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Steam Power Plant Cycle

  1. Oct 4, 2004 #1
    Carnot Cycle is the ideal heat engine cycle with maximum efficiency. We use Rankine Cycle as the working cycle for steam power plant which has significantly lower efficiency that Carnot Cycle.

    My question is why we don't adopt Carnot Cycle as ideal Steam Power Plant cycle?

    Thanks in advance!
  2. jcsd
  3. Oct 4, 2004 #2


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    The carnot cycle just plain doesn't work in practice due to isothermal (constant temperature) heat addition. As some websites I looked at noted, making the heat transfer isentropic requires slowing down the cycle so much your car would get great fuel economy while getting passed by people walking next to it.

    edit: right definition, wrong word - isothermal, not isentropic.

    More info HERE
    Last edited: Oct 5, 2004
  4. Oct 5, 2004 #3
    Thanks russ_watters!

    Even in Rankine Cycle heat addition and rejection processes are carried out in isothermal conditions, except for Economizer and Superheater where heat is added isentropically (ideally).

    So another question;

    "Why heat addition and rejection processes are preferred at constant temperature"?
  5. Oct 6, 2004 #4


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    It's impossible to add heat isentropically. It has no sense. It is the same thing like increasing temperature isothermally. In the Rankine cycle the water is converted into vapor, the heat is exchanged isothermally and isobarically due to the change of state constraint. It's an special case, when exchanging heat by an isothermal process is valid. But this is due to the change of state. During it P and T are constant. It's clearly an irreversible process, so that a iso-temperature line and a iso-pressure line are the same thing.

    In a thermal engine that does not use change of state of the fluid, it is practically impossible to exchange heat at a constant temperature varying the volume. You would need such a slow transfer process that engine economics would be the worst of the world. You would need a heat focus of constant temperature surrounding the cylinder. And the movement of the cylinder would be so slow as allowing internal gas to adapt itself at the focus temperature. After three or four years your engine will complete only one stroke. :surprised
    Last edited: Oct 6, 2004
  6. Oct 6, 2004 #5
    Thanks Clausius2 for the correction!

    Heat addition 'isentropically' --- really sounds stupid. Actually I was to say 'isobarically'. In Boiler (Evaporator) and Condenser heat addition and rejection processes are carried out both isothermally and isobarically (ideally of-course). But in Economizer and Superheater sections of the Plant, it's only the pressure which is kept constant not the temperature, which increases.

    While reading the topic of 'Reversibility', I found that heat transfer thru a finite temperature difference is a irreversible process. But we know that heat is the energy which is transferred thru a temperature difference. The question is "Is it possible to have a reversible heat transfer process at all?"

    Thanks for your time!
  7. Oct 7, 2004 #6


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    The answer is yes and no. Yes if the process is made extremely slow (quasi-statically). Then, as I have said above, the internal gas has time to adapt itself to the temperature of the focus. But any process of heat transfer not made quasi-statically (reversible) has the consequence of increasing the entropy. For a reversible process, the heat has to be transferred so slow as to accomodate substances to reach each infinitesimal increment of temperature each time. A reversible process would be an infinitely slow process, and there is not any machine able to do that.

    But probably you are thinking of the processes represented in the P-V or P-T diagrams. Have you ever calculated some figures with these diagrams?.
    Last edited: Oct 7, 2004
  8. Oct 7, 2004 #7
    Yes, this is exactly what I'm doing these days. For Rankine Cycle I'm using T-s diagram which is most commonly used.

    My engineering classes have just been started and Thermodynamics is the subject which I've found most interesting. But I have not yet been able to fully understand the concept of Reversibility. Reversibility is usually defined as:

    "A process is called reversible that once having taken place can be reversed and in doing so leave no change either on system or surroundings."

    But in practice we don't need to reverse a process. Take, for example, the compression process in a Gas Power Plant. We desire to achieve a reversible adiabatic compression process (isentropic) why? since in practice we'll never need to reverse this compression process.

    What thing I'm not being able to understand? So far all I have been able to understand about reversibility is that a reversible process is desirable only because it involves no losses. Am I right?

    Please help me out. Can you explain this topic and/or suggest some good readings?

    Thanks for your time!
  9. Oct 7, 2004 #8


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    Well, all that you can paint in a P-V diagram is generally a reversible process. Do you understand what does a quasi-static process mean?
    Do you understand which is the difference between a process made quasi-statically and another heat transfer process made in a finite time?

    We want to be near a reversible process in the compression process of a gas plant because the less entropy generation the less mechanic energy we have to supply to the compressor. There is a limit, where that work supplied to the compressor is equal to the heat added. This will be a reversible process.

    The reversibility is desirable because although we always are going to waste some energy (realise an efficiency of a Carnot engine is less than unity although is reversible), there is an extra consumption of energy due to internal irreversibilities. Thermal engineers want to eliminate this extra and chase the thresold of reversibility.

    Moreover a reversible process do produce losses. Do you think your Carnot cycle doesn't produce losses? Why the efficiency is less than one? What do you mean with losses?.
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