Solution Manual For Gas Dynamics, 3rd Edition

Name: Solution Manual For Gas Dynamics, 3rd Edition
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Price: 15 USD
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Author: Madison Taylor

Solution Manual For Gas Dynamics, 3rd Edition is your study companion, providing answers to textbook exercises and reinforcing key concepts.

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1
CChhaapptteerr OOnnee
BBAASSIICC EEQQUUAATTIIOONNSS
OOFF CCOOMMPPRREESSSSIIBBLLEE FFLLOOWW
Problem 1. – Air is stored in a pressurized tank at a pressure of 120 kPa (gage) and a temperature
of 27°C. The tank volume is 1 m3 . Atmospheric pressure is 101 kPa and the local acceleration
of gravity is 9.81 m/s2 . (a) Determine the density and weight of the air in the tank, and (b)
determine the density and weight of the air if the tank was located on the Moon where the
acceleration of gravity is one sixth that on the Earth.
Kkg/kJ728.0Rs/m81.9g
m1
C30027327T
kpa122101120PPP
2
3
atmgageabs
⋅==
=∀
°=+=
=+=+=
a) 3
m
kg
5668.2
)300)(287.0(
221
RT
P ===ρ
N1801.25)81.9)(1)(5668.2(gmgW ==∀ρ==
b) 3earthmoon m
kg
5668.2=ρ=ρ
N1967.4W
6
1
W
g
g
W earthearth
earth
moon
moon ===
Problem 2. – (a) Show that p/ρ has units of velocity squared. (b) Show that p/ρ has the same
units as h (kJ/kg). (c) Determine the units conversion factor that must be applied to kinetic
energy, V2 /2, (m2 /s 2 ) in order to add this term to specific enthalpy h (kJ/kg).
Air

2
a)
2
2
2
2
3
2
32
V
s
m
sN
mkg
1
kg
mN
kg
m
m
Np
m
kg
,
m
N
p
≈=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
⋅−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
≈
ρ
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ≈ρ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ≈
b) kg
kJ
1000
1
J0100
kJ1
mN
J
1
kg
mNP =⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⋅
⋅
≈
ρ
c)
c
2
2
22
g1000
1
factor
kg
kJ
J0100
kJ1
mN
J
1
mkg
sN
1
s
m
2
V
=∴
≈⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⋅⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⋅
⋅
≈
Problem 3. – Air flows steadily through a circular jet ejector, refer to Figure 1.15. The primary
jet flows through a 10 cm diameter tube with a velocity of 20 m/s. The secondary flow is through
the annular region that surrounds the primary jet. The outer diameter of the annular duct is 30
cm and the velocity entering the annulus is 5 m/s. If the flows at both the inlet and exit are
uniform, determine the exit velocity. Assume the air speeds are small enough so that the flow
may be treated as an incompressible flow, i.e., one in which the density is constant.
ei mm && =
ssppspi VAVAmmm ρ+ρ=+= &&&
eee VAm ρ=&
eesspp VAVAVA =+∴
So
e
sspp
e A
VAVA
V +
=
pse AAA +=
2
pp D
4
A π
= 2
p
2
os D
4
D
4
A π
−
π
= 2
oe D
4
A π
=
i e
s
p

3
( ) ( )sp2
o
2
p
s2
o
s
2
p
2
op
2
p
e
sspp
e VV
D
D
V
D
VDDVD
A
VAVA
V −+=
−+
=
+
=
( ) s/m6667.6520
30
10
5 2
2
=−+=
Problem 4. – A slow leak develops in a storage bottle and oxygen slowly leaks out. The volume
of the bottle is 0.1 m3 and the diameter of the hole is 0.1 mm. The initial pressure is 10 MPa and
the temperature is 20˚C. The oxygen escapes through the hole according to the relation
ee A
T
p
04248.0m =&
where p is the tank pressure and T is the tank temperature. The constant 0.04248 is based on the
gas constant and the ratio of specific heats of oxygen. The units are: pressure N/m2 , temperature
K, area m2 and mass flow rate kg/s. Assuming that the temperature of the oxygen in the bottle
does not change with time, determine the time it takes to reduce the pressure to one half of its
initial value.
3
m1.0=∀
MPa01p1 =
21 TK293T ==
MPa5p 2 =
Kkg
J
8219.259
32
3.314,8
R ⋅
==
From the continuity equation
em
dt
dm &−=
but
RT
p
m ∀
=
so
p
T
A80424.0
m
dt
dp
RTdt
dm e
e −=−=
∀
= &
Integrating we get,
O2
m(t)
d = 0.1mm
ee A
T
p
04248.0m =&

4
t
ATR80424.0
p
p
ln e
1
2
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∀
−=
hrs5979.12sec4076.713,46
2
1
ln
293)8219.259(
mm0100
m
mm1.0
4
)04248.0(
1.0
p
p
ln
TRA)04248.0(
t
2
1
2
e
==
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛ π
−=
∀
−=
Problem 5. – A normal shock wave occurs in a nozzle in which air is steadily flowing. Because
the shock has a very small thickness, changes in flow variables across the shock may be assumed
to occur without change of cross-sectional area. The velocity just upstream of the shock is 500
m/s, the static pressure is 50 kPa and the static temperature is 250 K. On the downstream side of
the shock the pressure is 137 kPa and the temperature is 343.3 K. Determine the velocity of the
air just downstream of the shock.
s/m050V1 = ?V2 =
kPa05p1 = kPa713p 2 =
K025T1 = K3.343T2 =
21 AA =
From the continuity equation
21 mm && =
So
222111 VAVA ρ=ρ
s/m5839.250)500(
250
3.343
137
50
V
T
T
p
p
V
RT/p
RT/p
VV 1
1
2
2
1
1
22
11
1
2
1
2 =⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
===
ρ
ρ
=
2
1

5
Problem 6. – A gas flows steadily in a 2.0 cm diameter circular tube with a uniform velocity of
1.0 cm/s and a density ρo . At a cross section farther down the tube, the velocity distribution is
given by V = Uo [1-( r/R) 2 ], with r in centimeters. Find Uo , assuming the gas density to be
ρo [1+( r/R)2
].
s/cm1V1 = ⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
−=
2
o2 R
r
1UV
o1 ρ=ρ ⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
+ρ=ρ
2
o2 R
r
1
21 mm && =
o
22
1o
R
o 111
R
o 11 RRVrdr2VdAVm ρπ=πρ=πρ=ρ= ∫∫&
( )
oo
22
oo
1
o
522
oo
2
2
o2
2R
o
R
o o222
UR
3
2
6
1
2
1
RU2
R
r
wheredR2U
rdr2
R
r
1U
R
r
1dAVm
ρπ=⎟
⎠
⎞
⎜
⎝
⎛ −πρ=
=ξξξ−ξπρ=
π
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +ρ=ρ=
∫
∫ ∫&
oo
2
o
2 UR
3
2
R ρ
π
=ρπ∴
so s/cm
2
3
Vo =
Problem 7. – For the rocket shown in Figure 1.6, determine the thrust. Assume that exit plane
pressure is equal to ambient pressure.
( ) ( ) ( )
ee
2
oH
ee
oH
oHeeeatme A
mm
V
mm
mm0VmApp ρ
+
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
ρ
+
++=+−= &&&&
&&&T
r
1 2

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6
Problem 8. – Determine the force F required to push the flat plate of Figure Pl.8 against the
round air jet with a velocity of 10 cm/s. The air jet velocity is 100 cm/s, with a jet diameter of 5.0
cm. Air density is 1.2 kg/m3 .
Figure P1.8
To obtain steady state add + Vp to all velocities
VmF &=
( ) ( ) ( )1.015.0
4
2.1AVm 2 +⎟
⎠
⎞
⎜
⎝
⎛ π
=ρ=& s/kg200259.0=
( )( ) N100285.01.1002592.0F ==
Problem 9. – A jet engine (Figure P1.9) is traveling through the air with a forward velocity of
300 m/s. The exhaust gases leave the nozzle with an exit velocity of 800 m/s with respect to the
nozzle. If the mass flow rate through the engine is 10 kg/s, determine the jet engine thrust. Exit
plane static pressure is 80 kPa, inlet plane static pressure is 20 kPa, ambient pressure surrounding
the engine is 20 kPa, and the exit plane area is 4.0 m2 .
F
x
V = -10 s
cm
Vj = 100 s
cm
Vj = 110 s
cm
x
V = 0
F

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7
Figure P1.9
( ) ( ) ( )( ) ( )( ) kN24552403008001042080VVmApp ieeatme =+=−+−=−+−= &T
Problem 10. – A high-pressure oxygen cylinder, typically found in most welding shops,
accidentally is knocked over and the valve on top of the cylinder breaks off. This creates a hole
with a cross-sectional area of 6.5 x 10-4 m2 . Prior to the accident, the internal pressure of the
oxygen is 14 MPa and the temperature is 27˚C. Based on critical flow calculations, the velocity
of the oxygen exiting the cylinder is estimated to be 300 m/s, the exit pressure 7.4 MPa and the
exit temperature 250 K. How much thrust does the oxygen being expelled from the cylinder
generate? What percentage is due to the pressure difference? What percentage due to the exiting
momentum? Atmospheric pressure is 101 kPa. Also note that 0.2248 lbf = 1 N.
Figure P1.10
s/m030Ve = 24
e m105.6A −
×=
MPa4.7p e = MPa110.0kPa110p atm ==
k025Te = ee
e
e
eee VA
RT
p
VAm =ρ=&
kkg
J
82.259R ⋅
= ( )( ) ( )( ) s/kg2.22300105.6
25082.259
m
N
104.7
4
2
6
=×
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ×
= −
300 m/s 800 m/s

Loading page 8...

8
( )
( ) ( )( )
( )
f
2
2
43
eeatme
lb3.548,2N0.336,11
N6.66644.4671
mkg
sN
1
s
mkg
3002.22N104.6101017400
VmApp
==
+=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⋅
⋅
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅
+×××−=
+−=
−
&T
The thrust due to the pressure is 41% of total and that due to momentum 59%.
Problem 11. – Air enters a hand held hair dryer with a velocity of 3 m/s at a temperature of 20°C
and a pressure of 101 kPa. Internal resistance heaters warm the air and it exits through an area of
20 cm2 with a velocity of 10 m/s at a temperature of 80°C. Assume that internal obstructions do
not appreciably affect the pressure between inlet and exit and that heat transfer to the
surroundings are negligible. Determine the power in kW needed to operate the hair dryer at
steady state.
( ) ( )( )
( )( ) ( ) ( ) ( ) s
kg
019939.010
100
m
20
353287
10101
VA
RT
p
VAm 2
23
22
2
2
222 ==⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=ρ=&
( ) ( ) kg
kJ
3.602080
Kkg
kJ
005.1TTchh 12p12 =−
⋅
=−=−
( )( )
kg
kJ
0455.0
2000
310310
g0100
1
2
VV
c
2
1
2
2 =
−+
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−
i
e

Loading page 9...

9
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+=− 2
V
hm
2
V
hmWQ
2
1
1
2
2
2 &&&&
( ) ( )( )
W2051.203,1
kg
kJ
203205.1
0455.03.60019939.0
2
VV
hhmW
2
1
2
2
12
==
+=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
+−=− &&
Problem 12. – Air is expanded isentropically in a horizontal nozzle from an initial pressure of
1.0 MPa, of a temperature of 800 K, to an exhaust pressure of 101 kPa. If the air enters the
nozzle with a velocity of 100 m/s, determine the air exhaust velocity. Assume the air behaves as
a perfect gas, with R = 0.287 kJ/kg · K and γ = 1.4. Repeat for a vertical nozzle with exhaust
plane 2.0 m above the intake plane.
(a) Horizontal nozzle
2
V
h
2
V
hh
2
2
2
2
1
1o +=+=
( ) ( )21
2
121p
2
12 TT
1
R2
VTTc2VV −
−γ
γ
+=−+=
kkg
J
287R ⋅
=
5194.0
1000
101
p
p
T
T 4.1
4.1
1
2
1
2 =⎟
⎠
⎞
⎜
⎝
⎛
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
= γ
−γ
K5.415T2 =
( ) ( )( )( ) ( ) s/m856.8845.460,772000,105.415800
4.0
2874.12
100V 2
2 =+=−+=
(b) Vertical nozzle
1 2
p2 = 101kPa
p1 = 1MPa
T 1 = 800K
V1 = 100m/s

Loading page 10...

10
2
2
2
21
2
1
1 gz
2
V
hgz
2
V
h ++=++
( ) ( ) ( )( )( )281.925.460,782zzg2TT
1
R2
VV 2121
2
12 +=−+−
−γ
γ
+=
s/m059.88474.499,782 ==
Problem 13. – Nitrogen is expanded isentropically in a nozzle from a pressure of 2000 kPa, at a
temperature of 1000 K, to a pressure of 101 kPa. If the velocity of the nitrogen entering the
nozzle is negligible, determine the exit nozzle area required for a nitrogen flow of 0.5 kg/s.
Assume the nitrogen to behave as a perfect gas with constant specific heats, mean molecular
mass of 28.0, and γ = 1.4.
2
V
h
2
V
hh
2
2
2
2
1
1o +=+=
( ) ( )21p212 TTc2hh2V −=−=
( ) K1.426
2000
101
1000
P
P
TT 4.1
4.01
1
2
12 =⎟
⎠
⎞
⎜
⎝
⎛
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
= γ
−γ
Kkg
J
9.296
28
3.8314
R ⋅
==
( ) ( ) ( )( )( ) s/m2.10921.42610009.2967TTR7TT
1
R
2V 21212 =−=−=−
−γ
γ
=
( )( ) 3
2
2
2 m
kg
798.0
1.4269.296
000,101
RT
p ===ρ
p1 = 2000kPa
T 1 = 1000K
V1 ~ small
1
2
p2 = 101kPa
m = 0.5kg/s = ρ2A 2V 2
•
A2 = ?
V 2 = ?

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11
( )( ) 22
22
2 cm473.5m4000573.0
2.1092798.0
5.0
V
m
A ===
ρ
= &
Problem 14. – Air enters a compressor with a pressure of 100 kPa and a temperature 20°C; the
mass flow rate is 0.25 kg/s. Compressed air is discharged from the compressor at 800 kPa and
50°C. Inlet and exit pipe diameters are 4.0 cm. Determine the exit velocity of the air at the
compressor outlet and the compressor power required. Assume an adiabatic, steady, flow and
that the air behaves as a perfect gas with constant specific heats; cp = 1.005 kJ/kg · K and
R = 0.287kJ/kg·K.
kg
kJ
005.1c p = kkg
kJ
287.0R ⋅
=
s/kg52.0mmm 21 === &&&
( ) 222
21 m60012.004.0
4
d
4
AA =
π
=
π
==
( )( ) 3
1
1
1 m
kg
189.1
2930.287
kPa0.10
RT
p ===ρ
( )( ) 3
1
2
2 m
kg
630.8
3230.287
800
RT
p ===ρ
( )( ) s
m
3.167
00126.189.1
25.0
A
m
V
11
1 ==
ρ
= &
( )( ) s
m
1.23
00126.63.8
25.0
A
m
V
22
2 ==
ρ
= &
p 1 = 100kPa
T1 = 293K
d = 4.cm
1
2
W = ?
•
m = 0.25 kg/s•
p 2 = 800KPa
T2 = 323K
V 2 = ?

Loading page 12...

12
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+=− 2
V
hm
2
V
hmWQ
2
1
11
2
2
22 &&&&
( ) ⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡ −
+−=− 2
VV
hhmW
2
1
2
2
12&&
( ) ( )( ) ( )( )
( )( ) ⎥
⎦
⎤
⎢
⎣
⎡ −+
+−= 10002
16731.231.233.167
293323005.125.0
( )( ) kW1.4
s
kJ
1.473.1315.3025.0 ==−=
Problem 15. – Hot gases enter a jet engine turbine with a velocity of 50 m/s, a temperature of
1200 K, and a pressure of 600 kPa. The gases exit the turbine at a pressure of 250 kPa and a
velocity of 75 m/s. Assume isentropic steady flow and that the hot gases behave as a perfect gas
with constant specific heats (mean molecular mass 25, γ = 1.37). Find the turbine power output
in kJ/(kg of mass flowing through the turbine).
Kkg
J
6.332
25
3.8314
R ⋅
== 37.1=γ ( )( )
Kkg
kJ
2314.1
37.
6.33237.1
1
R
c p ⋅
==
−γ
γ
=
K3.947
600
250
1200
p
p
TT 37.1
37.01
1
2
12 =⎟
⎠
⎞
⎜
⎝
⎛
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
= γ
−γ
1
2
W
•
V 1 = 50m/s
T1 = 1200K
p 1 = 600kPa
p 2 = 250 kPa
V 2 = 75 m/s

Loading page 13...

13
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+=− 2
V
hm
2
V
hmWQ
2
1
11
2
2
22 &&&&
( ) ⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡ −
+−= 2
VV
hhmW
2
2
2
1
21&&
( ) ( )( )
2000
VVVV
TTc
2
VV
hh
m
W
W 2121
21p
2
2
2
1
21
−+
+−=
−
+−== &
&
( )( ) ( )( )
kg
kJ
6.309563.118.311
2000
25125
3.94712002314.1W =−=⎟
⎠
⎞
⎜
⎝
⎛ −
+−=
Problem 16. – Hydrogen is stored in a tank at 1000 kPa and 30°C. A valve is opened, which
vents the hydrogen and allows the pressure in the tank to fall to 200 kPa. Assuming that the
hydrogen that remains in the tank has undergone an isentropic process, determine the amount of
hydrogen left in the tank. Assume hydrogen is a perfect gas with constant specific heats; the ratio
of specific heats is 1.4, and the gas constant is 4.124 kJ/kg · K. The tank volume is 2.0m3 .
K303TkPa1000p 11 ==
kPa200p 2 = ( ) K3.191
1000
200
303
p
p
TT 4.1
4.01
1
2
12 =⎟
⎠
⎞
⎜
⎝
⎛
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
= γ
−γ
( )( )
( )( ) kg507.0
3.191124.4
2200
RT
p
m
2
2
2 ==
∀
=

Loading page 14...

14
Problem 17. – Methane enters a constant-diameter, 3 cm duct at a pressure of 200 kPa, a
temperature of 250 K, and a velocity of 20 m/s. At the duct exit, the velocity reaches 25 m/s. For
isothermal steady flow in the duct, determine the exit pressure, mass flow rate, and rate at which
heat is added to the methane. Assume methane behaves as a perfect gas; the ratio of specific
heats is 1.32 (constant) and the mean molecular mass is 16.0.
222111 VAVA ρ=ρ
2
2
2
1
1
1 V
RT
p
V
RT
p
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
2211 VpVp =
( ) 2
2
1
12 m
N
160
25
20
200
V
V
pp =⎟
⎠
⎞
⎜
⎝
⎛
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
( ) 22 m000707.003.0
4
A =
π
=
Kkg
J
519.6
16
3.8314
R ⋅
==
( ) ( )
( )
( ) ( ) s
kg
02176.020
250
000707.0
6.519
200
VA
RT
p
VAm 11
1
1
111 ==⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=ρ=&
( ) ( )( ) W448.2
2
52025
02176.0
2
VV
mQ
2
1
2
2 =
+
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
= &&
Problem 18. – Air is adiabatically compressed from a pressure of 300 kPa and a temperature of
27 C to a pressure of 600 kPa and a temperature of 327 C. Is this compression actually possible?
1 2
T = constant
Q
•
p 1 = 200kPa
T1 = 250K
V 1 = 20 m/s
p 2 = ?
m = ?•
V 2 = 25m/s
Q = ?
•
d = 3cm
γ = 1.32
MW = 16

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15
kPa300p1 = kPa600p 2 =
K30027327T1 =+= K600273327T2 =+=
300
600
lnR
300
600
lnc
p
p
lnR
T
T
lncss p
1
2
1
2
p12 −=−=−
( ) 02lnc2lnRc vp >=−= possible∴
Problem 19. – Two streams of air mix in a constant-area mixing tube of a jet ejector. The
primary jet enters the tube with a speed of 600 m/s, a pressure of 200 kPa and a temperature of
400˚C. The secondary stream enters with a velocity of 30 m/s, a pressure of 200 kPa and a
temperature of 100˚C. The ratio of the area of the secondary flow to the primary jet is 5:1. The
air behaves as a perfect gas with constant specific heats, cp = 1.0045 kJ/kg· K. Using the iterative
numerical procedure described in Example 1.9 determine the velocity, pressure and temperature
of the air leaving the mixing tube.
g c 1
α 5
γ 1.4
R 287
cp 1004.5
Primary Secondary
V 600.00 30.00
T 673 373
P 200,000 200,000
A 43,122.5078
B 263,528.7595
C 706,538.5693
n V e (m/s) Pe (Pa) Te (K)
1 0.0000 101,000.0 293.1500
2 125.1620 244,722.8 695.5757
3 122.5671 245,112.7 695.8957
4 122.4284 245,133.6 695.9126
5 122.4210 245,134.7 695.9135
6 122.4206 245,134.7 695.9136
7 122.4206 245,134.7 695.9136

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Solution Manual For Gas Dynamics, 3rd Edition

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