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# ch08

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```\uf8f4\uf8ef \uf8fa\uf8f0 \uf8fb\uf8f3 \uf8fe
( ) ( ) 11 2 4conv,o o iR h D L 30.9 W / m K 1m 100m 1.03 10 K / W\u3c0 \u3c0 \u2212\u2212 \u2212\u2248 = \u22c5 × × × = ×
Hence, from Eq. (1)
( ) ( )o i 3 4 4 3
ins
ln D / D
2.96 10 3.33 10 1.03 10 K / W 2.52 10 K / W
2 k L\u3c0
\u2212 \u2212 \u2212 \u2212
= × \u2212 × \u2212 × = ×
( ) ( )3 2o i insD D exp 2 k L 2.52 10 K / W 1m exp 1.58 10 K / W 0.125 W / m K 100m 1.22m\u3c0 \u2212 \u2212= × × = × × × \u22c5 × =
Hence, the minimum insulation thickness is
( )min o it D D / 2 0.11m= \u2212 = <
COMMENTS: With Do = 1.22m, use of Di = 1m to evaluate the outer convection coefficient and
thermal resistance is a reasonable approximation. However, improved accuracy may be obtained by
using the calculated value of Do to determine conditions at the outer surface and iterating on the
solution.
PROBLEM 8.42
KNOWN: Flow rate, inlet temperature and desired outlet temperature of liquid mercury flowing
through a tube of prescribed diameter and surface temperature.
FIND: Required tube length and error associated with use of a correlation for moderate to large Pr
fluids.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible kinetic energy,
potential energy and flow work effects, (4) Fully developed flow.
PROPERTIES: Table A-5, Mercury T Km = 350c h: cp = 137.7 J/kg×K, m = 0.1309 ´ 10-2 N×s/m2,
k = 9.18 W/m×K, Pr = 0.0196.
ANALYSIS: The Reynolds and Peclet numbers are
( )D 2 2
4m 4 0.5 kg/s
Re 9727
D 0.05m 0.1309 10 N s/mp m p -
´= = =
´ ×
&
( )D DPe Re Pr 9727 0.0196 191.= = =
Hence, assuming fully developed turbulent flow throughout the tube, it follows from Eq. 8.66 that
( ) ( )0.8 0.8 2Dk 9.18 W/m Kh 5.0 0.025 Pe 5.0 0.025 191 1224 W/m K.D 0.05 m
×= + = + ´ = ×
From Eq. 8.42a, it follows that
( )
( )
p o
2i
m c 0.5 kg/s 137.7 J/kg KT 450 400
L n n 0.39 m.
Dh T 450 3000.05 m 1224 W/m Kp p
×D -
= - = - =
D -×
&
l l <
If the Dittus-Boelter correlation, Eq. 8.60, is used in place of Eq. 8.66,
( ) ( )
2
4/5 0.44/5 0.4 2
D
k 9.18 W/m K
h 0.023 Re Pr 0.023 9727 0.0196 1358 W/m K
D 0.05 m
×= = = ×
and the required tube length is
( )
( )
p o
2i
m c 0.5 kg/s 137.7 J/kg KT 450 400
L n n 0.35 m.
Dh T 450 3000.05 m 1358 W/m Kp p
×D -= - = - =
D -×
&
l l <
COMMENTS: Such good agreement between results does not occur in general. For example, if
ReD = 2 ´ 10
4
, h = 1463 from Eq. 8.66 and 2417 from Eq. 8.60. Large errors are usually associated
with using conventional (moderate to large Pr) correlations with liquid metals.
PROBLEM 8.43
KNOWN: Surface temperature and diameter of a tube. Velocity and temperature of air in cross
flow. Velocity and temperature of air in fully developed internal flow.
FIND: Convection heat flux associated with the external and internal flows.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform cylinder surface temperature, (3) Fully
developed internal flow.
PROPERTIES: Table A-4, Air (298K): n = 15.71 ´ 10-6 m2/s, k = 0.0261 W/m×K, Pr = 0.71.
ANALYSIS: For the external and internal flows,
4m
D -6 2
VD u D 30 m/s 0.05 m
Re 9.55 10 .
15.71 10 m / sn n
´= = = = ´
´
From the Zhukauskas relation for the external flow, with C = 0.26 and m = 0.6,
( ) ( ) ( ) ( )0.61/4 0.37 1 / 4m n 4D sDNu C Re Pr Pr/Pr 0.26 9.55 10 0.71 1 223.= = ´ =
Hence, the convection coefficient and heat flux are
2
D
k 0.0261 W/m K
h Nu 223 116.4 W/m K
D 0.05 m
×= = ´ = ×
( ) ( )2 3 2sq h T T 116.4W/m K 100 25 C 8.73 10 W/m .¥¢¢ = - = × - = ´o <
Using the Dittus-Boelter correlation, Eq. 8.60, for the internal flow, which is turbulent,
( ) ( )4/5 0.44/5 0.4 4D DNu 0.023 Re Pr 0.023 9.55 10 0.71 193= = ´ =
2
D
k 0.0261 W/m K
h Nu 193 101 W/m K
D 0.05 m
×= = ´ = ×
and the heat flux is
( ) ( )2 3 2s mq h T T 101 W/m K 100 25 C 7.58 10 W/m .¢¢ = - = × - = ´o <
COMMENTS: Convection effects associated with the two flow conditions are comparable.
PROBLEM 8.44
KNOWN: Diameter, length and surface temperature of condenser tubes. Water velocity and inlet
temperature.
FIND: (a) Water outlet temperature evaluating properties at Tm = 300 K, (b) Repeat calculations using
properties evaluated at the appropriate temperature, mT = (Tm,i + Tm,o)/2, and (c) Coolant mean
velocities for the range 4 \u2264 L \u2264 7 m which provide the same Tm,o as found in part (b).
SCHEMATIC:
ASSUMPTIONS: (1) Negligible tube wall conduction resistance, (2) Negligible kinetic energy,
potential energy and flow work changes.
PROPERTIES: Table A.6, Water ( mT = 300 K): \u3c1 = 997 kg/m3, cp = 4179 J/kg\u22c5K, µ = 855 × 10-6
kg/s\u22c5m, k = 0.613 W/m\u22c5K, Pr = 5.83.
ANALYSIS: (a) From Equation 8.42b
( ) ( )m,o s s m,i pT T T T exp DL mc h\u3c0\uf8ee \uf8f9= \u2212 \u2212 \u2212\uf8f0 \uf8fb\ufffd .
and evaluating properties at mT = 300 K, find
( )3m
D 6
997 kg m 1m s 0.0254 mu DRe 29,618
855 10 kg s m
\u3c1
µ \u2212
= = =
× \u22c5
The flow is turbulent, and since L/D = 197, it is reasonable to assume fully developed flow throughout
the tube. Hence, h \u2248 hfd. From the Dittus-Boelter equation,
( ) ( )4 /5 0.44 /5 0.4D DNu 0.023Re Pr 0.023 29,618 5.83 176= = =
( ) ( ) 2Dh Nu k D 176 0.613W m K 0.0254 m 4248 W m K= = \u22c5 = \u22c5 .
With
( ) ( ) ( )( )22 3mm u D 4 4 997 kg m 1m s 0.0254m 0.505kg s\u3c1 \u3c0 \u3c0= = =\ufffd .
Equation 8.42b yields
( ) ( ) ( )( )
2
m,o
0.0254m 5m 4248W m K
T 350K 60K exp 323K 50 C
0.505kg s 4179J kg K
\u3c0\uf8ee \uf8f9\u22c5\uf8ef \uf8fa
= \u2212 \u2212 = =\uf8ef \uf8fa
\u22c5\uf8ef \uf8fa\uf8f0 \uf8fb
\$ <
(b) Using the IHT Correlations Tool, Internal Flow, for fully developed Turbulent Flow, along with the
energy balance and rate equations above, the calculation of part (a) is repeated with mT = (Tm,i + Tm,o)/2
giving these results:
mT 307.3K= Tm,o = 51.7°C = 324.7 K <
(c) Using the IHT model developed for the part (b) analysis, the coolant mean velocity, um, as a function
of tube length L with Tm,o = 51.7°C is calculated and the results plotted below.
Continued...
PROBLEM 8.44 (Cont.)
4 5 6 7
Tube length, L (m)
0
1
2
3
4
5
6
Ve
lo
ci
ty
, u
m
(m
/s)
COMMENTS: (1) Using mT = 300 K vs. mT = (Tm,i + Tm,o)/2 = 307 K for this application resulted in
a difference of Tm,o = 50°C vs.Tm,o = 51.7°C. While the difference is only 1.7°C, it is good practice to
use the proper value for mT .
(2) Note that um must be increased markedly with increasing length in order that Tm,o remain fixed.
PROBLEM 8.45
KNOWN: Gas turbine vane approximated as a tube of prescribed diameter and length maintained at a
known surface temperature. Air inlet temperature and flowrate.
FIND: (a) Outlet temperature of the air coolant for the prescribed conditions and (b) Compute and plot
the air outlet temperature Tm,o as a function of flow rate, 0.1 \u2264 m\ufffd \u2264 0.6 kg/h. Compare this result with
those for vanes having passage diameters of 2 and 4 mm.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy changes.
PROPERTIES: Table A.4, Air (assume mT = 780 K, 1 atm): cp = 1094 J/kg\u22c5K, k = 0.0563 W/m\u22c5K, µ
= 363.7 × 10-7 N\u22c5s/m2, Pr = 0.706; (Ts = 650°C = 923 K, 1 atm): µ = 404.2 × 10-7 N\u22c5s/m2.
ANALYSIS: (a) For constant wall temperature heating, from Eq. 8.43,
s m,o
s m,i p
T T PLh
exp
T T mc
\uf8eb \uf8f6
\u2212
= \u2212\uf8ec \uf8f7\uf8ec \uf8f7
\u2212 \uf8ed \uf8f8\ufffd
(1)
where P = \u3c0D. For flow in circular passage,
( )
( )D 7 2
4 0.18kg h 1 3600s h4mRe 584
D 0.003m 363.7 10 N s m\u3c0 µ \u3c0 \u2212
×
= = =
× \u22c5
\ufffd
. (2)
The flow is laminar, and since L/D = 75 mm/3 mm = 25, the Sieder-Tate correlation including combined
entry length yields
0.141/3
DD
s
hD Re PrNu 1.86
k L D
µ
µ
\uf8eb \uf8f6\uf8eb \uf8f6
= = \uf8ec \uf8f7\uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8 (3)
0.141/3 7
2
7
0.0563W m K 584 0.706 363.7 10h 1.86 87.5 W m K
0.003m 25 404.2 10
\u2212
\u2212
\uf8eb \uf8f6
\u22c5 × ×\uf8eb \uf8f6 \uf8ec \uf8f7= = \u22c5\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 ×\uf8ed \uf8f8
.
Hence, the air outlet temperature is
( )
( )
( )
2
m,o650 T 0.003m 0.075m 87.5W m K
exp```