Question #104331

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?

Expert's answer

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"

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