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Unformatted text preview: ulus can be
treated as a noncircular duct with a hydraulic diameter of Dh Do Di. The
Nusselt number in this case can be determined from a suitable turbulent flow
relation such as the Gnielinski equation. To improve the accuracy of Nusselt
numbers obtained from these relations for annular flow, Petukhov and Roizen
(1964, Ref. 22) recommend multiplying them by the following correction factors when one of the tube walls is adiabatic and heat transfer is through the
other wall:
0.86 Di
Do 0.16 Fi 0.86 Di
Do 0.16 Fo (a) Finned surface Fin D2)/4
i
Di) Annular flow is associated with two Nusselt numbers—Nui on the inner
tube surface and Nuo on the outer tube surface—since it may involve heat
transfer on both surfaces. The Nusselt numbers for fully developed laminar
flow with one surface isothermal and the other adiabatic are given in
Table 8–4. When Nusselt numbers are known, the convection coefficients for
the inner and the outer surfaces are determined from Nuo 0
0.05
0.10
0.25
0.50
1.00 Dh (outer wall adiabatic) (877) (inner wall adiabatic) (878) Heat Transfer Enhancement
Tubes with rough surfaces have much higher heat transfer coefficients than
tubes with smooth surfaces. Therefore, tube surfaces are often intentionally roughened, corrugated, or finned in order to enhance the convection
heat transfer coefficient and thus the convection heat transfer rate (Fig. 8–27).
Heat transfer in turbulent flow in a tube has been increased by as much as cen58933_ch08.qxd 9/4/2002 11:29 AM Page 445 445
CHAPTER 8 400 percent by roughening the surface. Roughening the surface, of course,
also increases the friction factor and thus the power requirement for the pump
or the fan.
The convection heat transfer coefficient can also be increased by inducing
pulsating flow by pulse generators, by inducing swirl by inserting a twisted
tape into the tube, or by inducing secondary flows by coiling the tube.
EXAMPLE 8–4 Pressure Drop in a Water Pipe Water at 60°F (
62.36 lbm/ft3 and
2.713 lbm/ft h) is flowing steadily
in a 2in.diameter horizontal pipe made of stainless steel at a rate of 0.2 ft3/s
(Fig. 8–28). Determine the pressure drop and the required pumping power input for flow through a 200ftlong section of the pipe. 0.2 ft3/s
water 2 in. 200 ft FIGURE 8–28
Schematic for Example 8–4. SOLUTION The flow rate through a specified water pipe is given. The pressure
drop and the pumping power requirements are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The entrance effects
are negligible, and thus the flow is fully developed. 3 The pipe involves no components such as bends, valves, and connectors. 4 The piping section involves
no work devices such as a pump or a turbine.
Properties The density and dynamic viscosity of water are given by
62.36
lbm/ft3 and
2.713 lbm/ft h 0.0007536 lbm/ft s, respectively.
Analysis First we calculate the mean velocity and the Reynolds number to
determine the flow regime:
·
V
Ac
Re ·
V
0.2 ft3/s
9.17 ft/s
2
D /4
(2/12 ft)2/4
D (62.36 lbm/ft3)(9.17 ft/s)(2/12 ft) 3600 s
2.713 lbm/ft · h
1h 126,400 which is greater than 10,000. Therefore, the flow is turbulent. The relative
roughness of the pipe is /D 0.000007 ft
2/12 ft 0.000042 The friction factor corresponding to this relative roughness and the Reynolds
number can simply be determined from the Moody chart. To avoid the reading
error, we determine it from the Colebrook equation: 1 2.0 log f /D
3.7 2.51
Re f → 1
f 2.0 log 0.000042
3.7 2.51
126,400 f Using an equation solver or an iterative scheme, the friction factor is determined to be f 0.0174. Then the pressure drop and the required power input
become
2 P ·
Wpump 200 ft (62.36 lbm/ft3)(9.17 ft/s)2
1 lbf
L
0.0174
D2
2
2/12 ft
32.2 lbm · ft/s2
1700 lbf/ft2 11.8 psi
·
1W
V P (0.2 ft3/s)(1700 lbf/ft2)
461 W
0.737 lbf · ft/s
f cen58933_ch08.qxd 9/4/2002 11:29 AM Page 446 446
HEAT TRANSFER Therefore, power input in the amount of 461 W is needed to overcome the frictional losses in the pipe.
Discussion The friction factor also could be determined easily from the explicit
Haaland relation. It would give f
0.0172, which is sufficiently close to
0.0174. Also, the friction factor corresponding to
0 in this case is 0.0171,
which indicates that stainless steel pipes can be assumed to be smooth with
negligible error. EXAMPLE 8–5
·
qs = constant
15°C Water
D = 3 cm 65°C Heating of Water by Resistance Heaters in a Tube Water is to be heated from 15°C to 65°C as it flows through a 3cminternaldiameter 5mlong tube (Fig. 8–29). The tube is equipped with an electric resistance heater that provides uniform heating throughout the surface of the
tube. The outer surface of the heater is well insulated, so that in steady operation all the heat generated in the heater is transferred to the water in the tube.
If the system is to provide hot water at a rate of 10 L/min, determine the power
rating of the resistance heater. Also, estimate the inner surface temperature of
the pipe at the exit. 5...
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