62
CHAPTER 7
7.1 e2 – e1 =
p p
1 2
2
+
(v1 – v2)
1
−
1
−
T
1
p
1
1
−
−+
1
2
p
T
1
−
1
T
2
T
2
T
2
+
−
1
1
1
p
p
1
2
+
−+
+
−
+
1
1
1
11
2
1
2
1
p
p
p
p
Inverting,
1
2
11
1
2
1
++
−
p
p
64
up = 680 (1 – 0.376) = 424 m/sec
(b)
Stationary wave moving wave
Stationary Moving
a
2
a
2
or,
65
7.3 Behind the wave, the mass motion velocity relative to the laboratory is up = 423 m/sec.
Hence, the Mach number of this flow relative to the laboratory is
7.4 Assume W = 1000 m/sec.
a1 =
RT114 287 300=( . )( )( )
= 347.2 m/sec
W
1000
a
2
66
aM
2
2
5533
We want up/a2 = 1.5, so assume new values of W until up/a2 converges to 1.5.
W Ms T2/T1 M2 a2 up/a2
m/sec
7.5 From Table A.2, for p2/p1 = 1050, M1 =
W
a1
= 30.
a1 =
RT114 287 300=( . )( )( )
= 347.2 m/sec
67
From Eq. (7.23),
M
M
M
M
R
R
s
s
2 2
1 1−=−
( )
12 1
111
2
2
2
+−
+− +
( )
( )
MM
s
s
30
.
For MR = 2.638, T5/T2 = 2.28 and p5/p2 = 7.965
T5 = 2.28 (52770) = 120,300K
68
7.6 Assume WR = 5000 m/sec
W u
R p
+
5000 8671
+
WR = M5 a5 = (0.5005)(6953) = 3480 m/sec.
69
7.7 (a) From Table A.2 for p2/p1 = 40.4, Ms = 5.9, M2 = 0.405, T2/T1 = 7.709. Hence,
T2 = (7.709)(300) = 2313K
From Table A.1, for M = 1.72:
p
o2
T
o2
70
From Eq. (7.23), for Ms = 5.9
( )
M
R
2
59
.
)( . ) .
7.8 From Eq. (7.86):
u
a
3
4
2
1
=−
13
4
1
2
−
−
p
p
= 5[1 – (0.4)0.143] = 0.614
a4 =
RT414 287 2500=( . )( )( )
= 1002 m/sec
u3 = 0.614 a4 = 615 m/sec
T
T
3
4
=
p
p
3
4
1
−
(0.4)0.286 = 0.769
T3 = (0.769)(2500) = 1923
a3 =
( . )( )( )14 287 1923
= 879 m/sec
M3 = u3/a3 = 615/879 = 0.70
7.9
For the tail of the wave to remain in the driver section, u3 a3. The maximum strength wave that
reflects this criteria is the one where a3 = u3.
From Eq. (7.86):
p
u
3
3
2
1
−
p
4
p
4
p
a
3
3
2
1
−
p
4
1
2 67
.
7.10 Eq. (7.94) must be solved for
by trial-and-error:
Assume p2/p1 = 2. From Eq. (7.94)
7
−
( . )( )( )
73
7
−
( . )( )( . )
From Eq. (7.23)
p
5
M
R
2
14
.
.
p
p
3
4
p
p
1
4
7.11 For the incident shock, from problem 7.10, Ms = 1.4.
From Table A.2:
T
T
2
1
= 1.255, M2 = 0.7397.
dx
dx
dx
W u
p
−
From problem 7.10, the strength of the incident expansion wave is p3/p4 = 0.426. Hence
T
p
3
3
1
−
75
a3 =
( . )( )( . )14 287 2352
= 307.3 m/sec
From Eq. (7.84)
dx
dx
u
a
3
3
2
.
Also, from Eq. (7.74) for C– characteristics
J– = u –
2
1
a
−
And for C+ characteristics
J+ = u +
2
1
a
−
dx
76
dx
For points a, b, c and d:
(J+)a = (J+)b = (J+)c = (J+)d = 0 +
2347 2
0 4
( . )
.
= 1736 m/sec
dx
dx
dt dk
For point f:
(J+)f = (J+)e = 1596
77
dx
dt fc
For point g:
(J+)g = (J+)f = 1596
(J–)g = (J–)d = -1337
dt gd
For point h:
(J–)h = (J–)f = -1456
uh = 0
For point i:
(J+)i = (J+)h = 1456 m/sec
u6 = 0 and a6 = 267.4 m/sec
Since the flow is isentropic:
p
T
6
6
1
=
4
15 1 01389 16 1 14 1
16
+ − +
. ( ) .
= 0.5365
MR = 0.5365 MR2 – 0.5365
or,
81
M
R
54
.
2 0493
( . )
T
5