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Transcript of solucionario ccc 1
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8/3/2019 solucionario ccc 1
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1. INTRODUCTION
1.1 A composite wall consist of alternative layers of fir ( 5 cm thick ) , aluminum ( 1
cm thick ), lead ( 1 cm thick ), and corkboard ( 6 cm thick ). The temperature is 60
C of the outside of the fir and 10 C on the outside of the corkboard. Plot thetemperature gradient through the wall. Does the temperature profile suggest any
simplifying assumptions that might be made in subsequent analysis of the wall?
Solution:
Thermal Conductivities:
kfir= 0.12 W/m.K (Table A.2, Appendix A)
kalu = 237 W/m.K (Table A.1, Appendix A)
kld= 35 W/m.K (Table A.1, Appendix A)
kcb = 0.04 W/m.K (Table A.2, Appendix A).
Question No. 1: Plot the temperature gradient through the wall.
Answer:
Question No. 2: Does the temperature profile suggest any simplifying assumptions
that might be made in subsequent analysis of the wall?
Answer:
Yes, since the thermal conductivity of aluminum and lead are very high than fir and
corkboard, they are considered isothermal. Therefore consider only fir and corkboard.
fir+ Tcb = 60 C 10 C = 50 K
cbfir L
Tk
L
Tkq
=
=
Lfir= 5 cm = 0.05 m
Lcb = 6 cm = 0.06 m
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1. INTRODUCTION
Then,
( )( )
( )( )( )m
TKmWm
TKmWq cbfir
06.0/04.0
05.0/12.0 ==
Tcb = 3.6Tfir
Then,
fir+ 3.6fir= 50 K
fir= 10.87 K
( )
=
=
m
KKmW
L
Tkq
fir 05.0
87.10./12.0 = 26.09 W/m
2
Considering all walls:
Tfir+ Talu + Tld+ Tcb = 60 C 10 C = 50 K
cbldalufir L
Tk
L
Tk
L
Tk
L
Tkq
=
=
=
=
Lfir= 5 cm = 0.05 m
Lcb = 6 cm = 0.06 m
Lalu = 1 cm = 0.01 m
Lld= 1 cm = 0.01 m
=
alu
fir
firalu
L
k
L
k
TT
=
ld
fir
firld
L
k
L
k
TT
=
cb
fir
fircb
L
k
L
k
TT
Then
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1. INTRODUCTION
+
+
+
cbldalu
fir
fir
L
k
L
k
L
kL
k
T
111
1 = 50 K
+
+
+
06.0
04.0
1
01.0
35
1
01.0
237
1
05.0
12.01firT = 50
Tfir= 10.87 K
( )
=
=
m
KKmW
L
Tkq
fir 05.0
87.10./12.0 = 26.09 W/m
2
There it is equal to simplified solution.
1.2 Verify Equation (1.15).
Solution:
Equation (1.15)
TT
dt
dTbody
body
For verification only
Equation (1.3)
dt
dTmc
dt
dUQ ==
Equation (1.16)
TTQ body
Then
TT
dt
dTmc body
TT
dtdT
body
Then
TT
dt
dTbody
bodywhere mc is constant.
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1. INTRODUCTION
1.3 q = 5000 W/m2
in a 1 cm slab and T= 140 C on the cold side. Tabulate the
temperature drop through the slab if it is made of
Silver
Aluminum
Mild steel (0.5 % carbon)
Ice
Spruce
Insulation (85 % magnesia)
Silica aerogel
Indicate which situations would be unreasonable and why.
Solution:
L = 1 cm = 0.01 m
(a) Silver Slab
SiL
Tkq
= = 5000 W/m
2
Thermal conductivity of silver at 140 C, 99.99+ % Pure, Table A.1, Appendix A
ksi = 420 W/m.K
( )
=
m
TKmWq Si
01.0
/420 = 5000 W/m2
TSi = 0.12 K
(b) Alumium Slab
aluL
Tkq
= = 5000 W/m
2
Thermal conductivity of aluminum at 140 C, 99.99+ % Pure, Table A.1, App. A
Kalu = 237.6 W/m.K
( )
=
m
TKmWq alu
01.0/6.237 = 5000 W/m
2
Talu = 0.21 K
(c) Mild Steel Slab
msL
Tkq
= = 5000 W/m
2
Thermal conductivity of mild steel at 140 C, Table A.1, Appendix AKms = 50.4 W/m.K
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1. INTRODUCTION
( )
=
m
TKmWq ms
01.0/4.50 = 5000 W/m
2
Tms = 0.992 K
(d) Ice Slab
iceL
Tkq
= = 5000 W/m
2
Thermal conductivity of ice at 140 C, Table A.1, Appendix A
ice at 0 C, kice = 2.215 W/m.K
Note: there is no ice at 140 C, but continue calculation at 0 C.
( )
=
m
TKmWq ice
01.0
/215.2 = 5000 W/m2
Tice = 22.57 K
(e) Spruce Slab
SiL
Tkq
= = 5000 W/m
2
Thermal conductivity of spruce at 140 C, Table A.1, Appendix A
Ksp = 0.11 W/m.K @ 20 C (available)
( )
=
m
TKmWq
Sp
01.0/11.0 = 5000 W/m
2
TSp = 454.55 K
(f) Insulation (85 % Magnesia)
SiL
Tkq
= = 5000 W/m
2
Thermal conductivity of insulation at 140 C, Table A.1, Appendix A
Kin = 0.074 W/m.K @ 150 C (available)
( )
=
m
TKmWq in
01.0/074.0 = 5000 W/m
2
TSi = 675.8 K
(g) Silica Aerogel Slab
SiL
Tkq
= = 5000 W/m
2
Thermal conductivity of silica aerogel at 140 C, Table A.1, Appendix A
ksa = 0.022 W/m.K @ 120 C
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1. INTRODUCTION
( )
=
m
TKmWq sa
01.0/022.0 = 5000 W/m
2
Tsa = 2,273 K
Tabulation:
Slab Temperature Drop, K
Silver 0.12
Aluminum 0.21
Mild Steel (0.5 % Carbon) 0.992
Ice 22.57
Spruce 454.55
Insulation (85 % Magnesia) 675.8Silica Aerogel 2273
The situation which is unreasonable here is the use of ice as slab at 140 C, since ice
will melt at temperature of 0 C and above. Thats it.
1.4 Explain in words why the heat diffusion equation, eq. no. (1.13), shows that in
transient conduction the temperature depends on the thermal diffusitivity, , butwe can solve steady conduction problems using just k(as in Example 1.1).
Solution:
Equation (1.13)
xdt
dTcAx
dt
TTdcA
dt
dUQ
ref
net =
==
Answer: The application of heat diffusion equation eq. no. (1.13) depends on the
thermal diffusivity as the value oft
T
is not equal to zero as it I s under unsteady
state conduction. While in steady conduction depends only on kbecause the value of
t
T
= 0 for steady state conduction giving 2
2
x
T
= 0 , so dx
dT
kq = .
1.5 A 1-m rod of pure copper 1 cm2 in cross section connects a 200 C thermal
reservoir with a 0 C thermal reservoir. The system has already reached steady
state. What are the rates of change of entropy of (a) the first reservoir, (b) the
second reservoir, (c) the rod, and (d) the whole universe, as a result of the
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1. INTRODUCTION
process? Explain whether or not your answer satisfies the Second Law of
Thermodynamics.
Solution:
Equation (1.9)
L
Tkq
=
Thermal conductivity of copper at 100 C, Table A.1, Appendix A
k= 391 W/m.K
L = 1 m
T= 200 C 0 C = 200 K
( )
=
m
KKmWq
1
200/391 = 78,200 W/m2.K
Q = qA
A = 1 cm2
= 1 x 10-4
m2
Q = (78,200 W/m2.K)(1 x 10
-4m
2) = 7.82 W
(a)( )K
W
T
QS rev
273200
82.7
1
1+
=
=
& = - 0.01654 W/K
(b)( )K
W
T
QS rev
2730
82.7
2
2+
+==
& = + 0.02864 W/K
(c) =rS& = 0.0 W/K (see Eq. 1.5, steady state)
(d) =+= 21 SSSUn&&& = - 0.01654 W/K + 0.02864 W/K = + 0.0121 W/K
Since 0UnS& , therefore it satisfied Second Law of Thermodynamics.
1.6 Two thermal energy reservoirs at temperatures of 27 C and 43 C, respectively,
are separated by a slab of material 10 cm thick and 930 cm2
in cross-sectional
area. The slab has a thermal conductivity of 0.14 W/m.K. The system is operating
at steady-state conditions. What are the rates of change of entropy of (a) the
higher temperature reservoir, (b) the lower temperature reservoir, (c) the slab, and
(d) the whole universe as a result of this process? (e) Does your answer satisfy the
Second Law of Thermodynamics?
Solution:
Equation (1.9)
L
Tkq
=
Thermal conductivity , k= 0.14 W/m.K
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1. INTRODUCTION
A = 930 cm2
= 0.093 m2
L = 10 cm = 0.10 mT= 27 C (- 43 C) = 70 KT1 = 27 + 273 = 300 K
T2 = -43 + 273 = 230 K
( )
=
m
KKmWq
10.0
70./14.0 = 98 W/m
2
Q = qA = (98 W/m2)(0.093 m
2) = 9.114 W
(a)( )K
W
T
QS rev
300
114.9
1
1
=
=& = - 0.03038 W/K
(b)( )K
W
T
QS rev
230
114.9
2
2
+==& = + 0.03963 W/K
(c) =rS& = 0.0 W/K (see Eq. 1.5, steady state)
(d) =+= 21 SSSUn&&& = - 0.03038 W/K + 0.03963 W/K = + 0.00925 W/K
Since 0UnS& , therefore it satisfied Second Law of Thermodynamics.
1.7 (a) If the thermal energy reservoirs in Problem 1.6 are suddenly replaced with
adiabatic walls, determine the final equilibrium temperature of the slab. (b) What
is the entropy change for the slab for this process? (c) Does your answer satisfythe Second Law of Thermodynamics in this instance? Explain. The density of the
slab is 26 lb/ft3
and the specific heat 0.65 Btu/lb-F.
Solution:
( )
=
3
33
/1
/018.16/26
ftlb
mkgftlb = 416.468 kg/m
3
( )
=
FlbBtu
KkgJFlbBtuc
./1
./8.4186./65.0 = 2721.42 J/kg.K
k= 0.14 W/m.KT= 27 C (-43 C) = 70 CT1 = 27 C + 273 = 300 K
T2 = - 43 C + 273 = 230 K
A = 0.093 m2
L = 0.10 m
(a) =2
1
T
TT
dQ
T
Q
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1. INTRODUCTION
=2
1
T
TTcVdT
TQ
( )
=
1
212 lnT
TcV
T
TTcV
( )
=
1
212 lnT
T
T
TT
( ) ( )
=
=
300
230ln
300230
ln1
2
12
T
T
TTT = 263.45 K
(b)( ) ( )
T
TTcAL
T
TTcV
T
QS 1212
=
==
( )( )( )( )( )
45.263
30023010.0093.042.2721468.416 =S = - 2801 J/K
(c) This will not satisfy the Second Law of Thermodynamic since this is not a rate of
entropy of production of the universe.
1.8 A copper sphere 2.5 cm in diameter has a uniform temperature of 40 C. The
sphere is suspended in a slow-moving air stream at 0 C. The air stream produces a
convection heat transfer coefficient of 15 W/m2.K. Radiation can be neglected.
Since copper is highly conductive, temperature gradients in the sphere willsmooth out rapidly, and its temperature can be taken as uniform throughout the
cooling process (i.e., Bi
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1. INTRODUCTION
( )( )
=
TTcV
Ah
dt
TTd
( ) C
Ah
cV
tTT +
=
ln
at T(t= 0) Ti,
( )
= TTC ln
( ) ( )
+
= TT
Ah
cV
tTT
ilnln
xi T
t
Ah
cV
t
TT
TT=
=
ln
=
Ah
cVTx
xT
t
i
eTT
TT
=
T = 0 C + 273 = 273 K
iT = 40 C + 273 = 313 K
=
Ah
cVTx
3
3
4rV =
r= (1/2)(2.5 cm) = 1.25 cm = 0.0125 m24 rA =
( ) hcr
rh
rc
Ah
cVTx
34
3
4
2
3
=
==
h = 15 W/m2.K
Properties of copper, Table A.1, App. A
= 8954 kg/m3
cp = 384 J/kg.K
= 11.57 x 10-5 m2/s
2
( )( )( )( )KmW
mKkgJmkgTx
./153
0125.0./3843/8954= = 955 sec
Then:
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1. INTRODUCTION
( ) xTt
ieTTTT
=
( )
+= TeTTT x
T
t
i
( ) KeTt
273273313 955 +=
KeT
t
27340 955 +=
95540
t
eT
= C
where tin seconds
Tabulation:
Time, t, seconds Temperature, T, C
0 40
10 39.6
20 39.2
40 38.4
60 37.6
80 36.9
100 36.2
200 32.7
300 29.6
400 26.8
600 22
800 18
1000 14.75000 0.3
10000 0.0
100000 0.0
1000000 0.0
0.0
Plot:
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1. INTRODUCTION
1.9 Determine the total heat transfer in Problem 1.8 as the sphere cools from 40 C to
0 C. Plot the net entropy increase resulting from the cooling process above, S vs
T(K).
Solution:
T = 0 C + 273 = 273 K
24 rA = ,3
3
4rV =
r= 0.0125 m
= 8954 kg/m3
cp = 384 J/kg.K
= 11.57 x 10-5 m2/s
2
T= 40 C 0 C = 40 K
Total Heat Transfer:
Q = cVT = (8954 kg/m3)(384 J/kg.K)(4/3)()(0.0125 m)
3(40 K)
Q = 1125 J - - - - Answer.
Plotting the net-entropy increase:
Equation (1.24)
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1. INTRODUCTION
b
T
Tb
dTTT
cVS
b
b
=
0
11
( )( ) ( )b
T
Tb
dTTT
S
b
b
=
0
110125.0
3
43848954
3
=
0
0lnln13.28
b
b
b
bT
T
TT
T
TS
=
0
0ln13.28
b
bbb
T
T
T
TTS
Tb0 = 40 C = 313 K
=313
ln273
31313.28
bbTT
S
Tb, C Tb, K S
40 313 0
35 308 0.0622
30 303 0.117
25 298 0.1642
20 293 0.2034
15 288 0.2344
10 283 0.2569
5 278 0.2707
0 273 0.2754
Plot:
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1. INTRODUCTION
T2 = 6 C
dxdTkAQ
=
x
DD
L
DD =
121
D1 = 15 cm = 0.15 m
D2 = 7.5 cm = 0.075 m
L = 30 cm = 0.30 m
x
Dm
m
mm =
15.0
30.0
075.015.0
D = 0.15 m 0.25x
2
4DA =
dx
dTDkQ
=
2
4
( )dx
dTxmkQ
225.015.0
4
=
( ) dTkdxxQ
=
425.015.0
2
( ) dTkdxxQ
m
=
425.015.0
3.0
0
2
Thermal Conductivity of Portland Cement, Table A.2, Appendix A.
k= 0.70 W/m.K
( ) ( )[ ] ( ) ( )6404
70.025.015.025.0
11
3.0
0
1
=
xQ
( ) ( )( ) ( )[ ] ( ) ( )344
70.015.03.025.015.0411
=
Q
( ) ( ) ( )344
70.015.0
1
075.0
14
=
Q
Q = -0.70 W Ans.