12-40 is to be determined. 12-72E

advertisement
12-40
12-72E A person extends his uncovered arms into the windy air outside. The rate of heat loss from the arm
is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas
with constant properties. 4 The arm is treated as a 2-ft-long and 3-in-diameter cylinder with insulated ends.
5 The local atmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the film temperature of (Ts + Tf)/2 = (86+54)/2 = 70qF are
(Table A-22E)
k
0.01457 Btu/h.ft.qF
Q
0.1643 u 10 -3 ft 2 /s
0.7306
Pr
Air
V = 20 mph
Tf = 54qF
Analysis The Reynolds number is
Re
>(20 u 5280/3600) ft/[email protected](3/12) ft
VD
Q
0.1643 u 10 3 ft 2 /s
4.463 u 10 4
Arm
D = 3 in
Ts = 86qF
The Nusselt number corresponding to this Reynolds number is determined to be
hD
k
Nu
5/8
0.62 Re 0.5 Pr 1 / 3 ª § Re · º
«1 ¨¨
0.3 ¸
¸ »
1/ 4
«¬ © 282,000 ¹ »¼
ª § 0.4 · 2 / 3 º
«1 ¨
¸ »
Pr ¹ »
¼
¬« ©
4/5
ª
0.62(4.463 u 10 4 ) 0.5 (0.7306)1 / 3 « § 4.463 u 10 4
0.3 1 ¨
1/ 4
¨ 282,000
«
2
/
3
ª § 0.4 · º
¬ ©
«1 ¨
¸ »
«¬ © 0.7306 ¹ »¼
·
¸
¸
¹
5/8 º
4/5
»
»
¼
129.6
Then the heat transfer coefficient and the heat transfer rate from the arm becomes
h
k
Nu
D
As
Q conv
0.01457 Btu/h.ft.qF
(129.6)
(3 / 12) ft
7.557 Btu/h.ft 2 .qF
SDL S (3 / 12 ft)(2 ft) = 1.571 ft 2
hAs (Ts Tf )
(7.557 Btu/h.ft 2 .qF)(1.571 ft 2 )(86 - 54)qF = 380 Btu/h
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
Download