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# ch03

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```( )
( ) ( ) ( )
( )
,i s,o s,o ,o 4 4
o s,o sur
i i o i o o
s,o s,o
1 12 2
8 2
T T T T
D T T
1/ D h ln D / D / 2 k 1/ D h
523K T T 293K
ln 75/60
0.6m 500 W/m K 0.075m 25 W/m K
2 56.5 W/m K
+0.8 0.075m 5.67 10 W/m K
\u3b5\u3c0 \u3c3
\u3c0 \u3c0 \u3c0
\u3c0 \u3c0
\u3c0
\u3c0
\u221e \u221e
\u2212 \u2212
\u2212
\u2212 \u2212
= + \u2212
+
\u2212 \u2212
=
× × \u22c5 + × × \u22c5
× \u22c5
× × × \u22c5 4 4 4 4s,o
s,o s,o 8 4 4
s,o
T 293 K
523 T T 293
1.07 10 T 293 .
0.0106+0.0006 0.170
\u2212
\uf8ee \uf8f9
\u2212\uf8ef \uf8fa\uf8f0 \uf8fb
\u2212 \u2212 \uf8ee \uf8f9
= + × \u2212\uf8ef \uf8fa\uf8f0 \uf8fb
From a trial-and-error solution, Ts,o \u2248 502K. Hence the heat loss is
( ) ( )4 4o o s,o ,o o s,o surq = D h T T D T T\u3c0 \u3b5\u3c0 \u3c3\u221e\u2032 \u2212 + \u2212
( ) ( ) ( )2 8 4 4 42 4
W
q = 0.075m 25 W/m K 502-293 0.8 0.075m 5.67 10 502 243 K
m K
\u3c0 \u3c0 \u2212\u2032 \u22c5 + × \u2212
\u22c5
\uf8ee \uf8f9\uf8f0 \uf8fb
q =1231 W/m+600 W/m=1831 W/m.\u2032 <
COMMENTS: The thermal resistance between the outer surface and the surroundings is
much larger than that between the outer surface and the steam.
PROBLEM 3.50
KNOWN: Temperature and convection coefficient associated with steam flow through a pipe of
prescribed inner and outer radii. Emissivity of outer surface magnesia insulation, and convection
coefficient. Temperature of ambient air and surroundings.
FIND: Heat loss per unit length \u2032q and outer surface temperature Ts,o as a function of insulation
thickness. Recommended insulation thickness. Corresponding annual savings and temperature
distribution.
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): ks = 56.5 W/m\u22c5K. Table A-3, Magnesia,
85% (T \u2248 365 K): km = 0.055 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
T T T T T T
R R R R R
\u221e \u221e
\u2212 \u2212 \u2212
= +
\u2032 \u2032 \u2032 \u2032 \u2032+ +
or from Eqs. 3.9, 3.28 and 1.7,
( ) ( ) ( ) ( ) ( ) ( )( )
,i s,o s,o ,o s,o sur
12 21 i 2 1 s 3 2 m 3 o
3 s,o sur s,o sur
T T T T T T
1 2 r h ln r r 2 k ln r r 2 k 1 2 r h
2 r T T T T
\u3c0 \u3c0 \u3c0 \u3c0
\u3c0 \u3b5\u3c3
\u221e \u221e
\u2212
\u2212 \u2212 \u2212
= +
+ +
+ +\uf8ee \uf8f9\uf8f0 \uf8fb
This expression may be solved for Ts,o as a function of r3, and the heat loss may then be determined by
evaluating either the left-or right-hand side of the energy balance equation. The results are plotted as
follows.
Continued...
PROBLEM 3.50 (Cont.)
0.035 0.045 0.055 0.065 0.075
0
400
800
1200
1600
2000
H
ea
t l
os
s,
q
pr
im
e(W
/m
)
q1

0.035 0.045 0.055 0.065 0.075
0
0.5
1
1.5
2
Th
er
m
al
re
si
st
an
ce
, R
pr
im
e(K
/m
.W
)
Insulation conduction resistance, Rcond,m
Outer convection resistance, Rconv,o
The rapid decay in q\u2032 with increasing r3 is attributable to the dominant contribution which the insulation
begins to make to the total thermal resistance. The inside convection and tube wall conduction
resistances are fixed at 0.0106 m\u22c5K/W and 6.29×10-4 m\u22c5K/W, respectively, while the resistance of the
insulation increases to approximately 2 m\u22c5K/W at r3 = 0.075 m.
The heat loss may be reduced by almost 91% from a value of approximately 1830 W/m at r3 = r2
= 0.0375 m (no insulation) to 172 W/m at r3 = 0.0575 m and by only an additional 3% if the insulation
thickness is increased to r3 = 0.0775 m. Hence, an insulation thickness of (r3 - r2) = 0.020 m is
recommended, for which q\u2032 = 172 W/m. The corresponding annual savings (AS) in energy costs is
therefore
( )[ ] 9
\$4 h s
AS 1830 172 W m 7000 3600 \$167 / m
y h10 J
= \u2212 × × = <
The corresponding temperature distribution is
0.038 0.042 0.046 0.05 0.054 0.058
300
340
380
420
460
500
Lo
ca
l t
em
pe
ra
tu
re
, T
(K
)
Tr
The temperature in the insulation decreases from T(r) = T2 = 521 K at r = r2 = 0.0375 m to T(r) = T3 =
309 K at r = r3 = 0.0575 m.
Continued...
PROBLEM 3.50 (Cont.)
COMMENTS: 1. The annual energy and costs savings associated with insulating the steam line are
substantial, as is the reduction in the outer surface temperature (from Ts,o \u2248 502 K for r3 = r2, to 309 K for
r3 = 0.0575 m).
2. The increase in radR\u2032 to a maximum value of 0.63 m\u22c5K/W at r3 = 0.0455 m and the subsequent decay
is due to the competing effects of hrad and ( )3 3A 1 2 r\u3c0\u2032 = . Because the initial decay in T3 = Ts,o with
increasing r3, and hence, the reduction in hrad, is more pronounced than the increase in 3A\u2032 , radR\u2032
increases with r3. However, as the decay in Ts,o, and hence hrad, becomes less pronounced, the increase in
3A\u2032 becomes more pronounced and radR\u2032 decreases with increasing r3.
PROBLEM 3.51
KNOWN: Pipe wall temperature and convection conditions associated with water flow through the pipe
and ice layer formation on the inner surface.
FIND: Ice layer thickness \u3b4.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Negligible pipe wall thermal
resistance, (3) negligible ice/wall contact resistance, (4) Constant k.
PROPERTIES: Table A.3, Ice (T = 265 K): k \u2248 1.94 W/m\u22c5K.
ANALYSIS: Performing an energy balance for a control surface about the ice/water interface, it follows
that, for a unit length of pipe,
conv condq q\u2032 \u2032=
( )( ) ( )s,i s,oi 1 ,i s,i 2 1
T T
h 2 r T T
ln r r 2 k
\u3c0
\u3c0
\u221e
\u2212
\u2212 =
Dividing both sides of the equation by r2,
( )
( ) ( )( )
s,i s,o2 1
22 1 i 2 ,i s,i
T Tln r r k 1.94 W m K 15 C 0.097
r r h r T T 3 C2000 W m K 0.05m\u221e
\u2212
\u22c5
= × = × =
\u2212
\u22c5
\$
\$
The equation is satisfied by r2/r1 = 1.114, in which case r1 = 0.050 m/1.114 = 0.045 m, and the ice layer
thickness is
2 1r r 0.005m 5mm\u3b4 = \u2212 = = <
COMMENTS: With no flow, hi \u2192 0, in which case r1 \u2192 0 and complete blockage could occur. The
pipe should be insulated.
PROBLEM 3.52
KNOWN: Inner surface temperature of insulation blanket comprised of two semi-cylindrical shells of different
materials. Ambient air conditions.
FIND: (a) Equivalent thermal circuit, (b) Total heat loss and material outer surface temperatures.
SCHEMATIC:
resistance between materials, (4) Constant properties.
ANALYSIS: (a) The thermal circuit is,
conv,A conv,B 2R R 1/ r h\u3c0\u2032 \u2032= =
( ) ( )2 1cond A
A
ln r / r
R
k\u3c0
\u2032 = <
( ) ( )2 icond
B
ln r / r
R
k\u3c0
\u2032 =B
The conduction resistances follow from Section 3.3.1 and Eq. 3.28. Each resistance is larger by a factor of 2 than
the result of Eq. 3.28 due to the reduced area.
(b) Evaluating the thermal resistances and the heat rate ( )A Bq =q q ,\u2032 \u2032 \u2032+
( ) 12convR 0.1m 25 W/m K 0.1273 m K/W\u3c0 \u2212\u2032 = × × \u22c5 = \u22c5
( ) ( ) ( ) ( )cond A cond B cond Aln 0.1m/0.05mR 0.1103 m K/W R 8 R 0.8825 m K/W2 W/m K\u3c0\u2032 \u2032 \u2032= = \u22c5 = = \u22c5× \u22c5
( ) ( )
s,1 s,1
conv convcond A cond B
T T T T
q =
R R R R
\u221e \u221e\u2212 \u2212
\u2032 +
\u2032 \u2032 \u2032 \u2032+ +
( )
( )
( )
( ) ( )
500 300 K 500 300 K
q = 842 198 W/m=1040 W/m.
0.1103+0.1273 m K/W 0.8825+0.1273 m K/W
\u2212 \u2212
\u2032 + = +
\u22c5 \u22c5
<
Hence, the temperatures are
( ) ( )s,1 As,2 A cond A W m KT T q R 500K 842 0.1103 407Km W
\u22c5
\u2032 \u2032= \u2212 = \u2212 × = <
( ) ( )s,1 Bs,2 B cond B W m KT T q R 500K 198 0.8825 325K.m W
\u22c5
\u2032 \u2032= \u2212 = \u2212 × = <
COMMENTS: The total heat loss can also be computed from ( )s,1 equivq = T T / R ,\u221e\u2032 \u2212
where ( )( ) ( ) 11 1equiv conv,A cond(B) conv,Bcond AR R R R R 0.1923 m K/W.\u2212\u2212 \u2212\u2032 \u2032 \u2032 \u2032= + + + = \u22c5\uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb
Hence ( )q = 500 300 K/0.1923 m K/W=1040 W/m.\u2032 \u2212 \u22c5
PROBLEM 3.53
KNOWN: Surface temperature of a circular rod coated with bakelite and adjoining```