Question

A water pump lifts water from 10 m below the ground. Water is pumped at a rate of 30 kg/minute with negligible velocity. Calculate the minimum horsepower that the engine should have to do this.

Answer 1

The minimum horsepower is therefore 5.36 x $10^{-2}$ hp.

Explanation:

Step 1:

Given values

Mass, pump Flow rate = 30 kg / min. = 30 kg / 60 s. = 0.5 kg/s.

Height h = 10 m.

We know that pump water has an insignificant speed, then K.E is zero.  

Pumping work to increase water level by more than 10 m.  

Now water weight= mg

= 0.5 × 9.8  

= 4.9 N.

Step 2:

Now power = mgh

P = 4.9 × 10 m

P = 49 W.

Now we know 1 hp = 746 W.

Step 3:

So minimum power = 49/746

= 5.36 ×$10^{-3}$

= 0.0536

= 5.36 × 10⁻² hp.

The minimum horsepower is therefore 5.36 x $10^{-2}$ hp.

Answer 2

$\huge{\boxed{\mathcal\pink{\fcolorbox{red}{yellow}{Answer}}}}$

$5.36 \times {10}^ - {2} hp$

hope it help ☺☺