I am trying to show how a heat pump works by roughly identifying one set of temperatures and pressures for each stage of the heat pump cycle. I want to understand how propane can be useful as a refrigerant when it boils at minus 43°C at one-atmosphere pressure.
I am considering:
. an air-source, hot-water, heat pump,
. with a propane refrigerant,
. the heat source is the air outside the house,
. the hot water tank is also outside the house,
. the target hot water tank temperature = 60°C
. the current hot water tank temperature = 50°C, and
. the air temperature outside the house used as the heat source = 17°C.
The pressure-volume diagram suggests there are only a few temperatures and pressures to work out. In the cold, low-pressure space of the heat pump, there is:
. one low pressure (Pc), and
. one low temperature (Tc), which is the boiling point of propane at this pressure.
In the hot, high-pressure space, there is:
. one high-pressure (Ph)
. one high-temperature (Th), which is the boiling point of propane at this pressure, and
. a second superheated temperature (Ts) which is higher than Th.
****** Propane Boiling points
Here are propane boiling points from the internet calculator. Below I use the rounded values in brackets to keep it simpler.
Pressure Propane boiling point
1 Atm Minus 43°C
5.2 Atm (say 5 Atm) 14.9°C, (say 15°C)
16 Atm 74.47°C (say 74°C}
https://www.omnicalculator.com/chemistry/boiling-point#boiling-point-definition
***** The low-pressure space
To estimate the temperature in the heat pump's cold, low-pressure space, consider the temperature after evaporation and before compression. The pressure-volume diagram shows that the evaporated propane is a saturated gas. Since the gas is at its boiling point, the pressure will be the vapour pressure of the propane at this temperature.
This refrigerant has passed through a heat exchanger and is warmed by air at 17°C. I guess that the refrigerant will be cooler than the air, So I assume that Tc = 15°C.
** Is this reasonable, or should it be 17°C?
From my propane boiling point table, the propane vapour pressure at 15°C is 5 atm, meaning that Pc = 5 atm.
** Is this reasonable?
A controller keeps this refrigerant slightly superheated to prevent liquid from entering the compressor and damaging the heat pump. I will ignore this.
** Would it be easy to include this in my figures?
The propane after the expansion valve and before the evaporator will be at the same Tc = 15°C and Pc = 5 atm, still at boiling point but with a mixture of propane liquid and vapour.
****** The high-pressure space
To estimate the temperature in the heat pump's hot, high-pressure space, I consider the temperature after condensation and before expansion. Here, the pressure-volume diagram shows that the condensed propane is a saturated liquid. This liquid is at its boiling point, so the pressure will be the vapour pressure of the propane at this temperature.
This refrigerant needs to heat the hot water to 60°C, so I have assumed that the propane needs to get a good bit hotter than that, so Th = 74°C.
** Is this reasonable?
From my propane boiling point table, the propane vapour pressure at 74°C is 16 atm, meaning that Pc = 16 atm.
** Is this reasonable?
The pressure-volume diagram shows that, after compression, the propane is at the same pressure as after condensation, but it is now a superheated vapour. Its temperature will be higher than that of the saturated liquid after compression, Th = 74°C. Assume Ts = 80°C.
** Is this reasonable?
*******Is this real or an ideal
** Is this pressure-volume representation of a heat pump an ideal?
** Does a real propane heat pump come close to this?
Now I wonder how a real propane heat pump could behave like this ideal. (1) In vaporisation, the heat pump transfers exactly the latent heat of vaporisation into the propane, and (2) the compression does superheat the propane, so condensation transfers to the hot water its superheat and then exactly the latent heat of vaporisation.
** Any references to help here?