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Cycle (thermodynamics)

  1. Nov 28, 2009 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    One mole of an ideal gas is used as the working substance of an engine that operates on the cycle shown in the figure.
    B-C and D-A are reversible adiabatic processes.
    1) Is the gas monoatomic or polyatomic?
    2)What is the efficiency of the engine?




    2. Relevant equations
    None given. This is the exercise 48, page 530 of Fundamentals of physics, 5th edition.


    3. The attempt at a solution
    1)I used the formula [tex]PV^{\gamma}=\text{a constant}[/tex] which gave me that the gas is monoatomic.
    2)I'm stuck here. I think that [tex]\varepsilon=\frac{W_{\text{out}}}{Q_{\text{in}}}[/tex]. I think that there's work done by the system in A-B while W is done on the system in C-D.
    For A-B, [tex]\Delta E =\frac{3R\Delta T}{2}[/tex] but I didn't succeed in getting rid of the temperature...
    I also found the work in A-B to be [tex]RT \ln 2[/tex], where T is the temperature in point A... I still have the same problem with the temperature.

    Feel free to help me, I truly appreciate it.
     

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  3. Nov 28, 2009 #2

    Andrew Mason

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    In an actual engine, the work done on the gas is done by the engine itself: eg the output work of the A-B and B-C cycles in one cylinder is actually used to perform the work in the C-D and D-A cycles of another cylinder. So the net output work of the engine is what you want in order to measure efficiency.

    The net work done by the gas is the area under A-B and B-C LESS the area under C-D and D-A. So, the net work is the area inside the graph.

    AM
     
  4. Nov 28, 2009 #3

    ehild

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    This is a monoatomic ideal gas, só

    [tex]PV=nRT [/tex]

    where n is the number of moles, and the internal energy U is

    [tex] U= \frac{3}{2}nRT [/tex].

    Express all temperatures with T0 (the temperature at A).

    Both A-->B and C-->D are at constant pressure so the work done by the gas is

    [tex] W=P \Delta V [/tex].

    From the first law of Thermodynamics, you know that

    [tex] \Delta U = Q-W [/tex].

    You can express all work done and heat absorbed by the gas with nRT0. The efficiency is the work done / heat absorbed during the whole cycle.

    ehild
     
  5. Nov 29, 2009 #4

    Andrew Mason

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    You have to calculate Q and W for each part of the cycle.

    For A-B, the Qin is:

    [tex]Q_{in} = C_v\Delta T + P\Delta V = C_p\Delta T[/tex]

    where [tex]\Delta T = P\Delta V/R[/tex]. Since [itex]C_p = 5R/2[/itex],

    [tex]Q_{in} = \frac{5}{2}P\Delta V[/tex]

    For B-C, there is no heat flow. The change in temperature from B-C is determined by the adiabatic condition:

    [tex]TV^{\gamma - 1} = K[/tex]

    Since it is adiabatic, the work done is the same as the change in internal energy: [itex]W = -\Delta U = -C_v\Delta T[/itex]

    AM
     
  6. Nov 29, 2009 #5

    fluidistic

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    Thanks to both. I realized that the net work was done by the machine and it was the area between the curves. I also realize that if the sense was reversed, then the net work would have been done on the machine, also the area between the curves. However I don't think it helps me to calculate it, unless the graph shows what is the area between the curves.

    Ok, that's what I had started but stopped because of the temperature I couldn't get rid of.

    Ok here I understand that [tex]Q_{in}=C_p\Delta T[/tex] since it's a constant pressure process. But why it is also worth [tex]C_v\Delta T + P\Delta V[/tex]? I've no idea here, I'd love to know the explanation.
    Also, why does [tex]\Delta T = P\Delta V/R[/tex]? I'm also at a loss here.


    Ok, I can follow you if I assume the anterior.

    I can follow you here. I'd have to work out [tex]\Delta T[/tex] I think, but I'll try.
     
  7. Nov 29, 2009 #6

    Andrew Mason

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    The area inside the path is (the area under B-C + area under C-D) minus (the area under D-A + area under A-B). Just calculate the area under each part of the curve and then do the arithmetic.

    You don't want to get rid of the temperature.

    The first law: dQ = dU + dW. Change in internal energy is: [itex]\Delta U = nC_v\Delta T[/itex] and here n = 1. Since P is constant, [itex]W = P\Delta V[/itex]. So [itex]Q = C_v\Delta T + P\Delta V[/itex]. But by definition, [itex]Q = C_p\Delta T[/itex] for a constant pressure process. So [itex]C_p\Delta T = C_v\Delta T + P\Delta V[/itex].

    It is just the ideal gas law: PV = nRT. Since P is constant from B-C and from D-A, applying PV = nRT at the B and at C (or at D and at A) and subtracting gives [itex]P\Delta V = nR\Delta T[/itex].


    You have to work out the temperature changes to solve this.

    AM
     
  8. Nov 29, 2009 #7

    fluidistic

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    Are you sure? There's no area under C-D, and I'm tempted to say that the area between the curves is the area under A-B + area under B-C - area under D-A.
    I don't see how it could be "area under B-C + area under C-D) minus (the area under D-A + area under A-B)"

    Ok, thanks.
    Ok, it makes perfectly sense to me now. Thanks you a million.
    I'd appreciate if you could say a word about my thoughts for the area between the curves. That is, the net work done by the system.
     
  9. Nov 29, 2009 #8

    Andrew Mason

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    The area within the path is what you have to quantify. Divide the path into 4 segments and calculate the area under each segment.

    I may have confused you. I thought A was at the lower left corner when I wrote post #6. My post #2 had it right. The area within the path is [itex](A_{A-B} + A_{B-C}) - (|A_{C-D}| + |A_{D-A}|)[/itex].

    I am not sure why you do not think there is an area under C-D:

    [tex]A_{C-D} = \frac{P_0}{32}(8-16)V_0 = -P_0V_0/4[/tex]

    AM
     
    Last edited: Nov 30, 2009
  10. Dec 1, 2009 #9

    fluidistic

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    Thank you very much once again, you've helped me a lot. I'm sorry for the area under C-D, I guess I was so tired I was thinking about another problem involving the Diesel cycle...
     
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