Problem 13.1
[2]
Given:
Air extracted from a large tank
Find:
Mass flow rate
Solution:
Basic equations:
m
rate
ρ
V
⋅
A
⋅
=
h
1
V
1
2
2
+
h
2
V
2
2
2
+
=
p
ρ
k
const
=
T p
1 k
−
(
)
k
⋅
const
=
Given or available data
T
0
70
273
+
(
) K
⋅
=
p
0
101 kPa
⋅
=
p
25 kPa
⋅
=
D
15 cm
⋅
=
c
p
1004
J
kg K
⋅
⋅
=
k
1.4
=
R
286.9
J
kg K
⋅
⋅
=
The mass flow rate is given by
m
rate
ρ
A
⋅
V
⋅
=
A
π
D
2
⋅
4
=
A
0.0177m
2
=
We need the density and velocity at the nozzle.
In the tank
ρ
0
p
0
R T
0
⋅
=
ρ
0
1.026
kg
m
3
=
From the isentropic relation
ρ
ρ
0
p
p
0
⎛
⎜
⎝
⎞
⎟
⎠
1
k
⋅
=
ρ
0.379
kg
m
3
=
We can apply the energy equation between the tank (stagnation conditions) and the point in the nozzle to find the velocity
h
0
h
V
2
2
+
=
V
2
h
0
h
−
(
)
⋅
=
2 c
p
⋅
T
0
T
−
(
)
⋅
=
Fot T we again use insentropic relations
T
T
0
p
0
p
⎛
⎜
⎝
⎞
⎟
⎠
1 k
−
(
)
k
⋅
=
T
230.167K
=
T
43.0
−
°C
⋅
=
Then
V
2 c
p
⋅
T
0
T
−
(
)
⋅
=
V
476
m
s
=
The mass flow rate is
m
rate
ρ
A
⋅
V
⋅
=
m
rate
3.18
kg
s
=
Note that the flow is supersonic at this point
c
k R
⋅
=
c
304
m
s
=
M
V
c
=
M
1.57
=
Hence we must have a converging-diverging nozzle

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