Respostas_Livro FT
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Respostas_Livro FT


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be determined. 
Assumptions 1 Steady operating conditions exist. 2 The heat 
transfer coefficient is constant and uniform over the entire exposed 
surface of the person. 3 The surrounding surfaces are at the same 
temperature as the indoor air temperature. 4 Heat generation 
within the 0.5-cm thick outer layer of the tissue is negligible. 
Qrad
Tskin
 
Qconv
Properties The thermal conductivity of the tissue near the skin is 
given to be k = 0.3 W/m\u22c5°C. 
Analysis The skin temperature can be determined directly from 
 
C35.5°=°\u22c5\u2212°=\u2212=
\u2212=
)m C)(1.7 W/m3.0(
m) 005.0 W)(150(
C37
21
1
kA
LQTT
L
TT
kAQ
skin
skin
&
&
 
 
 
 
 
 
3-28 Heat is transferred steadily to the boiling water in an aluminum pan. The inner surface temperature of 
the bottom of the pan is given. The boiling heat transfer coefficient and the outer surface temperature of the 
bottom of the pan are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional since the thickness of 
the bottom of the pan is small relative to its diameter. 3 The thermal conductivity of the pan is constant. 
Properties The thermal conductivity of the aluminum pan is given to be k = 237 W/m\u22c5°C. 
Analysis (a) The boiling heat transfer coefficient is 
 2
22
m 0491.0
4
m) 25.0(
4
=== \u3c0\u3c0DAs 
95°C 
 108°C 
600 W 0.5 cm
 
C. W/m1254 2 °=°\u2212=\u2212=
\u2212=
\u221e
\u221e
C)95108)(m 0491.0(
 W800
)(
)(
2TTA
Qh
TThAQ
ss
ss
&
&
 
(b) The outer surface temperature of the bottom of the pan is 
 
C108.3°=°\u22c5°=+=
\u2212=
)m C)(0.0491 W/m237(
m) 005.0 W)(800(+C108
21,,
,,
kA
LQTT
L
TT
kAQ
innersouters
innersouters
&
&
 
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. 
 3-10
3-29E A wall is constructed of two layers of sheetrock with fiberglass insulation in between. The thermal 
resistance of the wall and its R-value of insulation are to be determined. 
Assumptions 1 Heat transfer through the wall is one-dimensional. 2 Thermal conductivities are constant. 
Properties The thermal conductivities are given to be 
ksheetrock = 0.10 Btu/h\u22c5ft\u22c5°F and kinsulation = 0.020 Btu/h\u22c5ft\u22c5°F. 
 
R1 R2 R3
L1 L2 L3 
Analysis (a) The surface area of the wall is not given and thus 
we consider a unit surface area (A = 1 ft2). Then the R-value of 
insulation of the wall becomes equivalent to its thermal 
resistance, which is determined from. 
F.h/Btu.ft 30.34 2 °=+×=+=
°=°===
°=°====
17.29583.022
F.h/Btu.ft 17.29
F)Btu/h.ft. 020.0(
ft 12/7
F.h/Btu.ft 583.0
F)Btu/h.ft. 10.0(
ft 12/7.0
21
2
2
2
2
2
1
1
31
RRR
k
L
RR
k
L
RRR
total
fiberglass
sheetrock
 
(b) Therefore, this is approximately a R-30 wall in English units. 
 
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. 
 3-11
3-30 The roof of a house with a gas furnace consists of a concrete that is losing heat to the outdoors by 
radiation and convection. The rate of heat transfer through the roof and the money lost through the roof 
that night during a 14 hour period are to be determined. 
Assumptions 1 Steady operating conditions exist. 2 
The emissivity and thermal conductivity of the roof 
are constant. 
Properties The thermal conductivity of the concrete 
is given to be k = 2 W/m\u22c5°C. The emissivity of both 
surfaces of the roof is given to be 0.9. 
Analysis When the surrounding surface temperature is different 
than the ambient temperature, the thermal resistances network 
approach becomes cumbersome in problems that involve radiation. 
Therefore, we will use a different but intuitive approach. 
 In steady operation, heat transfer from the room to the 
roof (by convection and radiation) must be equal to the heat 
transfer from the roof to the surroundings (by convection and 
radiation), that must be equal to the heat transfer through the roof 
by conduction. That is, 
Q&
Tsky = 100 K 
 
Tair =10°C 
Tin=20°C 
L=15 cm
 rad+conv gs,surroundin toroofcond roof,rad+conv roof, toroom QQQQ &&&& ===
Taking the inner and outer surface temperatures of the roof to be Ts,in and Ts,out , respectively, the quantities 
above can be expressed as 
[ ]4,44282 ,
224
,
4
, rad+conv roof, toroom
K) 273(K) 27320()K W/m1067.5)(m 300)(9.0( 
C))(20m C)(300 W/m5()()(
+\u2212+\u22c5×+
°\u2212°\u22c5=\u2212+\u2212=
\u2212
ins
insinsroominsroomi
T
TTTATTAhQ \u3c3\u3b5&
 
 
m 15.0
)m 300)(C W/m2( ,,2,,cond roof,
outsinsoutsins TT
L
TT
kAQ
\u2212°\u22c5=\u2212=& 
[ ]44,4282 ,
2244
,, rad+conv surr, toroof
K) 100(K) 273()K W/m1067.5)(m 300)(9.0( 
C)10)(m C)(300 W/m12()()(
\u2212+\u22c5×+
°\u2212°\u22c5=\u2212+\u2212=
\u2212
outs
outssurroutssurroutso
T
TTTATTAhQ \u3c3\u3b5&
 
Solving the equations above simultaneously gives 
 C1.2and , , out,, °\u2212=°== sins TTQ C7.3 W37,440&
The total amount of natural gas consumption during a 14-hour period is 
 therms36.22
kJ 105,500
 therm1
80.0
)s 360014)(kJ/s 440.37(
80.080.0
=\u239f\u239f\u23a0
\u239e\u239c\u239c\u239d
\u239b×=\u394== tQQQ totalgas
&
 
Finally, the money lost through the roof during that period is 
 $26.8== )therm/20.1$ therms)(36.22(lostMoney 
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. 
 3-12
3-31 An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat loss 
through that section of the wall by 90 percent. The thickness of the insulation that needs to be used is to be 
determined. Also, the length of time it will take for the insulation to pay for itself from the energy it saves 
will be determined. 
Assumptions 1 Heat transfer through the wall is steady and one-dimensional. 2 Thermal conductivities are 
constant. 3 The furnace operates continuously. 4 The given heat transfer coefficient accounts for the 
radiation effects. 
Properties The thermal conductivity of the glass wool 
insulation is given to be k = 0.038 W/m\u22c5°C. Insulation 
Ro 
T\u221e
Rinsulation 
Ts
L 
Analysis The rate of heat transfer without insulation is 
 2m 3m) m)(1.5 2( ==A
 W1500C)3080)(m 3(C) W/m10()( 22 =°\u2212°\u22c5=\u2212= \u221eTThAQ s&
In order to reduce heat loss by 90%, the new heat transfer rate and 
thermal resistance must be 
 
C/W 333.0
 W150
C)3080(
 W150 W150010.0
°=°\u2212=\u394=\u23af\u2192\u23af\u394=
=×=
Q
TR
R
TQ
Q
total
total
&
&
&
 
and in order to have this thermal resistance, the thickness of insulation must be 
 
cm 3.4==
°=°+°\u22c5=
+=+=
m 034.0
C/W 333.0
)m C)(3 W/m.038.0()m C)(3 W/m10(
1
1
222
conv
L
L
kA
L
hA
RRR insulationtotal
 
Noting that heat is saved at a rate of 0.9×1500 = 1350 W and the furnace operates continuously and thus 
365×24 = 8760 h per year, and that the furnace efficiency is 78%, the amount of natural gas saved per year 
is 
 therms4.517
kJ 105,500
 therm1
h 1
s 3600
0.78
h) kJ/s)(8760 350.1(SavedEnergy =\u239f\u239f\u23a0
\u239e\u239c\u239c\u239d
\u239b\u239f\u23a0
\u239e\u239c\u239d
\u239b=\u394=
Efficiency
tQsaved& 
The money saved is 
year)(per 1.569$)1.10/therm therms)($4.517(energy) oft Saved)(CosEnergy (savedMoney === 
The insulation will pay for its cost of $250 in 
 yr 0.44===
$569.1/yr
$250
savedMoney 
spent Money periodPayback 
which is equal to 5.3 months. 
 
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. 
 3-13
3-32 An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat loss 
through that section of the wall by 90 percent. The thickness of the insulation that needs to