two dimensional oblique shocks expansion flows

8/7/2019 Two Dimensional Oblique Shocks Expansion Flows

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Two-Dimensional Steady Oblique Shocks and Expansion

Flows

I. Differential Balance Equations and Uniform Flow

We consider two-dimensional flows governed by the Euler equations written below in

conservative form.

F G0

x y

(1.1)

2

2

u v

u P uv

F Guv v Pv v v v

u e Pu v e Pv2 2

(1.2)

We will study uniform flows that encounter shocks or expansion waves and remain

uniform after the encounter. The direction of the uniform flow is such that the flow is

tangent to a physical boundary. This is a special situation in which gradients of the flow

variables are zero and the Euler equations are trivially satisfied. Discussion of more

general flows will require computational methods and will be discussed later. The

uniform flow assumption will enable the study of fundamental flow situations that appear

in gas dynamics and will provide insight for more advanced approaches requiringcomputational or experimental approaches.

II. Disturbances and the Mach Angle

Mechanisms for the creation of shocks have previously been discussed and this chapter

highlights another interesting possibility. Figure 4.5 in the text illustrates the concept of a

Mach wave and the Mach angle. The Mach wave is an extremely weak oblique shock

with angle

1 1sinM

(2.1)Figure 4.5 is a striking photograph showing a bow shock in front of a supersonic

projectile traveling under a perforated plate. The fluid above the perforated plate feels the

disturbance moving at a supersonic speed and a Mach wave forms as the disturbances

coelesce.


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III. Integral Balance Equations for Oblique Shocks

The figure below shows an oblique shock caused by a boundary creating a flow turning

into itself. We seek to find conditions that hold across the shock similar to theconditions developed across a normal shock.

2 nu ,M

2

1 tw ,M

We begin with the control volume equations which we have shown hold when a shock is

within the control volume.

Continuity:

CV CS

dV v ndS 0t

(3.1)Momentum:

CV CS CS

vdV v v n dS Pn dSt

(3.2)Energy:

CV CS CS

1 1( e v v )dV ( e v v )v ndS Pv n dS

t 2 2 (3.3)

1 1V ,M

n

2 t 2w ,M 1

1 nu ,M

1

2 2V ,M

2

1


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We consider steady flows, uniform on either side of the shock. The text chooses a control

volume parallel to the velocity and tangent to the shock. If the flow is not uniform or

steady on either side of the stationary shock the same result would be obtained by

collapsing the control volume to the shock thus these results are valid for transient, non-

uniform flows with stationary oblique shocks.

The resulting conditions across the shock are:

Continuity:

1 1 2 2u u (3.4)

Normal Momentum:

(3.5)2 21 1 1 2 2 2u P u P

Tangential Momentum:

1w w2 (3.6)

Energy:

2

1 1 2 2

1( h u ) ( h u )

2 2 21 (3.7)

These conditions indicate that the conditions for an oblique shock are identical to the

conditions for a normal shock using the normal components of velocity. The new

condition is that the tangential component of velocity does not change across an oblique

shock.

The Mach number may be decomposed into normal and tangential components and the

normal component may be calculated from the earlier figure as

n1 1M M sin (3.8)

For calorically perfect gases the same myriad of algebraic manipulations that were

applied in chapter three may be undertaken to arrive at the expressions

2

n1

2n2

2

n1

11 M

2M 1M

2

(3.9)

2

n12 1

2

1 2 n

1 Mu

u 2 1 M

1

(3.10)


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22 n11P 2

1 MP 1

1 (3.11)

2

n122 2 1 2n1 2

1 1 2 1 n1

2 1 M T h P 21 M 1

T h P 1 1 M

(3.12)

The normal shock tables therefore apply to the normal direction for an oblique shock.

From the figure it is easy to see that

n22M

Msin (3.13)

Given equations (3.8) through (3.13) it is clear that changes across an oblique shock are

functions of both and1

M which contrasts to the case of a normal shock in whichchanges are only a function of . All that remains is to define a relationship between

the angle of the oblique shock1

M

and the turning (deflection) angle (usually theturning angle is known as it is given by the geometry and the shock angle must be

determined). The geometry of the figure as well as the jump conditions may be used to

find such a relationship.

2 2

1

2

1

M sin 1tan 2 cot

M cos 2 2

(3.14)

This relationship is graphed in figure 4.8 in the text may be used to arrive at thefollowing conclusions:

1. For any given1

M there is a maximum turning angle . If the geometry is suchthat the turning angle is greater that the turning angle predicted by equation (3.14)

then no oblique shock solution is possible. There will instead be a curved and

detached shock.

2. Given a turning angle there are two possible shock angles. The larger shock angleis termed the strong shock solution and the smaller shock angle is termed the

weak shock solution. Generally the weak shock solution is favored. The weak

solution results in 2M being supersonic. The strong shock solution almost always

results in2

M being subsonic.

3. 0 corresponds to2

which is a normal shock or to which is aMach wave.


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4. For a fixed as1

M ranges from high to low supersonic values, for weak

shocks increases. There is a 1M below which no solutions are possible. At thispoint is maximum and lower values of 1M result in detached, curved shocks asin figure 4.9 of the text.

Equation (3.14) may be used to calculate given and is implemented in theMATLAB file theta_calc.m.

function theta = theta_calc( M1, beta )

top = ( M1*sin(beta) )^2 - 1;bot = M1^2*( 1.4 + cos(2*beta) ) + 2;

term = 2*top/( bot*tan(beta));

theta = atan(term);

theta_calc.m

Equation (3.14) may also be iterated to calculate given and this calculation isimplemented in the MATLAB file beta_calc.m.

function beta = beta_calc( M1, theta, bguess )

% secant method to get mach # given area ratio

bo = bguess;

db = 1/1000.;delb = 1.0;

while abs(delb) > .0001,br = bo+db;bl = bo-db;dfdb = ( theta_calc(M1, br) - theta_calc(M1, bl) )/(2*db);delb = -(theta_calc(M1, bo)-theta)/dfdb;bo = bo + delb;

end

beta = bo;

beta_calc.m

Examples 4.1 through 4.3 are similar and may be examined using the following

MATLAB file. The conclusions of these examples are summarized in the paragraph in

the text following example 4.3 and we will read this in class.


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% example 4.1M1 = 3.0;

p1 = 1;

T1 = 288;theta = 20*pi/180;

beta_guess = 10*pi/180;beta = beta_calc(M1, theta, beta_guess);sprintf('beta = %0.5g',beta*180/pi)

Mn1 = M1*sin(beta)[presr1, densr1, tempr1, por1 , po21, Mn2 ] = shock(Mn1);p2 = presr1*p1T2 = tempr1*T1M2 = Mn2/sin( beta - theta )

[presr2,densr2,tempr2,arear2] = isentrop(M2);To2 = tempr2*T2po2 = presr2*p2

ex_4_1.m

IV. Regular Reflection of a Shock from a Solid Boundary

We consider next a case where a flow is first turned into itself and directed towards a

second boundary that must turn the flow again. In this case a solution may be found by

adding second shock reflected from the second boundary as shown below.

1 tw ,M

22 2V ,M1

2

1

1 1

V , M

1

1

2

1

3 3V ,M

3


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In this figure the turning angles1

and2

are the same but this is not generally true. The

method of calculation here is the same as for the single oblique shock case marching

from region 1 to region 3. Below is example 4.10 from the text

% example 4.10 - in this case theta1 = theta2

M1 = 2.8;

p1 = 1; T1 = 519;theta1 = 16*pi/180;

beta1 = beta_calc(M1, theta1, 10*pi/180);sprintf('beta1 = %0.5g',beta1*180/pi)

Mn1 = M1*sin(beta1);

sprintf('Mn1 = %0.5g',Mn1)

[presr1, densr1, tempr1, por1 , po21, Mn2 ] = shock(Mn1);

M2 = Mn2/sin( beta1 - theta1 );sprintf('M2 = %0.5g',M2)


Mn2 = M2*sin(beta2);sprintf('Mn2 = %0.5g',Mn2)



p3 = p1*presr1*presr2

T3 = T1*tempr1*tempr2phi = 180*( beta2 - theta1)/pi

ex_4_10.m

Note that in the case of a reflected shock we know the final flow direction but we do not

know the second shock direction. Note also that this solution is just a two steps of the

standard oblique shock calculation.

Example 4.11 is similar but in this case the second deflection angle is no the same as the

first. However, the calculation procedure is the same as in example 4.10.


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% example 4.11 - in this case theta1 is not equal to theta2% note that the second deflection is theta1 + theta2

M1 = 3.0;p1 = 1; T1 = 300;theta1 = 14*pi/180;theta2 = 10*pi/180;





beta2 = beta_calc(M2, theta1 + theta2, 10*pi/180);sprintf('beta2 = %0.5g',beta2*180/pi)


[presr2, densr2, tempr2, por2 , po32, Mn3 ] = shock(Mn2);sprintf('Mn3 = %0.5g',Mn3)phi = beta2 - theta1 - theta2;

M3 = Mn3/sin( phi );sprintf('M3 = %0.5g',M3)

p3 = p1*presr1*presr2

T3 = T1*tempr1*tempr2

ex_4_11.m

We will discuss example 4.12 in class. The Section on flow through multiple shock

systems relates to this example. The key point is that multiple shocks are a more efficient

way to decelerate a flow to a given Mach number. This means that the total pressure is

higher after multiple weak shocks than after one stronger shock. Following is example

4.12 and the text section 4.7 is important as a practical design consideration for the

engineer dealing with high speed gas dynamics.


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% example 4.12 calculate theta in degrees to get M3 = 1.4578% to compare efficiency of one oblique shock with two shock solution% from example 4.10 - want same M3 = 1.4578

% iterate

M2_given = 1.4578;M1 = 2.8;

p1 = 1;% guess theta = 20;theta = 20*pi/180;

func = 1;while abs(func) > 1.0e-6 % Newton loop% first get current M2beta_guess = 10*pi/180;beta = beta_calc(M1, theta, beta_guess);Mn1 = M1*sin(beta);

[presr1, densr1, tempr1, por1 , po21, Mn2 ] = shock(Mn1);M2 = Mn2/sin( beta - theta );

% get derivative for Newton methoddt = 1.0d-4*theta;thetar = theta + dt;

beta_guess = 10*pi/180;beta = beta_calc(M1, thetar, beta_guess);

Mn1 = M1*sin(beta);[presr1, densr1, tempr1, por1 , po21, Mn2 ] = shock(Mn1);Mr2 = Mn2/sin( beta - theta );fprime = (Mr2 - M2)/dt;func = M2 - M2_given;theta = theta - func/fprime;

end

sprintf('theta = %0.5g',theta*180/pi)p2 = presr1*p1;T2 = tempr1*T1;

[presr2,densr2,tempr2,arear2] = isentrop(M2);To2 = tempr2*T2;

po2 = presr2*p2;sprintf('po2 = %0.5g',po2)

ex_4_12.m

We will skip the section on pressure-deflection diagrams as they provide a graphical

solution and interpretation for oblique shock problems. As should have been mentioned

previously, some of the solutions in these notes differ from the text as the text uses charts

and interpolation to get data whereas the notes use the formulas directly in MATLAB.


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V. Intersection of Shocks of Opposite Families

We consider next a case where a flow is first turned into itself and directed towards a

second boundary that also turns the flow towards itself. In this case a solution may be

found by adding second shock reflected from the second boundary as shown in the figurefrom the text below.

In this section of the notes we seek to develop a method by which we can

compute solutions of this type without recourse to the pressure-deflection diagrams ( see

figure 4.22 ).

Consider the figure above from the text. This situation corresponds to a regular

reflection from a solid boundary on both the top and bottom walls. The added condition

is that the flow in regions 4 and 4 must be parallel and thus the pressure must be

continuous across the slip line. By a slip line we mean that there is a possibility that the

velocity tangential to the slip line is discontinuous. The condition that the pressure is the

same across a slip line means that there is no velocity component normal to the slip line.

We set up an iterative method by constructing the function:

(4.1)

This function is zero when the downstream deflection angle has the correct value. We

use Newtons (secant) method and the regular reflection problem discussed earlier to

write a simple program to implement this iteration. First we write a function to calculate

the regular reflection with pressure and Mach number as output. Essentially we write a

function to do the calculations of problems 4.10 through 4.12 but we are only interested

in the final pressure and Mach number. The function is shown below


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function [press3,M3] = reg_ref(M1, p1, theta1, theta2 )

beta1 = beta_calc(M1, theta1, 10*pi/180);

Mn1 = M1*sin(beta1);


M2 = Mn2/sin( beta1 - theta1 );

beta2 = beta_calc(M2, theta2, 10*pi/180);

Mn2p = M2*sin(beta2);[presr2, densr2, tempr2, por2 , po22, Mn3 ] = shock(Mn2p);

M3 = Mn3/sin( beta2 - theta2 );press3 = p1*presr1*presr2;

reg_ref.mIn order to solve example 4.10 with this function we would issue the MATLAB

command[press3,M3] = reg_ref(2.8, 1, 16*pi/180, 16*pi/180 )

In order to solve example 4.11 with this function we would issue the MATLAB

command

[press3,M3] = reg_ref(3.0, 1, 14*pi/180, 24*pi/180 )

We use this function to solve the regular reflections from the top and bottom boundaries

as part of an iteration. Consider the following example:

Example

A two-dimensional oblique shock diffuser ( ) operates in a

supersonic stream of air with and . Determine the pressures,

the deflection angle and the Mach numbers for the downstream regions 4 and 4.

The resulting solution is: (the initial guess was , positive guesses resulted in

divergence)


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Following is the program opp_shock.m to solve this

% input parameters - user provides a guess for phi

M1 = 2.3;p1 = 101.3;theta2 = 5*pi/180;theta3 = 15*pi/180;phi = -5phi = phi*pi/180;

dphi = 0.01;delp = 1.0;

while abs(delp) > 1.0e-12,phir = phi + dphi;theta4 = phir + theta3;theta4p = theta2 - phir;

[p4p, M4p] = reg_ref(M1, p1, theta2, theta4p );[p4 , M4 ] = reg_ref(M1, p1, theta3, theta4 );funcr = p4-p4p;

phil = phi - dphi;theta4 = phil + theta3;theta4p = theta2 - phil;

[p4p , M4p] = reg_ref(M1, p1, theta2, theta4p );[p4 , M4 ] = reg_ref(M1, p1, theta3, theta4 );funcl = p4-p4p;

dfdphi = ( funcr - funcl )/(2.0*dphi);

theta4 = phi + theta3;theta4p = theta2 - phi;[p4p , M4p] = reg_ref(M1, p1, theta2, theta4p );[p4 , M4 ] = reg_ref(M1, p1, theta3, theta4 );func = p4-p4p;

delp = -func/dfdphi;

phi = phi + delp;end

theta4 = phi + theta3;theta4p = theta2 - phi;

[p4p , M4p] = reg_ref(M1, p1, theta2, theta4p );[p4 , M4 ] = reg_ref(M1, p1, theta3, theta4 );

sprintf('p4 = %0.5g',p4)sprintf('p4p = %0.5g',p4p)sprintf('M4 = %0.5g',M4)sprintf('M4p = %0.5g',M4p)sprintf('phi = %0.5g',phi*180/pi)

opp_shock.m


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VI. Intersection of Shocks of the Same FamilyConsider the situation shown below. In this case we have two changes in geometry

that turn the flow into itself.

It is possible to find solutions for this geometry by enforcing the conditions4 5

P P and

4 5 . The solution procedure would be similar to that used in the previous section.

VII. Prandtl-Meyer Expansion WavesConsider the situation shown below where the geometry causes the flow to turn away

from itself.

Such a situation is termed an expansion wave and the behavior of expansion waves is

antithetical to the behavior of shock waves in the following ways:


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1. Expansion waves are continuous2.

2 1M M

3. 2 1 2 1 2 1P P T T 4. Streamlines through an expansion wave are smooth curves5. Expansion waves are isentropic

Consider the following figure depicting the expansion through the angle d andassuming the tangential component of velocity is continuous through the expansion:

From the law of sines we have

sinV dV 2

Vs in d

2

(5.1)

But

sin cos2

(5.2)And

s in d cos cos d sin sind 2 (5.3)

Equation (5.1) becomes

V dV dV cos 11

V V cos d sin 1 d ta

n

(5.4)


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Recall that for small x

2 311 x x x ... 1 x

1 x (5.5)

thus

dV1 1 d ta

Vn (5.6)

Recall that the Mach angle is defined as

1 1sinM

(5.7)or

2

1tan

M 1

(5.8)

This leads to

2 dVd M 1

V (5.9)

This expression is approximate for finite d and exact asd 0 . For a caloricallyperfect gas this may be integrated to obtain (note that there will be a change in sign

convention in later chapters reminder to me).

(5.10) 21

2 2d M

1M

Where is the Prandtl-Meyer function given byM

1 2 1 21 1M tan M 1 tan M 1 1

1 (5.11)

Given and1M 2 the calculation steps are

1. Calculate 1M 2. 2 1M M 2 3. Calculate

2M by iterating equation (5.11)

4. Use the isentropic relationships to find2

2

1

221

11 M

T 2

1T1 M

2

(5.12)


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12

2

1

221

11 M

P 2

1P 1 M

2

(5.13)

The Prandtl-Meyer function is tabulated in table A.5 and a MATLAB function called

p_m.m is provided below

function prandtl = p_m( M )

gamm = 1.4;

term1 = sqrt((gamm+1.0)/(gamm-1.0));term2 = sqrt((M^2 - 1.0)*(gamm-1.0)/(gamm+1.0));

result = term1*atan(term2) - atan(sqrt(M^2 - 1.0));

prandtl = result;

p_m.m

For the case where the Prandtl-Meyer function is known then equation (5.11) must be

iterated to calculate the Mach number. The Matlab function Inv_p_m.m does this

function mach = inv_p_m(prandtl, mguess)

% secant method to get mach # given area ratio

mo = mguess;dm = mo/1000.;delm = 1.0;

while abs(delm) > .0001,

mr = mo+dm;ml = mo-dm;dadm = ( p_m(mr) - p_m(ml) )/(2*dm);delm = -(p_m(mo)-prandtl)/dadm;mo = mo + delm;

end

mach = mo;

inv_p_m.m

Finally, the first part of example 4.13 is given in the MATLAB file ex_4_13.m

% example 4.13M1 = 1.5;mu1 = asin(1/M1);


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sprintf('mu1 = %0.5g',mu1*180/pi)theta2 = 20*pi/180;nu1 = p_m( M1 );

sprintf('nu1 = %0.5g',nu1*180/pi)nu2 = nu1 + theta2;M2 = inv_p_m(nu2,2.0);sprintf('M2 = %0.5g',M2)

mu2 = asin(1/M2);sprintf('mu2 = %0.5g',mu2*180/pi)

ex_4_13.m

two dimensional oblique shocks expansion flows

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