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I'm trying to get a sense of how much energy a $27$ horsepower engine outputs.

$$27 \mathrm{hp} = 20133\mathrm{W}$$

Potential energy can be calculated as

$$E = mgh$$

where $g \sim9.8\mathrm{ms}^{-2}$ on earth.

$$\frac{20133\mathrm{W}}{9.8\mathrm{ms}^{-2}} \sim 3000\mathrm{kgs}^{-1}$$

So for example if you use a $27\mathrm{hp}$ engine to pump hydraulic fluid. Would it output as much fluid as a $3000 \mathrm{kg}$ stone block pushing down a piston at $1\mathrm{ms}^{-1}$?

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This is not “might be”, this is exactly physics.SE material. :) –  theUg Aug 18 '13 at 2:48
You didn’t have to. You could’ve flagged for moderator and they would migrate it. –  theUg Aug 18 '13 at 12:02
@KristofferNolgren - it looks like the question no longer needs to be migrated. Would you like to delete the question here? Also, the answer to your final question is: yes, in perfect physics imaginary problem world, the two situations are the same at the instant that you let the block fall. In reality, though, pumps and pipes have more complicated flow loses and limitations. In effect, this is an instance of "assume a spherical chicken of uniform density on a frictionless plane in a vacuum...". –  Bob Cross Aug 19 '13 at 12:31
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migrated from mechanics.stackexchange.com Aug 19 '13 at 12:53

This question came from our site for mechanics and DIY enthusiast owners of cars, trucks, and motorcycles.

marked as duplicate by Dilaton, Manishearth Aug 19 '13 at 18:05

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2 Answers

Some terms have become a little mangled here.

$mgh$ is the defining expression you require. As this is going to be a rough calcualtion lets take $g=10\textrm{ms}^{-2}$. At the moment this is time independent, you are asking about power and $mgh$ is energy. So dividing bt $t$, time taken, we get power, $P$ as $$P= \frac{mgh}{t}$$ It is hard to work out exactly what you are asking but rearranging the above euqtion for the desired number, and filling values for the rest should give you what you need. HOWEVER, this is an idealised model, real world problems often require adjustment to take into account subtleties not bothered with by physicists.

So taking $P=20\mathrm{kW}$, $g=10\textrm{ms}^{-2}$, $m=1\textrm{kg}$ and $h=3\textrm{m}$; we rearrange for $t$ and we have the time taken to shift a litre of water a height of three metres. $$t=\frac{mgh}{P}$$ $$t= \frac{1*10*3}{20*10^3}$$ $$t= 1.5 \mathrm{ms}$$ Pretty quick and probably physically unreal but illustrates the basics of solving such problems.

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You know $m=1000$ and not $m=1$ in the original post. –  ja72 Aug 19 '13 at 15:19
Yeah, but I wanted to give a general approach to answering such problems, the 'physics' as I see it rather than the engineering of computing specific numbers... –  Nic Aug 19 '13 at 15:41
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Power is force times speed so you are correct to say

$$ 27\;{\rm hp} = 20130\,{\rm W} = \left( 2.05\,{\rm m s^{-1}} \right) \left(1000\,{\rm kg}\right) \left( 9.81\,{\rm m s^{-2}} \right) $$

Can can also say power is equal to momentum times acceleration

$$ P = \left(m \dot{x} \right) \left( \ddot{x}+g \right) $$ $$ \ddot{x} = \frac{P}{m \dot{x}} - g $$

So a $P=27\,{\rm hp}$ engine accelerating a $m=1000\,{kg}$ load at $\dot{x}=3\,{m/s}$ will accelerate at

$$ \ddot{x} = \frac{(20130\,{\rm W})}{(1000{\rm kg}) (3 {\rm m/s})} - 9.81{\rm m/s^2} = -3.1{\rm m/s^2} $$

which is a deceleration. So the speed is not sustainable with this horsepower provided.

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