Question 10.5: The 1750 rpm pump with 7 in. impeller of Fig. 10-11 is opera......
The 1750 rpm pump with 7 in. impeller of Fig. 10-11 is operating in a system as
shown as point 1 of Fig. 10-16. It is desired to reduce the pump speed until the flow rate is 100 gpm. Find the new pump head, shaft power, and efficiency.
From the system characteristic it may be observed that the pump must produce 25 ft of head at a flow rate of 100 gpm. This is one point on the new pump characteristic. The new pump speed can be found from either Eq. 10-22 or 10-23. Using Eq. 10-22,
\dot{Q}_{n} = \dot{Q}_{o} \frac{rpm_{n}}{rpm_{o}} (10-22)
H_{pn} = H_{po} \left[\frac{rpm_{n}}{rpm_{o}}\right]^{2} (10-23)
rpm_{n} = rpm_{o} (\dot{Q}_{n}/\dot{Q}_{o})
= 1750 (100/130) = 1346
The new shaft power is given by Eq. 10-24 with \dot{W}_{so} = 2.1 hp from Fig. 10-11:
\dot{W}_{sn} = \dot{W}_{so} \left[\frac{rpm_{n}}{rpm_{o}}\right]^{3} (10-24)
\dot{W}_{sn} = 2.1 (1346/1750)³ = 0.96 hp
The pump efficiency could be recalculated using Eq. 10-19. However, it may be deduced from the affinity laws that the efficiency will remain constant at about 69.4 percent. Thus
\eta_{p} = \frac{\dot{W}}{\dot{W}_{s}} = \frac{\dot{m} w}{\dot{W}_{s}} = \frac{ρ\dot{Q}w}{\dot{W}_{s}} (10-19)
\frac{\eta_{pn}}{\eta_{po}} = \frac{\dot{Q}_{n} H_{pn}/ \dot{W}_{sn}}{\dot{Q}_{o} H_{po}/ \dot{W}_{so}} = 1