7-81
7-83 A cylindrical electronic component mounted on a circuit board is cooled by air flowing across it. The surface
temperature of the component 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 We assume the film temperature to be 50C based on the problem
statement. The properties of air at 1 atm and at this temperature are (Table A-15)
7228.0Pr
/sm 10798.1
C W/m.02735.0
25–
=
=
=
k
Analysis The Reynolds number is
m) m/s)(0.003 (240/60
VD
Q
Resistor
Air
V = 240 m/min
T = 35C
7-82
7-84 A cylindrical hot water tank is exposed to windy air. The temperature of the tank after a 45-min cooling period is to be
estimated.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas with constant
properties. 4 The surface of the tank is at the same temperature as the water temperature. 5 The heat transfer coefficient on
the top and bottom surfaces is the same as that on the side surfaces.
Properties The properties of water at 80C are (Table A-9)
CJ/kg. 4197
kg/m 8.971 3
=
=
p
c
The properties of air at 1 atm and at the film temperature of 50C (based on the problem statement) are (Table A-15)
7228.0Pr
/sm 10798.1
C W/m.02735.0
25–
=
=
=
k
Analysis The Reynolds number is
5
m) (0.50m/s
3600
100040
VD
Water tank
D =50 cm
L = 95 cm
7-83
7-85 Prob. 7-84 is reconsidered. The temperature of the tank as a function of the cooling time is to be plotted.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
D=0.50 [m]
L=0.95 [m]
T_w1=80 [C]
T_infinity=18 [C]
Vel=40 [km/h]
time=45 [min]
“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_w_ave+T_infinity)
rho_w=Density(water, T=T_w_ave, P=101.3)
c_p_w=CP(Water, T=T_w_ave, P=101.3)*Convert(kJ/kg-C, J/kg-C)
T_w_ave=1/2*(T_w1+T_w2)
“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
A=pi*D*L+2*pi*D^2/4
Q_dot=h*A*(T_w_ave-T_infinity)
m_w=rho_w*V_w
V_w=pi*D^2/4*L
Q=m_w*c_p_w*(T_w1-T_w2)
Q_dot=Q/(time*Convert(min, s))
time
[min]
Tw2
[C]
30
73.06
45
69.86
60
66.83
75
63.96
90
61.23
105
58.63
120
56.16
135
53.8
150
51.54
165
49.39
180
47.33
195
45.36
210
43.47
225
41.65
240
39.91
255
38.24
270
36.63
285
35.09
300
33.6
050 100 150 200 250 300
30
35
40
45
50
55
60
65
70
75
time [min]
Tw2 [C]
7-84
7-86 A steam pipe is exposed to a light winds in the atmosphere. The amount of heat loss from the steam during a certain
period and the money the facility will save a year as a result of insulating the steam pipe are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The plant operates every
day of the year for 10 h a day. 4 The local atmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the film temperature of (Ts + T)/2 = (75+5)/2 = 40C are (Table A-15)
7255.0Pr
/sm 10702.1
C W/m.02662.0
25–
=
=
=
k
Analysis The Reynolds number is
4
m) (0.12m/s 1000/3600)(25
VD
m 12.0
D
The rate of heat loss by convection is
2
W9617=C5))(75m C)(4.524. W/m37.30()( 22 −=−=
TThAQssconv
The rate of heat loss by radiation is
W1870)K 2730()K 27375().K W/m10)(5.67m (4.524)8.0()( 4442-82
44 =+−+=−= surrssrad TTAQ
The total rate of heat loss then becomes
W487,1118709617
total =+=+= radconv QQQ
The amount of heat loss from the steam during a 10-hour work day is
kJ/day 104.135 5
=== )s/h 3600h/day 10)(kJ/s 487.11(tQQ total
The total amount of heat loss from the steam per year is
kJ/yr 10509.1)days/yr 365)(kJ/day 10135.4()days of no.( 85 === daytotal QQ
Noting that the steam generator has an efficiency of 80%, the amount of gas used is
therms/yr1788
kJ 105,500
therm1
80.0
kJ/yr 10509.1
80.0
8
=
== total
gas
Q
Q
Insulation reduces this amount by 90%. The amount of energy and money saved becomes
therms/yr1609=) therms/yr1788)(90.0()90.0(savedEnergy == gas
Q
$1690== erm))($1.05/th therms/yr(1609=energy) ofcost t saved)(UniEnergy (savedMoney
Wind
V = 25 km/h
T = 5C
Steam pipe
Ts = 75C
D = 12 cm
7-86
7-88 A cylindrical bottle containing cold water is exposed to windy air. The average wind velocity is to be estimated.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas with constant
properties. 4 Heat transfer at the top and bottom surfaces is negligible.
Properties The properties of water at the average temperature of (T1 + T2)/2=(3+11)/2=7C are (Table A-9)
CJ/kg. 4200
kg/m 8.999 3
=
=
p
c
The properties of air at 1 atm and the film temperature of (Ts + T)/2 = (7+27)/2 = 17C are (Table A-15)
7317.0Pr
/sm 10488.1
C W/m.02491.0
25–
=
=
=
k
Analysis The mass of water in the bottle is
23
2
D
Air
V
T = 27C
Bottle
D =10 cm
L = 30 cm
7-87
7-89 A 10-m tall exhaust stack discharging exhaust gases at a rate of 1.2 kg/s is subjected to solar radiation and convection at
the outer surface. The outer surface temperature of the exhaust stack is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties
are constant. 3 The surface temperature is constant.
Properties The properties of air at 80°C are k = 0.02953
W/m∙K,
= 2.097 10−5 m2/s, Pr = 0.7154 (from Table A-15).
Analysis The Reynolds number for the air flowing across the
exhaust stack is
5
)m 1)(m/s 10(
VD
7-88
7-90 Liquid NH3 flows in a pipe, which is insulated. The insulation thickness on the pipe that is necessary to keep
the liquid NH3 temperature below −35°C is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas with local
atmospheric pressureat 1 atm. 4 One-dimensional heat conduction through walls. 5 The thermal conductivities are constant. 6
The thermal contact resistance at the interface is negligible.
Properties The thermal conductivities of the pipe and the insulation are given to be kpipe= 25 W/mK and kins = 0.75 W/mK,
respectively.
The properties of air at 1 atm and Tf = (Ts + T)/2 = (10 + 20)/2 = 15C are k = 0.02476 W/m∙K, ν = 1.470 10−5 m2/s, and Pr
= 0.7323 (Table A-15).
Analysis The convection heat transfer coefficient on the outer surface can be determined using the Nusselt number relation
for flow across a cylinder. The Reynolds number and Nusselt number can be determined using
o
VD
5/4
8/5
4/13/2
3/12/1
air
000,282
Re
]Pr)/4.0(1[
PrRe62.0
+
Dh o
From Chapter 3, the thermal resistances of different layers are
LDhAh
R
ii
i
NH3NH3
conv,
11 ==
(liq. NH3 convection resistance)
Lk
DD
Ri
pipe
interface
pipe 2
)/ln(
=
(pipe layer resistance)
Lk
DD
Ro
ins
interface
ins 2
)/ln(
=
(insulation layer resistance)
LDhAh
R
oo
o
airair
conv,
11 ==
(air flow across cylinder convection resistance)
The total thermal resistance and the rate of heat transfer are
convinspipe conv,total RRRRR i+++=
and
o
os
R
TT
R
TT
Q
conv,
,
total
NH3 −
=
−
=
and the insulation thickness is
2
interface
ins
DD
to−
=
Solving for the insulation thickness yields
cm 4.26== m 0426.0
ins
t
Solved by EES Software. Copy-and-paste the following lines on a blank EES screen to verify the solutions.
“GIVEN”
h_NH3=100 [W/m^2-K] “liq. NH3 convection heat transfer coefficient”
7-89
Pr=Prandtl(Fluid$, T=T_film)
rho=Density(Fluid$, T=T_film, P=101.3)
mu=Viscosity(Fluid$, T=T_film)
nu=mu/rho
“Walls – pipe and insulation”
k_pipe=25 [W/m-K] “pipe thermal conductivity”
k_ins=0.75 [W/m-K] “insulation thermal conductivity”
“ANALYSIS”
Re=V*D_o/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_air=Nusselt*k/D_o
“Thermal resistances for diffenrent layers”
R_conv_i=1/(h_NH3*pi#*D_i*L) “liq. NH3 convection resistance”
R_pipe=ln(D_interface/D_i)/(2*pi#*k_pipe*L) “pipe layer resistance”
R_ins=ln(D_o/D_interface)/(2*pi#*k_ins*L) “insulation layer resistance”
R_conv_o=1/(h_air*pi#*D_o*L) “ambient air convection resistance”
R_total=R_conv_i+R_pipe+R_ins+R_conv_o
“Solving for the insulation thickness”
Q_dot=(T_infinity-T_s_i)/(R_total)
Q_dot=(T_infinity-T_s_o)/(R_conv_o)
t_ins=(D_o-D_interface)/2
Discussion To keep the liquid NH3 below −35°C, the pipe insulation thickness must be at least 4.26 cm thick.
7-91
“Walls – pipe and insulation”
k_pipe=15 [W/m-K] “pipe thermal conductivity“
k_ins=0.95 [W/m-K] “insulation thermal conductivity”
7-92
7-92 Air is flowing over a 5-cm diameter sphere, (a) the average drag coefficient on the sphere and (b) the heat transfer rate
from the sphere are to be determined.
Assumptions1 Steady operating conditions exist. 2 Properties are constant. 3 The surface temperature is constant.
Properties The properties of air (1 atm) at the free stream temperature T∞ = 20°C (Table A-15):
= 1.204 kg/m3, k = 0.02514
W/m∙K,
= 1.825 10−5 kg/m∙s, and Pr = 0.7309; at the surface temperature Ts = 80°C:
s = 2.096 10−5 kg/m∙s; at the film
temperature Tf = (80°C + 20°C)/2 = 50°C:
= 1.092 kg/m3 and
= 1.798 10−5 m2/s.
Analysis (a) The Reynolds number for air properties evaluated from the film temperature is
)m 05.0)(m/s 5.3(
VD
D
(b) The Reynolds number for air properties evaluated from the free stream temperature is
4
)05.0)(5.3)(204.1(VD
m m/s kg/m 3
7-94
7-94 Prob. 7-93 is reconsidered. The effect of air velocity on the average convection heat transfer coefficient and the
cooling time is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
D=0.15 [m]
T_1=125 [C]
T_2=75 [C]
T_infinity=30 [C]
P=101.3 [kPa]
Vel=6 [m/s]
rho_ball=8055 [kg/m^3]
c_p_ball=480 [J/kg-C]
“PROPERTIES”
Fluid$=’air’
k=Conductivity(Fluid$, T=T_infinity)
Pr=Prandtl(Fluid$, T=T_infinity)
rho=Density(Fluid$, T=T_infinity, P=P)
mu_infinity=Viscosity(Fluid$, T=T_infinity)
nu=mu_infinity/rho
mu_s=Viscosity(Fluid$, T=T_s_ave)
T_s_ave=1/2*(T_1+T_2)
“ANALYSIS”
Re=(Vel*D)/nu
Nusselt=2+(0.4*Re^0.5+0.06*Re^(2/3))*Pr^0.4*(mu_infinity/mu_s)^0.25
h=k/D*Nusselt
A=pi*D^2
Q_dot_ave=h*A*(T_s_ave-T_infinity)
Q_total=m_ball*c_p_ball*(T_1-T_2)
m_ball=rho_ball*V_ball
V_ball=(pi*D^3)/6
time=Q_total/Q_dot_ave*Convert(s, min)
Vel
[m/s]
h
[W/m2.C]
time
[min]
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
9.886
12.36
14.51
16.45
18.22
19.88
21.45
22.94
24.36
25.72
27.03
28.3
116.4
93.08
79.3
69.97
63.14
57.87
53.65
50.17
47.25
44.74
42.57
40.66
15
20
25
30
35
40
50
60
70
80
90
100
110
120
h (W/m2K)
time (s)
7-95
7-95 The average surface temperature of the head of a person when it is not covered and is subjected to winds 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 One-quarter of the heat the person generates is lost from the head. 5 The head can be approximated as a 30–cm–
diameter sphere. 6 The local atmospheric pressure is 1 atm.
Properties We assume the surface temperature to be 15C for viscosity based on the problem statement. The properties of air
at 1 atm pressure and the free stream temperature of 10C are (Table A-15)
7336.0Pr
kg/m.s 10802.1
kg/m.s 10778.1
/sm 10426.1
C W/m.02439.0
5
C15@
,
5
25–
=
=
=
=
=
−
−
s
k
AnalysisThe Reynolds number is
(12 1000/3600) m /s (0.3 m )
VD
éù
´
ëû
1/4
5
0.5 2/3 0.4
5
1.778 10
2 0 .4 ( 7 0 ,1 2 6 ) 0 .0 6 ( 7 0 ,1 2 6 ) (0.7336) 185.1
1.802 10
–
–
æö
´÷
ç
éù
÷
ç
= + + =
÷
ç
êú
÷
ëû
ç÷
´
èø
The heat transfer coefficient is
2
0.02439 W /m .C (185.1) 15.05 W /m .C
k
°
Air
V = 12 km/h
T = 10C
Head
Q = 21 W
D =0.3 m
7-96
7-96 A light bulb is cooled by a fan. The equilibrium temperature of the glass bulb is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The light bulb is in
spherical shape. 4 The local atmospheric pressure is 1 atm.
Properties We assume the surface temperature to be 100C for viscosity based on the problem statement. The properties of
air at 1 atm pressure and the free stream temperature of 30C are (Table A–15)
7282.0Pr
kg/m.s 10181.2
kg/m.s 10872.1
/sm 10608.1
C W/m.02588.0
5
C100@
,
5
25–
=
=
=
=
=
−
−
s
k
Analysis The Reynolds number is
4
m) m/s)(0.1 (2
VD
m 1.0
D
Noting that 90 % of electrical energy is converted to heat,
W90= W)100)(90.0(=Q
The bulb loses heat by both convection and radiation. The equilibrium temperature of the glass bulb can be determined by
iteration or by an equation solver:
222 m 0314.0)m 1.0( ===
C136.9==
K 9.409
s
s
T
Discussion This surface temperature is not close to the assumed surface temperature of 100C. For better accuracy, we can
repeat the calculations using a new viscosity value at 136.9C:
kg/m.s 10332.2 5
C9.136@
−
=
s
(Table A-15). It gives Ts =
412.6 K = 139.6C. The difference between the two results is 2.7C.
Lamp
100 W
= 0.9
Air
V = 2 m/s
T = 30C
7-97
7-97 Air flows over a spherical tank containing iced water. The rate of heat transfer to the tank and the rate at which ice melts
are 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 properties of air at 1 atm pressure and the free stream temperature of 25C are (Table A-15)
7296.0Pr
kg/m.s 10729.1
kg/m.s 10849.1
/sm 10562.1
C W/m.02551.0
5
C0@
,
5
25–
=
=
=
=
=
−
−
s
k
Analysis The Reynolds number is
(3 m /s)(0.4 m )
VD
Air
V = 3 m/s
T =25C
D =0.4 m
Iced water
0C
7-100
7-100 The temperature of H2 gas stream is to be measured by a spherical thermocouple junction. The time it takes to
register 99%of the initial Tand the convection heat transfer coefficient as functions of the free stream velocity are to be
evaluated.
Assumptions 1 The junction is spherical in shape. 2 The thermal properties of the junction are constant. 3 H2 gas behaves as
ideal gas at 1 atm. 4 Radiation effects are negligible. 5 The Biot number is Bi < 0.1 so that the lumped system analysis is
applicable (this assumption will be verified).
Properties The properties of the junction are given to be kTC = 35 W/mK, ρTC = 8500 kg/m3, and cp,TC = 320 J/kgK. The
properties of H2 gas are evaluated at T∞ = 200°C, and the dynamic viscosity μs of H2 gas at the sphere surface is evaluated at
Tave = (10°C + 200°C)/2 = 105°C
Analysis The problem is solved using EES, and the solution is given below:
“GIVEN”
T_infinity=200 [C]