Answer to Question #104331 in Mechanics | Relativity for SADINENI KARTHIKEYA

Question

Find the useful power used in pumping 3425m cube of water per hour from a well 8 metres deep to the surface,supposing 40 percent of the horse power during pumping is wasted.What is the horse power of the engine?

Answer 1

Net power used$=\frac{mgh}{t}$ ,where $m$ is mass of water uplifted,$h$ be height and $t$ be the time required.

$t=1 h=3600 sec$

Mass uplifted$=$ Density$\times$ Volume$=1000\times 3425=3.425\times 10^6 Kg$

Net useful power$=\frac{3.425\times 10^6 \times 9.8\times 8}{3600}=7.458 \times 10^4 W$ $=\frac{7.458\times 10^4}{746}=100 hp$

And it is $100-40=60$ % of total engine power.

So,$\frac{60}{100}\times P_{net}=100$

$P_{net}=166.67 hp$