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Work done by compressed air on the piston

  1. May 19, 2016 #1
    1. The problem statement, all variables and given/known data
    A closed cylinder of 0.25 m diameter is fitted with a light friction-less piston.The piston is retained in position by a catch in the cylinder wall and the volume on one side of the piston contains air at a pressure of 750 kN/m2. The volume on the other side of the piston is evacuated. A helical spring is mounted coaxially with the cylinder in this evacuated space to give a force of 120 N on the piston in this position. The catch is released and the piston travels along the cylinder until it comes to rest after a stroke of 1.2 m. The piston is then held in its position of maximum travel by ratchet mechanism. The spring force increases linearly with the piston displacement to a final value of 5kN. Calculate the work done by the compressed air on the piston.

    2. Relevant equations
    W= FxL
    ##W=\int p.dV ##
    F= KxL
    W- work done
    F- force
    K-spring constant
    L- piston displacement length = spring compression length
    p-pressure
    V-volume
    3. The attempt at a solution

    Total work done by compressed air = workdone against the spring + workdone by expansion right ?

    work done against the spring = ##\int F.L dF ## integrated between 120N and 5kN ??
    i got this as 14991360 joules ,this is way more than the final answer i was given.
    work done by expansion = ##\int pdV## = 750x103( (pi/4)0.252x1.2) = 44178.64 joules ??

    The final answer i was given is 3.07 kJ ,my answer is nowhere close to this ,any help will be appreciated.
     
  2. jcsd
  3. May 19, 2016 #2

    andrewkirk

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    You don't actually need to do any calculations with pressure, volume or area of piston. The work is simply that of the integral of force over distance, which is
    $$\int_0^{1.2}f(x)dx$$
    Where ##f## is the linear function that satisfies ##f(0)=120## and ##f(1.2)=5000##.
    When I integrate this I get 3,071J, which is the answer in the book.

    In your post you used force as integration variable rather than distance. That should give the same answer as mine. Can you post your working and maybe we can spot what went wrong?
     
  4. May 20, 2016 #3
    Oh did i forget how to integrate ?? did you get 50002/2 -1202/2 ?
     
  5. May 20, 2016 #4

    andrewkirk

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    No I didn't get that.

    First write out the function ##f## explicitly in terms of ##x##. It's pretty straightforward from there on.
     
  6. May 21, 2016 #5
    ##f(x)= Kx ## where ##K## is the spring constant and x is the change in length of spring ?
     
  7. May 21, 2016 #6

    andrewkirk

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    No that can't be it, because the starting value of ##x## is 0 but the starting force is 120N, not 0 (which will be because the spring is already compressed beyond its equilibrium point). See post 2. There is only one linear function that satisfies the requirements set out there. It will be ##f(x)=mx+b##. What are ##m## and ##b##?
     
  8. May 23, 2016 #7
    yea m=4066.666 and b=120 , i got it thanks !!

    Now the value of K is also 4066.666 N/m ,so for a pre-compressed spring we need to add a constant term to ##f(x)= Kx ## equal to the initial force ?

    How to do it with force as an integration variable ?
     
  9. May 24, 2016 #8

    andrewkirk

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    Yes, sort of. It's a different ##x##. In Hooke's spring law, the ##x## is the displacement of the spring from equilibrium, whereas in the above problem ##x## has been used as distance from the initial position, which is not the equilibrium point. Let's use ##x'## for distance from equilibrium, so that Hooke's law is then ##F=Kx'##. The difference between the two (the initial distance from equilibrium) is ##x'-x=\frac{120N}{ 4066.7 N/m}## metres.
    First express ##x## in terms of force ##F## as ##x=(F-b)/m##. However you can't just calculate the work as ##\int_{120}^{5000} x\,dF## because that leaves out part of the shape that needs to be integrated. Draw a diagram in the number plane of the integration over ##x## and the one I just wrote to see why. So the distance (##x##) function you need to integrate over ##F## is zero for ##F\in[0,120]## and equal to ##\frac{F-b}m## thereafter.
    This will make a lot more sense to you if you draw both diagrams. Let me know if it still is not clear.
     
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