Respostas_Livro FT
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Respostas_Livro FT


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aT2 + bT = C1x2 + C2 (e) None of them 
 
Answer (c) aT2 + bT = C1x + C2
 
 
2-161 Harvested grains, like wheat, undergo a volumetric exothermic reaction while they are being stored. 
This heat generation causes these grains to spoil or even start fires if not controlled properly. Wheat (k = 
0.5 W/m\u22c5K) is stored on the ground (effectively an adiabatic surface) in 5-m thick layers. Air at 20°C 
contacts the upper surface of this layer of wheat with h = 3 W/m2\u22c5K. The temperature distribution inside 
this layer is given by 
 
2
0
1 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212=\u2212
\u2212
L
x
TT
TT
s
s 
where Ts is the upper surface temperature, T0 is the lower surface temperature, x is measured upwards from 
the ground, and L is the thickness of the layer. When the temperature of the upper surface is 24oC, what is 
the temperature of the wheat next to the ground? 
(a) 39oC (b) 51oC (c) 72oC (d) 84oC (e) 91°C 
 
Answer (d) 84oC 
 
k=0.5 [W/m-K] 
h=3 [W/m2-K] 
L=5[m] 
Ts=24 [C] 
Ta=20 [C] 
To=(h*L/(2*k))*(Ts-Ta)+Ts 
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. 
 2-85
2-162 The conduction equation boundary condition for an adiabatic surface with direction n being normal 
to the surface is 
(a) T = 0 (b) dT/dn = 0 (c) d2T/dn2 = 0 (d) d3T/dn3 = 0 (e) -kdT/dn = 1 
 
Answer (b) dT/dn = 0 
 
 
 
2-163 Which one of the followings is the correct expression for one-dimensional, steady-state, constant 
thermal conductivity heat conduction equation for a cylinder with heat generation? 
(a) 
t
Tce
r
Trk
rr \u2202
\u2202=+\u239f\u23a0
\u239e\u239c\u239d
\u239b
\u2202
\u2202
\u2202
\u2202 \u3c1gen1 & (b) t
T
k
e
r
Tr
rr \u2202
\u2202=+\u239f\u23a0
\u239e\u239c\u239d
\u239b
\u2202
\u2202
\u2202
\u2202
\u3b1
11 gen& 
(c) 
t
T
r
Tr
rr \u2202
\u2202
\u3b1=\u239f\u23a0
\u239e\u239c\u239d
\u239b
\u2202
\u2202
\u2202
\u2202 11 (d) 01 gen =+\u239f\u23a0
\u239e\u239c\u239d
\u239b
k
e
dr
dTr
dr
d
r
&
 (e) 0=\u239f\u23a0
\u239e\u239c\u239d
\u239b
dr
dTr
dr
d 
 
Answer (d) 01 gen =+\u239f\u23a0
\u239e\u239c\u239d
\u239b
k
e
dr
dTr
dr
d
r
&
 
 
 
 
 
2-164 .... 2-167 Design and Essay Problems 
 
 
KJ 
 
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. 
 2-33
2-70 A compressed air pipe is subjected to uniform heat flux on the outer surface and convection on the 
inner surface. The mathematical formulation, the variation of temperature in the pipe, and the surface 
temperatures are to be determined for steady one-dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its 
thickness, and there is thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There 
is no heat generation in the pipe. 
Properties The thermal conductivity is given to be k = 14 W/m\u22c5°C. 
Analysis (a) Noting that the 85% of the 300 W generated by the strip heater is transferred to the pipe, the 
heat flux through the outer surface is determined to be 
 2
22
 W/m1.169
m) m)(6 (0.042
 W30085.0
2
=×=== \u3c0\u3c0 Lr
Q
A
Q
q sss
&&
& 
Noting that heat transfer is one-dimensional in the radial r direction and heat flux is in the negative r 
direction, the mathematical formulation of this problem can be expressed as 
 0=\u239f\u23a0
\u239e\u239c\u239d
\u239b
dr
dTr
dr
d 
and )]([)( 11 rTThdr
rdT
k \u2212=\u2212 \u221e 
 sqdr
rdT
k &=)( 2 
 (b) Integrating the differential equation once with respect to r gives 
Heater 
L=6 m 
Air, -10°C
r
r1
r2
 1Cdr
dTr = 
Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, 
 
r
C
dr
dT 1= 
 21 ln)( CrCrT += 
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
r = r2: k
rq
Cq
r
C
k ss
2
1
2
1 
&& =\u2192= 
r = r1: k
rq
hr
krTC
hr
krTCCrCTh
r
C
k s 2
1
11
1
12211
1
1 ln=ln )]ln([
&
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b \u2212\u2212\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b \u2212\u2212=\u2192+\u2212=\u2212 \u221e\u221e\u221e 
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 
 
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b ++\u2212=°\u22c5\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
°\u22c5
°\u22c5++°\u2212=
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b ++=\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b +\u2212+=\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b \u2212\u2212+= \u221e\u221e\u221e
61.12ln483.010
C W/m14
m) 04.0)( W/m1.169(
m) C)(0.037 W/m30(
C W/m14lnC10
lnlnlnlnln)(
1
2
2
1
2
11
1
1
11
1
11
r
r
r
r
k
rq
hr
k
r
rTC
hr
krrTC
hr
krTrCrT s
&
 
(c) The inner and outer surface temperatures are determined by direct substitution to be 
Inner surface (r = r1): ( ) C3.91°\u2212=++\u2212=\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b ++\u2212= 61.120483.01061.12ln483.010)(
1
1
1 r
r
rT 
Outer surface (r = r2): C3.87°\u2212=\u239f\u23a0
\u239e\u239c\u239d
\u239b ++\u2212=\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b ++\u2212= 61.12
037.0
04.0ln483.01061.12ln483.010)(
1
2
1 r
r
rT 
Note that the pipe is essentially isothermal at a temperature of about -3.9°C. 
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. 
 2-34
2-71 EES Prob. 2-70 is reconsidered. The temperature as a function of the radius is to be plotted. 
Analysis The problem is solved using EES, and the solution is given below. 
 
"GIVEN" 
L=6 [m] 
r_1=0.037 [m] 
r_2=0.04 [m] 
k=14 [W/m-C] 
Q_dot=300 [W] 
T_infinity=-10 [C] 
h=30 [W/m^2-C] 
f_loss=0.15 
 
"ANALYSIS" 
q_dot_s=((1-f_loss)*Q_dot)/A 
A=2*pi*r_2*L 
T=T_infinity+(ln(r/r_1)+k/(h*r_1))*(q_dot_s*r_2)/k "Variation of temperature" 
"r is the parameter to be varied" 
 
 
r [m] T [C] 
0.037 3.906 
0.03733 3.902 
0.03767 3.898 
0.038 3.893 
0.03833 3.889 
0.03867 3.885 
0.039 3.881 
0.03933 3.877 
0.03967 3.873 
0.04 3.869 
 
 
0.037 0.0375 0.038 0.0385 0.039 0.0395 0.04
-3.906
-3.897
-3.888
-3.879
-3.87
r [m]
T 
 [C
]
 
 
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. 
 2-35
2-72 A spherical container is subjected to uniform heat flux on the outer surface and specified temperature 
on the inner surface. The mathematical formulation, the variation of temperature in the pipe, and the outer 
surface temperature, and the maximum rate of hot water supply are to be determined for steady one-
dimensional heat transfer. 
Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and 
there is thermal symmetry about the mid point. 2 Thermal conductivity is constant. 3 There is no heat 
generation in the container. 
Properties The thermal conductivity is given to be k = 1.5 W/m\u22c5°C. The specific heat of water at the 
average temperature of (100+20)/2 = 60°C is 4.185 kJ/kg\u22c5°C (Table A-9). 
Analysis (a) Noting that the 90% of the 500 W generated by the strip heater is transferred to the container, 
the heat flux through the outer surface is determined to be 
 2
22
22
 W/m0.213
m) (0.414
 W50090.0
4
=×=== \u3c0\u3c0r
Q
A
Q
q sss
&&
& 
Noting that heat transfer is one-dimensional in the radial r direction and heat flux is in the negative r 
direction, the mathematical formulation of this problem can be expressed as 
 02 =\u239f\u23a0
\u239e\u239c\u239d
\u239b
dr
dTr
dr
d 
and C100)( 11 °== TrT
 sqdr
rdT
k &=)( 2 
(b) Integrating the differential equation once with respect to r gives 
 1
2 C
dr
dTr = 
Dividing both sides of the equation above by r2 and then integrating, 
r1 r2
T1k
Heater
r
Insulation
 
2
1
r
C
dr
dT = 
 2
1)( C
r
C
rT +\u2212= 
where C1 and C2 are arbitrary constants. Applying the boundary conditions give 
r = r2: k
rq
Cq
r
C
k ss
2
2
12
2
1 
&& =\u2192= 
r = r1: 
1
2
2
1
1
1
122
1
1
11 )( kr
rq
T
r
C
TCC
r
C
TrT s
&+=+=\u2192+\u2212== 
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 
\u239f\u23a0
\u239e\u239c\u239d
\u239b \u2212+=°\u22c5\u239f\u23a0
\u239e\u239c\u239d
\u239b \u2212+°=
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b \u2212+=\u239f\u239f\u23a0