Wednesday, December 7, 2011

Beyond-Carnot heat pump?

Figure 3 below shows a heat pump that has a Coefficient of Performance superior to a Carnot heat pump operating over the same temperature range.  But that contradicts the 2nd Law of Thermodynamics.  How is this paradox to be resolved?

Please read on …


  • The exhaust stream is hotter and drier than the inlet air stream.




COP = ma CaP (T4-T1) / W


Let’s remind ourselves about Carnot heat pumps, as shown in Figure 2.  This depicts a theoretical heat pump that takes heat energy QL from a reservoir at temperature TL and delivers heat energy QH to a reservoir at temperature TH.  This requires work W.

Figure 2:  Illustrating heat transfer and work requirement of a heat pump.

The COP of the heat pump is COP = QH / W.  But conservation of energy gives QL + W = QH and so

COP = QH / (QH – QL) =   1 / (1 – QL/QH) .

The 2nd Law of Thermodynamics gives that for a reversible process

QL/QH = TL/TH    (temperatures in Kelvin),

so that the COP can be written

COP_Carnot = 1 / (1 – TL/TH) = TH / (TH – TL).

This is the highest COP that a heat pump operating on a closed cycle between TL and TH can have.

Now let’s take the COP shown in Figure 1 (bottom left) for the case when T1 = 25°C and compare it to the Carnot COP (with TL = T1 and TH = T4).  I’ll now also explicitly take account of the water vapour that passes through my heat pump as follows

COP_BDE = { ma CaP + (mv - δmv )CvP } (T4 -T1) / W.

Here mv is the mass of vapour in the parcel of air at the inlet and δmv is the mass of vapour that condenses into water.  CvP is the specific heat capacity of water vapour at constant pressure.  (This expression for COP_BDE gives almost identical results to the previous COP expression used in Figure 1.)

Figure 3 has the results …

Figure 3: Plots of COP_BDE and COP_Carnot for the case when the inlet air is at 25°C and saturated.

But this says COP_BDE is greater than COP_Carnot, which is not possible according to the 2nd Law of Thermodynamics.  I have done similar plots for other saturated inlet conditions, with similar results:  COP_BDE   >  COP_Carnot.  Assuming I haven’t made any numerical mistakes (I have checked very carefully!), and assuming the 2nd Law of Thermodynamics isn’t wrong (it isn’t!), how can this be so?

The paradox is resolved with the aid of Figure 4.

Figure 4: Schematic diagram of the condensation heat pump.

In Figure 4, the BDE is a mechanical device indicated by the yellow box.  It takes in work W and moist air at temperature T1, and gives out condensed water at temperature T3 and drier air at temperature T4.  Let h denote specific enthalpy and subscripts a, v and w denote dry air, water vapour and water respectively.  The change in enthalpy of the water vapour that condenses is

Δ = δmv × { hv(T1) – hw(T3) } = δmv × { hv(T1) – [hv(T3) – L(T3)] }

in which L(T3) is the latent heat at T3, i.e. the specific enthalpy required to evaporate water.  Since we are assuming ideal gas theory with constant specific heats, it follows that

Δ = δmv × { CvP (T1 – T3) + L(T3) }.

Both terms in Δ have the same sign, and Δ is large and positive.

By conservation of energy, if the condensed water loses an amount Δ in enthalpy, then the gaseous components that pass through the heat pump must gain enthalpy (W + Δ).  Therefore the COP of the BDE heat pump is

            COP_BDE  =  (W + Δ) / W  =  1 + (Δ/W).

As an illustrative example, consider the following data from the loss-free computations published in [1]:

inlet temperature:                T1 = 25°C and saturated
lowest temperature:            T3 = 4.9°C
outlet temperature:             T4 = 52.7°C
expansion ratio:                    r = 1.6
water condensed:                 0.01067 kg/kg dry air
work requirement:               W = 1,765 J/kg dry air
latent heat:                             L(T3) = 2,489,600 J/kg water

With the above data, Δ/W evaluates to 15.3, and so COP_BDE is 16.3, which agrees exactly with the results shown in Figure 3 at r = 1.6.  (For the same temperature range,  COP_Carnot  =  (52.7 + 273.15) / (52.7 – 25) = 11.8, also as shown in Figure 3.)

Thus the gas mixture (air plus vapour) that passes through the BDE has an excellent COP, which is sustained by the large loss of enthalpy in the vapour that is condensed. 

Finally, to resolve the paradox in words …

A Carnot heat pump must operate on a closed-loop cycle, with the state of the device at the end of the cycle identical to its state at the beginning.  Also the internal workings of the device must not generate any entropy, so that everything is reversible.

It’s easy to see from Figure 4 that the BDE heat pump does not satisfy this condition, and its COP, as I have defined it, is therefore not subject to the Carnot constraint.  The expansion and compression processes are indeed reversible, but removal of condensate means that the conditions at the end of the cycle are not the same as those at the beginning.

To finish, a brief practical comment.  Theoretically the BDE has great promise.  I am currently evaluating how the performance is affected by inevitable losses in the device.



1 comment:

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