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Question regarding carnot cycle

  1. Jun 5, 2012 #1
    In Carnot cycle we assume that heat addition takes place at constant temperature(temperature of system and source being same) and same is the case for Heat rejection.
    But as we know that for heat flow (ie heat addition and heat rejection), temperature difference has to be there between system and heat reservoir. Then how is heat addition/rejection possible at constant temperature?
     
  2. jcsd
  3. Jun 5, 2012 #2
    The heat reservoirs are part of the system.
     
  4. Jun 5, 2012 #3

    AlephZero

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    The Carnot cycle is an idealized theoretical idea, to find the limit of the maximum possible efficiency of a heat engine. It can never be achieved in practice. If a real engine cycle was very close to the "perfect" Carnot cycle, the temperature difference between the source or sink and the working fluid would be very small, it would take a very long time to transfer a fixed amount of heat energy, and the power output (energy per unit time) from the engine would be close to zero.

    The "perfect" Carnot cycle represents the limit of this situation, where it takes "an infinite amount of time to transfer the heat energy infinitely slowly with no temperature difference".
     
  5. Jun 5, 2012 #4

    Andy Resnick

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    There are several mechanisms- a phase change (latent heat of melting, vaporization, etc), as well as volume expansion/contraction.
     
  6. Jun 5, 2012 #5
    I tried to post a longer answer but got kicked out by the software.
    Let me try again. I will provide a shorter answer and hope it gets through.
    One hypothesis in the Carnot cycle limit is that during the isothermal
    parts of the cycle, the temperature difference between container and
    heat reservoir is nonzero but much smaller than the temperature difference
    between the two heat reservoirs.
    For instance, suppose that the high temperature reservoir was 600 degrees
    while the low temperature reservoir was only 100 degrees (in any temperature
    unit that you like). The temperature difference between the two reservoir is
    500 degrees. The temperature difference between container and heat
    reservoir that it is in contact with has to be far less than 500 degrees.
    Suppose that the container is in contact with the low temperature reservoir.
    The container can be at only 101 degrees. That is only one degree higher than the
    temperature of the low temperature reservoir. However, heat flow will take place.
    Obviously, 500 degrees is much larger than 1 degree. So the engine will be very
    close to a Carnot cycle. That one degree makes the engine work, but doesn't
    cause the engine to work much different from a Carnot cycle.
     
  7. Jun 5, 2012 #6
    Although I think legend_xeon has left the party, please note his actual question
    which was not about efficiency.

     
  8. Jun 5, 2012 #7
    The assumption is that the system is at near equilibrium at every step.
    The system can't be at near equilibrium unless the container is at nearly
    the same temperature as the temperature reservoir that it is attached too.
    Therefore, the difference in temperature between the container with the
    ideal gas and the reservoir it is attached to has to be very small compared
    to the temperature difference between the reservoirs during the isothermal
    parts.
    There is another constraint that should be observed. The speed and
    power of the near Carnot cycle increases with the difference in temperature
    between container and the heat reservoir it is attached to. This is because the
    rate of heat flow is proportional to the difference in temperatures.
    Therefore, the Carnot cycle effectively stops moving in the limit of
    no temperature difference between container and reservoir. If the container
    is exactly the same temperature as the reservoir it is attached too, there
    can be no heat flow and the Carnot cycle stops. Therefore, one has to
    run an engine slowly before it even approximates a Carnot cycle.
    When an engine is running with an efficiency close to maximum, the
    temperature difference between reservoir and container has to be
    small. To decrease this distance, the engine has to run even slower.
    Efforts to speed up the engine will only result in a temperature difference
    increasing between container and reservoir. If the temperature difference
    between container and reservoir increases, the efficiency of the engine
    decreases. Entropy is created if the temperature difference between container and
    reservoir is nonzero.
    Therefore, there is a trade off between engine efficiency and speed. Running
    an engine too hard causes a decrease in engine efficiency. The hypothesis of
    isothermal expansion is self consistent only if the engine runs slowly.
     
  9. Jun 6, 2012 #8
    Studiot
    "Although I think legend_xeon has left the party, please note his actual question
    which was not about efficiency."
    LEGEND_XENON
    "Then how is heat addition/rejection possible at constant temperature?"
    I already gave two posts. However, I may not have answered his question
    directly. So, here is another try.
    In the near Carnot cycle, the temperature inside the container is never
    completely constant. The temperature of the container is slightly different
    from the temperature of the heat reservoir that it is in contact with during
    the isothermal parts of the cycle. The temperature of the container slowly
    changes until it is the same temperature as the reservoir.
    The OP's hypothesis, that the temperature is constant, is not precisely true
    for the container. It can be true for the reservoir, but not the container with
    the ideal gas. The hypothesis that the temperature of the container be
    constant is only self consistent in the limit of a cycle with an infinite period of
    time.
    Suppose that the container is placed at time,t=0, on the cold reservoir
    which is at a temperature of 100 degrees. Suppose, at time, t=0, the container
    is at a temperature 101 degrees.
    As heat leaves the container, the temperature of the container decreases.
    At time t=10 seconds, the temperature of the container can be 100.3 degrees.
    At time t=20 degrees, the temperature of the container can be 100.1 degrees.
    Hence, the temperature of the container is not constant during the isothermal
    parts of the cycle. It is just changing very slowly.
    As someone else pointed out, the Carnot cycle works in the limit where
    every cycle takes an infinite amount of time. The less the initial temperature
    difference between container and reservoir, the closer the engine is to the
    Carnot limit. However, the smaller the initial difference between container
    and reservoir the longer it takes to transfer a given amount of heat energy.
    I want to thank the OP for asking that question. Before he asked the
    question, I didn't know about the trade-off between efficiency and time.
    The key in understanding this trade off is realizing that the temperature
    of the container and the temperature of the reservoir is going to be nonzero
    in the real world.
    Carnot point out that if the container and the reservoir had a different
    temperature, the efficiency of the engine would be smaller than the maximum
    possible efficiency. I figured out the answer to the OP's question by going over
    Carnot's original monograph. Once I realized that the temperature of the
    container and the reservoir is going to be nonzero in the real world, I realized
    that there had to be a transfer of heat energy in the real world because of this
    difference. The use of formulas for heat conductivity helped me realize that
    the speed of the engine increases with this temperature difference. So sometimes
    it is useful to go back to the original sources, even if it was written in 1815 !-)
     
  10. Jun 6, 2012 #9
    Darwin, I already responded to this part of his post

    by noting that he has not drawn the system boundary correctly.
     
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