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Math Help - Couple of Carnot Cycle questions I can't get...

  1. #1
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    Couple of Carnot Cycle questions I can't get...

    These two questions are the only ones I can't get in this chapter. I don't even know where to start and would really appreciate help. Cheers.



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  2. #2
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    All you are trying to do is to find the area inside the curve (the shaded bit). You will then need to apply a correction factor as the axes are not in SI units (a simple x100 I think).

    For the first question the shaded area can be found by calculating four simpler areas and then summing them up. I suggest:

    Area1 defined by the points (1,0),(3,0),C,D
    Area2 defined by the points (1,0),(3,0),A,D
    Area3 defined by the points (3,0),(9,0),B,C
    Area4 defined by the points (3,0),(9,0),A,B

    Two of these areas can be calculated easily because they are rectangles, the other two can be calculated by integrating the equations that you are given over suitable limits.

    The total area is then Area1-Area2+Area3-Area4.

    The second problem can be done in a very similar way. To do this one you would define six areas and add/subtract them to get the required total.
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  3. #3
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    So would the top one be DCEF + VCFG - DAFE - ABGF?

    Why would the second one be 6 areas and what would they be?
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  4. #4
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    Actually you could also do the second one with just 4 areas in the same way. Assuming that you have made a typo (typed V instead of B) then yes you are looking at the correct areas.
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  5. #5
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    I tried using 4 but didn't get it and it's frustrating me! I think it's because of that small gap going up at the F. Would you be able to give me the the sides to use, as I don't know how to do it when only a small part of the curve is being used. Here's my working:



    I have:

    Area = CDEF + BCFG - DAFE - ABGF

    CDEF = 1.8 x 0.3 = 0.54

    ABGF = 0.9 x 0.3 = 0.27

    BCFG = f (0.72/V) limits of b = 2.5 and a = 0.5

    = 0.72ln(V)

    = [(0.72ln(2.5)) - (0.72ln(0.5))]

    = 1.258795297

    DAFE = f (1.8/V) limits of b = 0.4 and a = 0.2

    = 1.8ln(V)

    = [(1.8ln(0.4)) - (1.8ln(0.2))]

    = 1.247884925

    Area = 0.54 + 1.158795297 - 1.1247664925 - 0.27

    =0.181130372 * 100

    =18.1130372

    But the real answer is 118.7
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  6. #6
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    Quote Originally Posted by zapparage View Post

    Area = CDEF + BCFG - DAFE - ABGF

    BCFG = f (0.72/V) limits of b = 2.5 and a = 0.5

    DAFE = f (1.8/V) limits of b = 0.4 and a = 0.2
    The last area ABGF is not quite right. It would be more accurate to say ABG(F+0.2) or to mark the node to the right of F as F'. However, I think you have calculated the required area anyway.

    I am confused at how you have determined the limits of integration and this may be where your main error is.

    For BCFG the limits should be V= 1.0 to V = 3.0

    For DAFE the limits should be V= 0.4 to V = 1.2
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  7. #7
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    I really have no idea how I got those limits.

    How'd you get the 0.2 btw?
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