Chapter 8
Internal Forced Convection
Review Problems
861
Geothermal water is supplied to a city through stainless steel pipes at a specified rate. The electric
power consumption and its daily cost are to be determined, and it is to be assessed if the frictional heating
during flow can make up for the temperature drop caused by heat loss.
Assumptions
1
The flow is steady and incompressible.
2
The entrance effects are negligible, and thus the
flow is fully developed.
3
The minor losses are negligible because of the large lengthtodiameter ratio and
the relatively small number of components that cause minor losses.
4
The geothermal well and the city are
at about the same elevation.
5
The properties of geothermal water are the same as fresh water.
6
The fluid
pressures at the wellhead and the arrival point in the city are the same.
Properties
The properties of water at 110
C are
= 950.6 kg/m
3
,
= 0.255
10
3
kg/m
s, and
C
p
=
4.229
kJ/kg
C (Table A9). The roughness of stainless steel pipes is 2
10
6
m (Table 83).
Analysis
(
a
) We take point 1 at the wellhead of geothermal resource and point 2 at the final point of
delivery at the city, and the entire piping system as the control volume. Both points are at the same
elevation
(
z
2
=
z
2
) and the same velocity (
V
1
=
V
2
) since the pipe diameter is constant, and the same
pressure (
P
1
=
P
2
). Then the energy equation for this control volume simplifies to
2
2
u
pump,
turbine
2
2
2
2
u
pump,
1
2
1
1
L
L
h
h
h
h
z
g
g
P
h
z
g
g
P
V
V
That is, the pumping power is to be used to overcome the head losses
due to friction in flow. The mean velocity and the Reynolds number
are
7
3
3
2
3
2
10
186
.
1
s
kg/m
10
255
.
0
m)
m/s)(0.60
305
.
5
)(
kg/m
6
.
950
(
V
Re
m/s
305
.
5
4
/
m)
(0.60
/s
m
1.5
4
/
V
D
D
V
A
V
m
c
m
which is greater than 10,000. Therefore, the flow is turbulent. The
relative roughness of the pipe is
10
33
.
3
m
60
.
0
m
10
2
/
6
6
D
The friction factor can be determined from the Moody chart, but to avoid the reading error, we determine
it from the Colebrook equation using an equation solver (or an iterative scheme),
f
f
f
D
f
7
6
10
187
.
1
51
.
2
7
.
3
10
33
.
3
log
0
.
2
1
Re
51
.
2
7
.
3
/
log
0
.
2
1
It gives
f
= 0.00829. Then the pressure drop, the head loss, and the required power input become
kPa
2218
kN/m
1
kPa
1
m/s
kg
1000
kN
1
2
m/s)
305
.
5
)(
kg/m
6
.
950
(
m
0.60
m
000
,
12
00829
.
0
2
2
2
3
2
V
m
D
L
f
P
kW
5118
/s
m
kPa
1
kW
1
0.65
)
kPa
2218
)(
/s
m
(1.5
3
3
motor

pump
motor

pump
u
pump,
elect
P
V
W
W
Therefore, the pumps will consume 5118 kW of electric power to overcome friction and maintain flow.
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 Spring '08
 BENARD
 Heat Transfer

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