fluid mechanics kundu cohen 6th edition solutions sm ch (10)

Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.1. A thin flat plate 2 meters long and 1 meter wide is placed at zero angle of attack in a low speed wind tunnel in the two positions sketched below.

a) For steady airflow, what is the ratio: drag on the plate in position #1drag on the plate in position #2 ?

b) For steady airflow at 10 m/sec, what is the total drag on the plate in position #1? c) If the air flow is impulsively raised from zero to 10 m/sec at t = 0, will the initial drag on the plate in position #1 be greater or less than the steady-state drag value calculated for part b)? d) Estimate how long it will take for drag on the plate in position #1 in the impulsively started flow to reach the steady-state drag value calculated for part b)?

Solution 10.1. a) Here we need only consider the drag coefficient, CD, from the Blasius solution.

€

CD (L) =Drag per unit span on one side of the plate of length L( )

12 ρU

2L=

1.328ReL

where ReL= UL/ν. Therefore:

Total Drag on plate "j" = 2(span of plate "j")

€

⋅1.328ReL

⋅12ρU 2L where j = 1 or 2 , so

€

Drag on plate #1Drag on plate #2

=span of plate #1span of plate #2

⋅L1 L1

L2 L2

=2

2 2= 2 .

b) Total Drag on plate #1 = 2(span of plate #1)

€

⋅1.328ReL1

⋅12ρU 2L1

=

€

2(2m)1.328(10m /s)(1m) (1.5 ×10−5m2 /s)

⋅12(1.21kg /m3)(10m /s)2(1m)

= 0.394 N. c) When the flow is impulsively started, a temporally-developing boundary layer is produced on the top and bottom of plate #1. Initially, the shear stress is very high since:

€

τw ≅ µU(πνt)−1 2 so the initial drag will be greater than the steady-state drag. d) The flow will have reached steady-state when the temporally developing boundary layer skin friction has reached the Blasius boundary layer skin friction everywhere on the plate. The Blasius skin friction is lowest at the trailing edge, so an estimate for the drag-relaxation time can be based on matching skin friction coefficients at the plate's trailing edge between the temporally-developing boundary layer and the Blasius solution. Therefore, with ReL= UL/ν:

€

c f( )temporal BL≈ c f( )spatial BL

at x = L –>

€

2U

νπt

≈0.664ReL

, or

€

t ≈ 4Lπ (0.664)2U

≈ 3.0 LU

= 0.30s.

Interestingly, this result is independent of the fluid viscosity!


Exercise 10.2. Solve the Blasius equations (10.27) through (10.29) with a computer, using the Runge–Kutta scheme of numerical integration, and plot the results. What value of !!f at η = 0 leads to a successful profile? Solution 10.2. The Blasius equation is a non-linear third-order differential equation for f(η):

€

d3 fdη3

+12f d

2 fdη2

= 0.

The boundary conditions are:

€

df dη→1 as

€

η→∞, df/dη = f = 0 at η = 0. For a computer solution using a Runge-Kutta integration scheme, this equation must be reduced to a set of three first-order differential equations by defining:

€

g(η) = df dη , and

€

h(η) = d2 f dη2 . The above equation can then be written as three equations:

€

df dη = g(η) (A)

€

dg dη = h(η) (B)

€

dh dη = − fh 2 (C) subject to:

f(0) = g(0) = 0 and g(∞) = 1. The set of three equations are readily integrated via the Runge-Kutta method starting from η = 0 where the initial values of f and g are known. The initial value of h (= !!f at η = 0) is not known but it can be found by trial and error (0.33205) by looking for the value of h(0) that produces g = 1 at some suitably large value of η, perhaps 10 or 20. A simple MatlabTM code that does this based on a trial & error value of h(0) is: %Compute the Blasius Boundary Layer Profile clear; clc; f0 = 0; %The first known boundary condition g0 = 0; %The second known boundary condition h0 = 0.33205; %The value adjusted by trial & error eta_start = 0; %Starting point for eta eta_end = 10; %End point for eta %The next command invokes a Runga-Kutta integration scheme [eta,f] = ode45(@Blasius,[eta_start eta_end],[f0 g0 h0]); with the function function df = blasius(eta,f) % Solve the Blasius LBL profile equation. % f(1)=stream function, f(2)=velocity, f(3)=velocity gradient df = zeros(3,1); df(1) = f(2); %The equivalent of (A) df(2) = f(3); %The equivalent of (B) df(3) = -0.500*f(1)*f(3); %The equivalent of (C) end The resulting plot of df/dη vs. η is:


!"

!#$"

!#%"

!#&"

!#'"

!#("

!#)"

!#*"

!#+"

!#,"

$"

!" %" '" )" +" $!"

€

dfdη

=uU

€

η = y U νx


Exercise 10.3. A flat plate 4 m wide and 1 m long (in the direction of flow) is immersed in kerosene at 20°C, (v = 2.29 × 10−6 m2/s, ρ = 800 kg/m3) flowing with an undisturbed velocity of 0.5 m/s. Verify that the Reynolds number is less than critical everywhere, so that the flow is laminar. Show that the thickness of the boundary layer and the shear stress at the center of the plate are δ = 0.74 cm and τ0 = 0.2 N/m2, and those at the trailing edge are δ = 1.05 cm and τ0 = 0.14 N/m2. Show also that the total frictional drag on one side of the plate is 1.14 N. Assume that the similarity solution holds for the entire plate. Solution 10.3. ReL = UL/ν = (0.5m/s)(1m)/(2.29x106 m2/s) = 2.18x105 < Recr ~ 106. Thus, the flow is expected to be laminar everywhere. At x = 0.5 m, Rex = 1.09x105 so the 99% thickness from (10.30) and the shear stress from (10.31) are:

€

δ99 = 4.9x Rex1 2 = 4.9(0.5) 1.09 ×105 = 0.742cm, and

€

τ 0 = 0.332ρU 2 Rex1 2 = 0.332(800)(0.5)2 1.09 ×105 = 0.201N /m2 .

At x = 1.0 m, Rex = 2.18x105 so

€

δ99 = 4.9x Rex1 2 = 4.9(1.0) 2.18 ×105 =1.05cm , and

€

τ 0 = 0.332ρU 2 Rex1 2 = 0.332(800)(0.5)2 2.18 ×105 = 0.142N /m2 .

The total drag can be obtained from (10.33):

€

CD =1.33 Rex1 2 = 2.85 ×10−3, so

€

D =12ρU 2(Area)CD = 0.5(800)(0.5)2(4)(1)(0.00285) =1.14N .


Exercise 10.4. A fluid with constant density and viscosity flows with a constant horizontal speed U∞ over an infinite flat porous plate placed at y = 0 through which fluid is drawn with a constant velocity Vs. For this flow the steady two-dimensional zero-pressure-gradient boundary layer equations are (7.2) and (10.18) and the boundary conditions are u(y = 0) = 0, v(y = 0) = –Vs, and u = U∞ for

€

y→∞. a) Assuming u depends only on y, determine u(y) in terms of ν, Vs, U∞, and y. b) What is the wall shear stress τw? How does it depend on µ? c) What parametric change(s) decrease the boundary layer thickness? Solution 10.4. a) If u = u(y), then ∂u/∂x = 0 = ∂v/∂y and this means that v is a most a function of x. However, the boundary condition on v at y = 0 does not depend on x, thus v = const. = –Vs.

Therefore, the horizontal momentum equation simplifies to:

€

−Vs∂u∂y

= ν∂ 2u∂y 2

, which can be

integrated to find:

€

−Vsy ν = ln ∂u ∂y( ) +C , where C is a constant. This can be rearranged and integrated again to find:

€

u(y) = D+ E exp −Vsy ν( ) . The boundary condition, u(0) = 0 requires D + E = 0, while u(∞) = U∞ sets D = U∞. Thus, the velocity profile is:

€

u(y) =U∞ 1− exp −Vsy ν( )[ ] b) The wall shear stress is

€

τw = µ ∂u ∂y( )y= 0 = −µU∞ −Vs ν( ) = ρU∞Vs . It does not depend on µ! c) The boundary layer thickness decreases when Vs increases, and ν decreases. The horizontal flow speed does not influence the thickness of this boundary layer.


Exercise 10.5. A square-duct wind tunnel test section of length L = 1 m is being designed to operate at room temperature and atmospheric conditions. A uniform air flow at U = 1 m/s enters through an opening of D = 20 cm. Due to the viscosity of air, it is necessary to design a variable cross-sectional area if a constant velocity is to be maintained in the middle part of the cross-section throughout the wind tunnel. a) Determine the duct size, D(x), as a function of x. b) How will the result be affected if U = 20 m/s? At a given value of x, will D(x) be larger or smaller than (or the same as) the value obtained in a)? Explain. c) How will the result be affected if the wind tunnel is to be operated at 10 atm (and U = 1 m/s)? At a given value of x, will D(x) be larger or smaller than (or the same as) the value obtained in a)? Explain. [Hint: the dynamic viscosity of air (µ [N·s/m2]) is largely unaffected by pressure.] d) Does the airflow apply a net force to the wind tunnel test section? If so, indicate the direction of the force.

Solution 10.5. a) The duct length (1 m) and the flow speed (1 m/s) imply a Reynolds number of: ReL = UL/ν = (1 m/s)(1 m)/(1.5x10–5m2/s) = 67,000, which much larger than unity but still in the laminar boundary layer range. If δ* is the displacement thickness of the boundary layer at position x, then conservation of mass between the inlet of the duct and location x implies:

€

ρUD2(0) = ρU D(x) −δ*( )2.

For constant density and velocity in the duct, the pressure gradient must be zero. Therefore, the Blasius solution for a flat plate laminar boundary layer can be used, and the above equation reduces to:

€

D(x) = D(0) + δ* = D(0) +1.72 νx U = 0.20m + (1.5 ×10−5m2 /s)x (1m /s)

= 0.20 + 0.00387 x ,

where the final numbers provide D in meters when x is in meters. b) If U is 20 m/s instead of 1 m/s and the flow remains laminar, the displacement thickness of the boundary layer will be reduced by a factor of

€

20 ≈ 4.47. Thus, the required D(x) will be

€

D(x) = D(0) +1.72 νx U = 0.20m + (1.5 ×10−5m2 /s)x (20m /s) = 0.20 + 0.000866 x . Here, the requisite D(x) is smaller because the boundary layer will be thinner at the higher speed. c) The factor of ten increase in pressure causes a factor of 10 increase in density. Thus for constant µ, ν = µ/ρ will decrease by a factor of 10. This implies:

€

D(x) = D(0) +1.72 νx U = 0.20m + (1.5 ×10−6m2 /s)x (1m /s) = 0.20 + 0.00122 x .


So, the requisite D(x) will be smaller than the part a) result because the boundary layer will be thinner at the higher pressure because of the drop in kinematic viscosity (or equivalently, the increase in Reynolds number). d) Yes, the airflow applies a force to the wind tunnel via wall shear stress. This force points to the right.


Exercise 10.6. Use the control volume shown to derive the definition of the momentum thickness, θ, for flow over a flat plate:

ρU 2θ = ρU 2 uU

1− uU

"

#$

%

&'

0

h

∫ dy = Drag force on the plate from zero to xunit depth into the page

= τ w dx0

x

∫

The words in the figure describe the upper and lower control volume boundaries.

Solution 10.6. The CV is stationary, so the integral form of the continuity equation is:

€

ρUho = ρ u(y)dy0

h

∫ ,

where h is defined in the figure to be greater than the boundary layer thickness. The pressure is the same everywhere so the integral form of the x-momentum equation is:

−ρU 2ho + ρ u2 (y)0

h

∫ dy = − τ w dx0

x

∫ .

Here it is assumed that τw is positive (since µ and ∂u/∂y will be positive for the boundary layer flow inside the CV), so the right-side integral represents the force the plate applies to the fluid in the CV. Using the continuity equation, eliminate ho from the x-momentum equation:

−ρU u(y)dy0

h

∫ + ρ u2 (y)0

h

∫ dy = − τ w dx0

x

∫ .

Combine the two integrals, multiply by minus one, factor out ρU2, and recognize the definition of the momentum thickness θ:

ρ u(y) U −u(y)( )dy0

h

∫ = ρU 2 u(y)U

1− u(y)U

#

$%

&

'(dy = ρU 2θ =

0

∞

∫ τ w dx0

x

∫ .

Since u(y) = U for y > h, it is OK to extend the upper limit of the y-integration to +∞. The integral of the shear stress on the right side of the last equation corresponds to the force the flow exerts on the plate (per unit depth into the page). This force is positive and points in the downstream direction (increasing x) and is therefore a drag force.

U!

h > δ99!ho!

U!zero shear stress, !

constant pressure, no through flow!

y!

x!

no slip, τw ≠ 0!


Exercise 10.7. Estimate the 99% boundary layer thickness on: a) a paper airplane wing (length = 0.25 m, U = 1 m/sec), b) the underside of a super tanker (length = 300 m, U = 5 m/sec), and c) an airport run way on a blustery day (length = 5 km, U = 10 m/sec). d) Will these estimates be accurate in each case? Explain. Solution 10.7. Estimate the various thicknesses from the laminar zero-pressure-gradient (Blasius) boundary layer solution:

€

δ99 = 4.9 νx U . a)

€

δ99 = 4.9 (1.5 ×10−5m2s–1)(0.25m) (1ms−1) = 9.5mm b)

€

δ99 = 4.9 (1.0 ×10−6m2s–1)(300m) (5ms−1) = 3.8cm c)

€

δ99 = 4.9 (1.5 ×10−5m2s–1)(5000m) (10ms−1) = 0.42m d) The estimate for part a) may be fairly accurate, but the estimates for b) and c) will not be accurate. The downstream-distance-based Reynolds numbers in a), b), and c) are

€

1.7 ×104 ,

€

1.5 ×109 , and

€

3.3×109 respectively. At the larger two Reynolds numbers, the boundary layer will be turbulent and its 99% thickness will be much greater than the Blasius boundary layer estimate given above.


Exercise 10.8. Air at 20°C and 100 kPa (ρ = 1.167 kg/m3, ν = 1.5 × 10−5 m2/s) flows over a thin plate with a free-stream velocity of 6 m/s. At a point 15 cm from the leading edge, determine the value of y at which u/U = 0.456. Also calculate v and ∂u/∂y at this point. [Answer: y = 0.857 mm, v = 0.384 cm/s, ∂u/∂y = 3012 s–1.] Solution 10.8. From Table 10.1,

€

η = y U νx =1.4 when u/U = 0.456. Therefore,

y =1.4 νx U =1.4 (1.5×10−5 )(0.15) 6 = 0.857 mm. At this wall normal distance, the Blasuis BL formula for v and Table 10.1 produce:

v = 12

νUx

− f +η dfdη

"

#$

%

&'=121.5×10−5(6)

0.15−0.325+1.4 ⋅0.456( ) = 0.00384 m/s.

Again using Table 10.1, the slope of the velocity profile is: ∂u∂y

=U ∂(u /U)∂η

∂η∂y

=U !!f U νx = (6)(0.3074) 6 (1.5×10−5 )(0.15) = 3012 s–1.


Exercise 10.9. An incompressible fluid (density ρ, viscosity µ) flows steadily from a large reservoir into a long pipe with diameter D. Assume the pipe-wall boundary layer thickness is zero at x = 0. The Reynolds number based on D, ReD, is greater than 104. a) Estimate the necessary pipe length for establishing a parabolic velocity profile in the pipe. b) Will the pressure drop in this entry length be larger or smaller than an equivalent pipe length in which the flow has a parabolic profile? Why?

Solution 10.9. a) Fortunately, the exercise asks for an estimate. The flow in the entrance length of a round pipe will accelerate on the pipe's centerline, and the inner wall of the pipe is curved. Both of these features will cause the boundary layer growth inside the pipe to differ from that of a Blasius boundary. However, the Blasius solution does account for diffusive boundary layer growth, so it should be fine for producing an estimate of the entrance length L. The presence of the pipe walls is felt on the pipe centerline when the wall boundary layer has attained a thickness of D/2. Therefore, set δ99 ≈ D/2 and estimate δ99 from the Blasius solution:

D/2 ≈ 5(νL/U)1/2 = 5(LD/ReD)1/2 , and solve for L to find: L ≈ 10–2D(ReD). b) The pressure drop in the entry section will be larger because: 1) wall shear stresses are larger when BL's are thin, and 2) the central portion of the flow must accelerate to accommodate the displacement effect of the BL so the velocity at the edge of the boundary layer increases from the x = 0 to x = L.


Exercise 10.10. A variety of different dimensionless groups have been used to characterize the importance of a pressure gradient in boundary layer flows. Develop an expression for each of the following parameters for the Falkner-Skan boundary layer solutions in terms of the exponent n in Ue(x) = axn, Rex = Uex/ν, integrals involving the profile function !f , and !!f (0) , the profile slope at y = 0. Here u(x, y) =Ue(x) !f y δ(x)( ) =Ue !f (η) and the wall shear stressτ w = µ ∂u ∂y( )y=0 = µUe δ(x)( ) !!f (0) . What value does each parameter take in a Blasius boundary layer. What value does each parameter achieve at the separation condition? a) ν Ue

2( ) dUe dx( ) , an inverse Reynolds number

b) (θ2/ν)(dUe/dx), the Holstein and Bohlen correlation parameter

c) µ ρτ w3( ) dp dx( ) , Patel's parameter

d) δ* τ w( ) dp dx( ) , Clauser’s parameter Solution 10.10. a) Here Ue(x) = axn, so

€

dUe dx = naxn−1; thus

€

νUe2dUe

dx=νUe

naxn−1

axn= n ν

Uex=

nRex

.

The profile function does not enter here. This parameter is zero for the Blasius boundary layer, and is –0.0904/Rex at separation. b) This time the profile function enters through the definition of the momentum thickness.

€

θ 2

νdUe

dx=

1ν

uUe

1− uUe

%

& '

(

) *

0

∞

∫ dy%

& '

(

) *

2dUe

dx=δ 2

ν. f (η) 1− . f (η)( )

0

∞

∫ dη%

& '

(

) *

2dUe

dx

=δ 2

ν. f (η) 1− . f (η)( )

0

∞

∫ dη%

& '

(

) *

2dUe

dx=

1ν

νxUe

%

& '

(

) * . f 1− . f ( )

0

∞

∫ dη%

& '

(

) *

2nUe

x

= n . f 1− . f ( )0

∞

∫ dη%

& '

(

) *

2

.

This parameter is zero for the Blasius boundary layer, and is –0.0904

€

" f 1− " f ( )0

∞

∫ dη'

( )

*

+ ,

2

at

separation. c) This parameter involves the shear stress and pressure gradient directly. Differentiating the steady Bernoulli equation without a body force produces

€

dp dx + ρUe dUe dx = 0 , so

µ

ρτ w3

dpdx

=µρ

1µUe !!f (0) δ(x)( )3 2 −ρUe

dUe

dx#

$%

&

'(= −

µρ

νx Ue

µUe !!f (0)

#

$%%

&

'((

3 2

ρnUe

2

x

= −n µρν3 4

ρµ3 2

1!!f (0)[ ]3 2

x3 4

x1

Ue3 2Ue

3 4

#

$%

&

'(Ue

2 = −n µ1 4

ρ1 41

!!f (0)[ ]3 21x1 4

1Ue

1 4

= −n 1!!f (0)[ ]3 2

1Rex

1 4


This parameter is zero for the Blasius boundary layer, and it goes to +∞ at separation since

€

" " f = 0 at separation and for the Falkner-Skan laminar boundary layer solutions –n = +0.0904 > 0. d) This parameter involves the profile function through the displacement thickness and through the shear stress. Using the differentiated Bernoulli relation specified in part c) leads to:

δ*

τ w

dpdx

= 1−u U( )dy0

∞

∫$

%&

'

()

δ(x)µUe **f (0)

−ρUedUe

dx$

%&

'

()= − 1− *f( )dη

0

∞

∫$

%&

'

()

δ 2 (x)µUe **f (0)

ρnUe

2

x

= −n 1− *f( )dη0

∞

∫$

%&

'

()νx Ue

µUe **f (0)ρUe

2

x= −

n**f (0)

1− *f( )dη0

∞

∫$

%&

'

().

This parameter is zero for the Blasius boundary layer, and it goes to +∞ at separation since

€

" " f = 0 at separation and for the Falkner-Skan laminar boundary layer solutions –n = +0.0904 > 0.


Solution 10.11. Consider the boundary layer that develops as a constant density viscous fluid is drawn to a point sink at x = 0 on and infinite flat plate in two dimensions (x, y). Here Ue(x) = –UoLo/x, so set

€

η = y νx |Ue | and

€

ψ = − νx |Ue | f (η) and redo the steps leading to (10.36) to find

€

" " " f − " f 2 +1= 0. Solve this equation and utilize appropriate boundary conditions to find

€

" f = 3 1−αe− 2η

1+αe− 2η

&

' (

)

* +

2

− 2 where

€

α =3 − 23 + 2

.

Exercise 10.11. Use the specified Ue to develop the appropriate expressions for ψ and h.

€

ψ = νx |Ue | f (η) = νx UoLox

f (η) = νUoLo f (η) , and

€

η =y

νx |Ue |=

yνx 2 UoLo

=UoLoν

yx

.

The Cartesian velocity components are:

€

u =∂ψ∂y

= − νUoLodf (η)

dη∂η∂y

= − νUoLo ' f UoLo

ν1x

= −UoLo

x' f , and

€

v = −∂ψ∂x

= + νUoLodf (η)

dη∂η∂x

= νUoLo ' f UoLo

ν−

yx 2

(

) *

+

, - = −

νUoLo

xη ' f .


€

" f →1 as

€

η→∞ (far from the wall the boundary layer velocity matches the free-stream),

€

" f → 0 as

€

η→ 0 (no slip at the wall), and

€

η # f → 0 as

€

η→ 0 (no flow through the wall).

The horizontal boundary layer momentum equation is (10.9),

€

u∂u∂x

+ v ∂u∂y

= −1ρ∂p∂x

+ ν∂ 2u∂y 2

. The

differentiated Bernoulli equation,

€

dp dx + ρUe dUe dx = 0 , allows this to be simplified:

€

u∂u∂x

+ v ∂u∂y

=UedUe

dx+ ν

∂ 2u∂y 2

or

€

u∂u∂x

+ v ∂u∂y

= −Uo2Lo2

x 3+ ν

∂ 2u∂y 2

.

where the second version of the equation follows when Ue(x) = –UoLo/x. Substitute in the velocity components in the final equation to find:

€

−UoLo

x# f ∂∂x

−UoLo

x# f

%

& '

(

) * −

νUoLo

xη # f ∂

∂y−

UoLo

x# f

%

& '

(

) * = −

Uo2Lo2

x 3+ ν

∂ 2

∂y 2−

UoLo

x# f

%

& '

(

) * .

Perform the differentiations and simplify:

€

−UoLo

x# f +

UoLo

x 2# f +

UoLo

x# # f η

x%

& '

(

) * +

νUoLo

xη # f UoLo

x# # f UoLo

ν1x

= −Uo2Lo2

x 3−ν

UoLo

xUoLo

ν1x 2

# # # f ,

€

− # f + # f +η # # f ( ) +η # f # # f = −1− # # # f or

€

− # f + # f +η # # f ( ) +η # f # # f = − # f 2 = −1− # # # f . With a little rearrangement, the final equality then leads to the desired equation:

€

" " " f − " f 2 +1= 0 . This equation can be solved for f´ as follows. Multiply by

€

" " f and integrate to find:

€

12

" " f ( )2 − 13

" f 3 + " f + A = 0.

The constant A can be evaluated by considering the limit

€

η→∞. The first boundary condition provides

€

" f (∞) =1, and if f´ is constant then

€

" " f (∞) = 0 . Thus, the last equation becomes


0 – 1/3 + 1 + A = 0 as

€

η→∞, which requires A = –2/3. Therefore,

€

12

" " f ( )2 =23− " f +

13

" f 3 , or

€

" " f = ±232 − 3 " f + " f 3[ ]

1 2,

which can be separated:

€

d " f 2 − 3 " f + " f 3[ ]1 2

=d " f

(2 + " f )(1− " f )(1− " f )[ ]1 2=

d " f (1− " f ) 2 + " f [ ]1 2

=23

dη .

The possibility of the minus sign in front of the square root is dropped because η is always positive along with

€

1− # f and

€

2 + " f . The final equality can be integrated with the substitution

€

F 2 = 2 + " f :

€

2FdF(3− F 2)F

=2dF3− F 2 =

23dη

Now set

€

F = 3 tanh(γ) , so that

€

dF = 3 1− tanh2(γ)( )dγ , thus

€

2dF3− F 2∫ =

2 3 1− tanh2(γ)( )dγ3 1− tanh2(γ)( )∫ =

23γ =

23tanh−1 F

3%

& '

(

) * =

23η + C

Backtracking all the way to

€

" f produces:

€

23η + C =

23tanh−1 2 + $ f

3

%

& '

(

) * .

The constant C can be evaluated from

€

" f (0) = 0 :

€

C =23tanh−1 2

3

#

$ %

&

' (

Inverting for

€

" f yields:

€

" f = 3tanh2 η

2+

3C2

$

% &

'

( ) − 2 = 3tanh2 η

2+ tanh−1 2

3

$

% &

'

( ) − 2 .

To reach the desired form, use the sum formula for the hyperbolic tangent, and the definition of the hyperbolic tangent. Then manipulate the exponentials and the square roots.

€

" f = 3tanh η 2( ) + 2 3

1+ tanh η 2( ) 2 3

$

%

& &

'

(

) )

2

− 2 = 3

eη 2 − e−η 2

eη 2 + e−η 2+

23

1+eη 2 − e−η 2

eη 2 + e−η 2

+

, -

.

/ 0

23

$

%

& & & & &

'

(

) ) ) ) )

2

− 2 = 3

1− e− 2η

1+ e− 2η+

23

1+1− e− 2η

1+ e− 2η

+

, -

.

/ 0

23

$

%

& & & & &

'

(

) ) ) ) )

2

− 2

= 31− e− 2η +

23

1+ e− 2η( )1+ e− 2η + 1− e− 2η( ) 2

3

$

%

& & & &

'

(

) ) ) )

2

− 2 = 33 − 3e− 2η + 2 1+ e− 2η( )3 + 3e− 2η + 2 1− e− 2η( )

$

%

& &

'

(

) )

2

− 2

= 33 + 2 − 3 − 2( )e− 2η

3 + 2 + 3 − 2( )e− 2η

$

%

& &

'

(

) )

2

− 2 = 31− 3 − 2

3 + 2

+

, -

.

/ 0 e− 2η

1+3 − 23 + 2

+

, -

.

/ 0 e− 2η

$

%

& & & & &

'

(

) ) ) ) )

2

− 2.

The final equality is the desired form.


Exercise 10.12. Start from the boundary layer equations, (7.2), (10.9), and (10.10), and ψ =Ue(x)δ(x) f (η) , where η = y δ(x) , with δ(x) unspecified, to complete the following items. a) Show that the boundary-layer profile equation can be written:

!!!f +α f !!f +β(1− !f 2 ) = 0 , where α = δ ν( )d(Ueδ) dx , and β = δ 2 ν( )dUe dx .

b) The part a) equation will yield similarity solutions when α and β do not depend on x. Therefore, assume α and β are constants, set Ue = axn, and show that n = β/(2α – β). c) Deduce the values of α and β that allow the profile equation to simplify to the Falkner-Skan profile equation (10.36). Solution 10.12. a) The given stream function automatically satisfies the continuity equation, so the next step is to substitute it into (10.9):

u∂u∂x+ v ∂u

∂y=Ue

dUe

dx+ν

∂2u∂y2

.

Here, the surface-normal boundary-layer momentum equation (10.10), –∂p/∂y = 0, and the Bernoulli equation in the free stream have been used to replace the pressure gradient; − 1 ρ( ) ∂p ∂x( ) =Ue dUe dx( ) . For the given form of the stream function the various velocity components and derivatives are:

u = ∂ψ∂y

=Ueδ "f 1δ=Ue "f , v = −∂ψ

∂x= − "Ueδ f −Ue !δ f +Ueδ !f y

δ 2= − "Ueδ f −Ue !δ f +Ue !f y

δ ,

∂u∂x

= !Ue !f −Ue !!f yδ 2

!δ , ∂u∂y

=Uee !!f 1δ

, and ∂2u∂y2

=Ue """f 1δ 2

.

where a prime denotes differentiation of a function with respect to its argument. Equation (10.10) is then reconstructed:

Ue !f !Ue !f −Ue !!f yδ 2

!δ#

$%

&

'(+ − !Ueδ f −Ue !δ f +Ue !f y

δ

#

$%

&

'( Uee !!f 1

δ

#

$%

&

'(=Ue !Ue +νUe !!!f 1

δ 2,

and this can be simplified by performing the indicated multiplications on the left side, and canceling equal and opposite terms:

Ue !Ue !f 2 −Ue !Ue f !!f −Ue2 f !!f !δ

δ=Ue !Ue +νUe !!!f 1

δ 2.

Multiply through by δ2/νUe to reach: δ 2

ν!Ue !f 2 −

δ 2

ν!Ue f !!f −Ue

δ !δν

f !!f =δ 2

ν!Ue + !!!f ,

collect all the terms to one side of the equation,

!!!f +δ 2

ν!Ue 1− !f 2( )+ δ

νδ !Ue +Ue !δ( ) f !!f = 0 ,

and combine the terms in the last set of parentheses:

!!!f +δ 2

νdUe

dx1− !f 2( )+ δ

νddx

δUe( ) f !!f = 0 , or !!!f +β 1− !f 2( )+α f !!f = 0 ,

where the given definitions, α = δ ν( )d(Ueδ) dx & β = δ 2 ν( )dUe dx , have been used to reach the final equation.


b) When α and β are constants, their defining equations may be used to eliminate δ. Start with the α-equation:

α =δνd(Ueδ)dx

=δUe

νdδdx

+δ 2

νdUe

dx=Ue

2νdδ 2

dx+β .

Using the two ends of this extended equality, substitute for δ2 from the definition of β to find a non-linear equation for Ue:

α =Ue

2νdδ 2

dx+β =

Ue

2νddx

νβdUe dx

!

"#

$

%&+β = −

Ue

2β

dUe dx( )2d 2Ue

dx2+β .

Again using the two ends of this extended equality and simplifying leads to: α −ββ

dUe

dx"

#$

%

&'2

= −Ue

2d 2Ue

dx2 ,

and this equation has power-law solutions: Ue = axn. Substituting this trial solution into this equation produces:

α −ββ

a2n2x2n−2 = − axn

2an(n−1)xn−2 which simplifies to: α −β

βn = − 1

2(n−1) .

Solving this final algebraic equation for n produces the desired relationship: n = β/(2α – β).

c) The part b) result can be rearranged to find: α = β2

n+1n

!

"#

$

%& . Substitute this into the profile

equation from part a) to reach:

!!!f +β 1− !f 2( )+ β2n+1n

#

$%

&

'( f !!f = 0 ,

and this equation will match (10.36) when β = n. Thus, the values of α and β that produce the Falkner-Skan profile equation are:

α = (n + 1)/2 and β = n.


Exercise 10.13. Solve the Falkner-Skan profile equation (10.36) numerically for n = –0.0904, –0.654, 0, 1/9, 1/3, and 1 using boundary conditions (10.28) and (10.29) and the Runge–Kutta scheme of numerical integration. Plot the results and compare to Figure 10.8. What values of !!f at η = 0 lead to successful profiles at these six values of n? Solution 10.13. The Falkner-Skan profile equation is a non-linear third-order differential equation for f(η):

d3 fdη3

+n+12

f d2 fdη2

− n dfdη"

#$

%

&'

2

+ n = 0 .


€

df dη→1 as

€

η→∞, df/dη = f = 0 at η = 0. For a computer solution using a Runge-Kutta integration scheme, this equation must be reduced to a set of three first-order differential equations by defining:

f1(η) = f (η) , f2 (η) = df dη , and f3(η) = d2 f dη2 .

The above equation can then be written as three equations: df1 dη = f2 (η) (A) df2 dη = f3(η) (B)

df3 dη = −(1 2)(n+1) f1 f3 + nf22 − n (C)

subject to: f1(0) = f2(0) = 0 and f2(∞) = 1.

The set of three equations are readily integrated via the Runge-Kutta method starting from η = 0 where the initial values of f and g are known. The initial value of f3 (= !!f at η = 0) is not known but it can be found by trial and error by looking for the value of f3 that produces f2 = 1 at some suitably large value of η, perhaps 10 or 20. These initial values of !!f are : 0.00, 0.1640, 0.33205, 0.51181, 0.75741, 1.232533. A simple MatlabTM code that does this based on a trial & error value of h(0) is: %Compute the Falkner-Skan Boundary Layer Profile clear; clc; eta_start = 0; eta_end = 10; f0 = 0; %The first known boundary condition g0 = 0; %The second known boundary condition n = [-0.0904, -0.0654, 0, 1/9, 1/3, 1]; %Values of n h0 = [0.00, 0.1640, 0.33205, 0.51181, 0.75741, 1.232533]; %Values adjusted by trial & error figure; hold all; for i = 1:length(n); %The next command invokes a Runga-Kutta integration scheme [eta, f] = ode45(@(eta, f) FS(eta, f, n(i)), [eta_start eta_end],[f0 g0 h0(i)]); xlswrite(['FS_data_n=' num2str(n(i)) '.xlsx'],[eta f])


x=(.5*(n(i)+1)).^.5*eta; plot(x,f(:,2)); end with the function function df = FS( eta, f, n ) %UNTITLED Falkner-Skan LBL Profile %f(1)=stream function, f(2)=velocity, f(3)=velocity gradient %n=1; df = zeros(3,1); df(1) = f(2); %same as (A) df(2) = f(3); %same as (B) df(3) = -0.500*(n+1)*f(1)*f(3) + n*f(2)^2 - n; %same as (C) end The resulting plot of df/dη vs. η is:

The left most curve corresponds to n = 1, and the right-most curve corresponds to n = –0.0904 with the other monotonically arrayed in between. And – except for the aspect ratio and extent of the horizontal axis – this plot is identical to Figure 10.8.

0"

0.1"

0.2"

0.3"

0.4"

0.5"

0.6"

0.7"

0.8"

0.9"

1"

0" 2" 4" 6" 8" 10"

€

dfdη

=uU

€

η = y U νx


Exercise 10.14. By completing the steps below, show that it is possible to derive von Karman's boundary layer integral equation without integrating to infinity in the surface-normal direction using the three boundary layer thicknesses commonly defined for laminar and turbulent boundary layers: i) δ (or δ99) = the full boundary layer thickness that encompasses all (or 99%) of the region of viscous influence, ii) δ* = the displacement thickness of the boundary layer, and iii) θ = momentum thickness of the boundary layer. Here, the definitions of the later two involve the

first:

€

δ*(x) = 1− u(x,y)Ue (x)

$

% &

'

( )

y= 0

y=δ

∫ dy and

€

θ(x) =u(x,y)Ue (x)

1− u(x,y)Ue (x)

$

% &

'

( )

y= 0

y=δ

∫ dy , where Ue(x) is the flow

speed parallel to the wall outside the boundary layer, and δ is presumed to depend on x too. a) Integrate the two-dimensional continuity equation from y = 0 to δ to show that the vertical

velocity at the edge of the boundary layer is:

€

v(x,y = δ) =ddx

Ue (x)δ*(x)( ) −δ dUe

dx.

b) Integrate the steady two-dimensional x-direction boundary layer momentum equation from y =

0 to δ to show that:

€

τ 0ρ

=ddx

Ue2(x)θ(x)( ) +

δ*(x)2

dUe2(x)dx

.

[Hint: Use Leibnitz’s rule

€

ddx

f (x,y)dya(x )

b(x )

∫ = f (x,b) dbdx

#

$ % &

' ( − f (x,a) da

dx#

$ % &

' ( +

∂f (x,y)∂x

dya(x )

b(x )

∫ to handle

the fact that δ = δ(x)]

Solution 10.14. Start with

€

∂u∂x

+∂v∂y

= 0 , and integrate from y = 0 to δ to get:

€

∂u∂xdy +

0

δ

∫ v(x,y = δ) = 0

where v(x,y=0) = 0. Use Leibnitz’s rule to get the differentiation outside the integral:

€

v(x,y = δ) = −∂u∂xdy

0

δ

∫ = −ddx

udy0

δ

∫ +Ue (x)dδdx

Add and subtract Ue(x) within the integral and rearrange the result:

€

v(x,y = δ) = −ddx

u −Ue (x) +Ue (x)( )dy0

δ

∫ +Ue (x)dδdx

.

€

= −ddx

u −Ue (x)( )dy0

δ

∫ −ddx

Ueδ( ) +Ue (x)dδdx

€

=ddx

Ue (x) 1− uUe (x)

#

$ %

&

' ( dy

0

δ

∫#

$ %

&

' ( −δ

dUe

dx=ddx

Ueδ*( ) −δ dUe

dx

b) Start with

€

u∂u∂x

+ v ∂u∂y

= −1ρdpdx

+ ν∂ 2u∂y 2

, multiply the continuity equation by u and add it to this

equation to get:

€

∂u2

∂x+∂uv∂y

= −1ρdpdx

+ ν∂ 2u∂y 2

. Use the Bernoulli equation substitution for the

pressure and integrate this equation from y = 0 to δ to get:

€

∂u2

∂xdy

0

δ

∫ +Ue (x)v(x,y = δ) =UeδdUe

dx−τwρ


Use Leibnitz’s rule to get the differentiation outside the integral, substitute from part a) for the vertical velocity at the edge of the boundary layer, and add and subtract uUe within the integral:

€

ddx

u2 − uUe + uUe( )dy −Ue2 dδdx0

δ

∫ +Ue (x)ddx

Ueδ*( ) −δ dUe

dx%

& '

(

) * =Ueδ

dUe

dx−τwρ

Invoke the definition of the momentum thickness θ and then the definition of the displacement thickness δ*:

€

−ddx

Ue2θ( ) +

ddx

Ue u −Ue +Ue( )dy0

δ

∫&

' (

)

* + −Ue

2 dδdx

+Ue (x)ddx

Ueδ*( ) −δ dUe

dx&

' (

)

* + =Ueδ

dUe

dx−τwρ

€

−ddx

Ue2θ( ) +

ddx

−Ue2δ* +Ue

2δ( ) −Ue2 dδdx

+Ue (x)ddx

Ueδ*( ) −δ dUe

dx%

& '

(

) * =Ueδ

dUe

dx−τwρ

Now rearrange the result, and group like terms together:

€

τwρ

=ddx

Ue2θ( ) +

ddx

Ue2δ*( ) −Ue

ddx

Ueδ*( ) − dUe

2δdx

+Ue2 dδdx

+UeδdUe

dx+Ueδ

dUe

dx

The final four terms sum to zero, and the terms containing δ* can be expanded to obtain:

€

τwρ

=ddx

Ue2θ( ) + δ*

dUe2

dx+Ue

2 dδ*

dx−Ue

2 dδ*

dx−δ*Ue

dUe

dx=ddx

Ue2θ( ) +Ueδ

* dUe

dx,

and the final equality is identical to (10.43), von Karman's boundary layer integral equation.


Exercise 10.15. Derive the von Karman boundary layer integral equation by conserving mass and momentum in a control volume (C.V.) of width dx and height h that moves at the exterior flow speed Ue(x) as shown. Here h is a constant distance that is comfortably greater than the overall boundary layer thickness δ. Solution 10.15. Use the integral laws as applied to a moving control volume. Conservation of mass implies:

€

ddt

dx ρBdy0

h

∫$

% &

'

( ) + ρ u[ ]x−dx / 2 −Ue( )ex ⋅ (−exB)dy

0

h

∫

+ ρ u[ ]x+dx / 2 −Ue( )ex ⋅ (exB)dy0

h

∫ + ρ (u −Ue )ex + vey[ ]y= h⋅ eyBdx = 0

The four terms above correspond to unsteady mass addition to the CV, mass flux on the left side of the CV, mass flux on the right side of the CV, and mass flux on the top of the CV. The CV is a constant size and the density is presumed to be constant, so the first term is zero (the time derivative of a constant). Combine terms, perform the dot products, presume dx is small, and divide by ρBdx to reach:

€

1dx

u[ ]x+dx / 2 − u[ ]x−dx / 2( )dy0

h

∫ + v[ ]y= h = 0 , or

€

v[ ]y= h = −∂u∂xdy

0

h

∫ = −ddx

udy0

h

∫%

& '

(

) * , (*)

where the final equality follows from the fact that h is constant. Now conserve horizontal momentum using the same CV.

€

ddt

dx ρuBdy0

h

∫$

% &

'

( ) + ρ u[ ]x−dx / 2 u[ ]x−dx / 2 −Ue( )ex ⋅ (−exB)dy

0

h

∫

+ ρ u[ ]x+dx / 2 u[ ]x+dx / 2 −Ue( )ex ⋅ (exB)dy0

h

∫ + ρu (u −Ue )ex + vey[ ]y= h⋅ eyBdx

= −τ 0Bdx + p[ ]x−dx / 2 − p[ ]x+dx / 2( )hB

where τ0 is the wall shear stress, and p is the pressure. Simplify and combine terms, and divide by ρBdx to reach:

€

ddt

udy0

h

∫#

$ %

&

' ( +

1dx

u[ ]x+dx / 22

− u[ ]x−dx / 22{ }dy

0

h

∫ −Ue1dx

u[ ]x+dx / 2 − u[ ]x−dx / 2{ }dy0

h

∫ + uv[ ]y= h

= −τ 0

ρ+

1ρdx

p[ ]x−dx / 2 − p[ ]x+dx / 2( )h

Use the chain rule on the total time derivative of the unsteady term,

€

ddt

=dxdt

ddx

=Ueddx

, noting

that horizontal velocity of the CV (= dx/dt) is Ue; and presume dx is small:

€

Ueddx

udy0

h

∫#

$ %

&

' ( +

ddx

u2dy0

h

∫ −Ueddx

udy0

h

∫ + uv[ ]y= h = −τ 0ρ−1ρ∂p∂xh .

u(x+dx,y)!u(x,y)!

p(x+dx)!p(x)!

x+dx!x!

Ue!

C.V.!

Ue!

y = h!

!!


Two terms on the left cancel, and at y = h, u = Ue and v is given by (*). In addition, the steady Bernoulli equation implies:

€

dp dx = −ρUe dUe dx . So, with these substitutions, the last equation can be rewritten:

€

ddx

u2dy0

h

∫ −Ueddx

udy0

h

∫$

% &

'

( ) = −

τ 0ρ

+UedUe

dxh = −

τ 0ρ

+UedUe

dxdy

0

h

∫ .

Rearrange this equation so the skin friction term appears by itself with positive sign. Then manipulate the integrals to form the definitions of the momentum and displacement thicknesses.

€

τ 0

ρ= −

ddx

u2dy0

h

∫ +Ueddx

udy0

h

∫&

' (

)

* + +Ue

dUe

dxdy

0

h

∫

= −ddx

Ue2 u2

Ue2 dy

0

h

∫&

' (

)

* + +

ddx

Ue2 uUe

dy0

h

∫&

' (

)

* + −Ue

dUe

dxuUe

dy0

h

∫ +UedUe

dxdy

0

h

∫

=ddx

Ue2 uUe

1− uUe

&

' (

)

* + dy

0

h

∫&

' (

)

* + +Ue

dUe

dx1− u

Ue

&

' (

)

* + dy

0

h

∫ =ddx

Ue2θ( ) +Ueδ

* dUe

dx.

The final equality is identical to (10.43), von Karman's boundary layer integral equation.


Exercise 10.16. For the following approximate flat-plate boundary layer profile:

€

uU

=sin πy 2δ( ) for 0 ≤ y ≤ δ1 for y > δ

% & '

( ) *

, where δ is the generic boundary layer thickness, determine:

a) the displacement thickness δ*, the momentum thicknesses θ, and the shape factor

€

H = δ * θ . b) Use the zero-pressure gradient boundary layer integral equation to find:

€

δ x( )Rex1 2,

€

δ * x( )Rex1 2,

€

θ x( )Rex1 2 ,

€

c f Rex1 2 , and

€

CDReL1 2 for the approximate profile.

c) Compare these results to their equivalent Blasius boundary layer values.

Solution 10.16. a) The sinusoid boundary layer velocity profile,

€

uU

= sin πy2δ$

% &

'

( ) = sin

π2ζ

$

% &

'

( ) where

€

ζ =yδ

, is a reasonably accurate approximate laminar boundary layer profile. The momentum

thickness for this profile is:

€

θ =uU1− u

U$

% &

'

( )

0

δ

∫ dy = δ sin π2ζ

$

% &

'

( ) 1− sin

π2ζ

$

% &

'

( )

$

% &

'

( )

0

1

∫ dζ =2δπ

sin ψ( ) 1− sin ψ( )( )0

π 2

∫ dζ

The integration is not too complicated,

€

sin(ψ)dψ0

π 2∫ =1, and

€

sin2(ψ)dψ0

π 2∫ = π 4 , so

€

θ =2δπ1− π

4&

' (

)

* + = δ

4 −π2π

&

' (

)

* + = 0.1366δ

The displacement thickness for the approximate profile is:

€

δ* = 1− uU

$

% &

'

( )

0

δ

∫ dy =2δπ

1− sin ψ( )( )dψ =2δπ

π2−1

$

% &

'

( )

0

π 2

∫ = δπ − 2π

$

% &

'

( ) = 0.3634δ ,

so that the shape factor

€

H = δ* θ = 2.66 . Substitution into the zero-pressure gradient (ZPG) Von Karman boundary layer integral relationship produces:

€

τwρ

=U 2 dθdx

, or

€

c f =2τwρU 2 =

2 πµU 2δ( )ρU 2 =

πνUδ

= 2 dθdx

=4 −ππ

)

* +

,

- . dδdx

where

€

τw = µ ∂u ∂y( )y= 0 = πµU 2δ has been obtained directly from the approximate sinusoidal profile. Now use the fourth and sixth terms in the extended equality to form a differential equation for δ.

€

πνUδ

=4 −ππ

&

' (

)

* + dδdx

, or

€

δdδdx

=π 2

4 −π%

& '

(

) * νU

–>

€

δ 2 =2π 2

4 −π%

& '

(

) * νUx + const = 23.0 ν

Ux + const .

Here the implicit assumption is that the boundary layer thickness is zero at x = 0 so the constant

drops out. Thus:

€

δ = 23 νUx = 4.80 ⋅ x ⋅ Rex

−1 2 . Using the results above for the other length

scales and parameters produces:

b)

€

θxRex

1 2 = 0.654 (0.664)

€

δ*

xRex

1 2 =1.743 (1.721)

€

δxRex

1 2 = 4.80 (5.0)


€

c f Rex1 2 =

πνUδ

Rex1 2 =

πνU ⋅ 4.80xRex

−1 2 Rex1 2 = 0.654 (0.664)

€

CDReL1 2 =

1L

0.654Rex

1 20

L

∫ dx ⋅ ReL1 2 =

0.654L

νU

dxx0

L

∫ ⋅ ReL1 2 =

2 ⋅ 0.654ReL

1 2 ReL1 2 =1.309 (1.328)

c) The Blasius boundary layer results are listed above at the right in parentheses. The sinusoid velocity profile results are all within 4% of the Blasius values.


Exercise 10.17. An incompressible viscous fluid with kinematic viscosity ν flows steadily in a long two dimensional horn with cross sectional area A(x) = Aoexp{βx}. At x = 0, the fluid velocity in the horn is uniform and equal to Uo. The boundary layer momentum thickness is zero at x = 0. a) Assuming no separation, determine the boundary layer momentum thickness, θ(x), on the lower horn boundary using Thwaites method. b) Determine the condition on β that makes the no-separation assumption valid for 0 < x < L. c) If θ(x = 0) was nonzero and positive, would the flow in the horn be more or less likely to separate than the θ(x = 0) = 0 case with the same horn geometry?

Solution 10.17. a) As stated above, use Thwaites method to estimate the boundary layer momentum thickness. Start with conservation of mass to determine the U(x):

€

U(x)A(x) =UoAo , so

€

U(x) =UoAo A(x) =Uo exp −βx{ }. Thus, the Thwaites equation becomes:

θ 2 (x) = 0.45νU 6 (x)

U 5( !x )d !x0

x

∫ =0.45νUo

6 e+6βx Uo5e−5β !x d !x

0

x

∫

= 0.45νUo

e+6βx e−5βx

−5β+

15β

$

%&

'

()=

0.09νβUo

e+6βx 1− e−5βx( )

b) First compute

€

λ =θ 2

νdUdx

=0.09βUo

e+6βx 1− e−5βx( ) −βUoe−βx( ) = −0.09 e+5βx −1( ) . Separation will

have occurred if λ < λseparation = –0.090. Therefore, λ > λseparation at x = L is required to avoid separation. This means:

€

−0.09 e+5βL −1( ) > λseparation . Convert this requirement to a condition on β:

€

e+5βL −1< −λseparation0.09

, or

€

β <15Lln 1−

λseparation0.09

% & '

( ) * ≈0.139L

c) If the boundary layer starts with a non-zero momentum thickness it is thicker and therefore more likely to separate in an adverse pressure gradient.


Exercise 10.18. The steady two-dimensional velocity potential for a source of strength q located a distance b above a large flat surface located at y = 0 is:

φ(x, y) = q2π

ln x2 + (y− b)2 + ln x2 + (y+ b)2( )

a) Determine U(x), the horizontal fluid velocity on y = 0. b) Use this U(x) and Thwaites method to estimate the momentum thickness, θ(x), of the laminar boundary layer that develops on the flat surface when the initial momentum thickness θo is zero.

[Potentially useful information:

€

ξ 5dξ(ξ 2 + b2)50

x∫ =

x 6(x 2 + 4b2)24b4 (x 2 + b2)4

]

c) Will boundary layer separation occur in this flow? If so, at what value of x/b does Thwaites method predict zero wall shear stress? d) Using solid lines, sketch the streamlines for the ideal flow specified by the velocity potential given above. For comparison, on the same sketch, indicate with dashed lines the streamlines you expect for the flow of a real fluid in the same geometry at the same flow rate.

Solution 10.18. a) U(x) = ∂φ∂x!

"#

$

%&y=0

=q2π

xx2 + (y− b)2

+x

x2 + (y+ b)2!

"#

$

%&y=0

=qπ

xx2 + b2!

"#

$

%&

b) Use the Thwaites integral without the θo-term.

θ 2 (x) = 0.45νU 6 (x)

U 5(ξ )d0

x

∫ ξ =0.45νπ 6 x2 + b2( )

6

q6x6q5

π 5ξ

ξ 2 + b2"

#$

%

&'

0

x

∫5

dξ =0.45νπ x2 + b2( )

6

qx6ξ 5dξ

(ξ 2 + b2 )50

x

∫

Evaluate the integral using the given information:

θ 2 (x) = 0.45νπ (x2 + b2 )6

qx6x6 (x2 + 4b2 )24b4 (x2 + b2 )4

=0.45νπ (x2 + b2 )2

24b4q(x2 + 4b2 )

c) Compute the correlation parameter λ from the results of part b):

λ =θ 2

νdUdx

=0.45π (x2 + b2 )2

24b4q(x2 + 4b2 ) q

πb2 − x2

(x2 + b2 )2"

#$

%

&'=0.4524

4+ x2

b2"

#$

%

&' 1−

x2

b2"

#$

%

&' .

Clearly when x > b, λ becomes negative and its magnitude increases with increasing x, therefore at some x > b boundary layer separation will occur. Within Thwaites method, λ = λsep = –0.090 predicts the point, xzs, where the wall shear stress is zero; therefore, use the prior result and the quadratic formula to find

€

λsep =0.4524

4 +xzs2

b2#

$ %

&

' ( 1−

xzs2

b2#

$ %

&

' (

€

→

€

xzsb

= −32

+9 + 4(4 − 24λsep 0.45)

2

$

% & &

'

( ) )

1 2

=1.35 .

d) The streamlines for the ideal flow correspond to a point source above a flat surface. The real flow streamlines must include boundary layer separation at a value of x that is of order b. Furthermore, the thickness of the separated flow region will cause upward shifts in the ideal flow streamlines.

xsep x

y

b


Exercise 10.19. A fluid-mediated particle-deposition process requires a laminar boundary layer flow with a constant shear stress, τw, on a smooth flat surface. The fluid has viscosity µ and density ρ (both constant). The flow is steady, incompressible, and two-dimensional, and the flat surface extends from 0 < x < L. The flow speed above the boundary layer is U(x). Ignore body forces. a) Assume the boundary layer thickness is zero at x = 0, and use Thwaites’ formulation for the shear stress,

€

τw = µU θ( )l(λ) with

€

λ = θ 2 ν( ) dU dx( ) , to determine θ(x) and U(x) in terms of λ,

€

ν = µ ρ , x, and

€

τw µ = constant. [Hint: assume that

€

U θ = A and

€

l(λ) are both constants so that

€

τw µ = Al(λ) .] b) Using the Thwaites integral (10.50) and the results of part a), determine λ. c) Is boundary layer separation a concern in this flow? Explain with words or equations.

Solution 10.19. a) For Thwaites method, the shear stress is provided by the correlation

€

τw = µU θ( )l(λ) . Thus, the simplest way to achieve constant τw is for U to be proportional to θ, i.e.

€

U = Aθ where A is a constant, and l(λ) = constant, where

€

λ = θ 2 ν( ) dU dx( ) =constant. Eliminating U from this system of equations leaves:

€

τwµ

= Al(λ) , and

€

λ = A θ 2 ν( ) dθ dx( ) . (1,2)

Integrate (2) to find:

€

λνx = Aθ 3 3 where the constant of integration is zero because θ = 0 at x = 0. Now use equation (1) to eliminate A in the relationship for θ. Thus, the solution falls into an implicit form involving λ and l(λ).

€

θ(x) = 3λl(λ)νµx τw( )1 3 and

€

U(x) = Aθ(x) = 3τw2λνx µ2l2(λ)( )

1 3

b) Put the part a) results into the Thwaites integral:

€

3λl(λ)νµx τw( )2 3 =0.45ν

3τw2λνx µ2l2(λ)( )6 3

3τw2λνξ µ2l2(λ)( )

5 3dξ

0

x

∫ .

Perform the integration:

€

3λl(λ)νµx τw( )2 3 =0.45ν

3τw2λνx µ2l2(λ)( )6 3

3τw2λν µ2l2(λ)( )

5 3 x 8 3

8 3=0.458

3νx3τw

2λνx µ2l2(λ)( )1 3.

Use the two ends of the equality and simplify to find:

€

3λl(λ)νµx τw( )2 3 3τw2λνx µ2l2(λ)( )1 3

= 3νxλ =0.4583νx , or

€

λ =0.458

= 0.05625

c) Boundary layer separation is not a concern here because λ and τw are both positive and constant, so the separation condition λ = –0.090 is not approached.


Exercise 10.20. The steady two-dimensional potential for incompressible flow at nominal horizontal speed U over a stationary but mildly wavy wall is: φ(x, y) =Ux −Uε exp −ky( )cos kx( ) , where kε << 1. Here, ε is the amplitude of the waviness and k = 2π/Λ, where Λ = wavelength of the waviness. a) Use the potential to determine the horizontal velocity u(x, y) on y = 0. b) Assume that u(x, 0) from part a) is the exterior velocity on the wavy wall and use Thwaites’ method to approximately determine the momentum thickness, θ, of the laminar boundary layer that develops on the wavy wall when the fluid viscosity is µ, and θ = 0 at x = 0. Keep only the linear terms in kε and ε/x to simplify your work. c) Is the average wall shear stress higher for Λ/2 ≤ x ≤ 3Λ/4, or for 3Λ/4 ≤ x ≤ Λ. d) Does the boundary layer ever separate when kε = 0.01? e) In 0 ≤ x ≤ Λ, determine where the wall pressure is the highest and the lowest. f) If the wavy surface were actually an air-water interface, would a steady wind tend to increase or decrease water wave amplitudes? Explain.

Solution 10.20. a) The horizontal velocity u(x,y) is obtained from:

€

u = ∂φ ∂x = ∂ ∂x( ) Ux −Uεexp −ky( )cos kx( )[ ] =U +Ukεexp −ky( )sin kx( ) . when evaluated on y = 0, this becomes:

€

u(x,y = 0) =U 1+ kεsin(kx)( ) . b) Evaluate the Thwaites integral with θ = 0 at x = 0 to find:

€

θ 2 =0.45νU 6(x)

U 5( $ x )d $ x 0

x

∫ =0.45ν

U 1+ kεsin(kx)( )61+ kεsin(k $ x )( )5d $ x

0

x

∫

€

≈0.45ν

U1− 6kεsin(kx)( ) 1+ 5kεsin(k & x )( )d & x

0

x

∫ =0.45ν

U1− 6kεsin(kx)( ) & x − 5εcos(k & x )[ ]0

x

€

=0.45νU

1− 6kεsin(kx)( ) x − 5εcos(kx) + 5ε[ ] ≈ 0.45νxU

1− 6kεsin(kx) + 5(ε x)(1− cos(kx)( )

Thus,

€

θ ≈0.45νxU

1− 6kεsin(kx) + 5(ε x)(1− cos(kx)( )'

( ) *

+ ,

1 2

.

c) First compute the correlation parameter:

€

λ =θ 2

νdUdx

=0.45xU

1− 6kεsin(kx) + 5(ε x)(1− cos(kx)( ) Uk 2εcos(kx)( ).

This form can be made easier to look at if it is rearranged and the second order terms are dropped:

€

λ ≈ 0.45(kx)(kε)cos(kx) 1− 6kεsin(kx) + 5(ε x)(1− cos(kx)( ) ≈ 0.45(kx)(kε)cos(kx). Then note that

€

τw = µ U θ( )l(λ) and that l(λ) increases monotonically with increasing λ. Therefore, the shear stress will be highest where λ is highest. Now switch to considerations of

x!

y!

Λ/2! Λ!

U!

ε!


wavelength. Given that cos(kx) < 0 for Λ/2 ≤ x ≤ 3Λ/4 and that cos(kx) > 0 for 3Λ/4 ≤ x ≤ Λ, τw must be higher for 3λ/4 ≤ x ≤ λ. d) For separation, set

€

λsep = −0.090 ≈ 0.45(kx)(kε)cos(kx) . This can be solved numerically for kε = 0.01 to find kx ≈ 21.6, so separation is expected to occur when x is large enough. e) Use the steady flow Bernoulli equation with the reference condition taken far above the wavy surface:

€

12ρU 2 + P∞ =

12ρ U 2 1+ kεexp(−ky)sin(kx)[ ]2 +U 2 kεexp(−ky)cos(kx)[ ]2( ) + P(x) .

Evaluate on y = 0 and keep only the linear terms in kε to find:

€

P(x) − P∞ =12ρU 2 −2kεsin(kx) + ...( ) .

Thus, the minimum pressure occurs where sin(kx) = 1: kx = π/2 or x = Λ/4 (the wave crest), and the maximum pressure occurs where sin(kx) = –1: kx = 3π/2 or x = 3Λ/4 (the wave trough). f) Based on the results provided above, a steady wind will increase the wave amplitude by at least two mechanisms. The boundary layer shear tends to preferentially pull surface water away from wave troughs and toward wave crests (a shear stress effect), and the wind-induced pressure tends to suck wave crests up and push wave troughs down.


Exercise 10.21. Consider the boundary layer that develops in stagnation point flow:

€

Ue (x) =Uox L . a) With θ = 0 at x = 0, use Thwaites method to determine δ*(x), θ(x), and cf(x). b) This flow also has an exact similarity solution of the full Navier Stokes equations. Numerical evaluation of the final nonlinear ordinary differential equation produces:

€

c f Rex = 2.4652, where

€

Rex =Uex ν =Uo x2 Lν . Assess the accuracy of the predictions for cf(x) from the

Thwaites method for this flow.

Solution 10.21. a) The Thwaites equation is:

€

θ 2 =0.45νUe6(x)

Ue5(ξ)

0

x

∫ dξ + θo2 Uo

Ue (x)&

' (

)

* +

6

. Here, θ = 0

at x = 0 and

€

Ue (x) =Uox L , so only the integral term contributes:

€

θ 2 =0.45νUo6

L6

x 6Uo5

0

x

∫ ξ 5

L5dξ =

0.45νUo

L6

, or

€

θ = 0.2739 νLUo

.

Now compute the correlation parameter

€

λ =θ 2

νdUe

dx=θ 2

νUo

L=0.456

= 0.075

The wall shear stress is obtained from:

€

τw =µUe

θl(λ) =

µUo x L0.2739 νL Uo

(0.3267) =1.193µx Uo3

νL3&

' (

)

* +

1 2

,

where the numerical value for l(λ) has been obtained by interpolation in the tabulation of laminar boundary layer functions.

Thus, the skin friction coefficient is:

€

c f =τw

12ρUe

2 =2(1.193)µxρUo

2 x2 L2Uo3

νL3%

& '

(

) *

1 2

= 2.386 νLUox

2 .

Similarly,

€

δ* = θH(λ) = 0.2739 νLUo

(2.36) = 0.6464 νLUo

where again the value of H(λ) has been

obtained by interpolation. b) If

€

c f Rex = 2.4652 is the exact answer, then the Thwaites calculation is acceptably accurate (3% low) for most engineering purposes.


Exercise 10.22. A laminar boundary layer develops on a large smooth flat surface under the influence of an exterior flow velocity U(x) that varies with downstream distance, x. a) Using Thwaites method, find a single integral-differential equation for U(x) if the boundary layer is to remain perpetually right on the verge of separation so that the wall shear stress, τw, is zero. Assume that the boundary layer has zero thickness at x = 0. b) Assume

€

U(x) =Uo x L( )γ and use the result of part a) to find γ. c) Compute the boundary layer momentum thickness θ(x) for this situation. d) Determine the extent to which the results of parts b) and c) satisfy the von Karman boundary layer integral equation, (10.43), when τw = 0 by computing the residual of this equation. Interpret the meaning of your answer; is Von Karman’s equation well satisfied, or is the residual of sufficient size to be problematic? e) Can the U(x) determined for part b) be produced in a duct with cross sectional area

€

A(x) = Ao x L( )−γ ? Explain your reasoning.

Solution 10.22. a) When τw is zero, then

€

λsep = λ ≡ θ 2 ν( )dU dx . Now plug this into the Thwaites integral:

€

θ 2(x) =0.45νU 6(x)

U 5( $ x 0

x∫ )d $ x =

νλsep

dU dx, or

€

λsep =0.45

U 6(x)dUdx

U 5( # x 0

x∫ )d # x .

b) Plugging in

€

U(x) =Uo x L( )γ produces:

€

λsep =0.45

Uo6 x L( )6γ

γUo

LxL$

% & '

( ) γ −1 Uo

5L5γ +1

xL$

% & '

( ) 5γ +1

=0.45γ5γ +1

,

and for λsep = –0.090, this equation produces: γ = –0.10. [Note that this value is with in 0.01 of the exact Falkner-Skan exponent value, –0.0904 for a laminar boundary layer that is perpetually on the verge of separation.] c) Evaluate the Thwaites integral with θo = 0 using γ from part b):

€

θ 2 =0.45νU 6(x)

U 5( $ x 0

x∫ )d $ x =

0.45νUo6 x L( )−0.6

Uo5L

−0.5 +1xL'

( ) *

+ , −0.5+1

=0.90νL

Uo

xL'

( ) *

+ , 1.1

,

€

θ(x) =0.90νLUo

$

% &

'

( )

1 2xL$

% & '

( ) 0.55

.

d) Use τ wρ=ddx

U 2θ( )+Uδ* dUdx but set τw to zero and recognize that

€

δ* = 3.55θ from the

Thwaites shape-factor tabulation. Evaluate the two right side terms of the equation:

€

ddx

U 2θ( ) =ddx

Uo2 xL#

$ % &

' ( −0.2 0.90νL

Uo

#

$ %

&

' (

1 2xL#

$ % &

' ( 0.55#

$ % %

&

' ( ( = 0.35

Uo2

LxL#

$ % &

' ( −0.65 0.90νL

Uo

#

$ %

&

' (

1 2

€

Uδ* dUdx

=UoxL#

$ % &

' ( −0.1

3.55 0.90νLUo

#

$ %

&

' (

1 2xL#

$ % &

' ( 0.55

−0.1Uo

LxL#

$ % &

' ( −1.1#

$ % %

&

' ( ( = −0.355

Uo2

LxL#

$ % &

' ( −0.65 0.90νL

Uo

#

$ %

&

' (

1 2


Thus, we see that there is a residual =

€

0.005Uo2

LxL"

# $ %

& ' −0.65 0.90νL

Uo

"

# $

%

& '

1 2

which is only 1.4% of either

of the terms. This residual is acceptable given the approximate nature of Thwaites method. e) For an ideal two-dimensional duct with

€

A(x) = Ao x L( )−γ , the upper and lower walls would only diverge in proportion to

€

x L( )+0.1. However, the boundary layer displacement thickness grows like

€

x L( )+0.55 . Thus, adjustments would have to be made in the duct geometry to account for boundary layer growth. Such adjustments would involve greater divergence of the upper and lower walls, but increased wall divergence is likely to produce boundary layer separation. Thus, the part-b) flow speed

€

U(x) =Uo x L( )−0.10 cannot be produced in a duct with cross sectional area

€

A(x) = Ao x L( )−0.10 when viscous laminar boundary layers are present on the walls of the duct.


Exercise 10.23. Consider the boundary layer that develops on a cylinder of radius a in a cross flow a) Using Thwaites method, determine the momentum thickness as a function of ϕ, the angle from the upstream stagnation point (see drawing). b) Make a sketch of cf versus ϕ. c) At what angle does Thwaites method predict vanishing wall shear stress?

Solution 10.23. The potential for 2D ideal flow around a cylinder is:

€

φ(r,θ) =Urcosθ 1+ a2 r2( ) . For the specified geometry,

€

θ = π −ϕ and

€

cosθ = −cosϕ so

€

φ(r,ϕ) = −Urcosϕ 1+ a2 r2( ). Thus, the exterior velocity, Ue, outside the boundary layer that develops on the cylinder’s surface, is

given by:

€

Ue =1r∂φ∂ϕ

%

& '

(

) * r= a,x= aϕ

= 2U sinϕ[ ]x= aϕ = 2U sin x a( ) , so the Thwaites integral becomes:

€

θ 2 =0.45ν2U sinϕ( )6

32U 5

0

x

∫ sin5 ξ a( )dξ =0.45aν2U sin6ϕ

sin5 ' ϕ ( )0

ϕ

∫ d ' ϕ .

The initial condition term has been ignored since Ue(0) = 0. The integral of sin5(ϕ) may be looked up or performed via sin5(ϕ) = (1-cos2ϕ)2sin(ϕ) and an appropriate integration variable switch. The result is:

€

θ 2 =0.45aν2U sin6ϕ

815

− cosϕ +23cos3ϕ − 1

5cos5ϕ

&

' (

)

* + =

0.225a2

Resin6ϕ815


5cos5ϕ

&

' (

)

* + ,

where Re = Ua/ν (U is the velocity used in all the non-dimensional parameters in this problem).

b) Here

€

dUe

dx=2Uacosϕ , so for λ, we get:

€

λ =0.03cosϕsin6ϕ

8 −15cosϕ +10cos3ϕ − 3cos5ϕ( ) . Now,

evaluate the skin friction coefficient:

€

c f =µUe

θl(λ) 1

12 ρU

2 = 2l(λ)νUe

U 2θ= 8.43 l(λ)sin

4ϕRe

815


5cos5ϕ

(

) *

+

, - −1 2

.

Before plotting make a table of cf values. ϕ λ S(λ)

€

c f Re –––––––––––––––––––––––––––––––––––––––––––– 1° 0.075 0.33 0.11 15° 0.074 0.32 1.7 30° 0.072 0.32 3.1 60° 0.59 0.31 4.4 90° 0.0 0.22 2.5 100° -0.060 0.11 1.0


103° -0.089 ~0.0 ~0.0

The maximum surface shear stress is felt at ϕ ≈ 60°. c) The angle where Thwaites methog predicts a vanishing shear stress is ϕ ≈ 103°. Unfortunately, this value is not terribly useful as a prediction of the actual separation point (ϕs ≈ 80°-85°) because boundary layer separation causes significant changes to the outer flow around the cylinder that invalidates a portion of this computation. For this reason, reliable prediction of flows with separation remains a difficult task.

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

0 20 40 60 80 100 120

angle (degrees)

€

c f Re


Exercise 10.24. An ideal flow model predicts the following surface velocity for the suction (i.e. the upper) side of a thin airfoil with chord c placed in a uniform horizontal air stream of speed Uo:

Ue(x) = 2Uo x c[ ]1 5 exp −x c( ) . a) Assuming that x is the coordinate along the foil's suction surface, use Thwaites method to estimate the momentum thickness θ(x) of the laminar boundary layer that develops on this surface. b) Using the results of part a) show that the correlation parameter λ is given by:

λ =0.45

125 x c( )2e5x c −1− 5x

c"

#$%

&'1− 5x

c"

#$%

&'

c) Does Thwaites method predict boundary layer separation in the range 1/5 < x/c < 1? d) If a laminar boundary layer is predicted to separate from the surface of this airfoil, suggest at least two changes that could be made to the foil that would tend to prevent separation.

Solution 10.24. a) The boundary layer is launched from a stagnation point, so the basic Thwaites formula is:

θ 2 =0.45νUe6 Ue

5 dx0

x

∫ , or in this case: θ 2 = 0.45ν26Uo

6 x c( )6 5 e−6 x c25Uo

5 "x c( )e−5 "x c d "x0

x

∫ .

Simplify, change to an integration variable ξ = x´/c, and use the hint:

θ 2 =0.45νce−6 x c

2Uo x c( )6 5ξ exp −5ξ( )dξ

0

x c

∫ =0.45νce−6 x c

2Uo x c( )6 5−1+ 5ξ( )25

exp −5ξ( )#

$%

&

'(0

x c

.

Evaluate: θ 2 = 0.45νce−6 x c

2Uo x c( )6 5−1+ 5x c( )25

e−5x c + 125

"

#$

%

&'=

0.45νcex c

50Uo x c( )6 5e5x c − 1+ 5x c( )"# %& .

b) The correlation parameter is λ = (θ2/ν)(dUe/dx), and the missing ingredient from part a) is: dUe(x)dx

= 2Uo15c

x c[ ]−4 5 − x c[ ]1 5 1c

"

#$

%

&'e−x c .

So, λ = θ2

νdUe(x)dx

=0.45cex c

50Uo x c( )6 5e5x c − 1+ 5x c( )"# $%2Uo

15c

x c[ ]−4 5 − x c[ ]1 5 1c

&

'(

)

*+e−x c , or

0"

0.2"

0.4"

0.6"

0.8"

1"

1.2"

1.4"

0" 0.2" 0.4" 0.6" 0.8" 1"x/c!

Ue(x)Uo !

! !!

Uo!

c!


λ =0.45

125 x c( )2e5x c − 1+ 5x c( )"# $% 1− 5 x c( )( ) which is the desired form.

c) From part b), λ = 0 at x/c = 1/5, and λ = (0.45/125)(e5 – 6)(–4) = –2.05 at x/c = 1. d) Yes, the value of λ at x/c = 1 is less than λsep = –0.090. Thus, the Thwaites method predicts that a laminar boundary layer will separate downstream of x/c = 1/5 and upstream of x/c = 1. Four possible changes to the foil that might delay or prevent boundary layer separation are as follows. (i) The surface could be roughened or a boundary layer trip could be added to stimulate transition to a turbulent boundary layer. (ii) The thickness of the foil could be reduced so that the pressure gradient predicted by ideal flow for its upper surface is less adverse aft of x/c = 1/5. (iii) Compressed air could be jetted downstream along the foil's surface to speed up the slow moving portion of the boundary layer thereby keeping it attached. (iv) Suction could be applied to the aft portion of the foil to remove the lower (slow moving) portion of the boundary layer thereby keeping it attached.


Exercise 10.25. An incompressible viscous fluid flows steadily in a large duct with constant cross sectional area Ao and interior perimeter b. A laminar boundary layer develops on the duct’s sidewalls. At x = 0, the fluid velocity in the duct is uniform and equal to Uo, and the boundary layer thickness is zero. Assume the thickness of the duct-wall boundary layer is small compared to Ao/b. a) Calculate the duct-wall boundary layer momentum and displacement thicknesses, θ(x) and δ*(x) respectively, from Thwaites’ method when U(x) = Uo. b) Using the δ*(x) found for part a), compute a more accurate version of U(x) that includes boundary layer displacement effects. c) Using the U(x) found for part b), recompute θ(x) and compare to the results of part a). To simplify your work, linearize all the power-law expressions, i.e.

€

1− bδ* Ao( )n≅1− nbδ* Ao .

d) If the duct area expanded as the flow moved downstream, would the correction for the presence of the sidewall boundary layers be more likely to move boundary layer separation upstream or downstream? Explain. Solution 10.25. a) The Thwaites’ results with a constant velocity and zero momentum thickness at x = 0 are readily obtained:

€

θ 2 =0.45νUo6 Uo

5dx =0.45νxUo0

x

∫ , or

€

θ = 0.671 νx Uo .

Here, λ is zero because dU/dx = 0 so the shape factor is 2.61 which means that:

€

δ* = 2.61⋅ θ =1.75 νx Uo . b) At any downstream distance x in the duct, the displacement effect of the boundary layer will decrease the effective flow area of the duct. Therefore a simple mass balance between the duct entrance (x = 0) and the location x requires:

€

ρUoAo = ρU(x) Ao − bδ*( ) , or

€

U(x) =Uo 1− bδ* Ao( )

−1

c) Use the revised U(x) found in part b) and Thwaites’ method to find:

€

θ 2 =0.45ν

Uo6 1− bδ* Ao( )−6

Uo5 1− bδ* Ao( )

−5dx

0

x

∫ ≈0.45νUo

1− 6bδ* Ao( ) 1+ 5bδ* Ao( )dx0

x

∫

€

=0.45νUo

1− 6bδ* Ao( ) 1+ 5b(1.75 νx Uo Ao( )dx0

x

∫ =0.45νUo

1− 6bδ* Ao( ) x +10bδ*x 3Ao( )

€

≈0.45νxUo

1− 6bδ* Ao +10bδ* 3Ao( ) =0.45νxUo

1− 8bδ* 3Ao( )

where linearization has been used to make the integration and algebra easier. Here, we see that the revised value of θ(x) will be smaller than that estimated in part a). d) The displacement effect of the boundary layer acts to increase the flow speed. This means that dU/dx will be larger when the boundary layer’s displacement effect is accounted for, thus the correlation parameter λ will be larger and further from its “separation” value of –0.090. Therefore, the predicted location of separation will be further downstream in an expanding duct when the boundary layer displacement effect is accounted for. In the present situation, λ is zero without accounting for the boundary layer displacement effect but λ becomes positive because dU/dx is nonzero and positive. Thus, when the boundary layer displacement effect is included:


€

dU(x)dx

=ddxUo 1− bδ

* Ao( )−1≈ddxUo 1+ bδ* Ao( ) =

UobAo

dδ*

dx=Uob2Aox

δ*,

and

€

λ ≡θ 2(x)ν

dU(x)dx

≈0.45xUo

1− 8bδ* 3Ao( ) dU(x)dx≈0.452

1− 8bδ* 3Ao( ) bδ*

A0> 0 !


Exercise 10.26. Water flows over a flat plate 30 m long and 17 m wide with a free-stream velocity of 1 m/s. Verify that the Reynolds number at the end of the plate is larger than the critical value for transition to turbulence. Using the drag coefficient in Figure 10.12, estimate the drag on the plate. Solution 10.26. ReL = UL/ν = (1.0m/s)(30m)/(1.0x106 m2/s) = 3.0x105 > Recr ~ 106. Thus, the flow is expected to be turbulent over most of the plate. From Figure 10.12, CD ≈ 0.003 so that:

€

D =12ρU 2(Area)CD = 0.5(103)(1.0)2(17)(30)(0.003) = 765N .


Exercise 10.27. A common means of assessing boundary layer separation is to observe the surface streaks left by oil or paint drops that were smeared across a surface by the flow. Such investigations can be carried out in an elementary manner for cross-flow past a cylinder using a blow dryer, a cylinder 0.5 to 1 cm in diameter that is ~10 cm long (a common ball-point pen), and a suitable viscous liquid. Here, creamy salad dressing, shampoo, dish washing liquid, or molasses should work. And, for the best observations, the liquid should not be clear and the cylinder & liquid should be different colors. Dip your finger into the viscous liquid and wipe it over two thirds of the surface of the cylinder. The liquid layer should be thick enough so that you can easily tell where it is thick or thin. Use the dry one third of the cylinder to hold the cylinder horizontal. Now, turn on the blow dryer leaving the heat off and direct its outflow across the wetted portion of the horizontal cylinder to mimic the flow situation in the drawing for Exercise 10.23. a) Hold the cylinder stationary, and observe how the viscous fluid moves on the surface of the cylinder and try to determine the angle ϕs at which boundary layer separation occurs. To get good consistent results you may have to experiment with different liquids, different initial liquid thicknesses, different blow-dryer fan settings, and different distances between cylinder and blow dryer. Estimate the cylinder-diameter-based Reynolds number of the flow you’ve studied. b) If you have completed Exercise 10.23, do your boundary layer separation observations match the calculations? Explain any discrepancies between your experiments and the calculations. Solution 10.27. a) Using a dark shampoo on a white ball-point pen, the 3rd author of this textbook found that the separation point occurred on the upstream side of the cylinder near ϕ = 90°. The cylinder diameter was 8 mm and flow speed was probably about 10 m/s. Thus, an estimate of the Reynolds number is:

€

ReD =UDν

≈(10m /s)(0.008m)1.5 ×10−5m2 /s

≈ 5,000

b) The blow-drier-and-ball-point-pen separation point results, ϕs near to but less than 90°, match commonly quoted experimental values for the separation angle (ϕ ≈ 80°-85°). However, they do not match the Thwaites-method-calculated location of zero shear stress (ϕ ≈ 103°). The primary difference between the experiments and the Thwaites calculation is the separated flow in the wake of the cylinder. The experiment includes the separated flow while the potential flow surface velocity that is input to the Thwaites calculation does not include the separated flow. In this case the cylinder’s wake makes important changes to the surface flow on the cylinder so the experimental and calculated boundary layers develop under different flow fields and therefore separate at different points. In general, Thwaites method is only successful in predicting whether or not boundary layer separation will occur. Once boundary layer separation has occurred, a theory that accounts for flow in the separation zone is needed.


Exercise 10.28. Find the diameter of a parachute required to provide a fall velocity no larger than that caused by jumping from a 2.5 m height, if the total load is 80 kg. Assume that the properties of air are ρ = 1.167 kg/m3, ν = 1.5 × 10-5 m2/s, and treat the parachute as a hemispherical shell with CD = 2.3. [Answer: 3.9 m] Solution 10.28. The fall velocity from a 2.5 m height is [2gh]1/2 = [2(9.81)(2.5)]1/2 = 7.0 m/s. At steady state, the drag on the parachute equals the load, so that D = mg = (80)(9.81). Thus, using CD = 2.3, the parachute's cross sectional area should be

€

Area =D

CD12 ρU

2 =80(9.81)

2.3(0.5)(1.167)(7)2=11.93m2 , or a diameter = [4(11.93)/π]1/2 = 3.90 m.


Exercise 10.29. The boundary layer approximation is sometimes applied to flows that do not have a bounding surface. Here the approximation is based on two conditions: downstream fluid motion dominates over the cross-stream flow, and any moving layer thickness defined in the transverse direction evolves slowly in the downstream direction. Consider a laminar jet of momentum flux J that emerges from a small orifice into a large pool of stationary viscous fluid at z = 0. Assume the jet is directed along the positive z-axis in a cylindrical coordinate system. In this case, the steady, incompressible, axisymmetric boundary layer equations are:

€

1R∂(RuR )∂R

+∂w∂z

= 0 , and

€

w ∂w∂z

+ uR∂w∂R

= −1ρ∂p∂z

+νR∂∂R

R∂w∂R

&

' (

)

* + ,

where w is the (axial) z-direction velocity component, and R is the radial coordinate. Let r(z) denote the generic radius of the cone of jet flow. a) Let

€

w(R,z) = ν z( ) f (η) where

€

η = R z , and derive the following equation for f:

€

η # f + f ηfdη = 0η∫ .

b) Solve this equation by defining a new function

€

F = ηfdηη∫ . Determine constants from the

boundary condition: w → 0 as η → ∞, and the requirement:

€

J = 2πρ w2(R,z)RdRR= 0

R= r(z )

∫ = constant.

c) At fixed z, does r(z) increase or decrease with increasing J? [Hints: i) the fact that the jet emerges into a pool of quiescent fluid should provide information about ∂p/∂z, and ii)

€

f (η)∝ (1+ const ⋅η2)−2, but try to obtain this result without using it.] Solution 10.29. a) Start with the given boundary layer equations and use

€

w(R,z) = ν z( ) f (η) where

€

η = R z ,

€

∂ ∂R = 1 z( )∂ ∂η and

€

∂ ∂z = −η z( )∂ ∂η in the continuity equation to find:

€

uR = −1R

R∂w∂z

dR0

R

∫ = −1R

R −νηz2

' f − νz2

f(

) *

+

, - dR

0

R

∫ =νR

η2 ' f +ηf( )dη0

η

∫

Now start assembling terms of the momentum equation:

€

w ∂w∂z

=νz

f ⋅ − νz2

f − νηz2

f'

( )

*

+ , = −

ν 2

z3f 2 +ηf - f ( ),

€

uR∂w∂R

=νηz

η2 % f +ηf( )dη0

η

∫'

( )

*

+ , νz2

% f , and

€

νR∂∂R

R∂w∂R

$

% &

'

( ) =

νR∂∂R

R νz2

* f $

% &

'

( ) =

νηz

νz2

* f + R νz3

* * f $

% &

'

( ) =

ν 2

ηz3* f +η * * f ( ).

If the fluid far from the jet is still, then ∂p/∂z = 0 in the boundary layer approximation because ∂p/∂R = 0 too. Reassemble the momentum equation:

€

−ν 2

z3f 2 +ηf % f ( ) +

νηz

η2 % f +ηf( )dη0

η

∫'

( )

*

+ , νz2

% f =ν 2

ηz3% f +η % % f ( )

Divide out the common factor of

€

ν 2 z3 :

€

− f 2 +ηf $ f ( ) +1η

η2 $ f +ηf( )dη0

η

∫&

' (

)

* + $ f =

1η

$ f +η $ $ f ( )

This is a single equation for f without z or R appearing; thus the assumption of a similarity solution is successful. Multiply both sides by η, expand terms and integrate by parts.

€

−ηf 2 −η2 f $ f + $ f η2 f − 2 ηfdη0

η

∫&

' (

)

* + + $ f ηfdη

0

η

∫ = $ f +η $ $ f


Combine common terms to find:

€

−ηf 2 − $ f ηfdη0

η

∫ = $ f +η $ $ f . This compact form can be further

simplified by noting that:

€

ddη

f ηfdη0

η

∫$

% &

'

( ) = * f ηfdη

0

η

∫ +ηf 2 , and

€

ddη

η # f ( ) =η # # f + # f . Thus, one

integration with respect to η yields:

€

η # f + f ηfdη0

η

∫ = 0 . Here, the constant must be zero based

on an evaluation of the equation at η = 0. b) If

€

F = ηfdηη

∫ , then

€

" F =ηf , and

€

" " F =η " f + f , so the above equation becomes:

€

" " F − 1η

" F +1η

" F F = 0 , or

€

η2 # # F +η(F −1) # F = 0 . This last equation is equidimensional so the

substitution

€

η = et will lead to simplification. Here,

€

ηddη

=ddt

, and

€

η2d2

dη2=d2

dt 2−ddt

, so the

equation becomes:

€

d2Fdt 2

+ (F − 2) dFdt

= 0. This equation is autonomous; thus we can let

€

dFdt

= y(F) so that

€

d2Fdt 2

=dydF

dFdt

= y " y , and the last remnant of the momentum equation

becomes:

€

y " y − (F − 2)y = 0 . Therefore when y ≠ 0,

€

" y = −F + 2 , or

€

y = A + 2F − F 2 /2. Now go back to the definition of F to determine the constant A:

€

y(F) =dFdt

=ηdFdη

= A + 2F − F 2 /2 = A + 2 ηfdηη

∫ +12

ηfdηη

∫[ ]2

.

When η = 0, then F = 0, so A = 0. Thus,

€

y(F) = 2F − F 2 2 =ηdF dη . Separate and integrate

using a partial fraction decomposition:

€

dη2η

=dF

F 2 − 4F= −

dF4(F − 4)

+dF4F

to reach:

€

lnη1 2 + B =14ln F4 − F$

% &

'

( ) . Now exponentiate:

€

Cη1 2 =F

4 − F$

% &

'

( ) 1 4

and solve for F:

€

F =4C4η2

(1+ C4η2).

Recall that

€

" F =ηf , so

€

f =1η

# F =8C4

(1+ C4η2)2. The constant C4 can be evaluated from:

€

J2πρ

= w2(R,z)RdRR= 0

R= r(z)

∫ =ν 2

z2f 2(η)RdR

R= 0

R= r(z)

∫ = ν 2 ηf 2dη0

r(z ) / z

∫ = ν 264C8η

(1+ C4η2)4dη

0

r(z) / z

∫

The integral may be performed after a change of variable

€

γ =1+ C4η2 , and by noting that the upper limit corresponds to η → ∞. Thus:

€

J2πρ

= 32C4ν 2 γ−4dγ0

∞

∫ =32C4

3ν 2 , so

€

C4 =3J

64πν 2ρ=

364π

Re2, where

€

Re =J ρν

Therefore, the final solution is:

€

w(R,z) =νz

3 8π( )Re2

1+ 3 64π( )Re2(R /z)2( )2

c) From the result of part b), r(z) will decrease with increasing J (increasing Re). The laminar jet narrows as its Reynolds number increases.


Exercise 10.30. A simple realization of a temporal boundary layer involves the spinning fluid in a cylindrical container. Consider a viscous incompressible fluid (density = ρ, viscosity = µ) in solid body rotation (rotational speed = Ω) in a cylindrical container of diameter d. The mean depth of the fluid is h. An external stirring mechanism forces the fluid to maintain solid body rotation. At t = 0, the external stirring ceases. Denote the time for the fluid to spin-down (e.g. to stop rotating) by τ. a) Case I: h >> d. Write a simple laminar-flow scaling law for τ assuming that the velocity perturbation produced by the no-slip condition on the container’s sidewall must travel inward a distance d/2 via diffusion. b) Case II: h << d. Write a simple laminar-flow scaling law for τ assuming that the velocity perturbation produced by the no-slip condition on the container's bottom must travel upward a distance h via diffusion. c) Using partially-filled cylindrical containers of several different sizes (drinking glasses and pots & pans are suggested) with different amounts of water, test the validity of the above diffusion estimates. Use a spoon or a whirling motion of the container to bring the water into something approaching solid body rotation. You'll know when you're close to solid body rotation because the fluid surface will be a paraboloid of revolution. Once you have this initial flow condition set-up, cease the stirring or whirling and note how long it takes for the fluid to stop moving. Perform at least one test when d & h are several inches or more. Cookie or bread crumbs sprinkled on the water surface will help visualize surface motion. The judicious addition of a few drops of milk after the fluid starts slowing down may prove interesting. d) Compute numbers from your scaling laws for parts a) and b) using the viscosity of water, the dimensions of the containers, and the experimental water depths. Are the scaling laws from parts a) and b) useful for predicting the experimental results? If not, explain why. (The phenomena investigated here have some important practical consequences in atmospheric and oceanic flows and in IC engines where swirl and tumble are exploited to mix the fuel charge and increase combustion speeds.) Solution 10.30. For all simple unsteady diffusion problems, the length scale of “diffusion-penetration” is proportional to the square root of the product of the diffusion constant and time. For momentum diffusion in fluid flows, ν is the diffusion constant. In the following, let τ be the time it takes for the swirling fluid to come to rest. a) If h >> d, it should be possible to ignore the bottom of the container. Therefore, the "stopping" signal must penetrate a distance of d/2 inward from the walls of the container. So, for an approximate unsteady diffusion analysis, we have: d/2 ≈

€

ντ , or τ ≈ d2/4ν . b) If d >> h, it should be possible to ignore the walls of the container. Therefore, the "stopping" signal from the no-slip boundary condition on the bottom of the container must penetrate upward a distance of h from the bottom of the container. So, for an approximate unsteady diffusion analysis, we have: h ≈

€

ντ , or τ ≈ h2/ν c) For all standard-kitchen-size containers with water depths of a several cm, the experimental stopping time was less than a minute.


d) Using the formulae from part a) or b), a length scale of 5 cm (~two inches), and the kinematic

viscosity of water: τ ≈

€

(0.05m)2

1.0 ×10−6m2 /sec= 2500 seconds ≈ 40 minutes!

The expectation here is that the estimated and actual times would match within a factor of two or so. Clearly something is wrong with the simple diffusion analysis for these simple swirling flows. The milk drops in the swirling water show that the flow is very three-dimensional and that the interaction of the flow with the bottom and sides of the pan or cup is very important. In particular, the viscous boundary layer on the bottom of the container causes inward radial flow along the bottom and the fluid stops rotating because of a phenomenon called Ekman pumping.


Exercise 10.31. Mississippi River boatmen know that when rounding a bend in the river, they must stay close to the outer bank or else they will run aground. Explain in fluid mechanical terms the reason for the cross-sectional shape of the river at the bend. Solution 10.31. The Reynolds number based on mean channel width or radius of curvature is very large. Thus, the riverbed boundary layer is thin compared with the width or depth of the river, and most of the flow is inviscid. The primary turning flow around the bend is governed by

€

∂p ∂r = ρvθ2 r , but vθ decreases rapidly to zero in the thin boundary layer on the riverbed. Thus,

€

∂p ∂r in the boundary layer, which is largely the same as that in the inviscid flow, is not balanced by

€

ρvθ2 r in the boundary layer since [vθ]boundary layer < [vθ]inviscid flow. Here,

€

∂p ∂r is too large in the boundary layer flow for it to remain parallel to the flow above it. The excess pressure gradient pushes the boundary layer flow toward the lower pressure which occurs at the riverbank on the inside of the river bend. This induces cross-flow in the boundary layer that moves boundary layer fluid toward the inner riverbank, and causes a downward secondary flow at the outer riverbank. Together these effects set up the secondary flow pattern shown below. In the Mississippi River, the water transports silt which is scoured from the outer bank and deposited on the inner bank. This transport accounts for the cross section of the river channel. This secondary flow phenomenon also explains why small river meanderings tend to grow over time as the river erodes its outer bank and deposits silt on its inner bank. The "goose necks" of the San Juan River is southeastern Utah provide an extreme (and impressive) example of river meandering.

r

!"#!$

boundary

layer

fluid mechanics kundu cohen 6th edition solutions sm ch (10)

Engineering