p' = -p0v0v'x .

Besides, / (dP\ i 2 / p = U J p=a°p-

From these relationships, using also the condition v' = grad ip', one obtains

b) dif/dn = n ■ grady = v0nx + n ■ gradip' = 0 on the surface of the aerofoil E0, n is the normal to Eo- c) Since the aerofoil is thin and the angle of attack is small, the normal n to Eo is approximately perpendicular to the x-axis, i.e., nx is small. Therefore, on Eo,

dw , . dip' dip' -T- = v0nx+n- gradtp = v0nx + — - T ^ + — - n z . on ay oz

Next, since the points of the surface Eo are close to the x-axis, it can be assumed that the boundary condition should not be met on Eo, but on the segment [0, /] of the x-axis. Thus, the boundary condition on the surface of the aerofoil is reduced to the condition

Vo°L + W d± + W_ dl = o dx dy dy dz dz

at 0 < x < /, y = z = 0. Here, f(x,y,z) — 0 is the equation of E0 (n = grad/ / |grad/ | ) . d) For a stream of an incompressible fluid, the equation for the potential is Aip = 0, and the Cauchy-Lagrange integral and boundary conditions have the same form as for the stream of a compressible fluid with small disturbances.

24.24 a) Use the relationships obtained in solution of Problem 24.23. In this stream, tp = ip(x,y), and the equation of the surface of the wing has the form y — /i2(i) = 0 for the lower side, y — hi(x) = 0 for the upper side. The equation for the potential is

( ! - M o ) - ^ - + ^ = 0.

The boundary conditions on the surface of the wing are dtp'/dy = VQ dh/dx at y = 0, 0 < x < I where h = /i2 on the lower side, h = hi on the upper side, b) Let M0 < 1, then introduce new coordinates x, y and potential Tp by the formulae

x = x , y = ysj\ -Ml , Tp = <p\/l - M£

The equation for Tp is the Laplace equation

dx* + dy2 '

and the boundary condition is:

dip dh = VQ — at 0 < i < /. ,„ dx

y—»±0

The formula for p' takes the form

PQVQ dip

2 dx P = - V^M? It is obvious that the problem for Tp in the coordinates x, y coincides with the problem for <p in the coordinates x, y corresponding to the flow around a wing by a stream of an incompressible fluid (see Problem 24.23). The pressure disturbance at every point of the surface of the wing in the stream of the compressible fluid is 1/yl - MQ times as much as that in the stream of the incompressible fluid. The total force acting on the wing is the same number of times as much, since it equals

- pndo = - / ( p - Po)n da = — / p'n da . £o Eo Eo

Here, So is the surface of the wing, n is the normal to Eo- For a stream of an incompressible fluid under the conditions considered, the d'Alembert paradox is known to be valid, i.e., the drag of the body vanishes. The same is true for a subsonic stream of a compressible fluid.

24.25 Use the relationships obtained in Problem 24.24a. If M0 > 1, the equation for <p' is the wave equation

dy2 dx2

where A = V/MQ — 1. The general solution of this equation is

<p' = fi{x-\y) + f2{x + \y) .

The function / i is constant on the lines x — Xy = const, and f2 is constant on the lines x + Xy = const. The sine of the angle of inclination of these lines to the direction of the velocity of the stream equals ±1/M. These lines are called Mach lines. Since <p' = 0 as x —► —oo, the values of the function f2 are zero in the upper half-plane (y > 0), and f\ = 0 in the lower half-plane (y < 0). Let us find f\ and f2, using the boundary conditions on the surface of the wing. If 0 < x < I,

-Xf^vo— , Vi = v0—.

Hence, /i(z) = - j M*) . Mx) = T h^x) X

if 0 < x < I. Thus, in the upper half-plane, Xtp' = —v0 hi(x — Xy) if 0 < x — Xy < I, ip' = 0 if x — Xy < 0. In the lower half-plane, X(p' = v0 hi (x + Xy) if 0 < x + Xy < I, <p' = 0 if x + Xy < 0. It follows from the condition for pressure and, therefore, velocities vx and vy to be continuous at the x-axes downstream from the wing, i.e. at y = 0,x>l, that f{(x) = f'2{x) and -X}[{x) = Xf^x) if x > Z, i.e., f[[x) = f'2{x) = 0 if x > /. Consequently, /i(x — Ay) = const if x — Ay > I, and f2(x + Xy) = const if x + Ay > I. Therefore, the stream is not perturbed downstream from the lines x - Ay = / and x + Ay = /. Thus, the stream is perturbed only in the region bounded by the corresponding Mach lines (see Figure a24.3). Using the Cauchy-Lagrange

Figure a24.3

integral (see Problem 24.24), one finds the pressure disturbance on the upper and


lower surfaces of the wing

/ _ PQVQ dhi , _ p0v0 dh2

X dx X dx The projection of the force acting on the wing onto the i-axis (the drag) is

- / p'nx da . So

Here, E0 is the surface of the wing, n is the normal to Eo external with respect to the wing. In the approximation used, nx = —dhi/dx on the upper side of the wing, nx = —dh2/dx on the lower side, and integration should be done over the corresponding part of the plane y = 0. The drag per unit length of the wing is

P _ [ _PoVo_ (dhA2 (dh2\2 dx

For a plane plate with a small angle of attack e,

Therefore, the drag Rx is

dhi\ _ (dh2\ _ 2

dx I \ dx

= 2p0v$le2

^/K:rl Thus, a wing in supersonic flow experiences drag. This type of drag is referred to as wave drag and is essentially connected with the fact that the region, where the velocities and pressures are perturbed, expands downstream to infinity and the disturbances in this region do not decay downstream from the body.

Propagation of finite disturbances in an ideal compressible fluid

24.26 a)

dp dp_ dv_ _ dt dx dx dv dv 1 dp dt dx p dx ds ds , . - + v - = 0 , s = s(p,p)


b) The third equation of the system of the item a) has the characteristic form, the speed of the corresponding characteristics is c = v; the two first equations do not have the characteristic form, c) First, rewrite the continuity equation so that it includes only derivatives of p and v, taking into account that p = p(p, s) and ds/dt+v ds/dx = 0:

? ( £ + - E ) + ' ! - 0 ' •--(!). «=const Multiply the continuity equation by l\, the equation of motion by l^ and sum them. As a result, one obtains

h (dp ( W\ dp\ , (dv ( hp\ dv\ , „

This combination has the characteristic form if

l2a2 hp v + -— =v + — = c ,

hP h

i.e., if I1/I2 = ±a/p. Thus, there are two values for the speed of characteristics:

C i = v + a , C2 = v — a

and two equations in the characteristic form:

/ dp dv , d d , .8 pa dt dt dt at ox

I dp dv , d d , . d 17 + "17 = F* ' w h e r e 17 = a7 + (u ~ a ) ^~ •

pa dt dt dt at ox These two equations together with the third equation of the original system compose a complete system in the characteristic form, d) For an adiabatic barotropic motion, s — const in the whole stream, a = a(p), and dp = a2 dp. Let us introduce J+ and J- by the formulae

J+ = v + -dp , J- = v — -dp . J p J p

Then the equations in the characteristic form are equivalent to the statements: J+ is constant along the characteristics dx/dt = v + a, and J_ is constant along the characteristics dx/dt = v — a.

24.27 v±a,v; J± = v ± 2a/('y - 1), s; here a = ^-yp/p, 7 = (cv + R)/cv.

Figure a24.4

24.28 The characteristics L~k and Lg, the equations of which are, respectively, dx/dt = v — a and dx/dt = v + a, passing through the points XA, 0 and XB, 0 (see Figure a24.4a) and the values of v, p, s in the regions located at the left with respect to L~K and at the right with respect to Lg, do not vary if the values of v, p, s at the instant t = t0 vary only in the segment [XA,XB]. This statement can be proved using the method of characteristics (applied to a numerical solution of the problem). The essence of the method is as follows. Let the solution for t — t0 be known, and the solution for x = xi at the instant tx = t0 + At is to be found. Draw characteristics through the point xi, ti (see Figure a24.4b), and replace the derivatives in the equations in the characteristic form by the ratios of finite differences. From the resulting relationships, knowing the values of v, p, s at the intersections of the characteristics and the axis t = to, obtain their values at the point xi for the instant t\. This way, the solution for all x at t = ti can be found. Next, similarly, one can construct the solution for the instant t2 = ti + At, etc. It is obvious that the values at the point X\ of v, p, s at the instant t depend on the values of v, p, s at the instant t0 only at Xj < x < xr where Xj, and xr are the values of x at the intersections of the characteristics L+ and L~, respectively, passing through the point x\, t (see Figure a24.4b). Continuity of the solution is essential, since only under the condition of continuity the characteristics of the same family do not intersect; and, for that reason, the solution obtained by the method of characteristics is single-valued.

24.29 For the motion considered, s = const. Substitute into the system presented in solution of Problem 24.26a v = v(8), p = p(0). As a result, one obtains

o2 [dt + v

30 dt + v

d0_ dx d0_ dx

dp 86 dv dl + Pdidl dv IdO^dp dd + pdx dO


This is a system for determination of dp/dd, dv/dO. In order that it has a nontrivial solution its determinant should equal zero, i.e.,

39 39 39 39 , , . 39 n

m+vd-x = *ad-x

or di + {v±a)d-x = 0-Substituting these relationships into the original system, one obtains

39 f I dp dv\ „ . fdp ,, 3x-[11ade+PM)=0' L e - ^ y - = M = const.

Thus, two solutions in the form of a Riemann wave are possible. They correspond to the signs + or — in the formulae. For the first of them, J_ = v — / dp/(pa) = const in the whole stream, and 9 = const along the characteristics L+ with the equation dx/dt = v + a; L+ are straight lines x — (v + a)t = const, since v = v(9), a = a(9), and, consequently, v + a = const along L+. If x — (v + a)t = const, v, p do not change, consequently, v = v(x — (v + a)t), p = p(x — (v + a)t). These formulae show that each value of v and p is transferred with the speed v + a, therefore the solution is a wave. The value of v + a in a Riemann wave equals / dp/(pa) +a + M and depends upon p. Consequently, different values of p are transferred with different speeds, and therefore, the shape of the wave varies. The second solution is a wave in which constant values of v, p are transferred with the speeds v — a.

24.30 a) Direct the x-axis along the tube. At t = 0, let the piston be at the coordinate origin and the gas occupy the region i > 0. The system of equations has the form (see Problems 24.26-24.27)

J+ = v + 2a/(-y - 1) = const at L+ (a24.15) J_ = v - 2a/(7 - 1) = const at L~ (a24.16)

where L* are the characteristics with equations dx/dt = v ± a, a = ■J'yp/p = Cp(-r-i)/27 j s t n e Speed 0f Sound (C — const = ao/po "')■ The boundary condition is v = vx = u at x = X(t) where X(t) is the coordinate of the piston. This condition is valid until the gas does not separate from the piston (see item d)) b) Supposing the solution to be continuous, let us apply the method of characteristics (see the solution of Problem 24.28). Using the relationships (a24.15) for the characteristics passing through the points of the x-axis (x > 0, t = 0), one obtains that v = 0, a — ao, p = po everywhere in the region x > aot (see Figure a24.5). Consequently, the boundary T of the perturbed region moves with the speed ao and coincides with the characteristic LQ . On all the characteristics L~ initiating from the points t = 0, x > 0 and crossing LQ", v-2a/(j— 1) = const = -200/ (7-1) , i-e-. a = 00 + 0.5(7 —l)u everywhere in the region adjacent to LQ. Therefore, v and a are constant on L+, and L+ are straight

Figure a24.5

lines. This is a Riemann wave (see Problem 24.29). c) Assuming that the region of the Riemann wave extends up to the piston and using the boundary condition D = UOH the piston, one obtains that v(x, t) = U(T) on the characteristics x—X(T) = c(r)(t—T) where r is the value of t at the intersection of the characteristic L+ passing through the point x, t with the line x = X(t), C(T) — (v + a ) I = x ( r ) = ao + 0.5(7 + 1)U(T).

t=T

Expressing r in terms of x and t, using the equations of the characteristic and substituting into U(T), one finds v(x,t) and, consequently, also a(x, t) and p(x,t). At T < ti, i.e. x>xi = X{ti) + c(ti)(t - ti):

v=- ^j(ao + ^ t i At) - 2A7(oot - x) - - (ao + '1~ At) .

If T > t\, U(T) = UI = const; consequently, at x < i;(t) = X(t\) + c(ti)(t — ti), there is a translatory stream with velocity v = u\, and

In the region x;(t) < x < oot, the inclination of the characteristics L+ to the t-axis decreases as T grows, since u < 0 and |u| increases. Therefore, the characteristics L+

do not intersect and the solution is unique. The density and pressure in the particles of the gas decrease with time in this region (rarefaction wave), and the wave itself extends in space, d) Since a > 0, consequently, v > —2ao/(7 — 1) everywhere up to the piston, and the obtained solution is valid if |u| < 2ao/(-y— 1) (= 5ao if 7 = 1.4). If u = —2a0/(7 — 1), the pressure on the piston vanishes. If |M| > 2ao/(-y — 1), then the piston separates from the gas, i.e., there is vacuum between the gas and the piston. The velocity of the gas adjucent to vacuum and that of the interface itself equals 2ao/(7 — 1) (the velocity of unsteady outflow into a vacuum). It is calculated from the condition p = 0 at the interface with the vacuum, e) If <i tends to zero, the set of characteristics on the plane (x, t) takes the form plotted on Figure a24.6. In the

Figure a24.6

region xi(t) < x < a^t, all the characteristics L+ pass through the coordinate origin, and their equations are

( 7 + 1 \ , x = I a0 H — v I t .

Hence,

7 + (x \ 2 / 7 - l x \

P = Po

7-

7 — 1 i 7 + IV 2 dot)

The boundary i ; is determined by the conditions v(x,t) = u\ if |«i| < 2ao/(7 — 1) and p(i t , t) = 0 if |ui| > 2ao/(7 - 1)-

24.31 Formally, the solution can be tried in the same form as in Problem 24.30. As a result, the solution is a wave in which the pressure and density in particles of the gas increase (compression wave). However, since u(t) > 0 and u(t) increases as t grows, the characteristics L+ originating from the points of the trajectory of the piston intersect (see Figure a24.7). As a result, the velocity and pressure lose their

»- x

Figure a24.7

uniqueness. Consequently, this solution becomes invalid beginning with the instant


of the first intersection of the characteristics L+ . If u(t) — At, the instant of the first intersection of the characteristics L+ is

_ 2oo_ ^ - ( T T T J A -

For this problem, a solution can be constructed which contains discontinuities of the velocity and pressure — shock waves.

24.32 a) As a result of the motion of the piston, a simple rarefaction wave (Riemann wave) adjacent to the piston arises in the gas. Let the piston move to the right from the plane x = 0; and the boundary of the rarefaction wave, being a weak discontinuity, moves to the left with the speed of sound ao = (7PO/A>)^2 (7 is the adiabatic exponent). Then the velocity distribution in the gas is expressed in terms of the pressure p (see Problem 24.30) by the formula

•-£(«-(£) ^

Using this relationship on the surface of the piston, one obtains its equation of motion

( 7 - 1 \ ^ mvp(t) = Spp = Sp0 [1 - — - vv(t)J

Integrating, one obtains

_( 1 + (7+i ) f tg ,y \ 2mao /

As m —> 0 (or t —► oo), vp —* 2a0/(7 — 1). b) If 7 —► 00, the condition for the motion to be adiabatic

dt\ 0 becomes

? = «• dt In this limit, vp = 0; i.e., the pressure is instantly discharged, and the piston remains at its initial place.

24.33 a) The functions describing motion of the gas have the form x = a(t)£, a(0) = 1, A = a/a, a = xp(t)/l, I = xp(0) = mg/Sp0. Using the equation of adiabatic motion of the gas, one obtains (7 is the adiabatic exponent)

Poa(t)Z + - 5 7 = ° . P= — > P= — r • #£ a a7

Separation of variables £ and t yields

da1 = A = const > 0 , PoiO = ~^Po

from which it follows that

a2 A _ ifloo) 2 ( 7 - l ) a ^ - 1 2Z2 '

Calculation of the conserved total energy of the system at t —> oo yields

-E-— (m + ros/3) or rj = I m + m.g/6 '9/

(the internal energy of the gas tending to zero, since a —» oo). Then

wj!(oo)(7 - 1) £0(7 - 1) A = 2/2 (m + m9/3)/2

1 With the equalities E0 = S fp0(x)dx/(y - 1) and p0 = f(x)p1 taken into account,

0 one obtains

(7 - l)Ep m + 0.5m8(l - x2/l2) Po~ SI m + mg/3

b) In this case, p = A)(0/°W< anc* t n e ° t n e r relationships are the same as in the item a);

m _ £0(7 ~ 1) ■, _ 1 ~ « ^ " m + X ' (m + ^ f ) / 2 ' ^ " S - a ™ 9

_ (7 ~ 1)^0 m + (1 + 1/(2 - a))M{l - (x/l)2~a) Po SI m + M

As Q —> 1, pa —► (7 — \)EQ/SI and 77 —»• 1. At the initial instant, almost the whole gas is concentrated near x = 0, and its energy is distributed uniformly from x = 0 up to x = /.

M o t i o n wi th shock waves

24.34 a) Prom the conditions on a shock wave (see Problems 17.10-17.11)

Pi(D-u) = p0D = j , ju=p1-p0

fu2 p\ Po \

(T+(T3T^-(^rT^J=^

If p0 and p0 ahead of the wave are given and v = u behind the wave, one can obtain Pi, p\, and the speed of the wave D. In particular

Pi. = 1 | 7(7 +l)u2 7U Po 4ag ao \ 16a(j

1 + ( 7 + l ) 2 « 2

b) One obtains for the difference of the values of entropy behind and ahead of the wave

Si - s0 = cv In I — I — \Po \Pi

Let us denote pi/po = t > 1. The equation of the Hugoniot adiabat (see Problem 17.11) yields

po _ 7 + 1 + (7 ~ 1)* Pi 7 - 1 + (7+1)*"

Then PifPoY > / ( 7 + l ) + ( 7 - l ) « X 7

SUJ =H(7 + Dt + ( 7 - l ) J > 1 ' a t i > 1

Therefore, si — s0 > 0 at pi > po- To examine evolution of the shock wave, it is necessary to note that the number Nj of characteristics going away from it equals two since there are three jump conditions (of mass, momentum and energy conservation). There are three types of characteristics at both sides of the shock wave; their velocities are ao, —ao, 0 in front of it and u+ai , u — a.\, u behind it. It follows from the equation of mass conservation

Pi{D - u) = p0D > 0

that D > u > u — ai; so characteristics dx/dt = u and dx/dt = u — ax "go away" from the shock wave. Then all the other characteristics must come toward it, i.e., the evolutionarity conditions (24.1) take the form

—ao < 0 < a o < D , u — a.i <u<D<u + a\

The validity of the first of these inequalities follows from the formula for D of the item a). Satisfaction of the second inequality follows, with the inequality p0 < pi taken into account, from the formula

( D_W i ) 2 = a?7ZJjJl_MWPi

which can be derived from the conditions on the shock wave, c) In this case pi/po < 1; therefore, the inequalities of the item b) are not satisfied (see also Problem 14.16).


24.35 a) Due to spherical symmetry, the equalities va — v^ = 0, vT = v(r, t), P = p(ri 0 a^d P = p(r, 0 are valid for the velocity components, pressure and density of the gas, respectively. The system of equations is

dp d(pv) 2 fdv dv\

I(J)+-|(J)-'-

9p n

b) Let m = An f p(ri, r)r2 dr\. Then, regarding r a s a function of m and t, one obtains o

_dr 1 dt ' 4irr2dr/dm

The required equations have the form

dv . ■> dp ~ P r, x

c) The energy equation

9 /pu2

( T ^ B ! H T + ^ I ) ' " * ) -takes the form

dt V 2 ^ (T - i)py + -z-(4Ttr2pv) = 0

am d) Let r = R(t) be the function describing motion of the discontinuity. In terms of the mass, m = M(t), R(t) = r(M(t),t). The radial velocity of the discontinuity is D = R(t). Then, if the medium is continuous, one can write the jump conditions in the form ( [a] denotes the jump of a )

[p(v-D)} = 47IT2

i (v_v_dL.M) dr/dm \ dm )

M r 1 47T

[p(v - D)v + p] 47T.R2

* -»>(T^)

dm

\-Mv + 4TT R2p] = 0,

+ pv

R?\ = 0 ,

4TT^2 H^(^W pv = 0


* \ 6 / 5 „J/5 2/5

24.36 a) The function describing the motion of gas has the form (use the II Theorem, Section 36)

(3mY/3 . , . tEWJ3

r = ( 4 ^ o j n(T '7) ' Where T=-^~ On the shock wave,

/ 1 , 6 / 5

The expression for the function / is

.EoMj)

b) dE/dt = -4nr2pv, dE/dm = v2/2 + p/(7 - \)p. Due to the jump condition connected with energy conservation law (Problem 24.35d) and the fact that the velocity and pressure ahead of the shock wave vanish, E(M(t), t) = 0. Since limt_+o E(M, t) = EQ in accordance with the statement of the problem, E(M, t) = EQ. According to dimensional analysis, E(m,t) = EQEI(T), EI(T,) = 1. Differentiating this relationship with respect to m and t, eliminating the derivative E[(T) and using the expressions for the derivatives dE/dt and dE/dm, one obtains

m (v2 p \ lOrr 2 U + F^J-r '^-

c) Expressing the density and velocity in terms of the function r(m, t) (see item (a)), one obtains

!/3 /IT„xl/2 EoV' ' , _ Po_ rlT

V - \ ^ ) \m) r'v P ~ r ? ( l - 2 . 5 u ) '

The energy integral (item (b)) yields the equation for ri(r)

1 (±\2/3 n = /i(2.57ti-l)

2WJ r i ( 7 - l ) r f 7 - 1 ) ( l - 2 . 5 u ) 7

which can be integrated by taking u as the parameter. Prom the conditions across the shock wave (Problem 24.35d), one can derive the relationships

7 + 1 v, P, = 7 A),

2

7 - 1 " " 2 (7-l)p.

from which the values

4 / 2 2 \ t 7 - 1 / 3 \ 2 / 3 / 7 - n 7 _ 1

5(7 + 1) V 5 7; 5 J ' A 2 UJ (7+1J ( r j

can be determined. The quantity rs can be obtained from the integral equation for the energy E(M,t) = EQ (the item (b)) having the form

"7(i(sf^rta-§.»- 6 dr_ 5 ~

The equation for rt has a power solution of the form rt = const • T 2 / ' 3 7 - 1 ' for u = 2/(37 — 1) coinciding with u3 = 4/(5(7 + 1)) at 7 = 7. This case should be analyzed separately. If 7 =£ 7, the solution can be presented in the parametric form. At 7 ^ 2

{ ( 7 - l ) u 2 ( 1 _ | w ) 7 ; • °^{Usy S-r 5-r

13-(2-7-, + 12 /5-V \ 2(1+ 27) / 5 \ 6(-»-2) f r i u _ 1 j I 1 u )

At 7 = 2 u 2 4

5u - 1 15 It follows from the condition for the function <p(u) to be monotonic that, if 1 < 7 < 7, then u e (O.4/7; 0.8/(7 + 1)] and, if 7 > 7, then u 6 [0.8/(7 + 1); 0.4). If 7 > 7, the asymptotes of the solution as m —► 0 (r —» co), since u RS 0.4, have the form r « const • t2/s > 0; if 7 < 7, u « O.4/7 and r ss const • £2/57rn(7-i)/37 _> Q Thus, if 7 > 7, there is an expanding cavity, d) If 7 = 7 then « = 1/10, r, = (r/r,)1/10 , / , = 6(3/4)7, rs = ( ^ ) (3/47T)1/3, R = (225E0/(2nPo))1/5t^,r = (3/An)1'3 El

0/10/pT (t/ra)^°m^.

24.37 a) It follows from the equations of motion for the gas (see solution to Problem 24.35b)

d2x dp „ „, , ,, (dx\~ ^ + £ = 0,p=/(m)p^,p=^j

(7 is the adiabatic exponent) that

1 P = 7 7 T T v 7 ' ti = l/vp i/(m)(t + ti)' y {t + Up'


where C is constant, b) Prom the jump conditions in Lagrangian descriptions (see solution to Problem 24.35d)

[v - pt's(m)} = 2 V p

~2~ + ( 7 - 1)P pvt's(m) = 0 ,

one obtains two equations for the functions v and ts

Ct' = 0,

Cv' (t. + U)-' ' 2 {i-l)(t,+t,)T

Solving these equations, one obtains

t, kT-'i-'-'i1 5)""1.-

r = o, t, = i/Vp

3(7 - \)P-lM

The initial mass distribution is obtained from the jump condition [x(m, t)] — 0:

or

x = v(m)(t3(m) + t,)-l = l(l-jj)

3 ( T - 1 )

m ( X\~ -r+i M = 1 - ( 1 + y)

- .7t' 3(7-1)

-I

from here, 4-r-2

" 1+1 . , dm 3 ( 7 - 1 ) Mr xy

c) The distribution of mass density of energy is

2 + ( 7 - l ) p 2 \\ M) +K+tt} V M) )

As t —> oo, the density of internal energy is negligible almost everywhere and equal to the kinetic energy density only on the shock wave. The kinetic energy density v2/2, in turn, tends to infinity as t —» oo and m —* M. The total energy transmitted to the gas equals | Mvi and is completely converted into kinetic energy of the gas. The speed of the shock wave is

D = xa(t) = l±±vp(l + t/tl)2^ ,

and the jump in temperature is

where cv is the specific heat capacity of the gas at constant volume.

24.38 a) Since the pressure on the boundaries of the layer vanishes, the initial pressure of the gas equals zero everywhere. Let po a n d lo be the initial density and thickness of the layer, respectively. In the first stage, the given problem is equivalent to the problem of moving a piston into a gas with constant velocity — VQ (Problem 24.34). At the instant of impact (t = 0) a shock wave arises in the gas and passes through the layer with constant speed D. The gas is at rest in the region between the wall and the shock wave. Its density p\ and pressure p\ are constant and can be obtained, together with D, from the jump conditions connected with the mass, momentum and energy conservation laws

7 + 1 7 + 1 2 ^ , 7 ~ 1 Pi = r Po , Pi = —7T- PoVo . D = —v0 .

7 — 1 I I In the second stage, after the instant t\ = 2/0/(7 + l)vo when the shock wave reaches the free surface of the layer, a rarefaction Riemann wave arises related with outflow of the gas into vacuum (Problem 24.30). The free surface begins to move in the opposite direction with the velocity Vb

2 Vb r ^ i , ai =

7 - 1 In the third stage, at the instant t2 = h + h/ai where lt = l0(-y — l) /(7 + 1), the rarefaction wave reflects from the wall. The boundary of the reflected wave (which is again of the rarefaction type) is a characteristic C and propagates with the speed v — a known from the problem of the preceding rarefaction wave. Let us show that it never reaches the surface of the layer, and, consequently, the velocity Vb does not change and equals — v0J2r)l{^ — 1). The distributions of v and a in the first rarefaction wave axe

2 / , i + M 2 / 7 - l i + l, v = -01 + -—— , a = —— cii + 7 + 1 \ t-tij 7 + l \ 2 t-h

( i = 0 corresponds to the position of the wall). The equation of the characteristic C \t > t2) is

dx 2 / 3 - 7 x + lA

its solution is

Comparing with

( 7 - l ) / i 7 - 1

2ai(t-U)

3-1

= _/ ( 1 + 2 a i ( < - < i ) 7 + 1 f a i ( * - t i ) V + I

xb = ~h 1 + ( 7 - l ) * i


one obtains xc > Xb. b) Within the framework of linear elasticity, the processes of loading and unloading during impact of the elastic plate are reversible. The distributions of velocities and stresses are piecewise constant. The discontinuities propagate with the speed of longitudinal waves. After impact, a region that is in a stressed state is formed near the wall. Next, after the discontinuity reaches the free surface, the non-stressed part of the plate starts into motion in the opposite direction with the constant velocity — v0. Also, after the unloading wave passes through the plate, that is then in a non-stressed state and rebounds from the wall with the velocity — VQ. The speed of the boundary of the expanded layer of gas is always greater than \/2 u0.

24.39 Let the x-axis be directed along the velocity v0 of the stream and tp be the angle made by the front of the shock wave and v0- The conditions across the shock wave following from the conservation laws (see Problem 17.10) can be written in the form

PlVln = PoVon

Pi + Plvln = PO + P0VL' ViT=V(yr (a24.17)

"in , ^ Pi = "0» i ? P0 2 7 - 1 pi 2 7 - 1 po

where the subscripts n and r mark respectively normal and tangential components of the velocities:

v\n = fix sin ip - viy cos ip , von = v0 sin tp "IT = f l i COS ip + V\v Sin tp , VQT = V0 COS (p .

The velocity behind the shock wave is directed along the surface of the wedge, i.e., v\y/v\x = tan 0. Prom these relationships, regarding po, po, v0 and 6 as known, one obtains p\, p\, V\x, Viy and tp. In particular, the relation between 8 and tp has the form

cot 9 = tan „ ( _ £ ± l M _ - l ) ^ \2(Mgsin2 tp-1) )

where M0 = v0/ao, of, = -yp0/po- From the conditions (a24.17), the following relationship can be obtained

2 uo«ii - a2, v\y = (v0 - vlx)2

2 2" '—^ (a24.18) 7+1

where a2, = ((7 — \)VQ + 2al)/(j + 1) is the square of the critical speed of sound (see Problem 24.45). From this relationship, with the formula vly = V\x tan 6 taken into account, one can obtain V\v and v\x. The dependence of v\y on v\x, determined by the formula (a24.18), is called the shock polar; it has the form shown on Figure a24.8.

O

Figure a24.8

The distance OD equals v0. The shock polar crosses the straight line Vxy = vix tan 9 at three points A, B and C if 8 < 0max. The solution corresponding to the point C has no physical meaning, since the entropy decreases in such a wave. The solutions corresponding to the points A and B do not contradict all the known conditions, and, for a wedge of finite sizes, which of these solutions is actually realized depends on conditions far downstream. The angle 0max is the angle of inclination of the tangent to the shock polar passing through the point O, and the angle <p is determined by the condition for the tangential component of the velocity to be continuous as the angle of inclination of the perpendicular to the line DB passing through the point O (for the solution corresponding to the point B).

Detonation and slow combustion 24.40 a) By introducing mass flux j the relation (6.11) in Section 6 at m = 0

can be written as vi=jVu v2=jV2

Here t>i, v2 are the gas velocities relative to the discontinuity surface on different sides of it. Then, using the component of the equation (6.12) in the direction of the normal to the shock front at R = 0 and pn = —pn, one can obtain

■2=P2- Pi 3 v,~v2

The absolute values of the wave velocities relative to gas are V\\j\ = Vi\/(P2 -Pi) /(Vi - V2) and V2\j\ = V2sJ{p2 - pi)/(Vi - V2). b) Operate in a manner similar to one of Problem 17.11. Eliminate values V\ and v2 from equation (6.14) at

W = 0, qni = 0, qn2 = 0 with the use of equalities given in item a) to obtain

«a(P2, V2) - «i(pi , Ki) + ^ (P2 + Pi)(V2 - Vi) = 0 (a24.19)

c) Substituting the expression u = pV/(-y — 1) + const into the equation (a24.19) one obtains the equation for the shock adiabat in the form

P2 + ~^Pl){V2 7

7 + 1 7 + 1 V, 47V1P1

( 7 + I ) 2 (a24.20)

Thus, for a perfect gas, the shock adiabat is the hyperbola that meets the point p\,V\ and has the asymptotes p = —Pi(7 - l ) / ( 7 + 1), V = Vi(j — l ) / ( 7 + 1). From (a24.20) the derivative dp2/dV2 at fixed pi, V\ can be calculated. At V2 = V\ it equals —7P1/V1 = — a\2/Vi2 since a2 = 7pV for a perfect gas.

24.41 a) By entering «i = piVi/f /y-l) + Cu u2 = p2V2/(-y-l) + C2 in the equation (a24.19), one obtains

^ 7 - 1 P2 + — - T P l

7 + 1 . , l - l v \ 4 7 K l P l , 2(7 -1)Q

7 + 1 ( 7 + I ) 2 (7 + 1)

where Q = C\ — C2. At fixed Pi,V\,Q and variable P2,V2 the last equality specifies a hyperbola meeting the point V = VI, p = p\ + Q(-y - 1)/V\. The detonation adiabat is shown in Figure a24.9. It is obvious that conditions p 2 > 0, V2 > 0 can be fitted only by the points of one hyperbola branch, b) If there are three jump conditions

p,

Px -

U " kp,

\ A \ -f -T \

i |v, A

\

k=(y-myH)

V

Figure a24.9

then, for the evolu t ionary of the jump, two characteristics should "go away" from it ( see conditions (24.1)). This requirement leads to inequalities (if «i > 0, v2 > 0)

«i > a i , v2 < a2 (a24.21)

where v is the gas velocity relative to the discontinuity, a is the speed of sound, and subscripts 1 and 2 refer to states ahead of and behind the discontinuity. The first of these inequalities means that there are no "going away" characteristics in front of the discontinuity; the second one means that behind it one family of characteristics "comes toward" the discontinuity and two others "go away" from it. It can be proved that conditions (a24.21) are satisfied by the points that lie on the detonation adiabat above the so-called Jouguet point J (the point of contact of detonation adiabat and the tangent drawn from the initial point A with coordinates pi, Vi, see Figure a24.10). The point J correspond to minimal value of j 2 , i.e. the minimal value of the speed of the detonation front relative to gas in front of it (see Solution to Problem 24.40). To

V ♦ ^

V, V Figure a24.10

prove this statement, let us examine the shock adiabat meeting the point A, besides the detonation adiabat. Both curves are hyperbolae with the same asymptotes. In the (p, V)-plane the point p2, Vi is the point of intersection of the detonation adiabat and the line of constant mass flux

(P - Pi)/(Vi ~V)= j 2 = const (a24.22)

The line (a24.22) meets the shock adiabat in a single point E (in addition to the point A). Let us consider the first evolutionarity condition ui > a\. Asai2 = — Vi2(dp/dV)i) and Vi2 = Vij2 (see Solution to Problem 24.40) then the condition vx > a\ means that — j 2 < (dp/dV)i, i.e., the angle to the V-axis for the line (a24.22) should not be more than that for the tangent to the shock adiabat at the point pi, Vi. It follows from here that the first evolutionarity condition is suited only by the points of the detonation adiabat p2, V2 such that the line (a24.22) meets the shock adiabat above the point p\, Vi (the points B and C on the Figure a24.10). The second evolutionarity condition can be suited only by the points lying above the point J. Really, the jumps A —► B and A —► C (see Figure a24.10) correspond to those that involve generation

P

P,

of the same amount of chemical energy and that move with the same velocity relative to gas(which is equal to v\ = jV\. So the jump B —> C corresponds to a shock wave (moving with the same velocity). It follows from evolutionarity of the shock wave B —» C that vB > oB, vc < ac (see a solution to the Problem 24.34b). So the second evolutionarity condition is fulfilled only for the points of type C, i.e. for the points lying on the detonation adiabat above the point J (marked by dashes on the Figure a24.10). At the point J the condition vj = a.j is satisfied.

24.42 Let us study detonation waves travelling over gas at a given state pi,Vi with a certain velocity i.e. with a certain mass flux j across the wave. The equations for a steady flow describing the structure of these waves (in the frames of reference connected with the wave) can be integrated to give

pv = const = piVi = j , pv2 + p = const = p\v\ + pi v2/2 + cvT = ifi/2 + CpTi+q (a24.23)

where q = C\ — uChem, 9 varies continuously from q = 0 immediately behind the front shock to q = Q = C1—C2 far behind it. Formally, the relations (a24.23) for every point of the flow coincide with the jump conditions for a certain detonation front. So one can find the parameters in the flow by studying a set of detonation fronts for all possible values of q at j = const. The states with prescribed pi, Vi and q are (on the p, V plane) the points of intersection of the line of constant mass flux (p—P\)/{V — Vi) = j 2

and the detonation adiabats corresponding to generated chemical energy q. For all q, corresponding detonation adiabats have the same asymptotes. It is evident that solutions describing the wave can exist only for such values of j 2 for which the line of constant mass flux intersect the detonation adiabat corresponding q = Q. At q = 0 the detonation adiabat coincides with the shock adiabat meeting the point pi,Vi. So the state immediately behind the front shock is represented by the point E on the Figure a24.10. As q increases the point representing the state of the gas moves along the line EA. In case (a) q increases continuously and monotonically up to the value Q. The flow is continuous and p, V vary monotonically from their values at the point E to those at the point C. Discontinuities cannot exist in the flow, for the only possible discontinuity should be a shock wave of the type C —* B. But such a shock would be a rarefaction shock which cannot exist in a perfect gas (see solution to Problem 24.34b,c). So in case (a) a structure exists only for such detonation waves for which the states behind them are represented by the points lying on the detonation adiabat above the point J or by point J itself. Note that this result is in accordance with the result obtained from the evolutionarity requirements in the solution to Problem 24.41. In the case (b) let us consider two detonation adiabats: for q = Q and for q = <7max = C\ - C^a (see Figure a24.11). Immediately behind the front shock we again have the state E. Then q increases and the point p, V


Figure a24.11

moves along EA to the point F. After that q decreases and the point moves back along FE to the point C that is the top point of intersection of the adiabat q = Q and the line (p — Pi)/{V — Vt) = j 2 . It is evident that this solution can exist only for j 2 > j£,jn where j m i n is the mass flux for which the line (p — Pi)/(V — V|) = j 2

is tangent to the detonation adiabat q = qmax (see the point Jqmmx at the Figure a24.11). Thus, the states behind the wave are represented by the points lying above the point G. In special case j = jmim besides the solution described above, another one exists: for decreasing q, the point p, V can move from the point Jq„^ upward to the point G or downward to the point H. So in the case (b) a structure exists for waves with P2, V2 represented by points of the detonation adiabat lying above the point G (marked by dashes at the Figure a24.11) and, besides, by the point H. The point H corresponds to "undercompressed" detonation. Note that conditions (a24.21) (see solution to Problem 24.41) are not satisfied at the point H. However, the corresponding detonation wave is evolutionary as there exists an extra condition 0 = jmin)- This condition fixes the velocity of discontinuity, which is the smallest possible velocity of the detonation front.

24.43 Using the dimensionality theory (see Sections 36, 37) one finds that the solution of the problem can depend only on one combination of the independent variables x/t. Thus, it can consist of self-similar Riemann waves, discontinuities and domains of constant parameters. Consider the forms of solutions for different values of the velocity of the piston vp. If the piston moves into the gas and vp (> 0) is sufficiently large, then both in cases (a) and (b) the solution is constructed of the detonation wave and the domain with constant parameters, bordering on the piston. The detonation wave velocity is determined from the condition that the velocity of gas must be equal to that of the piston. As vp is decreased the velocity of the detonation wave is also decreased. In the case (a) at a certain positive vp = v' the detonation wave corresponds to the point J on Figure a24.10 (Jouguet detonation).

If vp < v* (vp can be negative if the piston moves out of the gas) then the solution of the problem consists of the Jouguet detonation front and of the Riemann wave bordering it, with its intensity being determined by the value of piston velocity vp. Such a solution is possible because the gas velocity behind the Jouguet detonation front equals the speed of sound vj = aj (see solution to Problem 24.41). If vp is negative and its absolute value is sufficiently large there is vacuum between the gas and the piston; the Riemann wave is then bounded by a surface where p = 0 (see Problem 24.30). In case (b) the velocity of the detonation front also decreases as the piston velocity decreases, and at a certain vp = v" it achieves the minimum possible value corresponding to j m i n (see solution to Problem 24.42). At vp > v" the solution is a detonation front represented by the point lying above the point G (see Figure a24.10). Because of the continuity of the dependence of the solution on Vp, at Up = v" the solution is represented by the point G. The jump A —► G can be considered as a sum of two jumps: the undercompressed detonation A —► H and the shock wave H —> G (see the solution to Problem 24.42). If vp is further decreased then the detonation wave A —* H does not vary; the shock wave amplitude is decreased and at a certain value of vp it is replaced by the rarefaction Riemann wave. At sufficiently large negative vp, the Riemann wave is ended by p = 0.

24.44 In this case the equation expressing energy flux constancy in the frame connected to the wave has the form:

jrp

With the use of an equation for the chemical reaction

u^- = aT(Q-q),

and, using equalities u = jV, p = RT/V = const, for T > T» we arrive at the system of equations

S-£fe<r-iy-<,-<»i. ! - - > - « . Too = T0 + Q

Temperature T^ corresponds to the homogeneous flow after the end of chemical energy generation, at q = Q. Thus, we have a linear system with a saddle point singularity at T = T^, q = Q. There are two singular directions and two linear integral curves (separatrices) meeting the singular point. One of them is q = Q, and the other is

't+jS)<r-r->-£<« - « - a


The existence of the solution linking the initial state T = T0, q = 0 with the final one T = Too, Q = Q requires that the separatrix given above meets the point T = T„, g = 0, from where

.2 = akp TQQ-T. 3 RcviT.-To)

The wave propagation velocity is v = jV = jRTo/p. This relation determining the wave velocity should be taken into consideration as a supplementary jump condition ( the jump arises in the limit k —► 0, a —* oo with ka remaining finite). Due to the existence of one extra jump condition, the subsonic flame front is evolutionary.

Steady motion of a compressible fluid 24.45 For a perfect gas (p = pRT, R = Cp — cv, 7 = Cp/cv) in adiabatic motion,

p = Ap1. The Bernoulli integral takes the form

where C(C) is constant for a streamUne C. This constant can be determined in terms of the stagnation parameters po, Po, ao, T0

+ ■ _7_p = 7 Po - 1 p 7 - 1 A)

= r = Cpio 7 - 1

The maximum velocity v^n possible for a chosen streamline is reached if the gas flows out into the vacuum, i.e., p = 0, p/p — 0. From the Bernoulli integral, one obtains Umax = t/2cpT0 = ooy 2/(7 — 1). The critical velocity is reached at the point where v = a = vm. From the same equation, one obtains

For air at To = 288 °K, ao =* 340 m/s, !!„», ~ 850 m/s, v. ~ 310 m/s. If the outflow is unsteady, the velocity on the boundary with vacuum equals 2ao/(7 — 1) (see Problem 24.30). For air, it is y/E times as much as Vmax.

24.46 The Bernoulli integral for an incompressible fluid, if there are no body forces, has the form

^ + ? = P« or £ « ! - £ (a24.24) 2 p p Po 2po

where p0 is the stagnation pressure, p = po = const. From the Bernoulli integral for adiabatic motion of a perfect gas (see Problem 24.45)

v 7 P = 7 Po 2 7 - 1 p 7 - 1 po '

using the equation of adiabat p/po = (p/po)7, one obtains

= ( l - - ^ ^ l = [ l - - ^ r ^ f Po P_=(l_ 7 - 1 P o ^ \ ^ = L _ 7 - 1 ^ :

Po V 7 2 p 0 j ^ 2 agj

If v <C OQ, the right-hand side can be replaced by its approximate expression, using the expansion into power series:

L Po 2oo \ l 4 4 + - ) - L 2p0 V 4agJ

If u2/ao >C 1, then the second term within the parentheses can be neglected and the expression for p/p0 in a compressible gas becomes the same as in a incompressible fluid. If the formula for a incompressible fluid is applied to a gas, the error is of order 1% for v2/4a% < 0.01, i.e., for v < 0.2a0. For air CLQ ~ 340 m/s. Therefore, for the velocities v < 68 m/s = 240 km/h, compressibility can be neglected for steady motions (with an error less than 1%).

24.47 For a spherically symmetrical steady flow, the mass flux through any sphere with a center at the coordinate origin is the same and equals the mass flow rate of the source. The equation expressing the mass conservation law can then be written in the form Anr2pv = Q where v(r) = vT. Hence,

Q pv - inr2

In order to find v(r), express p{r) from the Bernoulli integral where, e.g., Vmax is used to determine the streamline constant (see Problem 24.45)

v2 . 7 P_*>Lx ^ - ^ ( P ^

where p0, po are stagnation parameters. As a result, one obtains

Substitution into the equation of mass conservation yields

-3T = / ( — ) r2 ._ QIVn

4 T A ) ^ 7

The function /(v/vm^) has asymptotes at v/^max = 0 and v/vmax = 1 and a minimum at the point where

v h — 1 Wmax V 7 + 1

At this point, i; = v„ and M = jj = 1 (see Problem 24.45). The flow is possible only in the region r > rmili where

r2- = ' min 4irpo(io

/ 7 + l \ ^ T T

In the region r > rmin, there may be a subsonic flow, in which v grows from zero at r = oo up to n, at r = rmin, or a supersonic flow, in which v grows from v, at r = r„un up to umax at r = oo.

24.48 The velocity field v = v^ = A/r (in the polar coordinate system) satisfies the continuity equation div pv = 0 and the condition curlv = 0 both for an incompressible fluid and for a compressible gas. The constant A specifies the intensity of the vortex. As the motion is steady, barotropic and irrotational, the integral of Cauchy-Lagrange is valid

^ + IP _ 7Po _ i>iLx 2 ( 7 - l ) p (7- l)A> 2

where p<j, po are the values of p, p at r —> oo. From here

(7-i)p0^r /(7_1) Po 1 27p0r2

The flow can take place only in the region r >r„

' nun — — -*M "max V 27p0

If r = rmin, v = Umax! M = oo. The velocity v = v, is reached at r = r , = A/v,. At r > r,, the velocity is subsonic, decreasing up to zero at infinity.

24.49 The continuity equation is obtained by writing the mass conservation law for the control volume shown on Figure a24.12:

— f pdV = J pvnda = (pvxQ)x+dx - {pvxn)x = d(pvQ.) = 0 ,

since vn = 0 on the lateral surface of the tube, vn = vx in the section x + dx, vn = —vx in the section i (later on, vx is denoted by v). Similarly, using the momentum and energy conservation laws, one can obtain the remaining equations.

x+dx

Figure a24.12

24.50 Write the continuity equation in the form

dp dv dil p v il

Next, from the equation of motion, using the condition for the motion to be reversible and adiabatic ds = 0, one obtains

1 a2

-dp = — dp = —vdv ; 9 9

hence,

Therefore

dp v 2 dv — = — - d v = - M 2 — , p aJ v

( M 2 - l ) ^ = d n

v

M

n ' Thus, as the tube becomes more narrow, the velocity increases in a subsonic (M < 1) steady stream and decreases in a supersonic (M > 1) one. In order to obtain a supersonic velocity at the outlet of the tube, the velocity at the inlet being subsonic, it is necessary to use a tube the cross-sectional area of which first decreases and then increases as shown on Figure a24.13. A tube of such a shape for the discharge of a gas is called a convergent-divergent nozzle (Laval nozzle).

CX3 Figure a24.13

24.51 Operating in the same manner as in solution to Problem 24.50, one obtains (taking into account that Cl = const and ds = 0 where s is the entropy)

( M 2 - l ) - = F,dx. v

Let the x-axis be directed so that v > 0. a) Fx = g > 0, a subsonic stream decelerates, a supersonic one accelerates, b) Fx = — g < 0, a subsonic stream accelerates, a supersonic one decelerates. The physical meaning of these results can be understood with using the following reasoning. Let us consider such small velocities that the quantity vdv in the equations of motion can be neglected. Then, the dependence of the pressure and density on x is determined by the equation of hydrostatics and by the condition s = const, from which it follows that the density decreases as the height grows. Due to conservation of the mass flux, pv = const, consequently, the velocity increases as the height grows. In the other limiting case, for which the velocity is very large and, consequently, the term {dp)/p can be neglected, the motion can be regarded as caused only by inertia, and the gravity force obviously decelerates the particles if they move upwards and accelerates them if they move downwards.

24.52 Use the equation of state for a perfect gas in the form (see Problem 15.8)

p = Ce'l^p1 , C = const .

Then , 2 J . Pds 2 IP Cp R + Cy dp = a'dp+ -— , az = — , 7 = -2- = .

Cy p Cy Cy

From the energy equation (see Problem 24.49) with Gibbs' identity du = Tds + (p/p2) dp taken into account, one obtains

dp a2 p — = — dp H ds = -vdv -Tds = -vdv + fdx p p pcv

Hence,

Using the continuity equation in addition, one obtains

v ' v R RT

Since the motion is adiabatic, the entropy of the particles increases due to friction: ds > 0 if v > 0. Consequently, a subsonic steady stream accelerates due to friction, and a supersonic one decelerates.

24.53 Using the method of Problem 24.52, one obtains

v pa'-Cv a*

since for the process under consideration ds = dq/T where dq is the quantity of heat transmitted to unit mass of the gas for the time dt = dx/v (v > 0). If heat is transmitted to the gas, then dq > 0, and, consequently, a subsonic stream accelerates, a supersonic one decelerates. If heat is conducted away from the gas (dq < 0), then a subsonic stream decelerates, and a supersonic one accelerates. It is possible to cause acceleration of the gas from a subsonic velocity to a supersonic one by transmitting heat, until the velocity becomes equal to the speed of sound, and conducting heat away on the subsequent part of the tube. A device using this phenomenon is called a thermal nozzle.

Chapter 6

Elasticity Theory

26 Linearly Elastic Solid Constitutive relations

26.1 e„ = p(A + A*)/M(3A + 2/x) = p/E, e22 = £33 = -pA/2/x(3A + 2/z) = - a e u , £ = H(3X + 2/i)/(A + /x), a = A/2(A + A*), A = Ea/{1 - 2a)(l + a), n = E/2(l + a).

26.2 p„ = - p , p12 = p13 = p23 = 0, p22 = P33 = -?A/(A + 2fi), en = -p/(A -I- 2AJ), £22 = £33 = 0.

26.3 p12 = P21 = r, the other py — 0, 7 = 2ei2 = T/AJ, T = G7, G = n = \E/{\ + a).

26.4 a) 6 = p/(3A + 2^), b) 0 = -p/(A + 2/z), c) 6 = 0.

26.5 e y = ip f l y / (A + §At), 6 = p/(A + f M), AT = (A + § M) = E/3(l - 2a).

26.7 poT = I KJ? + G(J2 - I Jf), AT = A + § At, G = At.

26.8 Use the expansion of the free energy of state into a series over small quantities Eij and T — To, assuming that there are no strains in the non-stressed state.

») H^T) = ( 4 f e ) o exjekl + (§)o (T-To) + ( 0 ) o ( T - T 0 ) ^ ( ^ ) o stj(T-T0), p" = A««£„ + B«(T - T0), ^ < = (gg-\, B* = ( ^ ) Q .

b) poFVu J2, J3, T) = \ AJ? + AtJ2 - aJ,(T -T0)-\ b(T - T0)2 + 7 ( T - T0), Pij = XJiSij + 2/iey - a{T - T0)«y, -A = ek

k, J2 = eye'7'-

26.9 b) p^ = X'JAj + 2n'Eij - b'(s - «„)%

26.10 a = QQK = ZaK, K = X+\n-

165

166 SOLUTIONS. ELASTICITY THEORY

26.11 Use the second law of thermodynamics; b = p0cy/T0.

26.12 eii = £ [py-9AJ15y/(3A+2/x)]+a(T-To)6y = ± [(l+a)pij-aJ1{P)6ij}+ a(T-T 0 )V

26.13 The expression for T — T0 is derived from the equation — ^ = s = SQ. T - T0 = - Q 6 T 0 ( 3 A + 2/i)/(A>Cv), Py = [A + 9a2Ar2To/(poCv)]<A(e) «y + 2/zey.

26.14 A^ = A + a2(3A + 2/i)2T0/(A,cv), * * = K[l + 9a2T0K/(pocv)}, Mad =

26.15 Use the thermodynamic equation of Gibbs

d f (e«, T) = - p y de« - a dT = d^^- - - e« dpy - a dT . P P P

As is obvious from this equation, it is possible to introduce a thermodynamic function $(p«,T) = ±p y e y - T(eij,T) such that £y = Pgpj- If Hooke's law is valid, F and $ are quadratic forms, i.e., ey ^- = 2.F, consequently, $(py) = ^(p i j) .

26.16 An equilibrium is stable if, to cause an arbitrary additional strain Aey = £y — £y, it is necessary to add forces which perform additional work for the change in strain. Obviously, this work equals the difference between the increase in energy and the work of the applied stresses jf'*\ /"(ey.T) - F(e'j,T) - ±py*Aey > 0. For small additional strain Aey = ey — ey-, neglecting the terms of order higher than two in the expansion of .F(ey), one obtains dc £c Aey Aeju > 0 unless all £y equal zero. (Such functions T are referred to as convex.) Thus, for stability it is necessary that the function T is convex. For the case of a linearly elastic medium, represent T as the sum of the energies due to volume variation and shear: p§T = \ KJ\ + fj,( J2 — | J2) , K = A + § \i (see Problem 26.7). Convexity of the function yields K > 0, fi > 0. Using results of the solutions of Problems 26.1, 26.3, 26.4, one obtains A > 0, — 1 < a < | .

26.17 p0^(ey, dy) = \ XieCf + /ieye« + m ( e y ^ ) 2 + ne^£kidkl + le^dj + kdijd:'kekl£mn9mn- To specify the material, 9 elastic constants are required: A, /x, m, n, I, k, di, d2, d3 where d\, di, d3 are the principal components of the symmetric tensor dij. The scalar invariants K\, K%, K3 of the tensor dij can be used instead of di, d2, d3.

26.18 pT = \ \J\ + M + me3je>3 + le33Ji + ne^,

Pn = AJi + 2/ien + Ze33 , P12 = 2M£I2 , P22 = AJi + 2/xe22 + l£33 , P13 = 2(/i + m)ei3 , P33 = AJi + 2/*e33 + 2(m + n +1)833 + * (en + £22) , P23 = 2(/x + m)e23

26. Linearly Elastic Solid 167

26.19 _ i (2/i +OP

2 (2m + 2n + 3/i)(A + n) - {n - I)2

_ = (M - O P , , 33 (2m + 2n + 3/i)(A + / z ) - ( / z - 0 2 '

26.20 Use the formulae for stresses obtained in solution of Problem 26.18.

/» \ 9 ,. . ,, . d2w3 SPwi

pa2 = (X + fj,)— div w + (i Aw2 + {I + m)-jr—-+ m-j—j- ,

/9a3 = (A + /z)-—div«; +(/x + m) Aw3 + (m + 2n + 2/) —-T-+ az oz'-

{m+2l)[o^d-z + dWz)

26.21 Apply the operation dzv to the equation of equilibrium for displacements.

26.22 Use the results of solution of Problem 26.21.

26.23 Write the displacement components Wi in the form w, = tpi + ViV> where (Pi are three harmonic functions satisfying Laplace's equation. Substituting w* into the Lame equations, one obtains Poisson's equation At/) + 2(i-a) ^V« = 0 f° r *l>- Its solution has the form tp = ô + i>p where t/)o is the solution of the homogeneous equation Aô = 0, i.e., tp0 is a harmonic function, and tpp can be sought in the form ^P = Acpkxk, k= 1,2,3.

1 /")

Wi = ^ " 4(1 -a)d^ (^ t X* + ^ ' *'fc = 1 ' 2 ' 3 '

Thus, since the solution is reduced to determination of the four harmonic functions 1>o, ît-

Equilibrium under extension and shear 26.24 exx = - ~ (I - x), ew = ezz = — pg(l - x), ey = 0, at i ^ j .

oal2 PI 2 6 . 2 5 . = ^ A V = - - ( 1 - 2a).

26.26 Direct the x-axis downward along the rope, letting x = 0 at the point of suspension. The equation of equilibrium projected onto the x-axis is dpn/dx + pg = 0. At x = I: pn = F/S where F is the weight of the load, S is the cross-sectional area. The stress distribution ispn(x) = F/S+pg(l-x). The maximal stress is (pn)max = F/S+pgl < ab where <rb is the allowed stress limit. F/S ~ 3185 kgf/cm2, I < g »~^ 5 ~ 400 m.

26.27 The extensional force P acts along the x-axis of a rod with cross-sectional area S. It causes a state of stress such that pn = ^, and the other p^ = 0. Let us find the maximum extensional and tangential stresses. The stress vector on any surface element whose normal makes an angle a with the x-axis (see Figure a26.1) is p n = pn cos a e i = | cos a e j , |ei| = 1. pnn = ^ cos2 a is the extensional

-^x

Figure a26.1

stress, and pnT = ^ sin 2a is the tangential stress. Their maximum values are (Pnn)max = P/S at a = 0, (p„ T ) m a x = 5 P/S at a = 7T/4.

f 1? I 'Zl^rl P< 30000 kgf. \ 2I - 6 0 ° kgf/cnr b

26.28 6 = i pw2Z3/£ ~ 0.027 cm.

26.29 For each section x

It follows from the equations of equilibrium that

P12 —

P22 = / ,

aP

(60 + a i ) '

« 2 P

The functions / and y> are obtained from the boundary conditions on the lateral surface pn = 0, n = {sin 6, cos 0}: / = tp = 0. Since the problem is solved for

stresses, one should verify that the compatibility conditions are satisfied. For a plane problem they are reduced to one equation (see Problem 26.53)

d2Pn d2p22 _ d2pi2 . , ,

^ + ^ = 2 a ^ + ffA(pil+P22)-The obtained solution satisfies this equation if the terms with a2 are neglected. To calculate <5, use the formula en = j^ (pn — CTP22) (Hooke's law). Since it is taken that P22 - 0,

6 = ^IPndx = W^bZ)Hb/bo) E(b - 60)

26.30 Seek an approximate solution of the problem, assuming that there are only stresses along the axis of the rod pxx. The stresses are the superposition of the stresses caused by the force P and those caused by weight of the part of the rod below the section of observation. Let the stress be pxx(x) = Q in the section S(x). Then pxx(x + Ax) = ^ f f i ^ 5 in the nearby section S(x + Ax). Assuming that pxx — const, one obtains the differential equation for S: (q + pgdx)S = q(S + dS). Its solution is S = Soe"3*/" with q = pxx(0) = P/S0. Thus, the form of the rod is determined by the formula S(x) = S0e^x^i.

26.31 See the solution of the preceding problem. Replace the gravity force by the centrifugal force F = uPx. Equality of the stresses in the sections x and x + dx yields the equation q = {<,~^dx)b. From here, b(x) = boe-^'1^".

26.32 en = | ( p i - <jp2), £22 = i (P2 - <rp\), £33 = - | (Pi + P2), £ij = 0, at i^h^r- = l (P i+P2) ( l -2 f f ) , Tmax = \ (p, - p 2 ) .

Bending of beams 26.33 Using St.Venant's principle, seek a solution in the form pxx = —ay,

a = const, with the other p^ = 0. The conditions on the lateral surfaces are satisfied. On the end-walls Tx = JpxxdS — 0, Mx = My = 0, Mz = al = M where / =

5 / y2 dS is the moment of inertia of a cross section of the rod relative to the 2-axis. s

M From here, a = M/I. The strains are calculated by Hooke's law: exx — —=-= y,

El M

Syy = ezz = a — y, e^ = 0, at i ^ j . Next, the displacements are obtained from El

them. It is found that there is a neutral axis on which exx = 0, and the sections of the rod rotate, remaining plane and orthogonal to the axis (see Figure a26.2). Let

Figure a26.2

us denote the deflection, i.e., the displacement of points of the neutral axis along the M

2EI x , with

1 M curvature — = —

H til

y-axis, with r\. The equation of the neutral axis is the parabola rj •■ d\_ dx2'

26.34 Suppose that there is a neutral axis whose material line elements conserve their lengths. Consider an element of the beam between two nearby sections. Before bending, the length S = So for all longitudinal material Unes of this element. After bending, S = So = R>p for the material line of the neutral axis where R is the radius of curvature of the neutral axis, and (p is the angle of rotation of the section due to bending. Also due to the hypothesis of plane sections, for a material line at a distance y from the axis, S = S0 + AS = (R - y)ip. Thus, en = A 5 / 5 0 = -y/R. From Hooke's law pn = —Ey/R. Such a stress distribution corresponds to the moment Mz = — JpnydE =^Jy2dS = ^ = M. From here, one derives the equation for

M where TJ(X) is the deflection of the beam

1 <PT) the bent axis of the beam — ^ —z „ T R dx2 El (vertical displacement of points on the neutral axis)

26.35 Express the bending moment in each section, considering equilibrium of the right part of the beam cut by this section, so that M(x) = —P{1 — x). Using results from the solution of Problem 26.34, one obtains

e" = — E T V = E1^1~x^y' Pxx = 7 ( l ~ x ^ y ' d?i± _ _ £ x x P_(,_ \ dx2~ y~ EI( X)-

From here, JJ = - j f j i2(3/ - x), and |»?|m»Jt = |T?(/)| = ^ .

26.36 The differential equation of the neutral axis of the beam has the form £$ = E7 M(X). The bending moment in a section x is calculated by the formula

26. Liaearly Elastic Solid 171

l-x M(x) = / q(x + £)£d£ (the x-axis is directed along the beam, the origin is at the

o left fixed end). Substitute this expression for M(x) into the differential equation of the bent axis, and differentiate it twice with respect to x. Then M" = q(x), and the equation for r)(x) takes the desired form.

26.37 The equation of the axis of the beam has the form ^jf = ^gp where M(x) is the bending moment in the section x. For the given fixation of the beam, unknown reaction forces R and moments M0 arise in its end sections. Due to symmetry, the forces and moments are identical for both ends. They also contribute to M(x). As a result,

M(x) = - f q£dZ + Mo + Rx = - - qx2 + M0 + Rx , q = const , o

and the equation of the bent axis is

1 " ( I ) El

^ qx* + i Rx3 + Mo y + Cx + Co

The unknown constants R, M0, C, Co are obtained from the boundary conditions for fixation of the beam rj(0) = r)(l) = 0, r)'(0) = rf(l) = 0. As a result, one obtains 1 = -5&7 x*(l ~ x) 2- R = \ql,Mfi = - i ql\ pxx = -*& = i q(x2 - Ix + | P)y. Remark: This solution can also be obtained by integrating the equation ^3 = — -gj = const (see Problem 26.36) with the given boundary conditions taken into account.

26.38 Use the differential equation of Problem 26.36 or the hint to Problem 26.37. V=-^Eix(x3-2lx2 + l3).

26.39 See the hint to the preceding problem.

ql* V = ~7^7 *(2z - 3Zx2 + P) , Mn™ = f?(0.42I) = 0.0054

26.40 For bending, the profile of the longitudinal strain has the form £u = —ky. The coefficient A; is obtained from the boundary conditions for temperature £„(£) = a{Ti - T0), £ n ( - | ) = a{T2 - T0): k = Q(T2 - Tx)/h, and the equation of the bent axis is 0 = k = a fSjj^, 77 = a ^ ^ - x{x - I).


l-x 26.41 The bending moment in a section x is M(x) = — / q£d4 =

o — |g(Z — i ) 2 where q = const is the weight per unit length of the beam. Then pn = —jM(x)y = -£jq(l — x)2y, and pi2 is obtained from the equation of equilibrium dpn/dx + dp\2/dy = 0. As a result of integration, one obtains P12 = 2j (I — x)y2 + f(x)- To determine f(x), the boundary conditions Pi2|y=±* = 0 are used. As a result, one obtains / = - ^ - (/ - x), pi2 = 2/ ( T — V) (' — x)> Pi2max = Pn{x = 0, y = 0) = ^ .

26.42 According to the hypothesis of plane sections, en = ky over the whole thickness of the beam; the i-axis extends along the neutral axis of the beam (en (0) = 0), the location of which, o, is required to be determined. By Hooke's law, p°^ = kEay, a = 1,2. The total bending moment M is given M = kE\ f y2 dS + kE2 f y2 dS =

st s2 k{EiJi + E2J2), Sa are the areas of the sections of the beams made of different materials, and Ia are their moments of inertia. Prom M, k and, consequently, p°j are obtained:

Jfc - M v° - MEaV

E\J\ + E2J2 E\J\ + E2J2 The location of the neutral axis a is obtained from the condition that the total force on the ends equals zero:

-(fca-o) a

kEi J ydy + kE2 j ydy = Q. -(ft2+/ii-a) - ( - a + / i s )

Hence, 1 Ejhl + E2hl + 2Elhlh2

a~2 Eihi + E2h2

Torsion of rods 26.43 Seek the displacements in the form 101 = —azy, io2 = azx, w3 = f(x, y)

where z is the coordinate along the axis of the rod, and a = const. From the displacements, the strains and stresses are obtained: pu = an i—y + g£j, P23 = otfi (x + -J-), and the other Py = 0. Thus, the problem is reduced to determination of the function f(x, y) by solution of the equation of equilibrium A / = 0 with the boundary condition on the lateral surface free from stresses pn\T=R = 0. This yields %\ _ = 0 for the case of a circular cylinder, i.e., f(x,y) = const = 0. The sections perpendicular

to the z-axis remain plane and rotate each relative to other by the angle A0 = a Az. The value of a is determined by the torsional moment M2 = M given on the ends:

M = j{xp32 - yp3i)dS = Q/X/ = - otfinR* , a = - — .

/ is the polar moment of inertia of the section.

26.44 It is found in the preceding problem that p13 = —afxy, P23 = a/xx. The stress pn on an arbitrary surface element with normal n = nii + n2 j + n3fc is Pn = Pi3n3* + P23«3j + (pi3™i +P23n2)k- Since the problem is axially symmetric, it is sufficient to consider the vector n lying in the planes of longitudinal sections passing through the axis of the rod, i.e., n2 = 0. Then pnn = pi3 sin 2d, and pnT = p13 cos 29 where ni = cos 6. The maximal tensile stress |pnn | = afiR is on the external surface of the rod and acts on the surface elements making an angle of 45° with the z-axis. Maximal tangential stresses acting on the surface elements perpendicular to the z-axis also are on the external surface of the rod, so that pTmax = af^R-

26.45 Use the hint to Problem 26.44.

P33± = 2 ^ p ( l ± \A + 16MVF^2) ,

P Mimic — sj\ + 16M*/P*R?

26.46 Use the results of solution of Problem 26.43.

<p = / a(z)az = / — n. az = — — . Y J K ' J uirR* uirR* 0 0 r n

26.47 Let the right end of the shaft B be free, so that rigid fixation is replaced by the reaction moment Mg, in opposite direction to MQ. The total angle ips, by which the right section rotates under the action of the moments MB and Af 0 is ipB = (M0 - MB)a/(nIa) - MBb/{nh) where I = J r2 dS is the polar moment of inertia

s of the section. Prom the condition <PB = 0 (the end is fixed), MB is obtained. Next, the angle cpo by which the interface section rotates are obtained from the condition Mo = MA + MB, and the reaction moment MA on the left end is calculated:

M0bRl , . M0aRt 2M0ab aRt + bR*' ° aR* + bR* ' ™ irn(aR4

b + bR4a)

26.48 -Wr ~ } (see Problem 26.43). IQ = *£■ for the solid shaft and / = *[& - (0.6)4]/2 for the hollow one.

TmaxALx = 1 1 5 > P/Po = 0-64.

2 6 4 9 « = c i f e - r. = aSgffe. n = c ? S v Here> ' = / ^ = "f is

the polar moment of inertia of the section.

26.50 AT = — 2ajx - Poisson equation, T\c = const = 0, A is the Laplacian.

States of plane stress and plane strain 26.51 P33 = A(EH + £22) = ff(Pll + P22)-

26.52 £33 = -A(en + £22)/(A + 2/i) = -\Ji(e)/(2n),

d2w „ E ,. E A P W = PF + 2(1^) * * * ( 5 n + £22 ) + 2 (1+^) A W

where grad, A are the two-dimensional operators.

26.53 The equations of compatibility for a state of plane strain have the form

d2£33 = ^ £ 3 3 = d2£33 = Q ^ £ 1 1 ^ £ 2 2 = 2 d 2£i2 dx2 dy2 dxdy ' dy2 dx2 dxdy

The first group of these equations yields £33 = Ax + By + C. Using Hooke's law, express pap in terms of E ^ , Q = 1,2, and substitute into the remaining equation of compatibility conditions. As a result, one obtains dPpn/dy2 + d^p^/dx2 = 2dPpn/dxdy + o A(pn +P22) where A is the Laplacian with respect to two variables.

26.54 a) AA„ = g + 2 ^ + 0 = 0. b) PH = §Jf, P22 = f £ + pgy, P12 = - § •

26.55 From f>(x,y) the stress components pxx = c, pm = a, p^ = —b are obtained. Such a state pap can be realized if it is given that pxx = c on the sides perpendicular to the z-axis, pm = a on the sides of the plate perpendicular to the y-axis, and the tangential stress is r = — b everywhere on the boundary.

26.56 Obtain the stresses from <p(x, y): a) Pxx = 6dy — pure bending, b) Pxx — 2cx, p w = 0, pxy = — 2cy — such stresses correspond to the following boundary conditions: pXI = 2ci, p ^ = =p2c/i at j / = ±/i, pxx = ±2cZ, p ^ = —2cy at x = ±l.

26.57 * , = $§? + £ £ , , * = & , , * = - £ ( ! £ ) ,

r dr \ drj r2 BO2 (p = 0

26.58 tp = A\n r + Br2\n r + Cr2 + D, prr = A/r2 + 2B In r + B + 2C, pee = -A/r2 + 2B]nr + 3B + 2C,Pre = 0.

Various problems of equilibrium of elastic bodies 26.59 The equation of equilibrium (A+/x) grad div to+/x Aw = 0 can be written

in the form (A + 2/z) grad div w — jicurl (curl to) = 0, using the formula of vector calculus grad div to = Aw + curl (curl to). The problem is spherically symmetric, i.e., we = wv = 0, wT = wr(r), and, consequently, curl to = 0. The equation of equilibrium then yields grad div to = 0, and div to = j j | ( r V ) = const = 3 A in the spherical coordinate system. Prom here, one obtains the expressions for the displacements wr = Ar + £, strains e„ = A — 2f, egg = e w = ^ = A + j% and stresses p r r = (3A + 2/i)A - 4B/i /r3 , p9g = p w = (3A + 2/i)i4 + 2B/x/r3. Prom the boundary conditions p„-|oo = 0, Prr|r=« = -p0 , one obtains A = 0, B = E ^ .

£rr = -

Prr =

Pofl3

2/xr3 ' p0R3

£$e = £< _ P o ^ 3

w - 4 / i r 3 •

Ps« = P w = Po-R3

2r3

26.60 The problem is spherically symmetric: to = to(r) . Note the solution of the preceding problem. The integration constants A and B are obtained from the boundary conditions on r = R\ and r = R2-

A = p2R*+plR3

1

3(A + 2 M ) ( f l 3 - f l 3 ) ' B P1+P2

4/4*5 - «?) R[R%

26.61 See solution of Problem 26.59. The constants are obtained from the boundary conditions.

p„ = p( l - a3/r3) , pe9 = p w = p ( l + o3 /2r3) .

so that |p9fl|max = | p is reached on the boundary of the cavity.

26.62 The problem is plane and axially symmetric, i.e., w = w(r). Similarly to Problem 26.59, one can obtain divw = £ -^(rwr) = const = 2A, r = y/x'2 + y2. Hence, wT = Ar + f, wg = wz = 0. Next, the expressions for the strains and stresses are found: £„• = A - ^ , eee = A + ^ , p„ = 2(A + (i)A — 2Bfj,/r2, pee = 2(A + p)A + 2Bfj,/r2. The integration constants A and B are determined by the boundary conditions on r = Ri and r = R2.

Prr =

Pzz =

_RlPi

RlPi -Rl{1 r2) R2-R2{ r>) '

7-2 J ^ - Ji? y + r2 J ' K, ,

R2 — Ri A RjPl - fl|p2

A + ix Rl-R\

26.63 The results of solution of Problem 26.62 can be used. The constants are determined by the boundary conditions:

- , ( . ♦ £ ) . Prr = V 1 - T I , P»fi = P 1 + - r I , Peffmax = 2p . r-

26.64 The gravity force acting on a unit mass of the spherical body is proportional to the distance from the center and directed to the center: FT = —gr/R. The problem is spherically symmetric: w = w(r), and, consequently, curlu; = 0. After substituting the density of the gravity force into the equation of equilibrium, one can follow the solution of Problem 26.59.

_ pgR ( r2 5A + 6// 10(A + 2/z) V R2 3A + 2M

pgR{\-2a){\ + a) (^f^_Z-o 10£(l-<7) \ R? \ + o

26.65 Solve this axially symmetric problem in a cylindrical coordinate system r, 0, z. The equations of equilibrium for the stresses projected onto the z- and 0- axes and the boundary conditions prz = prB = 0 on r = a yield prz = prg = 0. Those projected onto the r-axis yield

dr r i.e.,

r - j — +Prr = Pee - pu rz . dr


Solving this non-homogeneous equation for p r r , one obtains p r r = A(r)/r where dA/dr = pee — pu2r2. FYom Hooke's law, the expressions for the strains in terms of A(r) are obtained:

1 (A dA 2 , \ 1 (dA 2 , A £"=E\-r--°l[r--°l"r)' E w l t + , W r - f f ;

For axially symmetric problems, £„. = dwT/dr, eee = wr/r, and the compatibility condition has the form e r r = eg$ + r de.ee I dr. That yields the equation for A(r): A" + ;A'--±IA = - ( 3 + a)pu>2r, the solution of which is A{r) = Cxr + - - § ( 3 + a ) / w V . FYom the condition that the stresses are bounded at r = 0 and the boundary condition prr(a) = 0, one obtains the constants Ci and C2, so that

Prr = g (3 + o)pu2{a2 - r2) , pee = - (3 + a)pw2 (a2 - - ^± -^ r 2 ) .

26.66 The problem is planar. Let us choose the polar coordinate system r, 9 in the plane of the plate with origin at the point of application of the force (see the figure). The stress function f{r,9) satisfies the biharmonic equation (see Problem 26.57). The boundary conditions on the boundary of the half-plane (at 8 = a) have the form pe${r, a) = 0, pr${r, a) = 0. To satisfy these conditions, one should take if = A{9)r which leads to the equation for the function A(6): A™ + 2A" + A = 0. Its general solution is A(9) = a9 sin 9 + bO cos 6 + at sin 9 + bi cos 9. Only the first and second terms contribute to the stress component prr. The constants a and 6 are determined by the condition fpndS= -F where integration over the semicircle

s of radius g with the center at the point of application of the force as g —» 0 is implied. Projecting onto the direction of the force and onto that perpendicular to it, one obtains

a a I pTT cos 9rd9= -F , f prr sin 9rd9 = 0 .

a-ir Q-n

As a result, prr = —2F cos 9 /vr, pre = pee = 0, and, consequently, e r r = prr/E, £ee = -oprr/E, ere = 0.

26.67 The problem is planar: pQp = pO0(x,y), a = 1,2. The system of equations consists of the two equations of equilibrium and the equations of compatibility conditions for the stresses AApi ; = 0. It is sufficient to supplement the first two equations by one of the latter, e.g., the compatibility equation for p22:

dpu dpi2 dpi2,dp22 „ A . n

ox ay ox oy

http://de.ee


The boundary conditions on the surface y = 0 can be written in the form pa2(z, 0) = 00

Fa6(x), a = 1,2 where 6(x) is the Dirac delta function such that / 6(£)d£ = 1. —oo

Substitute the Fourier transforms with respect to the variable x

1 °° Pa/?(s. V) = -/S= / Pap(x, y)e"x dx

into the equations and boundary conditions. As a result, one obtains the system of ordinary differential equations for the images

^ _ i s p . i = 0 , ^ i _ i s p J 2 = o , ^ - 2 * ^ + 5 ^ = 0

with the conditions p„2(s>0) = Fa/\/2~w. After solution of this system for p*a0, one obtains the functions pap(x,y), using inverse Fourier transformations

1 °° pap{x,y) = -j= J P*a0(s,y) :ds

26.68 The conditions at infinity are pj? = T, pg = pg = 0. If there is no hole, the stresses would be the same in the whole plane, and the stress function (see Problem 26.55) would be tp° = i Ty2. Next, choose the polar coordinate system r, 6 with the origin at the center of the circular hole; then ip° = i Tr2 sin2 9 = j Tr2(l — cos 20). Taking into account that there is a hole in the plate, seek the stress function (p in the form tp = tp" + ^ ( r .cos 29) = (p° + /i(r) + /2(r) cos 29 where each term satisfies the biharmonic equation. From here, one obtains / i = AT2 In r+Br2+C In r, /2 = i4xr2 + B\r4 + C\/r2. The constants are determined by the boundary conditions on the contour of the hole, which is free from stresses, p„ = prg = 0 at r = R, and by the conditions at infinity.

Prr

Pee

Pre

p2 / D4\

^ ( l + 2«!-3£)sin20.

cos 20

^


Temperature strains 26.69 The rigid walls on the ends do not allow the plate to extend in the

direction of the i-axis; thus £n = 0. Since there are no stresses on the lateral faces, one can seek the stresses in the form P22 = P33 = Pij = 0, i ^ j . Prom Hooke's law for a thermoelastic solid (see Problems 26.8b, 26.10) pu = \J\ + 2/x£« — a(3A + 2fj.)(T - Taverage) where Q is the thermal expansion coefficient, one obtains Pn = —/i(3A + 2fi)a(T — Taverage)/(A + fi). For the given distribution of temperature, Average = §T 0 . Thus, p u = § aET0 - CcET0{\ - g ) .

^ d

26.70 (A + /x)grad div w + fj. Aw - a(3A 4- 2/x)grad T = 0,

1 d(ru) r

C)d~r Here u

dr

1 d{r2u)

3A + 2fi dT . , = a — — in the polar coordinate system,

r2 dr

1 d{ru)'

A + 2/x dr

3A + 2/i dT

r dr = wr.

= a

A + 2/i dr

3A + 2/i dT . 2(A + p) dr

— in the spherical coordinate system,

in the polar coordinate system.

26.71 The solution of the Lame equation for a state of plane stress in a polar coordinate system (see Problem 26.70c)

L dr

1 d{ru) r dr

3A + 2/x dT „ . dT 2(A + /x) dr ' dr

is u = (1 + o)a - J Tr dr + C\r + C2/r. It follows from Hooke's law that r 0

Prr = -, j[£rr + oeee - (1 + a)aT) , 1-a2

E 1

[eee + OE^ - (1 + a)aT] ,

£ r r — u £0$ — — r

Thus, one obtains

= -aE r2J Trdr +

l-a- ;[C,(l + a ) - C 2 ( l - a ) / r 2 ]

and a similar formula for pgg. The constant C^ is determined by the condition for the solution to be finite at the center of the disk (at r = 0) which yields C2 = 0. From the boundary condition Prr(R) = 0 on the external contour of the disk, one obtains

Prr = OcE

pee = aE

^JrT(r)dr-^frT(r)dr 7?

-T+-frfrT(r)dr-$IrT(r)dr 0 0

(a26.1)

26.72 First, find the temperature distribution in the disk, solving the heat conduction equation. Since the temperature on the boundary of the region is constant, dT/dt = 0 in the annulus. Thus, the heat conduction equation is reduced to the Laplace equation AT = 0. For the case of axial symmetry, the solution of the equation in the polar coordinate system AT = £ (r ^ J = 0 satisfying the given boundary condition is T = To In ^ / In j | . The general solution of the problem of elasticity is presented in the answer to Problem 26.71 (formula (a26.1)). The resulting temperature distribution T(r) = To at r < a, T — T0 In ^ / In j^ at a < r < R should be substituted into that solution; as a result one obtains

Prr = aET0a2 (4r j _

ft2 J . 2 l n f t - )

4 In

26.73 Displacements for a state of plane strain in an axially symmetric problem are described by Lame equation in the polar coordinate system of Problem 26.70a. Next, the solution is constructed in the same manner as in Problem 26.71.

£rr =

e$e

a( l + g) 1-CT

Q ( 1 + <T)

l - ( 7

1 r ft T - - y rT(r) dr + (1 - 2a)— jrT(r) dr

0 0 ■ T ft

^ J rT(r) dr + (l-2a)-jpj rT{r) dr

26.74 First, find the temperature field T(r) for stationary heat flow in the same manner as in Problem 26.72. Next, the solution is constructed in the same way as indicated in Problems 26.73 and 26.71, but integration is performed from the internal radius, and the constant Ci is determined by the condition on the internal contour Prr(a) = 0:

aET0

Prr =

a I n - -r ft2 — a2 r* n

2(1 -a) In i


Pzz =

a f f r 0 [ l - l n * - ^ ( l + £ ) l n * ] 2(1 - a) In &

2(1 - a) In |

26.75 The problem is spherically symmetric. The equation for displacements (see Problem 26.70b) is

d ( \ d(r2u)\ 3X + 2fidT = a ■

dr \r2 dr J A + 2fi dr

The solution is the same as for Problem 26.71:

£rr =

£ee =

g(3A 4- 2[i) A + 2/x

Q(3A + 2/x) A + 2/x

r r

/ Tr2 dr + 2fM 1 3A + 2/x fl3

n,

J Tr2dr

r3J Tr2 dr + 2/i 1 3A + 2/i R3J Tr2dr

26.76 The temperature distribution is determined by the equation

1 d ( 2dT\ r2 dr \ dr J

with the given boundary conditions T(a) = T0, T{b) = 0 : T = T0^ (£ - l ) . Next, use the equation of Problem 26.70b, constructing the solution by the method described in Problems 26.71, and 26.74.

aET0 ab 1 - a b3 - a3 a + b- -(b2 + ab + a2) + - ^ -

r r J

_ _ aET0 ab pee = Pw-J^-p^tf

. 1 ,L2 . 2, a2b2 a + b- — {b2 + ab + a2)- —r 2r 2H

Stability of equilibrium 26.77 In the small deflection approximation for bending of slender beams, the

equation for the deflection d^rj/dx2 = M{x)/EI is obtained; here, J is the moment of inertia of the section, M(x) is the moment of the applied force relative to the considered section x. In the linear approximation of the given problem, the moment

M would be calculated relative to a point x of the undeformed (unbent) axis and would be zero. On account of the deflection TJ(X), the moment of the forces acting on one side of the considered section is M(x) = — P 77(1). The solution of the equation for the deflection d?T}/dx2 = —Pr]/EI is 77 = C\ sin kx + C2 cos kx where k = ±JP/EI. The boundary condition 77(0) = 0 is satisfied if Ci = 0 and 77 = 770 sin kx. The boundary condition on the other end rj(l) = 0 is satisfied if 1) 770 = 0 — the shape of the rod is rectilinear, or 2) sin kx = 0, i.e., k = 7rn//, n = 1,2,..., giving rise to a discrete set of values of the force P = ^jf- n2 (the eigenvalues of P) causing a bent shape of the rod. The minimum (critical) value of the force P at which bifurcation arises is Pa = ^jf--

26.78 The problem is reduced to the equation of the preceding problem if the origin of the coordinate system is chosen at the point of application of the force P. The boundary conditions at the fixation point are r](—l) = T)'(—1) = 0. The critical value of the force is Pa = ^ 5 - .

26.79 Let the free end of the rectilinear rod compressed by the force P along its axis be deflected by a small distance in a transverse direction (see the figure of Problem 26.78) by a force Q. A stable equilibrium corresponds to the minimum of the energy, i.e., to disturb the rod from equilibrium, a positive work dA = Qdy should be performed. For the chosen coordinate system in which the deflection is regarded as positive and the origin is taken at the point of application of the force P, dy is negative. Consequently, dA > 0 requires Q < 0. Let us determine equilibrium of the bent shape of the rod under action of the forces P and Q. The differential equation for the bent axis is jgf = —£j y- -§j x. Its solution is y — C\ sin kx+C-i cos kx- ^ x where fc2 = £j. From the boundary conditions y(0) = 0, j/(—I) = 0, one obtains the

constants C\ and C2- The shape of the rod is y = ^ (—x + fc,^g

fc^I)- The maximum deflection at the point x = — I is ymax = ^a~p™ kl' > 0. The condition for stability Q < 0 yields tan kl > kl. Consequently, the shape is stable if kl < w/2. If kl > TT/2, the shape is instable. Hence, fcCT = 7r/2/ and Pa = ^J— If P > Pa, the rectilinear shape of the rod is instable.

26.80 The equation for the deflection of the beam is d^rj/dx4 = q/EI. The value of q is proportional to the total deflection — the sum of the initial and current ones, i.e., ^ 3 = J j (77 + 770 sin ™J. The solution of this nonhomogeneous equation is

77(1) = C\ sin fcx + C2 cos A;x +

C3 cosh kx + C\ sinh fcx + 4 ^ 770 s in — .

The boundary conditions 77(0) = 77(f) = 0, T/'(0) = rf'(l) = 0 determine the constants: C\ = d = C3 = C4 = 0. The solution has the form

n4EI . nx Vo sin ' ^El-cd*" I '

The amplification coefficient of the amplitude ^^!ali equals 1 at a = 0 and grows with increase of a. At a = acr = ^EI/l*, the amplification coefficient becomes infinite, loss of stability takes place. The limit value of a equals acr-

26.81 The solution is similar to that of the preceding problem; 77 = j ^ - , 7 = - « L < 1 48E/ ^ *'

Dynamical problems of isothermal linear elasticity theory 26.82 Direct substitution of Wi(x,t) into the equation of motion for a linear

elastic medium, in terms of displacements, yields

d2wx _ 2 d2wx d2wv<z _ 2 d2wViZ _ /A + 2/j _ jfj. _ c i " S I T > a,2 ~ °i »_2 > c i - \ / : > c 2 -at* * &r2 ' dt2 "> dx2 ' -1 V P ' w V />

One of the solutions of this system is wx = f(x ± c\t), wv = wz — 0 — longitudinal wave, curl to = 0. The other two solutions of this system wVtZ = ip(x ± c2t), wx = 0 are transverse waves, divti; = 0. If the processes are isothermal, the Lame coefficients A and fj. are used, but if they are adiabatic, A^ and fx^ are used instead (see Problem 26.14).

26.83 Substitute w = W\ + w2 into the Lame equation (see Problem 26.22). Next, apply sequentially the operations div and curl to this equation. As a result, one obtains

divT/>j = 0, curl 2 = 0,

where V>i = P - ^ - ( A + 2/i)Au>i, ij>2 = p - ^ - fj.Aw2

In accordance with this approximation, curl^j = 0, and div ij)2 = 0. A function ip for which simultaneously curlV> = 0 and div^> = 0 is equal to zero (if it is defined in the whole space and vanishes at infinity). Thus, one obtains the equations for u>i and w2

d2w\ 2 . cPw2 2A 2 A + 2u ~ a - w - = c2Aw1,— = c>Arv2,c2 = - ^ , c i = ^

26.84 In a thin rod with lateral boundaries free from stresses, pu = Ee\\, P12 = P13 = 0- The equation of motion projected onto the axis of the rod is

d2u _ dpu _ „ d2u P~W~~dx~~cW '

This is a wave equation; the speed of the wave is c = JE/p = J1^^2^ < C\ = J^^ p

where C\ is the speed of a longitudinal wave in an unbounded elastic medium.

26.85 The equation for longitudinal waves in the rod (Problem 26.84) is

&u_Ed?u dt2~ p dx2 '

To satisfy the condition for fixation u(0, t) = 0, seek the solution in the form of a standing wave u = A sin kx cos u)t, which satisfies the equation if w = k JE/p. The wave number k is determined by the condition on the other end where pn(/,t) = 0. Thus, k = n(l + 2n)/2l, and u = yJWfp | ( 1 + 2n), n = 1,2,....

26.86 The equation for transverse displacements of the points of the axis of the rod wv = t] (see Problem 26.36) has the form

d*V ^ q dx4 El '

Here, / is the moment of inertia of the section S, and q = FS where F is the body force per unit volume of the rod and 5 is the cross-sectional area. In non-stationary problems the inertia force is a body force — p ^? where p is the mass dencity per unit volume of the rod. Thus, the equation for displacements takes the form

EJ_ A = _A pS dx* ~ dt2 '

Seek the solution in the form 77 = rfce'^1-""' . Substitution into the equation yields the relation between w and k — the dispersion relation ^ fc4 = w2. The phase speed of the wave is c = ^ = Jg k and depends upon k (there is dispersion). In this case the group speed U = ^ = 2JMk does not coincide with the phase speed: U = 2c.

26.87 In reflection in an elastic medium, both longitudinal and transverse waves arise. The displacement vectors for these waves can be written in the form u^ = fi(x cos a+y sin a—cit)ni for the longitudinal wave and w2 = fo(x cos 0+y sin 0— C2<)T2 for the transverse wave where (see Figure a26.3) n.i = {cos a, sin a} and

Figure a26.3

n 2 = {cos P, sin P} are the wave vectors of the reflected waves, respectively, of the longitudinal and transverse ones, and T 2 = {—sin /?,cos 0} _L n 2 is a direction in the plane of the front of the transverse wave, and c2 = Jpi/p is the speed of the transverse wave. The amplitudes / i , / 2 and the angles a and P of the reflected waves are determined by the condition for the displacement vector on the rigid boundary to equal zero: w0 + ti?i + w2 = 0 at x = 0. Projecting onto the normal and tangent to the boundary, one obtains

fo(y sin c*o + c\t) cos c*o + f\{y sin a — C\t) cos a— -/2( t / sin 0 - c2t) sin /? = 0 , f0(y sin a0 + cit) sin a0 + /i(y sin a — cit) sin a + +/2(t/ sin 0 - c2<) cos P = 0 .

To satisfy these equations, it is necessary for all the functions to have identical arguments at any y and t, i.e.,

y sin ao + C\t = y sin a — C\t = y sin p — c2t .

From here, sin a = — sin a0, sin P = —f^ sin ao- Solving the system of equations for fit /21 ° n e finds the amplitudes of the reflected waves:

/o(c2 sin2 a0 - cx cos a0 cos /?) h =

h = c2 sin ao + c-i cos ao cos p

fpCi sin 2a0

c2 sin2 ao + Ci cos ao cos P'

cos /? = A 1 — ^

c\ sin2 ao c?

There is no transverse wave (/2 = 0), if the longitudinal wave t«o is incident along the normal to the boundary: QQ = 0. If the incident wave is transverse, then the similar

expression for the angle of the reflected longitudinal wave sin 0 = —a sin c*o makes sense only for the angles of incidence a0 for which | sin a0\ < = \ / / J / ( A + 2/x).

26.88 A one-dimensional motion with plane waves has displacement components Wi(x, t). Substituting them into the equation of motion of a transversely isotropic medium (see Problem 26.20), one obtains, as projections onto axes, three wave equations which yield, as the solution, one longitudinal and two transverse waves with speeds ciong = J{\ + 2/i)/p, oi = ^//x/p, c3 = ^/(/z + m)/p.

26.89 The problem is spherically symmetric: wT = u(r, t). Let us introduce the potential of displacements (p(r, t) so that u = dtp/dr. The function <p satisfies the wave equation

# V 2 A 2 * d ( 2 d<P\ ^c'A^c'- -^-] ,

_2 _ A + 2/x

Its solution is (p = tjj(ct — r)/r. The equation for the function ip is derived from the boundary condition Prr{R) = —Po-

Solving it, one obtains

<p = I je-T^-^) [A sin 0(ct - r) + B cos (ct - r)} - ^ j

where /? = J£ — I7 , 7 = 2 ^gj, c% = *. To determine the constants /4 and B, use the initial conditions u = 0, du/dt = 0, $„ = 0 at £ = 0. It is convenient to use the variable f = ct — (r — R) in the solution. The result is

r 4/x I ^ sin /?(ct - r + R) + cos /3(ct -r + R) }■

(a26.2)

26.90 Use the representation of the displacement vector in the form w = w^ + w^ where curlto'1 ' = 0, divti/2 ' = 0 (see Problem 26.84). For the given plane problem, that means

dw® dwM dyjW dwW

^ - + ^ - = 0' -W-^T'0- (a 2 6 3>

Each component satisfies its own wave equation

= «?.) - 5 ^ " + - 5 ^ - . a , / J = l , 2 . at2 (a) ^ ax2 5y:

Here, c\ = J^r^i ci = J*- are the speeds of the longitudinal and transverse waves.

Seek a solution of the form w^ = /^V***-1"0. The functions f^a)(y) satisfy the equations

Since we are interested only in a solution decaying as y —► —oo, the sign of the coefficient in the equation should be positive

2 2 fc2 - ^ - = n2 > 0 , k2 - ^j = K\ > 0 . (a26.4)

The solution has the form

wia) = AaeK°v+i{kx-ut) , u>la) = BaeK°v+i{kx-u*) .

To determine the four constants Aa, Ba, use the condition (a26.3) and the boundary

conditions on the free surface p w = Pxy = 0 at y = 0. The problem is solvable if the

following relationship holds between ui and k

which is called the Rayleigh equation. It is a cubic equation with respect to the quantity ui2/k2 = c2 = m:

tj){m) = m3 - 8(%m2 + &% (-5 - 3 1 m-16c§ ( 3 - 3 ) = ° -

According to the condition (a26.4) the quantity c2 can take only those values for which 0 < c2 < c2,. The cubic equation has a real root on that segment (since ip(0) < 0, T/»(C|) > 0, it being the only one (since ip'(m) does not have roots in that range). If A; and ui are real, then c = ui/k is the speed of propagation of the wave along the x-axis. It is proved that the surface wave (Rayleigh wave) exists, its speed is c < c^ < c\, and it has no dispersion.

26.91 Use the solution of Problem 26.90. The constants Aa, Ba are determined by the conditions on the two free boundaries pm = p^ = 0 at y = ±h. For the speed c = cj/k, one derives the equation

tanh mh Ak2; nm '"'^HH^H) tanh nh (m2 + A:2)2

a) A » h, i.e., mh € l , n A « l (long waves). The equation takes the form

m 4k2nm n (m 2 + Jfc2)2

and has two solutions: c ^ = c2 = Jl*fp — the speed of the shear wave; c^ =

2c2Jl — p = JptfLa) — the speed of the longitudinal wave in the plate.

b) A <C h (short waves). The equation for the speed 4k2nm = (m2 + k2)2 coincides with the Rayleigh equation (see the preceding problem), and c = CRayieigh-

27 Nonlinear Elasticity 27.1 TTW = ^ {6ik + ViWk).

27.2 The relative change in volume is determined by the formula (see Section 4)

■ - / 0= Vl + 2 / 1 + 4 / 2 + 8 / 3 - 1

where / i =ekk, I2 = § {l\— £y £ y) , Iz — det || £y||, £y are the strain components in

the initial Lagrangian basis. For an incompressible medium, 9 = 0 and I3 is expressed in terms of I\ and I2.

27.3 pd^(ey) = \\J2 + nh + PJ1J2 + 1J3 + vJl J\ = £kk, h = £y£y, h = £y£ ifc4.

27.4 p0T = I J2 + /xJ2 + /JJx J2 + 7J3 + J7J?, £y are the strain components in the Lagrangian basis of the initial state; a coordinate system is chosen in the initial undeformed state. The components ir/w are calculated by the formulae

VTfci = -Q— (Ojk + YjWfc) ,

7Tn = (A + 2n)u + 3au2 + -b{ul + u23) , o = - + / * + /3 + 7 + 7j,

27. Nonlinear Elasticity 189

3 TTIQ = nua + (n + (3 + - 7)uwa ,

3 7Tai = ixua + 6wwQ , b = X + 2fi + p+-'y,

dw\ dwa u = -r— ua = -5— a = 2,3 . ox ox

27.5 Let us suppose that, for the given case, displacements only in the direction of the i-axis are possible: vj\ = IOI(I, t), Wi = W3 = 0. Use the formula of the preceding problem for TTU- One obtains Viu>i = u, V1W2 = V1U/3 = 0, irn = (A + 2fi)u + 3au2 = p where a=^X + fx + P + -y + r}. I f a < 0 , the graph of ^U(u) is convex; if a > 0, the graph is concave.

27.6 Let the waves propagate along the direction of the x-axis: u>i = Wi(x,t). The equations of motion for the Lagrangian variables in a Cartesian coordinate system of the initial state have the form

dVi _ OTTn

dt dx

where po is the density in the initial state, and Uj = dwi/dt are the velocity components. For the components of the Piola stress tensor, take the expressions obtained in the solution of Problem 27.4. Let us introduce the notation Ui = duii/dx. Then the equations of motion take the form

*«- = (A + 2/i)-dx- + 3aUl ax- + 2 6 r ax" + U3 ax"] • dva dua ( dua duA

to-bT = fl-dx- + b{Ul-dx-+Ua-dx-) ' a = 2 ' 3 -

For a purely longitudinal wave, W2 = W3 = 0. The second and third equations are satisfied as identities, and, for the longitudinal components,

p 0 — = (A + 2M) — + 3aUl— , ~^ = -oT.

For purely transverse waves, W\ = 0, which results in the system of equations

/ du2 du3\ 2 H U 2 a x - + U 3 ^ j = 0>

dva dua

Only a transverse wave can satisfy this system for which -§^{u\ + ul) = 0. i-e-> u l + ul ~ const- Since there are no initial strains, u2 = 0, u3 = 0, and there are no such waves.

I o o o

27.7 In an incompressible medium, 6 = v l + 2 / 1 + 4 / 2 + 8 /3 — 1 = 0 (see Problem 27.2). For plane one-dimensional motions, /1 = £/ = u\ + £ (u2 + u\ + v%), / 2 = —j(«2 + u§) , /3 = 0. The incompressibility condition 0 = 0 yields ui = 0 for longitudinal strains, and, consequently, there are no longitudinal waves.

27.8 For the equations of plane longitudinal waves for Lagrangian variables (see Problem 27.6)

dv , , . du du du dv

* « - < A + ^ f e + 3 f l 0 a* ' m = dx where u = dwx/dx, v = dwx/dt, a=^X + fj. + 0 + ^ + Tj, seek a solution of the form v = tp{u) (Riemann wave). Substituting into the system, one obtains the equation |H = ±Jc + 3 for the function v = ip(u). The dependence u(x, t) is found as the solution of the equation

du du , . du Tt = m+c{u)d-x=0

along the line ^ = c(u), i.e., u = f[x — c(u)t]. Each state moves with its own speed c(u); c(u) is the characteristic speed given by c2 = ^±p + 3 u. Waves for which | j > 0 have the tendency to break if a > 0; waves for which |*j < 0 tend to break if a < 0 .

28 Couple Elasticity and Averaging in Media with Microstructure

28.1 In the linear approximation, u> = ^ cwlvectv = ^ = -^- ^L_l from w h i c n

it follows that | # | is the magnitude of the angle of rotation of a particle of the medium as a rigid body around the vector # / | # | . Due to the equations Jfl = u> x n(Q), the quantities (TILJ) form the matrix of the corresponding rotation, approximately, nU = 5a + *<*****. «mi" = em™V,$, = i V,(Vmu>n - Vnwm).

28.2 Use the algebraic properties of the Levi-Civita tensor and the fact that the second covariant derivatives are symmetric.

28.3 Since /cmjn = em,MV,*, where # = ± curl to, and V*^ = 0,

pF = ^ (eV)2 + /i£yey + 2vV<^ • V'<^ + 2 ( 5 ^ • Vj¥

28. Couple Elasticity and Averaging in Media with Microstructure 191

with A + | / x > 0 , / i > 0 , v + 6 > 0, and v — 6 > 0. Since K is a third-rank tensor, there are no cross terms between K and e. The equations of state are

2

The equations of motion are

P%) = J^'Vkgi* = 2i/e«'A*,

Ot2 = (A + ^V'VjW3 + fj.Aw' + vAiV'VjW3 - Aw') + pF< .

28.4 p-^ = (A + 2/*)A0 + pdivF, p — = M A # - i / A A # + ^ c u r l F .

28-5 "-flP" = ( A + 2/x) 9^?^-^ = /Jw~"w-Wl = ±V"V*'

±yfi*&£k. The computed wave has the form (for real A')

uP = 2Aj cos h(u/j - Jj)t - (k' - k")x) cos i((u^ + b/j)t - {k' + k")x) .

In the limit, *^~ ,' = %±. For longitudinal waves, ^ = ^; for transverse ones, dw _ fc(M+2i/A;a) I dull ^ l u l dk up ' IdJfcl * Ifcl-

28.6 to1 = tu3 = 0, and w2(xl) satisfies the equation / W 2 — i/w""2 = 0. a) If x = ±L, w"2 = £, and / W 2 - uw'"2 = 0. Taking into account the equality w2(0) = 0, one finds w2 = y ^ (cosh £ - 1). b) If x1 = ±L, uw'2 - uw"12 = p, w"2(L) = - g , u/ '2(-L) = g . w2 = £ I x1 — 8 I " 7" ) • As //L —► 0, the solutions tend effectively, with exclusion of the

boundary points (boundary layers), to the solutions of the corresponding problems of classical elasticity.

28.7 a) # = -eg(ze2-ye3), Q = -^(u-6)(e2e3-e3e2). If y2 + z2 = R2, Qn = -*§g (i/ - <5)(ze2 - ye3); if x = 0 and x = - L , Q n = 0. The load is self-equilibrated. Let us note that the coefficient 6, which is not contained in the equations of equilibrium explicitly, arises in the boundary conditions.

b) * = f, (aze2 + xe3), Q = f f {(6 + u<r)exe3 + {da + v)eie1). If y2 + z2 = R2, Qn = wm ((*+"")*«i; if * = 0, Q„ = - ^ (*a+i/)c3; if x = L, Qn = % {6a+u)e3. The load is self-equilibrated. c) * = - f (xei+ye2-2ze3), Q = -2a(v + 6){elei+e2e2-2e3e3). If x2 + y2 = rt2, Qn = " i f (v+«)(xe1+yc2), if z = 0, Q n = -4a(u+6)e3; if z = L, Qn = 4a{u+6)e3. The load is self-balanced.

28.8 It follows from the equations of equilibrium that p 1 1 ' ^ 1 ) = 0 p u = p =

const, t i ^ V ) = e11 = jfe, and w1 = p11 / ^ . If x1 = L, w\L) = L(en)L =

Lpn \x+2z) ' fr°m whichp11/(e11) t = (x+ty) • For any periodic function / ( i ) with

period /, (f)L = 1 J / ( * ) dx = a J /(x) dx = (/),. Let £ = [£] + £ where [£] = m

is the integral part, £ € [0; 1). Then wl(xl) = Lp" (f ( _ i - ) j + ^ - | L . j =

^P U I ( T — e v \ X+2£/i + e / X+l ) • The ^ast t w o t e r m s form a periodic function with period 1 if £ € R.

28.9 Assuming that p, p. are functions of f and choosing the scale L — 1, one obtains

^UK-f)^) dx (dp(KJ dN2\ / 0 d M d ^ N X a2(w) ^ , .

Hence, if | | ( M ( l + ^ ) ) = 0, Nl = ^ ( J f - ( | f } ) - * + 1. Averaging

the term the equation of order 1, one obtains

d2(w) _ 1 &>(w) (p) dt2 (l//i) dx2

or (to) = f(x — ct) + g(x + ct) where c = {{p)(\/p)) 1^2, and / , g are arbitrary functions. Compare the equation for (w) with the original equation at constant p, p.

28.10 Let the amplitudes of the transient and reflected waves equal A2 and B\, P\, p2 aud k\, k2 be the modules of elasticity and the wave numbers before and after the jump, and ca = Jpa/Pa (a = 1,2) be the speeds of sound where ka = wjca. Then A* = $$&,' Bl = ElS+ffS Al- W h e n m a t e r i a l P ° i n t s ^ t h masses m,, m2 and initial speeds v\, 0, colhde, the resulting speeds become v[ = ^1~^|a v\, v2 = ^"V ) .

28. Couple Elasticity and Averaging in Media with Microstructure 193

28.11 Due to the energy and momentum conservation laws, after passage of the wave of collisions, J2%Lo ^v'J + £<» = ^r11. T^=omnv'n = moV0. Due to the Cauchy-Bunyakovskii inequality, 2(M + mo) ££Lo ^ ^ n 2 ^ 02™=o mnv'n)2 = TTIQV2,, the equality holding when at v'n = ^ v " which corresponds to the maximal value

^oo = -^r"- (1 — 6) where 6 = m0/(M + mo). Eliminating the speeds vn, v'n from the formulae obtained in the solution of Problem 28.10, one finds the optimal mass distribution: mn = n+(^i)s)(l+ns) w ^h S = i m> = ^- ^n *ne framework of continuum mechanics, the corresponding motion of a discontinuity along the x-axis is described by the equations (see, e.g., Problem 18.16)

a = — - p2(D - v2) , (oD) = -p2v2{D-v2) ,

aD2\ 1 2 . — J =-^P2vi{D-v2) , x = D,

with initial data ao, D0, x0 = 0. For the best transmission of energy, the speed of the oo

medium behind the jump is v2 — D0S (here, 6 = jf^-, M = J p\(x)dx). Then the distribution of p\ is found in the implicit form

Pi = D26 -*(£->-0-ID(D - 6D0) '

28.12 Let the solution in the regions [- | ;0) and [0; | ) have the form

wa = R e ^ ^ e " ^ 1 + Baeik°x))

(respectively, a = 1,2). Then, since F(x) is periodic,

w3 = Re(e iwt- i*'(^1e-'* l ( l- i ) + B^*1'*-'*))

in the region [|;/). From the conditions on the contact discontinuities at x = 0 and x = | , one obtains the system of four linear homogeneous equations for the quantities A\, B\, A2, B2. Assuming that the determinant of the system equals zero, one obtains

- kxl k2l 1 fpici p2c2\ . kj . k2l k = ms rns I sin cos kl = cos —— cos — 1 sin -— sin n 2 2 2 \p2c2 pid) 2 2

where ca = y/na/pa, ka = u/ca. If c^ = c2 = c, cos kl = 1 - ^ * ^ sin2 fc. The zones of non-transparency corresponding to the effective reflection of the incident wave are determined by the inequality (pi + p2)2 sin2 |£ > Apxp2. In the long-wave approximation,

2 & ( k2 (P1/M2-P2/M1)2

LJ = . . . .—- I 1 — (l//i) V 384 (p)2(l/M)2

28.13 w = ±2y/?/m sin ^ , |fc| < *, since oscillations of the system with wavelength m less than 2a cannot be distinguished from those with indicated wn. For small \ka\, u2 = ^ - k2 ( l - *j|-J which yields the equation

dt2 m \dx2 + 12 dx*)

differing from the equations of couple elasticity by the sign of the fourth derivative term.

28.14 The equations of the oscillations have the form mw2n = —(3{2w2n — wtn+i — W2n-i)> Mwin+i = -/?(2u>2n+i - t/>2n+2 - w2n)- Solving these equations, one obtains

\ l / 2 u2

± "(^^^((^^-i^'") tj-(k) is the acoustic mode, uj+(k) is the optical one, Au = m™ u>+ — m " w _ . Also B± = 2g~oa fc* A, where A is an arbitrary complex number. As \ka\ —► 0, w2. = ^r^f,

_ / m (afc)2 M - m \ 1 + M/m ' + ~ ^"M + ~T~ M T ^ J

B = l - ( a f c ) 2 / 2

Finally, the equation

d2w na/l 1 \ 2/3a2 d2™ n -—- + 2/9 ( - + — )w+ H -g-j = 0 OT2 \ m M / m + M ax2

is of the elliptic type (cannot be valid for large |aA;|, as it follows from the process of construction). For acoustic oscillations, displacements of the neighboring particles are directed identically (a running wave), but for the optical ones, in the opposite directions (proper oscillations of the chain).

Chapter 7

Inelastic Solids

29 Plastic Flow Theory 29.1 Use the solution to Problem 8.19.

29.2 a) If only the spheric part of the stress tensor varies, the principal stress components p( change by the same amount. Then the differences pi — pj involved in the Mises and Tresca yield criteria do not vary. b) Values of any isotropic function of a symmetric tensor p can be expressed in terms of its three independent invariants, e.g., in terms of Jy = pkk, 4d\ jf. The values of the functions specifying the Mises and Tresca criteria do not depend upon J\. c) j \ ' is involved in the function specifying the Tresca criterion, since otherwise both criteria would coincide (if the ratio of the constants TS and as takes a certain value).

29.3 Use 1) the no-load condition: PijTij = 0 on the lateral surface (n, are the components of the unit normal), 2) one of the conditions for equilibrium of the half of the specimen cut off by a plane perpendicular to the z-axis: the resultant force should have zero projection on the z-axis, 3) one of the conditions for equilibrium of the half of the specimen cut off by a plane passing through z-axis (see Figure a29.1): the resultant force should have zero projection on the i3-axis.

PTT =PVHfi = PTZ = PrV = 0, pVz = M/2ira28, pzz = F/2na6

29.4 a) F = 0, M ? 0; b) F ? 0, M = 0.

29.5 The yield stress in pure tension <xs is the value of P33 at which the stress tensor with physical components py, where the only nonzero component is P33 = 0, satisfies the yield criterion. The yield stress can be found if the yield criterion is

195

196 SOLUTIONS. INELASTIC SOLIDS

P~

Figure a29.1

known. The Tresca yield criterion is completely determined by the constant rs which has the meaning of the yield stress in pure shear (see Problem 29.4) and, consequently, is given in the problem formulation.

Answer: <rs = 2rs = 46 kN/cm2.

b) The yield stress in pure shear is the value of pn at which the stress tensor with physical components p^, where the only nonzero components are p12 = P21 ¥" 0, satisfies the yield criterion. This value is given in the problem formulation. Substituting these pij in the Mises criterion, one can find the only constant determining this criterion (the meaning of this constant is the yield stress in for pure tension; see Problem 29.4).

Answer: aa = \/3TS « 40 kN/cm2.

29.6 Plastic flow is possible provided the yield criterion /(py, x) = 0 is satisfied. Differentiating it with respect to time and using then the equation determining x and the normality flow rule, one can obtain the expression for A.

In the case of a perfectly plastic body, plastic flow is possible at constant stress, while the plastic strain rate can vary. (Consider, for example, a stressed state with the same values of the stress components at all points of the body, the only nonzero components in the Cartesian coordinate system being pi2 = p2i. Then simple shear takes place at an arbitrary rate.) Therefore, the multiplier A in the normality rule for a perfectly plastic body cannot be expressed in terms of the stress rate.

Answer. For a body with the yield criterion f(pij,x) = 0, the multiplier in the normality flow rule is

A = — ^ - P i - /(—Pki —

29. Plastic Flow Theory 197

29.7 Use the expression for A given in the solution to Problem 29.6. Take into account that A > 0. In the case of a perfectly plastic body: unloading takes place if df/dt < 0; a process starting at /(py) = 0 at df/dt > 0 is impossible (the stresses cannot reach beyond the yield surface); if /(py) = 0 and df/dt = 0, both £? ^ 0 and £? = 0 are possible.

29.8 Use the fact that, in the case under consideration, the function specifying the yield criterion depends only on the deviatoric part of the stress: /(py) = F{p\i');

therefore g- = S y # .

29.9 a) The derivatives in the normality rule can be written as M- = J ^ jjj&- + I f If?" + I f %?-■ F ind Ir*- on the right hand side as the derivatives of the function 0P2 OPi] OP3 Opt) OPij °

given implicitly by the characteristic equation

p3 - IlP2 + hp - 73 = 0,

where h =Pkk , h = ^ l(Pkk)2 ~ PijPij] , 3 = det ||py|| .

(It is useful to remember that the derivative of a determinant with respect to its element equals the cofactor of this element.) Show that, if the matrices of the derivatives ||9pi/9py||, ||dp2/dpy||, ||dp3/3py|| are considered in the basis of the principal axes of the stress tensor, then their only nonzero components are dpi/dpu = dp2/dp22 = dp3/dp33 = 1. b) The equalities £12 = £13 = £23 = 0 are valid in the invariable basis of the principal axes of the strain tensor. Derive in this case from the constitutive relations of elastic-plastic body the equations pi2 = -2/iA ^ , p13 = -2/xA ^ , p23 = -2/xA ^ . These equations can be interpreted as a system of ordinary differential equations with respect to P12, P13, P23 (with regarding X(t), pn(i). P22(0> P33(*) as given). The right hand sides of these equations vanish at p12 = P13 = P23 = 0 (see item (a)), and, according to the problem formulation, Pi2(0) = Pi3(0) = P23(0) = 0. Use the uniqueness theorem for a solution of a system of ordinary differential equations. c) Use the result of item (a). Answer: e\ = A, e = 0, e§ = —A.

29.10 Take into account that the normality rule implies that £JJfc = 0 (see Problem 29.8), and, consequently, e£fc = 0.

29.11 Use the Prandtl-Reuss equations (see Problem 29.10). It is convenient to consider the equation of the yield surface written in the form resolved with respect to x : X — v(Pi)i here, p\ is the intensity of shear stresses p\ = (py p^ ))1/2 .

a) The Prandtl-Reuss equations for all stress components except eztfi = evz = e take the form £y = 0. The equation for ezv together with the equation of the yield surface and that for the parameter x evolution form the system i = j - a + ACT, x — A2cr2, X = v(y/2a) {ezlf = e, pzip = a). Eliminating A and x, obtain an ordinary differential equation for the function e(<r). Find its solution taking into account that there is no plastic deformation until the stress reaches the initial yield surface. The function e(o) is odd (i.e., e(—<j) = -e(cr)); e = a/2(i at 0 < a < fc, e =

Via ajl\i + -4 / ip'(£) f _ 1 d£ at a > k (here, k = CTS/v3, and k is the initial yield stress

v/2fc in pure shear). b) The diagram a — e is piecewise linear for the given hardening law.

29.12 Using Hooke's law (pn = 2fj£U), evaluate the elastic shear modulus from the elastic part of the diagram. Using the corner point in the diagram, find the initial yield stress in pure shear. To find the function specifying hardening, use the equation of the yield surface in the form x — y(v/2cr); derive the equation </>'(£) = l/(2/Xp) - l/(2/x) for the function ip, where 2/ip is the slope ACT/AC of the plastic part of the diagram. This equation allows to find expression for (p (the fact is also to be taken into account that the hardening parameter equals zero when the stress reaches the yield surface for the first time: a = k, i.e., <p(y/2k) = 0). Show that the yield stress in shear varies as T„ = [fc2 + 2x/{^pl — A4"1)]1^2 with growth of x-Answer. The elastic shear modulus is /x « 0.4 • 104 kN/cm2. The initial yield stress in pure shear is fc « 12 kN/cm2; the evolution of the yield stress in shear is given by the formula rs = (fc2 + a2x)1/2 where a2 « 2 • 102 kN/cm2.

29.13 Use the Prandtl-Reuss equations (see Problem 29.10). Using the constitutive relations (as in the solution to Problem 29.11), eliminate x ^ d A and obtain the equations efj — py /(2/x) + y'(pi)piPy /p2 . Note that this equations imply in the case of a simple loading (py ' = TS°J) that the tensors e y and s y are proportional, and, consequently, £y /£ i = Pij /Pi- Derive also the equation de\/dp\ = l/(2/x) + p'(pi)/pi-Solving this equation results in the relations between the intensities: £\ = pj/(2/x) at Pi < \/2fc, and

ei=pI/(2/i) + -^ yVtor 1 ^

at pi > y/2k.

29.14 a) Use the equation of the yield surface, and show that, if the stress vary at pi = const, then the hardening parameterf x is constant, and, consequently, A = 0.


Therefore, the strain evolution follows Hooke's law if the process is considered within the framework of the plastic flow theory. The prediction of deformation plasticity theory is different.

29.15 Answer. Plastic deformation begins when the stress reaches the initial yield surface. Unloading begins when the load passes through its first maximum. Under the repeated loading cycles, the strain evolution obeys the elasticity law (see Figure a29.2).

Figure a29.2

29.16 Take into account that, under the plane strain condition, e.g., e13 = 0. Derive the formula pi3/(2/i) + Xp\3 = 0 from the Prandtl-Reuss equations. Using this formula, show that p13 = 0. Similarly verify that p ^ = 0. Derive from the Prandtl-Reuss equations and the incompressibility condition, the equation (pfj +P22)7(2A0 + A(pft -I- P22) = 0; this formula allows to show that p33 = (pn + pn)/^-

29.17 It is convenient to solve the problem using the cylindrical coordinate system r, ip, z. Due to the symmetry, the physical components of the velocity are of the form vT = vT(r), vv = 0, vz = 0. Using the incompressibility condition, show that the radial component of the velocity can be written as vr = A(t)/r. Note that the displacement has the similar property: u(r) = B{t)/r, B = A. As in the preceding problem, derive from the Prandtl-Reuss equations the following formulae for the physical components of the stress tensor

Pr* = PTZ =P,pz=0 , Pzz = ( P W + P T ) / 2 •

a) If deformation is elastic, the problem reduces to solving the system of equations consisting of the equilibrium equations and Hooke's law, with the boundary conditions

prr(r) = — p at r = a, prr = 0 at r = 6. Using the above-mentioned properties of the displacement and stress, verify that some of the equations are identities, and the rest are of the form dprr/dr + (prr - pw)/r = 0, dtir/dr = {p„ - p w ) /4 / i . Solve this system with the given boundary conditions, looking for the displacement of the form Ur = B/r, where B is constant.

Answer: If the deformation is purely elastic, the displacement and stress distributions are described by the formulae

pa2b2

^ = 2 ^ - a 2 ) ' u » = tt' = 0 ' _ a?p . b2 __ a2p b2

Pnp = Prz =P<pz = 0 , Pzz = (PW + Prr)/2 .

b) Show that the above-mentioned properties of the stresses reduce the Mises yield criterion to the equation |prr — p w | = 2k. Using the already known stress distribution under purely elastic deformation, find the value of the pressure at which the stress reaches the yield surface for the first time. Similarily, find the place in the cross-section where this happens.

Answer: Plastic deformation starts at p0 = (1 — a2/b2)k. The plastic zone propagates from the inner surface inside the pipe.

c) Since the problem is axially symmetric, the plastic zone is a cylinder (adjacent to the inner surface of the tube). The problem of finding stresses in this zone is statically determined, i.e., it can be solved only using the equilibrium equation dpTT/dr + (pTT — p w ) / r = 0 (the only one which is not an identity) and the yield criterion Ipr,. — p w | = 2k\ the boundary condition Prr(r) = —p at r = a is to be taken into account.

The sign of p r r — p w can be found from the non-negativeness of the multiplier A in the normality rule. For this purpose, derive from the Prandtl-Reuss equations the formula A = puey/2/c2. The incompressibility condition and the above-mentioned properties of the stress and velocity reduce this formula to A = e w (p r r —pvv)/{2k2) = vT(pTT — Pw)/(2fc2r). It remains to make use of the condition vr > 0.

Answer: The stress distribution in the plastic zone is described by the formulae p r r = - p -I- 2fc In I, p w = - p + 2fc (In I +1), prip = pTZ = pvx = 0, and pzz = ( p w +p r r ) /2 .

d) Let R be the radius of the cylindrical interface between the elastic and plastic zones. Then the elastic zone can be considered as a pipe (with the inner radius R and outer radius b) subjected to internal pressure. This pressure is determined by the known solution in the plastic zone; it equals —p + 2fc In jj. Use the solution already found for the pure elastic deformation of the pipe.

Answer: If the radius of the cylindrical interface between the elastic and plastic zones is R, the stress distribution in the elastic zone is described by the formulae Prr = (p-2k In f X l - f t V r W / i P - l J . P w = (p-2k In Z)(l + P/r*)/(P/ff-l), Vr<p = Vrz = P<pi = 0 , a n d pzz = ( p w + P r r ) / 2 .

e) The components pTT and prv, are continuous on the interface between the elastic and plastic zones, the stress fields in both zones are already found. The condition for p w continuity results in the equation for the radius R of the interface: kR2 + b2(p — k) = 2kb2 In *.

a f) To analyze solvability of the equation In f + (1 — R2/b2)/2 = p/2k (see item (e)), find on the interval [a, b] the maximum and minimum values of the function on the left hand side, and check that the function is monotone.

29.18 a) In the case of purely elastic deformation, the system of equations describing behavior of the sheet consists of Hooke's law and the equilibrium equations. Take into account the plane strain conditions and the equalities p\2 = P22 = P13 = P23 = 0; this results in the following form of Hooke's law:

e 2 2 = = "AT2^ £ l 1 ' £ l 2 _ 0 ' Pn~ A + 2M £ l" P 3 3 = 2 ( A T ^ ) e n -

These formulae together with the equilibrium equation (the only one of the equilibrium equations that is not an identity), implies that the components of the displacement are of the form U\ = 0,1X1X2 + doX\\ cio,cii = const; u2 = —a\x\/2 + A(aji2 + aoX2)/(A + 2/i). Show that the no-load boundary condition on the upper and lower surfaces of the sheet are satisfied by virtue of the formulation of the problem. The requirement of vanishing of the resultant force at the sides of the sheet can be written

b as / pn dx2 = 0. Write the boundary condition for the resultant moment at the sides

-b of sheet in the similar form, and find the constants ao and ai using these conditions. Answer: In the case of purely elastic deformation, the displacement distribution is given by the above formulae with the constants ao = 0, a,\ = 3M(A + 2/z)/8/i3/i(A+/i); the stress distribution is given by the formulae pn = 3A/x2/263, P33 = \pu/2(\ + fi), P12 = P13 = P23 = P22 = 0. The stress reaches the yield surface for the first time at M = M0 = 4TS / I2 /3; this happens on both upper and lower surfaces of the sheet. Two plastic zones propagate from these surfaces inside the sheet.

b) Using the results of item (a), show that the stress reaches the yield surface for the first time in its point where |pi — p2\ = 2TS, with pi = pn, p2 = P22 = 0. The processes in the plastic zones are described by the system of equations that consists of the equilibrium equation and constitutive relations for the elastic-plastic material. The latter can be written, similarly to the Prandtl-Reuss equations (see

Problem 29.10), as a formula expressing strain rate in terms of the stresses p^ , their rates py, and the multiplier A. Write these equations taking into account that

1) the deformation takes place under the plain strain conditions,

2) Pl2 = Pl3 = P23 = P22 = 0,

3) the stresses belong to the part |pi — P2I = 2r„ of the yield surface; also use the constitutive relations written in the coordinate system whose axes coincide with the principal axes of the stress tensor (see Problem 29.9). Show that these relations are reduced to the equalities |pn| = 2r8, pM = 0, en + £22 = 0, £i2 = 0, e n = A.

Since the components of the stress tensor are continuous on the interface between the elastic and plastic zones, the condition P33 = 0 means that, at a point currently in the plastic zone, the value of the component P33 is the same as at the moment when the interface passed through this point. Using the equalities P33 = Apn/2(A + fj,) at the elastic zone side of the interface and |pn | = 2rg at its plastic zone side, find P33 in the plastic zone.

Answer: The stress is distributed uniformly, and pu = 2rs, p33 = ATS/2(A -I- /x) P12 = P13 = P23 = P22 = 0 in the plastic zone adjacent to the upper surface of the sheet. In the plastic zone adjacent the lower surface, the stresses have the same magnitudes and the opposite signs. c) The plastic zones are adjacent to the upper and lower surfaces of the sheet, the elastic zone is located in its central part. Let the thickness of the elastic zone be 2/i. Show that the elastic zone can be considered as a sheet of the thickness 2/i free from load on the upper and lower surfaces (take into account the above-found values of the stress components in plastic zones and the stress continuity condition on the interfaces between the plastic and elastic zones). Then the stress distribution in the elastic zone is of the same form as the solution in item (a): pu = AX2, P33 = Apn/2(A -I- p), the constant A being unknown.

Using the stress continuity condition on the interface between the elastic and plastic zones and the boundary condition for the bending moment, find the constant A and the thickness h of the elastic zone. Answer: The elastic zone occupies the domain |x2| < h, h = \Z3yb2 — M/2rg. The stress distribution in the elastic zone is described by the formulae given in item (a) with the constant A = 2r„//i. The distribution of the component pu is shown in Figure a29.3 d) The limit load is the one at which the thickness of the elastic zone vanishes. In this case the plane X2 = 0 separates two plastic zones with pu = 2T, and pu = — 2T„. There is no solution to the problem under consideration at bigger loads.

30. Rate-Dependent Effects in Solids 203

Figure a29.3

OX i

Answer. The limit value of the bending moment M is M, = 2TS62.

e) Velocity field in the plastic zone is described by the system of equations

dvi dv2 n dv\ dx\ 8x2 dx2

(see item (b)) and the boundary conditions. The latter consist of the continuity of the velocity components on the interfaces separating elastic and plastic zones. To write these conditions, find the velocity fields in the elastic zones. Note that the displacement fields in the elastic zones are of the form found in item (a), but with a different value of the coefficient ai. To find ai, use Hooke's law and the known stress distributions in the elastic zones (see item (c)). Answer: The velocity field in the plastic zone adjacent to the upper surface of the sheet is given by the formulae

vi = Axix2 , v2 = A -I ? + I ? 2 3A + 2M .

2(A + 2/i) ) ■

where h = V3jb2-M/2Ts, A = 3(A + 2p)

8/x(A + fi)h: M .

30 Rate-Dependent Effects in Solids 30.1 Use one of the constitutive relations of the Maxwell model, izv = pzv +

PZV/T, under the condition ezv = const. Answer: If the material is described by the Maxwell model, the stress relaxes: pz^{t) = pzv(0) e~'/T. If the material is described by the Voigt model or by the plastic flow model, the stress does not vary.

30.2 Consider the constitutive relations of the Maxwell model in the case the only nonzero stress component is pzz. Derive the equation

izz = Ei + 2 ^ ( £ = M 3 A + 2M)/(A + /X))

and find stress relaxation law at ezz = const. Similarly derive the equation (£„. — £w>) = 0- I* implies that the equality £„. = e w , being satisfied at the initial instant, remains valid. Using this equality, the relationship pkk = 3K£kk and the obtained stress relaxation law, find dependence e„. and e w upon time. The known e„., , ezz determine relative elongation of any material line element and, thus, the evolution of the cross-sectional area too. Answer, a) The stress relaxation is described by the formula pzz(t) = pzz(0) e~Et^3,1T\ b) Evolution of the cross-sectional area is described by the formula S(t) = 5(0) [1 + pzz(t)/3K-ezz(0)).

30.3 Use the constitutive relation that relates the components ezifi and pzifl. Solve the corresponding differential equation for ezip taking account the conditions PZ<P = Po = const and ezv(0) = e0-Answer: a) If the material is described by the Voigt model, the strain evolution is given by the formula

£(t) = ^ + ( ^ o - ^ J e , •

b) If the material is described by the Maxwell model, the strain linearly increases: £zV = £o+ p0t/2fir.

30.4 The equation of the diagram is e = (a + a2/2Ar)/2fj, + e0, where e0 is the value of the strain at the initial moment. The fundamental difference between the case under consideration and the case of plastic deformation is dependence of the diagram upon the rate of loading (given by the parameter A).

30.5 In the case of an elastic-plastic specimen, unloading takes place; therefore the strain instantaneously follows the stress. In the case of the Maxwell material, when the stress rate is large, it determines the strain rate (and the strain rate is large too). The strain remains constant at zero stress. In the case of the Voigt material, when the stress decreases from its initial value to zero, the strain also decreases, but it cannot change too much since the process takes not too long. Strain relaxation takes place at zero stress. Answer: The first diagram in Figure a30.1 shows the strain evolution in the case of elastic-plastic material; the second in the case of the Maxwell material; the third in the case of the Voigt material (a0 and e0 are the initial values of the stress and strain).

30. Rate-Dependent Effects in Solids 205

e,=e,- 2u

Figure a30.1

30.6 Rigid visco-plastic medium is determined by the constitutive relations

vf = any satisfying the condition P<£j){d)mn < 2k2 if ey = 0 .

30.7 a) Show that the incompressibility condition implies that the velocity component i>i does not depend on the coordinate i i . Using the constitutive relations of the rigid visco-plastic model (see Problem 30.6), show that the only nonzero components of the stress deviator are p12 = fct>'i/luil + V^i (prime denote derivative with respect to X2). Derive the required properties of the pressure from the equilibrium equations. b) The velocity component Vi is a symmetric function of X2 due to symmetry of the problem. If the domains where v[ = 0 are adjacent to the plane x2 = 0 from above and below, then pi2(+0) = — pn{—0) ^ 0, and the condition \p^]vi = 0, representing the momentum conservation law on the discontinuity surface, is not satisfied (here, brackets denote the jump of a variable across a discontinuity surface). c) Find the values of the stresses on the upper and lower boundaries of the undeforming layer, using the inequality Vij v\j < 2fc2 inside the undeforming layer and the inequality p\},'p-j > 2k2 in the deforming layers adjacent to the undeforming one. Using the constitutive relations show that the derivative v[ vanishes on the boundaries of the undeforming layer. d) Consider the equilibrium condition for a part of of the undeforming layer contained between its two cross-sections orthogonal to the X-axis. Take into account the results of item (c). Answer: B = kh

e) Behavior of the medium in the deforming layer is described by the system of equations consisting of the equilibrium equations and constitutive relations. Most of them are identities with the exception of the constitutive relation mentioned in the solution to item (a) and the equilibrium equation ^ 1 + ^ 2 = 0. Here, pn = Ax\ + const; and A equals the constant B (see item (d)). dolve these equation taking into account the given boundary conditions: vanishing of the velocity on the wall of the channel and the equality dvi/dx2 = 0 on the interface between the deforming and undeforming layers (see item (c)).

Answer: The component of the velocity v^fa) is a symmetric function, ^1(12) = -^1(^2); and

l{-2)~\MH2-h2)-^H'-h) at 0<x2<h. The thickness h of the undeforming layer is related to the flow rate Q by the formula 3r)Q = kH2(H/h - 1)(2 - h/H - h2/H2).

Chapter 8

Basic Notions of Relativistic Kinematics and Dynamics. General Properties of Electromagnetic Field

31 Lorentz Transformations. Minkowski Space 31.1 It obviously follows from the condition a) that the transformation is (dis

placements of the coordinate system origin are not taken into account)

f = Af , rf = fir) (f = ct - x, 77 = ct + x) . (a31.1)

Hence, x> = i(A + /i)x - |(A - n)ct , ct' = - i ( A - p ) x + i ( A + /*)tf . (Ml^>

One can obtain the transformation inverse to (a31.1), by replacing quantities with primes by those without prime, as well as A and \i by 1/A and l//x. The speed v of the system x'.i ' relative to x,t is obtained from the condition x' = 0 which yields x = vt, v = c(X — /i)/(A + p). The speed v' of the system x, t relative to x', tf is v' = c(A-1 —/z-1)/(A_1+/i_1) = —v. It is obvious that the speed of the system (-x) , t relative to the system (—x'),1? equals v. Thus, on the first hand, the transformation of ( -x ' ) , f into (—x),t can be obtained from (a31.2) by replacing A and n with A-1

and /i"1; and, on the other hand, according to condition b), it should be given by the same matrix as the transformation (a31.2). It is possible if and only if A/x = 1. Using the expression for v in terms of A and (i, one obtains A = J(l + v/c)/{\ — v/c), H = 1/A. Substituting these equations into (a31.2), one finds x' = a,uX — a^ct, ct? = -ai2x + a22ct where an = 022 = 1/^/(1 - v2/c2), a12 = (v/c)/^(l - v2/c2).

207

208 SOLUTIONS. RELATIVITY THEORY

31.2 It is verified by direct calculation.

31.3 An orthogonal transformation of two variables (without reflections which can be considered separately) is fj = fi cos 9 — £2 sin 9, £2 = £i sin 9 + f2 cos 9. If £[ = 0, one obtains £x = £2 tan 9 = id tan 9, from which ic tan 9 = v, tan 9 = -«;/<:, cos 0 = (1 + tan2^)-1 /2 = 1/^/(1 - v2/c?), sin 0 = cos 0 tan 0 =

—i(v/c)/J(l — v2/c?). Introducing the real variable <p = —i9, theLorentz transformation can be written in the form x' = coship-x — siohip-ct, elf = — sinhyj-x+cosh^-c£.

31.4 The transformations of problem 31.3 — "rotations by an imaginary angle" — obviously, form a group. As a real parameter, incorporating successive transformations of the group, <p = arctanh (v/c) can be chosen, —00 < ip < 00, — c < v < c; ip = 0 corresponds to the identity transformation, (p = ±00 corresponds to v = ±c. Let V\= c tanh <p1 and v2 = c tanh < 2 be given. Then, execution of the corresponding Lorentz transformations is a Lorentz transformation with tp = ip^ + ip2 (which one can verify immediately). The value of v corresponding to ip is determined by the equality

. , vi + v2 v = c-t&nhifi, sov = ———— 1 -I- v'/cr

This equality is called the relativistic law of velocity addition.

31.5 Use the identity 2gijx,yj = 5yi*i J ;+flyyV l _Sy^' -J/ ')^—V*)- According to the definition of Lorentz transformations, they preserve the values of each of the three terms in the right side of the identity. Since <7y is diagonal, space and time are orthogonal.

31.6 Let a general Lorentz transformation transform the variables x, y, z, t into x', y,, z7, f. The axis tf should he inside the light cone x2 + y2 + z2 < c*t2, which follows from conservation of the value ffyx'y5 (and, consequently, its sign). Let the positive directions of the axes t and t' can be brought into coincidence by continuous movement. Otherwise, the direction of the axis tf should be changed by reflection. Consider the two-dimensional plane passing through the axes t and tf. Let Z1 denote the spatial axis in that plane orthogonal (in terms of the scalar product defined in Problem 31.5) to t and f2, £3 denote spatial axes orthogonal to f *. The transition from the variables x, y, z to £*, f2, £3 is represented by an orthogonal transformation. For the particular Lorentz transformation in the t,^1

plane such that the t-axis coincides with tf, t1 is transformed into f and f2, f3 are not transformed. The subspace passing through the axes £', £2, £3 coincides with the subspace of, j / , z\ since both they are orthogonal to the axis if (this fact can be easily checked); and, consequently, the transformation from some variables to others is orthogonal. Thus, the general Lorentz transformation is represented as the sequence of the transformations enumerated above.

31. Lorentz Transformations. Minkowski Space 209

31.7 Determine a tensor gxi which has the enumerated components gij in a coordinate system x1 = x, x2 = y, x3 = z, x4 = t. In an arbitrary Lorentz coordinate system x", <7ijX'xJ = 5^x"x'J = <fyx"x'J. Since x" are arbitrary, g1^ = p^.

31.8 a) A transformation of basis vectors is known to be represented by the inverse conjugated matrix with respect to the matrix of the coordinate transformation. Since the inverse Lorentz transformation is obtained with replacing v by — v (see solution of Problem 31.1), the transformation of the basis vectors e* directed along the axes x1 = x, x2 = y, x3 = z, x4 = ct is e\ = Quei + a12e4, e4 = ai2ei + a22e4, e'2 = e2, e'3 = e3 where an = a22 = l / y ( l - v2/c2), a12 = (v/c)/^(l - v2/c2). It is obvious that the vectors e[ and e'4 on the plane x, ct are located symmetrically relative to the straight line x = ct, and, as v varies, the ends of these vectors move along the hyperbolas crossing the coordinate axes at x = 1 and ct = 1 with asymptotes x = ct and x = — ct. b) If, for a pair of points xi,ti and x2,t2, the slope of the segment Ax, At (Ax = x2 — Xi, At = t2 — t\) on the x, t plane relative to the x-axis is less than c, then, according to the above, it is possible to choose v so that the vector e'j is parallel to the vector {Ax,cAt} (i.e., At' = 0); alternatively is possible to have the vector e'4 be parallel to that vector (i.e., Ax' = 0). c) The Jacobian of Lorentz transformations holding orientation constant equals 1, since, according to the principle of relativity, coordinate systems are equivalent. For a particular Lorentz transformation, this can be easily verified by direct calculation, d) Let us analyze the identity obtained by integration (Gauss formula)

r£dv=irds< where S is the hypersurface bounding Vj4), and dSi are the projections of hypersurface elements onto the coordinate hypersurfaces. Note that dSi (as well as dV(4) — see above) is assumed to have the same meaning as in calculus, i.e., as the products of the differentials of an independent variables, so that the integral is taken over a set of points that are projections of points belonging to dS onto the coordinate hypersurfaces. Let / ' be components of an arbitrary contravariant vector which are nonzero on S only on an element dS with projections dSi. Considering / ' as arbitrary constants we conclude that dSi are components of a covariant vector, since the left side of the equality written above is a scalar.

31.9 Let the coordinates of the ends of the rod in the proper coordinate system be x[ = 0 and x'2 = I. Then (see Problem 31.1) i i = vt, x2 = ly/(l - v2/c2) + vt. It is obvious (see Figure a31.1) that L/l = J(l — v2/c2).

31.10 Let the events take place at x\ = 0, t\ = 0 and x'2 = 0, t'2 = r in the "moving" coordinate system. Then according to 31.1 we have xt = 0, t\ = 0 and


Figure a31.1

x2 = vt2, t2 = T = r/y/(l - v2/c>), T/T = 1/^(1 - v2/c2) in the system of the observer.

31.11 According to 31.10, the time r = 7 y (1 — v2/c2) elapses in the rocket. While the rocket moves with constant speed, the both observers are equivalent, and, from the viewpoint of each, the time elapsing in the other's system is less than that in his own (see problem 31.10). Non-equivalence arises at the moment when the speed

Figure a31.2

of the rocket changes. At that moment, the time of the unmoving observer from the viewpoint of moving one (t at x = 0), as a function of the proper time elapsing in the rocket has the jump denoted on Figure a31.2 by At. If v = v(t), then r = B , / J(l — v2/^) dt. The quantity r is an invariant (does not depend on the coordinate A v

system in which it is calculated), since dr = ds/c where ds2 = c2 dt — dx2 — dy2 — dz2

(see Problem 31.2). r reaches its maximum if integration is carried out over the straight line in the x,t plane connecting the points A and B. This statement is obvious for the coordinate system in which the straight line AB is the time axis. Non-equivalence of the observers (the "unmoving" one and the one in the rocket)

32. Concepts of Relativistic Kinematics and Dynamics 211

mentioned above can be explained by the fact that according to the initial supposition about the form of the metric, one of them is on the same geodesic during the whole time and the other is not.

32 Concepts of Relativistic Kinematics and Dynamics

32.1 a) Let us denote va = dxa/dt where xa are Cartesian coordinates, a = 1,2,3; then

ds = y/{c2 dt2 - dx2 - dy2 - dz2) = cy/(l ~ v2/c2) dt, v2 = (^ ) 2 + (i;2)2 + (a3)2,

ua = dxa/ds = vQ/cv/(l - v2/c2), u4 = dt/ds = l/cy/(l - v2/c2),

pa = mva/J(l - v2/c2),p4 = m/y/{l - v2/c2).

b) - ELi("a)2 + ^(«4)2 = (" E3Q=i(^)2 + c W ) Ids2 = 1.

32.2

pa = mcua = mvalsj(\ - v2/c2) = mva{l + v2/2c2 + 3v4/8c4 + ...),

p4 = mcu4 = m/y/(l - v2/c2) = -^{mc2 + mv2/2 + 3mv*/8c2 + . . . ) .

32.3 The new values of m and va are obtained from the equations

y/(l ~ V2/C2)

, m = PS + A P 4 ^ P 4

where p{, is the initial four-momentum, and Ap* is the increase in four-momentum. They are va = pa/pi, m = p 4 y( l — v2/c2). For m to be real it is necessary that v2 < c2 or Da=i(pa/p4)2 < c2, i.e. that the new four-momentum is a time-like vector.

32.4 No, it is not. When the bomb explodes, the four-momentum of the products of explosion remains equal to the four-momentum of the bomb before explosion (conservation of momentum and energy), although the mass at rest of the particles which are the products of explosion is less than the original mass of the bomb. We have here the Problem 32.3 with Apa = 0. The velocity and the mass of the system do not change.

32.5 Let the derivative of the four-momentum with respect to the proper time be known: dp'/dr = / ' , p* = m dx'/dr, f* being a four-dimensional vector. Neglecting the quantities of the order v2/c2 as compared with those of the order 1, one obtains d(mva)/dt = fa, d(mc2 + mv2/2)/dt = c2 /4 from which it follows that, for the same order of approximation, dm/dt = f* - (v1/1 + v2f2 + v3f3)/c2. At va = 0 (in the proper coordinate system), dm/dt = dm/dr = / '4 . The prime means that / 4 is calculated in the proper coordinate system.

32.6 Let the acceleration of the particle in the proper coordinate system equal o. For an interval of proper time dr, the particle velocity increases by adr in the proper coordinate system. According to the relativistic rule of addition of velocities (Problem 31.4), the velocity of the particle in an unmoving system gets the increase

v + adr ( v2\ dv = T~, TT1 -v=[l--^)adT .

1 + va drjc1 \ cr) Integrating, one obtains v/c = tanh (a(r — T0)/C). By definition, v = dx/dt and dr =

1 — v2/c?) dt. Integrating the last relationships, one obtains t — to = c sinh (a(r — r0)/c)/a, x—Xo = (c2/a)(cosh (a(r—r0)/c) — 1). In the x, t plane the particle describes the hyperbola <?{t — to)2 — (x - x0 + c2/a)2 — —c4/a2 as the proper time r varies.

32.7 Let us define the amount of four-momentum passing through a hyper-surface element with projections dSi onto the coordinate hypersurfaces (see Problem 31.8) by means of the equality dp' = 7*' dSj. This equality is obvious if the vector dSj is directed along one of the spatial axes. Then dp1 = T*" dSa (no summation over a) defines the corresponding flux of four-momentum as it follows from the statement of the problem. If the vector dSj is directed along the time axis, then the formula dp' = T*4 dSi defines the fluxes of momentum from the past to the future through the corresponding hypersurface element. To obtain the formula presented above for the general case, let us consider the inputs of four-momentum to a small four-dimensional pentahedron, the four faces of which lie in the coordinate hypersurfaces, and the fifth one closes these four ones bounding a small four-dimensional volume. Supposing that there is no volume input of four-momentum, or it is proportional to the four-volume so that it can be neglected as compared with the influx through the faces for a sufficiently small pentahedron, and setting the flux of four-momentum through the fifth face equal to the sum of the four first ones, one obtains the formula dp' = T^'dS] in general case. There are components of a four-vector on the left side of the obtained equality; dSj are also components of a vector (see Problem 31.8). Therefore, the matrix T*J, providing a linear transformation of vectors, is a tensor. The whole reasoning conducted above repeats the well-known proof that stresses in a continuum are characterized by a tensor.

V(

33. Maxwell Equations 213

32.8 In the proper Lorentz coordinate system for an ideal gas, TQ/3 = —p6a0, Ta* = QT40 = 0 ) T44 = p(! + ui^ w h e r e v = [ / ( P ) p) i s ^ g j j S i n c e t h e v e c t o r o f

four-velocity in the same coordinate system has the components u4 = 1, ua = 0, one can write

7* = pgv + (P(u + c2) - p ) u v / c2 , II^II = H^ii"1

These expressions are valid in an arbitrary coordinate system.

32.9 The momentum and energy conservation laws are dT%:>/dxj = 0, i = 1,2,3,4. The equation expressing conservation of mass at rest is written in the form d(pul)/dx% = 0, where p is the density of the mass at rest. This equation together with the preceding one forms a closed system of equations for ua, p, p, a = 1,2,3. For the case of dust, T*J = puluj.

33 Maxwell Equations 33.1 a) Use the formulae of Stokes and Gauss.

b) Direct the x-axis along the normal to the discontinuity surface n at its arbitrary point, and consider the small surface E (shaded on Figure a33.1) in the shape of a plane rectangle with sides of lengths el and / parallel to the x— and y— axes, respectively. Let e —» 0 and / —► 0 so that the discontinuity surface does not cross

Figure a33.1

the sides of the rectangle parallel to the y-axis. Then, from the left equations (33.1 33.2), one obtains

&? - El"-±W{BV> - B?) = 0 ,


ET]

\BT]

+ -Wn c

\--Wn c

X[BT)

x[ET)

[Bn] [En]

=

=

= =

o, 4iri

c 0 , —Ana

where the superscripts 1 and 2 mean that the hmiting values of the corresponding quantities, approaching the discontinuity surface from side 1 or 2, are taken; the x-axis is directed to the side 1, W is the velocity of the discontinuity surface movement along the z-axis, and iz is the surface density of current in the direction of the 2-axis. Similar relationships can be obtained for the surface element lying in the x — z plane. Take the volume V for the second pair of the equations (33.1 - 33.2) in the shape of a cylinder with its element parallel to the i-axis. Let the diameter of the cylinder tend to zero and the length of the cylinder tend to zero more quickly. Then one obtains

B (2)_ B (D = 0 , £(2)_£(D = _4 7 R 7

where a is the surface density of charge. The obtained equalities can be written in the vector form

x n

where the square brackets denote the jumps of the corresponding functions: [A] = A{2) _ ^ ( 1 ) .

Let us note that one can obtain the relationship for the values of surface current and charge from the electric charge conservation law. However, it is shown to be a consequence of relationships already obtained, as it is found by staring from the original Maxwell equations (see Problem 33.7). If W = 0 (this can be often achieved by the appropriate choice of a moving coordinate system), the first two equalities are reduced to: [ET] = 0, [BT] = —4iri/c x n.

33.2 In the stationary case, E = grad tp. Then, if ET is discontinuous on a surface, the jump of potential (f on the surface does not equal zero. For that case, the integral / E • dx over a segment crossing the surface does not tend to zero as

Xl

x2 — x\ —► 0. This fact shows that the assumptions taken in Problem 33.1 are not always valid.

33.3 The motion and energy equations in a Lorentz coordinate system for a particle of a mass m with charge e can be written in the form

dPk r-ki i k k k d%k

-^— = -eF JUj , Uj = u gij , p = mcu , u = —— . dr as


These equations can be rewritten in the form

dpa („ vxB\a dp4 „ a dxa

They naturally generalize the non-relativistic motion and energy equations for a charged particle. Thus, the Lorentz force is precisely the derivative of the first three components of four-momentum of the particle with respect to time. The charge e is a scalar and dpk/dr, Uj are vector components, hence F*J' are tensor components.

33.4 It is verified by direct calculation.

33.5 Existence of the required coordinate system is an obvious consequence of Problem 33.4. The Larmor frequency is obtained from the equation muiv = evB/c: w = eB/mc.

33.6 a) The first and second pairs of Maxwell equations are written in a Lorentz coordinate system

~ = - J \ Ja=ja, J4 = Pe, (a33.1) ox> c dF* dFki dF{i _ dx{ dxj dxk

when —— = gmn ——, the matrix gmn is inverse to g« (and diagonal), g11 = g22 = oxm oxn

g33 = — 1, and g44 = 1/c2. Each term in equations (a33.2) is a component of a third-order tensor, and there are four independent equations. It follows from the form of the equations (a33.1) and from the fact that ir 'J are tensor components that J* are vector components (four-dimensional density of current). b) Let us consider the case of motion of several charged fluids moving, in general, with different velocities. For each of them, the four-dimensional density of current can be defined as follows: Jk = cp'euk, ua = va/(c^l — v2/c2), and u4 = \/{cJ\ — v2/c2), p* being the electric charge density in the proper coordinate system, so that pe = pl/Jl — v2/c2, and j a = peva. The denominator in the expression for pe can be interpreted as arising due to decrease of lengths in a moving medium and corresponding increase of the charge density. For the case of several charged fluids, the four-current is a sum each term of which is a four-vector.

c) The transformation of the quantities j a , pe is the four-dimensional vector transformation, the same as that of x' (see Problem 31.1). It is usually assumed for the non-relativistic case that j ' a = j a — peva, p'e = pe. These formulae are valid if

v2/c2 <£. 1. For the equality p'e = pe to be valid, it is also necessary that the inequality pe » vj/c2 be valid. It is necessarily invalid if pe = 0, j ^ 0. The fact that, for pe = 0, the charge density p'e 0 arises on transition to another coordinate system is a relativistic effect connected with different Lorentz contraction of lengths and a corresponding change in densities of charged media moving with different velocities.

33.7 Taking the four-divergence of the first Maxwell equation, one obtains the equation of charge conservation 6Jk/dxk = 0 or dpe/dt + dja/dxa = 0, since d2Fii/dxtdxj = 0 due to antisymmetry of the tensor F , J .

33.8 Taking the divergences of the vector Maxwell equations and taking into account the equation of charge conservation, one obtains

— ( d i v B ) = 0 , - ( 4 7 r p e - d i v . E ) = 0 .

33.9 To obtain the force and energy input sum, over all the charged particles contained in a unit volume, the forces and powers applied to every particle (see Problem 33.3):

/ = ^SJE+^XB) , Q \ C /

a

Introducing the densities of charge and current pe = £ Q eQ, j = £ Q eavQ, one obtains f = peE + \jxB,N = j-E.

33.10 It is verified by direct calculation that Sa0 = {BaB0 + EaEp)/4n -6a0{B2 + E2)l%iT, Sai = (Ex B/47rc)°, S4" = {E x B/4irc)", S44 = (B2 + E2)/8ivc2, a,0 = 1,2,3, in Lorentz coordinate systems, Sal3 is called the tensor of Maxwell stresses, ga = Sai is called the momentum density, W = c2S44 is called the energy density, and S0 = c?Sil3 is called the density of electromagnetic energy flux (Poynting vector). Using Maxwell equations, the divergence of the tensor Stk can be transformed to the form

dSak

dxk

dS4k

dxk

The last relationship is called the Poynting theorem. The right hand sides in the equalities obtained are, respectively, the momentum and energy divided by c2 transmitted to the field by the charged particles (being represented by pe and j).

dSa0 dga

dx0

1 (dS0

+ 1 r = -^Q -(j x B)a

c2 \dx0 dt + — 1 = " J ■ E

33.11 a) Let us consider two solutions of the Maxwell equations satisfying identical initial and boundary conditions and corresponding to identical distributions of electric current j(xa,t). Since the equations are linear, the difference of these solutions E,, B, satisfies the same equations with j = 0 and zero initial and boundary conditions. Using the Poynting theorem (see Problem 33.10) for E,, B. and the condition E,T = 0, one obtains

d_ r(E2, + B2m)dV ,{E.xBm)ndE

dtJ 8TT J 4TT V dV

from where §-J(El + Bl)dV = 0,

v i.e., E, = 0, B. = 0. b) The solution of the Cauchy problem with initial data given at t = 0, evaluated at any point and at an instant t,, can depend only on the distribution of the initial field in the sphere of the radius ct, with center at that point. The initial electromagnetic field can be set equal to zero at t = 0 outside of that sphere without any changes in the solution. Then, the electromagnetic field at 0 < t < t, vanishes at r > 2ct,. Uniqueness of the solution of the Cauchy problem follows from uniqueness of the initial-boundary-value problem with zero boundary conditions given on the closed surface containing the sphere of radius r = 2ct, inside of it and with initial conditions coinciding with the original ones inside the sphere of radius r = ct, and vanishing out of it.

33.12 The solution is analogous to that for problem 33.11

33.13 Since the required transformation does not depend upon the pre-history, one may consider transformation of the electromagnetic field caused at zero initial conditions by a four-current that is nonvanishing only at ( > to- The electromagnetic field under consideration can be regarded as equal to zero at infinity, because the speed of light is finite. If the electromagnetic field is transformed according to the tensor rule, then, in the new coordinate system, one obtains the solution of the Maxwell equations with the same four-current on the right hand side and with zero initial and boundary conditions. Since the solution of the Maxwell equations is unique under the enumerated conditions, then the transformation of an electromagnetic field satisfying the conditions (l)-(3) is also unique.

Chapter 9

Models of Media Interacting with Electromagnetic Field

34 Magnetohydrodynamics 34.1 The equation of internal energy for an element of the fluid is dU = —p dV+

(j' ■ E'/p)dt. Here, V = 1/p, and it is assumed that pn = —pn (there is no viscosity), dq = 0 (there is no heat conduction), and the amount of energy transmitted to the medium from the electromagnetic field per unit mass in the proper coordinate system of the medium element is dq* = - f • E' dt (see Problem 33.9). Comparison of Gibbs' identity dU = —pdV+Tds with the equation internal energy yields pTds/dt = j'E'. The last formula makes it obvious that the energy f ■ E' manifests itself under the suppositions made above in the form of heat, called the Joule heat (which causes the same change in entropy). If the electromagnetic field and the fluid are considered as an integral system, then entropy variation caused by the electric current is an internal process for it, and one may write

ds _ dvs _ j ' • E' dt dt pT

d\s/dt being the internal entropy production in the system. The quantity dq' = T dxs is called the uncompensated heat (in general case, the uncompensated heat includes only that part of TdiS that is not connected with heat conduction).

34.2 a) According to the preceding problem, dxs/dt = j ' ■ E'/(pT). In the thermodynamics of irreversible processes, entropy production is represented as the sum of products of multipliers characterizing deviation from thermodynamic equilibrium. Some of these multipliers are called "fluxes" the others are called "forces" (of course, the names are not important). For the processes slightly deviating from equilibrium

219

220 SOLUTIONS. MEDIA INTERACTING WITH ELECTROMAGNETIC FIELD

ones, linear relations between the "fluxes" and "forces" are postulated. Thus,

ji = oyE- (a34.1)

(the factor pT is included into <ry). This relation is Ohm's law. The coefficients (7y are components of a tensor. This tensor is called the electric conductivity tensor. In general, it depends on the scalar, vector and tensor parameters characterizing the thermodynamic state of the medium. The vector and tensor parameters can cause anisotropy of the relation (a34.1). It is more convenient to analyze <7y dependence upon the axial vector B' as dependence upon the tensor tijkB'k- ft follows from the theorem of Hamilton-Cayley (see Section 5) that any analytical relationship between three-dimensional second rank tensors can be represented in the form of a quadratic polynomial with scalar coefficients. In the case under consideration, one obtains

ay = a{Sii + peijkB'k/\B'\ + aBty/B*)

in a Cartesian coordinate system. It is obvious that, in the case of isotropy, when the influence of B' is negligible, Ohm's law (a34.1) takes the form

f = oE' where the scalar a is called the electric conductivity coefficient, b) To estimate the quantities a, a and /? in the proper coordinate system of the element of the medium, write the equation motion for the "electron fluid", neglecting forces of inertia and the gradient of electronic pressure: 0 = neeE' + j ' x B'/c + /ft.. Suppose that the friction force experienced by the electron fluid is /ft- = —nemev/T where ne is the concentration of electrons, me is the mass of electron, r is the average time between collisions of electrons with particles of the medium, and v is the average velocity of electrons relative to the medium related to j ' by the equality j ' = neev (the current is assumed to be caused only by motion of electrons). With the introduction of the notation eB'/cme = u>, the equality given above can be rewritten in the form J'+UJTXJ' = aE', a = nee2T/me. The quantity e\B'\/cme is the Larmor frequency of electrons (see Problem 33.3). That relation represents generalized Ohm's law which is used just in this form. Comparison of the last equality with the expression for <7y-obtained before shows that a = U)2T2/(\+IM2T2), 0 = |U>T|/(1+U;2T2) . Appearance of the current j ' component perpendicular to B' and E' is called the Hall effect. That component is usually small in dense media (if W T C I ) and can be neglected.

34.3 The input equations are dpjdx = n^eE, dE/dx = Ai:e(n\ — ne), dpe/dx = —neeE, pi = n\kTt pe = nekT where nj, ne are the numbers of ions and electrons, A; is the Boltzmann constant, T = const, the ion charge being assumed definiteness to be equal in magnitude to the electron charge e. From here,

dnx n^eE dne neeE dE , . & = l ^ ' ^ = -^F'^ = 47re(ni-ne)-

34. Magnetohydrodynamics 221

The solution of these equations can be found in an explicit form. To calculate the Debye length, consider the equations linearized in the vicinity of infinity. As a result, one obtains cPtp/dx2 = 8nn°e2ip/kT, LD ~ JkT/8Tm°e2 (n° is the density of electrons at infinity).

34.4 The input equations are j = CJE, dpe/dt + divj = 0, divE = 4irpe, from here, pe = p%t~^at. Consequently, div j > 0 everywhere, and the charge flows out to infinity.

34.5 Let us denote the characteristic value of a quantity with the subscript "0" and its characteristic interval of variation in the problem under consideration with "*". a) Assuming that a&ao, write the first pair of Maxwell equations taking Ohm's law into account, and estimate the orders of their terms:

dE curl B - i — - | ( (JE+ a^xB+ pev) = 0 , div E = 4ir pe

B E aEo <7v0B0 E L cT c c L

Here, under each term of an equation, its estimation is written, L and T being the typical length and time. 1) Let

^ « 1 , ^ « 1 , (.34.2)

so that 1/oT, v0/aL can be neglected compared with 1. Then, taking the second equation into account, one can neglect the second and last terms in the first equation compared with the third one. As a result,

E = - (vmcuilB--xB\ , E0<maxl^-, — Bo\ (a34.3)

where um = C2/4TT(7 is called the magnetic viscosity. The second and last terms in the Maxwell equation can be estimated in terms of B: - dE/dt ~ max{B/LaT,v0B0/c2T}, pev/c ~ max{v0B/aL2,vlB0/c2L}. If, besides (a34.2), the relations

are supposed to be valid, then the terms indicated above are small as compared with curlB ~ B/L. If it is supposed that

| » | , ( . 3 U )

then the relations (a34.4) are valid due to (a34.2). Thus, if (a34.2) and (a34.4) (or (a34.2) and (a34.5)) are valid, then

ATC v curlB = —j, j=a(E + -xB). (a34.6)

c c

2) With accuracy up to the terms of the order v2/c*, the transformation of the magnetic field on transition to a moving coordinate system is B' = B — " x E. The

last term, according to (a34.3) is of the order max < ——, ° 2 > and, due to (a34.2)

and (a34.5) is small compared with the first one and can be neglected. 3) The electric and magnetic forces have the following orders of magnitude \peE\ ~ EEQ/LI ~ El/L ~ max{c4B2/o2L3,v%B%/c2L}, \j x B/c\ ~ BBQ/L. For the electric force to be neglected compared with the magnetic one it is necessary that

The first relation coincides with (a34.5), and the second one can be a stronger restriction of the magnitude of conductivity than (a34.2). Since, according to (a34.6), the convection current is small compared with the conduction current, the Joule heat can be written in the form

f .E' = -jr2 = -j2=1^ (curl B)2 . (a34.8)

b) From the second group of Maxwell equations and the relationships (a34.6), one obtains

BB — curl (v x B) + curl (i/m curl B) = 0 , div B = 0 (a34.9)

(the first of these equations is called the "induction equation"). The equations of momentum, continuity and energy are written in the form

dv j 1 , „ „ dp n dS vm{rotB)2 . „ , , „ . P - = - ^ P + -curlBxB,^t+P<livv = 0 , - = ^ . (a34.10)

The equations (a34.9), (a34.10) form the system of equations of magnetohydrody-namics. Similar to the complete system of Maxwell equations, the second equation (a34.9) is a restriction on initial conditions for B, since it follows from the first equation (a34.9) that d(div B)/dt = 0.

34.6 a) Rem = v,L/vm where v. and L are the characteristic speed and length. b) The integral Maxwell equation describing variation in time of the magnetic field, on substitution into it of the expression for E used in magnetic hydrodynamics (the

equality (a34.3) in solution of Problem 34.5), takes the form (integral form of the induction equation)

— f BndZ- j{v x B) ■ dl + jumcwlB ■ dl = 0 Ho L L

where E0 is an unmoving surface, and L is its boundary. It is obvious (as in the general case) that, for a closed surface S, the equality § Bn dE = 0 is always valid if it is valid

s at a single instant. The integral induction equation can be modified, considering variation of the magnetic flux through the surface E(t) moving together with the medium and taking into account that, for an arbitrary vector field B(x, y, z, t) with zero divergence,

J BndZ = j f BndZ + JB ■ (v x dl) dt

£(t) £o L

where Eo is the unmoving surface coinciding with E(i) at the instant under consideration. Then, the integral equations corresponding to the induction equation take the form

— j BndY, + J umcui\B-dl = 0 , JBndZ = Q E(t) L E

where E(t) is a material surface, L is its boundary.

34.7 If i/mcurlB = 0 or according to ( 34.2) E = -v x B/c, then, for an unmoving surface E0 and for a material surface E(t), the following equivalent integral forms of the induction equation exist:

±jBndE = J(vxB)-dl, j t j 5ndE = 0. dt Eo L E(t)

The last equality is also called the theorem "of freezing in".

34.8 According to the theorem "of freezing in" of a magnetic field (see Problem 34.7), the scalar triple product [dl(i)(t) x dl^)(t)\ ■ B(t) = const, where <2i(i)(<) and dl(2){t) are two vectors linked with the medium, i.e., dll^(t) = (dx'/dxfydl^fi), a = 1,2, remains invariant during on deformation of the medium. If one more vector dl{t) linked with the medium and parallel to the vector B at an instant is considered, then

dV{i) = [dl{l)(t) x dl{2){t)\ ■ dl(t) = dV0 • A(t)

where A(t) is the ratio of the current and initial volumes. Since di(i)(0) and df(2)(0) are arbitrary, comparison of the two presented equalities yields B(t) =


const • dl(t)/A(t) which makes it obvious that the vector lines of B are linked with the medium. Using the rule for calculation of the components of a vector dl(t) linked with a medium, one obtains Bx = B^dx1 / dx^) / A(t) where BQ are the components of the vector B at the initial instant. The continuity equation yields A(t) = det {dx{/dxk) = Po/p(t)-

34.9 a) It is verified by computation with taking into account that curlB = 4TTJ/C. In the Cartesian coordinate system with x-axis directed along the vector B, Txx = —B2/8-K, Tm = Tzz = B2/8n, and the other components are equal to zero. b) Tlk =Tik-p6ik = BiBk/A-K - 6ik(p + B2/8n). Since in perfectly conductive media (i.e., when a = oo) the change of the vector B can be expressed in terms of the tensor of displacement gradients dwk/dxl

0 (w is the displacement vector, calculation of the components and differentiation are carried out in the non-deformed coordinate system) (see Problem 34.8), and the pressure is p = p(p,s) = p(po/A(t),s), s = const, the total stresses are also expressed in terms of the tensor of displacement gradients and entropy s. Therefore a gas with a magnetic field frozen into it can be regarded as an elastic medium. The equations of ideal magnetohydrodynamics (i.e., magnetohydrodynamics of perfectly conductive media) can be written in the form of Piola-Kirchhoff

d2wk _ _d_ ( dU. \ ds _ p0 dt2 ~ dxi \d{dwkldxi)) ' dt

where U, = p0(U(p, s) + B2/8np) = U,(dwk/dxl0, s) is the total energy per unit initial

volume, c) The flux of energy through a surface element connected with the medium is represented by the normal component of the Poynting vector (see Problem 33.10) 5 n = c(E x B) ■ n/4ir. For the case of perfect conduction, when E = — v x B/c, S„ can be written in the form

Sn = -Vk (B{Bk 6ikB2\ „ 2 l n

Here, the summand expressing the work of the stress tensor is isolated, and the rest term represents the flux of energy of magnetic field if that energy is regarded as moving with velocity of the medium due to freezing in of the magnetic field.

34.10 For a discontinuity having zero velocity in a certain coordinate system, one obtains, using the results of solution of the preceding problems,

[(»>, 'n] = 0 , C . ^ i k B{Bk 6ikB2\ = o ,

. . . v2\ B2\ Bn(B-v) P ^ + T + P + 7 - P n -2) 4n J 47r

Bn] = 0 , [v x B]T = 0 .

= 0 ,

As usual, the square brackets denote the jump of the corresponding quantity. The last two equalities follow from the general relationships on a discontinuity of electromagnetic field (Problem 33.1) for E = — v x B/c. They are obtained immediately from the integral form of the induction equation (Problem 34.6) for vm = 0.

34.11 The linearized system is

IT- 0 - £ "> • <■*»>

f - B ^ = 0 , (.34.13, dp du ,

+ A.7T = 0 , (a34.14) dt ' ™ dx du J_ dt po dv B° 6BV

dt 47r/?o dx dw Bl dBz

\dp) dx po \dsj dx 4irp0 dx

= 0 , (a34.16)

= 0 , (a34.17) dt 47rpo dx ds to = 0 (a34.18)

where u = ux, v = vv, w = vz, the index "0" corresponds to the undisturbed uniform state, and the coordinate system is chosen so that B° = 0, u° = v" = w° = 0. From equations ( a34.11) it follows that BT can be chosen equal to zero. The system of equations (a34.12)-(a34.18) has constant coefficients, is uniform with respect to the order of differentiation, is hyperbolic, and its general solution can be sought as a superposition of travelling waves, i.e., of solutions dependent on arguments of the form x — Xt, X = const. The general form of such systems is dui/dt + ay &Uj/dx = 0. Seeking the travelling wave solution leads to the equation (ay — A5y)u' = 0 from which it follows that A satisfies the characteristic equation |ay — A(5y| = 0, and u\ are proportional to the right eigenvectors of the matrix a'-' so that, for the studied wave, Ui = Ti ■ f(x — Xt) + const where / is an arbitrary function of its argument. Such a solution can be found for each root of the characteristic equation. To simplify the calculation, note that there are three isolated independent subsystems in the system (a34.12)-(a34.18). The first one is the equation (a34.18) for which A = 0, and the eigenvector is

{ B„ = 0 , 5 2 = 0 , p = - f ^ J -s , u = 0 , v = 0, w = 0 , s} .


The corresponding wave (moving with the velocity of the gas) is referred to as entropic. The second one is formed by the equations (a34.13) and (a34.17). For this subsystem, A = ±B°/^/Anpo. The corresponding eigenvectors have the form

{Bv = 0,Bz,p = 0,u = 0,v = 0,w = T-7f^= , s = 0 } ,

and the corresponding waves are called the Alfven waves. For the rest equations (a34.12), (a34.14), (a34.15), (a34.16), for Bz = 0 if s = 0 (all the disturbances for which Bz =ft 0, s =/= 0 are already considered above), the characteristic equation is

\ Airpo j 47rpo

(al = {dp/dp)°s). To any root A of that equation, the eigenvector

_47rp(A2-ag) Xp B°xp(a2-X2) {Bv~ B°v ' P ' U - 7 o ' V - B-A.A '

Bz = 0 , w = 0 , s = 0 }

corresponds. These disturbances are referred to as magnetosonic (fast or slow depending on the chosen value of A).

34.12 For an unmoving medium with a = const, the induction equation is reduced to the heat conduction equation dB/dt = umd2B/dx2. The solution sought belongs to the set of solutions of the form Re(exp i(kx — wt)). Substituting the solution into the equation, one obtains iw = vmk? or k = ±(l+i)Ju/2i/m. The desired solution corresponds to the sign "+": B — Bo exp(—xJu/2i>m) cos (Ju)/2vmx+u)t). Hence, 6 ~ Jvm/u.

34.13 The eigenfunctions of the heat conduction equation given in Problem 34.12 are sin kx where kl = 7m, n is an integer, the corresponding "frequencies" are u = —ii/mk2. Arbitrary continuous data B = Bo(x) can be expanded in a series in terms of the eigenfunctions. The most slowly decaying disturbance is represented by the term corresponding to the first eigenfunction C exp(—ti/mn2/l2) • sin irx/l, while

i C = (2/1) / B0(x) sia(wx/l)dx, i.e., the time of the disturbance decay can be eval-

o uated as T = l2/n2i/m. If C equals zero, decay of the field is more quick; and, to estimate the rate of decay, the first nonzero term of the expansion should be found.

34.14 a) Due to the symmetry of the problem, j x = odyjdx = 0; therefore, everywhere Bv = 0, Bz = Bo- To find u(y, z) and Bx(y, z), two equations should be

34. Magnetohydrodynainics 227

used — projections of the motion and induction equations onto the x-axis: dp/dx = -B0(dBx/dz)/4ir + fj, Au, 0 = B0 du/dz + um ABX, A = £, + £ . Combining the equations linearly, one obtains

-JT± = aAvi+C, (a34.19)

2 --£ = aAv2 + C, (a34.20)

where v\ = u - f3Bx, v2 = u + (5BX, a = y/4n(u/m/B0, /? = Jvm/4irfj.. It will be assumed later that the value of C = — J4num/fj,-(dp/dx)/B0 is positive. Let the flow of the fluid take place in the region Z\(y) < z < z2(y). On the boundary, the no-slip condition u| r = 0 and the condition of impenetrability for electric current j n | r = 0 are satisfied. Since j = (C/4IT)CUI\B, the last equality yields Bx\r = const = 0; since By = 0, Bz = Bo, J = 0 and hence Bx = const = BXao = 0 outside the tube. Thus, for v\ and v2 zero boundary conditions are valid. If the Hartman number is Ha = B0L/y/4irfivm = L/a » 1, then the first terms in the right side of the equations for Vi and v2 are significant only in the narrow regions near the walls (boundary layers). Near the wall, A can be replaced by the second derivative along the normal to the wall d2/dn2 = (1/ cos2 9)(d2/dz2) where 6 is the angle made by the z-axis (i.e., by the magnetic field) with the wall. Thus, the equations for vi and v2 are reduced to ordinary differential equations. Their solutions show that v^ vanishes on the "lower" wall z = z\(y) and increases linearly on the receding from the wall: v\ = C(z — zi(y)). It then sharply (over distances of the order y/4ir/j.i/m/B0 cos2 6) decreases to zero on the "upper" wall z = z2(y). Similarly, v2 = C(z2(y) — z) everywhere far from the lower wall, and there is a sharp decrease to zero on the lower wall. It follows from the expressions for vx and v2 that, everywhere out of the narrow boundary layers with thicknesses of the order 6 = y/4~nJw^/B0 near the walls, the equality u= (vi+ v-i)/2 = | C(z2(y) — z\(y)) is valid, i.e., the velocity does not depend upon z, and its magnitude u(y) is proportional to the length of the segment parallel to the z-axis lying inside the tube. The expression obtained is not valid for values of y for which cos B < 1. Next, sJum/4nnBx = (v2 - v{)/2 = C(z - (z^y) + z2(y))/2), i.e., the lines of electric current Bx = const are parallel to the mean line of the cross section of the tube z = (zi(y) + z2(y))/2. For Ha » 1, it is easy to obtain the solution inside the boundary layers as well if their thicknesses are small as compared with the distance between the walls and with the length over which cos 0 changes essentially. b) Taking into account the zero boundary conditions at z = ± i / for u and Bz (here, the fact that there is no total current in the direction of the y-axis is used) and, as a consequence, the zero boundary conditions for v\ and v2, one obtains from the

equations ( a34.19)-( a34.20) of item a)

exp(±az) - cosh(aif) vh2 = ±Cz-CH

u = CH

Bx =

sinh(a//) cosh(aH) — cosh(az)

sinh(a//) C(z + H sinh(az)

47T/X

If the Hartman number is large, then the presented formulae make it obvious that the quantities sharply change near the walls, i.e., there are boundary layers referred to as Hartman's.

34.15 The induction equation (um = 0) is satisfied for v = kB. The rest equations take the form

/ fc2 \ B2

p 11 - - — 1 (v ■ V)t> = -gradp. + fiAv , p»=p+ — , divv = 0 y 4irpJ 07T

and, for fc = const, p = const, are reduced to the Navier-Stokes equations with the pressure p\ = p , / ( l — k2/4irp) + const and viscosity /zi = /x/(l - k2/4np). a) If p. = 0, then, for fc2 = 4irp, there exists a class of solutions: v(x*) is an arbitrary solenoidal vector, B = ±y/4npv, and p . = const. If B = B0 = const at infinity, then, performing a Galilean transformation, one obtains the solution describing the wave propagating over the fluid, which is at rest at infinity. For the wave, v = ± ( B — Bo)/y/4Trp; its speed equals ±Bo/y/4np. This is an Alfv6n wave, b) Flow around bodies in a magnetic field corresponds to flow at B = 0 with the Reynolds number calculated with the "viscosity" p\. If fc2 > 47rp, then, to obtain equations with a positive viscosity, it is necessary to change the sign of the velocity, i.e., to introduce «i = —v. For that case, the thickness of the boundary layer increases on moving to the side of the stream running over, and the viscous track is directed to the same side.

35 Electrohydrodynamics 35.1 a) In the coordinate system connected with the charge, E? = er/r3. In

the initial system, if v2/<? < 1, E - & + o(v2E'/c?), B = -v x E'/c, \B\ = ev\/y2 + z2/r3 (the charge moves along the i-axis).

35. Electrohydrodynamics 229

b) Let the charged particles occupy a volume with a characteristic length L, and the characteristic density of their charge be p*. Estimate the order of the electric field E+, starting from the equation div E = 4npe from which it follows that E, ~ 47rp*L. If the charged particles move with a characteristic speed v,, then, regarding the process as quasi-stationary and neglecting the term (1/c) dE/dt in the Maxwell equation (which has importance only for quickly alternating processes with the characteristic times T^L/c), one obtains curll? = 4irpev/c or B* ~ vplL/c; therefore , B, ~ E,v,/c as in item a). c) According to the Lorentz transformation rule, E1 = E + v x B/c = E + o(Ev2/c2).

35.2 The average velocity bE of the charge carriers relative to the medium is reached for the time interval r between two sequential collisions due to effect of the electric field, therefore bE = ar/2 where a = eE/m. Prom here, b = er/2m. If \bE\ <C \v\, i.e., e£'r/2m <C v, then the term pebE is small as compared with pev.

35.3 a) In general case the density of electromagnetic force / can be represented (see Problem 32.5) as follows:

where - ExB T°0 - Ea£P + B°B0 - ^ ( ^ + g 2 )

4irc ' 4TT Sir As is obvious, if E2 3> B2, all the terms with B in the expression for T 0 " can be neglected. The magnitude of dga/dt has the order LB/cTE with respect to peE. Assuming that, in problems of mechanics, L/T has the order of a certain velocity much less than the light velocity, and regarding B/E as a small quantity (in motion of charges of an identical sign, B/E ~ v/c — Problem 35.1), one deduces that the term dga/dt can be omitted. b) The system of equations of electrohydrodynamics is presented in the introduction to this section. The equation pT ds/dt = pebE2 follows from the fact that the density of energy input from the electromagnetic field to the medium in the proper coordinate system is j ' • E = pebE2 similarly to the shown in Problem 34.1. According to the second law of thermodynamics, peb > 0. The equation curl E = 0 is obtained from the corresponding Maxwell equation on neglecting the term (l/c)dB/dt which has the order LB/cTE with respect to the remaining one. The other equations do not need any comments.

35.4 Using the equality dE/dx = 4npe and supposing j = bpeE, write the charge conservation law in the form

form which

dt dx (a35.2)

To derive the last equality, the condition at infinity is used. The equality (a35.1) can be written also in the form dpe/dt = —bpl/4ir where d/dt denotes a derivative along the line dx/dt = bE which, according to (a35.2), is a straight line, on which E = const. Integration yields p~l — p& = 4irbt. Since, according to the statement of the problem, bpe » a at t = 0, the term p& can be neglected. Then p~l = 4nbt. Thus, the condition j > aE is satisfied on the straight lines x = x0 + bE0t (x0 is the initial point, EQ = E(x0,0)) at t < a /An.

35.5 For a stationary one-dimensional flow, dj/dx = 0, j = const. Using Ohm's law, one obtains (v + bE) dE/dx = 4irj = const. Integration yields E(x) = —(v — y/C + 8irbjx)/b. Here, C = (bEo + v)2 is the integration constant obtained from the condition E = EQ at x = 0. The sign at the radical is chosen with taking into account that j < 0, b < 0, dE/dx < 0. Such dependence E(x) is valid if

i EQI > y>i > —oo. Actually, for that case there always exists a j such that / Edx = ip\

o (see Figure a35.1) where the functions E(x) are shown for j = 0, ji, J2, h < ji < 0). If ipi/l > E0, then E(x) = tprfl, j = 0.

Figure a35.1

Let the device be a pump. The (efficient) work done on the fluid for unit time i

equals the integral of the product of the force and the velocity vJpeEdx. The o

i necessary electric power, power loss and efficiency equal, respectively, j J Edx = j<p\,

o i i / ( j - pev)Edx,vf pKEdx/jipv 0 0

Chapter 10

Dimensional Analysis and Modelling

37 Examples of Application of Dimensional Analysis

37.1 Considering the motion as steady (the reservoir is large) and supposing that only gravity forces and inertia of the fluid are relevant to this problem, write the dependence sought for in the form

G = f{p,g,h).

Here, p is the density of the fluid, and g is the gravitational acceleration. Next, the standard sequence of operations in dimensional analysis is apphed. Choose a class of systems, e.g., {LMT}. The dimensions of the determined and determining parameters in that class are written as follows

[G] = MT-3 , [p] = ML~3 , [g] = LT~2 ,[h] = L.

Deduce that the dimensions of p, g and h are independent (the particular case of the IT-theorem k = n). Compose the dimensionless combination II for G. For that purpose, the expression of the dimension [G] is written in the form of a monomial composed of powers of the dimensions [p], [g] and [h] with unknown exponents a, /3 and 7, i.e.,

[G] = [p}°[9f[h]-< • The powers of M, L and T are equated:

M : 1 =a L : 0 = - 3 a + 0 + 7 T : - 3 = -2/?

231

232 SOLUTIONS. DIMENSIONAL ANALYSIS AND MODELLING

From this system the only solution obtained is

a = l , /? = - , 7 = ^

Therefore, II = G/{pg3/2h3/2). According to the Il-theorem, II = C where C is constant. Consequently,

G = Cpg3l2h3'2 .

37.2 G = / ( a , p, g, h) (see Problem 37.1). In the class of systems {LMT}

[G] = MLT~3 , [p] = ML'3 , [g] = LT~2 , [h] = L .

According to the Il-theorem,

Therefore G = <p(a) pg3l2h^2. To determine the value of the coefficient <p(a) at a selected value of a a single experiment is sufficient.

37.3 Let us suppose that the paths of the particles of the fluid with mass m in the jet are determined mainly by the action of gravity force. Then

£ = f(vcosa, usina, g, m) .

Following the method of Huntley, one can choose two different units of length in the horisontal direction (the symbol Lx) and in the vertical direction (symbol L„). Use the class of measuring systems {Lx, Lv, M, T}. In this class of systems

[v cos a] = LXT~X , [v sin a] = LvT~l ,

[g] = LVT~2 , [m] = M , [£] = Lx .

All the arguments of the sought function are dimensionally independent (the case k = n). Obtain the dimensionless combination for £:

[£] = [v cos a]m[v sin a]n[g]p[m]«, i.e., Lx = L"T- m LnT~n LIT-2* M" .

Equating the powers of Lx, Lv, T and M, one obtains

Lx 1 = m Lv : 0 =n + p T : 0 = - m - n - 2p

M : 0 = g

37. Examples of Application of Dimensional Analysis 233

Prom here, m = 1, n = 1, p = — 1, q = 0. Then

£ n = (v cos ot)(v sin a)<?-1 t>2 cos a sin ap _ 1

According to the IT-theorem, n = C where C is constant. Then

_, v2 cos a sin a „ u2

£ = C = Ci sin 2a —

Similarly, one obtains v2

H = C2sm2a — 9

where Ci is constant. (Note: the exact solution is C\ = 1, C2 = 5.) If the method of Huntley is not used, then, in the class of systems {LMT},

[£] = L , [m\ = M , [v cos a] = LT~l ,

[v sin a] = LT'1 , [5] = LI1"2 .

Only two arguments of those in the initial relationship are dimensionally independent, e.g., v sin a and g. The dimensionless combinations for £ and v cos a have the form

_ £ _ v cos a II = 7 : T5 r , FIi = : = cot a .

(v sin ayg~l v sin a

According to the II-theorem, II = <p(Hi). Then

£ = (f(cot a)

where <p(cot a) is an unknown function. The amount of obtained information is essentially less than in the case of application of the method of Huntley. The success achieved by application of the method of Huntley is connected, for the given case, with the fact that the motions of each particle of the jet in the horisontal and vertical directions are independent. This makes irrelevant the dimensional parameter that indicates how the length units chosen for different directions differ from each other. Indeed, the equations of motion of a particle and the initial conditions have the form

m i = 0 . - \ x = v cos a , x = 0 at t = 0 < . . my = — mg [ y = v sin a , y = 0

The first and second equations can be solved independently, and the initial conditions turn out to be independent as well. These relationships makes it obvious that the solution is also independent of the mass of the particle.


37.4 The range of a jet located at a height h above the bottom of the vessel equals C = f(v, g, h) where v is the speed of outflow of the jet. Applying the method of Huntley in the class of systems {Lx, Ly,T} (see Problem 37.3) one obtains

[C] = Lx , [v] = LxT~l , [g] = LVT~2 , [h] = Lv .

All the determining parameters are dimensionally independent. The dimensionless combination for £ has the form

n - £

vg-VW2 ' According to the Il-theorem, U = C where C is constant. Consequently,

,vh1'2 C = C gl/2 ■

It can be assumed that the velocity of steady outflow from a small hole located on a depth 6 under the free surface is close to the value following from the Bernoulli integral, i.e,

v=yj2gl. Then one obtains for hole 1

£ i = c gi/2 = CyJ2Hh\.

and for hole 2

j2g(H + hi- h2)h2 , £ 2 = c 1 - ^ T72 — =C v /2 ( i? + / i 1 - / l 2 ) / i 2 .

The condition L\ = C2 is satisfied at H = h2.

37.5 Let us suppose that, if the motion is steady,

H = f(Q,d,g).

Use the method of Huntley in the class of systems {Lx, Lv, T}. In this class of systems,

[H] = Lv , [Q] = LlLyT-1 ,[d\ = L,, [g] = LyT~2 .

All the arguments of the initial relationship are dimensionally independent. The dimensionless combination for H has the form

H ~ Q2d-*g~i '

According to the Il-theorem, II = C where C is a constant. Consequently,

Q2

gd*

37.6 The relationship for the speed V{ at any point on the surface of the body with coordinates Xi can be written in the form

Vi = f [d,^,-j,v0,Po,p,ak\ .

Here, the shape and size of the body are given by the characteristic linear dimension d and ratios of the other linear dimensions to that one, i.e., ln/d; Xi/d are the dimen-sionless coordinates of the point on the surface of the body. Show that, in the case of non-separated flow around the body, the parameter po may be not included in the set of the arguments. Let us start from the mathematical statement of the problem. The pressure p arises only in the equation of motion and boundary condition at infinity

p-ir = - g r a d p , p|oo=Po-

If p is replaced by the new parameter p7 = p — Po where po = const, then the form of the equation of motion is not changed:

dv , p-^- = -gradp ,

and the boundary condition at infinity takes the form p' = 0. Thus, the parameter po does not appear in the statement of the problem, Q.E.D. Then the initial relationship takes the form

Vi = f{d,^,ak,-j,v0,p) .

In the class of systems {LMT},

[m] = LT'1 ,[d\=L, M = LT-1 , [p] = ML'3 .

Taking d, v0 and p as dimensionally independent arguments, one obtains from the fl-theorem

Similarly, one derives

C.=

Vi (k Xi\

2(P-P<>) .(In n Xt\ P — 2

37.7 If, for example, the drag force X is the parameter to be determined, then its dependence upon the determining parameters can be represented in the form

X = f [p,v0,po,Pd,-i,d,ak


Similar relations can be written also for the length of the cavity £ and its maximum diameter D. It can be easily shown that the arguments of that function p0 and pd should be replaced by the single argument — their difference pQ — pd. Indeed, using, as in Problem 37.6, the substitution p' = p — po, one obtains the condition p' = 0 at infinity and the condition p' = pd — p0 on the boundary of the cavity, the form of the equation of motion remaining the same. Besides, the following relationship is valid for the force X

X = pnx da = I (p- p0)nx da = I p'nx da . E E E

Here, £ is the surface of the body, nx is the projection of the normal vector onto the direction of the velocity of the stream, and the relationship fp0nxda = 0 is used.

E Thus,

X = f [d>2'ak'p'v°'Po~Pd) ' In the class of systems {LMT},

[X] = MLT-2 , [p] = ML'3 , [p0 - pd] = ML~lT~2 , [d] = L .

Choosing p, v0 and d as the dimensionally independent parameters, one obtains

n X n li n n Po ~ Pd pvod? d pvfi

According to the II-theorem,

X pvld2 f

l{ 2(p0 - pd) d pvfi

where 2(po — Pd)/(pVo) = O'N is the dimensionless parameter called the natural cavitation number. For dimensionless values of the cavity length C/d and its maximum diameter D/d, similar relationships are derived. The similarity relations are: 1) (li/d)F = {li/d)M — the condition for geometrical similarity of the prototype body and its model; 2) a% — a " — the condition of identical orientations of the bodies with respect to the stream; 3) [2(po — Pd)/(pVo)] = [2(Po - Pd)/(pVo)] — equality of the natural cavitation numbers (it can be easily realized in a water tunnel by choice of necessary values of (po — pd)M or v™). The formula for transformation of data taken from the "model" test to those for the "prototype", in the case of using the same liquid has the form

X1 -XM[i) [h]

37. Ex&mples of Application of Dimensional Analysis 237

If the indicated relations are maintained, the following formulae for transformation of £ and D are also valid:

It is obvious that the forms of the cavities are also geometrically similar if the similarity relations are maintained.

37.8 Natural cavitation should be expected to arise at the point where the maximum value of the velocity, and, consequently, the minimum value of the pressure Pmin, is reached. Such a point, if the stream flowing around the body is potential, should be situated on the boundary of the flow region (on the surface of the body for the case under consideration). In that point,

r _ 2(pmip - Po) """" " pvl

(see Problem 37.6). Natural cavitation at the point where p = pmin arises at p™,, = p,*, i.e., if

The left-hand side of the relationship, i.e. CPmin, depends neither upon the speed v0 nor upon the pressure po and size of the body, and the cavitation number <7N depends upon v0, po. Satisfaction of the condition CPmlll = —crN in experiments in a water tunnel can be reached by varying CTN by virtue of varying of p0 and vo-

37.9 v = Cy/gX where C is constant.

37.10 Assuming that c = /(p,P,7) .

one obtains from the II-theorem that

c = ^ ( 7 )

Using the Clapeyron equation of state p = pET where R is the gas constant, one obtains

c=Vfip(-y,R) .

37.11 h = f(pg, (7,0). In the class of systems {LMT},

[pg] = ML~2T-2 , [a] = MT~2 ,[h] = L.

The parameters pg and a are dimensionally independent, 6 is a dimensionless parameter. It follows from the II-theorem that

where <p(6) is an unknown function of the wetting angle.

37.12 H = f(6,pg,a,9). Let us use the method of Huntley. In the class of systems {Lx, Ly, M, T},

[H] = Ly , [S] = Lx , [pg] = ML~2T-2 , [a] = ML?LVT~2

(Lx is the symbol of the length unit in the horizontal direction, Lv is that in the vertical direction). The arguments 6, pg and a are dimensionally independent. Applying the II-theorem, one obtains

where rjj{0) is an unknown function of the wetting angle.

37.13 Assuming that Pi = f(d, v, p., p, I), one obtains from the II-theorem that

Pi _ , (pod l\ pvW W \ li'd) '

i.e.,

Here, Re = pvd/p. = vd/u is the Reynolds number, and v = p,/p is the kinematic viscosity. The drag force caused by viscous friction, if the velocity of the fluid is constant, is balanced by the force connected with the pressure drop pi — p^, i.e., Pi = 7r(pi — P2) d2/4. The pressure drop p\ — p% is known from experiments to be proportional to I. Consequently, Pj is also proportional to /, from where it follows that

Pi = ip{Re) l- pv2d2 . d

37.14 Let us suppose that

Pi = f(d,v,p,,p,l) .

Use the method of Huntley, choosing the class of system {LX,LV,M,T} where Lx and Ly are the symbols of the length measuring units respectively along the tube and across it:

[Pi] = MLXT~2 , [d] = Ly , [v] = LxT~l ,

[p] = ML^T-1 , [p] = ML-xlL-2 , [l] = Lx .

Choose the arguments d, v, p., p as the dimensionally independent parameters. Then it follows from the Il-theorem that

Pi ( ¥ \ <Pv2p \(Pvp)

Since inertial forces, if the flow is steady and laminar, can be neglected compared with friction forces, the formula for P; obviously cannot depend upon the density p, and, consequently, the function ^(^/((Pvp)) should have the form

( l» \ =c fa \cPvpJ cPvp

where C is constant. Then one obtains the formula for Pt

Pi = Cl/iv .

37.15 The first variant of solution. According to solution to Problem 37.13,

a

Since the flow is steady, the drag force should be balanced by the force caused by the pressure drop p\ — pi. Then, writing Pj in the form

Pi = (Pi ~ P2) - j -

and expressing the pressure drop j>\ - p 2 in terms of the drag coefficient A, one obtains

I pv2 ir<P P , = A d ~ 2 " X -

Comparison of the expressions obtained for P; yields

A = V(Re) .

Since, if the flow is steady and laminar, the density p is not relevant and, consequently, should not be present in the expression for Pj, it is obvious that A = C/Re where C is a constant.

The second variant of solution. Supposing that the dependence

Pi ~ V2 = f(l, v, p, d, p)

is valid and using the method of Huntley, one obtains for the class of systems {Lx,Ly,M,T} (Lx and Ly are the symbols of the length measuring units along the tube and across it)

bi -Pa] = MLxLy2T~2 , [Z] = Lx , [v] = LxT~l ,

[p] = ML'2LXX , [d] = Ly , [p] = ML-lT~l .

Choose I, p, d, p, as the dimensionally independent parameters. According to the Il-theorem,

Pi - ? 2 = (pv<P\ Pp^d-'p,2 V\lp ) '

If the flow in the tube is steady and laminar, the pressure drop cannot depend on the density p, i.e., <p (pvd?/lp) = C\ pvd?/lp where C\ is constant. Then (px — p2) = C\ Ipv/d2 and, consequently,

, _ Pi ~P2 _ C_ (l/d) (pv2/2) Re '

37.16 Let Q = f{d, p, i). All the arguments of the function / are dimensionally independent in the class of systems {LMT}. Consequently, Q = Cdti/p where C is constant.

37.17 Let us suppose that X = f(d,v,p,p). Then, in the class of systems {LMT},

[X] = MLT~2 , [d] = L , [v] = LT'1 , [p] = ML~3 , [p.] = ML^T'1 .

Choosing the arguments d, p, v as the dimensionally independent parameters and applying the Il-theorem, one obtains

x = J^^d2 = Ci(Re) !d2 where Re = pvd/p = vd/u is the Reynolds number. If Re 4; 1 (as, e.g., in slow motion for which inertial forces can be neglected as compared with friction forces), the density p is not relevant and should be absent from the formula for X. Therefore Cx(Re) = C/Re; and, consequently, X = Cipdv, i.e., the drag depends linearly upon the speed (the Stokes law). For very large values of the Reynolds number Re, it can be assumed that the drag force is caused mainly by inertial forces, and viscosity is not relevant. Hence, Cx(Re) = Ci = const, and, consequently, X = C? pv2 d2/2, i.e., the dependence of the drag on speed is described by a quadratic function.

37.18 If the rate of submersion is constant, the weight of the sphere P is balanced by Archimedes' force A and by the force of viscous hydrodynamic drag X, i.e., P = A + X. Using the expression for the force X obtained in Problem 37.17

X = C , ( R e ) ^ ,

write the balance equation in the form

4 3 ^ , _ ,_ . pv2 ncP d Pi 3 *r3g = - nr3pg + Cx(Re) — — , r = -

The speed t; can be obtained from this expression if Cx(Re) is known. Similarly to Problem 37.17, one deduces that if Re <g. 1, i.e., the rate of submersion is slow,

v = C ^ - ( ^ - l ) , C = const,

and, if submersion is very quick, Cx = const, and, consequently,

H T - 1

where C\ = const

37.19 1) If the penetration is quick, then, obviously, X = f(t,v,p,a) from which, due to the II-theorem, X = C(a) pv*t2. 2) If the penetration is slow, then X = f(t, v, fj,,a), therefore X = C(a) tv2fi.

37.20 The initial relationship is

tQ = J{dA,K,Q,p,n,g) ■

The dimensions of the determined and determining parameters in the class of systems {LMT} are

[*«] = T , [d] = [ho] = L , [Q] = L3 , \p] = ML'3 ,

M = ML-'T-1 , [g] = LT-2 .

Choosing d, g and p as the dimensionally independent parameters and applying the II-theorem, one obtains

fd Ik ho Q v tQ = \-f\ g^\d' d ' d3' gl/2d3'2

where v = \ijp is the kinematic viscosity. The similarity criteria are : 1) (U/d) = {h/d)M — this relation requires geometrical similarity between the prototype and model vessels and mouthpieces. 2) {ho/d) = (ho/d) — it follows from this relation that the initial depth of the liquid in the model vessel should be equal to hg = hp (dM/dp) = hp/n, where n = dp/dM. 3) (Q/d3)P = (Q/d3)M — this relation implies that one finds in the model experiment the time for which the following amount of the liquid pours out from the vessel: QM ~ Qp (dM/dpj = Qp/n3.

4) [i//(ff1/2d3/2)]P = [v/{g1/2d3/2)]M — iigp = gM, then it follows from this relation

that i/M = vp (dM/dpJ = fp/n3/2, i.e., the kinematic viscosity of the liquid used in the model experiment should be less by a factor of Vn? than the prototype kinematic viscosity. If all the similarity criteria are satisfied, then in the case for which <;p = gM

the model test data are transformed to prototype data according to the formula

jdp lQ lQ \ "3x7 V n £Q

where tq is the time during which the amount QM of the hquid pours out from the model vessel. All the requirements following from the similarity relations are in many cases easily satisfied, and, therefore, it is possible to achieve modelling with complete similarity of the phenomena in the prototype and model. Let a model vessel geometrically similar to a prototype have size one-fifth as large, i.e., n = 5. Then the initial depth of the hquid in the model experiment should be one-fifth as much as for the prototype. What should be found is the time £g for which one-125th as much amount of the hquid pours out. The viscosity of the hquid used in the model should be less by a factor of \/125 than that of the prototype one. If all the similarity relations are satisfied, then the time, for which Q liters of the hquid pour out from the prototype vessel, is determined by the formula

If there is a possibility to realize centrifugal modelling (i.e., to carry out the model experiment on a centrifuge allowing one to achieve any necessary value of "gravitational" acceleration g), then it would be possible to use the same liquid in the model experiment as in the prototype vessel. Indeed, it follows from the similarity relation (4) that the necessary gravitational acceleration equals gM = gp (dp/dMJ . The model test data should be transformed to prototype values according to the formula

tp -tM

dM gp Q \dM) "

For the particular case dM/dp = ±, gM = 125gp and tfj = 2 5 $ .


37.21 Write the desired relationship for W in the form

W = f(l,±,D,vtg,vL,p} .

Choosing a class of systems, e.g., {LMT}, and, next, choosing I, v and p as the dimensionally independent arguments in that class of systems, one obtains, applying the II-theorem,

W (D h gl fi\ pv2P J \l3' l'v2' pvl) '

Independence of the dimensionless arguments of the function / allows one to change the arguments by means of multiplying them by each other, dividing, raising to a power and other operations of a similar kind. As a result, various possible equivalent relationships can be obtained, e.g., of the following form

W _ ( I lA _v_ pvl\ \po*P~'P{yD'l'y/gl' n) "

The arguments written in the presented form are often encountered when applying dimensional analysis to problems dealing with motion of bodies on the surface of a heavy viscous liquid or under that surface. The dimensionless combination l/y/~D is called the delicacy ratio. Let us denote it by ip. It is obvious that the delicacy ratios of ships with larger length for a given volume displacement have a larger numerical value. The dimensionless parameter v/y/gl = Fr often called the FVoude number is of great importance in analysis of phenomena for which taking the weight of the fluid into account is essential. The parameter pvl/fM = vl/v = Re called the Reynolds number is one of the basic parameters in problems for which account of viscous friction is important. Determining the drag of a ship by virtue of modelling faces certain difficulties. In fact, the necessity for the similarity relations Rep = ReM

and Fi^ = FrM to be satisfied simultaneously leads at gp = gM and vp = uM to mutually exclusive requirements on the speed of the model. Namely, it follows from equality of the Reynolds numbers that vM = vp (lp/lM), and equality of the Proude numbers leads to the relation i.e., it turns to be impossible to realize complete similarity under these conditions. The indicated difficulties are overcome by the following way. The function ip(ijj, li/l, Fr, Re) is represented in the form

p(tf, It/l, Fr, Re) = Vl(Re) + ipaik/l, i>, Fr) ,

i.e., it is voluntarily assumed that the drag force caused by viscous friction can be singled out in the form of an additive term in the formula for the total drag of a ship. Besides, that force can be assumed to be approximately independent of the shape of the surface on which it acts and calculated using experimental data for the drag


coefficient C/(Re) of a plane plate in a longitudinal stream of flowing fluid. Thus, the formula for the drag W is written in the form

W = C'f(Re)^-l2 + C'w(li/l^,Fr)^-l2

2

= C/(Re) ^ - Sw + Cw(k/l,4, Fr) pgD .

Here, Sw is the wetted area of the surface of the hull (it is taken equal to the area of the part of the hull below the waterline). The coefficient Cw{li/l,ij), Fr) is called the coefficient of remainder drag. A model experiment is conducted in order to determine the form of the dependence Cw = Cw(li/l,ip,Fr), the following similarity relations being satisfied: (U/l) = (h/l) - the requirement of geometric similarity; xpp = ^ M

- easily realized condition; Fr1* = FrM - from which it follows that On the basis of the measurement of Wu, the function

is found. And, since it can now be assumed that C£ = Cjjj due to identity of the arguments of these functions, the formula

Wr = C / ( R e P ) E^fl si + C%(lj, A Fr) pgDp

is applied to determine Wp. Of course, the number of the strong assumptions allows us to obtain the drag of the ship only within a certain error; however, as experience shows, applying the described method is quite acceptable in practice.

37.22 Let us consider, e.g., the drag X overcome by the flying body as the determined parameter. The shape and size of the body are specified by a characteristic linear dimension d and a set of ratios of other linear dimensions and d, i.e., k/d. The orientation of the body in the stream is given by angles a*. Then

X- = f(d,li/d,ak,p0,po,H,f,v) ■

Using the class of systems {LMT}, choose d, p0 and v as the dimensionally independent arguments. Applying the II-theorem, one obtains

X = A j ^ _ _po_\ Pov'd2 \d' ' ' pvd' pov2)

The arguments of the function ip can be transformed. In fact, p,/(pvd) = 1/Re where Re is the Reynolds number. Multiplying the last argument by 7, one obtains the new


parameter rypo/{poV2) in place of it. Noting that 7P0/P0 = ^o where OQ is the speed of sound in the undisturbed gas, one can write this argument in the form

v2 M2 '

Here, M = w/ao is the Mach number — one of the most essential parameters in problems for which taking compressibility of the gas into account is essential. As a result of the transformations carried out, one obtains

Consequently, the similarity relations have the form

( y ) P = ( y ) M , a f cp = a f c

M ,7p = 7

M , R e p = R e M , M p = M M .

It is necessary to note that, as the Reynolds number grows, dependence upon it becomes weaker. For moderate supersonic speed the main similarity relations, besides those connected with the shape of the body and its orientation in the stream, are equalities of adiabatic exponents and the Mach numbers for the prototype and model flows: 7P = 7M, Mp = MM. It is essential to note that, since there is no characteristic linear dimension in these relations, the choice of the size of a model is quite arbitrary. Equality of the Mach numbers is provided by supersonic wind tunnels where the prototype speed can be reached.

37.23 The flow is obviously axially symmetrical. The expression for the speed at a point (x, y) is written in the form

v{x,y) = f(x,y,p,/J.,Jo) ■

Using the class of systems {LMT} and choosing x, p and J0 as the dimensionally independent parameters, one obtains, applying the fl-theorem

At y = 0, i.e., on the axis of the jet,

. .. 1 /Jo" (vpll2\


Consequently, the speed decreases along the axis of the jet varying inversely with the first power of the coordinate. The ratio of the speeds v(x, y) and v(x, 0) is

«(»,») * V Jl'2) _,(y VP1/2\

* $ )

which is in the form of a universal dependence on y/x.

37.24 Writing the relationship sought in the form r2 = f{E0, po, 7, t) and applying the Il-theorem, one obtains

- c w ( f ) 1/5

,2/5

37.25 At t = 0 at the origin of the spherical coordinate system (r, 6, ip), a finite amount of energy Eo instantly appeares, i.e., a point explosion takes place (see Problem 37.24). The equations of gas dynamics for determining the flow of the gas in the region bounded by the spherical shock wave of the radius r2 = r2(t) have the form 1) the continuity equation

dp d{pv) 2pv dt dr r

where p is the density v is the velocity, t is the time; 2) the equation of motion

dv dv 1 dp _ dt dr p dr

where p is the pressure; 3) the condition for motion to be adiabatic

d_ dt teKte)-

where 7 = Cp/cv is the adiabatic exponent. The unknowns in this system are p, v, p. The boundary conditions have the form l ) v = 0 a t r = 0 (the condition following from the fact that the flow is symmetrical); 2) at r = r2(t) = C(-y) (£0/po)1/512/5

(see Problem 37.24) the conditions across the strong shock wave should be satisfied, i.e.,

2 „ 7 + I 2 „ , v2 = ——r D , p2 = po , Pa = ——7 A>£>

7 + 1 7 - 1 7 + 1


where D = dr2/dt is the speed of the shock wave, po is the density ahead of it, and the subscript "2" marks the parameters behind the shock wave. The assumption that the explosion is strong means that the pressure p0 in front of the wave can be neglected in the conditions across the shock wave, i.e., that po can be excluded from the set of the determining parameters. It obviously follows from the mathematical statement of the problem that

v = v(t, r, E0, po, 7) , P = P{t, r, E0, po, 7) , p = p{t, r, E0, p0,7) •

Let us apply dimensional analysis. In the class of systems {LMT}

[v] = LT-1 , \p0] = [p] = ML"3 , [p] = ML-'T-2 ,

\t\ = T , [r] = L , [E0] = ML2T'2 .

Choosing t, E0 and po as the dimensionally independent variables and applying the ri-theorem, one obtains

v= ( ^ V 3 ^ ^ ) , P = A>*(A,7) , p=£02/5Po/5r6/5P(A,7)

where A = r(po/E0)l/it~2^ is a dimensionless parameter, and V(A,7), R(\,-y), P(A,7), are dimensionless functions. Substituting these expressions into the system of gas dynamics equations and deleting dimensional multipliers, one obtains 1) 5XVR - Afl'(2A - 5V) + IQRV = 0; 2) (2A - bV)RV + 3flV - 5P' = 0; 3) (2A — 5V) {P'/P - 7 K/R) = 5. This is a system of ordinary differential equations (the primes mean differentiation with respect to A). Substituting the expressions for v, p and p into the conditions across the shock wave, one obtains the boundary conditions for V, R and P

V 4 C ^ R ~>+l P 8 C^) Vi = — rr , n-2 = 7 . "1 = 5 ( 7 + 1 ) ' 7 - I ' J 2 5 ( 7 + 1 ) '

These conditions should be satisfied at r = r%, i.e., for A = C(7). Thus, the problem is reduced to seeking for the solution of the system of ordinary differential equations corresponding to a value of C(7) for which the condition V = 0 at the center of the explosion is satisfied, i.e., A = 0. L.I. Sedov showed that such a solution exists and is unique, and he found it in explicit form.

37.26 The determining parameters of the point explosion problem with counter-pressure taken into account are

t, r, EQ, po. 7> Po,


i.e., there is the parameter p0 in addition to the parameters of the problem of strong point explosion (see Problem 37.24). The parameters to be determined are v, p, p and the radius r2 of the shock wave. Choosing t, E0 and po as the dimensionally independent parameters in the class of systems {LMT} (as in Problem 37.24) and applying the Il-theorem, one obtains, e.g., the expression for the pressure

p 2 /5 3/5

where , r Pot6/b _cL

(E0/p0y/HV* ' T $*$/* ' T Cv

The equality of the prototype and model-phenomenon values of these dimensionless parameters is the condition of their physical similarity; i.e., the similarity relations have the form

/ = ^ , Ap = AM , r p = rM .

If they are satisfied, the following relation between the determined and determining parameters is obviously valid

pp(tp)6 /5 pM(tM)6/5

(£0P)V5(pP)3/5 ( £ M ) 2 / 5 ( p M ) 3 / 5

whence

p ~p [W) UJ le) ■ To determine the value of pf at a given distance rf at an instant tp with the method of modelling, let us write the similarity relations in details

/ = 7M

rM

(EE/p%y/5(tP)VS ( £M / pM ) l /5 ( tM )2/5 •

pP(tf)6/5 = pM(tM)6/5

( £ P ) ^ ( p P ) 3 / 5 " (£M)2/5(pM)3/5 •

In modelling with the use of an explosion chamber, it is possible to use the prototype medium (air) with the same 7, with the same initial density $ = pp and with the same pressure pjj1 = p p . Choosing an acceptable amount of an explosive, on explosion of which the energy EQ1 is extracted, one can use the two last relations for determining the values of rM and iM at which the pressure p14 should be measured


in the laboratory experiment. After the pressure pM is measured, the value of pf is found by the formula mentioned above for transformation of pressures:

37.27 v = CJE/p where C is constant.

37.28 To obtain the similarity relations, the determining parameters should be written out. The elastic properties of the material are specified by Young's modulus E and Poisson's ratio a. The shape and size of the beam are given by the characteristic linear dimension I — length of the beam — and the set of parameters U/l. The specific weight of the material pg (p is the density, g is the gravitational acceleration) is taken into account. It is obvious that

h = f(a,E,l,li/l,pg).

In the class of systems {LMT}, U/l and a are dimensionless quantities. As it follows from the Il-theorem,

The similarity relations have the form

If a model and the prototype beam are made of the same material and have an identical shape, then the condition

(gif = (gi)M

should be satisfied for modelling. If lM ^ lF, it is necessary to carry out centrifugal modelling, placing a model in a centrifuge or in an elevator to provide artificial increase in acceleration "g" by as many times as the size of the model is less than the size of the prototype beam. In this case, the displacement of the end of the prototype beam is greater than that of the model by as many times as the length of the prototype beam is greater than the length of the model.

Index

A adiabatic exponent 62 adiabatic Poisson's coefficient 64 adiabatic Young's modulus 64 Alfven waves 226, 228 apparent mass 101 Archimedes force 50

B Bernoulli integral 158

C characteristic form of a differential

equation 138 charge conservation 216 circulation of velocity 125 compatibility conditions for strain com

ponents 174 concomitant coordinate system 40 concurrent stream 48 conformal mapping method 93 constitutive equations 122 continuity equation 121 continuity equation in cylindrical coor

dinates 36 continuity equation in spherical coordi

nates 36 Coulomb's law 37 cylindrical coordinates, Christoffel sym

bols 12 cylindrical coordinates, metric tensor 9

detonation adiabat 153 deviator of a tensor 39 dimensionally independent quantities

239 discontinuity surface 65 dispersion relation 113, 128 Doppler effect 130 drag 137 drag force 48, 235

E electric conductivity coefficient 220 electric conductivity tensor 220 electromagnetic energy 216 electromagnetic Maxwell stress tensor

216 energy equation 121 entropic wave 226 entropy 152 entropy equation 122 entropy increase in a shock wave 79,

145 entropy production 219 equations of motion 121 equations of motion for an ideal fluid

(Euler equations) 43 Euler equations 43 Euler equations in cylindrical coordi

nates 43 Euler equations in spherical coordi

nates 44 evolutionarity conditions 145, 153

F

251

D d'Alembert paradox 48, 135

flow around a circular cylinder 101 flow around a sphere 101 Proude number 124

G group speed 184

H Hamilton-Cayley theorem 220 hardening parameter 198 heat conduction equation 226 Huntley method 232 hydraulic jump 117

I induction equation 223 intensity of shear stresses 197 invariants of a tensor 8

J Jaumann derivative 51 Jouguet detonation 157 Joule heat 219 jump conditions for an ideal fluid 78

L Lame parameters 95 Laplace equation 135 Laval nozzle 161 length scale 123 lift force 48 longitudinal wave 183, 189

M Mach cone 130 Mach lines 136 Mach number 123 magnetic viscosity 221 magnetohydrodynamics, ideal 224 magnetohydrodynamics, system of equa

tions 222 Maxwell visco-elastic solid 204 method of characteristics 139

method of conformal mappings 93 method of mirror reflections 94

N Neumann problem 37, 90

O Ohm's law 220

P Peclet number 124 perfectly conductive media 224 phase speed 114, 128, 184 physical components 43 Pi-theorem (II-theorem) 60 Piola-Kirchhoff stress tensor 40 Poisson adiabat 56 Poisson equation 174 Poynting vector 216 pressure coefficient 99 progressive wave 128

R Rayleigh waves 187 relativistic law of velocity addition 208 Reynolds number 123 Riemann wave 140, 190

S second law of thermodynamics 65, 80,

229 self-similar problem 60, 105 shock adiabat 153 shock polar 151 speed of sound 123 speed of sound in a mixture 133 spherical coordinates, Christoffel sym

bols 12 spherical coordinates, metric tensor 9 standing wave 184 Stokes approximation 109 Strouhal number 124

252

T tangential discontinuity 77 tangential stresses, extremal values 39 time scale 123 transverse wave 183, 189

U uncompensated heat 219

V Voigt visco-elastic solid 204

W wave drag 137 wave equation 127, 128, 186 waveguide 132

Y yield stress in shear 198

253

continuum mechanics via problems and exercises part ii answers and solutions ed by m e eglit d h...

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