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Chapter 7
External Forced Convection
For this staggered arrangement tube bank, the friction coefficient corresponding to Re
D
= 5294 and
S
L
/
D
=
1.5/0.8 = 1.875
is, from Fig. 727b,
f
= 0.44. Also,
χ
= 1 for the square arrangements. Then the pressure
drop across the tube bank becomes
Pa
638.2
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⋅
=
ρ
χ
=
Δ
2
2
3
2
max
m/s
kg
1
N
1
2
m/s)
571
.
8
)(
kg/m
316
.
1
(
)
1
)(
44
.
0
(
30
2
V
f
N
P
L
Discussion
The arithmetic mean fluid temperature is (
T
i
+
T
e
)/2 = (0 15.6)/2 = 7.8
°
C, which is fairly close
to the assumed value of 5
°
C. Therefore, there is no need to repeat calculations.
772
Air is heated by hot tubes in a tube bank. The average heat transfer coefficient is to be determined.
Assumptions
1
Steady operating conditions exist.
2
The surface temperature of the tubes is constant.
Properties
The exit temperature of air, and thus the mean temperature, is not known. We evaluate the air
properties at the assumed mean temperature of 70
°
C and 1 atm (Table A15):
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 Fall '10
 Dr.DanielArenas
 Thermodynamics, Convection, Force, Friction, Mass

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