gas dynamics summer project 2012 final

7/29/2019 Gas Dynamics Summer Project 2012 Final

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Gas Dynamics SS 2012

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GAS DYNAMICS

SUMMER

PROJECT 2012

Prof Dr.-Ing Ernst von Lavante

BY

NISHANT KUMAR

Matrikelnummer :-ES0227948700

M.Sc (Computational Mechanics)


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CONTENTS

Sr.No Chapter Name Page Number

1 Problem statement -----

2 Abstract 3

3 Introduction & theory 4-7

4 Calculations 8-22

5 Comparison of Analytical & Simulation Results 23-25

6 Steps of Simulation in Software Star CMM+ 26-28

7 Simulation Results

8 Conclusion

List of Graphs & Charts

Aim Graph or charts

To find angle Shock Wave angle() From Graph of Shock wave angle ()

VsDeflection angle

To find Mach Number, Pressure ratio,

Temperature ratio, Prandtl-MeyerFunction

Chart of Supersonic flowfor

compressible flow


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Chapter No:-1

Abstract

Given problem is case of oblique shock. It is a flow over wedge. It consist of 2

compression & 4 expansion corners. The fluid flowing through wedge is air.We are using ideal gas with isentropic expansion & compression.

For the 2 compression corners, 60 & 100 we split the wave front into 2

components of Mach Number, the Normal component Mn & the tangential

component Mt

For rest of the 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 alldecrease. The analysis of expansion corner is different as compared to

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

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

turned isentropically to achieve the indicated Mach number.


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Chapter No:-02

Introduction & Theory

Normal ShockWavesWheretherearesupersonicflows,thereareusuallyalsoshockwaves.

Afundamentaltypeofshockwave isthenormal shockwavethe

shockwavenormaltotheflowdirection.

1.Basic RelationsConsiderarectangularpieceofair(the system)aroundanormal

shockwave,asisshowninfigure1. Totheleftofthisshockwavearetheinitialpropertiesof theflow(denotedbythesubscript1). To the rightaretheconditionsbehindthe

wave.

Figure1: Anormalshockwave.

We can already note a few things about the flow. It is a steady flow(theproperties stay constant in time). Itisalsoadiabatic, sincenoheatisadded.Noviscouseffectsarepresentbetweenthesystemand its

boundaries. Finally,therearenobody forces. Usingthecontinuityequation,wecan find thatthemassflowthat entersthe systemontheleftis1 u1 A1 ,with u thevelocityoftheflowinx-direction. Themass

flowthatleavesthe systemonthe rightis2 u2 A2 . However,sincethe

systemisrectangular,wehave A1 =A2 . Sowe find that

..........(Continuity Equation)

......................................... (1)


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The stagnation enthalpy remains constant over the flow

but since enthalpy is a function of temperature only

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

Using the Momentum Equation

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

Equations 1,2,3 are the governing equations for the Normal shock. Expressingall the relations in terms of M1 we get the Normal Shock tables

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

strength of shock is given by

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


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h= Cpt............................................................... (5)

Oblique Shocks Waves:-Inrealitynormalshockwavesdontoftenoccur. Obliqueshockwavesaremorecommon.1 ShockWaveAngles Whenanaircraftisflying,itcreatesdisturbancesintheflow. Thesedisturbancespreadaroundwiththe speedofsounda. Figure1 visualizesthese disturbances foranairplanetravelingfrompoint Atopoint B.

Figure1: Visualization ofthedisturbancesinaflow.

When the airplane flies at a subsonic velocity (V a), the disturbances cannot. In fact, they all stay within a cone and stackup at the edge, forming aMach wave. This conehas an angle , whereis

calledtheMach angle. Fromfigure1itcanbederivedthat,

sin=1/M

Suppose we have an airflow moving along a wall, which suddenly makesan angle away from the flow

We then get an expansion wave. In this expansion wave, the airflow

bends around the wall edge

While the airflow changes direction, its velocity also changes.we can derive an expression for , being

= (M2) (M1).the flow is isentropic, so the entropy s stays constant.

In a shock wave the pressure, density and temperature increase. In an

expansion wave it is exactly


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The 3 basic governing equations are;

.................................................(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

......................................(eqn 7)

Where M1 is the Normal Component of the Upstream Flow


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Chapter No:-3

Analytical Calculations

Part 1. For Mach 3

Given Data:-Given free stream conditions are

M1 = 3

P1 = 1.01325 bar

T1 = 300 K

Compression Curves = 6o and 10o

Expansion Curves = four 4o curves

Solution:-1) Compression Corner of 6o

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

Location of Shock and Normal Mach Number

Refer the Oblique Shock Charts ( Reference 1). We obtain for mach

number 3 and = 6o the value of the shock wave angle as 1 as 23o

Mn1 = M1 * sin n1 = 3*sin (23)

Mn1 = 1.1721

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

number 1.22 the following values of pressure ratio, temperature ratio

and downstream mach number

p2/p1 = 1.430; T2/T1 = 1.109; Mn2 = 0.8615

p2 = p1 * 1.430; T2 = T1 * 1.109; M2 = Mn2/sin(- ) p2 = 1.0325 * 1.570; T2 = 300 * 1.141; M2 = 0.8300/sin(23-6)

p2 = 1.4489 bar; T2 = 332.7 K; M2 = 2.946

Strength of Shock is given by

(p2 - p1)/p1 or (p2/p1)1


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(1.44891.01325)/1.01325

= 0.429

Change in Entropy is given by

s = Cp * ln(T2/T1) - R * ln(p2/p1)

s = 1005 * ln(332.7/300)287 * ln(1.4489/1.01325)s = 1.3328 J/Kg K

Change in Enthalpy is given by

h= Cpt

h = 1005 * (T2T1)h= 1005 * (332.7300)h = 32863.5 J/Kg K

Calculated

Values

T2

(K)

P2(bar)

M2 s

(J/KgK)

h

(J/KgK)

Compression

Corner 60

332.7 1.4489 2.946 1.3328 32863.5

2) Compression Corner of 10oM2= 2.946;P2 = 1.4489 bar; T2 = 332.7 K, =10

0



number 3 and = 28o

Mn2 = M2 * sin

n2 = 2.946* sin (28)

Mn2 = 1.383

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

number 1.383 the following values of pressure ratio, temperature

ratio and downstream mach number

p3/p2 = 2.055; T3/T2 = 1.242; Mn2 = 0.7483

p3 = p2 * 2.055; T3 = T2 * 1.242; M3 = Mn2/sin ( - )

p3 = 2.977 bar; T3 = 413.21 K; M3 = 2.422


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(p3- p2)/p2 or (p3/p2)1

(2.9771.4489)/1.4489

= 1.054662


s = Cp * ln(T3/T2) - R * ln(p3/p2)

s = 1005 * ln(1.242)287 * ln(2.055)

s = 6.257 J/Kg K


h= Cpth = 1005 * (T3T2)h= 1005 * (413.21332.7)h = 80912.55J/Kg K

Calculated

Values

T3

(K)

P3(bar)

M3 s

(J/KgK)

h

(J/KgK)

Compression

Corner 100

413.21 2.997 2.422 6.257 80912.55

3) Expansion Corner of 40M3 = 2.422;P3 = 2.997 bar; T3 = 413.21 K; 3 = 4

o

Prandtl-Meyer Function

Refer the Isentropic Charts

( Reference 2). We obtain for mach number 2.422 the value of

Prandtl- Meyer Function 3 = 37.229

4 = 3 + 34 = 33.018 + 4

4 = 41.229o

Refer the Normal Shock tables (Reference 2). We obtain for 4 =

41.229o the value of corresponding mach numberM4 as 2.59.


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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 (Tt = constant). Also there are no losses inthe flow, hence the stagnation pressure is also constant (pt =

constant)

p4/p3 = 0.5090 * 1 * (0.6630)-1; T4/T3 = 0.4271 * 1 *(0.4606)

-1

p4 = 2.2855 bar; T4 = 383.15 K


(p4/p3)1

0.7625 - 1

= -0.2374


s = Cp * ln(T4/T3) - R * ln(p4/p3)

s = 1005 * ln(0.9272)287 * ln(0.7677)

s = -0.09367 J/Kg K


h= Cpt

h = 1005 * (T4T3)

h = 1005 * (383.15413.21)

h = -30210.3 J/Kg K

Calculated

Values

T4

(K)

P4(bar)

M4 s

(J/KgK)

h

(J/KgK)

Expansion

Corner 40

383.15 2.2855 2.59 -0.09367 -30210.3

4) Expansion Corner of 4oM4 = 2.59;P4 = 2.2855 bar; T4 = 383.15 K; 4 = 4

o


Refer the Isentropic Charts ( Reference 2). We obtain for mach

number 2.59 the value of Prandtl- Meyer Function 4 = 41.229


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5 = 4 + 4

5 = 45.229o

Refer the Normal Shock tables (Reference 2). We obtain for 5 = 41.018o the

value of corresponding mach numberM5 as 2.78. Referring the same followingvalues of pressure ratio and temperature ratio are obtained as follows


p5/p4 = 0.3799 * 1 * (0.5090)-1; T5/T4 = 0. 3928* 1 * (0.4271)

-1

p5 = 1.70 bar; T5 = 352.97 K


(p5/p4)1

= 0.2561


s = Cp * ln(T5/T4) - R * ln(p5/p4)

s = 1005 * ln(0.9196)287 * ln(0.7463)

s = -0.25144 J/Kg K


h= Cpt

h = 1005 * (T5T4)

h = 1005 * (352.37383.15)

h = -30933.9 J/Kg K

Calculated

Values

T5

(K)

P5(bar)

M5 s

(J/KgK)

h

(J/KgK)

Expansion

Corner 40

352.37 1.70 2.78 -0.25144 -30933.9

5) Expansion Corner of 4oM5 = 2.78;p5 = 1.70 bar; T5 = 352.37 K; 5 = 4

o



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Refer the Isentropic Charts/ Normal Shock Charts ( Reference 2). We

obtain for mach number 2.54 the value of Prandtl- Meyer Function 5 =

41.018

6 = 5 + 5

6 = 49.229o

Refer the Normal Shock tables (Reference 2). We obtain for 6 = 45.018o

the value of corresponding mach numberM6 as 2.98. Referring the same

following values of pressure ratio and temperature ratio are obtained as

follows


p6/p5 = 0.2895 * 1 * (0.3799)-1; T6/T5 = 0.3602 * 1 * (0.3928)

-1

p6 = 1.29 bar; T6 = 323.125 K


(p6/p5)1

= -0.2411


s = Cp * ln(T6/T5) - R * ln(p6/p5)

s = 1005 * ln(0.9170)287 * ln(1.55172)

s = -0.257 J/Kg K


h= Cpt

h = 1005 * (T6T5)

h = -29391.22 J/Kg K

CalculatedValues

T6(K)

P6(bar)

M6 s(J/KgK)

h(J/KgK)

Expansion

Corner 40

323.125 1.29 2.78 -0.257 -29391.22


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6)Expansion Corner of 4o

M6 = 2.78;p6 = 1.29 bar; T6 = 323.125 K; 6 = 4o


7 = 6 + 67 = 53.229

o

Refer the Normal Shock tables (Reference 2). for 7 = 53.229o, mach number

M7 as 3.0123. values of pressure ratio and temperature ratio are obtained as

follows


Tt = constant and pt = constant

p7/p6 = 0.2682 * 1 * (.2895)-1; T7/T6 = 0.3556 * 1 * (0.3602)

-1

p7/p6 = 0.9264; T7/T6 = 0.917p7 = 1.195 bar; T7 = 318.99 K


(p7/p6)1

= -0.0736


s = Cp * ln(T7/T6) - R * ln(p7/p6)

s = - 0.6514J/Kg K


h= Cpt

h = 1005 * (T7T6)

h = -4155.67 J/Kg K

CalculatedValues

T7(K)

P7(bar)

M7 s(J/KgK)

h(J/KgK)

Expansion

Corner 40

318.99 1.195 3.0123 - 0.65140J -4155.67


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The Total Change in Entropy

s = Cp * ln(T7/T1) - R * ln(p7/p1)

=1005*ln(318.99/300)-287*ln(1.195/1.0135)

= 0.01414 J/Kg K

The Total Change in Enthalpy

h =32863.5+80912.5-30210.3-30933.9-29391.22-4155.67 = 1.908 KJ/Kg

K

Part 2. For Mach 4:-

Given Data:-Given free stream conditions are

M1 = 4

P1 = 1.01325 bar

T1 = 300 K

Compression Curves = 6o and 10o

Expansion Curves = four 4o curves

Solution:-1) Compression Corner of 6o

M1 = 4;P1 = 1.01325 bar; T1 = 300 K; 1 = 6o



number 3 and = 19o

Mn1 = M1 * sin

n1 = 4*sin(19)

Mn1 = 1.3022


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

number 1.22 the following values of pressure ratio, temperature ratio

and downstream mach number

p2/p1 = 1.805; T2/T1 = 1.191; Mn2 = 0.7860

p2 = p1 * 1.805; T2 = T1 * 1.191; M2 = Mn2/sin( - )p2 = 1.828 bar; T2 = 357.3 K; M2 = 3.49


(p2 - p1)/p1 or (p2/p1)1

= 0.8045


s = Cp * ln(T2/T1) - R * ln(p2/p1)s = -0.342 J/Kg K


h= Cpt

h = 1005 * (T2T1)h = 57586.5 J/Kg K

CalculatedValues

T2(K)

P2(bar)

M2 s(J/KgK)

h(J/KgK)

Compression

Corner 60

357.3 1.828 3.49 -0.342 57586.5

2) Compression Corner of 10oM2= 3.49;P2 = 1.828 bar; T2 = 357.3 K, =10

0


Refer the Oblique Shock Charts ( Reference 1). We obtain for machnumber 4 and = 25.5o

Mn2 = M2 * sin

n2 = 3.49* sin (25.5)

Mn2 = 1.5024


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Refer the Normal Shock tables (Reference 2). We obtain forMach

number 0.7011 the following values of pressure ratio, temperature

ratio and downstream mach number

p3/p2 = 2.458; T3/T2 = 1.320; Mn2 = 1.5024

p3 = p2 * 2.458; T3 = T2 * 1.320; M3 = Mn2/sin( - )p3 = 4.493bar; T3 = 471.636 K; M3 = 2.623


(p3- p2)/p2 or (p3/p2)1


s = Cp * ln(T3/T2) - R * ln(p3/p2)s = 1005 * ln(1.320)287 * ln(2.458)

s = 0.209 J/Kg K


h= Cpt

h = 1005 * (T3T2)h = 114907.68J/Kg K

Calculated

Values

T3

(K)

P3(bar)

M3 s

(J/KgK)

h

(J/KgK)

Compression

Corner 100

471.636 4.493 2.623 0.209 114907.68

3) Expansion Corner of 40M3 = 2.623;P3 = 4.493 bar; T3 = 471.636K; 3 = 4o


Refer the Isentropic Charts/ Normal Shock Charts ( Reference 2).

We obtain for mach number 2.623 the value of Prandtl- Meyer

Function 3 = 41.868


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4 = 3 + 34 = 45.868

o

Refer the Normal Shock tables (Reference 2). We obtain for 4 =

41.229o the value of corresponding mach numberM4 as 2.81.

Referring the same following values of pressure ratio andtemperature ratio are obtained as follows


Since the flow is isentropic and no work is done, the stagnation

temperature is constant (Tt = constant). Also there are no losses in

the flow, hence the stagnation pressure is also constant (pt =

constant)

p4/p3 = 0.3629* 1 * (0.4259)-1; T4/T3 = 0.3877 * 1 *(0.4196)

-1

p4 = 3.3517 bar; T4 = 435.77 KStrength of Shock is given by

(p4/p3)1

= 0.254


s = Cp * ln(T4/T3) - R * ln(p4/p3)

s = -0.33523 J/Kg K


h= Cpt

h = 1005 * (T4T3)

h = -36039.3J/Kg K

Calculated

Values

T4

(K)

P4(bar)

M4 s

(J/KgK)

h

(J/KgK)

Expansion

Corner 40

435.77 3.3517 2.81 -0.33523 -36039.3

4) Expansion Corner of 4oM4 =2.81;P4 =3.3517 bar; T4 =435.77 K; 4 = 4

o



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Refer the Isentropic Charts/ Normal Shock Charts ( Reference 2).

We obtain for mach number 2.81 the value of Prandtl- Meyer

Function 4 = 45.868

5 = 4 + 4

5 = 49.868o

Refer the Normal Shock tables (Reference 2). We obtain for 5 = 49.868o the

value of corresponding mach numberM5 as 3.00. Referring the same following

values of pressure ratio and temperature ratio are obtained as follows


p5/p4 = 0.2722 * 1 * (0.3629)-1; T5/T4 = 0. 3571* 1 * (0.3877)

-1

p5 = 2.514 bar; T5 = 401.37 K


(p5/p4)1

= -0.2499


s = Cp * ln(T5/T4) - R * ln(p5/p4)

s = -0.2978 J/Kg K


h= Cpt

h = 1005 * (T5T4)

h = -34572 J/Kg K

Calculated

Values

T5

(K)

P5(bar)

M5 s

(J/KgK)

h

(J/KgK)

Expansion

Corner 40

401.37 2.514 3.00 -0.2978 -34572


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5) Expansion Corner of 4oM5 =3.00;p5 = 2.514bar; T5 =401.37 K; 5 = 4

o


Refer the Isentropic Charts/ Normal Shock Charts ( Reference 2). Weobtain for mach number 3 the value of Prandtl- Meyer Function 5 =

49.868

6 = 5 + 5

6 = 53.868o

Refer the Normal Shock tables (Reference 2). We obtain for 6 = 45.018o

the value of corresponding mach numberM6 as 3.22. Referring the same

following values of pressure ratio and temperature ratio are obtained asfollows


p6/p5 = 0.1964 * 1 * (0.2722)-1; T6/T5 = 0.3253 * 1 * (0.3571)

-1

p6 = 1.813 bar; T6 = 365.6 K


(p6/p5)1

= -0.2788


s = Cp * ln(T6/T5) - R * ln(p6/p5)

s = -0.0622 J/Kg K


h= Cpt

h = 1005 * (T6T5)

h = -35948.85 J/Kg K

Calculated

Values

T6

(K)

P6(bar)

M6 s

(J/KgK)

h

(J/KgK)

Expansion

Corner 40

365.60 1.813 3.22 -0.0622 -35948.85


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6)Expansion Corner of 4o

M6 =3.22;p6 = 1.813 bar; T6 = 365.60 K; 6 = 4o

Prandtl-Meyer Function7 = 6 + 6

7 = 57.868o

Refer the Normal Shock tables (Reference 2). for7 = 53.229o, mach number

M7 as 3.0123. values of pressure ratio and temperature ratio are obtained as

follows


Tt = constant and pt = constant

p7/p6 = 0.1388 * 1 * (.1964)

-1

; T7/T6 = 0.2946 * 1 * (0.3253)

-1

p7/p6 = 0.706; T7/T6 = 0.905

p7 = 1.281 bar; T7 = 331.09 K


(p7/p6)1

= -0.2934


s = Cp * ln(T7/T6) - R * ln(p7/p6)

s = -0.001723J/Kg K


h= Cpt

h = 1.005 * (T7T6)

h = -34682.55 J/Kg K

Calculated

Values

T7

(K)

P7(bar)

M7 s

(J/KgK)

h

(J/KgK)

Expansion

Corner 40

331.09 1.281 3.0123 -0.00172 -34682.55


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The Total Change in Entropy

s= Cp * ln(T7/T1) - R * ln(p7/p1)= 0.82995 J/Kg K

The Total Change in Enthalpy

h = -34682.55-35948.85-34572-36039.3+114907.98+57586.5 = 3.125KJ/Kg K


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CHAPTER NO:-04

Comparison of Analytical & Simulation results

1) Temperature

Sr No. Mach Number

Temperature(k)

Shock wave Number

1 2 3 4 5 6 7

1

3(Analytical

Values) 300 332.7 413.21 383.15 352.37 323.125 318.99

2

4(Analytical

Values) 300 357.3 471.636 435.77 401.37 365.6 331.09

3

3(simulation

results) 229.86 331.27 410 385 350 325 320

4

4(simulation

results) 299.78 355 465 430 400 360 331

X axis:-Shock Wave Number

Y axis:-Temperature (k)

0

50

100

150

200

250

300

350

400

450

500

1 2 3 4 5 6 7

Mach 3

Mach 4

Mach 3-Simulation values

Mach 4-simulation values


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2) Absolute Pressure(bar)

Sr No. Mach Number

Absolute Pressure(bar)

Shock wave Number1 2 3 4 5 6 7

1

3(Analytical

Values) 1.01325 1.4489 2.977 2.2855 1.7 1.29 1.195

2

4(Analytical

Values) 1.01325 1.828 4.493 3.3517 3 1.813 1.281

3

3(simulation

results) 1.0118 1.4407 3 2.27 1.7 1.3 1.2

4

4(simulation

results) 1.0035 1.725 4.267 3.225 2.3 1.657 1.25


Y axis:-Absolute Pressure (bar)


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3) Mach Number

Sr No. Mach Number

Mach Number

Shock wave Number

1 2 3 4 5 6 7

1

3(Analytical

Values) 3 2.946 2.422 2.59 2.78 2.98 3.0123

2

4(Analytical

Values) 4 3.49 2.623 2.81 3 3.22 3.0123

3

3(simulation

results) 3.0011 2.8 2.2 2.5 2.8 2.9 3

4

4(simulation

results) 4.002 3.5 2.8 2.1 3.2 3.3 3.01


Y axis:-Mach Number


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CHAPTER NO:-05

Steps of Simulation in Software Star CMM+

Step 1:- Initial Environment settingStar CMM+ is a server based software, all calculations & simulations

are done through remote server of company CD-adapco, so every time

for running the software we have to check for licensing of the software

for that we follow these steps;

File Menu-New Simulation-Run Mode-Serial-Click OK

After following these steps server will check for licensing of the software,

after that the work can be started in software.

Step 2:- Importing a meshed wedge fileFollow the following sequence for importing a meshed model

File Menu-import mesh-volume mesh-select the .cas file-click OK

Scaling the Mesh file;

Mesh menu-scale mesh-enter value 0.01-apply-close

Step 3:- Pre-Processing DataSelect the data for input for the problem

Continua-Physics-Implicit unsteady (as its easy to solve from implicit

unsteady equations)-Gas-Coupled Flow-ideal Gas-Laminar flow

Regions-Fluid-Velocity inlet-Free stream-physics value-mach number-

constant (change value to 3 & 4)

Step 4:- SolversNow after putting initial data the software has to be given a command ofsolving the problem for that follow the steps;

Solvers-Implicit unsteady-Change time step to 0.0001 sec-Change

Temporal Discritisation to 2nd Order (as these equations leads to more

converging solution)

Stopping criteria (Up to limits to which calculations has to be done)

Maximum inner iterations (the iterations carried out for calculations)-


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Disenabling maximum physical time-Maximum steps (Steps upto which

iterations has to be performed)

Step 5:- Post ProcessingAfter solving the data needs to be post proceeds for visulations

Scene-New Scene-Scalar-Change function to Mach Number, Absolute

Pressure & Temperature-For improving the colour quality of results-

Displayers-Scalar 1-Counter Style-smooth filled


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Chapter No:-06

Simulation Results


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Page 29

Chapter No:-07

Conclusion

It is observed that there is slight variation between the analytical &

simulation results it is due to following reasons;

1) As simulation (solving in software) is based on approximation ofgoverning algebraic equation which leads to deviation from actual

value.

2) The type of mesh used on component, if the descritization i.e meshcreation is not proper the results would deviate from actual value.

Also the mesh connectivity is of prime importance, if mesh

connectivity specifically near the curvature is not proper then solver

cannot compute the right result.

3) The higher the number of iterations selected would result into higherdegree of accurate results but it has a disadvantage of more iteratingtime & power.

4) The entropy change in each region and over the entire is region yieldsa very small value since the entropy change is directly affected by the

cube of flow deflection angle.

5) Coding behind the software is also one of the reasons behinddifferences in values.


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References

1) Zucker gas Dynamics2)Naca Report

gas dynamics summer project 2012 final

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