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Understanding Carnot Efficiency

  1. Nov 2, 2015 #1

    KTF

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    I'm struggling to understand the Carnot efficiency. I can follow proofs that prove efficiency= 1- (Th/Tc), and I understand the nature of the processes that make up a carnot cycle - however I still can't make sense of it in my head.
    I've looked online and most people try and explain it intuitively using a water wheel analogy - such that if you take some water at the top of the wheel and release it halfway, you have only utilized half of the available potential energy ( apparently how carnot understood it. ) The equivalent in terms of heat is explained such that if a 600k heat engine has to exhaust at 300k, it can be max 50% efficient.

    However in a carnot cycle the working fluid going from the source to the sink temperature doesn't represent the utilization of the energy you have put into the system, right? Because the working fluid is already at the source temperature - we didnt add energy to get it to this temperature. All the heat added is transformed into work?

    Looking at the cycle, the loss of efficiency comes from the work needed to return the piston to the initial position. Is it such that the work needed to return the piston is related to the temperature of the sink, and that it would take less work to return the piston if the working fluid was a lower temperature?And Qout is equivalent to this work? This would explain why lowering the source temperature would increase the efficiency? I'm unsure on why increasing the source temperature would increase the efficiency, though?

    Thanks!
     
  2. jcsd
  3. Nov 2, 2015 #2

    DrClaude

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    It is a question of entropy. The hotter the hot reservoir, the less entropy you get for the same amount of heat, since ##\Delta S = Q/T##. That entropy has to be dumped into the cold reservoir (since the working substance must return to its initial state), such that the amount of heat dumped in the cold reservoir will be greater if there is more entropy coming in from the hot reservoir. The first law imposes ##Q_h = Q_c + W##, so for the same for the same ##Q_h##, maximizing ##T_h## minimizes ##\Delta S## which minimizes ##Q_c##, which maximizes ##W##, which increases the efficiency.
     
  4. Nov 2, 2015 #3
    Think of Carnot's Engine in terms of a state diagram (P-V). Then you understand that every closed transformation would necessarily waste some energy (energy is represented as area).
     
  5. Nov 3, 2015 #4

    Andrew Mason

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    Welcome to Physicsforums KTF!

    All of the heat flow into the system during the isothermal expansion is converted to work. i.e. since ##\Delta U = 0, W = Q##. In the adiabatic expansion internal energy loss is converted entirely into work: i.e. since ##Q = 0, W = \Delta U##. Neither of these parts of the cycle violates the second law. However, these expansions cannot continue indefinitely. To create a repeatable cycle, one has to either compress the system's working gas to bring it up to the temperature of the hot reservoir or expel the system working gas and bring in gas from the cold surroundings and heat it up to the temperature of the hot reservoir. In either case, energy has to be put back into the system in order to keep it going. It is that energy in these latter parts of the cycle that causes the system to be less than 100% efficient.

    Increasing the source temperature results in greater expansion and more work being done during expansion (area below graph on the PV diagram increases). The amount of energy required to bring the working gas back to the temperature of the hot reservoir is larger too but is proportionately smaller than the amount of work done in the forward part of the process. You have to work this out: efficiency is W/Qh = (Qh-Qc)/Qh = 1-Qc/Qh and since Qc/Qh is proportional to Tc/Th for a Carnot cycle (##\Delta S_{sys} = -\Delta S_{surr} \text{ so } Q_h/T_h = -Q_c/T_c## ), the higher Qh means that efficiency increases.

    AM
     
    Last edited: Nov 3, 2015
  6. Nov 3, 2015 #5

    KTF

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    That's what i thought regarding the source temperature. In regards to the reverse cycle, is the adiabatic compression equal to the Adiabatic expansion ( in terms of work), as one is the reverse of the other?

    This would mean the net work is gained due to the Isothermal Expansion stage producing more work than the isothermal compression stage takes, right? And this is because the compression occurs at a lower temperature, thus pressure? ( the PV diagram suggests this.) The thing is i would assume the Qc=Qh, because the expansion and compression occur for the same time period, and the temp difference for heat transfer is the same?
     
  7. Nov 3, 2015 #6

    Andrew Mason

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    Yes. The work done is equal to the change in internal energy which is a function of the change in temperature (for an ideal gas). Since the adiabatic expansion is from Th to Tc and the compression is from Tc to Th, the ##\Delta U## in the expansion is equal and opposite to the ##\Delta U## in the compression regardless of the starting and ending volumes and pressures.

    Correct.
    No. The gas does more work in expanding at high constant temperature than the work that is required to compress it at low T (ie. net positive work is done by the system): This is because if T is constant: ##W = \int PdV = nRT\int dV/V = nRT\ln{\frac{V_f}{V_i}}## - i.e. W is proportional to T since ln(Vf/Vi) for expansion = - ln(Vf/Vi) for the compression.

    AM
     
  8. Nov 10, 2015 #7

    KTF

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    Thanks for the explanations so far Andrew, they have been very clear!I only have a few more ( I hope!)

    In terms of heat addition, why is reversible heat transfer more efficient than irreversible heat transfer? I understand why heat transfer through a finite temperature difference is irreversible(heat can't flow from cold to hot without energy), just not why it would produce less work?

    In a steady flow Carnot cycle, there is no work done during the isothermal heat transfers, however in the piston example, this is where the net work occurred? And the 2 adiabaic processes cancelled eachother out? I'm assuming in this case the compression requires less work that the expansion produces?

    Is the Rankine cycle less efficient than the Carnot efficiency because part of the heat addition isn't at a constant temperature? And the super heat Rankine cycle is closer to the Carnot efficiency because even though there is more non-constant temperature heat addition, Th is increased? This only confuses me because textbooks just say 'the average temperature of heat addition is higher/lower, so the cycle is more/less efficient' for both cases.

    Thanks again!
     
  9. Nov 11, 2015 #8

    Andrew Mason

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    If you compare a Carnot engine and a real engine operating on an irreversible cycle drawing the same amount of heat from the hot reservoir (Qh), the real engine will deliver more heat to the cold reservoir.

    If that was not the case, the ##\Delta S## of the surroundings (Qh/Th + Qc/Tc where Qh<0 and Qc>0) for the real engine would be less than the ##\Delta S## for the Carnot engine, which is necessarily 0 (##\Delta S## for the engine is always 0 after complete cycles in which the engine returns to its initial state). So ##\Delta S = |Qc|/Tc - |Qh|/Th > 0## for the irreversible engine. With Qh, Th and Tc being equal for both engines, Qc has to be greater for the irreversible engine than the Carnot engine,

    Since both engines return to the same state after complete cycles, ##\Delta U = 0## so W = Q. Q is just the total heat flow with respect to the engine, which is Qh+Qc in one full cycle (first law), and since relative to the engine Qh is positive and Qc is negative, W = |Qh|-|Qc|. Since |Qh| is the same for both but |Qc| is greater for the irreversible cycle, ##W_{irrev} < W_{Carnot}##

    Why do you say that no work is done during the isothermal heat transfers? The only way you can transfer heat into an ideal gas (Q>0) but maintain constant temperature (##\Delta U = 0##) is for the gas to do work equal to Q (from the first law).

    The ideal Rankine cycle would be reversible. It is less efficient than the Carnot engine in practice because the ideal isentropic processes are not achieved in the real engine.

    AM
     
  10. Nov 11, 2015 #9

    KTF

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    I was envisioning the boiler and turbine to be completely separate components in the steady flow system, and the only work output you get is when the working fluid expands adiabatically in the turbine. I think my confusion is due to moving onto vapour cycles where steam is the working fluid and not an ideal gas, where the heat addition occurs when the working fluid is changing state.

    If the ideal Rankine cycle is reversible , how come it is less efficient than the Carnot cycle? I thought the loss of efficiency in the ideal Rankine cycle comes from the heat addition that is not at the maximum temperature?
     
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