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7-121
7-121 The heat generated by four transistors mounted on a thin vertical plate is dissipated by air blown over the plate on both
surfaces. The temperature of the aluminum plate is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5105. 3 Radiation effects are
negligible. 4 The entire plate is nearly isothermal. 5 The exposed surface area of the transistor is taken to be equal to its base
area. 6 Air is an ideal gas with constant properties. 7 The pressure of air is 1 atm.
Properties Assuming a film temperature of 40C, the properties of air are evaluated to be (Table A-15)
7255.0Pr
/sm 10702.1
C W/m.02662.0
25-
=
=
=
k
Analysis The Reynolds number in this case is
4
25 10463.6
/sm 10702.1
m) m/s)(0.22 5(
Re =
== −
VL
L
which is smaller than the critical Reynolds number. Thus we
have laminar flow. Using the proper relation for Nusselt
number, heat transfer coefficient is determined to be
C. W/m35.18)7.151(
m 22.0
C W/m.02662.0
7.151)7255.0()10463.6(664.0PrRe664.0
2
3/15.043/1
5.0
=
==
====
Nu
L
k
h
k
hL
Nu L
The temperature of aluminum plate then becomes
C42.5=
+=+=⎯→⎯−= ])m 22.0(2)[C. W/m35.18(
W)104(
C20)( 22
s
sss hA
Q
TTTThAQ
Discussion In reality, the heat transfer coefficient will be higher since the transistors will cause turbulence in the air. Also,
the film temperature is (20 + 42.5)/2 = 31.3C, which is not close to the assumed value of 40C. Therefore, calculations
should be repeated with a lower film temperature.
Ts
10 W
V = 5 m/s
T = 20C
L= 22 cm
7-122
7-122 Oil flows over a flat plate that is maintained at a specified temperature. The rate of heat transfer is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5105. 3 Radiation effects are
negligible.
Properties The properties of oil are given to be ρ = 880 kg/m3, μ = 0.005 kg/m.s, k = 0.15
W/mK, and cp = 2.0 kJ/kgK.
Analysis The Prandtl and Reynolds numbers are
C)J/kg s)(2000kg/m 005.0(
cp
Ts = 20C
Oil
V = 20 cm/s
7-124
7-124 Air flows over the top and bottom surfaces of a thin, square plate. The total heat transfer rate are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5105. 3 Radiation effects are
negligible.
Properties The properties of air at the film temperature ofTf= (Ts + T)/2 = (54+10)/2 = 32C are (Table A-15)
K W/m02603.0
7276.0PrKJ/kg 1007
/sm 10627.1kg/m 156.1 253
=
==
== −
k
cp
Analysis The Reynolds number is
6
m) m/s)(1.2 48(
VL
Ts = 54C
Air
V = 48 m/s
T = 10C
7-125
7-125 Wind is blowing parallel to the walls of a house with windows. The rate of heat loss through the window is to be
determined.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5105. 3 Radiation effects are
negligible. 4 Air is an ideal gas with constant properties. 5 The pressure of air is 1 atm.
Properties Assuming a film temperature of 5C based on the problem
statement, the properties of air at 1 atm and this temperature are
evaluated to be (Table A-15)
7350.0Pr
/sm 10382.1
C W/m.02401.0
25-
=
=
=
k
Analysis Air flows along 1.8 m side. The Reynolds number in this case is
6
m) (1.8m/s )3600/100035(
VL
Air
V = 35 km/h
T2 = -2C
WINDOW
T1 = 22C
7-126
7-126 A car travels at a velocity of 60 km/h. The rate of heat transfer from the bottom surface of the hot automotive engine
block is to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5105. 3 Air is an ideal gas with
constant properties. 4 The pressure of air is 1 atm. 5 The flow is turbulent over the entire surface because of the constant
agitation of the engine block. 6 The bottom surface of the engine is a flat surface.
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
5
25 10855.6
/sm 10702.1
m) (0.7m/s )3600/100060(
Re =
== −
VL
L
which is greater than the critical Reynolds number. Thus we have
combined laminar and turbulent flow. But we will assume turbulent flow
because of the constant agitation of the engine block.
C. W/m97.58)1551(
m 7.0
C W/m.02662.0
1551)7255.0()10855.6(037.0PrRe037.0
2
3/18.053/1
8.0
=
==
====
Nu
L
k
h
k
hL
Nu L
W1734=C5)(75m) m)(0.7 (0.6C). W/m97.58()( 2−=−= ssconv TThAQ
The heat loss by radiation is then determined from Stefan-Boltzman law to be
)(
44428-
44
−= surrssrad TTAQ
Ts = 75C
= 0.92
Air
V = 60 km/h
T = 5C
L = 0.7 m
Engine block
Ts = 10C
7-127
7-127E A 15-ft long strip of sheet metal is being transported on a conveyor, while the coating on the upper surface is being
cured by infrared lamps. The surface temperature of the sheet metal is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat conduction through the sheet metal is negligible. 3 Thermal
properties are constant. 4 The surrounding ambient air is at 1 atm. 5 The critical Reynolds number is Recr = 5105.
Properties The properties of air at 180°F are k = 0.01715 Btu/h∙ft∙R,
= 2.281 10−4 ft2/s, Pr = 0.7148 (from Table A-15E).
Analysis The Reynolds number for L = 15 ft is
6
24 10052.1
/sft 10281.2
)ft 15)(ft/s 16(
Re =
== −
VL
L
Since 5 105< ReL< 107, the flow is a combined
laminar and turbulent flow. Using the proper
relation for Nusselt number, the average heat
transfer coefficient on the sheet metal is
1395)7148.0](871)10052.1(037.0[Pr)871Re037.0(Nu 3/18.063/18.0 =−=−== L
k
hL
RftBtu/h 595.1
ft 15
RftBtu/h 01715.0
13951395 2=
== L
k
h
From energy balance, we have
7-129
7-129 The top surface of a hot block is to be cooled by forced air. The rate of heat transfer is to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr = 5105. 3 Radiation effects are
negligible. 4 Air is an ideal gas.
Properties The properties k, , cp, and Pr of ideal gases are independent of pressure, while the properties and are
inversely proportional to density and thus pressure. The properties of air at the film temperature of Tf= (Ts+ T
)/2 =(140 +
20)/2 = 80°C and 1 atm pressure are (Table A–15):
k = 0.02953 W/m · K, Pr = 0.7154, @ 1 atm = 2.09710-5 m2/s
The atmospheric pressure in Denver is P = (83.4 kPa)/(101.325
kPa/atm) = 0.823 atm. Then the kinematic viscosity of air in Denver
becomes: = @ 1 atm /P = (2.097 10-5 m2/s)/0.823 = 2.548 10-5 m2/s
Analysis (a) When air flow is parallel to the long side,
we have L= 6 m, andthe Reynolds number at the end of
the plate becomes
6
25 10884.1
/sm 10548.2
)m 6)(m/s 8(
Re =
== −
VL
L
which is greater than the critical Reynolds number.
Thus, we have combinedlaminar and turbulent flow, and
the average Nusselt number for the entire plateis
determined to be
2687)7154.0](871)10884.1(037.0[Pr)871Re037.0(Nu 3/18.063/18.0 =−=−== L
k
hL
K W/m2.13
m 6
K W/m02953.0
)2687(2=
== L
k
Nuh
𝐴𝑠 = wL = (1.5 m)(6 m) = 9 m2
W10 1.43 4
−=−= =C20)(140)m)(9 W/m2.13()( 22 KTThAQss
Note that if we disregarded the laminar region and assumed turbulent flow overthe entire plate, we would get Nu = 3466 from
Eq. 7–22, which is 29 percenthigher than the value calculated above.
(b) When air flow is along the short side, we have L = 1.5 m, and the Reynoldsnumber at the end of the plate becomes
5
)m 5.1)(m/s 8(
VL
7-130
7-130 A silicon chip is mounted flush in a substrate that provides an unheated starting length. The maximum allowable power
dissipation is to be determined such that the surface temperature of the chip cannot exceed 75°C.
Assumptions 1 Steady operating conditions exist. 2 Thermal properties are constant. 3 The flow is turbulent. 4 Only the
upper surface of the chip is conditioned for heat transfer. 5 Heat transfer by radiation is negligible. 6 Heat dissipated from the
chip is uniform.
Properties The properties of air at Tf = (75°C + 25°C)/2 = 50°C are k = 0.02735 W/m∙K,
= 1.798 10−5 m2/s, Pr = 0.7228
(from Table A-15).
Analysis For uniform heat flux on the chip surface, the maximum surface temperature occurs at the trailing edge, where the
convection heat transfer coefficient is at minimum. The Reynolds number at the trailing edge (x = 0.040 m) is
4
)m 040.0)(m/s 25(
Vx
])40/20(1[
)m 040.0(
9/110/9
−
Hence, the maximum allowable power dissipation on the chip surface is
7-131
7-131 Airstream flows in parallel over a 3-m long flat plate where there is an unheated starting length of 1 m, (a) the local
convection heat transfer coefficient at x = 3 m and (b) the average convection heat transfer coefficient for the heated section
are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Surface temperature is uniform throughout the heated section. 3 Thermal
properties are constant. 4 The critical Reynolds number is Recr = 5105.
Properties The properties of air at Tf = (80°C + 20°C)/2 = 50°C are k = 0.02735 W/m∙K,
= 1.798 10−5 m2/s, Pr = 0.7228
(from Table A-15).
Analysis (a) The Reynolds number at x = 1 m is
5
)m 1)(m/s 15(
Vx
−
])3/1(1[
m) 3(
(b) The average convection heat transfer coefficient over the heated section is
7-134
7-134E A cylindrical transistor mounted on a circuit board is cooled by air flowing over it. The maximum power rating of the
transistor 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 pressure of air is 1 atm.
Properties The properties of air at 1 atm and the film temperature of
F1502/)120180(=+=
f
T
are (Table A-15E)
7188.0Pr
/sft 10099.2
FBtu/h.ft. 01646.0
24-
=
=
=
k
Analysis The Reynolds number is
ft) /12ft/s)(0.22 (600/60
VD
Air
600 ft/min
120F
Power
transistor
D = 0.22 in
L = 0.32 in
