Supersonic Flow Over a Wedge(Oblique Shock Problem)

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-By

Sameer Kadam

M.Sc Computational Mechanics

Problem State

Problem Statement

ment

Consider a supersonic flow over a

Consider a supersonic flow over a wedge (jet

wedge (jet

engine intake?) is to be analyzed. The shape,

given in Fig. 1,

given in Fig. 1, consists of

consists of two compression

two compression

corners of 6 and

corners of 6 and 10 degrees and 4 4

10 degrees and 4 4 degree

degree

expansion corners. The unit of t

expansion corners. The unit of the x-coordinate

he x-coordinate

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

1. 1. Mach number M =

Mach number M = 3, the a

3, the ambient pressure

mbient pressure

pa = 0.10132

pa = 0.101325 MPa and T

5 MPa and T = 300 K .

= 300 K .

2.

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Problem State

Problem Statement

ment

Consider a supersonic flow over a

Consider a supersonic flow over a wedge (jet

wedge (jet

engine intake?) is to be analyzed. The shape,

given in Fig. 1,

given in Fig. 1, consists of

consists of two compression

two compression

corners of 6 and

corners of 6 and 10 degrees and 4 4

10 degrees and 4 4 degree

degree

expansion corners. The unit of t

expansion corners. The unit of the x-coordinate

he x-coordinate

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

1. 1. Mach number M =

Mach number M = 3, the a

3, the ambient pressure

mbient pressure

pa = 0.10132

pa = 0.101325 MPa and T

5 MPa and T = 300 K .

= 300 K .

2. 2. Mach number M = 4

Mach number M = 4, the ambient

, the ambient pressure

pressure

pa = 0.10132

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Sr.No

Sr.No Topic Topic Page Page No.No.

1

1 Approach Approach 11

2

2 CalculatiCalculations ons for for Flow Flow With With Mach Mach Number Number = = 3 3 55

3

3 CalculatiCalculations ons for for Flow Flow With With Mach Mach Number Number = = 4 4 1111

4

4 Calculation Calculation of of Entropies Entropies and and Enthalpies Enthalpies 1717

5 5

Verification of Results Using

Verification of Results Using Flow Simulating SoftwareFlow Simulating Software Star

Star CCM CCM 1818

6

6 Comparison Comparison of of Analytical Analytical and and Numerical Numerical Results Results 1919

7

7 AppendiAppendix x 1 1 2020

6

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Approach

The given problem is a case of oblique shock as the wavefront is at an angle other than 90 The given problem is a case of oblique shock as the wavefront is at an angle other than 90oo toto the approaching flow. It can be divided into two parts basically,

the approaching flow. It can be divided into two parts basically, 1.

1. Analysis of Compression CornersAnalysis of Compression Corners 2.

2. Analysis of Expansion CornersAnalysis of Expansion Corners

1.

1. Analysis Analysis of of Compression Corners:Compression Corners:

There are two compression corners having angles 6

There are two compression corners having angles 6ooand 10and 10oorespectively. Let us denote theserespectively. Let us denote these

compression angle or deflection angles as

compression angle or deflection angles as ϴ. ϴ. Consider a flow Consider a flow with Machwith Machnumber; M1 is approachingnumber; M1 is approaching a compression corner. It results in a

a compression corner. It results in a oblique shock which can be visualized with the oblique shock which can be visualized with the help of the givenhelp of the given figure:

figure:

Properties of an Oblique Shock Wave Properties of an Oblique Shock Wave

Here we can see the properties of

Here we can see the properties of an oblique shock wave. The approaching flow is deflan oblique shock wave. The approaching flow is defl ected by angleected by angle

ϴ .

ϴ . Here β Here β is called the wave is called the wave angle and for aangle and for a Normal ShockNormal Shock it is 90it is 90oo. We split the approaching flow. We split the approaching flow into two components and hence we obtain

into two components and hence we obtain the Normal Component of mach number Mthe Normal Component of mach number Mn1n1& the& the

tangential component as M

tangential component as Mt1t1. Similarly we have. Similarly we have MMn2n2MMt2t2as theas the NormalNormal and tangential components of and tangential components of

the mach number of the flow after shock. the mach number of the flow after shock. Where

Where MMn1n1is given by M1/sinis given by M1/sin

In order to find angle β we r

In order to find angle β we r efer theefer the Oblique Shock Charts ( Appendix 1) which give the relationOblique Shock Charts ( Appendix 1) which give the relation between the shock wave angle and flow deflection angle for various values of upstream mach between the shock wave angle and flow deflection angle for various values of upstream mach numbers.

numbers. As we know the values of upstream Mach number MAs we know the values of upstream Mach number M11and the flow deflection angle ϴ weand the flow deflection angle ϴ we

can

canfind the corresponding value of β .find the corresponding value of β . Now, it can be solved as problem of

Now, it can be solved as problem of Normal Shock.Normal Shock.

Using the basic equations for continuity, energy and momentum we can obtain the relations 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.

for between the various parameters of the upstream and downstream flows. We start with the continuity equation

We start with the continuity equation

ρ

ρ11vv1=1= ρρ22vv22

From which we get the relation From which we get the relation

ρ

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The stagnation enthalpy remains constant over the flow The stagnation enthalpy remains constant over the flow

h ht1t1=h=ht2t2

but since enthalpy is a function of temperature only but since enthalpy is a function of temperature only

T Tt1t1=T=Tt2t2 T Ttt= T*(1+ (γ= T*(1+ (γ-1)*M-1)*M22 /2) /2) T T11*(1+ (γ-1)*M*(1+ (γ-1)*M1122 /2) = T /2) = T22*(1+ (γ-1)*M*(1+ (γ-1)*M2222 /2) /2) ... ... (2)(2)

Using the Momentum Equation Using the Momentum Equation

p

p11*(1+ γ*M*(1+ γ*M1122) = p) = p22*(1+ γ*M*(1+ γ*M2222)) ... ... (3)(3)

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

Equations 1,2,3 are the governing equations for the Normal shock. EmpressinEmpressing all tg all thehe relations in terms of M

relations in terms of M11 we get thewe get the Normal Shock tables( Appendix 2)Normal Shock tables( Appendix 2)

These tables provide the various values of pressure ratios, temperature ratios, density These tables provide the various values of pressure ratios, temperature ratios, density ratios, mach number downstream, etc

ratios, mach number downstream, etc. Using these Normal Shock tables (Appendix 2) we. Using these Normal Shock tables (Appendix 2) we obtain the values for pressure ratio (p

obtain the values for pressure ratio (p2/ 2/ pp11), temperature ratio (T), temperature ratio (T22 /T /T11) and Mach number after) and Mach number after

shock. shock.

Note that the static pressure and temperature are the same whether we are talking Note that the static pressure and temperature are the same whether we are talking about Normal Shock or Oblique Shock.

about Normal Shock or Oblique Shock.

Moreover the value of Mach number which we obtain for the flow after the shock is the Moreover the value of Mach number which we obtain for the flow after the shock is the Normal Component of M

Normal Component of M22 i.e Mi.e Mn2.n2.M can be obtained by using the relation.M can be obtained by using the relation.

M

M22= M= M2n2n/sin(β/sin(β--ϴ)ϴ)

The strength of shock can be calculated by using the pressure ratio. The strength of shock is The strength of shock can be calculated by using the pressure ratio. The strength of shock is given by

given by

(p

(p22-p-p11)/p)/p11

The change in Entropy for any ideal gas in terms of pressure ratio and density ratio is given The change in Entropy for any ideal gas in terms of pressure ratio and density ratio is given by the relation:

by the relation:

∆s

∆s = C= Cpp * ln(T* ln(T22 /T /T11) - R * ln(p) - R * ln(p22 /p /p11)) ... ... (4)(4)

Note that the entropy changes tend to be very small for oblique shocks as the entropy changes 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 are directly proportional to the cube of the flow deflection angle which is very small in this case.

case.

The change in Enthalpy is given by The change in Enthalpy is given by

∆h= C

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2.

2. Analysis Analysis of Exof Expansion Corners:pansion Corners:

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

There are four expansion corners each of them having an angle of 4oo. In a shock wave the. In a shock wave the pressure, density and temperature increase. In an expansion wave it is exactly opposite: they pressure, density and temperature increase. In an expansion wave it is exactly opposite: they all decrease. The analysis of expansion corner is different

all decrease. The analysis of expansion corner is different as compared to compressionas compared to compression corner. Here we use the

corner. Here we use the Prandtl-Prandtl-Meyer Function(ν).Meyer Function(ν). It is defined as

It is defined as the angle through which a flow with a Mthe angle through which a flow with a Mach number = 1 is turnedach number = 1 is turned isentropically to achieve the indicated Mach number.

isentropically to achieve the indicated Mach number.

Illustration of Prandtl Meyer Function Illustration of Prandtl Meyer Function

The Prandtl-Meyer Function can be obtained from the Isentropic Charts or the Normal The Prandtl-Meyer Function can be obtained from the Isentropic Charts or the Normal Shock Charts(Appendix 2).

Shock Charts(Appendix 2).

Otherwise it can be calculated by using the relationship Otherwise it can be calculated by using the relationship

Thus for calculating the Prandtl-Meyer Function we require only the upstream mach number. 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:

Based on this we can calculate the Mach number of the downstream flow as follows: 

 Find νFind ν11 for the upstream mach number from isefor the upstream mach number from isentropic chartsntropic charts



 TThe flow turns by the flow deflection angle ϴ. Add this value of ϴ to νhe flow turns by the flow deflection angle ϴ. Add this value of ϴ to ν11whichwhich

would indicate the

would indicate the Prandtl Meyer Function, νPrandtl Meyer Function, ν22for the downstream machfor the downstream mach

number. number. 

 i.e νi.e ν22-- ν ν11== ϴϴ



 From νFrom ν22wewe can obtain the corresponding down stream mach number from thecan obtain the corresponding down stream mach number from the

isentropic charts. isentropic charts.

Similarly using the Isentropic charts we can find the other values for the pressure ratio and Similarly using the Isentropic charts we can find the other values for the pressure ratio and temperature ratio.

temperature ratio.

The Cahnge in Enthalpy and Entropy can also be found out similarly as in case of The Cahnge in Enthalpy and Entropy can also be found out similarly as in case of Compression Corners using eqns 4 and 5

Compression Corners using eqns 4 and 5

Note that the change in entropy and enthalpy is negative in this case Note that the change in entropy and enthalpy is negative in this case However, we are also required to find the

However, we are also required to find the shock wave angleshock wave angle ββ, in order to , in order to determine thedetermine the location of shock.

location of shock.

ν ν

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

...

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

...

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

...

...(from ...(from eqn eqn 3)3) Substituting the last two eqns in the first eqn we get a relation between M

Substituting the last two eqns in the first eqn we get a relation between M11 and Mand M22 as follows:as follows:

...

...(eqn ...(eqn 6)6) Substituting the value of M

Substituting the value of M22 in eqn 3 we have a relation between the pressure in terms of Min eqn 3 we have a relation between the pressure in terms of M11

p

p22 /p /p11= 2γ/(γ= 2γ/(γ-1) * M-1) * M1122 – – (γ(γ--1) 1) /(γ+1).../(γ+1)...(eqn ...(eqn 7)7)

Where M

Where M11 is the Normal Component of the Upstream Flowis the Normal Component of the Upstream Flow

Thus for Expansion Corner it gets modified to Thus for Expansion Corner it gets modified to

p

p22 /p /p11= 2γ/(γ= 2γ/(γ-1) * M-1) * M1122sinsin22ββ – – (γ(γ--1) 1) /(γ+1)..../(γ+1)...(eqn 8)...(eqn 8)

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

calculate the wave angle ββ.. Note that after taking sin

Note that after taking sin-1-1 we will have to consider the negative value of the angle, sincewe will have to consider the negative value of the angle, since it is a case of expansion.

it is a case of expansion.

Wave Angle for an Expansion Shock Wave Angle for an Expansion Shock Negative Angle is Traced Clockwise Negative Angle is Traced Clockwise

--_β_β M

M₁₁

-M -M₂₂

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Analytical Calculations for Mach 3

Given free stream conditions are Given free stream conditions are M M11= 3= 3 P P11= 1.01325 bar= 1.01325 bar T T11 = 300 K= 300 K Compression Curves = 6

Compression Curves = 6ooand 10and 10oo Expansion Curves = four 4

Expansion Curves = four 4oocurvescurves

1.

1. Compression Corner of 6Compression Corner of 6oo M

M11 = 3;= 3; PP11 = 1.01325 bar; T= 1.01325 bar; T11= 300 K= 300 K; ϴ; ϴ11 = 6= 6oo

Location of Shock and Normal Mach Number Location of Shock and Normal Mach Number Refer the Oblique Shock Charts ( Appendix 1)

Refer the Oblique Shock Charts ( Appendix 1). We obtain for mach number 3 and ϴ =. We obtain for mach number 3 and ϴ = 6

6oo the value of the shock wave angle as βthe value of the shock wave angle as β11 as 24as 24oo

 MMn1n1 = M= M11* sin ϴ* sin ϴ  n1n1 = 3sin(24)= 3sin(24)  MMn1n1 = 1.22= 1.22

Refer the Normal Shock tables (Appendix 2). We obtain for mach number 1.22 the 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 following values of pressure ratio, temperature ratio and downstream mach number p p22 /p /p11= = 1.570; 1.570; TT22 /T /T1 =1 =1.141; M1.141; Mn2n2= 0.8300= 0.8300  pp22= p= p11* * 1.570; 1.570; TT22= T= T11* * 1.141; 1.141; MM22= M= Mn2n2/sin(β/sin(β--ϴ)ϴ)  pp22= = 1.0325 * 1.0325 * 1.570; 1.570; TT22 = = 300 300 * * 1.141; 1.141; MM22 = = 0.8300/sin0.8300/sin(24-6)(24-6)  pp22= = 1.59 1.59 bar; bar; TT22= = 342.3 342.3 K; K; MM22 = 2.685= 2.685

Strength of Shock is given by Strength of Shock is given by (p (p22- p- p1)1) /p /p11 or (por (p22 /p /p11)) – – 11  1.5701.570 – – 11 = = 0.5700.570

Change in Entropy is given by Change in Entropy is given by ∆ ∆s = Cs = Cpp* ln(T* ln(T22 /T /T11) - R * ln(p) - R * ln(p22 /p /p11))  ∆∆s = 1005 * ln(1.141)s = 1005 * ln(1.141) – – 287 * ln(1.570)287 * ln(1.570)  ∆s = 3.10 J/Kg K ∆s = 3.10 J/Kg K

Change in Enthalpy is given by Change in Enthalpy is given by ∆h= C ∆h= Cpp∆t∆t ∆h ∆h = 1005 * (T= 1005 * (T22 – – TT11)) ∆h ∆h = 1005 * (342.3= 1005 * (342.3 – – 300)300) ∆h = 42511.5 J/Kg K ∆h = 42511.5 J/Kg K

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2.

M22 = 2.685;= 2.685; PP22= 1.59 bar; T= 1.59 bar; T22= 342 K= 342 K; ϴ; ϴ22 = 10= 10oo

Location of Shock and Normal mach Number Location of Shock and Normal mach Number

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

ϴ = 1010oothe value of the shock wave angle as βthe value of the shock wave angle as β11 as 30as 30oo

 MMn2n2 = M= M11* sin ϴ* sin ϴ  n2n2 = 2.685sin(30)= 2.685sin(30)  MMn2n2 = 1.3425= 1.3425

Refer the Normal Shock tables (Appendix 2). We obtain for mach number 1.3425 the 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 following values of pressure ratio, temperature ratio and downstream mach number p p33 /p /p22= = 1.928; 1.928; TT33 /T /T22== 1.216; M1.216; Mn3n3 = 0.7664= 0.7664  pp33= p= p22* * 1.928; 1.928; TT33= T= T22* * 1.216; 1.216; MM33= M= Mn3n3/sin(β/sin(β--ϴ)ϴ)  pp33= = 1.59 1.59 * * 1.928; 1.928; TT33 = = 342 342 * * 1.216; 1.216; MM33 = = 0.7664/sin0.7664/sin(30-10)(30-10)  pp33= = 3.06 3.06 bar; bar; TT33 = = 416.23 416.23 K; K; MM33 = 2.25= 2.25

Strength of Shock is given by Strength of Shock is given by

(p (p33 /p /p22)) – – 11  1.9281.928 – – 11 = = 0.9280.928

Change in Entropy is given by Change in Entropy is given by ∆ ∆s = Cs = Cpp* ln(T* ln(T33 /T /T22) - R * ln(p) - R * ln(p33 /p /p22))  ∆∆s = 1005 * ln(1.216)s = 1005 * ln(1.216) – – 287 * ln(1.928)287 * ln(1.928)  ∆s =∆s = 8.14 J/Kg K8.14 J/Kg K

Change in Enthalpy is given by Change in Enthalpy is given by ∆h= C ∆h= Cpp∆t∆t ∆h ∆h = 1005 * (T= 1005 * (T33 – – TT22)) ∆h ∆h = 1005 * (416.23= 1005 * (416.23 – – 342.3)342.3) ∆h = ∆h = 74290 J/Kg K74290 J/Kg K

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3.

3. Expansion Corner of 4Expansion Corner of 4oo M

M33 = 2.25;= 2.25; PP33= 3.06 bar; T= 3.06 bar; T33 = 416.23 K= 416.23 K; ϴ; ϴ33 = 4= 4oo

Prandtl-Meyer Function Prandtl-Meyer Function

Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach number 2.25 the value of Prandtl- Meyer Function

number 2.25 the value of Prandtl- Meyer Function ν ν33 = 33.018= 33.018

  ν ν44= ν= ν33+ ϴ+ ϴ33   ν ν44= 33.018 + 4= 33.018 + 4   ν ν44 = 37.018= 37.018oo

Refer the Normal Shock tables (Appendix 2). We obtain for

Refer the Normal Shock tables (Appendix 2). We obtain for ν ν44 = 37.018= 37.018oothe value of the value of

corresponding mach number M

corresponding mach number M44 as 2.41. Referring the same following values of as 2.41. Referring the same following values of

pressure ratio and temperature ratio are obtained as follows pressure ratio and temperature ratio are obtained as follows p

p44 /p /p33= p= p44 /p /p4t4t* p* p4t4t /p /p3t3t * p* p3t3t /p /p33; ; TT44 /T /T33= T= T44 /T /T4t4t * T* T4t4t /T /T3t3t* T* T3t3t /T /T33;;

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

constant). Also there are no losses in the flow, hence the stagnation pressure is also constant constant). Also there are no losses in the flow, hence the stagnation pressure is also constant (p (ptt = constant)= constant)  pp44 /p /p33= 0.06734 * 1 * (0.08648)= 0.06734 * 1 * (0.08648)-1-1; ; TT44 /T /T33= 0.46262 * 1 * (0.49689)= 0.46262 * 1 * (0.49689)-1-1  pp44 /p /p33= = 0.7786; 0.7786; TT44 /T /T33 = 0.931= 0.931  pp44= = 0.7786 0.7786 * * 3.06; 3.06; TT44= 0.931 * 416.23= 0.931 * 416.23  pp44= = 2.382516 2.382516 bar; bar; TT44= 387.52 K= 387.52 K

(p (p44 /p /p33)) – – 11  0.7786 - 10.7786 - 1 = = -0.2214-0.2214

Change in Entropy is given by Change in Entropy is given by ∆ ∆s = Cs = Cpp* ln(T* ln(T44 /T /T33) - R * ln(p) - R * ln(p44 /p /p33))  ∆∆s = 1005 * ln(0.931)s = 1005 * ln(0.931) – – 287 * ln(0.7786)287 * ln(0.7786)  ∆s =∆s = -0.02948 J/Kg K-0.02948 J/Kg K

Change in Enthalpy is given by Change in Enthalpy is given by ∆h= C ∆h= Cpp∆t∆t ∆h ∆h = 1005 * (T= 1005 * (T44 – – TT33)) ∆h ∆h = 1005 * (387.52 - 416.23)= 1005 * (387.52 - 416.23) ∆h = ∆h = -28853 J/Kg K-28853 J/Kg K Location of Shock β Location of Shock β p

piiii /p /pii = 2γ/(γ= 2γ/(γ-1) * M-1) * Mii22 sinsin22β – β – (γ(γ--1) /(γ+1)1) /(γ+1)...(eqn ...(eqn 8)8)

0.7786 = 2*1.4/(1.4-1) * 2.25

0.7786 = 2*1.4/(1.4-1) * 2.2522sinsin22β – β – (1.4-1)/(1.4+1)(1.4-1)/(1.4+1) 

β = sinβ = sin-1-1(0.4);(0.4); β = +23.58β = +23.58ooor -23.58or -23.58oo 

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4.

M44 = 2.41;= 2.41; PP44= 2.382 bar; T= 2.382 bar; T44= 387.52 K= 387.52 K; ϴ; ϴ44= 4= 4oo

Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach number 2.41 the value o

number 2.41 the value of Prandtl-f Prandtl- Meyer Function νMeyer Function ν44 = 37.018= 37.018

  ν ν55= ν= ν44+ ϴ+ ϴ44   ν ν44= 37.018 + 4= 37.018 + 4   ν ν44 = 41.018= 41.018oo

constant). Also there are no losses in the flow, hence the stagnation pressure is also constant constant). Also there are no losses in the flow, hence the stagnation pressure is also constant (p (ptt = constant)= constant)  pp55 /p /p44= 0.0550 * 1 * (0.06734)= 0.0550 * 1 * (0.06734)-1-1; ; TT55 /T /T44 = 0. 43662 * 1 * (0.46262)= 0. 43662 * 1 * (0.46262)-1-1  pp55 /p /p44= = 0.8167; 0.8167; TT55 /T /T44= 0.9437= 0.9437  pp55= = 0.8167*2.3820.8167*2.382; ; TT44= 0.9437 * 387.52= 0.9437 * 387.52  pp55= = 1.94 1.94 bar; bar; TT55= 365.74 K= 365.74 K

(p (p55 /p /p44)) – – 11  0.8167 - 10.8167 - 1 = = -0.1833-0.1833

Change in Enthalpy is given by Change in Enthalpy is given by ∆h= C ∆h= Cpp∆t∆t ∆h ∆h = 1005 * (T= 1005 * (T55 – – TT44)) ∆h ∆h = 1005 * (365.74 - 387.52)= 1005 * (365.74 - 387.52) ∆h = ∆h = -21889 J/Kg K-21889 J/Kg K Location of Shock β Location of Shock β p

0.8167 = 2*1.4/(1.4-1) * 2.41

0.8167 = 2*1.4/(1.4-1) * 2.4122sinsin22β – β – (1.4-1)/(1.4+1)(1.4-1)/(1.4+1) 

β = sinβ = sin-1-1(0.3809);(0.3809); β = +2β = +22.392.39ooor -22.39or -22.39oo 

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5.

M55 = 2.54;= 2.54; pp55= 1.94 bar; T= 1.94 bar; T55= 365.74 K= 365.74 K; ϴ; ϴ55 = 4= 4oo

  ν ν66= ν= ν55+ ϴ+ ϴ55   ν ν66= 41.018 + 4= 41.018 + 4   ν ν66 = 45.018= 45.018oo

constant). Also there are no losses in the flow, hence the stagnation pressure is also constant constant). Also there are no losses in the flow, hence the stagnation pressure is also constant (p (ptt = constant)= constant)  pp66 /p /p55= 0.03858 * 1 * (0.0550)= 0.03858 * 1 * (0.0550)-1-1; ; TT66 /T /T55 = 0.3954 * 1 * (0.4366)= 0.3954 * 1 * (0.4366)-1-1  pp66 /p /p55= = 0.7014; 0.7014; TT66 /T /T55= 0.9036= 0.9036  pp66= = 0.7014*1.940.7014*1.94; ; TT66= 0.9036 * 365.74= 0.9036 * 365.74  pp66= = 1.36 1.36 bar; bar; TT66= 330.4 K= 330.4 K

(p (p66 /p /p55)) – – 11  0.7014 - 10.7014 - 1 = = -0.2986-0.2986

Change in Enthalpy is given by Change in Enthalpy is given by ∆h= C ∆h= Cpp∆t∆t ∆h ∆h = 1005 * (T= 1005 * (T66 – – TT55)) ∆h ∆h = 1005 * (330= 1005 * (330 – – 365.74)365.74) ∆h = ∆h = -35516.7 J/Kg K-35516.7 J/Kg K Location of Shock β Location of Shock β p

0.7014 = 2*1.4/(1.4-1) * 2.54

0.7014 = 2*1.4/(1.4-1) * 2.5422sinsin22β – β – (1.4-1)/(1.4+1)(1.4-1)/(1.4+1) 

β = sinβ = sin-1-1(0.34054);(0.34054); β = +β = +19.9119.91oo or -19.91or -19.91oo 

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6.

6. Expansion Corner of 4Expansion Corner of 4oo M M66 = 2.77;= 2.77; pp66= 1.36 bar; T= 1.36 bar; T66= 330.4 K= 330.4 K; ϴ; ϴ66 = 4= 4oo Prandtl-Meyer Function Prandtl-Meyer Function   ν ν77= ν= ν66+ ϴ+ ϴ66 

 ν ν77= 45.018 + 4= 45.018 + 4………(From Appendix 2)………(From Appendix 2)



 ν ν77 = 49.018= 49.018oo

Refer the Normal Shock tables (Appendix 2). for

Refer the Normal Shock tables (Appendix 2). for ν ν77= 49.018= 49.018oo, mach number M, mach number M77as 2.97.as 2.97.

values of pressure ratio and temperature ratio are obtained as follows values of pressure ratio and temperature ratio are obtained as follows p

T

Ttt = constant and p= constant and ptt = constant= constant

 pp77 /p /p66= 0.02848 * 1 * (0.03858)= 0.02848 * 1 * (0.03858)-1-1; ; TT77 /T /T66= 0.36177 * 1 * (0.39454)= 0.36177 * 1 * (0.39454)-1-1  pp77 /p /p66= = 0.7382; 0.7382; TT77 /T /T66 = 0.917= 0.917  pp77= = 0.7382*1.360.7382*1.36; ; TT66= 0.917 * 330.4= 0.917 * 330.4  pp77= = 1.004 1.004 bar; bar; TT77= 302.95 K= 302.95 K

(p (p77 /p /p66)) – – 11  0.7382 - 10.7382 - 1 = = -0.2618-0.2618

Change in Enthalpy is given by Change in Enthalpy is given by ∆h= C ∆h= Cpp∆t∆t ∆h ∆h = 1005 * (T= 1005 * (T77 – – TT66)) ∆h ∆h = 1005 * (302.95 - 330)= 1005 * (302.95 - 330) ∆h = ∆h = -27185.25 J/Kg K-27185.25 J/Kg K Location of Shock β Location of Shock β p

piiii /p /pii = 2γ/(γ= 2γ/(γ-1) * M-1) * Mii22 sinsin22β – β – (γ(γ--1) /(γ+1)1) /(γ+1)...(eqn ..(eqn 8)8)

0.7382 = 2*1.4/(1.4-1) * 2.77

0.7382 = 2*1.4/(1.4-1) * 2.7722sinsin22β – β – (1.4-1)/(1.4+1)(1.4-1)/(1.4+1) 

β = sinβ = sin-1-1(0.3380);(0.3380); β = +19.β = +19.7676ooor -19.76or -19.76oo 

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Analytical Calculations for Mach 4

Given free stream conditions are Given free stream conditions are M M11= 4= 4 P P11= 1.01325 bar= 1.01325 bar T T11 = 300 K= 300 K Compression Curves = 6

Compression Curves = 6ooand 10and 10oo Expansion Curves = four 4

Expansion Curves = four 4oocurvescurves

1.

M11 = 4;= 4; PP11 = 1.01325 bar; T= 1.01325 bar; T11= 300 K= 300 K; ϴ; ϴ11 = 6= 6oo

Location of Shock and Normal Mach Number Location of Shock and Normal Mach Number Refer the Oblique Shock Charts ( Appendix 1)

Refer the Oblique Shock Charts ( Appendix 1). We obtain for mach number 3 and ϴ =. We obtain for mach number 3 and ϴ = 6

6oo the value of the shock wave angle as βthe value of the shock wave angle as β11 as 19as 19oo

 MMn1n1 = M= M11* sin ϴ* sin ϴ  n1n1 = 4sin(19)= 4sin(19)  MMn1n1 = 1.3022= 1.3022

Refer the Normal Shock tables (Appendix 2). We obtain for mach number 1.3022 the 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 following values of pressure ratio, temperature ratio and downstream mach number p p22 /p /p11= = 1.806; 1.806; TT22 /T /T1 =1 =1.191; M1.191; Mn2n2= 0.7866= 0.7866  pp22= p= p11* * 1.806; 1.806; TT22= T= T11* * 1.191; 1.191; MM22= M= Mn2n2/sin(β/sin(β--ϴ)ϴ)  pp22= = 1.0325 * 1.0325 * 1.806; 1.806; TT22 = = 300 300 * * 1.191; 1.191; MM22 = = 1.3022/sin1.3022/sin(19-6)(19-6)  pp22= = 1.83 1.83 bar; bar; TT22= = 357.3 357.3 K; K; MM22 = 3.5= 3.5

Strength of Shock is given by Strength of Shock is given by (p (p22- p- p1)1) /p /p11 or (por (p22 /p /p11)) – – 11  1.8061.806 – – 11 = = 0.8060.806

Change in Entropy is given by Change in Entropy is given by ∆ ∆s = Cs = Cpp* ln(T* ln(T22 /T /T11) - R * ln(p) - R * ln(p22 /p /p11))  ∆∆s = 1005 * ln(1.191)s = 1005 * ln(1.191) – – 287 * ln(1.806)287 * ln(1.806)  ∆∆s = 6.0174 J/Kg Ks = 6.0174 J/Kg K

Change in Enthalpy is given by Change in Enthalpy is given by ∆h= C ∆h= Cpp∆t∆t ∆h ∆h = 1005 * (T= 1005 * (T22 – – TT11)) ∆h ∆h = 1005 * (357.3= 1005 * (357.3 – – 300)300) ∆ ∆h = 57580 J/Kg Kh = 57580 J/Kg K

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2.

M22 = 3.5;= 3.5; PP22= 1.83 bar; T= 1.83 bar; T22= 357.3 K= 357.3 K; ϴ; ϴ22 = 10= 10oo

Location of Shock Location of Shock

Refer the Oblique Shock Charts (

Refer the Oblique Shock Charts ( Appendix 1). We obtain for mach number 3.5Appendix 1). We obtain for mach number 3.5 and ϴand ϴ = 10

= 10oo the value of the shock wave angle as βthe value of the shock wave angle as β22 as 24.5as 24.5oo

 MMn2n2 = M= M11* sin ϴ* sin ϴ  n2n2 = 3.5sin(24.5)= 3.5sin(24.5)  MMn2n2 = 1.45= 1.45

Refer the Normal Shock tables (Appendix 2). We obtain for mach number 1.45 the 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 following values of pressure ratio, temperature ratio and downstream mach number p p33 /p /p22= = 2.286; 2.286; TT33 /T /T22== 1.287; M1.287; Mn3n3 = 0.7196= 0.7196  pp33= p= p22* * 2.286; 2.286; TT33= T= T22* * 1.287; 1.287; MM33= M= Mn3n3/sin(β/sin(β--ϴ)ϴ)  pp33= = 1.83 1.83 * * 2.286; 2.286; TT33 = = 357.3 357.3 * * 1.287; 1.287; MM33= 1.45/sin(24.5-10)= 1.45/sin(24.5-10)  pp33= = 4.1 4.1 bar; bar; TT33= = 459.87 459.87 K; K; MM33= 2.874= 2.874

(p (p33 /p /p22)) – – 11  2.2862.286 – – 11 = = 1.2861.286

Change in Entropy is given by Change in Entropy is given by ∆ ∆s = Cs = Cpp* ln(T* ln(T33 /T /T22) - R * ln(p) - R * ln(p33 /p /p22))  ∆∆s = 1005 * ln(1.287)s = 1005 * ln(1.287) – – 287 * ln(2.286)287 * ln(2.286)  ∆s =∆s = 16.28 J/Kg K16.28 J/Kg K

Change in Enthalpy is given by Change in Enthalpy is given by ∆h= C ∆h= Cpp∆t∆t ∆h ∆h = 1005 * (T= 1005 * (T33 – – TT22)) ∆h ∆h = 1005 * (459.87= 1005 * (459.87 – – 357.3)357.3) ∆h = ∆h = 103068 J/Kg K103068 J/Kg K

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3.

M33 = 2.874;= 2.874; PP33= 4.1 bar; T= 4.1 bar; T33 = 459.87 K= 459.87 K; ϴ; ϴ33 = 4= 4oo

Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach number 2.874 the value of

Prandtl-number 2.874 the value of Prandtl- Meyer Function νMeyer Function ν33= 47.18523= 47.18523

  ν ν44= ν= ν33+ ϴ+ ϴ33   ν ν44= 47.18523 + 4= 47.18523 + 4   ν ν44 = 51.18523= 51.18523oo

constant). Also there are no losses in the flow, hence the stagnation pressure is also constant constant). Also there are no losses in the flow, hence the stagnation pressure is also constant (p (ptt = constant)= constant)  pp44 /p /p33= 0.02453 * 1 * (0.03312)= 0.02453 * 1 * (0.03312)-1-1; ; TT44 /T /T33= 0.34810 * 1 * (0.3773)= 0.34810 * 1 * (0.3773)-1-1  pp44 /p /p33= = 0.7426; 0.7426; TT44 /T /T33 = 0.918= 0.918  pp44= = 0.7426 0.7426 * * 4.10; 4.10; TT44= 0.918 * 459.87= 0.918 * 459.87  pp44= = 3.035000 3.035000 bar; bar; TT44= 424.279 K= 424.279 K

(p (p44 /p /p33)) – – 11  0.7426 - 10.7426 - 1 = = -0.2626-0.2626

Change in Enthalpy is given by Change in Enthalpy is given by ∆h= C ∆h= Cpp∆t∆t ∆h ∆h = 1005 * (T= 1005 * (T44 – – TT33)) ∆h ∆h = 1005 * (424.279= 1005 * (424.279 – – 459.87)459.87) ∆h = ∆h = -35591 J/Kg K-35591 J/Kg K Location of Shock β Location of Shock β p

0.7426 = 2*1.4/(1.4-1) * 2.874

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

β = sinβ = sin-1-1(0.3067);(0.3067); β = +β = +17.8617.86ooor -17.86or -17.86oo 

β =β = -17.86-17.86oo………(consider –ve value, since it’s a case o………(consider –ve value, since it’s a case of expansion)f expansion) 4.

M44 = 3.07;= 3.07; PP44= 3.035 bar; T= 3.035 bar; T44= 424.279 K= 424.279 K; ϴ; ϴ44= 4= 4oo

  ν ν55= ν= ν44+ ϴ+ ϴ44   ν ν44= 51.18523 + 4= 51.18523 + 4   ν ν44 = 55.18523= 55.18523oo

corresponding mach number M55 as 3.9. Referring the same following values of pressureas 3.9. Referring the same following values of pressure

ratio and temperature ratio are obtained as follows ratio and temperature ratio are obtained as follows p

constant). Also there are no losses in the flow, hence the stagnation pressure is also constant constant). Also there are no losses in the flow, hence the stagnation pressure is also constant (p (ptt = constant)= constant)  pp55 /p /p44= 0.01773 * 1 * (0.02452)= 0.01773 * 1 * (0.02452)-1-1; ; TT55 /T /T44= 0. 31597 * 1 * (0.34810)= 0. 31597 * 1 * (0.34810)-1-1  pp55 /p /p44= = 0.72308; 0.72308; TT55 /T /T44 = 0.907= 0.907  pp55= = 0.72308*3.070.72308*3.07; ; TT44= 0.907 * 424.279= 0.907 * 424.279  pp55= = 2.19455 2.19455 bar; bar; TT55= 385.11 K= 385.11 K

(p (p55 /p /p44)) – – 11  0.72308 - 10.72308 - 1 = = -0.277-0.277

Change in Enthalpy is given by Change in Enthalpy is given by ∆h= C ∆h= Cpp∆t∆t ∆h ∆h = 1005 * (T= 1005 * (T55 – – TT44)) ∆h ∆h = 1005 * (385.11= 1005 * (385.11 – – 424.279)424.279) ∆h = ∆h = -39364.845 J/Kg K-39364.845 J/Kg K Location of Shock β Location of Shock β p

0.72308 = 2*1.4/(1.4-1) * 3.07

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

β = sinβ = sin-1-1(0.2843);(0.2843); β = +β = +16.5216.52ooor -16.52or -16.52oo 

β =β = -16.52-16.52oo………(consider –ve value, since it’s a case o………(consider –ve value, since it’s a case of expansion)f expansion) 5.

M55 = 3.29;= 3.29; pp55= 2.19455 bar; T= 2.19455 bar; T55= 385.11 K; ϴ= 385.11 K; ϴ55 = 4= 4oo

  ν ν66= ν= ν55+ ϴ+ ϴ55   ν ν66= 55.18523 + 4= 55.18523 + 4   ν ν66 = 59.18523= 59.18523oo

constant). Also there are no losses in the flow, hence the stagnation pressure is also constant constant). Also there are no losses in the flow, hence the stagnation pressure is also constant (p (ptt = constant)= constant)  pp66 /p /p55= 0.01239 * 1 * (0.01773)= 0.01239 * 1 * (0.01773)-1-1; ; TT66 /T /T55= 0.28520 * 1 * (0.31597)= 0.28520 * 1 * (0.31597)-1-1  pp66 /p /p55= = 0.7000; 0.7000; TT66 /T /T55 = 0.9026= 0.9026  pp66= = 0.7000*2.1940.7000*2.19455; 55; TT66= 0.9026 * 385.11= 0.9026 * 385.11  pp66= = 1.46 1.46 bar; bar; TT66= 347.06 K= 347.06 K

(p (p66 /p /p55)) – – 11  0.7000 - 10.7000 - 1 = = -0.3000-0.3000

Change in Enthalpy is given by Change in Enthalpy is given by ∆h= C ∆h= Cpp∆t∆t ∆h ∆h = 1005 * (T= 1005 * (T66 – – TT55)) ∆h ∆h = 1005 * (347.06= 1005 * (347.06 – – 384.11)384.11) ∆h = ∆h = -38240.25 J/Kg K-38240.25 J/Kg K Location of Shock β Location of Shock β p

0.7000 = 2*1.4/(1.4-1) * 3.29

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

β = sinβ = sin-1-1(0.2619);(0.2619); β = +1β = +15.1875.187oo or -15.187or -15.187oo 

β =β = -15.187-15.187oo………(consider – ………(consider – ve valuve value, since it’s a case e, since it’s a case of expansion)of expansion) 6.

6. Expansion Corner of 4Expansion Corner of 4oo M M66 = 3.54;= 3.54; pp66= 1.46 bar; T= 1.46 bar; T66= 347.06 K= 347.06 K; ϴ; ϴ66 = 4= 4oo Prandtl-Meyer Function Prandtl-Meyer Function   ν ν77= ν= ν66+ ϴ+ ϴ66 

 ν ν77= 59.18523+ 4= 59.18523+ 4………(Ap………(Appendix pendix 2)2)



 ν ν77 = 64.18523= 64.18523oo

Refer the Normal Shock tables (Appendix 2). for

Refer the Normal Shock tables (Appendix 2). for ν ν77= 64.18523= 64.18523oo, mach number M, mach number M77asas

3.81. pressure ratio and temperature ratio are obtained as follows 3.81. pressure ratio and temperature ratio are obtained as follows p

T

Ttt = = constant; constant; pptt = constant= constant

 pp77 /p /p66= 0.00851 * 1 * (0.01239)= 0.00851 * 1 * (0.01239)-1-1; ; TT77 /T /T66= 0.25620 * 1 * (0.28520)= 0.25620 * 1 * (0.28520)-1-1  pp77 /p /p66= = 0.6868; 0.6868; TT77 /T /T66 = 0.8983= 0.8983  pp77= = 0.6868*1.460.6868*1.46; ; TT66= 0.8983 * 347.06= 0.8983 * 347.06  pp77= = 1.00279 1.00279 bar; bar; TT77= 308.88 K= 308.88 K

(p (p77 /p /p66)) – – 11  0.6868 - 10.6868 - 1 = = -0.3132-0.3132

Change in Enthalpy is given by Change in Enthalpy is given by ∆h= C ∆h= Cpp∆t∆t ∆h ∆h = 1005 * (T= 1005 * (T77 – – TT66)) ∆h ∆h = 1005 * (302.95 - 330)= 1005 * (302.95 - 330) ∆h = ∆h = -38370.9J/Kg K-38370.9J/Kg K Location of Shock β Location of Shock β p

0.6868 = 2*1.4/(1.4-1) * 3.54

0.6868 = 2*1.4/(1.4-1) * 3.5422sinsin22β – β – (1.4-1)/(1.4+1)(1.4-1)/(1.4+1) 

β = sinβ = sin-1-1(0.2415);(0.2415); β = +1β = +13.983.98ooor -13.98or -13.98oo 

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The Total Change in Entropy for Mach 3 The Total Change in Entropy for Mach 3 ∑∆s = 3.10 + 8.14

∑∆s = 3.10 + 8.14 -0.2948-0.2948 – – 0.12390.1239 – – 0.083050.08305 – – 0.03504 =0.03504 = 10.703 J/Kg K10.703 J/Kg K

The Total Change in Enthalpy for Mach 3 The Total Change in Enthalpy for Mach 3 ∑∆h = 42.511 + 74.290 –

∑∆h = 42.511 + 74.290 – 28.85328.853 – – 21.8821.88 – – 35.516735.5167 – – 27.185 =27.185 = 3.716 KJ/Kg K3.716 KJ/Kg K

The Total Change in Entropy for Mach 4 The Total Change in Entropy for Mach 4 ∑∆s = 6.0

∑∆s = 6.0174 + 16.28174 + 16.28 – – 0.575120.57512 – – 5.045.04 – – 0.622460.62246 – – 0.2937 =0.2937 = 15.76612 J/Kg K15.76612 J/Kg K

The Total Change in Enthalpy Mach 4 The Total Change in Enthalpy Mach 4 ∑∆h = 57.58 + 103.68 –

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Verification of Results Using Flow Simulating Software Star

CCM

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For Mach 3 For Mach 3

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For Mach 4 For Mach 4

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Comparison of Analytical and

Comparison of Analytical and Numerical Results

Numerical Results

For Mach 3

At

At points points 1 1 2 2 3 3 4 4 5 5 6 6 77

Mach Number

Mach Number Analytical Analytical 3 3 2.685 2.685 2.25 2.25 2.41 2.41 2.54 2.54 2.77 2.77 2.972.97

Numerical

Numerical 3 3 2.74 2.74 2.3195 2.5227 2.3195 2.5227 2.698 2.698 2.92 2.92 2.96322.9632

Pressure(bar)

Pressure(bar) Analytical 1.0132 Analytical 1.0132 1.59 1.59 3.06 3.06 2.382 2.382 11.95 .95 1.36 1.36 1.0041.004

Numerical

Numerical 1.0132 1.0132 1.612 1.612 2.98 2.98 2.46 2.46 1.869 1.869 1.269 1.269 1.01181.0118

Temperature(K)

Temperature(K) Analytical Analytical 300 300 342 342 416.23 416.23 387.52 387.52 365.94 365.94 330.4 330.4 302.95302.95

Numerical Numerical 300 300 343.46 343.46 411.98 411.98 393.3 393.3 362.15 362.15 324.78 324.78 299.86299.86

For Mach 4

At At points points 1 1 2 2 3 3 4 4 5 5 6 6 77 Mach Number

Mach Number _Numerical_NumericalAnalytical Analytical 4 ₄₄4 _3.57_3.573.5 3.5 _2.874_2.8742.874 2.874 _3.035_3.0353.07 3.07 _3.35_3.353.29 3.29 _3.62_3.623.54 3.54 _3.948_3.9483.813.81

Pressure(bar)

Pressure(bar) Analytical Analytical 1.013 1.013 1.83 1.83 4.1 4.1 3.035 3.035 2.1945 2.1945 1.46 1.46 1.0021.002

Numerical

Numerical 1.013 1.013 1.786 1.786 4.13 4.13 3.08 3.08 2.04 2.04 1.394 1.394 1.0031.003

Temperature(K)

Temperature(K) Analytical Analytical 300 300 357.3 357.3 459.8 459.8 424.27 424.27 385.11 385.11 347.06 347.06 308.8308.8

Numerical

Numerical 300 300 361.9 361.9 459.7 459.7 415.3 415.3 379.76 379.76 344.22 344.22 299.7299.7

Conclusion Conclusion

From the above results we can see that the analytical and numerical results almost 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. comply with each other although there are small differences between the two values. These differences are mainly because the fact that numerical simulation approximates These differences are mainly because the fact that numerical simulation approximates the governing equations to algebraic equations. This approximation leads to deviation the governing equations to algebraic equations. This approximation leads to deviation from the analytical values.

from the analytical values.

Other factor resulting into deviation is fineness of meshing of the component. 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 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 domain is discretized into number of small domains. The discontinuities of the numerical solution over these number of small domains results in deviation. 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 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 higher the number of inner iterations the more the numerical solution is close to analytical solution

analytical solution

Also the entropy change in each region and over the entire is region yields a very small 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. 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. This was found to be correct after obtaining the values for the change in entropy.

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Appendix 1

Fig: Variation of Wave Deflection

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Appendix 2 Appendix 2

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