978-0072424430 Chapter 3

subject Type Homework Help
subject Pages 9
subject Words 1209
subject Authors John Anderson

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page-pf1
CHAPTER 3
3.1 From Table A.1, for M = 0.7; po/p = 1.387 and To/T = 1.098. Hence,
p
o
3.2
=
15 10
510
6
4
. x
x
= 30
page-pf2
12
3.3 a =
RT
=
( . )( )( )14 1716 500
= 1096 ft/sec
M = V/a = 3000/1096 = (2.74)
3.4 From Table A.2, for M1 = 3; p2/p1 = 10.33, 2/1 = 3.857,
p p
o o
2 1
/
= 0.3283, and
M2 =0.4752. Thus,
p2 = p1 (p2/p1) = (1)(1.01 x 105)(10.33) = 1.043 x 106 N/m2 (or 10.33 atm)
Thus,
p
2
page-pf3
13
T T T T
T
p
R
T
T
x
o o
o o
2 1
1 1
1
1
1
1 1
5
101 10
287 123
= =
=
=
.
()( . )
(2.8) = 801K
3.5 (a) po/p =
122 10
101 10
5
5
.
.
x
x
= 1.21. From Table A.1: M = 0.53
7222
o21
1020
3.6 For the shock compression, use M1 as a parameter, along with Table A.2. For the
isentropic compression, use p2/p1 = (v1/v1)-
page-pf4
Note that:
3.7 From Table A.1 for M = 38, To/T = 289.9; To = (289.8)(270) = 78246K
This temperature is almost 8 times the surface temperature of the sun. Long before this
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15
3.8 (a) From Table A.3 for M1 = 2.0:
p
p
1
*
= 0.363,
T
T
1
*
= 0.5289,
T
T
o
o
1
*
= 0.7934
From Table A.1 for M1 = 2.0:
T
T
o1
1
= 1.8
Hence:
To* =
T
T
o
o
*
1
T
T
o1
1
T1 =
1
07934.
(1.8)(288) = 653.4K
This is the total temperature at the exit, since for choked conditions at the exit, M2 = 1 and hence
= To*, p2 = p*, etc. At the inlet,
= 1.8 T1 = (1.8)(288) = 518.4K.
q = cp (
-
) = 1005 (653.4 - 581.4)
q = 1.357 x 105 joule/kg
p2 = p* = p1/0.3636 =
1
03636
atm
.
= 2.75 atm
T2 = T* = T1/0.5289 =
288
05289.
= 544.4K
(b) From Table A.3, for M1 = 0.2:
p1/p* = 2.273, T1/T* = 0.2066,
To1
/To* = 0.1736
From Table A.1, for M1 = 0.2:
page-pf6
16
T
o1
02066.
3.9 1 slug air + 0.06 slug fuel = 1.06 slug of mixture
page-pf7
page-pf8
Also,
T
o*
page-pf9
page-pfa
20
Eq. (3.95) becomes
M2
4fdx
T
Substitute (A4) into (A1).
dT
M2
4fdx
page-pfb
21
M
2
Substitute (A6) and (A8) into (A5)
dM
1
dM
1
3.16 q = cp (T0,2 To,1)
page-pfc
22
q
c T
T
T
p 0 1
0 2
0 1,
,
,
=
- 1 =
T
T
o2
0*
T
T
o*
,0 1
- 1
At M = 2.5, from Table A.3,
T
T
0 1
0
,
*
= 0.7101.
Since cp T0,1 is the total enthalpy of the gas entering the duct, and q is given as 30% of this total
enthalpy, we have
q
c T
p 0 1,
=
0.3
Hence

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