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 May 20th 2009, 11:18 AM #1 Junior Member   Join Date: Feb 2009 Posts: 20 Power and Work Problem A farmer is trying to decide whether to buy a traditional windmill or an 84-hp (horsepower) electric pump to provide water from a well that is 45 m deep. (1 hp  746 W) a. Calculate the work necessary to bring 1.0 kg of water to the surface. b. Calculate the flow rate of the water from the electric pump. (1.0 kg of water  1000 cm3 volume) c. It takes 98 kW of electricity to run the electric pump for 1.0 h. What is the efficiency of this pump? d. With a steady wind, the windmill is guaranteed to pump water at a rate of 0.050 m3/s. What horsepower electric pump would be equivalent to this windmill?
 May 20th 2009, 11:37 AM #2 Physics Team   Join Date: Feb 2009 Posts: 1,425 a. Calculate the work necessary to bring 1.0 kg of water to the surface. The work done is equal to the increase in P.E.
 May 20th 2009, 11:54 AM #3 Physics Team   Join Date: Feb 2009 Posts: 1,425 b. Calculate the flow rate of the water from the electric pump. (1.0 kg of water 1000 cm3 volume) . 84 HP = 84 x 746 W or joules/sec .Thus the energy expended per sec is 84 x 746 .The energy required to lift I kg of water thru 45m = 1 x 9.8 x 45 So the no of kgs lifted per sec = 84 x 746 / 1 x 9.8 x 45 1kg corresponds to a volume of 1000 cm^3. Find the vol for 84 x 746 / 1 x 9.8 x 45 . This volume per sec = flow rate.
 May 20th 2009, 12:10 PM #4 Physics Team   Join Date: Feb 2009 Posts: 1,425 c. It takes 98 kW of electricity to run the electric pump for 1.0 h. What is the efficiency of this pump. i.e. 98000 W to run for 60 x 60 = 3600 secs i.e. actual energy required = 98000 x 3600. Ideally the energy required would have been 86 x 746 x 3600 So efficiency = ideal / actual = 86 x 746 x 3600 / 98000 x 3600 = 0.655
 May 20th 2009, 12:15 PM #5 Physics Team   Join Date: Feb 2009 Posts: 1,425 d. With a steady wind, the windmill is guaranteed to pump water at a rate of 0.050 m3/s. What horsepower electric pump would be equivalent to this windmill? Vol flow rate x density = mass /sec (mass / sec) x g x h will give you the power in watts. Then use 1hp = 746 W to find the value in hp.
May 20th 2009, 04:10 PM   #6
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 Originally Posted by physicsquest a. Calculate the work necessary to bring 1.0 kg of water to the surface. The work done is equal to the increase in P.E.
I dont understand. can u explain how to get to that?
Thanks

May 20th 2009, 04:47 PM   #7
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 Originally Posted by physicsquest b. Calculate the flow rate of the water from the electric pump. (1.0 kg of water 1000 cm3 volume) . 84 HP = 84 x 746 W or joules/sec .Thus the energy expended per sec is 84 x 746 .The energy required to lift I kg of water thru 45m = 1 x 9.8 x 45 So the no of kgs lifted per sec = 84 x 746 / 1 x 9.8 x 45 1kg corresponds to a volume of 1000 cm^3. Find the vol for 84 x 746 / 1 x 9.8 x 45 . This volume per sec = flow rate.

My answer is 142 kg. How do I calculate for flow rate? I am a bit confused.

May 20th 2009, 09:25 PM   #8
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 Originally Posted by angelica My answer is 142 kg. How do I calculate for flow rate? I am a bit confused.
142 kg/sec =142*1000 cm^3/sec

 May 20th 2009, 10:49 PM #9 Physics Team   Join Date: Feb 2009 Posts: 1,425 The work done is equal to the increase in P.E. I dont understand. can u explain how to get to that? Thanks Increase in P.E. = m g h where you treat the P.E. at the bottom as zero. I think you should revise and brush up on your fundamental concepts a bit. I deliberately did not go into detail here though it is implicit in one of my subsequent posts on the same question.
May 20th 2009, 11:22 PM   #10
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 Originally Posted by angelica I dont understand. can u explain how to get to that? Thanks
The work done force times distance. In order to raise a bucket of mass m you have to exert a force to the bucket which is equal to the weight of the bucket. The distance is the height you have to raise the bucket. Let h be depth of well. let F = W = mg = weight of bucket. Then the work = force * distance = (Mg)(h) = mgh. But this is simply the increase in the potential energy of the bucket!

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