ch03
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ch03


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K 20 0 C
\u3c1
\u221e
×
\u2212
= =
\u2212
\u22c5 \u2212
$
mt 11,690s 3.25 h= = <
COMMENTS: The tube radius r1 exceeds the critical radius rcr = k/h = 0.4 W/m\u22c5K/100 W/m2\u22c5K = 0.004
m, in which case any frost formation will reduce the performance of the coil.
PROBLEM 3.46
KNOWN: Conditions associated with a composite wall and a thin electric heater.
FIND: (a) Equivalent thermal circuit, (b) Expression for heater temperature, (c) Ratio of outer and inner
heat flows and conditions for which ratio is minimized.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Constant properties, (3) Isothermal
heater, (4) Negligible contact resistance(s).
ANALYSIS: (a) On the basis of a unit axial length, the circuit, thermal resistances, and heat rates are as
shown in the schematic.
(b) Performing an energy balance for the heater, in outE E=\ufffd \ufffd , it follows that
( )
( ) ( ) ( ) ( )
h ,i h ,o
h 2 i o
1 12 1 3 2
i 1 o 3
B A
T T T T
q 2 r q q
ln r r ln r r
h 2 r h 2 r
2 k 2 k
\u3c0
\u3c0 \u3c0
\u3c0 \u3c0
\u221e \u221e
\u2212 \u2212
\u2212 \u2212
\u2032\u2032 \u2032 \u2032= + = +
+ +
<
(c) From the circuit,
( )
( )
( ) ( )
( ) ( )
1 2 1
i 1h ,oo B
1 3 2i h ,i
o 3
A
ln r r
h 2 rT Tq 2 k
ln r rq T T h 2 r
2 k
\u3c0
\u3c0
\u3c0
\u3c0
\u2212
\u221e
\u2212\u221e
+
\u2212\u2032
= ×
\u2032
\u2212
+
<
To reduce o iq q\u2032 \u2032 , one could increase kB, hi, and r3/r2, while reducing kA, ho and r2/r1.
COMMENTS: Contact resistances between the heater and materials A and B could be important.
PROBLEM 3.47
KNOWN: Electric current flow, resistance, diameter and environmental conditions
associated with a cable.
FIND: (a) Surface temperature of bare cable, (b) Cable surface and insulation temperatures
for a thin coating of insulation, (c) Insulation thickness which provides the lowest value of the
maximum insulation temperature. Corresponding value of this temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r, (3)
Constant properties.
ANALYSIS: (a) The rate at which heat is transferred to the surroundings is fixed by the rate
of heat generation in the cable. Performing an energy balance for a control surface about the
cable, it follows that gE q=\ufffd or, for the bare cable, ( )( )2 e i sI R L=h D L T T .\u3c0 \u221e\u2032 \u2212 With
( ) ( )22 4eq =I R 700A 6 10 / m 294 W/m,\u2212\u2032 \u2032 = × \u2126 = it follows that
( ) ( )s 2i
q 294 W/mT T 30 C+
h D 25 W/m K 0.005m\u3c0 \u3c0
\u221e
\u2032
= + =
\u22c5
$
sT 778.7 C.=
$ <
(b) With a thin coating of insulation, there exist contact and convection resistances to heat
transfer from the cable. The heat transfer rate is determined by heating within the cable,
however, and therefore remains the same.
( )
s s
t,c
t,c
i i i
i s
t,c
T T T Tq= 1 R 1R
h D L
 D L h D L
 D T T
q =
R 1/ h
\u3c0 \u3c0 \u3c0
\u3c0
\u221e \u221e
\u221e
\u2212 \u2212
=
\u2032\u2032
+ +
\u2212
\u2032
\u2032\u2032 +
and solving for the surface temperature, find
( )
2 2
s t,c
i
q 1 294 W/m m K m KT R T 0.02 0.04 30 C
 D h 0.005m W W\u3c0 \u3c0\u221e
\uf8ee \uf8f9\u2032 \u22c5 \u22c5\uf8ee \uf8f9
\u2032\u2032= + + = + +\uf8ef \uf8fa\uf8ef \uf8fa\uf8f0 \uf8fb \uf8ef \uf8fa\uf8f0 \uf8fb
$
sT 1153 C.= $ <
Continued \u2026..
PROBLEM 3.47 (Cont.)
The insulation temperature is then obtained from
s i
t,c
T Tq=
R
\u2212
or
( )
2
t,c
i s t,c
i
W m K294 0.02R m WT T qR 1153 C q 1153 C
 D L 0.005m\u3c0 \u3c0
\u22c5
×
\u2032\u2032
= \u2212 = \u2212 = \u2212
$ $
iT 778.7 C.=
$ <
(c) The maximum insulation temperature could be reduced by reducing the resistance to heat transfer
from the outer surface of the insulation. Such a reduction is possible if Di < Dcr. From Example 3.4,
cr 2
k 0.5 W/m K
r 0.02m.
h 25 W/m K
\u22c5
= = =
\u22c5
Hence, Dcr = 0.04m > Di = 0.005m. To minimize the maximum temperature, which exists at
the inner surface of the insulation, add insulation in the amount
( )o i cr i 0.04 0.005 mD D D Dt=
2 2 2
\u2212
\u2212 \u2212
= =
t = 0.0175m. <
The cable surface temperature may then be obtained from
( )
( )
( )
( ) ( )
s s
2t,c cr i
i cr
2
T T T 30 C
q =
R ln D / D 1 ln 0.04/0.0050.02 m K/W 1
 D 2 k h D W0.005m 2 0.5 W/m K 25 0.04m
m K
\u3c0 \u3c0 \u3c0 \u3c0 \u3c0 \u3c0
\u221e
\u2212 \u2212
\u2032 =
\u2032\u2032
\u22c5+ + + +
\u22c5
\u22c5
$
Hence,
( )
s sT 30 C T 30 CW294
m 1.27+0.66+0.32 m K/W 2.25 m K/W
\u2212 \u2212
= =
\u22c5 \u22c5
$ $
sT 692.5 C= $
Recognizing that q = (Ts - Ti)/Rt,c, find
( )
2
t,c
i s t,c s
i
W m K294 0.02R m WT T qR T q 692.5 C
 D L 0.005m\u3c0 \u3c0
\u22c5
×
\u2032\u2032
= \u2212 = \u2212 = \u2212
$
iT 318.2 C.=
$ <
COMMENTS: Use of the critical insulation thickness in lieu of a thin coating has the effect of
reducing the maximum insulation temperature from 778.7°C to 318.2°C. Use of the critical insulation
thickness also reduces the cable surface temperature to 692.5°C from 778.7°C with no insulation or
from 1153°C with a thin coating.
PROBLEM 3.48
KNOWN: Saturated steam conditions in a pipe with prescribed surroundings.
FIND: (a) Heat loss per unit length from bare pipe and from insulated pipe, (b) Pay back
period for insulation.
SCHEMATIC:
Steam Costs:
 $4 for 109 J
Insulation Cost:
 $100 per meter
Operation time:
 7500 h/yr
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3)
Constant properties, (4) Negligible pipe wall resistance, (5) Negligible steam side convection
resistance (pipe inner surface temperature is equal to steam temperature), (6) Negligible
contact resistance, (7) Tsur = T\u221e.
PROPERTIES: Table A-6, Saturated water (p = 20 bar): Tsat = Ts = 486K; Table A-3,
Magnesia, 85% (T \u2248 392K): k = 0.058 W/m\u22c5K.
ANALYSIS: (a) Without the insulation, the heat loss may be expressed in terms of radiation
and convection rates,
( ) ( )( )
( ) ( )
( ) ( )
4 4
s sur s
8 4 4 4
2 4
2
q = D T T h D T T
Wq =0.8 0.2m 5.67 10 486 298 K
m K
W
 +20 0.2m 486-298 K
m K
\u3b5\u3c0 \u3c3 \u3c0
\u3c0
\u3c0
\u221e
\u2212
\u2032
\u2212 + \u2212
\u2032 × \u2212
\u22c5
×
\u22c5
( )q = 1365+2362 W/m=3727 W/m.\u2032 <
With the insulation, the thermal circuit is of the form
Continued \u2026..
PROBLEM 3.48 (Cont.)
From an energy balance at the outer surface of the insulation,
( ) ( ) ( )
( )
( )
( )
( )( )
( )( )
cond conv rad
s,i s,o 4 4
o s,o o s,o sur
o i
s,o
s,o2
-8 4 4 4
s,o2 4
q q q
T T
h D T T D T T
ln D / D / 2 k
486 T K W20 0.3m T 298K
ln 0.3m/0.2m m K
2 0.058 W/m K
W
 +0.8 5.67 10 0.3m T 298 K .
m K
\u3c0 \u3b5\u3c3\u3c0
\u3c0
\u3c0
\u3c0
\u3c0
\u221e
\u2032 \u2032 \u2032= +
\u2212
= \u2212 + \u2212
\u2212
= \u2212
\u22c5
\u22c5
× × \u2212
\u22c5
By trial and error, we obtain
Ts,o \u2248 305K
in which case
( )
( )
( )
486-305 K
q = 163 W/m.
ln 0.3m/0.2m
2 0.055 W/m K\u3c0
\u2032 =
\u22c5
<
(b) The yearly energy savings per unit length of pipe due to use of the insulation is
( ) 9
Savings Energy Savings Cost
Yr m Yr. Energy
Savings J s h $43727 163 3600 7500
Yr m s m h Yr 10 J
Savings $385 / Yr m.
Yr m
= ×
\u22c5
= \u2212 × × ×
\u22c5 \u22c5
= \u22c5
\u22c5
The pay back period is then
Insulation Costs $100 / mPay Back Period = 
Savings/Yr. m $385/Yr m=\u22c5 \u22c5
Pay Back Period = 0.26 Yr = 3.1 mo. <
COMMENTS: Such a low pay back period is more than sufficient to justify investing in the
insulation.
PROBLEM 3.49
KNOWN: Temperature and convection coefficient associated with steam flow through a pipe
of prescribed inner and outer diameters. Outer surface emissivity and convection coefficient.
Temperature of ambient air and surroundings.
FIND: Heat loss per unit length.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3)
Constant properties, (4) Surroundings form a large enclosure about pipe.
PROPERTIES: Table A-1, Steel, AISI 1010 (T \u2248 450 K): k = 56.5 W/m\u22c5K.
ANALYSIS: Referring to the thermal circuit, it follows from an energy balance on the outer
surface that
,i s,o s,o ,o s,o sur
conv,i cond conv,o rad
T T T T T T
R R R R
\u221e \u221e\u2212 \u2212 \u2212
= +
+
or from Eqs. 3.9, 3.28 and 1.7,
( ) ( ) ( )