Question

Sewage at a certain pumping station is raised vertically by 5.48 m at the rate of 1 850 000 liters each day. The sewage, of density 1 050 kg/m³, enters and leaves the pump at atmospheric pressure and through pipes of equal diameter.

(a) Find the output mechanical power of the lift station.
W

(b) Assume an electric motor continuously operating with average power 4.10 kW runs the pump. Find its efficiency.
%

Answer 1

Given is:-

height h=5.48m, density of sewage

volume of the sewage is = 1850000 liters/day

now we know that

Power = Energy/ Time

and

Energy(Work) = Force Distance

The force is the weight of the water. Weight of the water is density () times gravity times the flow rate (Q) (converted to m^3 per s) and the distance moved is the vertical height (h) the water is lifted.

Thus Power is

W OR

Therefore the output mechanical power of the lift station is 1209 W (part-a)

Part-b

now the efficiency is

we know that output power is 1209W or 1.209KW

and the input power is 4.10KW

Thus the efficiency is

Or 29%

Therefore the efficiency of the electric motor is 29% which is not very good.