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Ideal Carnot Engine

  1. May 27, 2009 #1
    1. The problem statement, all variables and given/known data

    An ideal Carnot engine produces a power output of 5kW at a thermal efficiency of 35% while rejecting heat reversibly to a reservoir at 30degrees. It achieves this output while executing 10 cycles per second.

    a) What temperature is the engine recieving its heat supply?
    b)What are the rates of heat supply to and heat rejection from the engine?
    c)Hence determine the change in the total entropy of the working fluid during each of the two heat transfer process.

    2. Relevant equations

    Not really given but I got first part n_carnot=(Th-TL) / TH

    3. The attempt at a solution

    I use that about equation for part a to find Th which is where the temperature recieving the heat supply also which is process 1-2 isothermal heat addition.
    But im totally stoned about part b and c. How can I calculate the rates of heat supply etc, would someone be able gimme ahint on how to even start it?

    Thanks
     
  2. jcsd
  3. May 27, 2009 #2

    Mapes

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    Power is energy transferred per unit time. The relationship between the rates of heat supply, heat rejection, and power output are exactly the same as the relationship between heat transferred in (Q) and out and work done (W) per cycle. And this relationship is in turn related to the efficiency. Look for one more equation to tie these variables together.
     
  4. May 27, 2009 #3
    n carnot = Wcycle / Qin
    this would help me solve rate for power supply into system, given i know thermal efficency as .35 / 35% but I do not know the W cycle, I am only given the Work output which is 5KW. No volume is given so I cant go through the first law and calculate the total work for the cycle. What should I be doing? Thanks
     
  5. May 27, 2009 #4

    Mapes

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    From this equation, you can calculate the rate of heat supply immediately. Apply the first law, and you can calculate the rate of heat rejection.
     
  6. May 27, 2009 #5
    How do I calculated the rate heat supplied? I thought the 5KW output is not the total Work Cycle?
     
  7. May 27, 2009 #6

    Mapes

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    If you take a look at my post #2 again, the ratio between the work per cycle and the heat input per cycle is the same as the ratio between the (average) work per second and the heat input per second, which is the same as the ratio between the work power (which you're given) and the heat supply rate.
     
  8. May 27, 2009 #7
    ahh, I get it, so I do 5kw/.35= q in?
    another question how could i calculate heat reject. It is process 3-4 in a carnot cycle which is isothermal heat rejection. If I apply First Law it wud be typically that q3-4 or qout= RT3*ln(v4/v3) but I can not get this volume , or should I be using somthing else instead of doing this? Thanks

    edit: cant use pV=mRT as I dont have the mass or do I use pv=RT where its gonna be a Pressure vs Specific volume graph in that case I could where T=the temperature I found in part a, but lol I dont have the pressure.

    Thank you
     
    Last edited: May 27, 2009
  9. May 27, 2009 #8

    Mapes

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    Yet another way of looking at it is that, out of the heat energy that comes in, what isn't output as work must be rejected.
     
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