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 …

In 2004 I invented a new heat pump which I grandiosely called the Barton Drying Engine (BDE). This has the following features:

- Mechanically expand a parcel of moist air; this lowers the temperature of the air as well as the partial pressures and densities of its component gases.
- Condensation forms immediately when dew point is reached, and any further expansion takes place along a moist adiabat.
- After the expansion is complete, collect the condensed droplets and remove them from the parcel of air.
- Re-compress the parcel of air back to ambient pressure; this involves an increase in temperature and partial pressures and densities. Re-compression takes place along a dry adiabat. Overall, work must be supplied to complete the cycle.
- The exhaust stream is hotter and drier than the inlet air stream.

I published a theoretical analysis of the BDE cycle in 2008 [1], with the main assumptions that air and water vapour are ideal gases with constant specific heat capacities. The principal results are shown in Figure 1 (click to expand).

Figure 1: Results from [1] for the loss-free condensation heat pump. The inlet air stream is saturated and at the temperature shown. The results also depend on the expansion ratio. Clockwise from top left: specific mass change in water vapour during cycle [kg/kg dry air], specific work output per cycle [J/kg dry air], temperature increase from inlet to outlet [°C],

*COP*as a heat pump [dimensionless].Here the Coefficient of Performance (bottom left diagram in Figure 1) was defined by

*COP = m*

_{a}C_{aP}(T_{4}-T_{1}) / Win which

*C*is the specific heat capacity of dry air at constant pressure,_{aP}*m*is the mass of dry air in the parcel,_{a}*T*is the outlet temperature,_{4}*T*the inlet temperature and_{1}*W*is the work required to complete the cycle. In words, the*COP*is the desired output (heat increase in the parcel of air) divided by the necessary input (work to complete the cycle).Let’s remind ourselves about Carnot heat pumps, as shown in Figure 2. This depicts a theoretical heat pump that takes heat energy

*Q*from a reservoir at temperature_{L}*T*and delivers heat energy_{L}*Q*to a reservoir at temperature_{H}*T*. This requires work_{H}*W*.The

*COP*of the heat pump is*COP*=*Q*. But conservation of energy gives_{H}/ W*Q*and so_{L}+ W = Q_{H}*COP = Q*1

_{H}/ (Q_{H}– Q_{L}) =*/ (*1

*– Q*

_{L}/Q_{H}) .The 2nd Law of Thermodynamics gives that for a reversible process

*Q*(temperatures in Kelvin),

_{L}/Q_{H}= T_{L}/T_{H}so that the

*COP*can be written*COP_*Carnot

*=*1

*/ (*1

*– T*

_{L}/T_{H}) = T_{H}/ (T_{H}– T_{L}).This is the highest

*COP*that a heat pump operating on a closed cycle between*T*and_{L}*T*can have._{H}Now let’s take the

*COP*shown in Figure 1 (bottom left) for the case when*T*= 25°C and compare it to the Carnot_{1}*COP*(with*T*=_{L}*T*and_{1}*T*=_{H}*T*). I’ll now also explicitly take account of the water vapour that passes through my heat pump as follows_{4}*COP_*BDE

*= { m*

_{a}C_{aP}+ (m_{v }- δm_{v})C_{vP}} (T_{4 }-T_{1}) / W.Here

*m*is the mass of vapour in the parcel of air at the inlet and_{v }*δm*is the mass of vapour that condenses into water._{v}*C*is the specific heat capacity of water vapour at constant pressure. (This expression for_{vP}*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*T*, and gives out condensed water at temperature_{1}*T*and drier air at temperature_{3}*T*. Let_{4}*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 = δm*=

_{v }× { h_{v}(T_{1}) – h_{w}(T_{3}) }*δm*

_{v}× { h_{v}(T_{1}) – [h_{v}(T_{3}) – L(T_{3})] }in which

*L(T*is the latent heat at_{3})*T*,_{3}*i.e.*the specific enthalpy required to evaporate water. Since we are assuming ideal gas theory with constant specific heats, it follows that*Δ = δm*

_{v}× { C_{vP}(T_{1}– T_{3}) + L(T_{3}) }.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:

*T*= 25°C and saturated_{1}lowest temperature:

*T*= 4.9°C_{3}outlet temperature:

*T*= 52.7°C_{4}expansion ratio:

*r*= 1.6water condensed: 0.01067 kg/kg dry air

work requirement:

*W*= 1,765 J/kg dry airlatent heat:

*L(T*= 2,489,600 J/kg water_{3})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.

Reference

[1] N G Barton, “An evaporation heat engine and condensation heat pump”,

*ANZIAM Journal*,**49**(2008), 503-524.
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