a 3600-r/min speed, the specific gravity of the liquid is 0.52, and the total available ex-
haust head is 20 ft (6.1 m). Analyze the cavitation potential and operating characteristics
at an 80 percent flow rate.
Calculation Procedure:
- Choose the number of stages for the pump
Search of typical centrifugal-pump data shows that a head of 1290 ft (393.1 m) is too
large for a single-stage pump of conventional design. Hence, a two-stage pump will be
the preliminary choice for this application. The two-stage pump chosen will have a design
head of 645 ft (196.6 m) per stage.
- Compute the specific speed of the pump chosen
Use the relation Ns = pump rpm(Q)°^5 ///°^75 , where Ns = specific speed of the pump; rpm =
r/min of pump shaft; Q = pump capacity or flow rate, gal/min; H = pump head per stage,
ft. Substituting, we get N 3 = 3600(1500)°^5 /(645)^075 = 1090. Note that the specific speed
value is the same regardless of the system of units used—USCS or SI.
- Convert turbine design conditions to pump design conditions
To convert from turbine design conditions to pump design conditions, use the pump
manufacturer's conversion factors that relate turbine best efficiency point (bep) per-
formance with pump bep performance. Typically, as specific speed Ns varies from 500
to 2800, these bep factors generally vary as follows: the conversion factor for capacity
(gal/min or L/min) CQ, from 2.2 to 1.1; the conversion factor for head (ft or m) C 119
from 2.2 to 1.1; the conversion factor for efficiency CE9 from 0.92 to 0.99. Applying
these conversion factors to the turbine design conditions yields the pump design condi-
tions sought.
At the specific speed for this pump, the values of these conversion factors are deter-
mined from the manufacturer to be CQ = 1.24; Cn = 1.42; Q = 0.967.
Given these conversion factors, the turbine design conditions can be converted to the
pump design conditions thus: Qp = QJCQ, where Qp = pump capacity or flow rate,
gal/min or L/min; Qt = turbine capacity or flow rate in the same units; other symbols are
as given earlier. Substituting gives Qp = 1500/1.24 = 1210 gal/min (4580 L/min).
Likewise, the pump discharge head, in feet of liquid handled, is Hp = HJCH. So Hp =
645/1.42 = 454 ft (138.4m).
- Select a suitable pump for the operating conditions
Once the pump capacity, head, and rpm are known, a pump having its best bep at these
conditions can be selected. Searching a set of pump characteristic curves and capacity ta-
bles shows that a two-stage 4-in (10-cm) unit with an efficiency of 77 percent would be
suitable.
- Estimate the turbine horsepower developed
To predict the developed hp, convert the pump efficiency to turbine efficiency. Use the
conversion factor developed above. Or, the turbine efficiency Et = Ep CE = (0.77)(0.967)
= 0.745, or 74.5 percent.
With the turbine efficiency known, the output brake horsepower can be found from
bhp = QfHfEfS/3960, where s = fluid specific gravity; other symbols as before. Substitut-
ing, we get bhp = 1500(1290)(0.745)(0.52)/3960 = 198 hp (141 kW).
- Determine the cavitation potential of this pump
Just as pumping requires a minimum net positive suction head, turbine duty requires a net
positive exhaust head. The relation between the total required exhaust head (TREH) and
turbine head per stage is the cavitation constant ar = TREH///. Figure 5 shows ay. vs. Ns
