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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?

cool_tech
15-10-2013, 10:26 AM
your theory looks good, an easy way is energy in KW electricity to KW refirgeration . eg york chiller at part load has 100 kw of electrical energy in to make 700 kw of refrigeration. COP 7

hookster
15-10-2013, 04:20 PM
:confused: Must be a phenomenal chiller!! Part load COP of 7 I think the salesman has been telling porky pies!

desA
18-10-2013, 07:27 PM
COP = (What you get) / (What you pay for) Do make sure that you inject ALL the energy (power) paid for - compressor, fans, pumps, etc.

moideen
18-10-2013, 07:47 PM
:confused: Must be a phenomenal chiller!! Part load COP of 7 I think the salesman has been telling porky pies!

It is real. The latest Mitsubishi centrifugal ETA series chiller claim full load cop is 6.4 and part load cop is 10.2 and maximum part load cop is 23.8!

hookster
19-10-2013, 10:00 AM
I understand theoretical part load COP sounds good on manufacturer blurbs! But the reality is that this is marketing bull. You have a massive chunk of unused capital equipment sitting unused to get these high COP figures.

They are usually achieved at almost impossible conditions to replicate (enthalpy of Air in / air out etc.) in the real world besides for a split second in time

There is no real point to part load conditions, if they are not a large percentage of operating time.
Then you need to question why this was designed for small peak loads anyway. The move to SEER rating gives a better true interpretation of operating conditions although this is not perfect either.

Peter_1
19-10-2013, 08:28 PM
Info abouth this wonderful machine?

7171
23-10-2013, 03:08 AM
Thanks for your replys. This is going slightly off topic so I would like to bring attention back to the essence of what I was asking.


COP = (What you get) / (What you pay for) Do make sure that you inject ALL the energy (power) paid for - compressor, fans, pumps, etc.

I understand that in the real world, there are losses. Some here are talking about whether some figure are or can be archived in the real world. One could add that the cost of the parts should be included in the practicability of such implementation, which is true, important to remember, but that does not address the matter in question.

Even though energy consumed by fans to speed and energy loss form the compressor or pump affect the overall COP of a system we would build, it has nothing to do with the efficiency of the "cooling/heating" itself.


Approaching 0 K (of temperature differential), there is nearly no energy required (edit: the COP approach infinity) 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.

Pump and compressor have their own efficiency, which will affect the overall efficiency of a device in the real world, but not the efficiency on the cooling or heating effect itself. Those are outside the scope of this thread.

My interrogation here is about theory, i.e. what really is the maximum theatrical amount of energy we can move with a given amount of... "work" (or electricity or energy, or whatever you want to call it)?

From my understanding, the Carnot cycle is defined as that "maximum", either using work to move heat, or converting a temperature differential into work. However, using the formula that theory proposes, I seem to prove that it can't describe the maximum it claims to describe, but at best what .

So, logic tells me that at least one of the following statement would be true:
1. I misunderstand the Carnot's theorem. (e.g. it does not describe the maximum efficiency)
2. There is a flaw in my analysis. (e.g. Step X does not lead to Y)
3. I miscalculated something.
4. The Carnot's theorem proves itself to be wrong.
5. I am missing something really important here.

Which option do you vote for? And how do you back your vote?

mad fridgie
23-10-2013, 05:04 AM
Just a simple one first, did you add the work done from the previous stage, to the work needed to be done by the next stage. Or does this extra energy within the the stages magically dis-appear.

7171
23-10-2013, 06:57 AM
Just a simple one first, did you add the work done from the previous stage, to the work needed to be done by the next stage. Or does this extra energy within the the stages magically dis-appear.

Using the wrong formula, adding the heat from the previous step may have affected cooling negatively, while for heating, it would just have increase the efficiency even further.

HOWEVER:

Over the fact that I made a mistake in:
COP (heating) = Th/(Th-+Tc) = 277/(297-277) = 277/20 = 13.85
which should have been
COP (cooling) = Tc/(Th-+Tc) = 277/(297-277) = 277/20 = 13.85
(which I couldn't edit,)

there is the fact that, even though the formula for the COP is correct, the Carnot's theorem that defines the maximum efficiency is 1-(Th/Tc), which is different from the one for the COP.

With the right formula, more energy is required as steps are added. What a surprise!

Thanks a lot for your pointer. The revision brought me to realise I wasn't using the right formula in the first place.

COP: http://fr.wikipedia.org/wiki/Coefficient_de_performance
Carnot's theorem: http://en.wikipedia.org/wiki/Carnot%27s_theorem_%28thermodynamics%29

7171
28-12-2013, 07:06 PM
The energy doesn't magically disappear. In the example, each stage require respectively 0.361 and 0.348 joules for a total of 0.709 joules, which compares to 1.444 joules when done in one step.


It seems that I didn't think much before writing my last message because both formulae are basically the same i.e. the COP (Th/(Th-Tc)) is the reciprocal of the efficiency (1-(Tc/Th)). I had also wrote 1-(Th/Tc) instead 1-(Tc/Th).

So I'm still in the same boat. Any ideas?