Respostas_Livro FT
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Respostas_Livro FT


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at the specified values. 2 Heat transfer is one-dimensional since any significant 
temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivity of 
the glass is constant. 4 Heat transfer by radiation is negligible. 
Properties The thermal conductivity of the glass is given to be kglass = 0.78 W/m\u22c5°C. 
Analysis Heat cannot be conducted through an evacuated 
space since the thermal conductivity of vacuum is zero (no 
medium to conduct heat) and thus its thermal resistance is 
zero. Therefore, if radiation is disregarded, the heat transfer 
through the window will be zero. Then the answer of this 
problem is zero since the problem states to disregard 
radiation. 
Discussion In reality, heat will be transferred between the 
glasses by radiation. We do not know the inner surface 
temperatures of windows. In order to determine radiation 
heat resistance we assume them to be 5°C and 15°C, 
respectively, and take the emissivity to be 1. Then 
individual resistances are 
Vacuum 
R1 Rrad R3 Ro Ri
T\u221e1 T\u221e2 2m 4.2m) 2(m) 2.1( =×=A
C/W 1426.0
0167.00810.0)0016.0(20417.02
C/W 0167.0
)m 4.2(C). W/m25(
11
C/W 0810.0
]278288][278288)[m 4.2().K W/m1067.5(1
1 
))((
1
C/W 0016.0
)m 4.2(C) W/m.78.0(
m 003.0
C/W 0417.0
)m 4.2(C). W/m10(
11
2,11,
22
2
2,o
3222428
22
2
1
1
glass31
22
1
1,i
°=
+++=+++=
°=°===
°=
++×=
++=
°=°====
°=°===
\u2212
convradconvtotal
conv
surrssurrs
rad
conv
RRRRR
Ah
RR
K
TTTTA
R
Ak
L
RRR
Ah
RR
\u3b5\u3c3
 
The steady rate of heat transfer through window glass then becomes 
 W203=°
°\u2212\u2212=\u2212= \u221e\u221e
C/W1426.0
C)]5(24[21
totalR
TT
Q& 
The inner surface temperature of the window glass can be determined from 
 C15.5°=°\u2212°=\u2212=\u23af\u2192\u23af\u2212= \u221e\u221e C/W) W)(0.0417203(C241,11
1,
11
conv
conv
RQTT
R
TT
Q && 
Similarly, the inner surface temperatures of the glasses are calculated to be 15.2 and -1.2°C (we had 
assumed them to be 15 and 5°C when determining the radiation resistance). We can improve the result 
obtained by reevaluating the radiation resistance and repeating the calculations. 
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and 
educators for course preparation. If you are a student using this Manual, you are using it without permission. 
 3-6
3-22 EES Prob. 3-20 is reconsidered. The rate of heat transfer through the window as a function of the 
width of air space is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
A=1.2*2 [m^2] 
L_glass=3 [mm] 
k_glass=0.78 [W/m-C] 
L_air=12 [mm] 
T_infinity_1=24 [C] 
T_infinity_2=-5 [C] 
h_1=10 [W/m^2-C] 
h_2=25 [W/m^2-C] 
 
"PROPERTIES" 
k_air=conductivity(Air,T=25) 
 
"ANALYSIS" 
R_conv_1=1/(h_1*A) 
R_glass=(L_glass*Convert(mm, m))/(k_glass*A) 
R_air=(L_air*Convert(mm, m))/(k_air*A) 
R_conv_2=1/(h_2*A) 
R_total=R_conv_1+2*R_glass+R_air+R_conv_2 
Q_dot=(T_infinity_1-T_infinity_2)/R_total 
 
Lair [mm] Q [W] 
2 307.8 
4 228.6 
6 181.8 
8 150.9 
10 129 
12 112.6 
14 99.93 
16 89.82 
18 81.57 
20 74.7 
 
2 4 6 8 10 12 14 16 18 20
50
100
150
200
250
300
350
Lair [mm]
Q
 [
W
]
 
 
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and 
educators for course preparation. If you are a student using this Manual, you are using it without permission. 
 3-7
3-23E The inner and outer surfaces of the walls of an electrically heated house remain at specified 
temperatures during a winter day. The amount of heat lost from the house that day and its cost are to be 
determined. 
Assumptions 1 Heat transfer through the walls is steady since the surface temperatures of the walls remain 
constant at the specified values during the time period considered. 2 Heat transfer is one-dimensional since 
any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal 
conductivity of the walls is constant. 
Properties The thermal conductivity of the brick wall is given to be k = 0.40 Btu/h\u22c5ft\u22c5°F. 
Analysis We consider heat loss through the walls only. The total heat transfer area is 
 2ft 1530)935950(2 =×+×=A
The rate of heat loss during the daytime is 
 Btu/h 6120
ft 1
F)4555(
)ft F)(1530ftBtu/h 40.0( 221day =°\u2212°\u22c5\u22c5=\u2212= L
TT
kAQ& 
T2T1
L 
Q& 
Wall 
The rate of heat loss during nighttime is 
 
Btu/h 240,12
ft 1
C)3555()ft F)(1530ftBtu/h 40.0( 2
21
night
=°\u2212°\u22c5\u22c5=
\u2212=
L
TT
kAQ&
 
The amount of heat loss from the house that night will be 
 
Btu 232,560=
+=+=\u394=\u23af\u2192\u23af\u394=
 
)Btu/h 240,12(h) 14()Btu/h 6120(h) 10(1410 nightday QQtQQt
QQ &&&&
 
Then the cost of this heat loss for that day becomes 
 $6.13== kWh)/09.0)($kWh 3412/560,232(Cost 
 
 
 
 
3-24 A cylindrical resistor on a circuit board dissipates 0.15 W of power steadily in a specified 
environment. The amount of heat dissipated in 24 h, the surface heat flux, and the surface temperature of 
the resistor are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat is transferred uniformly from all surfaces of the 
resistor. 
Analysis (a) The amount of heat this resistor dissipates during a 24-hour period is 
 Wh3.6==\u394= h) W)(2415.0(tQQ &
Resistor 
0.15 W 
Q& 
(b) The heat flux on the surface of the resistor is 
 2
22
m 000127.0m) m)(0.012 003.0(
4
m) 003.0(
2
4
2 =+=+= \u3c0\u3c0\u3c0\u3c0 DLDAs 
 2 W/m1179===
2m 000127.0
 W15.0
sA
Qq
&
& 
(c) The surface temperature of the resistor can be determined from 
 C171°=°\u22c5+°=+=\u23af\u2192\u23af\u2212= \u221e\u221e )m 7C)(0.00012 W/m(9
 W15.0C40)(
22
s
sss hA
QTTTThAQ
&& 
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and 
educators for course preparation. If you are a student using this Manual, you are using it without permission. 
 3-8
3-25 A power transistor dissipates 0.2 W of power steadily in a specified environment. The amount of heat 
dissipated in 24 h, the surface heat flux, and the surface temperature of the resistor are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat is transferred uniformly from all surfaces of the 
transistor. 
Analysis (a) The amount of heat this transistor 
dissipates during a 24-hour period is 
Air, 
30°C 
Power 
Transistor 
0.2 W 
 kWh 0.0048===\u394= Wh8.4h) W)(242.0(tQQ &
(b) The heat flux on the surface of the transistor is 
2
2
2
m 0001021.0m) m)(0.004 005.0(
4
m) 005.0(
2
4
2
=+=
+=
\u3c0\u3c0
\u3c0\u3c0 DLDAs
 
2 W/m1959===
2m 0001021.0
 W2.0
sA
Qq
&
& 
(c) The surface temperature of the transistor can be determined from 
C139°=°\u22c5+°=+=\u23af\u2192\u23af\u2212= \u221e\u221e )m 21C)(0.00010 W/m(18
 W2.0C30)(
22
s
sss hA
QTTTThAQ
&& 
 
 
 
 
3-26 A circuit board houses 100 chips, each dissipating 0.06 W. The surface heat flux, the surface 
temperature of the chips, and the thermal resistance between the surface of the board and the cooling 
medium are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer from the back surface of the board is 
negligible. 2 Heat is transferred uniformly from the entire front surface. 
Analysis (a) The heat flux on the surface of the circuit board is 
Chips 
Ts
Q& 
T\u221e 
 2m 0216.0m) m)(0.18 12.0( ==sA
 2 W/m278=×==
2m 0216.0
 W)06.0100(
sA
Qq
&
& 
(b) The surface temperature of the chips is 
 
C67.8°=°\u22c5
×°=+=
\u2212=
\u221e
\u221e
)m 0216.0)(C W/m10(
 W)06.0100(+C40
)(
22
s
s
ss
hA
QTT
TThAQ
&
&
 
(c) The thermal resistance is 
 C/W4.63°=°\u22c5== )m 0216.0(C) W/m10(
11
22
s
conv hA
R 
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and 
educators for course preparation. If you are a student using this Manual, you are using it without permission. 
 3-9
3-27 A person is dissipating heat at a rate of 150 W by natural convection and radiation to the surrounding 
air and surfaces. For a given deep body temperature, the outer skin temperature is to