Supersonic Flow Over a Wedge(Oblique Shock Problem)

-By

Sameer Kadam

M.Sc Computational Mechanics

Problem 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 x-coordinate

is 0.1 m (10 cm). The free streem conditions are:

1. Mach number M = 3, the ambient pressure

pa = 0.101325 MPa and T = 300 K .

2. Mach number M = 4, the ambient pressure

pa = 0.101325 MPa and T = 300 K .

Sr.No Topic Page No.

1 Approach 1

2 Calculations for Flow With Mach Number = 3 5

3 Calculations 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

6 Comparison of Analytical and Numerical Results

19

7 Appendix 1 20

6 Appendix 2 21

Analytical and Numerical Methods for Oblique Shock Problems Page 1

Approach

The given problem is a case of oblique shock as the wavefront is at an angle other than 90o 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:

There are two compression corners having angles 6o and 10

o 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 deflected 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

In order to find angle β we refer the Oblique Shock Charts ( Appendix 1) which give the relation

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 M1 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)


The stagnation enthalpy remains constant over the flow

ht1=ht2

but since enthalpy is a function of temperature only

Tt1=Tt2

Tt= T*(1+ (γ-1)*M2/2)

T1 *(1+ (γ-1)*M12/2) = T2 *(1+ (γ-1)*M2

2/2) ......................................... (2)

Using the Momentum Equation

p1*(1+ γ*M12) = p2*(1+ γ*M2

2) ......................................... (3)

Equations 1,2,3 are the governing equations for the Normal shock. Empressing all the

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 (p2/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-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(p2/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)


2. Analysis of Expansion Corners:

There are four expansion corners each of them having an angle of 4o. 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

corner. Here we use the Prandtl-Meyer Function(ν).

It is defined as the angle through which a flow with a Mach number = 1 is turned

isentropically to achieve the indicated Mach number.

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:

Find ν1 for the upstream mach number from isentropic charts

The flow turns by the flow deflection angle ϴ. Add this value of ϴ to ν1 which

would indicate the Prandtl Meyer Function, ν2 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.

ν


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

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

....................................(from eqn 2)

.............................................(from eqn 3)

Substituting the last two eqns in the first eqn we get a relation between M1 and M2 as follows:

.........................(eqn 6)

Substituting the value of M2 in eqn 3 we have a relation between the pressure in terms of M1

p2/p1 = 2γ/(γ-1) * M12 – (γ-1) /(γ+1)......................................(eqn 7)

Where M1 is the Normal Component of the Upstream Flow

Thus for Expansion Corner it gets modified to

p2/p1 = 2γ/(γ-1) * M12 sin

2β – (γ-1) /(γ+1)..................................(eqn 8)

The pressure ratio can be found out from the isentropic charts and thus we can

calculate the wave angle β.

Note that after taking sin-1

we will have to consider the negative value of the angle, since

it is a case of expansion.

Wave Angle for an Expansion Shock

Negative Angle is Traced Clockwise

-β

M1

-M2


Analytical Calculations for Mach 3

Given free stream conditions are

M1 = 3

P1 = 1.01325 bar

T1 = 300 K

Compression Curves = 6o and 10

o

Expansion Curves = four 4o curves

1. Compression Corner of 6o

M1 = 3; P1 = 1.01325 bar; T1 = 300 K; ϴ1 = 6o

Location of Shock and Normal Mach Number

Refer the Oblique Shock Charts ( Appendix 1). We obtain for mach number 3 and ϴ =

6o the value of the shock wave angle as β1 as 24

o

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; M2 = Mn2/sin(β-ϴ)

p2 = 1.0325 * 1.570; T2 = 300 * 1.141; M2 = 0.8300/sin(24-6)

p2 = 1.59 bar; T2 = 342.3 K; M2 = 2.685

Strength of Shock is given by

(p2 - p1)/p1 or (p2/p1) – 1

1.570 – 1

= 0.570

Change in Entropy is given by

∆s = Cp * ln(T2/T1) - R * ln(p2/p1)

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

∆s = 3.10 J/Kg K

Change in Enthalpy is given by

∆h= Cp∆t

∆h = 1005 * (T2 – T1)

∆h = 1005 * (342.3 – 300)

∆h = 42511.5 J/Kg K



M2 = 2.685; P2 = 1.59 bar; T2 = 342 K; ϴ2 = 10o

Location of Shock and Normal mach Number

Refer the Oblique Shock Charts ( Appendix 1). We obtain for mach number 2.685 and

ϴ = 10o the value of the shock wave angle as β1 as 30

o

Mn2 = M1 * sin ϴ

n2 = 2.685sin(30)

Mn2 = 1.3425



p3/p2 = 1.928; T3/T2 = 1.216; Mn3 = 0.7664

p3 = p2 * 1.928; T3 = T2 * 1.216; M3 = Mn3/sin(β-ϴ)

p3 = 1.59 * 1.928; T3 = 342 * 1.216; M3 = 0.7664/sin(30-10)

p3 = 3.06 bar; T3 = 416.23 K; M3 = 2.25


(p3/p2) – 1

1.928 – 1

= 0.928


∆s = Cp * ln(T3/T2) - R * ln(p3/p2)

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

∆s = 8.14 J/Kg K


∆h= Cp∆t

∆h = 1005 * (T3 – T2)

∆h = 1005 * (416.23 – 342.3)

∆h = 74290 J/Kg K


3. Expansion Corner of 4o

M3 = 2.25; P3 = 3.06 bar; T3 = 416.23 K; ϴ3 = 4o

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 ν3 = 33.018

ν4 = ν3 + ϴ3

ν4 = 33.018 + 4

ν4 = 37.018o

Refer the Normal Shock tables (Appendix 2). We obtain for ν4 = 37.018o the value of

corresponding mach number M4 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/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)

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; T4 = 0.931 * 416.23

p4 = 2.382516 bar; T4 = 387.52 K


(p4/p3) – 1

0.7786 - 1

= -0.2214


∆s = Cp * ln(T4/T3) - R * ln(p4/p3)

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

∆s = -0.02948 J/Kg K


∆h= Cp∆t

∆h = 1005 * (T4 – T3)

∆h = 1005 * (387.52 - 416.23)

∆h = -28853 J/Kg K

Location of Shock β

pii/pi = 2γ/(γ-1) * Mi2 sin

2β – (γ-1) /(γ+1)........................................................................(eqn 8)

0.7786 = 2*1.4/(1.4-1) * 2.252sin

2β – (1.4-1)/(1.4+1)

β = sin-1

(0.4); β = +23.58o or -23.58

o

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



M4 = 2.41; P4 = 2.382 bar; T4 = 387.52 K; ϴ4 = 4o




ν5 = ν4 + ϴ4

ν4 = 37.018 + 4

ν4 = 41.018o







(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; T4 = 0.9437 * 387.52

p5 = 1.94 bar; T5 = 365.74 K


(p5/p4) – 1

0.8167 - 1

= -0.1833


∆s = Cp * ln(T5/T4) - R * ln(p5/p4)

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

∆s = -0.1239 J/Kg K


∆h= Cp∆t

∆h = 1005 * (T5 – T4)

∆h = 1005 * (365.74 - 387.52)

∆h = -21889 J/Kg K



2β – (γ-1) /(γ+1)........................................................................(eqn 8)

0.8167 = 2*1.4/(1.4-1) * 2.412sin

2β – (1.4-1)/(1.4+1)

β = sin-1

(0.3809); β = +22.39o or -22.39

o




M5 = 2.54; p5 = 1.94 bar; T5 = 365.74 K; ϴ5 = 4o




ν6 = ν5 + ϴ5

ν6 = 41.018 + 4

ν6 = 45.018o







(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; T6 = 0.9036 * 365.74

p6 = 1.36 bar; T6 = 330.4 K


(p6/p5) – 1

0.7014 - 1

= -0.2986


∆s = Cp * ln(T6/T5) - R * ln(p6/p5)

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

∆s = -0.08305 J/Kg K


∆h= Cp∆t

∆h = 1005 * (T6 – T5)

∆h = 1005 * (330 – 365.74)

∆h = -35516.7 J/Kg K



2β – (γ-1) /(γ+1)........................................................................(eqn 8)

0.7014 = 2*1.4/(1.4-1) * 2.542sin

2β – (1.4-1)/(1.4+1)

β = sin-1

(0.34054); β = +19.91o or -19.91

o




M6 = 2.77; p6 = 1.36 bar; T6 = 330.4 K; ϴ6 = 4o


ν7 = ν6 + ϴ6

ν7 = 45.018 + 4………………………………………………………(From Appendix 2)

ν7 = 49.018o

Refer the Normal Shock tables (Appendix 2). for ν7 = 49.018o, mach number M7 as 2.97.

values of pressure ratio and temperature ratio are obtained as follows


Tt = constant and pt = constant

p7/p6 = 0.02848 * 1 * (0.03858)-1

; T7/T6 = 0.36177 * 1 * (0.39454)-1

p7/p6 = 0.7382; T7/T6 = 0.917

p7 = 0.7382*1.36; T6 = 0.917 * 330.4

p7 = 1.004 bar; T7 = 302.95 K


(p7/p6) – 1

0.7382 - 1

= -0.2618


∆s = Cp * ln(T7/T6) - R * ln(p7/p6)

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

∆s = -0.03504 J/Kg K


∆h= Cp∆t

∆h = 1005 * (T7 – T6)

∆h = 1005 * (302.95 - 330)

∆h = -27185.25 J/Kg K



2β – (γ-1) /(γ+1)........................................................................(eqn 8)

0.7382 = 2*1.4/(1.4-1) * 2.772sin

2β – (1.4-1)/(1.4+1)

β = sin-1

(0.3380); β = +19.76o or -19.76

o



Analytical Calculations for Mach 4

Given free stream conditions are

M1 = 4

P1 = 1.01325 bar

T1 = 300 K

Compression Curves = 6o and 10

o

Expansion Curves = four 4o curves


M1 = 4; P1 = 1.01325 bar; T1 = 300 K; ϴ1 = 6o

Location of Shock and Normal Mach Number

Refer the Oblique Shock Charts ( Appendix 1). We obtain for mach number 3 and ϴ =

6o the value of the shock wave angle as β1 as 19

o

Mn1 = M1 * sin ϴ

n1 = 4sin(19)

Mn1 = 1.3022



p2/p1 = 1.806; T2/T1 = 1.191; Mn2 = 0.7866

p2 = p1 * 1.806; T2 = T1 * 1.191; M2 = Mn2/sin(β-ϴ)

p2 = 1.0325 * 1.806; T2 = 300 * 1.191; M2 = 1.3022/sin(19-6)

p2 = 1.83 bar; T2 = 357.3 K; M2 = 3.5


(p2 - p1)/p1 or (p2/p1) – 1

1.806 – 1

= 0.806


∆s = Cp * ln(T2/T1) - R * ln(p2/p1)

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

∆s = 6.0174 J/Kg K


∆h= Cp∆t

∆h = 1005 * (T2 – T1)

∆h = 1005 * (357.3 – 300)

∆h = 57580 J/Kg K



M2 = 3.5; P2 = 1.83 bar; T2 = 357.3 K; ϴ2 = 10o

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 β2 as 24.5

o

Mn2 = M1 * sin ϴ

n2 = 3.5sin(24.5)

Mn2 = 1.45



p3/p2 = 2.286; T3/T2 = 1.287; Mn3 = 0.7196

p3 = p2 * 2.286; T3 = T2 * 1.287; M3 = Mn3/sin(β-ϴ)

p3 = 1.83 * 2.286; T3 = 357.3 * 1.287; M3 = 1.45/sin(24.5-10)

p3 = 4.1 bar; T3 = 459.87 K; M3 = 2.874


(p3/p2) – 1

2.286 – 1

= 1.286


∆s = Cp * ln(T3/T2) - R * ln(p3/p2)

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

∆s = 16.28 J/Kg K


∆h= Cp∆t

∆h = 1005 * (T3 – T2)

∆h = 1005 * (459.87 – 357.3)

∆h = 103068 J/Kg K



M3 = 2.874; P3 = 4.1 bar; T3 = 459.87 K; ϴ3 = 4o




ν4 = ν3 + ϴ3

ν4 = 47.18523 + 4

ν4 = 51.18523o





Since the flow is isentropic and no work is done, the stagnation temperature is constant (Tt =


(pt = constant)

p4/p3 = 0.02453 * 1 * (0.03312)-1

; T4/T3 = 0.34810 * 1 * (0.3773)-1

p4/p3 = 0.7426; T4/T3 = 0.918

p4 = 0.7426 * 4.10; T4 = 0.918 * 459.87

p4 = 3.035000 bar; T4 = 424.279 K


(p4/p3) – 1

0.7426 - 1

= -0.2626


∆s = Cp * ln(T4/T3) - R * ln(p4/p3)

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

∆s = -0.57512 J/Kg K


∆h= Cp∆t

∆h = 1005 * (T4 – T3)

∆h = 1005 * (424.279 – 459.87)

∆h = -35591 J/Kg K



2β – (γ-1) /(γ+1)........................................................................(eqn 8)

0.7426 = 2*1.4/(1.4-1) * 2.8742sin

2β – (1.4-1)/(1.4+1)


β = sin-1

(0.3067); β = +17.86o or -17.86

o



M4 = 3.07; P4 = 3.035 bar; T4 = 424.279 K; ϴ4 = 4o




ν5 = ν4 + ϴ4

ν4 = 51.18523 + 4

ν4 = 55.18523o


corresponding mach number M5 as 3.9. Referring the same following values of pressure

ratio and temperature ratio are obtained as follows




(pt = constant)

p5/p4 = 0.01773 * 1 * (0.02452)-1

; T5/T4 = 0. 31597 * 1 * (0.34810)-1

p5/p4 = 0.72308; T5/T4 = 0.907

p5 = 0.72308*3.07; T4 = 0.907 * 424.279

p5 = 2.19455 bar; T5 = 385.11 K


(p5/p4) – 1

0.72308 - 1

= -0.277


∆s = Cp * ln(T5/T4) - R * ln(p5/p4)

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

∆s = -5.04 J/Kg K


∆h= Cp∆t

∆h = 1005 * (T5 – T4)

∆h = 1005 * (385.11 – 424.279)

∆h = -39364.845 J/Kg K



2β – (γ-1) /(γ+1)........................................................................(eqn 8)

0.72308 = 2*1.4/(1.4-1) * 3.072sin

2β – (1.4-1)/(1.4+1)


β = sin-1

(0.2843); β = +16.52o or -16.52

o



M5 = 3.29; p5 = 2.19455 bar; T5 = 385.11 K; ϴ5 = 4o




ν6 = ν5 + ϴ5

ν6 = 55.18523 + 4

ν6 = 59.18523o







(pt = constant)

p6/p5 = 0.01239 * 1 * (0.01773)-1

; T6/T5 = 0.28520 * 1 * (0.31597)-1

p6/p5 = 0.7000; T6/T5 = 0.9026

p6 = 0.7000*2.19455; T6 = 0.9026 * 385.11

p6 = 1.46 bar; T6 = 347.06 K


(p6/p5) – 1

0.7000 - 1

= -0.3000


∆s = Cp * ln(T6/T5) - R * ln(p6/p5)

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

∆s = -0.62246 J/Kg K


∆h= Cp∆t

∆h = 1005 * (T6 – T5)

∆h = 1005 * (347.06 – 384.11)

∆h = -38240.25 J/Kg K



2β – (γ-1) /(γ+1)........................................................................(eqn 8)

0.7000 = 2*1.4/(1.4-1) * 3.292sin

2β – (1.4-1)/(1.4+1)


β = sin-1

(0.2619); β = +15.187o or -15.187

o

β = -15.187o …………………………(consider –ve value, since it’s a case of expansion)


M6 = 3.54; p6 = 1.46 bar; T6 = 347.06 K; ϴ6 = 4o


ν7 = ν6 + ϴ6

ν7 = 59.18523+ 4……………………………………………………………(Appendix 2)

ν7 = 64.18523o

Refer the Normal Shock tables (Appendix 2). for ν7 = 64.18523o, mach number M7 as

3.81. pressure ratio and temperature ratio are obtained as follows


Tt = constant; pt = constant

p7/p6 = 0.00851 * 1 * (0.01239)-1

; T7/T6 = 0.25620 * 1 * (0.28520)-1

p7/p6 = 0.6868; T7/T6 = 0.8983

p7 = 0.6868*1.46; T6 = 0.8983 * 347.06

p7 = 1.00279 bar; T7 = 308.88 K


(p7/p6) – 1

0.6868 - 1

= -0.3132


∆s = Cp * ln(T7/T6) - R * ln(p7/p6)

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

∆s = -0.2937 J/Kg K


∆h= Cp∆t

∆h = 1005 * (T7 – T6)

∆h = 1005 * (302.95 - 330)

∆h = -38370.9J/Kg K



2β – (γ-1) /(γ+1)........................................................................(eqn 8)

0.6868 = 2*1.4/(1.4-1) * 3.542sin

2β – (1.4-1)/(1.4+1)

β = sin-1

(0.2415); β = +13.98o or -13.98

o

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


The Total Change in Entropy for Mach 3

∑∆s = 3.10 + 8.14 -0.2948 – 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 – 21.88 – 35.5167 – 27.185 = 3.716 KJ/Kg K

The Total Change in Entropy for Mach 4

∑∆s = 6.0174 + 16.28 – 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 – 39.364 – 38.240 – 38.370 = 9.695 KJ/Kg K


Verification of Results Using Flow Simulating Software Star

CCM

For Mach 3

For Mach 4


Comparison of Analytical and Numerical Results

For Mach 3

At points 1 2 3 4 5 6 7

Mach Number Analytical 3 2.685 2.25 2.41 2.54 2.77 2.97

Numerical 3 2.74 2.3195 2.5227 2.698 2.92 2.9632

Pressure(bar) Analytical 1.0132 1.59 3.06 2.382 1.95 1.36 1.004

Numerical 1.0132 1.612 2.98 2.46 1.869 1.269 1.0118

Temperature(K) Analytical 300 342 416.23 387.52 365.94 330.4 302.95

Numerical 300 343.46 411.98 393.3 362.15 324.78 299.86

For Mach 4 At points 1 2 3 4 5 6 7

Mach Number Analytical 4 3.5 2.874 3.07 3.29 3.54 3.81

Numerical 4 3.57 2.874 3.035 3.35 3.62 3.948

Pressure(bar) Analytical 1.013 1.83 4.1 3.035 2.1945 1.46 1.002

Numerical 1.013 1.786 4.13 3.08 2.04 1.394 1.003

Temperature(K) Analytical 300 357.3 459.8 424.27 385.11 347.06 308.8

Numerical 300 361.9 459.7 415.3 379.76 344.22 299.7

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.


Appendix 1

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

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Appendix 2

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