Calculating Average Power from a Windmill Pumping Water - Homework Solution

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Homework Help Overview

The problem involves calculating the average useful power extracted from a windmill that raises water from a well, with specific parameters including the depth of the well and the mass of water raised daily.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of work done by the windmill and the need to clarify the time interval for power calculation. Some question the interpretation of units and the conversion of measurements to SI units.

Discussion Status

Participants are exploring different interpretations of the problem and discussing the importance of unit conversion. Some guidance has been offered regarding the need to convert time from days to seconds and to ensure all units are consistent.

Contextual Notes

There is a mention of potential confusion regarding SI engineering unit prefixes, and participants are considering the significance of the data provided in the problem.

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Homework Statement


A windmill is used to raise water from a well. The depth of the well is 5.0 m. The windmill raises 204 kg of water every day. What is the average useful power extracted from the wind?
A. 12 mW
B. 120 mW
C. 690 mW
D. 1700 mW

Homework Equations


P = W/t

The Attempt at a Solution


Work done by windmill daily = 204 kg x 9.81 x 5.0 = 10006.2 kJ
Question doesn't provide info about time interval though
 
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The time interval is in the 3rd sentence in the problem and you correctly stated it in your answer...use it!
 
Hint :
azs8t1 said:
The windmill raises 204 kg of water every day. What is the average useful power extracted from the wind?
 
I got 204 x 9.81 x 5 = 10006.2 kJ
Power = 10006.2 kJ / 1 = 10006.2 kW = 10.0062 mW which doesn't fall under any of the options. Did I misinterpret something here
 
you need to convert everything to SI units

Kilograms
metres
seconds
watts


You need to convert a day to seconds.A very useful thing to do here, when you have multi-choice and the choices are quite different in size, is to use a mental arithmetic trick called zequals, which we were taught at school 30 years ago but is becoming quite trendy again.

Zequals is the trick of rounding everything up or down to make the maths as easy as possible to do in your head. It does not give you an exact answer, but in engineering, it is very important to have an idea of the size of an answer before you go near the calculator.So you could say the depth of the well is 10m and the mass of water is 100kg and acceleration due to gravity is 10m/s^2. You could say the approximate number of a seconds in a day is about 100,000

so,

10 x 10 x 100 / 100,000

cancel all the zeroes

1 / 10

so I would estimate, using zequals, that the answer is about 0.1 WattsAnd I would say my answer is 'more' correct than available answers, because you can (usually) only give the answer to the same number of significant figures as the least accurate piece of data supplied - which in this case is one sig figure.

It looks like SI engineering unit prefixes might be confusing too (you have equated 10 MW with 10 mW which is an easy mistake to make, so make it all Watts, and if necessary use powers of 10 until you can remember the prefixes).

So convert the answers to base units (and round up and down if it makes it easier)
A ) 0.01 W
B ) 0.1 W
C ) 1 W
D ) 2W
 
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