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20 Jun 2008 12:16:38 IST

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532

frm work power energy

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water drawn from a well of depth 10.2 m below the open end of the tube of cross section 0.245cm²comes out of the tube with the speed of 40 m/s.calculate the minimum horse power of motor to be used for this purpose......

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Anshul 007

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Joined: 19 Jun 2008

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20 Jun 2008 21:04:35 IST

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Mass of water flowing out per second = Density x (area x speed)
work done per sec in joules =power in watts=mgh
=1000 x 40 x 0.0000245 x 10 x 10.2=99.96 watts =0.134Hp
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varun kinhal

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22 Jun 2008 13:47:24 IST

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no, the ans is 1.182hp

DEEPAK

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Joined: 20 Jun 2008

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23 Jun 2008 14:51:48 IST

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let 'm' be the mass per unit time taken out

from equation of continuity mass if water flowing per unit time=dAV

**where d=density of water(1000kg/m^3),a=area(0.245*10^-4),v=velocity(40m/sec)**

pump has to do work against gravitation&give velocity to the mass of water taken

work done=mgh=dAVtgh wheret=time

kinetic energy=1/2mv^2=1/2dAVtV^2

**total energy=P*t=work done+kinetic energy=dAVtGh+1/2dAV^3t (t cancels out)**

putting the value of all variables&unit correctly we get

p=99.96+784=883.96watt

in hp=883.96/746=1.184 hp

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