An electric motor drives a pump that raises 1,000 litres of water each minute, to a tank 20m above ground level. Calculate the power that the motor must provide if the pump is only:
50% efficient:
80% efficient:
+36916 Similar to the other water pumping problem
Mass of 1000 liters of water = 1000 kg
F = mg = 1000 kg * 9.81 m/s2 = 9810 N
W = F * d = 9810 N * 20 m = 196200 j
Power = work /time = 196200j / 60 sec = 3270 watts
50% efficient means 50% is wasted..... it will take twice as much power to pump the water = 2 * 3270 = 6540 watts
80% efficient : 3270 / .8 = 4087.5 watts
+36916 Similar to the other water pumping problem
Mass of 1000 liters of water = 1000 kg
F = mg = 1000 kg * 9.81 m/s2 = 9810 N
W = F * d = 9810 N * 20 m = 196200 j
Power = work /time = 196200j / 60 sec = 3270 watts
50% efficient means 50% is wasted..... it will take twice as much power to pump the water = 2 * 3270 = 6540 watts
80% efficient : 3270 / .8 = 4087.5 watts
