TEST BANK FOR Gas Dynamics 3rd Edition By James John, Theo Keith

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.
g 9.81m/ s R 0.287 kJ / kg K
1m
T 27 273 300 C
P P P 120 101 221 kpa
2
3
abs gage atm
= = ⋅
∀ =
= + = °
= + = + =
a) m3
2.5668 kg
(0.287)(300)
221
RT
ρ = P = =
W = mg = ρ∀g = (2.5668)(1)(9.81) = 25.1801N
b) moon earth 3
m
ρ = ρ = 2.5668 kg
W 4.1967N
6
W 1
g
g
W earth earth
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/s2) in order to add this term to specific enthalpy h (kJ/kg).
Air
2
a)
2
2
2
2
3
2
2 3
V
s
m
N s
1 kg m
kg
N m
kg
m
m
p N
m
, kg
m
p N
≈ = ⎟
⎟⎠
⎞
⎜ ⎜⎝
⎛
⋅
− ⋅
= ⎟
⎟
⎠
⎞
⎜ ⎜
⎝
⎛
⎟ ⎟⎠
⎞
⎜ ⎜⎝
⎛
≈
ρ
⎟ ⎟⎠
⎞
⎜ ⎜⎝
⎛
≈ ρ ⎟
⎟⎠
⎞
⎜ ⎜⎝
⎛
≈
b)
kg
kJ
1000
1
1000 J
1kJ
N m
1 J
kg
P N m = ⎟⎠
⎞
⎜⎝
⎛
⎟⎠
⎞
⎜⎝
⎛
⋅
⋅
≈
ρ
c)
c
2
2
2 2
1000g
factor 1
kg
kJ
1000 J
1 kJ
N m
1 J
kg m
1 N s
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.
m& i = m& e
m& i = m& p + m& s = ρApVp + ρAsVs
m& e = ρAeVe
∴ApVp + AsVs = AeVe
So
e
p p s s
e A
A V A V
V
+
=
Ae = As + Ap
2p
p D
4
A
π
= 2p
2
s o D
4
D
4
A
π
−
π
= 2
e Do
4
A
π
=
i e
s
p
3
( ) 2 ( p s )
o
2p
2 s
o
s
2p
2
p o
2p
e
p p s s
e V V
D
D
V
D
D V D D V
A
A V A V
V = + −
+ −
=
+
=
(20 5) 6.6667m/ s
30
5 102
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
e Ae
T
m& = 0.04248 p
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.
∀ = 0.1 m3
p1 = 10 MPa
T1 = 293K = T2
p2 = 5 MPa
kg K
259.8219 J
32
R 8,314.3
⋅
= =
From the continuity equation
me
dt
dm = − &
but
RT
m p
∀
=
so
p
T
0.04248 A
m
dt
dp
dt RT
dm e
= − e = −
∀
= &
Integrating we get,
O2
m(t)
d = 0.1mm
e Ae
T
m& = 0.04248 p
4
t
0.04248 R TA
p
p
ln e
1
2
⎟ ⎟
⎠
⎞
⎜ ⎜
⎝
⎛
∀
= −
46,713.4076sec 12.9759 hrs
2
ln 1
(259.8219) 293
1000 mm
0.1 mm m
4
(0.04248)
0.1
p
p
ln
(0.04248)A R T
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.
V1 = 500 m/ s V2 = ?
p1 = 50 kPa p2 = 137 kPa
T1 = 250 K T2 = 343.3 K
A1 = A2
From the continuity equation
m& 1 = m& 2
So
ρ1A1V1 = ρ2A2V2
(500) 250.5839m/ s