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Old May 20th 2009, 10:18 AM   #1
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Thumbs up 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?
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Old May 20th 2009, 10:37 AM   #2
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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.
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Old May 20th 2009, 10:54 AM   #3
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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.
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Old May 20th 2009, 11:10 AM   #4
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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
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Old May 20th 2009, 11:15 AM   #5
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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.
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Old May 20th 2009, 03:10 PM   #6
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Originally Posted by physicsquest View Post
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
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Old May 20th 2009, 03:47 PM   #7
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Originally Posted by physicsquest View Post
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.
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Old May 20th 2009, 08:25 PM   #8
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Originally Posted by angelica View Post
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
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Old May 20th 2009, 09:49 PM   #9
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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.
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Old May 20th 2009, 10:22 PM   #10
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Originally Posted by angelica View Post
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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