gas dynamics summer project 2012 final
TRANSCRIPT
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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
Location of Shock and Normal Mach Number
Refer the Oblique Shock Charts ( Reference 1). We obtain for mach
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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Strength of Shock is given by
(p3- p2)/p2 or (p3/p2)1
(2.9771.4489)/1.4489
= 1.054662
Change in Entropy is given by
s = Cp * ln(T3/T2) - R * ln(p3/p2)
s = 1005 * ln(1.242)287 * ln(2.055)
s = 6.257 J/Kg K
Change in Enthalpy is given by
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
Strength of Shock is given by
(p4/p3)1
0.7625 - 1
= -0.2374
Change in Entropy is given by
s = Cp * ln(T4/T3) - R * ln(p4/p3)
s = 1005 * ln(0.9272)287 * ln(0.7677)
s = -0.09367 J/Kg K
Change in Enthalpy is given by
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
Prandtl-Meyer Function
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 = p5/p5t * p5t/p4t * p4t/p4; T5/T4 = T5/T5t * T5t/T4t * T4t/T4;
p5/p4 = 0.3799 * 1 * (0.5090)-1; T5/T4 = 0. 3928* 1 * (0.4271)
-1
p5 = 1.70 bar; T5 = 352.97 K
Strength of Shock is given by
(p5/p4)1
= 0.2561
Change in Entropy is given by
s = Cp * ln(T5/T4) - R * ln(p5/p4)
s = 1005 * ln(0.9196)287 * ln(0.7463)
s = -0.25144 J/Kg K
Change in Enthalpy is given by
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
Prandtl-Meyer Function
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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 = p6/p6t * p6t/p5t * p5t/p5; T6/T5 = T6/T6t * T6t/T5t * T5t/T5;
p6/p5 = 0.2895 * 1 * (0.3799)-1; T6/T5 = 0.3602 * 1 * (0.3928)
-1
p6 = 1.29 bar; T6 = 323.125 K
Strength of Shock is given by
(p6/p5)1
= -0.2411
Change in Entropy is given by
s = Cp * ln(T6/T5) - R * ln(p6/p5)
s = 1005 * ln(0.9170)287 * ln(1.55172)
s = -0.257 J/Kg K
Change in Enthalpy is given by
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
Prandtl-Meyer Function
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
p7/p6 = p7/p7t * p7t/p6t * p6t/p6; T7/T6 = T7/T7t * T7t/T6t * T6t/T6;
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
Strength of Shock is given by
(p7/p6)1
= -0.0736
Change in Entropy is given by
s = Cp * ln(T7/T6) - R * ln(p7/p6)
s = - 0.6514J/Kg K
Change in Enthalpy is given by
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
Location of Shock and Normal Mach Number
Refer the Oblique Shock Charts ( Reference 1). We obtain for mach
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
Strength of Shock is given by
(p2 - p1)/p1 or (p2/p1)1
= 0.8045
Change in Entropy is given by
s = Cp * ln(T2/T1) - R * ln(p2/p1)s = -0.342 J/Kg K
Change in Enthalpy is given by
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
Location of Shock and Normal Mach Number
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
Strength of Shock is given by
(p3- p2)/p2 or (p3/p2)1
Change in Entropy is given by
s = Cp * ln(T3/T2) - R * ln(p3/p2)s = 1005 * ln(1.320)287 * ln(2.458)
s = 0.209 J/Kg K
Change in Enthalpy is given by
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
Prandtl-Meyer Function
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
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 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
Change in Entropy is given by
s = Cp * ln(T4/T3) - R * ln(p4/p3)
s = -0.33523 J/Kg K
Change in Enthalpy is given by
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
Prandtl-Meyer Function
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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 = p5/p5t * p5t/p4t * p4t/p4; T5/T4 = T5/T5t * T5t/T4t * T4t/T4;
p5/p4 = 0.2722 * 1 * (0.3629)-1; T5/T4 = 0. 3571* 1 * (0.3877)
-1
p5 = 2.514 bar; T5 = 401.37 K
Strength of Shock is given by
(p5/p4)1
= -0.2499
Change in Entropy is given by
s = Cp * ln(T5/T4) - R * ln(p5/p4)
s = -0.2978 J/Kg K
Change in Enthalpy is given by
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
Prandtl-Meyer Function
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 = p6/p6t * p6t/p5t * p5t/p5; T6/T5 = T6/T6t * T6t/T5t * T5t/T5;
p6/p5 = 0.1964 * 1 * (0.2722)-1; T6/T5 = 0.3253 * 1 * (0.3571)
-1
p6 = 1.813 bar; T6 = 365.6 K
Strength of Shock is given by
(p6/p5)1
= -0.2788
Change in Entropy is given by
s = Cp * ln(T6/T5) - R * ln(p6/p5)
s = -0.0622 J/Kg K
Change in Enthalpy is given by
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
p7/p6 = p7/p7t * p7t/p6t * p6t/p6; T7/T6 = T7/T7t * T7t/T6t * T6t/T6;
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
Strength of Shock is given by
(p7/p6)1
= -0.2934
Change in Entropy is given by
s = Cp * ln(T7/T6) - R * ln(p7/p6)
s = -0.001723J/Kg K
Change in Enthalpy is given by
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
X axis:-Shock Wave Number
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
X axis:-Shock Wave Number
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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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