The annual cost of operation with an electric motor drive will be
81 0:746 0:05 60000:90 ¼ $20;142
(0.746 is the conversion factor from hp to kW.)
If steam is used, the annual cost of operation will be
3 81 2545 6000 0:85 0:90 106 ¼ $4595
(2545 Btu=h ¼ 1 hp; 0.85 is the boiler efficiency; 0.95 is the mechanical efficiency.) Hence the savings in cost of operation is 20,142 7 4545 ¼ $15,547=year.
Depending on the difference in investment between the two drives, payback can be worked out. In the calculation above it was assumed that the backpressure steam was used for process. If it was wasted, the economics may not work out the same way.
9.09a
Q:
How does the specific gravity or density of liquid pumped affect the BHP, flow, and head developed?
A:
A pump always delivers the same flow in gpm (assuming that viscosity effects can be neglected) and head in feet of liquid at any temperature. However, due to changes in density, the flow in lb=h, pressure in psi, and BHP would change. A variation of Eq. (9) is
BHP ¼ |
q DP |
¼ |
W DP |
ð10Þ |
1715Zp |
857;000Zps |
where
q ¼ liquid flow; gpm
W ¼ liquid flow; lb=h s ¼ specific gravity
DP ¼ pressure developed; psi
H ¼ head developed; ft of liquid
Also,
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Example
If a pump can develop 1000 gpm of water at 40 F through 1000 ft, what flow and head can it develop when the water is at 120 F? Assume that pump efficiency is 75% in both cases.
Solution. s1 at 40 F is 1 (from the steam tables; see the Appendix). s2 at 120 F is 0.988.
DP1 ¼ 1000 2:131 ¼ 433 psi
From Eq. (11),
433
BHP1 ¼ 1000 0:75 1715 ¼ 337 hp
W1 ¼ 500q1s1 ¼ 500 1000 1 ¼ 500;000 lb=h At 120 F,
DP2 ¼ 1000 02::98831 ¼ 427 psi
427
BHP2 ¼ 1000 0:75 1715 ¼ 332 hp
W2 ¼ 500 0:988 1000 ¼ 494;000 lb=h
If the same W is to be maintained, BHP must increase.
9.09b
Q:
How does the temperature of water affect pump power consumption?
A:
The answer can be obtained by analyzing the following equations for pump power consumption. One is based on flow in gpm and the other in lb=h.
where
Q ¼ flow; gpm
H ¼ head; ft of water s ¼ specific gravity
Zp ¼ efficiency
Copyright © 2003 Marcel Dekker, Inc.
In boilers, one would like to maintain a constant flow in lb=h, not in gpm, and at a particular pressure in psi. The relationships are
Q ¼ |
W |
|
and |
H ¼ 2:31 |
DP |
|
500s |
|
s |
where |
|
|
|
|
W ¼ flow; lb=h |
|
|
|
DP ¼ pump differential; psi |
|
|
Substituting these terms into (1), we have |
|
|
BHP ¼ W |
DP |
|
ð13Þ |
857;000Zps |
|
As s decreases with temperature, BHP will increase if we want to maintain the flow in lb=h and head or pressure in psi. However, if the flow in gpm and head in ft should be maintained, then the BHP will decrease with a decrease in s, which in turn is lower at lower temperatures.
A similar analogy can be drawn with fans in boiler plants, which require a certain amount of air in lb=h for combustion and a particular head in in. WC.
9.10
Q:
A centrifugal pump delivers 100 gpm at 155 ft of water with a 60 Hz supply. If the electric supply is changed to 50 Hz, how will the pump perform?
A:
For variations in speed or impeller size, the following equation applies:
|
|
2 |
|
|
2 |
|
|
|
q1 |
|
N1 |
pH1 |
|
|
q |
|
¼ |
N |
|
¼ pH |
|
ð14Þ |
where |
|
|
|
|
|
|
|
|
|
|
|
|
|
2 |
|
q ¼ pump flow; gpm
H ¼ head developed; ft
N ¼ speed; rpm
Use of Eq. (14) gives us the head and the flow characteristics of a pump at different speeds. However, to get the actual operating point, one must plot the
Copyright © 2003 Marcel Dekker, Inc.
new head versus flow curve and note the point of intersection of this curve with the system resistance curve. In the case above,
q2 ¼ 100 5060 ¼ 83 gpm H2 ¼ 155 5060 2¼ 107 ft
In this fashion, the new H versus q curve can be obtained. The new operating point can then be found.
9.11
Q:
How does the performance of a pump change with the viscosity of the fluids pumped?
A:
The Hydraulic Institute has published charts that give correction factors for head, flow, and efficiency for viscous fluids when the performance with water is known (see Figs. 9.3a and 9.3b).
Example
A pump delivers 750 gpm at 100 ft head when water is pumped. What is the performance when it pumps oil with viscosity 1000 SSU? Assume that efficiency with water is 82%.
Solution. In Fig. 9.3b, go up from capacity 750 gpm to cut the head line at 100 ft and move horizontally to cut viscosity at 1000 SSU; move up to cut the various correction factors.
CQ ¼ 0:94; CH ¼ 0:92; CE ¼ 0:64
Hence the new data are
q ¼ 0:94 750 ¼ 705 gpm
H ¼ 0:92 100 ¼ 92 ft
Zp ¼ 0:64 92 ¼ 52%
The new H versus q data can be plotted for various flows to obtain the characteristic curve. The operating point can be obtained by noting the point of intersection of the system resistance curve with the H versus q curve. CQ, CH , and CE are correction factors for flow, head, and efficiency.
Copyright © 2003 Marcel Dekker, Inc.
FIGURE 9.3a Viscosity corrections. (Courtesy of Hydraulic Institute=Gould Pump Manual.)
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FIGURE 9.3b Determination of pump performance when handling viscous liquids. (Courtesy of Hydraulic Institute=Gould Pump Manual.)
Copyright © 2003 Marcel Dekker, Inc.
9.12
Q:
What is the temperature rise of water when a pump delivers 100 gpm at 1000 ft at an efficiency of 60%?
A:
The temperature rise of fluids through the pump is an important factor in pump maintenance and performance considerations and must be limited. The recirculation valve is used to ensure that the desired flow goes through the pump at low load conditions of the plant, thus cooling it.
From energy balance, the friction losses are equated to the energy absorbed by the fluid.
DT ¼ ðBHP theoretical powerÞ |
2545 |
ð15aÞ |
|
WCp |
where |
|
|
DT ¼ temperature rise of the fluid; F
BHP ¼ brake horsepower
W ¼ flow of the fluid; lb=h
Cp ¼ specific heat of the fluid; Btu=lb F
For water, Cp ¼ 1.
From Eq. (9),
H
BHP ¼ W Zp 3600 550
where Zr is the pump efficiency, fraction. Substituting into Eqs. (15a) and (9) and simplifying, we have
DT ¼ H |
1=Zp 1 |
|
ð15bÞ |
|
778 |
|
|
If H ¼ 100 ft of water and Zr ¼ 0.6, then |
T |
¼ |
1000 |
|
1:66 1 |
|
1 F |
D |
|
778 |
|
|
9.13
Q:
How is the minimum recirculation flow through a centrifugal pump determined?
Copyright © 2003 Marcel Dekker, Inc.
A:
Let us illustrate this with the case of a pump whose characteristics are as shown in Fig. 9.4. We need to plot the DT versus Q characteristics first. Note that at low flows when the efficiency is low, we can expect a large temperature rise. At 100 gpm, for example,
Zp ¼ 0:23 and H ¼ 2150 ft
Then
DT ¼ 2150 1=0:23 1 ¼ 9 F 778
In a similar fashion, DT is estimated at various flows. Note that DT is higher at low flows owing to the low efficiency and also because of the lesser cooling capacity.
The maximum temperature rise is generally limited to about 20 F, depending on the recommendations of the pump manufacturer. This means that at least 40 gpm must be circulated through the pump in this case. If the load is only 30 gpm, then depending on the recirculation control logic, 10–70 gpm could be recirculated through the pump.
FIGURE 9.4 Typical characteristic curve of a multistage pump also showing temperature rise versus capacity.
Copyright © 2003 Marcel Dekker, Inc.
9.14
Q:
What is net positive suction head (NPSH), and how is it calculated?
A:
The NPSH is the net positive suction head in feet absolute determined at the pump suction after accounting for suction piping losses (friction) and vapor pressure. NPSH helps one to check if there is a possibility of cavitation at pump suction. This is likely when the liquid vaporizes or flashes due to low local pressure and collapses at the pump as soon as the pressure increases. NPSH determined from pump layout in this manner is NPSHa (NPSH available). This will vary depending on pump location as shown in Fig. 9.5.
FIGURE 9.5 Calculation of system NPSH available for typical suction conditions.
Copyright © 2003 Marcel Dekker, Inc.
NPSHr (NPSH required) is the positive head in feet absolute required to overcome the pressure drop due to fluid flow from the pump suction to the eye of the impeller and maintain the liquid above its vapor pressure. NPSHr varies with pump speed and capacity. Pump suppliers generally provide this information.
NPSHa can be determined by a gauge reading at pump suction:
NPSHa ¼ PB VP PG þ VH |
ð16Þ |
where
VH ¼ velocity head at the gauge connection; ft
PG ¼ pressure gauge reading; converted to ft
VP ¼ vapor pressure; ft absolute
PB ¼ barometric pressure; ft ðif suction is atmosphericÞ
To avoid cavitation, NPSHa must be greater than NPSHr.
9.15
Q:
Does the pump suction pressure change NPSHa?
A:
NPSHa is given by
NPSHa ¼ Ps þ H VP Hf |
ð17Þ |
where
Ps ¼ suction pressure; ft of liquid
H ¼ head of liquid; ft
VP ¼ vapor pressure of the liquid at operating temperature; ft
Hf ¼ friction loss in the suction line; ft
For saturated liquids, VP Ps, so changes in suction pressure do not significantly change NPSHa.
Example
Determine the NPSHa for the system shown in Fig. 9.5b when H ¼ 10 ft, Hf ¼ 3 ft, and VP ¼ 0.4 psia (from the steam tables). Assume that the water has a density of 62 lb=cu ft.
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