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Efficiency is always what you get out of a system divided by what you put in. If something is 50% efficient, then we needed two times the energy to get half the work in return. Mathematically speaking the formula for 50% efficiency is ½. Which is also to say, if we got 500 Btu/hour from our air-conditioner and used 1000 Watts to get it, 500 / 1000 = ½ = 50% efficient.
Thermal efficiency can be viewed in the same way. A coil operates at 45 degrees Fahrenheit, and the condenser operates at 100 degrees Fahrenheit. The efficiency becomes: What we got divided by what it took, minus were we started. Got that? What we got – 45, what it took – 100 minus where we started – 45. Mathematically, 45 / (100-45) = .82 or 82%, that is thermal efficiency.
On the 410a diagram below I drew a cycle of the above conditions with the compression line following the entropy line, or line of perfect compression, which put the temperature at 135F. I then marked the enthalpy line for the heat of compression at 185 and 197 for a specific heat of compression of 12 Btu/pound of gas.
Since the thermal efficiency of this process is 82%, dividing 12 by .82 gives a new specific enthalpy of 14.6 Btu/pound.
Using 185 (the original starting point) we add 14.6 to get 199.6, or about 200 Btu/lb
Re-plotting the compression line below the new discharge temperature emerges. (Temperature lines are in 10 degree increments).
The process can be used in the field using the gauge pressures as previously described. Actual discharge temperature will be little higher depending on the amount of friction in the compressor. If the temperature is much lower, determine if liquid may be entering the compressor
I worked for over thirty years in the HVACR industry. I have designed, installed, serviced, and trouble shot units of various types throughout the years. The posts here are information based on that experience, I hope you find them useful. If you have a different experience, please comment.