7171

05-10-2013, 07:23 AM

Using the formula for calculating the maximum theoretical COP, I mathematically come to the conclusion that it is the maximum COP per stage, while adding stage seems to be much more efficient.

I've seen that formula in many places, here is one of them:

http://en.wikipedia.org/wiki/Coefficient_of_performance

So I used:

Th=24C=297K

Tc=4C=277K

COP (heating) = Th/(Th-+Tc) = 297/(297-277) = 297/20 = 14.85

COP (cooling) = Th/(Th-+Tc) = 277/(297-277) = 277/20 = 13.85

I'm cool with that.

I'll assume it takes a 1 joule to raise the temperature by 1 degree. I know it's no real world value, but that is a constant and therefore don't affect the end result which is a percentage.

While I cool 13.85 degrees for 1 joule of energy provided to the system, I calculate the amount of energy I need to provide for 20 degrees, which is 1j*20K/13.85K = 1.444 joule

Now I do the very same process for in two steps, i.e. (Th, Tc) of (4, 14) and (14, 24) instead of one (4, 24) step.

COP (cooling step 1) = Th/(Th-+Tc) = 277/(287-277) = 277/10 = 27.7

COP (cooling step 2) = Th/(Th-+Tc) = 287/(297-277) = 287/10 = 28.7

1j*10K/27.7K = 0.361

1j*10K/28.7K = 0.348

When I ad the energy required by both step, I get (0.361 + 0.348) = 0.709 joules.

That is 50.87% more efficient.

Four steps (5 Kelvin each) results in only 0.351 joules required, for the very same cooling.

As for heating, the same kind of calculation shoes slightly higher COP.

Approaching 0 K (of temperature differential), there is nearly no energy required for all the cooling, which would also mean that it would take an infinite amount of space for the equipment and an infinite amount of time to do the cooling.

I know that the real world isn't as perfect as the theory and that there are losses, but it seems that very significant energy saving could be archived. While also significantly increasing the price of the equipment, it looks like it could be worth it, at least for high temperature differentials, like a freezer in summer time, or heating indoor space during winter.

I've seen multi-stage compressors, with in-between pressure(s), where there is a theoretical gain in efficiency for each stage that is added. After 4 or 5 stages, though, it's becomes so very insignificant. Yet, it is significant enough to be manufactured and used in industries.

The only multi-stage heat pump I've seen seems to refer either to a dual-speed compressor; two independent heat pumps, one runs more continuously and only the other start and stop as the temperature requires it so as to save on start-up and stabilization of the system; and some with drying stage of some sort...

Therefore, my final assumption, or my assumption so far, is that I am missing something. I misunderstand the formula, its definition, or some basic concept, or perhaps my math is wrong somewhere.

May someone please either tell me how great of an idea this is or, more likely, point out what I am missing, where my analysis fails, or where it is actually implemented?

I've seen that formula in many places, here is one of them:

http://en.wikipedia.org/wiki/Coefficient_of_performance

So I used:

Th=24C=297K

Tc=4C=277K

COP (heating) = Th/(Th-+Tc) = 297/(297-277) = 297/20 = 14.85

COP (cooling) = Th/(Th-+Tc) = 277/(297-277) = 277/20 = 13.85

I'm cool with that.

I'll assume it takes a 1 joule to raise the temperature by 1 degree. I know it's no real world value, but that is a constant and therefore don't affect the end result which is a percentage.

While I cool 13.85 degrees for 1 joule of energy provided to the system, I calculate the amount of energy I need to provide for 20 degrees, which is 1j*20K/13.85K = 1.444 joule

Now I do the very same process for in two steps, i.e. (Th, Tc) of (4, 14) and (14, 24) instead of one (4, 24) step.

COP (cooling step 1) = Th/(Th-+Tc) = 277/(287-277) = 277/10 = 27.7

COP (cooling step 2) = Th/(Th-+Tc) = 287/(297-277) = 287/10 = 28.7

1j*10K/27.7K = 0.361

1j*10K/28.7K = 0.348

When I ad the energy required by both step, I get (0.361 + 0.348) = 0.709 joules.

That is 50.87% more efficient.

Four steps (5 Kelvin each) results in only 0.351 joules required, for the very same cooling.

As for heating, the same kind of calculation shoes slightly higher COP.

Approaching 0 K (of temperature differential), there is nearly no energy required for all the cooling, which would also mean that it would take an infinite amount of space for the equipment and an infinite amount of time to do the cooling.

I know that the real world isn't as perfect as the theory and that there are losses, but it seems that very significant energy saving could be archived. While also significantly increasing the price of the equipment, it looks like it could be worth it, at least for high temperature differentials, like a freezer in summer time, or heating indoor space during winter.

I've seen multi-stage compressors, with in-between pressure(s), where there is a theoretical gain in efficiency for each stage that is added. After 4 or 5 stages, though, it's becomes so very insignificant. Yet, it is significant enough to be manufactured and used in industries.

The only multi-stage heat pump I've seen seems to refer either to a dual-speed compressor; two independent heat pumps, one runs more continuously and only the other start and stop as the temperature requires it so as to save on start-up and stabilization of the system; and some with drying stage of some sort...

Therefore, my final assumption, or my assumption so far, is that I am missing something. I misunderstand the formula, its definition, or some basic concept, or perhaps my math is wrong somewhere.

May someone please either tell me how great of an idea this is or, more likely, point out what I am missing, where my analysis fails, or where it is actually implemented?