310
0
x
ua
au
t
10
01 =
∂
∂
⎥
⎦
⎤
⎢
⎣
⎡
+
∂
∂
⎥
⎦
⎤
⎢
⎣
⎡ww
where the dependent column vector is w = ⎥
⎦
⎤
⎢
⎣
⎡
R
u. Defining the coefficient matrices as
⎥
⎦
⎤
⎢
⎣
⎡
=
⎥
⎦
⎤
⎢
⎣
⎡
=ua
au
10
01 BA
the above equation may be written as
0
x
t
=
∂
∂
+
∂
∂
w
B
w
A
Note A is the identity matrix I. Therefore, A-1 = I. So
0
x
t
x
t
=
∂
∂
+
∂
∂
=
∂
∂
+
∂
∂1w
B
ww
BA
w–
The characteristic directions are obtained by determining the eigenvalues of the
coefficient matrix of ∂w/∂x, which in this case is matrix B, hence
0=λ− IB
that is
0
ua
au =
λ−
λ−
This is the same determinant as obtained in the previous problem. Expanding gives
(
)
0au 2
2=−λ−
Solution of this expression yields
au
d
t
dx ±==λ
To derive the compatibility equation, the left eigenvectors must first be determined.
Rather than obtaining results for each characteristic. The following applies to
characteristics of either family
(
)
0=λ− IBlT
or
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311
[]
0
ua
au
ll 21 =
⎥
⎦
⎤
⎢
⎣
⎡
λ−
λ−
Expanding gives two equations produces
(
)
()
0lula
0allu
21
21
=λ−+
=
+
λ
−
Hence,
12
12
l
u
a
l
l
a
u
l
⎟
⎠
⎞
⎜
⎝
⎛
−λ
=
⎟
⎠
⎞
⎜
⎝
⎛−λ
=
Using the fact that
au
±
=
λ
produces that
12 ll
±
=
Take l1 to be unity. The compatibility equation for characteristics is
0=wlTd
or
[] []
0dRdu
dR
du
11
dR
du
ll 21 =±=
⎥
⎦
⎤
⎢
⎣
⎡
±=
⎥
⎦
⎤
⎢
⎣
⎡
Problem 9. – (a) Combine Eq.(14.29) with both expressions in Eq.(14.27) to obtain
Eq.(14.30); (b) Substitute Eq.(14.30) into Eq.(14.29) to obtain
3
31
2
22
C
CCCC
du
dv −−
=m
(a) We begin with the simpler of the two expressions
(
)
0CC 2121
=
σ
+
σ
−
λ
Hence,
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc.,
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312
λ−=
σ
σ
12
1
2CC
But
1
31
2
22
C
CCCC −±
=λ
Substitution brings
31
2
2
1
31
2
22
12
1
2CCC
C
CCCC
CC −=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛−±
−=
σ
σm
Next the second expression is used, i.e.,
(
)
0CC 2132
=
λ
σ
+
σ
−
λ
or
()
()
31
2
22
31
2
22
31
2
22
31
2
22
31
2
22
31
2
22
2
31
2
2
31
2
22
31
2
22231
2
31
2
22
31
2
323
1
2
CCCC
CCCC
CCCC
CCCC
CCCC
CCCCCCC
CCCC
CCCCCCC
C
CCCC
CC
C
CCC
−=
−±
−±
−=
−±
−−−
=
−±
−±−
=−
−±
=
−
λ
=
λ
λ−
=
σ
σ
m
m
m
(b) From Eq.(14.29) we have
0dvCduC
1
2
21 =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
σ
σ
++
Substituting the results from part (a) gives
(
)
0dvCCCCduC 31
2
221 =−+ m
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc.,
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313
Rearranging this brings
()
()
3
31
2
22
31
2
2
2
2
31
2
221
31
2
22
31
2
22
31
2
22
1
31
2
22
1
C
CCCC
CCCC
CCCCC
CCCC
CCCC
CCCC
C
CCCC
C
du
dv
−±−
=
−−
−−
=
−
−
−±
−
=
−±
−
=
m
m
m
Problem 10. – In example 14.2 only one of four compatibility equations was determined.
Complete this example by determine the remaining three.
The complete left eigenvector is
() ()
() ()
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
ρρ−
−+
−
−+
ρ
−+−+
−
ρ
=
0010
va1va1a11
avu
vu
avu
vu
0v
avu
1
avu
1
0u
2
222222
222222
L
The compatibility equations are established by application of
0=
λwlT
id
for each left eigenvector corresponding to a particular eigenvalue.
Along the characteristic given by dy/dy = λ2
[]
0d
a
dp
d
dp
dv
du
1
a
1
00
d
dp
dv
du
lllld 22
2
4
2
3
2
2
2
12 =ρ+−=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
ρ
⎥
⎦
⎤
⎢
⎣
⎡−
=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
ρ
=
λwlT
So we obtain the speed sound expression
2
a
d
dp =
ρ
Along the characteristics given by dy/dy = λ3 and λ4 we have
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
314
[]
0
d
dp
dv
du
0
va
1
a
vu
a
1
d
dp
dv
du
lllld 4,3
4
4,3
3
4,3
2
4,3
14,3 =
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
ρ
⎥
⎦
⎤
⎢
⎣
⎡
ρβ
±
β
=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
ρ
=
λ
m
wlT
where β =
()
1avu1M 2222 −+=− . Expanding and canceling terms yields
0dpudvvdu =
ρ
β
+±m
Now according to Bernoulli’s equation
()
vdvudu
dp +−=
ρ
Uniting the expressions and rearranging produces
vu
uv
du
dv
β
β
±
=m
Multiply both numerator and denominator by u ± βv. The numerator simplifies as
follows
()()
()( )
() ()
[]
4222
2
22
222
222
avuauv
a
vu
vu1uv
uvvuuvvuuv
−+±
+
=+β±β+=
β+β±β±=β±β±
Whereas the denominator simplifies as follows
()()
(
)
(
)()
22
2
22
22
2
22
2222 va
a
vu
vv
a
vu
uvuvuuu −
+
=+
+
−=β−=β±βm
Hence,
(
)
22
4222
va
avuauv
du
dv
−
−+±
=
Problem 11. – Use the Method of Indeterminate Derivatives to determine the equations
of the characteristics for linearized, two-dimensional, supersonic flow described by
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
315
()
0
yx
M1 2
2
2
2
2=
∂
φ∂
+
∂
φ∂
−∞
There is no need to go through the entire analysis since this is a simple extension of
theory presented in the Chapter. Instead simply let
(
)
1C
0C
M1C
3
2
22
1
=
=
β−=−= ∞
Note for supersonic flow C1 is negative and the potential equation is actually the wave
equation, which is hyperbolic.
1M
11
00
C
CCCC
dx
dy
2
2
2
1
31
2
22
−
=
β
=
β−
β+±
=
−±
=
∞
mm
1M
1
00
C
CCCC
du
dv 2
2
2
3
31
2
22 −=β=
β+
=
−−
=∞
mm
m
m
Problem 12. – (a) Show that each dependent variable in the following pair of equations
must satisfy the wave equation and therefore the set is hyperbolic
0
x
v
y
u
0
y
v
x
u
=
∂
∂
−
∂
∂
=
∂
∂
−
∂
∂
(b) Use the Method of Linear Combination to determine the equations of the
characteristics for this set of equations.
(a) The wave equation is written in the x-y plane as
2
2
2
2
2
y
f
c
x
f
∂
∂
=
∂
∂
where c is the wave speed.
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
316
Now differentiate the first equation wrt to x and the second wrt y to obtain
0
xy
v
y
u
0
yx
v
x
u
2
2
2
2
2
2
=
∂∂
∂
−
∂
∂
=
∂∂
∂
−
∂
∂
Subtraction produces the wave equation with a wave speed of ±1
2
2
2
2
y
u
x
u
∂
∂
=
∂
∂
Now differentiate the first equation wrt to y and the second wrt x to obtain
0
x
v
yx
u
0
y
v
xy
u
2
22
2
22
=
∂
∂
−
∂∂
∂
=
∂
∂
−
∂∂
∂
Subtraction produces the wave equation with a wave speed of ±1
2
2
2
2
y
v
x
v
∂
∂
=
∂
∂
(b) Multiply the given equations
0
x
v
y
u
0
y
v
x
u
=
∂
∂
−
∂
∂
=
∂
∂
−
∂
∂
by σ1 and σ2, respectively, and add to get
0
x
v
y
u
y
v
x
u
21 =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
−
∂
∂
σ+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
−
∂
∂
σ
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
317
Rearrangement brings
0
y
v
x
v
y
u
x
u
2
1
2
1
2
1=
⎥
⎦
⎤
⎢
⎣
⎡
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
σ
σ
+
∂
∂
σ−
⎥
⎦
⎤
⎢
⎣
⎡
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
σ
σ
+
∂
∂
σ
Compare the group of terms within the square brackets to the following total derivatives
dx
dv
y
v
dx
dy
x
v
dx
du
y
u
dx
dy
x
u
=
∂
∂
+
∂
∂
=
∂
∂
+
∂
∂
From this comparison we may write the slope, dx/dt, denoted as λ, as
2
1
1
2
du
dv
σ
σ
=
σ
σ
=λ=
Expanding this pair of equations produces two equations for σ1 and σ2
0
0
21
21
=λσ−σ
=
σ
−
λσ
A unique solution for σ1 and σ2 will be obtained if and only if the determinant of the
coefficients vanishes, i.e.,
0
1
1=
λ−
−λ
Expanding and rearranging the result produces the quadratic equation,
01
2=+λ−
Solution of this expression yields
1
d
t
dx ±==λ
The eigenvalues yield the wave speed of the wave equation.
To derive the compatibility equation, incorporate the total derivative equations into the
combined equation and obtain
0dvdu 21
=
σ
−
σ
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
318
or
0dvdu
2
1=−
σ
σ
The relationship between σ1 and σ2 is obtained from use of either
(
)
()
0ua
0au
21
21
=σλ−+σ
=
σ
+
σ
λ
−
Using either of these and the expression for λ produces
1
a
uau
a
u
2
1±=
−±
=
−λ
=
σ
σ
Hence,
0dvdu
=
−
±
or
1
du
dv ±=
along lines with slopes
au
dt
dx ±=
Problem 13. – Use Eigenanalysis to determine the equations of the characteristics in
problem 12.
The pair of equations
0
x
v
y
u
0
y
v
x
u
=
∂
∂
−
∂
∂
=
∂
∂
−
∂
∂
can be written in vector matrix form as
0
x
01
10
t
10
01 =
∂
∂
⎥
⎦
⎤
⎢
⎣
⎡−
+
∂
∂
⎥
⎦
⎤
⎢
⎣
⎡
−
ww
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
319
where the dependent column vector is w = ⎥
⎦
⎤
⎢
⎣
⎡
v
u. Defining the coefficient matrices as
⎥
⎦
⎤
⎢
⎣
⎡−
=
⎥
⎦
⎤
⎢
⎣
⎡
−
=01
10
10
01 BA
the above equation may be written as
0
xt =
∂
∂
+
∂
∂
w
B
w
A
The inverse of A is
10
01 ⎥
⎦
⎤
⎢
⎣
⎡
−
=
-1
A
So
0
xtxt =
∂
∂
+
∂
∂
=
∂
∂
+
∂
∂1w
C
ww
BA
w–
where
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
⎥
⎦
⎤
⎢
⎣
⎡−
⎥
⎦
⎤
⎢
⎣
⎡
−
== 1
01
10
01
10
10
01
BAC –
The characteristic directions are obtained by determining the eigenvalues of C
0=λ− IC
Therefore,
0
1
1=
λ−−
−λ−
Expanding gives
0122 =−λ
Solution yields
1
dx
dy ±==λ
To derive the compatibility equation, the left eigenvectors must first be determined.
Rather than obtaining results for each characteristic. The following applies to
characteristics of either family
(
)
0=λ− ICl T
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
320
or
[]
0
1
1
ll 21 =
⎥
⎦
⎤
⎢
⎣
⎡
λ−−
−λ−
Expanding gives two equations produces
0ll
0ll
21
21
=λ+
=
+
λ
Hence,
112
112
ll
1
l
lll
m
m
=
λ
−=
=
λ
−
=
where the slope of the characteristics ( 1
±
=
λ
) has been used. So
12 ll
±
=
Take l1 to be unity. The compatibility equation for characteristics is
0=wlTd
or
[] []
0dvdu
dv
du
11
dv
du
ll 21 =±=
⎥
⎦
⎤
⎢
⎣
⎡
±=
⎥
⎦
⎤
⎢
⎣
⎡
Problem 14. – Complete the solution of Example 14.3 by determining the solution at
points 19 to 32. Check the accuracy of the results.
The numbering of the points is contained in the following figure
•2
•3
•4
•
5
•1
•
6
•
7
•
8
•
9
•
10
•
11
•
12
•
13
•
14
•
15
•
16
•
17
•
18
•
19
•
20
•
21
•
22
•
23
•
24
•
25
•
26
•
27
•
28
•
29
•
30
•
31
•
32
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
321
A set of tables that contains data for all of the labeled points follows:
point
α =
(CI+CII)/2
deg
ν =
(CI
− CII)/2
deg
CI =
α + ν
de
g
CII =
α − ν
deg M
µ
deg
α + µ
deg
α − µ
deg
1 6.00 26.3798 32.3798 20.3798 2.0000 30.0000 36.0000 -24.0000
2 4.00 26.3798 30.3798 22.3798 2.0000 30.0000 34.0000 -26.0000
3 2.00 26.3798 28.3798 24.3798 2.0000 30.0000 32.0000 -28.0000
4 0.00 26.3798 26.3798 26.3798 2.0000 30.0000 30.0000 -30.0000
5 5.00 27.3798 32.3798 22.3798 2.0365 29.4095 34.4095 -24.4095
6 3.00 27.3798 30.3798 24.3798 2.0365 29.4095 32.4095 -26.4095
7 1.00 27.3798 28.3798 26.3798 2.0365 29.4095 30.4095 -28.4095
8 6.00 28.3798 34.3798 22.3798 2.0733 28.8370 34.8370 -22.8370
9 4.00 28.3798 32.3798 24.3798 2.0733 28.8370 32.8370 -24.8370
10 2.00 28.3798 30.3798 26.3798 2.0733 28.8370 30.8370 -26.8370
11 0.00 28.3798 28.3798 28.3798 2.0733 28.8370 28.8370 -28.8370
12 5.00 29.3798 34.3798 24.3798 2.1106 28.2815 33.2815 -23.2815
13 3.00 29.3798 32.3798 26.3798 2.1106 28.2815 31.2815 -25.2815
14 1.00 29.3798 30.3798 28.3798 2.1106 28.2815 29.2815 -27.2815
15 6.00 30.3798 36.3798 24.3798 2.1483 27.7419 33.7419 -21.7419
16 4.00 30.3798 34.3798 26.3798 2.1483 27.7419 31.7419 -23.7419
17 2.00 30.3798 32.3798 28.3798 2.1483 27.7419 29.7419 -25.7419
18 0.00 30.3798 30.3798 30.3798 2.1483 27.7419 27.7419 -27.7419
19 5.00 31.3798 36.3798 26.3798 2.1864 27.2173 32.2173 -22.2173
20 3.00 31.3798 34.3798 28.3798 2.1864 27.2173 30.2173 -24.2173
21 1.00 31.3798 32.3798 30.3798 2.1864 27.2173 28.2173 -26.2173
22 6.00 32.3798 38.3798 26.3798 2.2251 26.7068 32.7068 -20.7068
23 4.00 32.3798 36.3798 28.3798 2.2251 26.7068 30.7068 -22.7068
24 2.00 32.3798 34.3798 30.3798 2.2251 26.7068 28.7068 -24.7068
25 0.00 32.3798 32.3798 32.3798 2.2251 26.7068 26.7068 -26.7068
26 5.00 33.3798 38.3798 28.3798 2.2642 26.2096 31.2096 -21.2096
27 3.00 33.3798 36.3798 30.3798 2.2642 26.2096 29.2096 -23.2096
28 1.00 33.3798 34.3798 32.3798 2.2642 26.2096 27.2096 -25.2096
29 6.00 34.3798 40.3798 28.3798 2.3039 25.7250 31.7250 -19.7250
30 4.00 34.3798 38.3798 30.3798 2.3039 25.7250 29.7250 -21.7250
31 2.00 34.3798 36.3798 32.3798 2.3039 25.7250 27.7250 -23.7250
32 0.00 34.3798 34.3798 34.3798 2.3039 25.7250 25.7250 -25.7250
The coordinates and slopes of the characteristics in the physical plane are contained in the
following table
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
322
point
α
deg µ α + µ α − µ mI mII x y
1 6 30 36 -24 9.5144 1.0000
2 4 30 34 -26 9.5435 0.6673
3 2 30 32 -28 9.5609 0.3339
4 0 30 30 -30 9.5668 0.0000
5 5 29.4095 34.4095 -24.4095 -0.4495 0.6797 9.8265 0.8597
6 3 29.4095 32.4095 -26.4095 -0.4922 0.6298 9.8505 0.5162
7 1 29.4095 30.4095 -28.4095 -0.5363 0.5821 9.8625 0.1722
8 6 28.8370 34.8370 -22.8370 0.1051 0.6905 10.1222 1.0639
9 4 28.8370 32.8370 -24.8370 -0.4750 0.6106 10.1563 0.7030
10 2 28.8370 30.8370 -26.8370 -0.5186 0.6401 10.1541 0.3588
11 0 28.8370 28.8370 -28.8370 -0.5457 0.0000 10.1779 0.0000
12 5 28.2815 33.2815 -23.2815 -0.4257 0.6509 10.4780 0.9124
13 3 28.2815 31.2815 -25.2815 -0.4676 0.6023 10.4768 0.5532
14 1 28.2815 29.2815 -27.2815 -0.5108 0.5557 10.5030 0.1806
15 6 27.7419 33.7419 -21.7419 0.1051 0.6622 10.8171 1.1369
16 4 27.7419 31.7419 -23.7419 -0.4351 0.6131 10.8201 0.7636
17 2 27.7419 29.7419 -25.7419 -0.4772 0.5660 10.8481 0.3760
18 0 27.7419 27.7419 -27.7419 -0.5208 0.0000 10.8497 0.0000
19 5.00 27.2173 32.2173 -22.2173 -0.4036 0.6244 11.1821 0.9896
20 3.00 27.2173 30.2173 -24.2173 -0.4448 0.5769 11.2153 0.5878
21 1.00 27.2173 28.2173 -26.2173 -0.4873 0.5313 11.2181 0.1957
22 6.00 26.7068 32.7068 -20.7068 0.1051 0.6361 11.5317 1.2120
23 4.00 26.7068 30.7068 -22.7068 -0.4134 0.5882 11.6028 0.8157
24 2.00 26.7068 28.7068 -24.7068 -0.4549 0.5421 11.6101 0.4082
25 0.00 26.7068 26.7068 -26.7068 -0.4978 0.0000 11.6112 0.0000
26 5.00 26.2096 31.2096 -21.2096 -0.3830 0.5999 11.9783 1.0410
27 3.00 26.2096 29.2096 -23.2096 -0.4236 0.5534 12.0240 0.6373
28 1.00 26.2096 27.2096 -25.2096 -0.4654 0.5086 12.0298 0.2129
29 6.00 25.7250 31.7250 -19.7250 0.1051 0.6120 12.4084 1.3042
30 4.00 25.7250 29.7250 -21.7250 -0.3933 0.5650 12.4265 0.8647
31 2.00 25.7250 27.7250 -23.7250 -0.4341 0.5198 12.4720 0.4428
32 0.00 25.7250 25.7250 -25.7250 -0.4763 0.0000 12.4767 0.0000
The accuracy of the calculations is assessed in the following table
M
Point Area A/A*
(exact
solution)
M
(MOC) % Error
4 A4 1.6875 2 2
11 1.06395A4 1.7953 2.0733 2.0733 0
18 1.1341A4 1.9138 2.1472 2.1483 0.0514
25 1.2137A4 2.0471 2.2239 2.2251 0.0506
32 1.3042A4 2.2008 2.3038 2.3039 0.0011
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
323
Problem 15. – Using the same number of points repeat Example 14.3 for:
(a) Minitial = 2.0, total wedge angle of 12˚ and γ = 1.3;
(b) Minitial = 4.0, total wedge angle of 12˚ and γ = 1.4;
(c) Minitial = 2.0, total wedge angle of 24˚ and γ = 1.4
(a) Minitial = 2.0, total wedge angle of 12˚ and γ = 1.3
point α deg ν C(+)I C(-)II M µ α + µ α − µ
1 6.00 28.6809 34.6809 22.6809 2.0000 30.0000 36.0000 -24.0000
2 4.00 28.6809 32.6809 24.6809 2.0000 30.0000 34.0000 -26.0000
3 2.00 28.6809 30.6809 26.6809 2.0000 30.0000 32.0000 -28.0000
4 0.00 28.6809 28.6809 28.6809 2.0000 30.0000 30.0000 -30.0000
5 5.00 29.6809 34.6809 24.6809 2.0324 29.4747 34.4747 -24.4747
6 3.00 29.6809 32.6809 26.6809 2.0324 29.4747 32.4747 -26.4747
7 1.00 29.6809 30.6809 28.6809 2.0324 29.4747 30.4747 -28.4747
8 6.00 30.6809 36.6809 24.6809 2.0649 28.9650 34.9650 -22.9650
9 4.00 30.6809 34.6809 26.6809 2.0649 28.9650 32.9650 -24.9650
10 2.00 30.6809 32.6809 28.6809 2.0649 28.9650 30.9650 -26.9650
11 0.00 30.6809 30.6809 30.6809 2.0649 28.9650 28.9650 -28.9650
12 5.00 31.6809 36.6809 26.6809 2.0978 28.4698 33.4698 -23.4698
13 3.00 31.6809 34.6809 28.6809 2.0978 28.4698 31.4698 -25.4698
14 1.00 31.6809 32.6809 30.6809 2.0978 28.4698 29.4698 -27.4698
15 6.00 32.6809 38.6809 26.6809 2.1309 27.9884 33.9884 -21.9884
16 4.00 32.6809 36.6809 28.6809 2.1309 27.9884 31.9884 -23.9884
17 2.00 32.6809 34.6809 30.6809 2.1309 27.9884 29.9884 -25.9884
18 0.00 32.6809 32.6809 32.6809 2.1309 27.9884 27.9884 -27.9884
point
α
deg µ α + µ α − µ mI mII x y
1 6 30 36 -24 9.5144 1.0000
2 4 30 34 -26 9.5435 0.6673
3 2 30 32 -28 9.5609 0.3339
4 0 30 30 -30 9.5668 0.0000
5 5 29.4747 34.4747 -24.4747 -0.4502 0.6806 9.8261 0.8597
6 3 29.4747 32.4747 -26.4747 -0.4929 0.6306 9.8501 0.5162
7 1 29.4747 30.4747 -28.4747 -0.5370 0.5829 9.8621 0.1721
8 6 28.9650 34.9650 -22.9650 0.1051 0.6929 10.1205 1.0637
9 4 28.9650 32.9650 -24.9650 -0.4764 0.6122 10.1551 0.7029
10 2 28.9650 30.9650 -26.9650 -0.5200 0.6425 10.1527 0.3589
11 0 28.9650 28.9650 -28.9650 -0.5479 0.0000 10.1763 0.0000
12 5 28.4698 33.4698 -23.4698 -0.4290 0.6548 10.4743 0.9120
13 3 28.4698 31.4698 -25.4698 -0.4709 0.6060 10.4732 0.5531
14 1 28.4698 29.4698 -27.4698 -0.5143 0.5593 10.4992 0.1806
15 6 27.9884 33.9884 -21.9884 0.1051 0.6677 10.8101 1.1362
16 4 27.9884 31.9884 -23.9884 -0.4396 0.6183 10.8129 0.7631
17 2 27.9884 29.9884 -25.9884 -0.4819 0.5711 10.8411 0.3758
18 0 27.9884 27.9884 -27.9884 -0.5257 0.0000 10.8429 0.0000
.
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
324
Point Area A/A*
M
(Exact Solution)
M
(MOC) % Error
4 A4 1.7732 2 2
11 1.0637A4 1.8862 2.0649 2.0649 0
18 1.1334A4 2.0097 2.1299 2.1309 0.0438
(b) Minitial = 4.0, total wedge angle of 12˚ and γ = 1.4
point α deg ν C(+)I C(-)II M µ α + µ α − µ
1 6.00 65.7848 71.7848 59.7848 4.0000 14.4775 20.4775 -8.4775
2 4.00 65.7848 69.7848 61.7848 4.0000 14.4775 18.4775 -10.4775
3 2.00 65.7848 67.7848 63.7848 4.0000 14.4775 16.4775 -12.4775
4 0.00 65.7848 65.7848 65.7848 4.0000 14.4775 14.4775 -14.4775
5 5.00 66.7848 71.7848 61.7848 4.0768 14.1991 19.1991 -9.1991
6 3.00 66.7848 69.7848 63.7848 4.0768 14.1991 17.1991 -11.1991
7 1.00 66.7848 67.7848 65.7848 4.0768 14.1991 15.1991 -13.1991
8 6.00 67.7848 73.7848 61.7848 4.1557 13.9238 19.9238 -7.9238
9 4.00 67.7848 71.7848 63.7848 4.1557 13.9238 17.9238 -9.9238
10 2.00 67.7848 69.7848 65.7848 4.1557 13.9238 15.9238 -11.9238
11 0.00 67.7848 67.7848 67.7848 4.1557 13.9238 13.9238 -13.9238
12 5.00 68.7848 73.7848 63.7848 4.2370 13.6516 18.6516 -8.6516
13 3.00 68.7848 71.7848 65.7848 4.2370 13.6516 16.6516 -10.6516
14 1.00 68.7848 69.7848 67.7848 4.2370 13.6516 14.6516 -12.6516
15 6.00 69.7848 75.7848 63.7848 4.3207 13.3822 19.3822 -7.3822
16 4.00 69.7848 73.7848 65.7848 4.3207 13.3822 17.3822 -9.3822
17 2.00 69.7848 71.7848 67.7848 4.3207 13.3822 15.3822 -11.3822
18 0.00 69.7848 69.7848 69.7848 4.3207 13.3822 13.3822 -13.3822
point
α
deg µ α + µ α − µ mI mII x y
1 6 14 20 -8 9.5144 1.0000
2 4 14 18 -10 9.5435 0.6673
3 2 14 16 -12 9.5609 0.3339
4 0 14 14 -14 9.5668 0.0000
5 5 14.1991 19.1991 -9.1991 -0.1555 0.3412 10.2041 0.8927
6 3 14.1991 17.1991 -11.1991 -0.1915 0.3026 10.2291 0.5361
7 1 14.1991 15.1991 -13.1991 -0.2279 0.2649 10.2416 0.1788
8 6 13.9238 19.9238 -7.9238 0.1051 0.3553 10.9225 1.1480
9 4 13.9238 17.9238 -9.9238 -0.1799 0.2905 10.9777 0.7536
10 2 13.9238 15.9238 -11.9238 -0.2162 0.3165 10.9073 0.3895
11 0 13.9238 13.9238 -13.9238 -0.2412 0.0000 10.9827 0.0000
12 5 13.6516 18.6516 -8.6516 -0.1457 0.3305 11.7892 1.0218
13 3 13.6516 16.6516 -10.6516 -0.1815 0.2922 11.7029 0.6219
14 1 13.6516 14.6516 -12.6516 -0.2178 0.2547 11.7722 0.2011
15 6 13.3822 19.3822 -7.3822 0.1051 0.3447 12.6965 1.3345
16 4 13.3822 17.3822 -9.3822 -0.1587 0.3061 12.5927 0.8942
17 2 13.3822 15.3822 -11.3822 -0.1947 0.2683 12.6522 0.4371
18 0 13.3822 13.3822 -13.3822 -0.2312 0.0000 12.6420 0.0000
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. © 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.