2
$\begingroup$

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

27 hp $=$ 20 133 watts (joules/second). Potential energy can be calculated as $E = mgh$ where $g = ~9.8\ m/s^2$ on earth. Therefore,

$m = \frac{20133\ W}{9.8\ m/s^2} = ~3000\ kg$.

So, for example, if you use a 27 hp engine to pump hydraulic fluid. Would it output as much fluid as a 3000 kg stone block pushing down a piston at 1 meter per second? Everything else equal.

$\endgroup$
  • $\begingroup$ Actually, IIRC the horsepower is actually defined along these lines, clearly thinking of horses working mills or lifting things: I seem to recall 33 000 English pounds lifted through 1 foot in one minute (this is 747.5W, so my numbers are vaguely right). Your interpretation is fine - although you would have to make sure that the volume of fluid shoved out of the piston and its pressure were such that the upward force on the block were just a little less than the downward force from the stone - otherwise work is simply wasted. $\endgroup$ – WetSavannaAnimal Aug 18 '13 at 11:34
  • $\begingroup$ Your analysis is correct, except that someone is bound to point out that 20133/9.81 isn't equal to 3000 :-) $\endgroup$ – John Rennie Aug 18 '13 at 14:30
  • $\begingroup$ Lets see, power can be expressed as (Force)*(Speed), (Acceleration)*(Momentum), (Torque)*(Angular Speed), (Pressure)*(Flow Rate), (Voltage)*(Current Flow), ... $\endgroup$ – ja72 Aug 18 '13 at 16:02
0
$\begingroup$

1 HP = 745.7 Watts. 1 Watt is a Joule/second, which corresponds to the measure of the rate of energy.

27 HP= 20133 Joules/second. The output of a 27 horse power engine is 20133 Joules per second. The engine of a car is rated a t 50- 120 KW. This engine can make work 672 light bulbs of 30 W. Your analogy is quite right if the pressure of the piston is independent of time.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.