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Homework Help: Engine Calcs

  1. Dec 9, 2012 #1
    Hello, this is my first post on the forum and I am just looking for some help solving a problem.

    T = 520Sin(Ө) + 260Sin(2Ө)
    The engine has two working strokes per revolution

    Graph produced from equation


    Working out for power produced at 1800rpm


    Now i need to work out the the greatest fluctuation in energy per revolution.
    Can anyone point me in the right direction on where to start?

    I can provide more info if needed.
    All help greatly appreciated.

  2. jcsd
  3. Dec 9, 2012 #2


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    Hello forbes-1234. Welcome to PF !

    Please give the complete problem as it was given to you, word for word.
  4. Dec 9, 2012 #3
    It has been established that the variation in output torque T (Nm) of an engine in relation to the crank angle is given by -

    T = 520Sin(Ө) + 260Sin(2Ө)

    If the are two working strokes per revolution you are asked to calculate -

    a) The power developed by the engine at 1800 rpm
    b) The greatest fluctuation in energy per revolution

    I have completed part a) and believe my answer is correct, I integrated the equation then put the limits (in radians 0 and Pi) into the equation and subtracted the answer from the lower limit from the answer from the upper limit.
    Fairly simple integration question.

    However this next question is more of a mechanical principles question and need a place to start from.
  5. Dec 9, 2012 #4

    Ray Vickson

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    What is the formula for power in terms of T and other quantities?
  6. Dec 9, 2012 #5
    I'm not quite sure what you mean. All the information provided above is what I have. Only the equation for torque is given in relationship to crank angle in the question.
    The area under the graph is the work done per cycle.
    The amount of cycles per second is the speed divided by 60, then multiplied by 2 because there is 2 cycles per revolution.
  7. Dec 9, 2012 #6

    Ray Vickson

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    I mean: type out the formula here; I cannot really make out the attached handwritten stuff on my computer screen.
  8. Dec 9, 2012 #7
    Ok the handwritten stuff is the working out for the power produced at 1800 rpm, first off is the integrated version of the equation for torque in relationship to crank angle.
    I have then used the limits to work out the area under the graph which is equal to the work done per cycle.
    This works out to be 1040 Joules.

    At 1800 rpm the number of cycles per second = (1800/60) x 2

    Cycles per second = 60

    Therefore power developed at 1800 rpm = 60 x 1040
    = 64200 W
    = 64.2 kW

    Does this add up to you?

    Now for the calculation for the greatest fluctuation of energy per revolution.
  9. Dec 10, 2012 #8
    I'm a bit confused by the graph, after 180 degrees the cycle commences again as at 0 degrees?

    If so, wouldn't the greatest fluctuation in energy be the region covered by the steepest slope? Which appears to be from 0 - 60 degrees.

    I don't know much mechanical engineering principles so I might not be talking any sense.
  10. Dec 10, 2012 #9
    I did it slightly differently to you, although I got the same answer I didn't quite understand your method.

    If integrating from 0 to 180 degrees (half a revolution) produced 1040 J/s, then one complete revolution (360 degrees) produces 2080 J/s which is 2080/60 = 34.6667 J/m

    Multiplying by 1800 = 62,400 = 62.4kW @ 1800rpm.

    This definitely can't be a car engine right lol? Seems way too powerful.
  11. Dec 10, 2012 #10
    Marine engine I believe.

    How would you calculate the greatest fluctuation in energy then?
  12. Dec 10, 2012 #11


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    I'm not at all sure what that would mean. The energy in each revolution is the same, so there is no "fluctuation in energy per revolution". So I assume it means fluctuation in power during a revolution, but it's still unclear. Does it mean peak to trough difference or rate of change?
  13. Dec 11, 2012 #12
    I'm inclined to agree with haruspex, based on what I saw on the graph I broke it down from [0 - 60] and [60 - 150] and [150 - 180], it's rather vague to me. Maybe it would make more sense to someone who deals with internal combustion engines specifically? Googled it?
  14. Dec 11, 2012 #13
    I googled the topic when i first started this problem, I came up with a lot of information about flywheels and internal combustion engines, this is were i got my steer for the answer to the first part of the question.

    I am guessing that it means the rate of change, a quick google define on fluctuation gives - an instance of change; the rate or magnitude of change.

    So if i differentiate the equation and then equate to when torque is a a maxima this may give me an answer.

    I think I will ask my tutor for more information on the question and report back.
    There is also another two parts to the question which I may ask for help on later too, but I don't think they are concerned with calculus.
  15. Dec 11, 2012 #14


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    No, you want max rate of change of torque. Which derivative of T will be zero there?
  16. Dec 11, 2012 #15
    Thanks for your help, I appreciate your time to help me.

    Ok, so differentiate then equate to zero, to find the max rate of change of torque?
  17. Dec 11, 2012 #16


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    No, that would be for finding the max or min torque. Try again.
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