Supersonic Flow Over a Wedge(Oblique Shock Problem)
description
Transcript of 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)
Analytical and Numerical Methods for Oblique Shock Problems Page 2
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)
Analytical and Numerical Methods for Oblique Shock Problems Page 3
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
Analytical and Numerical Methods for Oblique Shock Problems Page 4
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 and Numerical Methods for Oblique Shock Problems Page 5
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
Analytical and Numerical Methods for Oblique Shock Problems Page 6
2. Compression Corner of 10o
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
Refer the Normal Shock tables (Appendix 2). We obtain for mach number 1.3425 the
following values of pressure ratio, temperature ratio and downstream mach number
p3/p2 = 1.928; T3/T2 = 1.216; Mn3 = 0.7664
p3 = p2 * 1.928; T3 = T2 * 1.216; 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
Strength of Shock is given by
(p3/p2) – 1
1.928 – 1
= 0.928
Change in Entropy is given by
∆s = Cp * ln(T3/T2) - R * ln(p3/p2)
∆s = 1005 * ln(1.216) – 287 * ln(1.928)
∆s = 8.14 J/Kg K
Change in Enthalpy is given by
∆h= Cp∆t
∆h = 1005 * (T3 – T2)
∆h = 1005 * (416.23 – 342.3)
∆h = 74290 J/Kg K
Analytical and Numerical Methods for Oblique Shock Problems Page 7
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
Strength of Shock is given by
(p4/p3) – 1
0.7786 - 1
= -0.2214
Change in Entropy is given by
∆s = Cp * ln(T4/T3) - R * ln(p4/p3)
∆s = 1005 * ln(0.931) – 287 * ln(0.7786)
∆s = -0.02948 J/Kg K
Change in Enthalpy is given by
∆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)
Analytical and Numerical Methods for Oblique Shock Problems Page 8
4. Expansion Corner of 4o
M4 = 2.41; P4 = 2.382 bar; T4 = 387.52 K; ϴ4 = 4o
Prandtl-Meyer Function
Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach
number 2.41 the value of Prandtl- Meyer Function ν4 = 37.018
ν5 = ν4 + ϴ4
ν4 = 37.018 + 4
ν4 = 41.018o
Refer the Normal Shock tables (Appendix 2). We obtain for ν5 = 41.018o the value of
corresponding mach number M5 as 2.54. Referring the same following values of
pressure ratio and temperature ratio are obtained as follows
p5/p4 = p5/p5t * p5t/p4t * p4t/p4; T5/T4 = T5/T5t * T5t/T4t * T4t/T4;
Since the flow is isentropic and no work is done, the stagnation temperature is constant (T t =
constant). Also there are no losses in the flow, hence the stagnation pressure is also constant
(pt = constant)
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
Strength of Shock is given by
(p5/p4) – 1
0.8167 - 1
= -0.1833
Change in Entropy is given by
∆s = Cp * ln(T5/T4) - R * ln(p5/p4)
∆s = 1005 * ln(0.9437) – 287 * ln(0.8167)
∆s = -0.1239 J/Kg K
Change in Enthalpy is given by
∆h= Cp∆t
∆h = 1005 * (T5 – T4)
∆h = 1005 * (365.74 - 387.52)
∆h = -21889 J/Kg K
Location of Shock β
pii/pi = 2γ/(γ-1) * Mi2 sin
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
β = -22.39 o ……………………………(consider –ve value, since it’s a case of expansion)
Analytical and Numerical Methods for Oblique Shock Problems Page 9
5. Expansion Corner of 4o
M5 = 2.54; p5 = 1.94 bar; T5 = 365.74 K; ϴ5 = 4o
Prandtl-Meyer Function
Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach
number 2.54 the value of Prandtl- Meyer Function ν5 = 41.018
ν6 = ν5 + ϴ5
ν6 = 41.018 + 4
ν6 = 45.018o
Refer the Normal Shock tables (Appendix 2). We obtain for ν6 = 45.018o the value of
corresponding mach number M6 as 2.77. Referring the same following values of
pressure ratio and temperature ratio are obtained as follows
p6/p5 = p6/p6t * p6t/p5t * p5t/p5; T6/T5 = T6/T6t * T6t/T5t * T5t/T5;
Since the flow is isentropic and no work is done, the stagnation temperature is constant (T t =
constant). Also there are no losses in the flow, hence the stagnation pressure is also constant
(pt = constant)
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
Strength of Shock is given by
(p6/p5) – 1
0.7014 - 1
= -0.2986
Change in Entropy is given by
∆s = Cp * ln(T6/T5) - R * ln(p6/p5)
∆s = 1005 * ln(0.9036) – 287 * ln(0.7014)
∆s = -0.08305 J/Kg K
Change in Enthalpy is given by
∆h= Cp∆t
∆h = 1005 * (T6 – T5)
∆h = 1005 * (330 – 365.74)
∆h = -35516.7 J/Kg K
Location of Shock β
pii/pi = 2γ/(γ-1) * Mi2 sin
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
β = -19.91 o ……………………………(consider –ve value, since it’s a case of expansion)
Analytical and Numerical Methods for Oblique Shock Problems Page 10
6. Expansion Corner of 4o
M6 = 2.77; p6 = 1.36 bar; T6 = 330.4 K; ϴ6 = 4o
Prandtl-Meyer Function
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
p7/p6 = p7/p7t * p7t/p6t * p6t/p6; T7/T6 = T7/T7t * T7t/T6t * T6t/T6;
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
Strength of Shock is given by
(p7/p6) – 1
0.7382 - 1
= -0.2618
Change in Entropy is given by
∆s = Cp * ln(T7/T6) - R * ln(p7/p6)
∆s = 1005 * ln(0.917) – 287 * ln(0.7382)
∆s = -0.03504 J/Kg K
Change in Enthalpy is given by
∆h= Cp∆t
∆h = 1005 * (T7 – T6)
∆h = 1005 * (302.95 - 330)
∆h = -27185.25 J/Kg K
Location of Shock β
pii/pi = 2γ/(γ-1) * Mi2 sin
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
β = -19.76 o ……………………………(consider –ve value, since it’s a case of expansion)
Analytical and Numerical Methods for Oblique Shock Problems Page 11
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
1. Compression Corner of 6o
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
Refer the Normal Shock tables (Appendix 2). We obtain for mach number 1.3022 the
following values of pressure ratio, temperature ratio and downstream mach number
p2/p1 = 1.806; T2/T1 = 1.191; Mn2 = 0.7866
p2 = p1 * 1.806; T2 = T1 * 1.191; 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
Strength of Shock is given by
(p2 - p1)/p1 or (p2/p1) – 1
1.806 – 1
= 0.806
Change in Entropy is given by
∆s = Cp * ln(T2/T1) - R * ln(p2/p1)
∆s = 1005 * ln(1.191) – 287 * ln(1.806)
∆s = 6.0174 J/Kg K
Change in Enthalpy is given by
∆h= Cp∆t
∆h = 1005 * (T2 – T1)
∆h = 1005 * (357.3 – 300)
∆h = 57580 J/Kg K
Analytical and Numerical Methods for Oblique Shock Problems Page 12
2. Compression Corner of 10o
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
Refer the Normal Shock tables (Appendix 2). We obtain for mach number 1.45 the
following values of pressure ratio, temperature ratio and downstream mach number
p3/p2 = 2.286; T3/T2 = 1.287; Mn3 = 0.7196
p3 = p2 * 2.286; T3 = T2 * 1.287; 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
Strength of Shock is given by
(p3/p2) – 1
2.286 – 1
= 1.286
Change in Entropy is given by
∆s = Cp * ln(T3/T2) - R * ln(p3/p2)
∆s = 1005 * ln(1.287) – 287 * ln(2.286)
∆s = 16.28 J/Kg K
Change in Enthalpy is given by
∆h= Cp∆t
∆h = 1005 * (T3 – T2)
∆h = 1005 * (459.87 – 357.3)
∆h = 103068 J/Kg K
Analytical and Numerical Methods for Oblique Shock Problems Page 13
3. Expansion Corner of 4o
M3 = 2.874; P3 = 4.1 bar; T3 = 459.87 K; ϴ3 = 4o
Prandtl-Meyer Function
Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach
number 2.874 the value of Prandtl- Meyer Function ν3 = 47.18523
ν4 = ν3 + ϴ3
ν4 = 47.18523 + 4
ν4 = 51.18523o
Refer the Normal Shock tables (Appendix 2). We obtain for ν4 = 51.18523o the value of
corresponding mach number M4 as 3.07. Referring the same following values of
pressure ratio and temperature ratio are obtained as follows
p4/p3 = p4/p4t * p4t/p3t * p3t/p3; T4/T3 = T4/T4t * T4t/T3t * T3t/T3;
Since the flow is isentropic and no work is done, the stagnation temperature is constant (Tt =
constant). Also there are no losses in the flow, hence the stagnation pressure is also constant
(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
Strength of Shock is given by
(p4/p3) – 1
0.7426 - 1
= -0.2626
Change in Entropy is given by
∆s = Cp * ln(T4/T3) - R * ln(p4/p3)
∆s = 1005 * ln(0.918) – 287 * ln(0.7426)
∆s = -0.57512 J/Kg K
Change in Enthalpy is given by
∆h= Cp∆t
∆h = 1005 * (T4 – T3)
∆h = 1005 * (424.279 – 459.87)
∆h = -35591 J/Kg K
Location of Shock β
pii/pi = 2γ/(γ-1) * Mi2 sin
2β – (γ-1) /(γ+1)........................................................................(eqn 8)
0.7426 = 2*1.4/(1.4-1) * 2.8742sin
2β – (1.4-1)/(1.4+1)
Analytical and Numerical Methods for Oblique Shock Problems Page 14
β = sin-1
(0.3067); β = +17.86o or -17.86
o
β = -17.86 o ……………………………(consider –ve value, since it’s a case of expansion)
4. Expansion Corner of 4o
M4 = 3.07; P4 = 3.035 bar; T4 = 424.279 K; ϴ4 = 4o
Prandtl-Meyer Function
Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach
number 2.41 the value of Prandtl- Meyer Function ν4 = 51.18523
ν5 = ν4 + ϴ4
ν4 = 51.18523 + 4
ν4 = 55.18523o
Refer the Normal Shock tables (Appendix 2). We obtain for ν5 = 55.18523o the value of
corresponding mach number M5 as 3.9. Referring the same following values of pressure
ratio and temperature ratio are obtained as follows
p5/p4 = p5/p5t * p5t/p4t * p4t/p4; T5/T4 = T5/T5t * T5t/T4t * T4t/T4;
Since the flow is isentropic and no work is done, the stagnation temperature is constant (T t =
constant). Also there are no losses in the flow, hence the stagnation pressure is also constant
(pt = constant)
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
Strength of Shock is given by
(p5/p4) – 1
0.72308 - 1
= -0.277
Change in Entropy is given by
∆s = Cp * ln(T5/T4) - R * ln(p5/p4)
∆s = 1005 * ln(0.907) – 287 * ln(0.72308)
∆s = -5.04 J/Kg K
Change in Enthalpy is given by
∆h= Cp∆t
∆h = 1005 * (T5 – T4)
∆h = 1005 * (385.11 – 424.279)
∆h = -39364.845 J/Kg K
Location of Shock β
pii/pi = 2γ/(γ-1) * Mi2 sin
2β – (γ-1) /(γ+1)........................................................................(eqn 8)
0.72308 = 2*1.4/(1.4-1) * 3.072sin
2β – (1.4-1)/(1.4+1)
Analytical and Numerical Methods for Oblique Shock Problems Page 15
β = sin-1
(0.2843); β = +16.52o or -16.52
o
β = -16.52 o ……………………………(consider –ve value, since it’s a case of expansion)
5. Expansion Corner of 4o
M5 = 3.29; p5 = 2.19455 bar; T5 = 385.11 K; ϴ5 = 4o
Prandtl-Meyer Function
Refer the Isentropic Charts/ Normal Shock Charts ( Appendix 2). We obtain for mach
number 3.29 the value of Prandtl- Meyer Function ν5 = 55.18523
ν6 = ν5 + ϴ5
ν6 = 55.18523 + 4
ν6 = 59.18523o
Refer the Normal Shock tables (Appendix 2). We obtain for ν6 = 59.18523o the value of
corresponding mach number M6 as 3.54. Referring the same following values of
pressure ratio and temperature ratio are obtained as follows
p6/p5 = p6/p6t * p6t/p5t * p5t/p5; T6/T5 = T6/T6t * T6t/T5t * T5t/T5;
Since the flow is isentropic and no work is done, the stagnation temperature is constant (T t =
constant). Also there are no losses in the flow, hence the stagnation pressure is also constant
(pt = constant)
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
Strength of Shock is given by
(p6/p5) – 1
0.7000 - 1
= -0.3000
Change in Entropy is given by
∆s = Cp * ln(T6/T5) - R * ln(p6/p5)
∆s = 1005 * ln(0.9026) – 287 * ln(0.7000)
∆s = -0.62246 J/Kg K
Change in Enthalpy is given by
∆h= Cp∆t
∆h = 1005 * (T6 – T5)
∆h = 1005 * (347.06 – 384.11)
∆h = -38240.25 J/Kg K
Location of Shock β
pii/pi = 2γ/(γ-1) * Mi2 sin
2β – (γ-1) /(γ+1)........................................................................(eqn 8)
0.7000 = 2*1.4/(1.4-1) * 3.292sin
2β – (1.4-1)/(1.4+1)
Analytical and Numerical Methods for Oblique Shock Problems Page 16
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)
6. Expansion Corner of 4o
M6 = 3.54; p6 = 1.46 bar; T6 = 347.06 K; ϴ6 = 4o
Prandtl-Meyer Function
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
p7/p6 = p7/p7t * p7t/p6t * p6t/p6; T7/T6 = T7/T7t * T7t/T6t * T6t/T6;
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
Strength of Shock is given by
(p7/p6) – 1
0.6868 - 1
= -0.3132
Change in Entropy is given by
∆s = Cp * ln(T7/T6) - R * ln(p7/p6)
∆s = 1005 * ln(0.8983) – 287 * ln(0.6868)
∆s = -0.2937 J/Kg K
Change in Enthalpy is given by
∆h= Cp∆t
∆h = 1005 * (T7 – T6)
∆h = 1005 * (302.95 - 330)
∆h = -38370.9J/Kg K
Location of Shock β
pii/pi = 2γ/(γ-1) * Mi2 sin
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)
Analytical and Numerical Methods for Oblique Shock Problems Page 17
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
Analytical and Numerical Methods for Oblique Shock Problems Page 18
Verification of Results Using Flow Simulating Software Star
CCM
For Mach 3
For Mach 4
Analytical and Numerical Methods for Oblique Shock Problems Page 19
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.
Analytical and Numerical Methods for Oblique Shock Problems Page 20
Appendix 1
Fig: Variation of Wave Deflection Angle wrt to Flow Deflection Angle for Various Mach Numbers