# 12-49 of the person is to be determined for two cases.

```12-49
12-80E A fan is blowing air over the entire body of a person. The average temperature of the outer surface
of the person is to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas
with constant properties. 4 The average human body can be treated as a 1-ft-diameter cylinder with an
exposed surface area of 18 ft2. 5 The local atmospheric pressure is 1 atm.
Properties We assume the film temperature to be 100qF.
V = 6 ft/s
The properties of air at this temperature are (Table A-22E)
Person, Ts
T
f = 85qF
k 0.01529 Btu/h.ft.qF
300 Btu/h
Q
1.809 u 10 - 4 ft 2 /s
Pr 0.7260
D = 1 ft
Analysis The Reynolds number is
(6 ft/s)(1 ft)
VD
3.317 u 10 4
Re
4
2
Q
1.809 u 10 ft /s
The proper relation for Nusselt number corresponding to this Reynolds number is
Nu
hD
k
0.3 0.62 Re 0.5 Pr 1 / 3
&gt;1 (0.4 / Pr) @
2 / 3 1/ 4
&ordf; &sect; Re &middot; 5 / 8 &ordm;
&laquo;1 &uml;&uml;
&cedil;&cedil; &raquo;
&laquo;&not; &copy; 282,000 &sup1; &raquo;&frac14;
4/5
&ordf;
0.62(3.317 u 10 4 ) 0.5 (0.7260)1 / 3 &laquo; &sect;&uml; 3.317 u 10 4
1
0.3 1/ 4
&laquo; &uml;&copy; 282,000
1 (0.4 / 0.7260) 2 / 3
&not;
&gt;
@
&middot;
&cedil;
&cedil;
&sup1;
5/8 &ordm;
4/5
&raquo;
&raquo;
&frac14;
107.8
The heat transfer coefficient is
k
0.01529 Btu/h.ft.qF
h
Nu
(107.8) 1.649 Btu/h.ft 2 .qF
D
1 ft
Then the average temperature of the outer surface of the person becomes
Q
300 Btu/h
Q hAs (Ts Tf ) o Ts Tf 85qF +
hAs
(1.649 Btu/h.ft 2 .qF)(18 ft 2 )
95.1qF
If the air velocity were doubled, the Reynolds number would be
(12 ft/s)(1 ft)
VD
6.633 u 10 4
Re
4
2
Q
1.809 u 10 ft /s
The proper relation for Nusselt number corresponding to this Reynolds number is
Nu
hD
k
0.3 0.62 Re 0.5 Pr 1 / 3
&gt;1 (0.4 / Pr) @
2 / 3 1/ 4
&ordf; &sect; Re &middot; 5 / 8 &ordm;
&laquo;1 &uml;&uml;
&cedil;&cedil; &raquo;
&laquo;&not; &copy; 282,000 &sup1; &raquo;&frac14;
4/5
&ordf;
0.62(6.633 u 10 4 ) 0.5 (0.7260)1 / 3 &laquo; &sect;&uml; 6.633 u 10 4
0.3 1
1/ 4
&laquo; &uml;&copy; 282,000
1 (0.4 / 0.7260) 2 / 3
&not;
&gt;
@
&middot;
&cedil;
&cedil;
&sup1;
5/8 &ordm;
Heat transfer coefficient is
k
0.01529 Btu/h.ft.qF
h
Nu
(165.9) 2.537 Btu/h.ft 2 .qF
D
1 ft
Then the average temperature of the outer surface of the person becomes
Q
300 Btu/h
85qF +
Q hAs (Ts Tf ) o Ts Tf hAs
(2.537 Btu/h.ft 2 .qF)(18 ft 2 )
&raquo;
&raquo;
&frac14;
4/5
165.9
91.6qF
PROPRIETARY MATERIAL. &copy; 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.
```