Calculating the Required Power for a Pump

1 ) pump horse power calculation

I will repeat the same answer for the same question asked around one month before

HP = ω × Q × H / 75 ×ξ

Where; ω is water density = 1000 Kg / CU meter

Q is pump discharge in Cubic meter /second

H is total head In meter

ξ is pump efficiency

example: for water pump with discharge of 60 CU. meter per hour at a total head 40 meter

HP = ω × Q × H / 75 ×ξ

ω = 1000 Kg / Cu meter

Q = 60 / 3600 =0.01667 Cu meter / second

H = 40 m

Consider pump efficiency 60 %

HP = 1000 ×0.01667 × 40 × 100 / 75 × 60 = 14.8 HP

2) For air receiver may I differ with bottom connection for outgoing air line?

This because compressed air contain water vapor which tends to condensate in air receiver due to expansion and big tank surface. Also some oil vapor escape from compressor to air receiver.

Condensate water with oil drops rejected out from the lowest point for tank bottom either manually or through automatic drain.

In case of connecting out going pipe to this bottom point, all condensate water, oil with any other deposits will discharged to air service line. In such case water separator, oil separator filter and air filter will be all under worse working condition, more frequently change for filters and overloading for air cooler.

Almost air receiver has incoming air line at the down third of the tank not at bottom also out going line at the upper third.

In all condition I think it will not affect air pressure but quality of product outgoing air will affected.