Respostas_Livro FT
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Respostas_Livro FT


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A plate with variable conductivity is subjected to specified 
temperatures on both sides. The rate of heat transfer through the plate 
is to be determined. 
T2
x
k(T) 
 L 
T1
Assumptions 1 Heat transfer is given to be steady and one-
dimensional. 2 Thermal conductivity varies quadratically. 3 There is 
no heat generation. 
Properties The thermal conductivity is given to be 
. )1()( 20 TkTk \u3b2+=
Analysis When the variation of thermal conductivity with temperature 
k(T) is known, the average value of the thermal conductivity in the 
temperature range between can be determined from 21 and TT
 
( ) ( )
( )\u23a5\u23a6\u23a4\u23a2\u23a3\u23a1 +++=
\u2212
\u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1 \u2212+\u2212
=\u2212
\u239f\u23a0
\u239e\u239c\u239d
\u239b +
=\u2212
+
=\u2212=
\u222b\u222b
2
121
2
20
12
3
1
3
2120
12
3
0
12
2
0
12
avg
3
1
33
)1()(
2
1
2
1
2
1
TTTTk
TT
TTTTk
TT
TTk
TT
dTTk
TT
dTTk
k
T
T
T
T
T
T
\u3b2
\u3b2\u3b2\u3b2
 
This relation is based on the requirement that the rate of heat transfer through a medium with constant 
average thermal conductivity kavg equals the rate of heat transfer through the same medium with variable 
conductivity k(T). Then the rate of heat conduction through the plate can be determined to be 
 ( )
L
TT
ATTTTk
L
TT
AkQ 212121
2
20
21
avg 3
1
\u2212
\u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1 +++=\u2212= \u3b2& 
Discussion We would obtain the same result if we substituted the given k(T) relation into the second part 
of Eq. 2-76, and performed the indicated integration. 
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. 
 2-53
2-103 A cylindrical shell with variable conductivity is subjected to specified temperatures on both sides. 
The variation of temperature and the rate of heat transfer through the shell are to be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies 
linearly. 3 There is no heat generation. 
Properties The thermal conductivity is given to be )1()( 0 TkTk \u3b2+= . 
r2
T2
r 
r1
T1
k(T)
Solution (a) The rate of heat transfer through the shell is 
expressed as 
 
)/ln(
2
12
21
avgcylinder rr
TT
LkQ
\u2212= \u3c0& 
where L is the length of the cylinder, r1 is the inner 
radius, and r2 is the outer radius, and 
 \u239f\u23a0
\u239e\u239c\u239d
\u239b ++==
2
1)( 120avgavg
TTkTkk \u3b2 
is the average thermal conductivity. 
(b) To determine the temperature distribution in the shell, we begin with the Fourier\u2019s law of heat 
conduction expressed as 
 
dr
dTATkQ )(\u2212=& 
where the rate of conduction heat transfer is constant and the heat conduction area A = 2\u3c0rL is variable. 
Separating the variables in the above equation and integrating from r = r
Q&
1 where to any r where 
, we get 
11 )( TrT =
TrT =)(
 \u222b\u222b \u2212= TTrr dTTkLrdrQ 11 )(2\u3c0& 
Substituting )1()( 0 TkTk \u3b2+= and performing the integrations gives 
 ]2/)()[(2ln 21
2
10
1
TTTTLk
r
rQ \u2212+\u2212\u2212= \u3b2\u3c0& 
Substituting the expression from part (a) and rearranging give &Q
 02)(
)/ln(
)/ln(22
1
2
121
12
1
0
avg2 =\u2212\u2212\u2212++ TTTT
rr
rr
k
k
TT \u3b2\u3b2\u3b2 
which is a quadratic equation in the unknown temperature T. Using the quadratic formula, the temperature 
distribution T(r) in the cylindrical shell is determined to be 
 1
2
121
12
1
0
avg
2
2)(
)/ln(
)/ln(211)( TTTT
rr
rr
k
k
rT \u3b2\u3b2\u3b2\u3b2 ++\u2212\u2212±\u2212= 
Discussion The proper sign of the square root term (+ or -) is determined from the requirement that the 
temperature at any point within the medium must remain between . 21 and TT
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. 
 2-54
2-104 A spherical shell with variable conductivity is subjected to specified temperatures on both sides. 
The variation of temperature and the rate of heat transfer through the shell are to be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies 
linearly. 3 There is no heat generation. 
Properties The thermal conductivity is given to be )1()( 0 TkTk \u3b2+= . 
r1 r2
T1k(T) 
r 
T2Solution (a) The rate of heat transfer through the shell is expressed as 
 
12
21
21avgsphere 4 rr
TT
rrkQ \u2212
\u2212= \u3c0& 
where r1 is the inner radius, r2 is the outer radius, and 
 \u239f\u23a0
\u239e\u239c\u239d
\u239b ++==
2
1)( 120avgavg
TTkTkk \u3b2 
is the average thermal conductivity. 
(b) To determine the temperature distribution in the shell, we begin with the Fourier\u2019s law of heat 
conduction expressed as 
 
dr
dTATkQ )(\u2212=& 
where the rate of conduction heat transfer is constant and the heat conduction area A = 4\u3c0rQ& 2 is variable. 
Separating the variables in the above equation and integrating from r = r1 where to any r where 
, we get 
11 )( TrT =
TrT =)(
 \u222b\u222b \u2212= TTrr dTTkrdrQ 11 )(42 \u3c0& 
Substituting )1()( 0 TkTk \u3b2+= and performing the integrations gives 
 ]2/)()[(411 21
2
10
1
TTTTk
rr
Q \u2212+\u2212\u2212=\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b \u2212 \u3b2\u3c0& 
Substituting the expression from part (a) and rearranging give Q&
 02)(
)(
)(22
1
2
121
12
12
0
avg2 =\u2212\u2212\u2212\u2212
\u2212++ TTTT
rrr
rrr
k
k
TT \u3b2\u3b2\u3b2 
which is a quadratic equation in the unknown temperature T. Using the quadratic formula, the temperature 
distribution T(r) in the cylindrical shell is determined to be 
 1
2
121
12
12
0
avg
2
2)(
)(
)(211)( TTTT
rrr
rrr
k
k
rT \u3b2\u3b2\u3b2\u3b2 ++\u2212\u2212
\u2212\u2212±\u2212= 
Discussion The proper sign of the square root term (+ or -) is determined from the requirement that the 
temperature at any point within the medium must remain between . 21 and TT
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. 
 2-55
2-105 A plate with variable conductivity is subjected to specified temperatures on both sides. The rate of 
heat transfer through the plate is to be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies 
linearly. 3 There is no heat generation. 
Properties The thermal conductivity is given to be )1()( 0 TkTk \u3b2+= . 
T2
k(T) 
T1 
 L 
Analysis The average thermal conductivity of the medium in 
this case is simply the conductivity value at the average 
temperature since the thermal conductivity varies linearly with 
temperature, and is determined to be 
 
K W/m24.34
2
K 350)+(500
)K 10(8.7+1K) W/m25(
2
1)(
1-4-
12
0avgave
\u22c5=
\u239f\u23a0
\u239e\u239c\u239d
\u239b ×\u22c5=
\u239f\u239f\u23a0
\u239e\u239c\u239c\u239d
\u239b ++== TTkTkk \u3b2
 
Then the rate of heat conduction through the plate becomes 
 kW 30.8==\u2212×\u22c5=\u2212= W30,820
m 15.0
0)K35(500m) 0.6 m K)(1.5 W/m24.34(21avg L
TT
AkQ& 
Discussion We would obtain the same result if we substituted the given k(T) relation into the second part 
of Eq, 2-76, and performed the indicated integration. 
 
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. 
 2-56
2-106 EES Prob. 2-105 is reconsidered. The rate of heat conduction through the plate as a function of the 
temperature of the hot side of the plate is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
A=1.5*0.6 [m^2] 
L=0.15 [m] 
T_1=500 [K] 
T_2=350 [K] 
k_0=25 [W/m-K] 
beta=8.7E-4 [1/K] 
 
"ANALYSIS" 
k=k_0*(1+beta*T) 
T=1/2*(T_1+T_2) 
Q_dot=k*A*(T_1-T_2)/L 
 
 
T1 [W] Q [W] 
400 9947 
425 15043 
450 20220 
475 25479 
500 30819 
525 36241 
550 41745 
575 47330 
600 52997 
625 58745 
650 64575 
675 70486 
700 76479 
 
 
400 450 500 550 600 650 700
0
10000
20000
30000
40000
50000
60000
70000
80000
T1 [K]
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