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Work of a Carnot machine

  1. May 28, 2005 #1
    integration of

    Dear people of this section.

    Suppose you have two indepenfent functions x and y and we need to find the integral of the following expression y dx ,and y does not depends upon x, but y varies between 4 and 8 in the interval of integration. ¿ Do you think that the value of the integration is (( 4+8)/2 ) x ?
     
  2. jcsd
  3. May 28, 2005 #2
    [tex]\int y dx= yx + h(y)[/tex]

    does that make sense? as far as y varying from 4 to 8... umm... i guess you would just plug those at the end...
     
  4. May 28, 2005 #3
    we need to find the integral of the following expression y dx ,and y does not depends upon x

    This statement shows that the integral will evaluate to [itex] yx + C [/tex]

    y varies between 4 and 8 in the interval of integration
    If y varies as plotted against x, then y does depend on x. If it varies when plotted against a different variable, then this information has no meaning for the given problem.
     
  5. May 28, 2005 #4

    Hurkyl

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    In other words, you just use the fundamental theorem of calculus like you always would:

    [tex]
    \int_a^b f_x(x, y) \, dx = f(b, y) - f(a, y)
    [/tex]
     
  6. May 29, 2005 #5
    In order to be clear my qustion to you, I am going to explain better the origin of this thread. I am interested to find the work of a carnot machine that operates between a hot reservoir at 800k (constant temperature) an a series of cold reservoirs whose temperature varies from 400 t0 800K. From thermodynamics the work, W, will be

    W = Integral (1-Tb/Ta) dQ
    where Tb is the variable temperature of the cold reservoir, Ta is the constant temperature of the hot reservoir and Q is the heat input. Solving we get

    W= Integral dQ - integral (Tb/Ta)dQ = Q- (1/Ta)integral Tb dQ

    Since Tb and Q are independent, we obtain

    W= Q-Tb Q

    To get a number a need to introduce a value for Q and Tb. ¿ What value do I choose for Tb if it varies between 400 and 800?. Q is equal to 2000 calories. Because of this I said, well, Tb is the arithmetic mean of 400 and 800 K, or Tb= (400 + 800)/2. ? Is this reasonable, honest?
     
  7. May 29, 2005 #6

    Hurkyl

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    Ack, I just noticed the section. This isn't even a number theory question to begin with! (Should be calculus & analysis, if you wanted to put it in the math forum)


    However, your problem appears to be your modelling of the scenario. Q and Tb cannot be functionally independent, since both depend on time. (Unless one of them is a constant) I'm going to punt this over to the physics section -- should be someone over there who knows exactly what you should be doing.
     
  8. May 31, 2005 #7
    Dear Hurkyl, since I don´t have any answer, yet. Let me to present a different solution to the original question of the thread. In other words, let us find the integral of YdX using the basic definition of integration. As you suggest Y and X are not functionally independent, but we don´t know how they depend. We know that X varies between a and b. Then , we can define an infinite set of equals intervals delta X = (b-a)/n, where n is the number of intervals. n tends to infinite. then,

    integral Y d X = Summatory (Y1 delta X+ Y2 deltaX + Y3 delta X+ ....Yn delta X)

    Y1, Y2, Y3,.............Yn, are the Y values in each interval delta X.

    Since each delta X = (b-a)/n, we get

    Integral Y dX = (b-a) Summmatory ( Y1+ Y2+ Y3+......Yn) /n
    where n tends to infinite

    But Summatory (Y1+ Y2+Y3+ .......+ Yn) / n is the arithmetic mean or the average of an infinite set of values , as I will prove now.
    By the mean value theorem

    Summatory ( Y1+Y2+Y3+......+Yn) = 1/ ( Yb -Ya) Integral YdY = (Yb +Ya) / 2

    where Ya and Yb are the initial and final values of Y.
    Therefore,

    Integral YdX= (b-a)(Ya+Yb)/2
    This final result coincides with the empirical value I initially suggested.

    Now, since I am not an expert mathematician, I have doubts about the validity of this demonstration, and I appreciate very much your comments and suggestions. If this is correct I think I can apply it to the Carnot machine.
    Thanks.
     
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