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Heat transfer through finite temperature difference

  1. Oct 5, 2015 #1
    Hello,

    Heat transfer through finite temperature difference is known a irreversible process because heat cannot be transferred from cold to hot temperature without doing any additional work. But, how this transfer affect the efficiency of the system? How this heat transfer decrease the amount of work for irreversible process?

    Another question, why heat in reversible process is more than a irreversible one? Does heat transfer through finite temperature difference decreases it?

    Thank you
     
  2. jcsd
  3. Oct 5, 2015 #2
    Based on the Carnot process:

    Imagine you want to heat your room in winter. You'd like to have it at a constant temperature T using a heat pump. To be able to transfer the heat to your room, the compressor has to increase the temperature of the fluid to T+ΔT.

    The heat your room receives is Q = T ⋅ ΔsR
    The heat your heat pump provides is Q = (T + ΔT) ⋅ ΔsHP

    Q must be the same for both cases, but the entropy change is different, the process is irreversible.

    The work input in the compressor is Win = ((T + ΔT) - Tamb) ⋅ ΔsHP

    reversible heating (without ΔT): COPrev = (T ⋅ ΔsHP) / ((T - Tamb) ⋅ ΔsHP)
    irreversible heating (with ΔT): COPirev = (T ⋅ ΔsR) / (((T + ΔT) -Tamb) ⋅ ΔsHP)

    → COPirev < COPrev

    As the benefit of heating is to increase the temperature, the benefit's larger, if the temperature is higher. As the temperature is lower (- ΔT) the benefit decreases.

    Imagine now you want to use the heat you put into the room to run the heat pump and produce electricity. You transfer the heat at T to run the cycle which gives you an output of work of Wout = (T - Tamb) ⋅ ΔsR

    The exergy is the part of the heat, which can be transformed into work again (in a Carnot process).

    E = Q ⋅ (1 - Tamb / T)

    Wout - Win = Q ⋅ [(1 - Tamb / T) - (1 - Tamb / (T + ΔT)] = Q ⋅ Tamb ⋅ (1 / T - 1 / (T + ΔT))

    → Q ⋅ Tamb ⋅ (1 / T - 1 / (T + ΔT)) = Tamb ⋅ ΔsHP ⋅ ΔT/T = Tamb ⋅ ΔsR ⋅ (1 - T / (T + ΔT)) ... "loss of work"

    As the energy after the heat transfer is at lower temperature level, it lost part of his capability to work (you can do "more" with higher temperature level).

    Could you please specify, I don't really understand what you mean. Are you referring te exergy/anergy?
     
    Last edited: Oct 5, 2015
  4. Oct 5, 2015 #3
    Ah sorry, maybe I had it wrong, does Q in reversible process is higher then Q irreversible? Maybe I am wrong about this information
     
  5. Oct 5, 2015 #4
    Do you mean the irreversible increase of heat due to friction? Or do you mean that a larger heat flux can be realized with a heat source of determined temperature and no necessary temperature difference ΔT?

    PS.: I edited my last post, I found a mistake
     
  6. Oct 5, 2015 #5
    Hello again,

    I mean the value of Qirr for a process is it superior to the same process but the reversible one. if yes does only the friction cause it or there is another cause? For exemple, a heat transfer through a finite temperature difference is it more than a reversible heat transfer? and if yes, why? Can we say because there's increase of entropy?
     
    Last edited: Oct 5, 2015
  7. Oct 5, 2015 #6
    So can we say; in irreversible process, some of the heat generated cannot make useful work right?
     
  8. Oct 6, 2015 #7
    Generally heat cannot transformed into work entirely. The amount of energy, that can be transformed depends on the temperature of the ambiance - It's called exergy E (contrarily to anergy B).

    E = Q ⋅ (1 - Tamb / T) ... for T > Tamb
    E = Q ⋅ (Tamb / T - 1) ... for T < Tamb

    Now E represents the amount of heat, that can be converted into work in a Carnot process. As it can converted into work (and back again into heat and back again into work,...) it's the reversible part of Q.

    Considering irreversibilty due to the ΔT at the heat transfer. Q transferred must be the same (due to energy consistency), but as the temperature will be lower after the transfer, part of the Q is lost, if you wanted to convert it back.

    So Q = Qrev + Qirr, whereas

    Qrev = Q ⋅ (1 - Tamb / T)

    The larger ΔT, the smaller T and the amount of Qirr increases. So the irreversible part of the heat is larger compared to a process with perfect heat transfer.

    And as already indicated in my first post, if the same heat is exchanged at different temperatures, the entropy change is different; it is larger at the part with the smaller temperature. This entropy increase is the measure for the irreversibility.

    Also friction causes irreversibilty. Part of the work of the heat pump has to be done to overcome the friction. This additional force is irreversibly converted into heat. In case of the heat pump it is not that bad, because you can use this heat to warm up the room, but if you imagine a refrigerator, which is basically the same machine, the transformed work into heat by friction isn't available any more for compression and has to be discharged additionally.

    Yes, part of the work transformed into heat cannot be transformed back and is lost for other purposes.
     
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