Question

a water pump pumps 3 Kg of water every second from the bottom of the well to the top of the well 50 m above if the pump uses 2000w of electrical power to accomplish the task what is the efficiency of the pump

Answer 1

70% efficiency

Step-by-step explanation:

Power out:

P=W/t

P=mgh/t

P=[(3kg)*(9.8m/s^2)*(50m)]/1sec

P=1470W

Power used: 2000W

efficiency=work out/work in

eff=1470/2000W

eff=0.735=73.5%

When accounting for significant figures:

eff=70%

Answer 2

m = mass of water pumped = 3 kg

t = time taken to pump mass "m" of water = 1 sec

h = depth from which the water is pumped = 50 m

P = electrical power used by the pump = Input power = 2000 Watt

P' = output power

U = potential energy gained by water

η = efficiency of the pump

potential energy gained by water is given as

U = mgh

output power is given as

Output power = Potential energy gained by water / time

P' = mgh/t

inserting the values

P' = (3) (9.8) (50)/1

P' = 1470 Watt

efficiency of the pump is given as

efficiency = output power/Input power

η = P'/P

η = 1470/2000

η = 0.735