7-61
7-62 Air flows in parallel over a flat plate where the first-half length has a constant surface temperature and the second-half
length is subjected to uniform heat flux.The local convection heat transfer coefficients at x = 1 and 3 m are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Local atmospheric pressure is 1 atm. 3 The critical Reynolds number is
Recr = 5105. 4 The boundary layer over the second portion of the plate with uniform heat flux has not been affected by the
first half of the plate with constant surface temperature.
Properties The properties of air at Tf =30°C are k = 0.02588 W/m∙K, ν = 1.608 10−5 m2/s, and Pr = 0.7282 (Table A-15).
Analysis The Reynolds numbers at x = 1 m and 3 m are
124378
m) m/s)(12(Vx
Re 25
x=
=
=−
(flow is laminar at x = 1 m)
7-63
x [m] Rex Ts [°C] Tf [°C] hx [W/m2∙K]
0.2 24875 50 30 6.092
0.3 37313 50 30 4.974
0.4 49751 50 30 4.308
0.5 62188 50 30 3.853
0.6 74626 50 30 3.517
0.8 99501 50 30 3.046
1.0 124377 50 30 2.725
1.2 149252 50 30 2.487
1.4 174128 50 30 2.303
1.6 199003 50 30 2.154
1.8 223878 50 30 2.031
2.0 248754 50 30 1.927
2.0 254093 42.70 26.35 2.630
2.2 278198 44.30 27.15 2.507
2.6 325924 47.29 28.65 2.306
3.0 373058 50.07 30.03 2.146
3.4 419648 52.66 31.33 2.016
3.8 465736 55.11 32.56 1.906
4.0 488603 56.29 33.14 1.858
0 0.5 1 1.5 2 2.5 3 3.5 4
1
2
3
4
5
6
7
hx [W/m2·K]
7-65
7-65 A circuit board is cooled by air. The surface temperatures of the electronic components at the leading edge and the end
of the board are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5105. 3 Radiation effects are
negligible. 4 Any heat transfer from the back surface of the board is disregarded. 5 Air is an ideal gas with constant
properties.
Properties We assume a film temperature of 35C based on the problem statement, the properties of air are evaluated at this
temperature to be (Table A-15)
7268.0Pr
/sm 10655.1
C W/m.0265.0
25–
=
=
=
k
Analysis (a) The convection heat transfer coefficient
at the leading edge approaches infinity, and thus the
Air
Circuit board
20 W
15 cm
7-66
Flow across Cylinders and Spheres
7-67C Turbulence moves the fluid separation point further back on the rear of the body, reducing the size of the wake, and
7-67
7-70 The flow of a fluid across an isothermal cylinder is considered. The change in the drag force and the rate of heat transfer
when the free-stream velocity of the fluid is doubled is to be determined.
Analysis The drag force on a cylinder is given by
2
V
Pipe
7-68
7-71 A steam pipe is exposed to windy air. The rate of heat loss from the steam is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas with constant
properties.
Properties The properties of air at 1 atm and the film temperature of (Ts + T)/2 = (90+7)/2 = 48.5C are (Table A-15)
7232.0Pr
/sm 10784.1
C W/m.02724.0
25–
=
=
=
k
Analysis The Reynolds number is
5
25 10214.1
/sm 10784.1
m) (0.12]s/h) 0m/km)/(360 1000(km/h) (65[
Re =
== −
VD
The Nusselt number corresponding to this Reynolds number is
( )
( )
4/5
5/8
0.5 1/3
1/4
2/3
5/8
5 0.5 1/3 5
1/4
2/3
0.62Re Pr Re
0.3 1 282,000
1 0.4 /Pr
0.62(1.214 10 ) (0.7232) 1.214 10
0 .3 1 282,000
1 0.4 /0.7232
hD
Nu
k
éù
æö
êú
÷
ç
= = + + ÷
ç
êú
÷
ç÷
èø
êú
éù
ëû
+
êú
ëû
éù
æö
êú
´´
÷
ç÷
ç
= + +
êú
÷
ç÷
êç÷
éù
èø
ê
+ëû
êú
ëû
4/5
247.5=
ú
ú
The heat transfer coefficient and the heat transfer rate become
C W/m.02724.0 2=
k
Air
V = 65 km/h
T = 7C
Pipe
D = 12 cm
Ts = 90C
7-70
7-73 A heated long cylindrical rod is placed in a cross flow of air. The rod surface has an emissivity of 0.95 and its surface
temperature is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 The surface temperature is constant. 4 Heat
flux dissipated from the rod is uniform.
Properties The properties of air (1 atm) at 70°C are given in Table A-15: k = 0.02881 W/m∙K,
= 1.995 10−5 m2/s, and Pr
= 0.7177.
Analysis The Reynolds number for the air flowing across the rod is
)m 005.0)(m/s 10(
VD
7-71
7-74E 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 + T)/2 = (84+54)/2 = 69F are (Table A-15E)
7308.0Pr
/sft 101638.0
FBtu/h.ft. 01455.0
23–
=
=
=
k
Analysis The Reynolds number is
4
ft (3/12)ft/s /3600)5280(30
VD
Air
V = 30 mph
T = 54F
7-73
Vel
[mph]
Qconv
[Btu/h]
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
234.8
261.3
286.5
310.5
333.6
356
377.7
398.9
419.6
439.8
459.7
479.3
498.6
517.6
536.3
554.8
10 15 20 25 30 35 40
200
250
300
350
400
450
500
550
600
Vel [mph]
Qconv [Btu/h]
7-74
7-76 The wind is blowing across a geothermal water pipe. The average wind velocity 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 local atmospheric pressure is 1 atm.
Properties The specific heat of water at the average temperature of 75ºC is 4193 J/kg.ºC. The properties of air at the film
temperature of (75+15)/2=45ºC are (Table A-15)
7241.0Pr
/sm 1075.1
C W/m.02699.0
25–
=
=
=
k
Analysis The rate of heat transfer from the pipe is the
energy change of the water from inlet to exit of the pipe,
and it can be determined from
Wind
V
T = 15C
Water
7-77
7-79 Prob. 7-78 is reconsidered. The effect of the wind velocity on the surface temperature of the wire is to be
investigated.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
D=0.005 [m]
L=1 [m] “unit length is considered”
I=50 [Ampere]
R=0.002 [Ohm]
T_infinity=10 [C]
Vel=50 [km/h]
“PROPERTIES”
Fluid$=’air’
k=Conductivity(Fluid$, T=T_film)
Pr=Prandtl(Fluid$, T=T_film)
rho=Density(Fluid$, T=T_film, P=101.3)
mu=Viscosity(Fluid$, T=T_film)
nu=mu/rho
T_film=1/2*(T_s+T_infinity)
“ANALYSIS”
Re=(Vel*Convert(km/h, m/s)*D)/nu
Nusselt=0.3+(0.62*Re^0.5*Pr^(1/3))/(1+(0.4/Pr)^(2/3))^0.25*(1+(Re/282000)^(5/8))^(4/5)
h=k/D*Nusselt
W_dot=I^2*R
Q_dot=W_dot
A=pi*D*L
Q_dot=h*A*(T_s-T_infinity)
Vel
[km/h]
Ts
[C]
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
14.08
13.32
12.87
12.56
12.32
12.14
12
11.88
11.78
11.69
11.61
11.54
11.48
11.43
11.38
10 20 30 40 50 60 70 80
11
11.5
12
12.5
13
13.5
14
Vel [km/h]
Ts [C]
