Given, the volume of the tank = 30 
Mass of the water, m = 30*1000 kg = 30000kg 
Time, t = 15 min = 15*60 = 900s 
Height, h = 40 m 
Total work done by the pump, W = F*h= mgh = 30000*9.8*40 = 11760000 J 
Power, P = W/t = 11760000/900 = 13066.67 watt 
But efficiency = 30% = 3/10 
Required power, P = 13033.67*10/3 = 43.56 kW.

Problem:
A pump on the ground floor of a building can pump up water to fill a tank of volume 30m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30 %, how much electric power is consumed by the pump?

Solution:

To find the electric power consumed by the pump, we need to consider the work done by the pump to raise the water to the tank.

Step 1: Calculate the work done:

The work done by the pump is equal to the change in potential energy of the water. The potential energy is given by the formula:

Potential Energy = mass * gravitational acceleration * height

Since the volume of the tank is given as 30m3, we can calculate the mass of the water using the density of water, which is 1000 kg/m3.

Mass of water = volume * density
= 30 * 1000
= 30000 kg

The gravitational acceleration can be taken as 9.8 m/s2.

The height is given as 40m.

Potential Energy = 30000 * 9.8 * 40
= 11760000 J

Step 2: Calculate the power:

The power is given by the formula:

Power = Work done / Time

Since the time taken to fill the tank is given as 15 min, we need to convert it to seconds.

Time = 15 * 60
= 900 s

Power = 11760000 / 900
= 13066.67 W

Step 3: Calculate the electric power consumed:

The efficiency of the pump is given as 30%. Efficiency is defined as the ratio of useful output energy to the input energy.

Efficiency = (Useful output energy / Input energy) * 100

Since the useful output energy is the work done by the pump, and the input energy is the electric power consumed, we can rewrite the formula as:

Efficiency = (Work done / Electric power consumed) * 100

Rearranging the formula, we get:

Electric power consumed = Work done / (Efficiency / 100)

Electric power consumed = 11760000 / (30 / 100)
= 11760000 / 0.3
= 39200000 W

Therefore, the electric power consumed by the pump is 39200000 W.