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Does the pressure take into consideration the area of the cylinder?

  1. Jul 15, 2009 #1
    Not a question to be solved, just an understanding of interpretation required.

    In thermodynamics when a question is asked regarding cylinder pressures, so say a cylinder contained a gas pressure of 600Kpa, when a question asks for you to conclude say How Much A Spring is Compressed, does the pressure take into consideration the area of the cylinder?

    Now the above may not sound briliant, but if the compressed gas was pushing on a piston which had a spring attached, so the spring constant was say 4.8 kN/m, does the following solution seem to be worked out correctly.

    PA - P atm A = K spring change in x

    (600 000 - 100 000)pie x 0.05 squared = 4800 change in x = 0.818m

    I know the maths is ok, I am concerned about my interpretation of the equation above, and whether PA should have the area of the cylinder included or is already included in the pressure?

    Please advise if you can

    David
     
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  3. Jul 15, 2009 #2

    tiny-tim

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    Hi David! :wink:

    I'm not sure what you're asking about, but if you mean how do we convert the pressure in a cylinder to the force at the end of the cylinder (which can push a piston which can compress a spring, for which we need to know the force), then it's force = pressure x cross-section area. :smile:
     
  4. Jul 16, 2009 #3
    Re: Thermodynamics

    Hi tiny-tim, very much appreciated for your response to my question, and I appreciate that I may not have worded it very well.

    The question asks how much would a spring be compressed if a gas pressure in a cylinder 600Kpa was acting on a piston, the spring K constant being 4.8 KN/m. The equation given is; PA - P atm A = Spring constant x change in distance X

    Using this equation P = Pressure and "A" I think = area subtract pressure of the atmosphere multiplied by the area, which then = spring constant "4.8 KN/m".

    The part I cannot see being used is any area calculation, which is why I thought that in this subject "Thermodynamics", just maybe the "Pressure" only is used, as the cylinder only gives the "Diameter" and no other measurements?

    Thanks

    David
     
  5. Jul 16, 2009 #4

    tiny-tim

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    Hi David! :wink:
    Yes it is … it's a cylinder, so you know its cross-section is circular, so the given diameter tells you what the area is. :smile:
     
  6. Jul 16, 2009 #5
    Re: Thermodynamics

    Hi tiny-tim, thanks for your replies.

    Using the equation as given; PA - P atm A = K spring x change in X. Can I justify just cancelling the A's on the left hand side to read; P - P atm = K spring x change in X.

    The example given shows; (600 000 - 100 000)pie x 0.05squared = 4800 change in X

    The answer is = 0.818m, but if I include the areas I get 0.2m which is 0.6m different?

    Is it the correct procedure therefore to cancel like terms?

    Many thanks

    David
     
  7. Jul 16, 2009 #6

    tiny-tim

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    Hi David! :smile:

    Don't use the word "cancel"!!

    Except in actual fractions (like 4A/3A = 4/3), say what you're actually doing, and add "on both sides" …

    in this case, say "dividing by A on both sides", and then you'll immediately see that you get P - P atm = K spring x change in X ÷ A
     
  8. Jul 16, 2009 #7
    Re: Thermodynamics

    P is pressure.....look at the units/dimensions..its obvious that u include A??
    Also...
    Whether u subtract Patm from 600kpa depends on whether the given pressure is gauge or absolute pressure...that should b mentioned in th problem itself...I dont think u can guess that unless sumthings provided on that....rest should b ok...
     
    Last edited: Jul 16, 2009
  9. Jul 16, 2009 #8
    Re: Thermodynamics

    Hi tiny-tim, thanks for that input, if what you say I divide both sides by A, then the text book answer of 0.818m would be incorrect because using the method of division where the right hand side of the equation is divided by A, the result shows; 0.643m

    Have I understood you correctly?

    David:smile:
     
  10. Jul 16, 2009 #9
    Re: Thermodynamics

    Hi nanunath, my limited understanding of Thermodynamics is that unless the question specifically says the pressure is "gauge", then it will be absolute, which in my case is absolute.

    Thanks

    David:smile:
     
  11. Jul 16, 2009 #10
    Re: Thermodynamics

    Ok....

    but the other thing..
    The answer is .2m ..not .6..reason :The same as u said earlier..
    but I think this question isnt a part of thermodynamics...
    is that all given in question?
     
  12. Jul 16, 2009 #11

    tiny-tim

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    Hi David! :smile:

    I don't understand what you think is wrong …

    (600 000 - 100 000)pie x 0.05squared = 4800 change in X does give 0.818m,

    and the area (π 0.052) is in there. :confused:
     
  13. Jul 16, 2009 #12
    Re: Thermodynamics

    Hi tiny-tim, and nanunath,

    Because one person is saying I am right with an answer of 0.2 and the text book answer says 0.818, then I must conclude that the equation;

    PA - P atm A = K spring muliplied by change in length X I am somewhere not completing understanding the idea being used. So in my endevour please allow me to make sure we are all singing from the same hymm sheet?

    The question asks;

    A 10 cm diameter cylinder contains a gas pressurised to 600Kpa. A frictionless piston is held in position by a stationary spring with a spring constant of 4.8 KN/m.

    How much is the spring compressed?

    The equation used is;

    PA - P atm A = K spring change in X.

    So my understanding of the above equation is as follows;

    P = Pressure in pascals
    A = Area in metres squared
    atm = atmospheric pressure
    K = the spring constant
    x = springs change in length.

    so the equation says;

    PA - P atm A = K spring change in X. Putting in the data

    (600 000 - 100 000)pie x 0.05 squared = 4800 multipled by X
    500 000 x 7.85 x 10 -3 (power three) = 4800 multipled by X

    divide both sides by 4800 = 0.818m

    Now, if I complete the solution this way;

    PA - P atm A = K spring change in X

    (600 000 x pie x 0.05 squared) - (100 000 x pie x 0.05 squared) = 4800 x change in length.

    This way I get a completely different answer?

    Now there is also;

    PA - P atm A = K spring chnage in X. Divide both sides by A, this gives;

    P - P atm A = K spring change in X / A

    which also gives a completely different answer?

    How do I know which method is the right one to use?

    Thanks

    David:smile:
     
  14. Jul 16, 2009 #13

    tiny-tim

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    Hi David! :smile:

    (have a pi: π and try using the X2 tag just above the Reply box :wink:)
    But that gives the same answer …

    show us how you get it to be different. :confused:
     
  15. Jul 16, 2009 #14
    Re: Thermodynamics

    Hi tiny-tim

    if you divide both sides by A, you will end up with 4800/pie x A on the right hand side, which then shows a different solution.

    Thanks

    David:smile:
     
  16. Jul 16, 2009 #15
    Re: Thermodynamics

    Hi tiny-tim, just having to nip out now for about an hour, will check your response when I return. Thank you very much for both your patience and your help, much appreciated.

    David
     
  17. Jul 16, 2009 #16

    tiny-tim

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    Yes, but it should still give the same solution. :confused:

    Show us how exactly you get a different solution.
     
  18. Jul 16, 2009 #17
    Re: Thermodynamics

    Hi tiny - tim, this is how I get the different answer.

    PA - P atm A = K spring change in X (divide both sides by A)

    P - P atm A = K spring/A multiply by change in X (now put in data)

    (600 000 - 100 000)pi x 0.05 squared = 4800 / pi x 0.05 squared. multiplied by change in X

    500 000 x 7.85 x 10 -3 (minus three is a superscript) = 611154.98. change in X

    3926.9908 = 911154.98 change in X (divide both sides by 911154.98)

    X = 6.43 x 10 -3 (minus three superscript) answer in standard form.

    Hope this helps

    David:smile:
     
  19. Jul 16, 2009 #18

    tiny-tim

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    But you still have A on the LHS. :confused:
     
  20. Jul 16, 2009 #19
    Re: Thermodynamics

    yes I agree, but my understanding is that when using transposition of formula, what you do to one side you do to the other, so as I invented A on the right hand side, then I must invent "A" on the left hand side when I divided both sides, but when cancelling like terms on the left hand side, one can only cancel one "A" with one "A" which leaves a remainder of one "A" on the left hand side?

    I don't know of an alternative way?

    Thanks

    David:smile:
     
  21. Jul 16, 2009 #20

    tiny-tim

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    I don't understand :confused:

    you had (a + b)A = X, or aA + bA = X

    so you divide both sides by A, to give a + b = X/A

    (and what do you mean by "invent"?)
     
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