45
CHAPTER 5
5.1 From Table A.1 for Me = 2.4: Ae/A* = 2.403 ,
po/pe = 14.62 and To/Te = 2.152.
p
e
T
e
5.2
p
p
o
o
2
1
=
0627
10
.
= 0.627.
5.3 (a)
p
p
o
=
5
4
= 1.25
46
5.4
For M1 = 0.3, from Table A.1: A1/A* = 2.035 and po/p1 = 1.064
p
p
o
A
A
A
t t
.
10 7
p
p
o
1
1094.
5.5
47
u
1
250
5.6 At the throat:
m
•
= p*A*u*
u* =
RT *
and * = p*/RT*
Hence,
•
p
*
p A
* *
T
*
2
2
2
m
•
p
o
+
−
1
1/
m
•
T
o
R
1
+
It is important to note that
m
•
p A
T
o
o
*
5.7 Since
m
•
p A
T
o
o
*
m
p
T
o
o
2
2
1
20
750
•
5.8
From Table A.1: For Ae/A* = 25, Me = 5.
49
The pressure and temperature at the nose of the model are the stagnation pressure and
temperature behind a normal shock at Me = 5. Hence, from Table A.2:
p
o
2
5.9
p
2
However,
50
5.10
p
o
2
0 6172
.
p
e
T
e
A
*
Hence,
RT
()( . )
287 833
51
5.11 We will solve this problem two ways: (a) first by trial-and-error, and then (b) by a direct
method.
(a) Trial and Error. We assume the A/A1* value for the shock location, and check to see if pe =
0.6 atm for the assumed location. If not, repeat.
Assume
52
Assumed
p
p
o
o
2
1
A
A2*
A
A
e
2*
p
p
o
e
e
atm( )
1.9 2.14 0.555 0.6557 1.24 1.305 0.52 1.202 0.5455
1.8 2.08 0.5643 0.6835 1.24 1.377 0.48 1.171 0.584
1.75 2.04 0.5707 0.7022 1.226 1.401 0.47 1.163 0.604
close enough
53
m
•
= uA.
This can also be expressed as
m
•
=
p A
TR
o
o
*
2
1
1
1
+
+
−
(1)
as derived in Problem 5.6. Now consider flow across a normal shock in a duct.
For any isentropic flow, A* is constant; however, it changes across a shock
m
•
54
p
oe
2
and
p
oe
A
e
A
e
2
2
1
2 1
+
−
( )
55
p
o
2
p
o
2
p
o
e
p
o
e
p
e
0 6
.
5.12
A
A
t
t
2
1
=
p
p
o
o
1
2
(Eq. 5.36 in text)
For starting the tunnel, the diffuser throat should be large enough to pass the mass flow through a
normal shock in the test section. If
At2
is this large, or larger, then the shock will be swallowed
by the diffuser, and the tunnel will start. Considering a normal shock in the test section (at the
entrance to the diffuser),
p
o
2
A
t
2
A
t
1
56
5.13 Since the diffuser exhausts to the atmosphere, and the exit Mach number is very low, pd =
pdo
= 1 atm. Also, from Table A.2 for Me = 2.6,
p
p
o
o
2
1
= 0.4601. Hence,
5.14
p
o
po
10
p
2
004
57
5.15 From the –-M diagram, across the shock wave,
()1 = 2 – 1 = 32.2
Mn1
= M1 sin = 4 sin 50 = 3.06
p
o
2
p
o
2
5.16 cp =
Rjoule
kg K−= =
1
122 519 6
022 28814
( . )( . )
..
ho = constant: cp Te +
Ve
2
2
= cp To
59
T
o
5.18 (a) From Eq. (3.28)
T
TM
o
e
e
= + −= + −
11
2114 1
2
2
.
(10)2 = 21
Te = To/21 = (3000)/21 = 142.9K
60
5.19 Maximum velocity will occur when the test section Mach number is infinite. Since
T
TM
e
e
= + −
11
2
o
2
5.20
T
TM
o
e
e
= + −
11
2
2
= 1 + 0.2 (20)2 = 81
5.21
T
TM
o
e
e
= + −
11
2
2
= 1 +
167 1
2
.−
(20)2 = 135
61
5.22 From Eq. (5.20), for helium, we have
A
AMM
e
e
e
*
( )/( )
=++−
+ −
2
2
2
1 1
1 2
111
2
=
1
20
2
167 11167 1
220
2
2
267
067
( ) .
.( )
.
.
++−
=
1
400 0 749 135 3985
. ( ) .
= 243,854
A
A
e
*
= 493.8
For air at Mach 20, from Table A.1,
At M = 20, Ae/A* = 0.1538 x 105 = 15,380
The area ratio required for Mach 20 in air is much larger, by a factor of 31, than that for helium.