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# Respostas_Livro FT

DisciplinaFenômenos de Transporte I12.652 materiais111.757 seguidores
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```are ,
the single equivalent heat transfer coefficient is
radconveqv hhh += when the medium and the surrounding
surfaces are at the same temperature. Then the equivalent thermal resistance will be . )/(1 AhR eqveqv =

3-9C The thermal resistance network associated with a five-layer composite wall involves five single-layer
resistances connected in series.

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.
3-2
3-10C Once the rate of heat transfer Q is known, the temperature drop across any layer can be determined
by multiplying heat transfer rate by the thermal resistance across that layer,
&
layerlayer RQT &=\u394

3-11C The temperature of each surface in this case can be determined from

)(/)(
)(/)(
22222222
11111111
\u221e\u2212\u221e\u221e\u2212\u221e
\u2212\u221e\u221e\u2212\u221e\u221e
+=\u23af\u2192\u23af\u2212=
\u2212=\u23af\u2192\u23af\u2212=
ssss
ssss
RQTTRTTQ
RQTTRTTQ
&&
&&
where is the thermal resistance between the environment iR \u2212\u221e \u221e and surface i.

3-12C Yes, it is.

3-13C The window glass which consists of two 4 mm thick glass sheets pressed tightly against each other
will probably have thermal contact resistance which serves as an additional thermal resistance to heat
transfer through window, and thus the heat transfer rate will be smaller relative to the one which consists of
a single 8 mm thick glass sheet.

3-14C Convection heat transfer through the wall is expressed as . In steady heat transfer,
heat transfer rate to the wall and from the wall are equal. Therefore at the outer surface which has
convection heat transfer coefficient three times that of the inner surface will experience three times smaller
temperature drop compared to the inner surface. Therefore, at the outer surface, the temperature will be
closer to the surrounding air temperature.
)( \u221e\u2212= TThAQ ss&

3-15C The new design introduces the thermal resistance of the copper layer in addition to the thermal
resistance of the aluminum which has the same value for both designs. Therefore, the new design will be a
poorer conductor of heat.

3-16C The blanket will introduce additional resistance to heat transfer and slow down the heat gain of the
drink wrapped in a blanket. Therefore, the drink left on a table will warm up faster.

3-17 The two surfaces of a wall are maintained at specified temperatures. The rate of heat loss through the
wall is to be determined.
Assumptions 1 Heat transfer through the wall is steady since the surface
temperatures remain constant 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 is constant.
2°C14°C
L= 0.3 m
Q&
Wall
Properties The thermal conductivity is given to be k = 0.8 W/m\u22c5°C.
Analysis The surface area of the wall and the rate of heat loss
through the wall are
2m 18m) 6(m) 3( =×=A
W576=°\u2212°\u22c5=\u2212=
m 3.0
C)214(
)m C)(18 W/m8.0( 221
L
TT
kAQ&
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.
3-3
3-18 A double-pane window is considered. The rate of heat loss through the window and the temperature
difference across the largest thermal resistance are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer coefficients are constant.
Properties The thermal conductivities of glass and air are given to be 0.78 W/m\u22c5K and 0.025 W/m\u22c5K,
respectively.
Analysis (a) The rate of heat transfer through the window is determined to be
[ ]
[ ] W210=++++
°×=
°\u22c5+°\u22c5+°\u22c5+°\u22c5+°\u22c5
°×=
++++
\u394=
05.0000513.02.0000513.0025.0
C(-20)-20)m 5.11(
C W/m20
1
C W/m78.0
m 004.0
C W/m025.0
m 005.0
C W/m78.0
m 004.0
C W/m40
1
C(-20)-20)m 5.11(
11
2
22
2
og
g
a
a
g
g
i hk
L
k
L
k
L
h
TAQ&

(b) Noting that the largest resistance is through the air gap, the temperature difference across the air gap is
determined from
C28°=×°\u22c5===\u394 )m 5.11(C) W/m025.0(
m 005.0 W)210(
2Ak
L
QRQT
a
a
aa
&&

3-19 The two surfaces of a window are maintained at specified temperatures. The rate of heat loss through
the window and the inner surface temperature are to be determined.
Assumptions 1 Heat transfer through the window is steady since the surface temperatures remain constant
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 is constant. 4 Heat transfer
Properties The thermal conductivity of the glass is given to be k = 0.78 W/m\u22c5°C.
Analysis The area of the window and the individual resistances are
2m 4.2m) 2(m) 2.1( =×=A
C/W06155.001667.000321.004167.0
C/W 01667.0
)m 4.2(C). W/m25(
11
C/W 00321.0
)m 4.2(C) W/m.78.0(
m 006.0
C/W 04167.0
)m 4.2(C). W/m10(
11
2,1,
22
2
2,o
2
1
glass
22
1
1,i
°=++=
++=
°=°===
°=°==
°=°===
convglassconvtotal
conv
conv
RRRR
Ah
RR
Ak
LR
Ah
RR

T1
Q&
Glass
Ri Rglass Ro
T\u221e1 T\u221e2
L
The steady rate of heat transfer through
window glass is then
W471=°
°\u2212\u2212=\u2212= \u221e\u221e
C/W 06155.0
C)]5(24[21
totalR
TT
Q&
The inner surface temperature of the window glass can be determined from
C4.4°=°\u2212°=\u2212=\u23af\u2192\u23af\u2212= \u221e\u221e C/W) 7 W)(0.0416471(C241,11
1,
11
conv
conv
RQTT
R
TT
Q &&
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.
3-4
3-20 A double-pane window consists of two layers of glass separated by a stagnant air space. For specified
indoors and outdoors temperatures, the rate of heat loss through the window and the inner surface
temperature of the window are to be determined.
Assumptions 1 Heat transfer through the window is steady
since the indoor and outdoor temperatures remain constant 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 conductivities of
the glass and air are constant. 4 Heat transfer by radiation is
negligible.
Properties The thermal conductivity of the glass and air are
given to be kglass = 0.78 W/m\u22c5°C and kair = 0.026 W/m\u22c5°C.
Air
R1 R2 R3 Ro Ri
T\u221e1 T\u221e2
Analysis The area of the window and the
individual resistances are
2m 4.2m) 2(m) 2.1( =×=A
C/W 2539.0
0167.01923.0)0016.0(20417.02
C/W 0167.0
)m 4.2(C). W/m25(
11
C/W 1923.0
)m 4.2(C) W/m.026.0(
m 012.0
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
2,211,
o
2o2
2
2,o
2
2
2
2
2
1
1
glass31
22
1
1,i
°=
+++=+++=
====
°=°===
°=°====
°=°===
convconvtotal
conv
air
conv
RRRRR
Ah
RR
Ak
LRR
Ak
LRRR
Ah
RR

The steady rate of heat transfer through window
glass then becomes
W114=°
°\u2212\u2212=\u2212= \u221e\u221e
C/W2539.0
C)]5(24[21
totalR
TT
Q&
The inner surface temperature of the window glass can be determined from
C19.2°°\u2212=\u2212=\u23af\u2192\u23af\u2212= \u221e\u221e =C/W) W)(0.0417114(C24o1,11
1,
11
conv
conv
RQTT
R
TT
Q &&
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.
3-5
3-21 A double-pane window consists of two layers of glass separated by an evacuated space. For specified
indoors and outdoors temperatures, the rate of heat loss through the window and the inner surface
temperature of the window are to be determined.
Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures
remain constant```