Calculating Average Power from a Windmill Pumping Water - Homework Solution

In summary: WA ) 10 mWB ) 100 mWC ) 1000 mWD ) 2000 mWso it looks like B is the best answerNote if you had done the whole question in SI units, and done it with zequals, then the answer would have been 0.1 W, which would have been exactly the same as answer B in mW.
  • #1
azs8t1
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0

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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  • #2
Welcome to PF!

The time interval is in the 3rd sentence in the problem and you correctly stated it in your answer...use it!
 
  • #3
Hint :
azs8t1 said:
The windmill raises 204 kg of water every day. What is the average useful power extracted from the wind?
 
  • #4
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
 
  • #5
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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1. What is the definition of average useful power?

Average useful power is a measure of the rate at which energy is converted into a useful form, such as electrical or mechanical energy. It is calculated by dividing the total energy output by the time it takes to produce that output.

2. How is average useful power different from instantaneous power?

Instantaneous power is the power at a specific moment in time, while average useful power is the average power over a period of time. Average useful power takes into account fluctuations in power output, while instantaneous power only reflects the power output at a single point in time.

3. What are some common units of measurement for average useful power?

The most common unit for average useful power is watts (W). Other units include horsepower (hp) and kilowatts (kW), depending on the context and the size of the power being measured.

4. How is average useful power used in real-world applications?

Average useful power is used in a wide range of applications, from measuring the efficiency of engines and machines to evaluating the performance of renewable energy sources. It is also used in the design and development of electrical systems and appliances.

5. How can average useful power be increased?

There are several ways to increase average useful power, such as optimizing the design and efficiency of a system, using more powerful and efficient components, and reducing energy losses. In some cases, changing the type of energy source being used can also increase average useful power.

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