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Thermodynamics - Need help with inclined plane problem

  1. Feb 9, 2017 #1
    1. The problem statement, all variables and given/known data
    A train engine climbs a hill. The engine is 30% efficient. The train has a mass of 12000kg. The hill is 750m in height, with an incline of 20° form the horizontal. Friction exerts a force opposing motion of 8000N throughout the climb. Find w, Qin and Qout for the engine.

    2. Relevant equations
    We have only covered the first law, and some equations for various types of work:
    dE/dt = Qin - Qout - w
    w = Fd

    3. The attempt at a solution
    I assume I have to first solve for the forces as if it were an inclined plane problem. I'm used to the friction being the opposite way, so I am a little confused about how to resolve these forces. I think the work will be:
    F|| * hypotenuse of the incline.
    But would this work be the work done by the engine? It just seems strange to me since I'm thinking that is due to the train itself, and not the engine within the train. Here's what I have so far:
    IMAG1298_zpsg3xpibr7.jpg
     
  2. jcsd
  3. Feb 9, 2017 #2

    Doc Al

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    Find the force that the train must exert to overcome friction and gravity. And the distance it must exert this force to get to the top of the hill. And thus the work it must do.

    Note that the mass is 12000 kg, not 12000 x 103 kg.
     
  4. Feb 9, 2017 #3
    If its 12000kg that means its 12,000 kg * 1000g/1kg = 12,000,000g = 12000x103 g, right?
    Also, I assumed that the force would be the F||. Now I am little confused, since I am used to having the friction force on the other side, and so F|| and F are just the components of the gravitation force. Would this still be so?
     
  5. Feb 9, 2017 #4

    kuruman

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    This doesn't look dimensionally correct to me. You have power on the left side and energy on the right side.
     
  6. Feb 9, 2017 #5

    Doc Al

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    Why in the world are you converting to grams?

    Not sure what you mean. Friction and gravity both act down the incline; the train must exert an equal force up the incline to climb the hill.
     
  7. Feb 9, 2017 #6
    For some reason I though I needed to. And gravity acts directly downwards, but I tilted the incline onto the x-axis to make it easier to work with. Is my diagram wrong? I am treating this as an incline plane problem, is that wrong?
     
  8. Feb 9, 2017 #7

    Doc Al

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    No, don't do that.

    It is an incline plane problem. So what's the component of gravity parallel to the incline?
     
  9. Feb 9, 2017 #8
    Oh I believe I know what I'm doing wrong. I didn't realize that the friction force is equal to the horizontal component of gravity. Here is a new FBD:
    IMAG1300_zpssa1gvmfn.jpg
    If this is correct, then Fpush (force required for the train to go up) is equal to 8kN?
    Also, where would the 30% efficiency come in? Would be applied to the work I will find?
     
  10. Feb 9, 2017 #9

    Doc Al

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    It's not. There are two downward forces that must be overcome by the engine: gravity & friction. Add them to find the total force. (The friction force is given; you need to calculate the component of gravity.)
     
  11. Feb 9, 2017 #10
    Oh oops, okay so there are two horizontal components acting to the left. Friction and mg*sin(20). So Fpush = 8kN + 40.26kN = 48.2kN.
    For work, the distance would be:
    d = 750/sin(20°) = 2192.85m
    And so the work would be:
    w = Fd = (48.2)*(2192.85) = 105.83kJ

    Now where does the efficiency come in?
     
  12. Feb 9, 2017 #11

    Doc Al

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    Good, but be careful with units. (I think you dropped a factor of 103.)

    What's the definition of efficiency?
     
  13. Feb 9, 2017 #12
    Oh okay so it should be 105.83MJ.
    Efficiency of a heat engine is Wout/Qin
    So would it be:
    .30 = Wout/Qin
    ?
     
  14. Feb 9, 2017 #13

    Doc Al

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    Good. Now apply the 1st law to find Qout.
     
  15. Feb 9, 2017 #14
    So Qin = Wout/0.30 = 3.53GJ
    dE/dt = Qin - Qout - Wout
    0 = 3.53GJ - Qout - 105.83MJ
    Qout = 3.53GJ - 105.83MJ = 3.42GJ
     
  16. Feb 9, 2017 #15

    Doc Al

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    Right idea, but careful with the decimal point.

    As kuruman pointed out, that equation is dimensionally inconsistent. Better to use:
    Qin = Qout + Wout
     
  17. Feb 9, 2017 #16
    Oops, should be:
    Qout = 352.77MJ - 105.83MJ = 246.94MJ
    And that equation came from my notes. Why is it plus and not minus?
     
  18. Feb 9, 2017 #17

    Doc Al

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    That looks better.
    We're using the same equation. (Just not equal to dE/dt, but dE alone.)
     
  19. Feb 9, 2017 #18
    Right, so the equation is the same, and we get the same answer. So why exactly are you guys pointing out my equation and saying it is dimensionally inconsistent?. What does that mean?
     
  20. Feb 9, 2017 #19

    Doc Al

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    The equation you quoted:
    Cannot be true as is, since the left side expresses power (J/sec) and the right expresses energy (J). Equations must have the same dimensions (units) on both sides.

    But the equation you actually used was this one:
    Qin = Qout + Wout
    Which is fine.
     
  21. Feb 9, 2017 #20
    Oh okay that makes sense, thank you both for your help. I appreciate it!
     
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