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PROBLEM 10.1
KNOWN:
Flow of air through a pipe
U(r) = 25[1 - (r/r
1
)
2
] cm/s
p = 1 bar abs = 100,000 N/m
2
abs
T = 5
o
C = 278 K
d
1
= 2r
1
= 5 cm
FIND:
mass flow rate
ASSUMPTIONS:
Steady, incompressible, axisymmetric flow of a perfect gas.
SOLUTION
Conservation of mass gives
which for steady, incompressible, axisymmetric flow becomes
The velocity can be written in vector form as
resulting in
For a perfect gas, the density can be estimated by
ρ
= p/RT = 1.16 kg/m
3
so that (with 1 m = 100 cm):
0
dA
n
ˆ
U
ρ
ρ
d
t
CV
CS
=
•
+
∀
∂
∂
∫∫∫
∫∫
∫ ∫
=
•
2
π
0
r1
0
ρ
U(r)rdrd
θ
m
z
e
2
1
)
r
r
25(1
U
−
=
→
/s
cm
39.3
ρ
ρπ
r
2
25
θ
d
dr
r
)
)
r
r
(
25(1
ρ
m
3
2
1
2
π
2
0
r1
0
1
=
=
−
=
∫ ∫
•
kg/hr
0.16
kg/s
10
4.5
m
5
=
×
=
−
•

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PROBLEM 10.2
KNOWN:
Air flow through a pipe
diameter = 2r
1
= 20 cm
N = 5 velocity measurements per cross-sectional traverse
M = 3 cross-sectional traverses
B
Q
/Q = ±2%
FIND:
Q'
ASSUMPTIONS:
Steady, incompressible flow of a perfect gas.
Flow rate remains perfectly controlled (constant) during all
measurements (equivalent to the assumption that all
measurements are taken simultaneously).
SOLUTION
The flow rate along each traverse line (j = 1, 2, 3) can be approximated by
1
5
1
0
2
2
r
j
ij
i
Q
Urdr
U r
r
π
π
=
=
≈
∆
∑
∫
i = 1,2,3,4,5
where
∆
r = 2 cm and j = 1, 2, 3. Then,
[
]
1
2
(25.31)(1)(2)
(22.48)(3)(2)
(21.66)(5)(2)
(15.24)(7)(2)
(5.12)(9)(2)
Q
π
=
+
+
+
+
= 4446 cm
3
/s
Similarly, Q
2
= 4421 cm
3
/s, Q
3
= 4400 cm
3
/s. The mean flow rate is
Q
= (1/3)[4446 + 4421 + 4400]cm
3
/s = 4423 cm
3
/s
with standard deviation,
1/ 2
3
2
1
1
(
)
2
Q
j
j
S
Q
Q
=
⎡
⎤
=
−
⎢
⎥
⎣
⎦
∑
= 23 cm
3
/s
and
1/ 2
/3
Q
Q
S
S
=
= 13.3 cm
3
/s
with
ν
= 2