Question

Water tank measuring 5 metre into 5 metre into 4 metre is at an average height of 10 metre above the ground level how much working has to be done in filling the time from a Reservoir at ground level what must be the power of an engine working at 80% efficiency if it fills the tank in 2 hours given density of water is equals to 1000 kg metre cube acceleration due to gravity G is equals to 10 metre per second square

Answer 1

Work done in pumping water = 10^7 J

the power of the engine must be 1.74 kW

-Work done in pumping the water is equal to the potential energy attained by the water when pumped fully.

-the volume of water being pumped, V = 5 * 5 * 4 =100 cubic metre

-if d is the density of water i.e., d=1000 Kg/m^3

-Work done, W = m.g.h = d.V.g.h = 1000 * 100 * 10 * 10 = 10^7 J

-power used to pump the water for 2 hours, Po= W/t =10^7 / (2*3600)

(Note that time is in seconds)

we get Po=1.39 kW

-if efficiency is n = 80%, power of the engine P=Po/n

P = (1.39 * 10^3) / 0.80

we get P=1.74 kW