Solve the following examples.

An electric pump has 2 kW power. How much water will the pump lift every minute to a height of 10 m?

Note: Here, electric pump will consume electrical energy , convert it to mechanical energy which will do work on water in raising its height and increasing potential energy of water, i.e. finally all energy is converted to potential energy of water.

Power of the pump is 2 kW i.e. 2000 W of Electrical energy is being consumed every second by motor, we have to find water lifted each minute i.e. energy consumed by motor in 1 minute will be equal to potential energy gained by water in raising to a height of 10 m

We know energy consumed is given as

E = P × t

Where E is the energy consumed by Pump in J, P is the power of Pump in Watt (W), t is the time for which Pump is operated in seconds (s)

Here power of the pump is

P = 2000 W

We want to obtain energy consumed in one 1 minute, so time is

t = 1 min = 60 s

so the energy consumed is

E = 2000 W × 60 s = 120000 J

Now this much energy will be transferred as potential energy of some amount of water

We know the Relation for Potential energy of an object of mass m due to earth’s gravity(g is acceleration due to gravity) held at an height h above ground is

P.E. = mgh

Value of acceleration due to gravity,

g = 9.8 m/s²

Suppose mass of water transferred is m

Height to which water is transferred is

h = 10m

so potential energy of transferred water is

P.E. = m × 9.8 m/s² × 10m = 98m J

Now equating potential energy of water of mass m equal to energy supplied by pump

P.E. = E

98m = 120000

Solving we get mass of water transferred as

m = 120000/98 Kg = 1224.489 Kg

i.e. around 1224.5 Kg of water is being lifted to height of 10 m by the pump.