Homework Sourse1 A pump storage system generates electricity by draining wa
(1) A pump storage system generates electricity by draining water from an upper lake to a lower one when electricity is expensive and then pumping the water back up at a time when electricity is cheaper. A pump storage system is operated between two lakes with an elevation difference of H=50 m. The pump and turbine each have mechanical efficiencies of = 0.92 and the head loss in the pipes is h-3 m. For a flow rate Q-0.3 m3/s calculate the: a. power output when generating b. power required to pump water back up c. Total efficiency of the system assuming that half the time is spent generating and the other half spent pumping. 
Solution
1) POWER GENERATING:
Total head = 50 m
head loss = 3 m
head available = 50-3 = 47 m
Q = 0.3 cumec
Power generated = (yQH) x efficiency
where y = unit weight of water = 9.81 kN/m3
H = head avialable
Q = discharge
Power = 9.81 x 3 x 47 x 0.92 = 1272.55 kilo Watt
2) Back pump
Total head required = elevation difference + head loss
head loss = 3 m
Total head = 50+3 = 53 m
Q = 0.3 cumec
Power generated = (yQH) / efficiency
where y = unit weight of water = 9.81 kN/m3
H = head required required
Q = discharge
Power = (9.81 x 3 x 53) / 0.92 = 1695.42 kilo Watt
3) TOTAL EFFICIENCY
Total efficiency = Power generated / Power required to back pump
= 1272.55 / 1695.42
= 0.75
= 75 %
