9-101
9-115 The significance of natural convection to the heat transfer process on a horizontal rod with water flowing across its
outer surface is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Properties are constant. 3 The rod is orientated such that the characteristic
length is its diameter.
Properties The properties of water at Tf = (Ts + T∞)/2 = 80°C are ρ = 971.8 kg/m3, μ = 0.355 × 10−3 kg/m·s, and β = 0.653 ×
10−3 K-1 (Table A-9).
Analysis The Reynolds number for the cross flow is
4
3
)m 150.0)(m/s 0.2)(kg/m 8.971(
VD
9-103
9-118 A circuit board is cooled by a fan that blows air upwards. The average temperature on the surface of the circuit board is
to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The atmospheric pressure at
that location is 1 atm.
Properties Based on the problem statement, the properties of air at 1 atm and 1 atm and the anticipated film temperature of
(Ts+T)/2 = (60+35)/2 = 47.5C are (Table A-15)
1–
25
K 00312.0
K)2735.47(
11
7235.0Pr
/sm 10774.1
C W/m.02717.0
=
+
==
=
=
=
−
f
T
k
Analysis We assume the surface temperature to be 60C. We will check this
assumption later on and repeat calculations with a better assumption, if necessary.
The characteristic length in this case is the length of the board in the flow (vertical)
(b) The Rayleigh number is
6
3-12
3
)m 12.0)(K 3560)(K 00312.0)(m/s 81.9(
)( =
−
−
LTTg
Air
T = 35C
V = 0.5 m/s
PCB, Ts
1000.05 W
L = 12 cm
9-104
9-119 A horizontal square plate is being cooled by blowing air over the surface of the plate. Determine the Nusslet number
for this case.
Assumptions1 Steady state conditions. 2 Air is an ideal gas. 3 The local atmospheric pressure is 1 atm. 4 Thermal radiation is
negligible.5 Backside of the plate is insulated.
Properties The properties of air at the film temperature of Tf = (T + Ts )/2 = (29 +125)/2 =77°C are (Table A–15) Pr =
0.7161, k = 0.02931 W/m∙K, ν = 2.066× 10−5 m2/s, β = 1/Tf = 1/(77 + 273 K) = 0.002857 K−1
AnalysisThe characteristic length of the plate is
9-105
Special Topic: Heat Transfer through Windows
9-120C Windows are considered in three regions when analyzing heat transfer through them because the structure and
properties of the frame are quite different than those of the glazing. As a result, heat transfer through the frame and the edge
section of the glazing adjacent to the frame is two-dimensional. Even in the absence of solar radiation and air infiltration, heat
transfer through the windows is more complicated than it appears to be. Therefore, it is customary to consider the windows in
heat transfer through the window will be a minimum for the window with 10–mm air gab, as can be seen from Fig. 9-37.
9-122C In an ordinary double pane window, about half of the heat transfer is by radiation. A practical way of reducing the
9-108
Wall
9-131E The R-value of the common double door windows that are double pane with 1/4-in of air space and have aluminum
frames is to be compared to the R-value of R-13 wall. It is also to be determined if more heat is transferred through the
windows or the walls.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the window is one-dimensional. 3 Thermal
properties of the windows and the heat transfer coefficients are constant. 4 Infiltration heat losses are not considered.
Properties The U-factor of the window is given in Table 9-6 to be
4.550.176 = 0.801 Btu/h.ft2.F.
Analysis The R-value of the windows is simply the inverse of its U-factor,
and is determined to be
11 2
R-13
9-109
9-132 The rate of heat loss through a double-door wood framed window and the inner surface temperature are to be
determined for the cases of single pane, double pane, and low-e triple pane windows.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the window is one-dimensional. 3 Thermal
properties of the windows and the heat transfer coefficients are constant. 4 Infiltration heat losses are not considered.
Properties The U-factors of the windows are given in Table 9-6.
Analysis The rate of heat transfer through the window can be determined from
window
i
where hi is the heat transfer coefficient on the inner surface of the window which is determined from Table 9-5 to be hi = 8.3
W/m2.C. Then the rate of heat loss and the interior glass temperature for each case are determined as follows:
(a) Single glazing:
W337=−−= C)8(20)m 16.2)(C W/m57.5( 22
window
Q
C1.2=
−=−= )m 16.2)(C W/m29.8(
W337
C20 22
window
window
glass Ah
Q
TT
i
i
(b) Double glazing (13 mm air space):
W173=−−= C)8(20)m 16.2)(C W/m86.2( 22
window
Q
C10.3=
−=−= )m 16.2)(C W/m29.8(
W173
C20 22
window
window
glass Ah
Q
TT
i
i
(c) Triple glazing (13 mm air space, low-e coated):
W88.3=−−= C)8(20)m 16.2)(C W/m46.1( 22
window
Q
C15.1=
−=−= )m C)(2.16. W/m(8.3
W88.3
20 22
glass
windowi
window
iAh
Q
TT
Discussion Note that heat loss through the window will be reduced by 49 percent in the case of double glazing and by 74
percent in the case of triple glazing relative to the single glazing case. Also, in the case of single glazing, the low inner glass
surface temperature will cause considerable discomfort in the occupants because of the excessive heat loss from the body by
radiation. It is raised from 1.2C to 10.3C in the case of double glazing and to 15.1C in the case of triple glazing.
Double-door
window
Wood frame
9-110
9-133 The overall U-factor for a double-door type window is to be determined, and the result is to be compared to the value
listed in Table 9–6.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through
the window is one-dimensional.
Properties The U-factors for the various sections of windows are given in
Table 9-6.
Analysis The areas of the window, the glazing, and the frame are
2
glazingwindowframe
2
glazing
2
window
m 42.038.480.4
m .384m) 14.m)(1 92.1(2WidthHeight2
m .804m) 4.m)(2 2(WidthHeight
=−=−=
===
===
AAA
A
A
The edge-of-glass region consists of a 6.5-cm wide band around the
perimeter of the glazings, and the areas of the center and edge sections
of the glazing are determined to be
2
center m .623m) 13.014.m)(1 13.092.1(2Width)Height(2 =−−==A
2
centerglazingedge m .76062.338.4 =−=−= AAA
The U-factor for the frame section is determined from Table 9-4 to be Uframe = 2.8 W/m2.C. The U-factor for the center and
edge sections are determined from Table 9-6 to be Ucenter = 2.78 W/m2.C and Uedge =3.40 W/m2.C. Then the overall U-factor
of the entire window becomes
C W/m2.88 2=
++=
++=
80.4/)42.08.276.040.362.378.2(
/)( windowframeframeedgeedgecentercenterwindow AAUAUAUU
Discussion The overall U-factor listed in Table 9-6 for the specified type of window is 2.86 W/m2.C, which is sufficiently
close to the value obtained above.
Center of
glass
Edge of
glass
Frame
2 m
1.92m
2.4 m
1.14 m
1.14 m
9-111
Review Problems
9-134 An electric resistance space heater filled with oil is placed against a wall. The power rating of the heater and the time it
will take for the heater to reach steady operation when it is first turned on are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 Heat transfer from the
back, bottom, and top surfaces are disregarded. 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+25)/2 = 50C are (Table A–15)
1–
25
K 003096.0
K)27350(
11
7228.0Pr
/sm 10798.1
C W/m.02735.0
=
+
==
=
=
=
−
f
T
k
Analysis Heat transfer from the top and bottom surfaces are said to be negligible,
and thus the heat transfer area in this case consists of the three exposed side
surfaces. The characteristic length is the height of the box, Lc = L = 0.5 m. Then,
8
3-12
3
)m 5.0)(K 2575)(K 003096.0)(m/s 81.9(
)( =
−
−
LTTg
Air
T =25C
= 0.8
Ts = 75C
15 cm
80 cm
50 cm
9-114
9-137 A plate inclined at 30° with the top surface of the plate well insulated. The rate of heat loss from the plate is to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 Thermal properties are constant. 4 Radiation heat
transfer is negligible.
Properties The properties of air at Tf = (Ts + T∞)/2 = 30°C are k = 0.02588 W/m∙K,
= 1.608 × 10−5 m2/s, Pr = 0.7282 (from
Table A-15). Also, β = 1/Tf = 0.0033 K-1.
Analysis The Rayleigh number (Lc = L) is
31–2
2
3
)m 5.0(K)060)(K 0033.0)(30)(cosm/s 81.9(
Pr
)(cos
Ra
−
−
=
LTTg s
L
9-116
9-139 A group of 25 transistors are cooled by attaching them to a square aluminum plate and positioning the plate
horizontally in a room. The required size of the plate to limit the surface temperature to 50C is to be determined for two
cases.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The local atmospheric
pressure is 1 atm. 4 Any heat transfer from the back side of the plate is negligible.
Properties The properties of air at 1 atm and the film temperature of
(Ts+T)/2 = (50+30)/2 = 40C are (Table A–15)
1–
25
K 003195.0
K)27340(
11
7255.0Pr
/sm 10702.1
C W/m.02662.0
=
+
==
=
=
=
−
f
T
k
Analysis The characteristic length and the Rayleigh number for
the horizontal case are determined to be
2L
L
A
rad 3.125K])27330()27350)[(.K W/m1067.5()9.0()( LLTTAQ skyss =+−+=−= −
(a) Hot surface facing up: We assume Ra < 107 and thus L <0.74 m so that we can determine the Nu number from Eq. 9-22.
Then the Nusselt number and the convection heat transfer coefficient become
4/34/1374/1 0.38)10454.2(54.054.0 LLRaNu ===
(b) Hot surface facing down: The Nusselt number in this case is determined from
4/34/1374/1 0.19)10454.2(27.027.0 LLRaNu ===
Transistors,
251.5 W
= 0.9
Ts = 50C
Room
30C
Plate
L L
9-118
9-141 A flat-plate solar collector placed horizontally on the flat roof of a house is exposed to the calm ambient air. The rate
of heat loss from the collector by natural convection and radiation are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The local atmospheric
pressure is 1 atm.
Properties The properties of air at 1 atm and the film temperature of (Ts+T)/2 = (42+8)/2 = 25C are (Table A-15)
1–
25
K 003356.0
K)27325(
11
7296.0Pr
/sm 10562.1
C W/m.02551.0
=
+
==
=
=
=
−
f
T
k
Analysis The characteristic length in this case is determined from
)m 5.4)(m 5.1( === p
A
Air
T = 8C
Tsky = –15C
Solar collector
Ts = 42C
= 0.85
L = 1.5 m
9-119
9-142 A horizontal skylight made of a single layer of glass on the roof of a house is considered. The rate of heat loss through
the skylight is to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The local atmospheric
pressure is 1 atm.
Properties Based on the problem statement, the properties of air at 1 atm
and the anticipated film temperature of (Ts+T)/2 = (-4-10)/2 = −7C are
(Table A-15)
1–
25
K 003759.0
K)2737(
11
738.0Pr
/sm 10278.1
C W/m.02311.0
=
+−
==
=
=
=
−
f
T
k
Analysis We assume radiation heat transfer inside the house to be negligible. We start the calculations by “guessing” the
glass temperature to be -4C for the evaluation of the properties and h. We will check the accuracy of this guess later and
repeat the calculations if necessary. The characteristic length in this case is determined from
m) m)(2.5 1( === p
A
Tin = 20C
Outdoors
T = –10C
Tsky = –30C
Skylight
2.5 m 1 m
= 0.9
t = 0.5 cm
9-120
W115
C. W/m90.6
1
C W/m.78.0
m 005.0
C. W/m998.1
1
11
22
glass
,,,
,,
=
+
+
=
++
=
++
−
=
hk
t
h
RRR
TT
Q
glassi
outrooms
ocombinedglascondiconv
ois
skylight
Using the same heat transfer coefficients for simplicity, the rate of heat loss through the roof in the case of R-5.34
construction is determined to be
W5.12
C. W/m90.6
1
C/W.m 34.5
C. W/m998.1
1
C)]10(20)[m 5.2(
11
)(
2
2
2
,,
,,
=
++
−−
=
++
−
=
++
−
=
h
R
h
TTA
RRR
TT
Q
glass
i
outrooms
ocombinedcondiconv
ois
roof
Therefore, a house loses 115/12.5 9 times more heat through the skylights than it does through an insulated wall of the
same size.