Sewage at a certain pumping station is raised vertically by 5.49 m at the rate of 1 890 000 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. (b) assume an electric motor continuously operating with average power 5.90 kw runs the pump. find its efficiency

a)the output mechanical power of the lift station is 1.237kW

b)the eficiency is 21%

Explanation:

Hello!

To solve this exercise we must follow the following steps.

1. Find the hydraulic power, this refers to the ideal energy required to move the indicated amount of wastewater in the indicated time.

It is calculated as the product of the flow, by height, by gravity by the density of the fluid. remember to use conversion factors

the output mechanical power of the lift station is 1.237kW

2. Remember that efficiency is the ratio between ideal hydraulic power and actual engine power.

the eficiency is 21%

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j