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Homework Help: Carnot Cycle, how ideal is it?

  1. Dec 8, 2007 #1
    1. The problem statement, all variables and given/known data
    We have stressed that the Carnot engine does not exist in real life: It is a purely theoretical device, useful for understanding the limitations of heat engines. Real engines never operate on the Carnot cycle; their efficiency is hence lower than that of the Carnot engine. However, no attempts to build a Carnot engine are being made. Why is that?
    a. A Carnot engine would generate too much thermal pollution.
    b. Building the Carnot engine is possible but is too expensive.
    c. The Carnot engine has zero power.
    d. The Carnot engine has too low an efficiency.

    and also,
    A real heat engine operates between temperatures Tc and Th. During a certain time, an amount Qc of heat is released to the cold reservoir. During that time, what is the maximum amount of work (Wmax) that the engine might have performed?

    2. Relevant equations

    for question 2.
    Wmax = eff*Qc

    3. The attempt at a solution
    for question 1.
    I think (d) and (c) are not correct, because carnot engine was made because its efficiency can reach up to 100%. That's why it is an ideal engine. so, it also can't have 0 power.
    then I don't know the answer is (a) or (b) ..

    for question 2.
    I calculated but I got wrong
    here is the calculation:
    Wmax = ((Th-Tc)/Th)*Qc

    Please help me..
    thank you
  2. jcsd
  3. Dec 8, 2007 #2
    You are right, C and D are wrong. As for A and B, ask yourself this, does a more efficient engine produce more or less pollution? A real life example will help here--think about SUVs compared with hybrid cars.

    As for your second question, if the real engine is performing the maximum amount of work possible it will behave according to the Carnot cycle. Use conservation of energy (heat in = work + heat out and be careful with signs), and Qc*Th=Qh*Tc (this follows from the Carnot cycle).
  4. Dec 9, 2007 #3
    for question I. is the answer B? since efficient engine produce less pollution.
  5. Dec 9, 2007 #4
    for question 2.
    work = heat in (Qh) - Heat out (Qc)
    then use Qc*Th = Qh*Tc --> Qh = (Qc*Th)/Tc
    then work = (Qc*Th)/Tc - Qc

    is that the answer?

    thanks for helping me..
  6. Dec 9, 2007 #5
    I got it..
    thanks for helping me..
  7. Mar 19, 2008 #6

    I have a question on Why the Carnot cycle is not used with real engines? Ive got to refer to real cycles in my explanation.
  8. Mar 19, 2008 #7

    Andrew Mason

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    D is wrong but not C. C is the right answer.

    Think about the Carnot engine. It uses a reversible cycle, which means that the engine is always in a state of thermodynamic equilibrium. The heat is transferred from the hot reservoir and dispersed to the cold reservoir reversibily - infinitessimal temperature differences mean it occurs over time scales approaching infinity. The work done divided by time gives the power, so the power of a Carnot engine is arbitrarily close to 0.

  9. Mar 19, 2008 #8

    Andrew Mason

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    Efficiency is thermodynamic efficiency which has nothing to do with the chemical efficiency and completeness of combustion. It may not use any chemical combustion at all. You could run a near-Carnot engine using electricity to produce the heat and it would have no pollution whatsoever.

  10. Mar 19, 2008 #9

    Andrew Mason

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    A Carnot engine may be an ideal engine, but that ideal is not 100% efficient.
    A Carnot engine can only have 100% efficiency if the cold reservoir is at absolute 0.

  11. Jan 22, 2009 #10
    see as the engine is real we can no use carnot efficiency to calculate the work. we shud use 2nd law of thermodynamics.solution is as
    let the temp of cold and hot reservoir be T2 and T1 resp.
    then to hav a feasible process dS>= 0
    this implies dat (QC/T2)-[(W+Qc)/T1]>=0
    here change in entropy of source and sink are taken into account. the working fluid doesnt has any chnge of entropy as it undergoes a cycle.
    for maximum work done possible we equate to zero the above equation(to hav a reversible process)
    this gives W max = (Qc/T2)-(Qc/T1)*T1
    =Qc[T1/T2 -1]
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