Unformatted text preview: erial, determine (a) the temperature of the oil when the pipe
leaves the lake, (b) the rate of heat transfer from the oil, and (c) the pumping
power required to overcome the pressure losses and to maintain the flow of the
oil in the pipe. Icy lake, 0°C
20°C Oil
2 m/s D = 0.3 m Te 0°C
200 m FIGURE 8–23
Schematic for Example 8–3. SOLUTION Oil flows in a pipeline that passes through icy waters of a lake at
0°C. The exit temperature of the oil, the rate of heat loss, and the pumping
power needed to overcome pressure losses are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of
the pipe is very nearly 0°C. 3 The thermal resistance of the pipe is negligible.
4 The inner surfaces of the pipeline are smooth. 5 The flow is hydrodynamically
developed when the pipeline reaches the lake.
Properties We do not know the exit temperature of the oil, and thus we cannot
determine the bulk mean temperature, which is the temperature at which the
properties of oil are to be evaluated. The mean temperature of the oil at the
inlet is 20°C, and we expect this temperature to drop somewhat as a result
of heat loss to the icy waters of the lake. We evaluate the properties of the oil
at the inlet temperature, but we will repeat the calculations, if necessary,
using properties at the evaluated bulk mean temperature. At 20°C we read
(Table A14)
k 888 kg/m3
0.145 W/m °C Cp
Pr 901 10 6 m2/s
1880 J/kg °C
10,400 Analysis (a) The Reynolds number is
m Dh Re (2 m/s)(0.3 m)
901 10 6 m2/s 666 which is less than the critical Reynolds number of 2300. Therefore, the flow is
laminar, and the thermal entry length in this case is roughly Lt 0.05 Re Pr D 0.05 666 10,400 (0.3 m) 104,000 m which is much greater than the total length of the pipe. This is typical of fluids
with high Prandtl numbers. Therefore, we assume thermally developing flow
and determine the Nusselt number from Nu hD
k
3.66
37.3 3.66 1 1 0.065 (D/L) Re Pr
0.04 [(D/L) Re Pr]2/3 0.065(0.3/200)
0.04[(0.3/200) 666
666 10,400
10,400]2/3 cen58933_ch08.qxd 9/4/2002 11:29 AM Page 440 440
HEAT TRANSFER Note that this Nusselt number is considerably higher than the fully developed
value of 3.66. Then, h k
Nu
D 0.145 W/m
(37.3)
0.3 m 18.0 W/m2 °C Also, As
·
m pL
Ac m DL
(0.3 m)(200 m) 188.5 m2
(888 kg/m3)[1 (0.3 m)2](2 m/s) 125.5 kg/s
4 Next we determine the exit temperature of oil from ·
Ti) exp ( hAs /m Cp)
(18.0 W/m2 · °C)(188.5 m2)
0°C [(0 20)°C] exp
(125.5 kg/s)(1880 J/kg · °C)
19.71°C Te Ts (Ts Thus, the mean temperature of oil drops by a mere 0.29°C as it crosses the
lake. This makes the bulk mean oil temperature 19.86°C, which is practically
identical to the inlet temperature of 20°C. Therefore, we do not need to reevaluate the properties.
(b) The logarithmic mean temperature difference and the rate of heat loss from
the oil are Tln
·
Q Ti Te
20 19.71
19.85°C
Ts Te
0 19.71
ln
ln
0 20
Ts Ti
hAs Tln (18.0 W/m2 °C)(188.5 m2)( 19.85°C) 6.74 104 Therefore, the oil will lose heat at a rate of 67.4 kW as it flows through the pipe
in the icy waters of the lake. Note that Tln is identical to the arithmetic mean
temperature in this case, since Ti
Te.
(c) The laminar flow of oil is hydrodynamically developed. Therefore, the friction
factor can be determined from f 64
Re 64
666 0.0961 Then the pressure drop in the pipe and the required pumping power become P
·
Wpump 3
2
2
200 m (888 kg/m )(2 m/s)
m
L
0.0961
1.14
D2
0.3 m
2
5
2
·
m P (125.5 kg/s)(1.14 10 N/m )
16.1 kW
3
888 kg/m f 105 N/m2 Discussion We will need a 16.1kW pump just to overcome the friction in the
pipe as the oil flows in the 200mlong pipe through the lake. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 441 441
CHAPTER 8 8–6 I TURBULENT FLOW IN TUBES We mentioned earlier that flow in smooth tubes is fully turbulent for Re
10,000. Turbulent flow is commonly utilized in practice because of the higher
heat transfer coefficients associated with it. Most correlations for the friction
and heat transfer coefficients in turbulent flow are based on experimental
studies because of the difficulty in dealing with turbulent flow theoretically.
For smooth tubes, the friction factor in turbulent flow can be determined
from the explicit first Petukhov equation [Petukhov (1970), Ref. 21] given as
Smooth tubes: f (0.790 ln Re 1.64) 2 104 Re 106 (865) The Nusselt number in turbulent flow is related to the friction factor through
the Chilton–Colburn analogy expressed as
Nu 0.125 f RePr1/3 (866) Once the friction factor is available, this equation can be used conveniently to
evaluate the Nusselt number for both smooth and rough tubes.
For fully developed turbulent flow in smooth tubes, a simple relation for the
Nusselt number can be obtained by substituting the simple power law relation
f 0.184 Re 0.2 for the friction factor into Eq. 8–66. It gives
Nu 0.7
Re 0.023 Re0.8 Pr1/3 Pr 160
10,000 (867) which is known as the Colburn equation. The accuracy of this equation can be
improved by...
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 Spring '10
 Ghaz
 Fluid Dynamics, Heat Transfer, TI, tube

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