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ch08


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C r n r T .
4k 2k 4k 2k
= = = - + + = - +
& && &
l l
The temperature distribution and the maximum wall temperature (r = ro) are
( ) ( )
2
2 2 o
si
i
q rq r
T r r r n T
4k 2k r
= - - + +
&&
l
( ) ( )
2
2 2 o o
w,max o o si
i
q r rq
T T r r r n T
4k 2k r
= = - - + +
&&
l
where Ts, the inner surface temperature of the wall at the exit, follows from
( ) ( ) ( ) ( )
2 2 2 2
o oi i
s s m,o
i i
q /4 D D L q D D
q h T T
 D L 4 D
p
p
- -
¢¢ = = = -
& &
where h is the local convection coefficient at the exit. With
( )D 6 2i
4 m 4 0.1 kg/s
Re 7928
 D 0.02m 803 10 N s/mp m p -
´
= = =
´ ×
&
the flow is turbulent and, with (L/Di) = (8.87 m/0.02m) = 444 >> (xfd/D) » 10, it is also fully
developed. Hence, from the Dittus-Boelter correlation, Eq. 8.60,
( ) ( )4 / 54/5 0.4 0.4 2D
i
k 0.617 W/m K
h 0.023 Re Pr 0.023 7928 5.45 1840 W/m K.
D 0.02 m
×
= = = ×
Hence, the inner surface temperature of the wall at the exit is
( ) ( ) ( )
( )
2 26 32 2
o i
s m,o 2i
10 W/m 0.04m 0.02mq D D
T T 40 C 48.2 C
4 h D 4 1840 W/m K 0.02m
é ù-- ê úë û= + = + =
´ ×
o o
&
and ( ) ( )
6 3
2 2
w,max
10 W/m
T 0.02m 0.01m
4 15 W/m K
é ù= - -ê úë û´ ×
( )26 310 W/m 0.02m 0.02
n 48.2 C 52.4 C.
2 15 W/m K 0.01
+ + =
´ ×
o ol <
COMMENTS: The physical situation corresponds to a uniform surface heat flux, and Tm
increases linearly with x. In the fully developed region, Ts also increases linearly with x.
PROBLEM 8.37
KNOWN: Dimensions and thermal conductivity of concrete duct. Convection conditions of ambient
air. Flow rate and inlet temperature of water flow through duct.
FIND: (a) Outlet temperature, (b) Pressure drop and pump power requirement, (c) Effect of flow rate
and pipe diameter on outlet temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Fully developed flow throughout duct, (3) Negligible pipe
wall conduction resistance, (4) Negligible potential energy, kinetic energy and flow work changes for
water, (5) Constant properties.
PROPERTIES: Table A-6, water ( )mT 360 K :\u2248 3967 kg / m ,\u3c1 = pc 4203 J / kg K,= \u22c5 6324 10µ \u2212= ×
2N s / m ,\u22c5 wk 0.674 W / m K, Pr 2.02.= \u22c5 =
ANALYSIS: (a) The outlet temperature is given by
( ) ( )m,o m,i pT T T T exp UA / m c\u221e \u221e= + \u2212 \u2212 \ufffd
where
( ) ( ) 11tot cnv,w cnd cnv,aUA R R R R \u2212\u2212= = + +
( ) ( )
( )
4
cnd
ln 1.08w / D ln 1.08 0.30m / 0.15m
R 8.75 10 K / W
2 kL 2 1.4 W / m K 100m\u3c0 \u3c0
\u2212
×
= = = ×
\u22c5
( ) ( ) 11 2 4cnv,aR 4 w L h 4 0.3m 100m 25W / m K 3.33 10 K / W\u2212\u2212 \u2212= = × × × \u22c5 = ×
With ( ) ( )6 2DRe 4 m / D 4 2 kg / s / 0.15m 324 10 N s / m 52, 400,\u3c0 µ \u3c0 \u2212= = × × × × \u22c5 =\ufffd
( ) ( )4 / 5 0.3 24 / 5 0.3ww fd Dk 0.674 W / m K 0.023h h 0.023 Re Pr 52, 400 2.02 761W / m KD 0.15m
\u22c5 ×
\u2248 = = = \u22c5
( ) ( ) 11 2 5cnv,w wR D L h 0.15m 100m 761W / m K 2.79 10 K / W\u3c0 \u3c0 \u2212\u2212 \u2212= = × × × \u22c5 = ×
( ) 15 4 4UA 2.79 10 8.75 10 3.33 10 K / W 809 W / K\u2212\u2212 \u2212 \u2212\uf8ee \uf8f9= × + × + × =\uf8ef \uf8fa\uf8f0 \uf8fb
m,o
809 W / KT 0 C 90 C exp 81.7 C
2kg / s 4203J / kg K
\uf8eb \uf8f6
= ° + ° \u2212 = °\uf8ec \uf8f7
× \u22c5\uf8ed \uf8f8 <
Continued \u2026..
PROBLEM 8.37 (Cont.)
(b) With f = 0.0206 from Fig. 8.3 and 2mu m / D / 4 0.117 m / s,\u3c1\u3c0= =\ufffd
( )232 2 4m 967 kg / m 0.117 m / sup f L 0.0206 100m 91N / m 8.98 10 bars
2D 2 0.15m
\u3c1
\u2212\u2206 = = = = ×
×
<
With 3 3m / 2.07 10 m / s,\u3c1 \u2212\u2200 = = ×\ufffd \ufffd the pump power requirement is
( )2 3 3P p 91N / m 2.07 10 m / s 0.19 W\u2212= \u2206 \u2200 = × =\ufffd <
(c) The effects of varying the flowrate and duct diameter were assessed using the IHT software, and
results are shown below.
Although cnv,wR , and hence totR , decreases with increasing m\ufffd , thereby increasing UA, the effect is
significantly less than that of m\ufffd to the first power, causing the exponential term, ( )pexp UA / m c ,\u2212 \ufffd to
approach unity and m,oT to approach m,iT . The effect can alternatively be attributed to a reduction
in the residence time of the water in the pipe (um increases with increasing m\ufffd for fixed D). With
increasing D for fixed m\ufffd and w, m,oT decreases due to an increase in the residence time, as well as a
reduction in the conduction resistance, cndR .
COMMENTS: (1) Use of mT 360 K= to evaluate properties of the water for Parts (a) and (b) is
reasonable, and iteration is not necessary. (2) The pressure drop and pump power requirement are
small.
0 1 2 3 4 5
Ma s s flow ra te , m d o t(kg /s )
6 0
6 5
7 0
7 5
8 0
8 5
9 0
O
u
tle
t t
e
m
pe
ra
tu
re
,
 
Tm
o
(C
)
0 .0 5 0 .1 0 .1 5 0 .2 0 .2 5
D u ct d ia m e te r, D (m )
7 5
7 8
8 1
8 4
8 7
9 0
Ou
tle
t t
e
m
pe
ra
tu
re
,
 
Tm
o
(C
)
PROBLEM 8.38
KNOWN: Water flow through a thick-walled tube immersed in a well stirred, hot reaction tank
maintained at 85°C; conduction thermal resistance of the tube wall based upon the inner surface area
is 2cdR 0.002 m K / W.\u2032\u2032 = \u22c5
FIND: (a) The outlet temperature of the process fluid, Tm,o; assume, and then justify, fully
developed flow and thermal conditions within the tube; and (b) Do you expect Tm,o to increase or
decrease if the combined thermal entry condition exists within the tube? Estimate the outlet
temperature of the process fluid for this condition.
SCHEMATIC:
ASSUMPTIONS: (1) Flow is fully developed, part (a), (2) Constant properties, (3) Negligible
kinetic and potential energy changes and flow work, and (4) Constant wall temperature heating.
PROPERTIES: Table A-6, Water (Tm = (Tm,o + Tm,i)/2 = 337 K): cp = 4187 J/kg\u22c5K, µ = 4.415 ×
10-4 N\u22c5s/m2, k = 0.6574 W/m\u22c5K, Pr = 2.80; (Ts = 358 K): µs = 3.316 × 10-4 N\u22c5s/m2.
ANALYSIS: (a) The outlet temperature is determined from the rate equation, Eq. 8.46a, written as
s m,o s
s m,i p
T T UA
exp
T T m c
\uf8eb \uf8f6
\u2212
= \u2212\uf8ec \uf8f7\uf8ec \uf8f7
\u2212 \uf8ed \uf8f8\ufffd
(1)
where the overall coefficient, based upon the inner surface area of the tube is expressed in terms of
the convection and conduction thermal resistances,
cd,i
1 1 R
U h
\u2032\u2032= +
(2)
To estimate h, begin by characterizing the flow
DRe 4 m / D\u3c0 µ= \ufffd (3)
( ) 4 2DRe 4 33/ 3600 kg / s / 0.012 m 4.415 10 N s / m 2210\u3c0 \u2212= × × × \u22c5 =
Consider the flow as laminar, and assuming fully developed conditions, estimate h with the
correlation of Eq. 8.55,
DNu hD / k 3.66= = (4)
2h 3.66 0.6574 W / m K / 0.012 m 201 W / m K= × \u22c5 = \u22c5
From Eq. (2),
12 2 2U 1/ 201 W / m K 0.002 m K / W 143.1 W / m K
\u2212\uf8ee \uf8f9
= \u22c5 + \u22c5 = \u22c5\uf8ef \uf8fa\uf8f0 \uf8fb
and from Eq. (1), with As = \u3c0DL, calculate Tm,o.
Continued \u2026..
PROBLEM 8.38 (Cont.)
2
m,o85 T 143.1 W / m K 0.012 m 8 m
exp
85 20 33/ 3600 kg / s 4187 J / kg K
\u3c0\uf8eb \uf8f6\u2212 \u22c5 × × ×\uf8ec \uf8f7= \u2212\uf8ec \uf8f7
\u2212 × \u22c5\uf8ed \uf8f8
m,oT 64 C= ° <
Fully developed flow and thermal conditions are justified if the tube length is much greater than the
fully developed length xfd,t. From Eq. 8.23,
fd,t
D
x
0.05 Re Pr
D
=
fd,tx 0.012 m 0.05 221.0 2.41 3.20 m= × × × =
That is, the length is only twice that required to reach fully developed conditions.
(b) Considering combined entry length conditions, estimate the convection coefficient using the
Sieder-Tate correlation, Eq. 8.56,
( )
( )D
D
2/3
D
0.0668 D / L Re Pr
Nu 3.66
1 0.04 D / L Re Pr
= +
\uf8ee \uf8f9+ \uf8f0 \uf8fb
(5)
substituting numerical values, find
D
2Nu 4.05 h 222 W / m K= = \u22c5
which is a 10% increase over the fully developed analysis result. Using the foregoing relations, find
2
m,oU 154 W / m K T 65.5 C= \u22c5 = ° <
COMMENTS: (1) The thermophysical properties for the fully developed correlation are evaluated at
the mean fluid temperature Tm = (Tm,o + Tm,i)/2. The values are shown above in the properties
section.
(2) For the Sieder-Tate correlation, the properties are also evaluated at Tm, except for µs, which is
evaluated at Ts.
(3) For this case where the tube length is about twice xfd,t, the average heat transfer