8-41
8-63 A circuit board is cooled by passing cool air through a channel drilled into the board. The maximum total power of the
electronic components is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat flux at the top surface of the channel is uniform, and heat
transfer through other surfaces is negligible. 3 The inner surfaces of the channel are smooth. 4 Air is an ideal gas with
constant properties. 5 The pressure of air in the channel is 1 atm. 5 Flow is fully developed in the channel.
Properties The properties of air at 1 atm and estimated average temperature of 25C based on the problem statement are
(Table A-15)
7296.0Pr
CJ/kg. 1007
/sm 10562.1
C W/m.02551.0
kg/m 184.1
25–
3
=
=
=
=
=
p
c
k
Analysis The cross-sectional and heat transfer
surface areas are
2
m 00028.0)m 14.0)(m 002.0(
==
c
A
Air
15C
4 m/s
Electronic components,
50C
L = 20 cm
Air channel
0.2 cm 14 cm
8-42
8-64 A circuit board is cooled by passing cool helium gas through a channel drilled into the board. The maximum total power
of the electronic components is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat flux at the top surface of the channel is uniform, and heat
transfer through other surfaces is negligible. 3 The inner surfaces of the channel are smooth. 4 Helium is an ideal gas. 5 The
pressure of helium in the channel is 1 atm. 6 Flow is fully developed in the channel.
Properties Use the following properties for helium:
6867.0Pr
CJ/kg. 5193
/sm 10214.1
C W/m.1502.0
kg/m 1636.0
24–
3
=
=
=
=
=
p
c
k
Analysis The cross-sectional and heat transfer
surface areas are
2
m 00028.0)m 14.0)(m 002.0(
==
c
A
Helium
15C
4 m/s
Electronic components,
50C
L = 20 cm
Air channel
0.2 cm 14 cm
8-44
Ts
[C]
Q
[W]
30
35
40
45
50
55
60
65
70
75
80
85
90
10.58
14.1
17.62
21.13
24.64
28.15
31.65
35.14
38.64
42.13
45.61
49.09
52.56
30 40 50 60 70 80 90
10
15
20
25
30
35
40
45
50
55
Ts [C]
Q [W]
8-45
8-66 A computer is cooled by a fan blowing air through its case. The flow rate of the air, the fraction of the temperature rise
of air that is due to heat generated by the fan, and the highest allowable inlet air temperature are to be determined.
Assumptions 1 Steady flow conditions exist. 2 Heat flux is uniformly distributed. 3 Air is an ideal gas with constant
properties. 4 The pressure of air is 1 atm.
Properties We assume the bulk mean temperature for air to be 25C.
The properties of air at 1 atm and this temperature are (Table A–15)
0.7296Pr
CJ/kg. 1007
/sm 10562.1
C W/m.02551.0
kg/m 184.1
25–
3
=
=
=
=
=
p
c
k
Analysis (a) Noting that the electric energy consumed by the fan is converted to
thermal energy, the mass flow rate of air is
+
+
W)10 12(8
WQ
Cooling
air
8-46
8-67 Water is flowing between two parallel 1-m wide plates with 12.5-mm spacing. Hydrogen gas flows width-wise in
parallel over the upper and lower surfaces of the two plates. The outlet mean temperature of the water, the surface
temperature of the plates, and the total rate of heat transfer are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 Isothermal parallel plates. 4 The thermal
resistance of the plates is negligible (thin plates). 5 The bulk mean fluid temperature of the water is 30°C (this will be
validated). 6 The film temperature of the H2 gas is 100°C (this will be validated).
Properties The properties of liquid water at 30°C are cp = 4178 J/kg∙K, k = 0.615 W/m∙K, μ = 0.798 × 10−3 kg/m∙s, and Pr =
5.42 (Table A-9). The properties of H2 gas at 100°C are kH2 = 0.2095 W/m∙K, νH2 = 1.582 10−4 m2/s, and PrH2 = 0.7196
(Table A-16)
Analysis The Reynolds number, hydrodynamic and thermal entry lengths can be determined to be
e
s
8-47
Discussion The bulk mean fluid temperature is Tb = (Ti + Te)/2 = 30.1°C, thus 30°C is an appropriate temperature for
evaluating the properties of glycerin. The film temperature of the H2 gas is Tf = (T∞ + Ts)/2 = 100.28°C, thus 100°C is an
appropriate temperature for evaluating the properties of H2 gas.
Equations (1) to (3) can be solved using the EES software with the following lines:
“GIVEN”
c_p=4178 [J/kg-K]
h=187.81 [W/m^2-K]
h_H2=22.171 [W/m^2-K]
A_s=20.25 [m^2]
m_dot=0.58 [kg/s]
T_i=20 [C]
T_infinity=155 [C]
V_infinity=5 [m/s]
L=10 [m]
width=1 [m]
“ANALYSIS”
8-49
ṁ [kg/s] V∞ [m/s] Ts [°C] Q
̇ [W]
0.01 0.001348 40 836.7
0.05 0.0337 40 4183
0.10 0.1348 40 8367
0.12 0.1942 40.01 10040
0.14 0.2644 40.03 11713
0.16 0.3456 40.08 13387
0.18 0.4379 40.14 15060
0.20 0.5415 40.23 16733
0.22 0.6566 40.35 18407
0.24 0.7834 40.50 20080
0.26 0.9222 40.67 21753
0.28 1.073 40.87 23427
0.30 1.237 41.08 25100
0.32 1.413 41.32 26773
0.35 1.702 41.70 29283
0.40 2.252 42.41 33467
0.45 2.891 43.20 37650
0.50 3.625 44.04 41833
0.55 4.459 44.93 46017
0.58 5.009 45.48 48527
0 1 2 3 4 5
0
10000
20000
30000
40000
50000
V¥ [m/s]
Q [W]
0 0.1 0.2 0.3 0.4 0.5 0.6
40
41
42
43
44
45
46
m [kg/s]
Ts [C]
8-50
8-69 Oil flows through a pipeline that passes through icy waters of a lake. The exit temperature of the oil and the rate of heat
loss are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the pipe is very nearly 0C. 3 The thermal
resistance of the pipe is negligible. 4 The inner surfaces of the pipeline are smooth. 5 The flow is hydrodynamically
developed when the pipeline reaches the lake.
Properties The properties of oil at 10C are (Table A-13)
750,28Pr C,J/kg. 1839
/sm 10592.2 kg/m.s, 326.2
C W/m.1460.0 ,kg/m 6.893
23–
3
==
==
==
p
c
k
Analysis (a) The Reynolds number in this case is
m) m/s)(0.4 (0.5
avg =
h
DV
m 4.0
D
Next we determine the exit temperature of oil
m 1885=m) m)(1500 4.0(
2
==
DLA
s
Oil
10C
0.5 m/s
(Icy lake, 0C)
L = 1500
D = 0.4 m
8-52
8-71 Liquid glycerin is flowing through a 25-mm diameter and 10-m long tube, the constant surface temperature of the tube
is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 Constant tube surface temperature.
Properties The properties of glycerin at Tb = (Ti + Te)/2 = 30°C are cp = 2447 J/kg∙K, k = 0.2860 W/m∙K,
= 0.6582 kg/m∙s,
and Pr = 5631 (Table A-13).
Analysis The Reynolds number, hydrodynamic and thermal entry lengths are
)kg/s 5.0(44
m
−−
s
where
)m 10)(m 025.0()K W/m152(2
s
hA
8-53
8-72 Liquid water entering at 60°C is heated in a circular tube with a constant surface temperature. The mass flow rate
of the flow is 11 g/s. The surface temperature of the tube is to be determined and whether a HNBR o-ring attached to the tube
surface is suitable. The maximum temperature permitted for the o-ring is 150°C.
Assumptions 1 The flow is steady and incompressible. 2 Uniform surface temperature. 3 Inner surface of the tube is smooth.
Properties The properties of water at 100°C are (Table A–9) cp = 4217 J/kg∙K, k = 0.679 W/m∙K, μ = 0.282 × 10−3 kg/m∙s,
and Pr = 1.75
Analysis The Reynolds number for the flow in circular tube is
8-54
8-73 Air at 20°C (1 atm) enters into a 5-mm diameter and 10-cm long circular tube, the convection heat transfer coefficient
and the outlet mean temperature are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 Constant tube surface temperature.
Properties The properties of air at 50°C: cp = 1007 J/kg∙K, k = 0.02735 W/m∙K,
= 1.092 kg/m3,
= 1.963 10−5 kg/m∙s,
= 1.798 10−5 m2/s, and Pr = 0.7228; at Ts = 160°C:
s = 2.420 10−5 kg/m∙s (Table A–15).
Analysis The Reynolds number, hydrodynamic and thermal entry lengths are
)m 005.0)(m/s 5(
avg =
DV
8-56
8-75 Glycerin is being heated by flowing between two parallel 1-m wide plates with 12.5-mm spacing. The outlet mean
temperature and the total rate of heat transfer are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 Isothermal parallel plates. 4 Bulk mean fluid
temperature is 30°C (this will be validated).
Properties The properties of glycerin at 30°C are cp = 2447 J/kg∙K, k = 0.2860 W/m∙K, μ = 0.6582 kg/m∙s, and Pr = 5631
(Table A-13).
Analysis The Reynolds number, hydrodynamic and thermal entry lengths can be determined to be
8-58
0 1 2 3 4 5 6
0
20000
40000
60000
80000
100000
120000
140000
160000
Q [W]
8-59
8-77 Glycerin is being heated by flowing between two parallel 1-m wide plates with 12.5-mm spacing. Hydrogen gas
flows width-wise in parallel over the upper and lower surfaces of the two plates. The outlet mean temperature of the glycerin,
the surface temperature of the plates, and the total rate of heat transfer are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 Isothermal parallel plates. 4 The thermal
resistance of the plates is negligible (thin plates). 5 The bulk mean fluid temperature of the glycerin is 30°C (this will be
validated). 6 The film temperature of the H2 gas is 100°C (this will be validated).
Properties The properties of glycerin at 30°C are cp = 2447 J/kg∙K, k = 0.2860 W/m∙K, μ = 0.6582 kg/m∙s, and Pr = 5631
(Table A-13). The properties of H2 gas at 100°C are kH2 = 0.2095 W/m∙K, νH2 = 1.582 10−4 m2/s, and PrH2 = 0.7196 (Table
A-16)
Analysis The Reynolds number, hydrodynamic and thermal entry lengths can be determined to be
m 025.2m )012501(2 =+= .p
,
2
m 0125.0)m 0125.0(m) 1( ==
c
A
,
m 02469.0/4 == pAD ch
)kg/s 7.0(4
4
m
Therefore the flow is laminar and hydrodynamically developed but still thermally developing. The appropriate equation to
determine the Nusselt number is from Edwards et al. (1979):
)5631)(101.2)(10/02469.0(03.0
PrRe)/(03.0
LD
h
e
s
8-60
“GIVEN”
c_p=2447 [J/kg-K]
h=96.156 [W/m^2-K]
h_H2=17.166 [W/m^2-K]
A_s=20.25 [m^2]
m_dot=0.7 [kg/s]