Supersonic Flow Over a Wedge(Oblique Shock Problem)

-By Sameer Kadam M.Sc Computational Mechanics

 Problem Statement Statement Consider a supersonic flow over a wedge (jet engine intake?) is to be analyzed. The shape,  given in Fig. 1, consists of two compression  corners of 6 and 10 degrees and 4 4 degree expansion corners. The unit of the t he x-coordinate is 0.1 m (10 cm). The free streem conditions are: 1. Mach 1.  Mach number M = 3, the ambient ambient pressure  pa = 0.101325 0.101325 MPa and T = 300 K .  2. Mach  Mach number M = 4, 4, the ambient pressure  pa = 0.101325 0.101325 MPa and T = 300 K .

 Problem Statement Statement Consider a supersonic flow over a wedge (jet engine intake?) is to be analyzed. The shape,  given in Fig. 1, consists of two compression  corners of 6 and 10 degrees and 4 4 degree expansion corners. The unit of the t he x-coordinate is 0.1 m (10 cm). The free streem conditions are: 1. Mach 1.  Mach number M = 3, the ambient ambient pressure  pa = 0.101325 0.101325 MPa and T = 300 K .  2. Mach  Mach number M = 4, 4, the ambient pressure  pa = 0.101325 0.101325 MPa and T = 300 K .

Sr.No

Topic

Page No.

1

Approach

1

2

Calculations Calculati ons for Flow With Mach Number = 3

5

3

Calculations Calculati ons for Flow With Mach Number = 4

11

4

Calculation of Entropies and Enthalpies

17

5

Verification of Results Using Flow Simulating Software Star CCM

18

Comparison of Analytical and Numerical Results

19

7

Appendix Appendi x 1

20

6

Appendix Appendi x 2

21

6

Approach o

The given problem is a case of oblique shock as the wavefront is at an angle other than 90 to the approaching flow. It can be divided into two parts basically, 1. Analysis of Compression Corners 2. Analysis of Expansion Corners

1. Analysis of Compression Corners: o

o

There are two compression corners having angles 6 and 10 respectively. Let us denote these compression angle or deflection angles as ϴ. Consider a flow with Mach number; M1 is approaching a compression corner. It results in a oblique shock which can be visualized with the help of the given figure:

Properties of an Oblique Shock Wave

Here we can see the properties of an oblique shock wave. The approaching flow is defl ected by angle ϴ . Here β is called the wave angle and for a Normal Shock it is 90o. We split the approaching flow into two components and hence we obtain the Normal Component of mach number Mn1 & the tangential component as Mt1. Similarly we have Mn2 Mt2 as the Normal and tangential components of  the mach number of the flow after shock. Where Mn1 is given by M1/sin efer the Oblique Shock Charts ( Appendix 1) which give the relation In order to find angle β we r efer between the shock wave angle and flow deflection angle for various values of upstream mach numbers. As we know the values of upstream Mach number M 1 and the flow deflection angle ϴ we can find the corresponding value of β .

Now, it can be solved as problem of  Normal Shock. Using the basic equations for continuity, energy and momentum we can obtain the relations for between the various parameters of the upstream and downstream flows. We start with the continuity equation ρ1v1= ρ2v2

From which we get the relation ρ1M1 /sqrt(T1)= ρ2M2 /sqrt(T2).................................... ......................................... ..... (1) Analytical and Numerical Methods for Oblique Shock Problems

Page 1

The stagnation enthalpy remains constant over the flow ht1=ht2 but since enthalpy is a function of temperature only Tt1=Tt2 (γ-1)*M2 /2) Tt= T*(1+ (γ-1)*M 2 2 T1 *(1+ (γ-1)*M (γ-1)*M1  /2) = T2 *(1+ (γ-1)*M (γ-1)*M2  /2) ................................. ......................................... ........ (2) Using the Momentum Equation p1*(1+ γ*M 12) = p2*(1+ γ*M 22) .............................. ......................................... ........... (3)

Equations 1,2,3 are the governing equations for the Normal shock. Empressing Empressing all the t he relations in terms of M1 we get the Normal Shock tables( Appendix 2) These tables provide the various values of pressure ratios, temperature ratios, density ratios, mach number downstream, etc. Using these Normal Shock tables (Appendix 2) we obtain the values for pressure ratio (p 2/ p1), temperature ratio (T2 /T1) and Mach number after shock. Note that the static pressure and temperature are the same whether we are talking about Normal Shock or Oblique Shock. Moreover the value of Mach number which we obtain for the flow after the shock is the Normal Component of M2 i.e Mn2. M can be obtained by using the relation. M2= M2n/sin(β/sin(β-ϴ) The strength of shock can be calculated by using the pressure ratio. The strength of shock is given by (p2-p1)/p1 The change in Entropy for any ideal gas in terms of pressure ratio and density ratio is given by the relation: ∆s = Cp * ln(T2 /T1) - R * ln(p 2 /p1) ................................... ......................................... ...... (4) Note that the entropy changes tend to be very small for oblique shocks as the entropy changes are directly proportional to the cube of the flow deflection angle which is very small in this case. The change in Enthalpy is given by ∆h= Cp∆t..................................... ......................................................... .......................... ...... (5)

Analytical and Numerical Methods for Oblique Shock Problems

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2. Analysis of Expansion Expansion Corners: o

There are four expansion corners each of them having an angle of 4 . In a shock wave the pressure, density and temperature increase. In an expansion wave it is exactly opposite: they all decrease. The analysis of expansion corner is different as compared to compression Prandtl-Meyer Function(ν). corner. Here we use the Prandtl-Meyer It is defined as the angle through which a flow with a Mach M ach number = 1 is turned isentropically to achieve the indicated Mach number.

 ν

Illustration of Prandtl Meyer Function

The Prandtl-Meyer Function can be obtained from the Isentropic Charts or the Normal Shock Charts(Appendix 2). Otherwise it can be calculated by using the relationship

Thus for calculating the Prandtl-Meyer Function we require only the upstream mach number. Based on this we can calculate the Mach number of the downstream flow as follows: 

ise ntropic charts Find ν1 for the upstream mach number from isentropic



The flow turns by the flow deflection angle ϴ. Add this value of ϴ to ν 1 which would indicate the Prandtl Meyer Function, ν 2 for the downstream mach number.



i.e ν2- ν1 = ϴ



From ν2 we can obtain the corresponding down stream mach number from the isentropic charts.

Similarly using the Isentropic charts we can find the other values for the pressure ratio and temperature ratio. The Cahnge in Enthalpy and Entropy can also be found out similarly as in case of  Compression Corners using eqns 4 and 5 Note that the change in entropy and enthalpy is negative in this case However, we are also required to find the shock wave angle β, in order to determine the location of shock. Analytical and Numerical Methods for Oblique Shock Problems

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For this we derive an equation based on eq 1,2,3

...................................... .................................................(from ...........(from eqn 1)

....................................(f ....................................(from rom eqn 2)

.................................... .............................................(from .........(from eqn 3) Substituting the last two eqns in the first eqn we get a relation between M 1 and M2 as follows:

.........................(eqn .........................(eqn 6) Substituting the value of M 2 in eqn 3 we have a relation between the pressure in terms of M 1 p2 /p1 = 2γ/(γ-1) 2γ/(γ-1) * M12 – (γ  – (γ--1) /(γ+1).............................. /(γ+1)......................................(eqn ........(eqn 7) Where M1 is the Normal Component of the Upstream Flow Thus for Expansion Corner it gets modified to 2

2

p2 /p1 = 2γ/(γ-1) 2γ/(γ-1) * M1 sin β – (γ  – (γ--1) /(γ+1)..................... /(γ+1)..................................(eqn .............(eqn 8) The pressure ratio can be found out from the isentropic charts and thus we can calculate the wave angle β. -1 Note that after taking sin we will have to consider the negative value of the angle, since it is a case of expansion.

M1 -M2

-β

Wave Angle for an Expansion Shock  Negative Angle is Traced Clockwise

Analytical and Numerical Methods for Oblique Shock Problems

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Analytical Calculations for Mach 3 Given free stream conditions are M1 = 3 P1 = 1.01325 bar T1 = 300 K Compression Curves = 6o and 10o Expansion Curves = four 4 o curves 1. Compression Corner of 6o o M1 = 3; P1 = 1.01325 bar; T 1 = 300 K; K; ϴ1 = 6 Location of Shock and Normal Mach Number Refer the Oblique Shock Charts ( Appendix 1). 1). We obtain for mach number 3 and ϴ = 6o the value of the shock wave angle as β 1 as 24o 

Mn1 = M1 * sin ϴ

n1 

= 3sin(24)

Mn1 = 1.22

Refer the Normal Shock tables (Appendix 2). We obtain for mach number 1.22 the following values of pressure ratio, temperature ratio and downstream mach number p2 /p1 = 1.570; T2 /T1 = 1.141; Mn2 = 0.8300 

p2 = p1 * 1.570;

T2 = T1 * 1.141;



p2 = 1.0325 * 1.570; T 2 = 300 * 1.141;



p2 = 1.59 bar;

T2 = 342.3 K;

M2 = Mn2/sin(β/sin(β-ϴ) M2 = 0.8300/sin(24-6) 0.8300/sin(24-6) M2 = 2.685

Strength of Shock is given by  – 1 (p2 - p1) /p  /p1 or (p2 /p1) – 1

1.570 –  1.570 – 1 1 = 0.570



Change in Entropy is given by ∆s = Cp * ln(T2 /T1) - R * ln(p 2 /p1) 

∆s = 1005 * ln(1.141)  – 287  – 287 * ln(1.570)



∆s = 3.10 J/Kg K 

Change in Enthalpy is given by ∆h= Cp∆t  – T1) ∆h = 1005 * (T2 – T (342.3 – 300) 300) ∆h = 1005 * (342.3 –  ∆h = 42511.5 J/Kg K 

Analytical and Numerical Methods for Oblique Shock Problems

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o

2. Compression Corner of 10 o M2 = 2.685; P2 = 1.59 bar; T 2 = 342 K; K; ϴ2 = 10 Location of Shock and Normal mach Number Refer the Oblique Shock Charts ( Appendix 1). We obtain for mach number 2.685 and o o ϴ = 10 the value of the shock wave angle as β 1 as 30 

Mn2 = M1 * sin ϴ

n2 

= 2.685sin(30)

Mn2 = 1.3425

Refer the Normal Shock tables (Appendix 2). We obtain for mach number 1.3425 the following values of pressure ratio, temperature ratio and downstream mach number p3 /p2 = 1.928; T3 /T2 = 1.216; Mn3 = 0.7664 

p3 = p2 * 1.928;

T3 = T2 * 1.216;

/sin(β-ϴ) M3 = Mn3/sin(β-



p3 = 1.59 * 1.928;

T 3 = 342 * 1.216;

M3 = 0.7664/sin(30-10) 0.7664/sin(30-10)



p3 = 3.06 bar;

T3 = 416.23 K;

M3 = 2.25

Strength of Shock is given by  – 1 (p3 /p2) – 1

1.928 –  1.928 – 1 1 = 0.928



Change in Entropy is given by ∆s = Cp * ln(T3 /T2) - R * ln(p 3 /p2) 

∆s = 1005 * ln(1.216)  – 287  – 287 * ln(1.928)



∆s = 8.14 J/Kg K

Change in Enthalpy is given by ∆h= Cp∆t

∆h = 1005 * (T3 – T  – T2) (416.23 – 342.3) 342.3) ∆h = 1005 * (416.23 –  ∆h = 74290 J/Kg K

Analytical and Numerical Methods for Oblique Shock Problems

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3. Expansion Corner of 4 o M3 = 2.25; P3 = 3.06 bar; T3 = 416.23 K; K; ϴ3 = 4 Prandtl-Meyer Function Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach number 2.25 the value of Prandtl- Meyer Function ν Function ν3 = 33.018  ν4 =

ν3 + ϴ3

 ν4  ν

= 33.018 + 4

 ν4

= 37.018

o



Refer the Normal Shock tables (Appendix 2). We obtain for ν for ν4 = 37.018o the value of  corresponding mach number M 4 as 2.41. Referring the same following values of  pressure ratio and temperature ratio are obtained as follows p4 /p3 = p4 /p4t * p4t /p3t * p3t /p  /p3; T4 /T3 = T4 /T4t * T4t /T  /T3t * T3t /T3; Since the flow is isentropic and no work is done, the stagnation temperature is constant (T t = constant). Also there are no losses in the flow, hence the stagnation pressure is also constant (pt = constant) 

p4 /p3 = 0.06734 * 1 * (0.08648) -1;

T4 /T3 = 0.46262 * 1 * (0.49689) -1



p4 /p3 = 0.7786;

T4 /T3 = 0.931



p4 = 0.7786 * 3.06;

T 4 = 0.931 * 416.23



p4 = 2.382516 bar;

T4 = 387.52 K

Strength of Shock is given by (p4 /p3) – 1  – 1

0.7786 - 1 = -0.2214



Change in Entropy is given by ∆s = Cp * ln(T4 /T3) - R * ln(p 4 /p3) 

∆s = 1005 * ln(0.931)  – 287  – 287 * ln(0.7786)



∆s = -0.02948 J/Kg K

Change in Enthalpy is given by ∆h= Cp∆t  – T3) ∆h = 1005 * (T4 – T

∆h = 1005 * (387.52 - 416.23) ∆h = -28853 J/Kg K Location of Shock β 2γ/(γ-1) * Mi2 sin2β – (γ – (γ--1) /(γ+1)............... /(γ+1).................................... pii /pi = 2γ/(γ-1) ......................................... ....................................(eqn ................(eqn 8) 2 2 – (1.4-1)/(1.4+1) 0.7786 = 2*1.4/(1.4-1) * 2.25 sin β – (1.4-1)/(1.4+1) 

β = sin-1(0.4);



β = -23.58 ……………………………(consider –ve value, since it’s a case of o f expansion)



β = +23.58o or -23.58o

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Analytical and Numerical Methods for Oblique Shock Problems

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4. Expansion Corner of 4 o M4 = 2.41; P4 = 2.382 bar; T 4 = 387.52 K; K; ϴ4 = 4 Prandtl-Meyer Function Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach number 2.41 the value of o f Prandtl- Meyer Function ν 4 = 37.018  ν5 =

ν4 + ϴ4

 ν4  ν

= 37.018 + 4

 ν4

= 41.018

o



Refer the Normal Shock tables (Appendix 2). We obtain for ν for ν5 = 41.018o the value of  corresponding mach number M 5 as 2.54. Referring the same following values of  pressure ratio and temperature ratio are obtained as follows p5 /p4 = p5 /p5t * p5t /p4t * p4t /p  /p4; T5 /T4 = T5 /T5t * T5t /T  /T4t * T4t /T4; Since the flow is isentropic and no work is done, the stagnation temperature is constant (T t = constant). Also there are no losses in the flow, hence the stagnation pressure is also constant (pt = constant) 

p5 /p4 = 0.0550 * 1 * (0.06734) -1;

T5 /T4 = 0. 43662 * 1 * (0.46262) -1



p5 /p4 = 0.8167;

T5 /T4 = 0.9437



p5 = 0.8167*2.382; 0.8167*2.382 ;

T4 = 0.9437 * 387.52



p5 = 1.94 bar;

T5 = 365.74 K

Strength of Shock is given by (p5 /p4) – 1  – 1

0.8167 - 1 = -0.1833



Change in Entropy is given by ∆s = Cp * ln(T5 /T4) - R * ln(p 5 /p4) 

∆s = 1005 * ln(0.9437) –  ln(0.9437)  – 287 287 * ln(0.8167)



∆s = -0.1239 J/Kg K

Change in Enthalpy is given by ∆h= Cp∆t  – T4) ∆h = 1005 * (T5 – T

∆h = 1005 * (365.74 - 387.52) ∆h = -21889 J/Kg K Location of Shock β 2γ/(γ-1) * Mi2 sin2β – (γ – (γ--1) /(γ+1)............... /(γ+1).................................... pii /pi = 2γ/(γ-1) ......................................... ....................................(eqn ................(eqn 8) 2 2 – (1.4-1)/(1.4+1) 0.8167 = 2*1.4/(1.4-1) * 2.41 sin β – (1.4-1)/(1.4+1) 

β = sin-1(0.3809);



β = -22.39 ……………………………(consider –ve value, since it’s a case of o f expansion)



β = +22.39 +22.39o or -22.39o

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Analytical and Numerical Methods for Oblique Shock Problems

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5. Expansion Corner of 4 o M5 = 2.54; p5 = 1.94 bar; T 5 = 365.74 K; K; ϴ5 = 4 Prandtl-Meyer Function Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach number 2.54 the value of o f Prandtl- Meyer Function ν 5 = 41.018  ν6 =

ν5 + ϴ5

 ν6  ν

= 41.018 + 4

 ν6

= 45.018

o



Refer the Normal Shock tables (Appendix 2). We obtain for ν for ν6 = 45.018o the value of  corresponding mach number M 6 as 2.77. Referring the same following values of  pressure ratio and temperature ratio are obtained as follows p6 /p5 = p6 /p6t * p6t /p5t * p5t /p  /p5; T6 /T5 = T6 /T6t * T6t /T  /T5t * T5t /T5; Since the flow is isentropic and no work is done, the stagnation temperature is constant (T t = constant). Also there are no losses in the flow, hence the stagnation pressure is also constant (pt = constant) 

p6 /p5 = 0.03858 * 1 * (0.0550) -1;

T6 /T5 = 0.3954 * 1 * (0.4366) -1



p6 /p5 = 0.7014;

T6 /T5 = 0.9036



p6 = 0.7014*1.94; 0.7014*1.94 ;

T6 = 0.9036 * 365.74



p6 = 1.36 bar;

T6 = 330.4 K

Strength of Shock is given by (p6 /p5) – 1  – 1

0.7014 - 1 = -0.2986



Change in Entropy is given by ∆s = Cp * ln(T6 /T5) - R * ln(p 6 /p5) 

∆s = 1005 * ln(0.9036) –  ln(0.9036)  – 287 287 * ln(0.7014)



∆s = -0.08305 J/Kg K

Change in Enthalpy is given by ∆h= Cp∆t  – T5) ∆h = 1005 * (T6 – T (330 – 365.74) 365.74) ∆h = 1005 * (330 –  ∆h = -35516.7 J/Kg K Location of Shock β 2γ/(γ-1) * Mi2 sin2β – (γ – (γ--1) /(γ+1)............... /(γ+1).................................... pii /pi = 2γ/(γ-1) ......................................... ....................................(eqn ................(eqn 8) 2 2 – (1.4-1)/(1.4+1) 0.7014 = 2*1.4/(1.4-1) * 2.54 sin β – (1.4-1)/(1.4+1) 

β = sin-1(0.34054);



β = -19.91 ……………………………(consider –ve value, since it’s a case of o f expansion)



β = +19.91 +19.91o or -19.91o

o

Analytical and Numerical Methods for Oblique Shock Problems

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6. Expansion Corner of 4 o M6 = 2.77; p6 = 1.36 bar; T 6 = 330.4 K; K; ϴ6 = 4 Prandtl-Meyer Function  ν7 =

ν6 + ϴ6

 ν7  ν

= 45.018 + 4………………………………………………………(From 4 ………………………………………………………(From Appendix 2)

 ν7

= 49.018

o



Refer the Normal Shock tables (Appendix 2). for ν for ν7 = 49.018o, mach number M 7 as 2.97. values of pressure ratio and temperature ratio are obtained as follows p7 /p6 = p7 /p7t * p7t /p6t * p6t /p  /p6; T7 /T6 = T7 /T7t * T7t /T  /T6t * T6t /T6; Tt = constant and p t = constant -1

p7 /p6 = 0.02848 * 1 * (0.03858) ;

T7 /T6 = 0.36177 * 1 * (0.39454)



p7 /p6 = 0.7382;

T7 /T6 = 0.917



p7 = 0.7382*1.36; 0.7382*1.36 ;

T6 = 0.917 * 330.4



p7 = 1.004 bar;

T7 = 302.95 K



-1

Strength of Shock is given by  – 1 (p7 /p6) – 1

0.7382 - 1 = -0.2618



Change in Entropy is given by ∆s = Cp * ln(T7 /T6) - R * ln(p 7 /p6) 

∆s = 1005 * ln(0.917)  – 287  – 287 * ln(0.7382)



∆s = -0.03504 J/Kg K

Change in Enthalpy is given by ∆h= Cp∆t  – T6) ∆h = 1005 * (T7 – T

∆h = 1005 * (302.95 - 330) ∆h = -27185.25 J/Kg K Location of Shock β 2γ/(γ-1) * Mi2 sin2β – (γ – (γ--1) /(γ+1)............... /(γ+1)................................ pii /pi = 2γ/(γ-1) ................................... ...................................... ......................(eqn ..(eqn 8) 2 2 – (1.4-1)/(1.4+1) 0.7382 = 2*1.4/(1.4-1) * 2.77 sin β – (1.4-1)/(1.4+1) -1

o

β = sin (0.3380);



β = -19.76 ……………………………(consider –ve value, since it’s a case of o f expansion)



β = +19.76 +19.76 or -19.76

o



o

Analytical and Numerical Methods for Oblique Shock Problems

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Analytical Calculations for Mach 4 Given free stream conditions are M1 = 4 P1 = 1.01325 bar T1 = 300 K Compression Curves = 6o and 10o Expansion Curves = four 4 o curves 1. Compression Corner of 6o o M1 = 4; P1 = 1.01325 bar; T 1 = 300 K; K; ϴ1 = 6 Location of Shock and Normal Mach Number Refer the Oblique Shock Charts ( Appendix 1). 1). We obtain for mach number 3 and ϴ = 6o the value of the shock wave angle as β 1 as 19o 

Mn1 = M1 * sin ϴ

n1 

= 4sin(19)

Mn1 = 1.3022

Refer the Normal Shock tables (Appendix 2). We obtain for mach number 1.3022 the following values of pressure ratio, temperature ratio and downstream mach number p2 /p1 = 1.806; T2 /T1 = 1.191; Mn2 = 0.7866 

p2 = p1 * 1.806;

T2 = T1 * 1.191;



p2 = 1.0325 * 1.806; T 2 = 300 * 1.191;



p2 = 1.83 bar;

T2 = 357.3 K;

M2 = Mn2/sin(β/sin(β-ϴ) M2 = 1.3022/sin(19-6) 1.3022/sin(19-6) M2 = 3.5

Strength of Shock is given by  – 1 (p2 - p1) /p  /p1 or (p2 /p1) – 1

1.806 –  1.806 – 1 1 = 0.806



Change in Entropy is given by ∆s = Cp * ln(T2 /T1) - R * ln(p 2 /p1) 

∆s = 1005 * ln(1.191)  – 287  – 287 * ln(1.806)



∆s = 6.0174 J/Kg K

Change in Enthalpy is given by ∆h= Cp∆t  – T1) ∆h = 1005 * (T2 – T (357.3 – 300) 300) ∆h = 1005 * (357.3 –  ∆h = 57580 J/Kg K

Analytical and Numerical Methods for Oblique Shock Problems

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2. Compression Corner of 10 o M2 = 3.5; P2 = 1.83 bar; T 2 = 357.3 K; K; ϴ2 = 10 Location of Shock Refer the Oblique Shock Charts ( Appendix 1). We obtain for mach number 3.5 and ϴ = 10o the value of the shock wave angle as β 2 as 24.5o 

Mn2 = M1 * sin ϴ

n2 

= 3.5sin(24.5)

Mn2 = 1.45

Refer the Normal Shock tables (Appendix 2). We obtain for mach number 1.45 the following values of pressure ratio, temperature ratio and downstream mach number p3 /p2 = 2.286; T3 /T2 = 1.287; Mn3 = 0.7196 

p3 = p2 * 2.286;

T3 = T2 * 1.287;

/sin(β-ϴ) M3 = Mn3/sin(β-



p3 = 1.83 * 2.286;

T 3 = 357.3 * 1.287;

M3 = 1.45/sin(24.5-10)



p3 = 4.1 bar;

T3 = 459.87 K;

M3 = 2.874

Strength of Shock is given by  – 1 (p3 /p2) – 1

2.286 –  2.286 – 1 1 = 1.286



Change in Entropy is given by ∆s = Cp * ln(T3 /T2) - R * ln(p 3 /p2) 

∆s = 1005 * ln(1.287)  – 287  – 287 * ln(2.286)



∆s = 16.28 J/Kg K

Change in Enthalpy is given by ∆h= Cp∆t

∆h = 1005 * (T3 – T  – T2) (459.87 – 357.3) 357.3) ∆h = 1005 * (459.87 –  ∆h = 103068 J/Kg K

Analytical and Numerical Methods for Oblique Shock Problems

Page 12

3. Expansion Corner of 4o o M3 = 2.874; P3 = 4.1 bar; T3 = 459.87 K; K; ϴ3 = 4 Prandtl-Meyer Function Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach number 2.874 the value of Prandtl- Meyer Function ν3 = 47.18523  ν4 =

ν3 + ϴ3

 ν4  ν

= 47.18523 + 4

 ν4

= 51.18523o o

Refer the Normal Shock tables (Appendix 2). We obtain for ν for ν4 = 51.18523 the value of  corresponding mach number M 4 as 3.07. Referring the same following values of  pressure ratio and temperature ratio are obtained as follows p4 /p3 = p4 /p4t * p4t /p3t * p3t /p3; T4 /T3 = T4 /T4t * T4t /T3t * T3t /T3; Since the flow is isentropic and no work is done, the stagnation temperature is constant (T t = constant). Also there are no losses in the flow, hence the stagnation pressure is also constant (pt = constant) -1



p4 /p3 = 0.02453 * 1 * (0.03312) ;

T4 /T3 = 0.34810 * 1 * (0.3773)



p4 /p3 = 0.7426;

T 4 /T3 = 0.918



p4 = 0.7426 * 4.10;

T 4 = 0.918 * 459.87



p4 = 3.035000 bar;

T4 = 424.279 K

-1

Strength of Shock is given by (p4 /p3) – 1  – 1

0.7426 - 1 = -0.2626



Change in Entropy is given by ∆s = Cp * ln(T4 /T3) - R * ln(p 4 /p3) 

∆s = 1005 * ln(0.918)  – 287  – 287 * ln(0.7426)



∆s = -0.57512 J/Kg K

Change in Enthalpy is given by ∆h= Cp∆t

∆h = 1005 * (T4 – T  – T3) (424.279 – 459.87) 459.87) ∆h = 1005 * (424.279 –  ∆h = -35591 J/Kg K Location of Shock β pii /pi = 2γ/(γ-1) 2γ/(γ-1) * Mi2 sin2β – (γ – (γ--1) /(γ+1)............... /(γ+1)................................ ................................... ...................................... ......................(eqn ..(eqn 8) 2 2 – (1.4-1)/(1.4+1) 0.7426 = 2*1.4/(1.4-1) * 2.874 sin β – (1.4-1)/(1.4+1) Analytical and Numerical Methods for Oblique Shock Problems

Page 13



β = sin-1(0.3067);



β = +17.86 +17.86o or -17.86o

o

o f expansion) β = -17.86 ……………………………(consider –ve value, since it’s a case of o 4. Expansion Corner of 4 o M4 = 3.07; P4 = 3.035 bar; T 4 = 424.279 K; K; ϴ4 = 4



Prandtl-Meyer Function Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach number 2.41 the value of o f Prandtl- Meyer Function ν 4 = 51.18523  ν5 =

ν4 + ϴ4

 ν4  ν

= 51.18523 + 4

 ν4

= 55.18523o o

Refer the Normal Shock tables (Appendix 2). We obtain for ν for ν5 = 55.18523 the value of  corresponding mach number M 5 as 3.9. Referring the same following values of pressure ratio and temperature ratio are obtained as follows p5 /p4 = p5 /p5t * p5t /p4t * p4t /p4; T5 /T4 = T5 /T5t * T5t /T4t * T4t /T4; Since the flow is isentropic and no work is done, the stagnation temperature is constant (T t = constant). Also there are no losses in the flow, hence the stagnation pressure is also constant (pt = constant) -1



p5 /p4 = 0.01773 * 1 * (0.02452) ;

T5 /T4 = 0. 31597 * 1 * (0.34810)



p5 /p4 = 0.72308;

T 5 /T4 = 0.907



p5 = 0.72308*3.07; 0.72308*3.07 ;

T4 = 0.907 * 424.279



p5 = 2.19455 bar;

-1

T5 = 385.11 K

Strength of Shock is given by  – 1 (p5 /p4) – 1

0.72308 - 1 = -0.277



Change in Entropy is given by ∆s = Cp * ln(T5 /T4) - R * ln(p 5 /p4) 

∆s = 1005 * ln(0.907)  – 287  – 287 * ln(0.72308)



∆s = -5.04 J/Kg K

Change in Enthalpy is given by ∆h= Cp∆t

∆h = 1005 * (T5 – T  – T4) (385.11 – 424.279) 424.279) ∆h = 1005 * (385.11 –  ∆h = -39364.845 J/Kg K Location of Shock β pii /pi = 2γ/(γ-1) 2γ/(γ-1) * Mi2 sin2β – (γ – (γ--1) /(γ+1)............... /(γ+1)................................ ......................................... ........................................(eqn ................(eqn 8) 2 2 – (1.4-1)/(1.4+1) 0.72308 = 2*1.4/(1.4-1) * 3.07 sin β – (1.4-1)/(1.4+1) Analytical and Numerical Methods for Oblique Shock Problems

Page 14



β = sin-1(0.2843);



β = +16.52 +16.52o or -16.52o

o

o f expansion) β = -16.52 ……………………………(consider –ve value, since it’s a case of o 5. Expansion Corner of 4 o M5 = 3.29; p5 = 2.19455 bar; T 5 = 385.11 K; K; ϴ5 = 4



Prandtl-Meyer Function Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach number 3.29 the value of o f Prandtl- Meyer Function ν 5 = 55.18523  ν6 =

ν5 + ϴ5

 ν6  ν

= 55.18523 + 4

 ν6

= 59.18523o o

Refer the Normal Shock tables (Appendix 2). We obtain for ν for ν6 = 59.18523 the value of  corresponding mach number M 6 as 3.54. Referring the same following values of  pressure ratio and temperature ratio are obtained as follows p6 /p5 = p6 /p6t * p6t /p5t * p5t /p5; T6 /T5 = T6 /T6t * T6t /T5t * T5t /T5; Since the flow is isentropic and no work is done, the stagnation temperature is constant (T t = constant). Also there are no losses in the flow, hence the stagnation pressure is also constant (pt = constant) -1



p6 /p5 = 0.01239 * 1 * (0.01773) ;

T6 /T5 = 0.28520 * 1 * (0.31597)



p6 /p5 = 0.7000;

T6 /T5 = 0.9026



p6 = 0.7000*2.19455; 0.7000*2.194 55;

T6 = 0.9026 * 385.11



p6 = 1.46 bar;

T6 = 347.06 K

-1

Strength of Shock is given by  – 1 (p6 /p5) – 1

0.7000 - 1 = -0.3000



Change in Entropy is given by ∆s = Cp * ln(T6 /T5) - R * ln(p 6 /p5) 

∆s = 1005 * ln(0.9026)  – 287  – 287 * ln(0.7000)



∆s = -0.62246 J/Kg K

Change in Enthalpy is given by ∆h= Cp∆t

∆h = 1005 * (T6 – T  – T5) (347.06 – 384.11) 384.11) ∆h = 1005 * (347.06 –  ∆h = -38240.25 J/Kg K Location of Shock β pii /pi = 2γ/(γ-1) 2γ/(γ-1) * Mi2 sin2β – (γ – (γ--1) /(γ+1)............... /(γ+1)................................ ................................... ...................................... ......................(eqn ..(eqn 8) 2 2 – (1.4-1)/(1.4+1) 0.7000 = 2*1.4/(1.4-1) * 3.29 sin β – (1.4-1)/(1.4+1) Analytical and Numerical Methods for Oblique Shock Problems

Page 15



β = sin-1(0.2619);



β = +15.187 +15.187o or -15.187o

o

β = -15.187 …………………………(consider – ve ve value, value, since it’s a case of expansion) o 6. Expansion Corner of 4 o M6 = 3.54; p6 = 1.46 bar; T 6 = 347.06 K; K; ϴ6 = 4



Prandtl-Meyer Function  ν7 =

ν6 + ϴ6

 ν7  ν

pendix 2) = 59.18523+ 4……………………………………………………………(Ap 4 ……………………………………………………………(Appendix

 ν7

= 64.18523o

Refer the Normal Shock tables (Appendix 2). for ν for ν7 = 64.18523o, mach number M 7 as 3.81. pressure ratio and temperature ratio are obtained as follows p7 /p6 = p7 /p7t * p7t /p6t * p6t /p  /p6; T7 /T6 = T7 /T7t * T7t /T  /T6t * T6t /T6; Tt = constant; pt = constant -1



p7 /p6 = 0.00851 * 1 * (0.01239) ;

T7 /T6 = 0.25620 * 1 * (0.28520)



p7 /p6 = 0.6868;

T 7 /T6 = 0.8983



p7 = 0.6868*1.46; 0.6868*1.46 ;

T6 = 0.8983 * 347.06



p7 = 1.00279 bar;

T7 = 308.88 K

-1

Strength of Shock is given by (p7 /p6) – 1  – 1

0.6868 - 1 = -0.3132



Change in Entropy is given by ∆s = Cp * ln(T7 /T6) - R * ln(p 7 /p6) 

∆s = 1005 * ln(0.8983)  – 287  – 287 * ln(0.6868)



∆s = -0.2937 J/Kg K

Change in Enthalpy is given by ∆h= Cp∆t

∆h = 1005 * (T7 – T  – T6) ∆h = 1005 * (302.95 - 330) ∆h = -38370.9J/Kg K Location of Shock β pii /pi = 2γ/(γ-1) 2γ/(γ-1) * Mi2 sin2β – (γ – (γ--1) /(γ+1)............... /(γ+1).................................... ......................................... ....................................(eqn ................(eqn 8) 0.6868 = 2*1.4/(1.4-1) * 3.54 2sin2β – (1.4-1)/(1.4+1) – (1.4-1)/(1.4+1) 

β = sin-1(0.2415);



o f expansion) β = -13.98 ……………………………(consider –ve value, since it’s a case of



β = +13.98 +13.98o or -13.98o

o

Analytical and Numerical Methods for Oblique Shock Problems

Page 16

The Total Change in Entropy for Mach 3 ∑∆s = 3.10 + 8.14 -0.2948 –   – 0.08305  – 0.03504 -0.2948 – 0.1239 0.1239 –  0.08305 –  0.03504 = 10.703 J/Kg K

The Total Change in Enthalpy for Mach 3 ∑∆h = 42.511 + 74.290 –  28.853 –   – 35.5167  – 27.185 28.853  – 21.88 21.88 –  35.5167 –  27.185 = 3.716 KJ/Kg K

The Total Change in Entropy for Mach 4 ∑∆s = 6.0174 6.0174 + 16.28 –   – 5.04  – 0.62246  – 0.2937 16.28  – 0.57512 0.57512 –  5.04 –  0.62246 –  0.2937 = 15.76612 J/Kg K

The Total Change in Enthalpy Mach 4 ∑∆h = 57.58 + 103.68 –  35.591 –   – 38.240  – 38.370 35.591  – 39.364 39.364 –  38.240 –  38.370 = 9.695 KJ/Kg K

Analytical and Numerical Methods for Oblique Shock Problems

Page 17

Verification of Results Using Flow Simulating Software Star CCM

Analytical and Numerical Methods for Oblique Shock Problems

Page 18

For Mach 3

For Mach 4

Comparison of Analytical and Numerical Results For Mach 3 At points Analytical Mach Number Numerical Analytical Pressure(bar) Numerical Analytical Temperature(K) Numerical

1 3 3 1.0132 1.0132 300 300

2 2.685 2.74 1.59 1.612 342 343.46

3 2.25 2.3195 3.06 2.98 416.23 411.98

4 2.41 2.5227 2.382 2.46 387.52 393.3

5 2.54 2.698 1.95 1 .95 1.869 365.94 362.15

6 7 2.77 2.97 2.92 2.9632 1.36 1.004 1.269 1.0118 330.4 302.95 324.78 299.86

For Mach 4 At points Analytical Mach Number Numerical Pressure(bar) Temperature(K)

Analytical Numerical Analytical Numerical

1 4 4

2 3.5 3.57

3 2.874 2.874

4 3.07 3.035

5 3.29 3.35

6 3.54 3.62

7 3.81 3.948

1.013 1.013 300 300

1.83 1.786 357.3 361.9

4.1 4.13 459.8 45 9.8 459.7

3.035 3.08 424.27 415.3

2.1945 2.04 385.11 379.76

1.46 1.394 347.06 344.22

1.002 1.003 308.8 299.7

Conclusion From the above results we can see that the analytical and numerical results almost comply with each other although there are small differences between the two values. These differences are mainly because the fact that numerical simulation approximates the governing equations to algebraic equations. This approximation leads to deviation from the analytical values. Other factor resulting into deviation is fineness of meshing of the component. Discretization of the domain is a part of numerical preprocessing. Thus the entire domain is discretized into number of small domains. The discontinuities of the numerical solution over these number of small domains results in deviation. Another important factor is the number of inner iterations selected by the user. The higher the number of inner iterations the more the numerical solution is close to analytical solution Also the entropy change in each region and over the entire is region yields a very small value since the entropy change is directly affected by the cube of flow deflection angle. This was found to be correct after obtaining the values for the change in entropy.

Analytical and Numerical Methods for Oblique Shock Problems

Page 19

Appendix 1

Fig: Variation of Wave Deflection Angle wrt to Flow Deflection Angle Angl e for Various Mach Numbers

Analytical and Numerical Methods for Oblique Shock Problems

Page 20

Appendix 2