1537 pág.

# Respostas_Livro FT

DisciplinaFenômenos de Transporte I11.511 materiais110.436 seguidores
Pré-visualização50 páginas
```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.
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
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
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
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
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```