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%

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.

grav potential energy