The cooling capacity of air-conditioner

[superkaho]

I would like to ask why the cooling capacity of air-conditioner is 2-3 time more than the work done by the compressor??

According to the conservation of energy, the amount of work input should be equal to the amount of heat absorbed(cooling capacity), isn't it??

Can anyone explain it, thanks in advance.

 

[rock.freak667]

The total energy input, some goes into cooling while some heat is rejected (that is how it must operate else it would violate the second law of thermodynamics).

The refrigerant/coolant used has a low boiling point usually, so that requires more energy to convert from liquid to gas while all the compressor does is basically compress the gas.

 

[Drakkith]

The only thing the compressor does is compress the warm gas. This heats it up further, and when it gets sent through the condenser coil, it loses that excess heat from the compressor and the room into the outside air. After that the now cool gas is actually a liquid. This liquid flows into an expansion valve where the pressure is reduced, causing the part of the liquid to boil and turn into a gas. This removes more heat from the remaining liquid, and the mix of cold gas/liquid coolant flows into the evaporator coil to absorb warm air from the room.

From wikipedian on AC:

Air conditioner equipment power in the U.S. is often described in terms of "tons of refrigeration". A "ton of refrigeration" is approximately equal to the cooling power of one short ton (2000 pounds or 907 kilograms) of ice melting in a 24-hour period. The value is defined as 12,000 BTU per hour, or 3517 watts.[14] Residential central air systems are usually from 1 to 5 tons (3 to 20 kilowatts (kW)) in capacity.

This is different from the electrical power used by the AC unit. In fact, they have an efficiency rating called SEER (Seasonal Enery Efficiency Rating) for AC units.
From wikipedia on SEER:

The SEER rating of a unit is the cooling output in Btu (British thermal unit) during a typical cooling-season divided by the total electric energy input in watt-hours during the same period. The higher the unit's SEER rating the more energy efficient it is.

For example, consider a 5,000-British-thermal-unit-per-hour (1,500 W) air-conditioning unit, with a SEER of 10 BTU/W·h, operating for a total of 1000 hours during an annual cooling season (e.g., 8 hours per day for 125 days).

The annual total cooling output would be:

5000 BTU/h × 8 h/day × 125 days/year = 5,000,000 BTU/year
With a SEER of 10, the annual electrical energy usage would be about:

5,000,000 BTU/year / 10 BTU/W·h = 500,000 W·h/year
The average power usage may also be calculated more simply by:

Average power = (BTU/h) / (SEER) = 5000 / 10 = 500 W
If your electricity cost is 20¢/kW·h, then your cost per operating hour is:

0.5 kW * 20¢/kW·h = 10¢/h

As you can see the amount of cooling done by an AC unit will change as more efficient ones are designed.

 

[superkaho]

The total energy input, some goes into cooling while some heat is rejected (that is how it must operate else it would violate the second law of thermodynamics).

The refrigerant/coolant used has a low boiling point usually, so that requires more energy to convert from liquid to gas while all the compressor does is basically compress the gas.

but the point is that the output(cooling capacity) is even more than the input(work done by the compressor), not the opposite. why? energy lost to the surroundings or by itself seems difficult to explain it...

 

[rock.freak667]

The work input to the compressor as Drakkith pointed out is only used to compress the gas (which is finite depending on the temperatures of the gas before and after). Where the cooling takes place is the space to be cooled. This space has nothing to do with the compressor's work input.

So you could be cooling 5 kW of space, but your compressor only requires 1 kW to compress the gas.

COP = 5/1 = 5.

 

[superkaho]

The work input to the compressor as Drakkith pointed out is only used to compress the gas (which is finite depending on the temperatures of the gas before and after). Where the cooling takes place is the space to be cooled. This space has nothing to do with the compressor's work input.

So you could be cooling 5 kW of space, but your compressor only requires 1 kW to compress the gas.

COP = 5/1 = 5.

When the problem is considered for each part such as (1) compressor (2) cooling part. I understand the functions and the amount of energy transfer to be defined. For example, the work done by the compressor is W while the heat absorbed by the cooling part is Qc. I also understand that the work input by compressor is independent of the heat transfer by the cooling part.

However, when I consider the air-conditioner as a whole system. I still cannot figure out the reason why the COP = Qc/W is larger than 1 in terms of the conservation of energy. Could you explain more about its mechanism? thanks~

 

[russ_watters]

but the point is that the output(cooling capacity) is even more than the input(work done by the compressor), not the opposite. why? energy lost to the surroundings or by itself seems difficult to explain it...

Related example: A pump used to lift water has an efficiency based on an input of electrical energy and an output of pressure and flow...but what if you're not interested in the pressure and flow of the pump but rather are interested in the temperature (energy) of the water that it is pumping?

Now consider the the walls of your house: energy spontaneously flows from an area of high temperature to an area of low temperature without an input of energy. So Eout/Ein is infinite. So is that a violation of conservation of energy?

Put the two together and you have COP. There's a reason they call it COP: It is not "efficiency".

 

[rock.freak667]

However, when I consider the air-conditioner as a whole system. I still cannot figure out the reason why the COP = Qc/W is larger than 1 in terms of the conservation of energy. Could you explain more about its mechanism? thanks~

In order to analyze the system as a whole, you need to get the heat/work transfers within each individual section.

As russ_waters pointed out COP is not a measure of efficiency but how much cooling you can get in the space for each unit of power to the compressor.

You are not violating conservation of energy since the energy you input to the compressor only does work in compressing the gas.

 

[TerryW1]

I agree with superkaho. The math doesn't make sense. It is a terminology problem but it should be explainable but none of the above answers explain it.
Look at "5,000,000 BTU/year / 10 BTU/W·h = 500,000 W·h/year
The average power usage may also be calculated more simply by:"
Hello. 1 WH = 3.412 BTU However, the above equation says 10 BTU = 1 wh. There is the problem.

To understand the true energy efficiency we should be looking at the amount of cooling as energy output as % of the energy input. It must be much less than one. i would guess less than 20%. But that is just a guess.
To use an opposite analogy is a power station. The electrical energy output is about 33% of the net neat value input for a coal fired power station.

 

[Drakkith]

From wikipedia on thermodynamic efficiency which should explain this.

Their efficiency (of a heat pump, AC, ETC) is measured by a coefficient of performance (COP). Heat pumps are measured by the efficiency with which they add heat to the hot reservoir, COPheating; refrigerators and air conditioners by the efficiency with which they remove heat from the cold interior, COPcooling:

(Insert some equations i couldn't copy over)


The reason for not using the term 'efficiency' is that the coefficient of performance can often be greater than 100%. Since these devices are moving heat, not creating it, the amount of heat they move can be greater than the input work. Therefore, heat pumps can be a more efficient way of heating than simply converting the input work into heat, as in an electric heater or furnace.

 

[russ_watters]

I agree with superkaho. The math doesn't make sense. It is a terminology problem but it should be explainable but none of the above answers explain it.

Welcome to PF.

Such confusion on terminology may come from a somewhat careless mixing of the concepts of COP and efficiency by websites such as the one you quoted. I think that is done to "dumb down" the concept (treat COP as if it were the same as efficiency) for a certain audience and when someone looks deeper, they uncover the flaw. The flaw is simply that they shouldn't be using the word "efficiency" at all.

Look at "5,000,000 BTU/year / 10 BTU/W·h = 500,000 W·h/year
The average power usage may also be calculated more simply by:"
Hello. 1 WH = 3.412 BTU However, the above equation says 10 BTU = 1 wh. There is the problem.

You missed the part about where the 10 btu/w-h came from: that's a hypothetical example performance of an air conditioner and it most definitely is a reasonable value because...

To understand the true energy efficiency we should be looking at the amount of cooling as energy output as % of the energy input. It must be much less than one. i would guess less than 20%. But that is just a guess.
To use an opposite analogy is a power station. The electrical energy output is about 33% of the net neat value input for a coal fired power station.

...the amount of cooling energy as a % of input is COP and you are very much wrong in your guess: typically, it is around 3 (or by your way of looking at it, 300%). So in the example they gave, it is COP=10/3.41=2.93 (or 293%).

Understanding how a COP can be >1 is the OP's question - your question was slightly different based on your guess that it couldn't be >1.

 

[superkaho]

Thanks all your replies, here is to summarize the key points we've made.

(1) COP is not the same as efficiency, but it can indicate how effective the air-conditioner to absorb heat by each unit of work input.
(2) Air conditioner is not an energy converter, but just a device to transfer heat from low temperature to high temperature.
(3) The law of conservation of energy should be considered in an isolated system, but NOT in air conditioner. As there is heat flow by the medium of refrigerant.
i.e. Qh = Qc + W

From wikipedia on thermodynamic efficiency which should explain this....
The reason for not using the term 'efficiency' is that the coefficient of performance can often be greater than 100%. Since these devices are moving heat, not creating it, the amount of heat they move can be greater than the input work...

(1) Can we claim that the air conditioner is NOT an energy converter? so the term 'efficiency' is not avaliable for it?

(2) The COP can be greater than 1... Is it correct to say that" As COP is NOT a measure of energy transformation, but the amount of heat transfer by each unit of input, so it is possible to be greater than 1."?

 

[rock.freak667]

(1) Can we claim that the air conditioner is NOT an energy converter? so the term 'efficiency' is not avaliable for it?

Yes as the air conditioning unit does not produce a "useful" power output (something that does mechanical work)

Efficiency = useful power output/power input.

(2) The COP can be greater than 1... Is it correct to say that" As COP is NOT a measure of energy transformation, but the amount of heat transfer by each unit of input, so it is possible to be greater than 1."?

More accurate to say that COP is the heat transferred (Cooling) for each unit of work input. But essentially yes.

 

[superkaho]

Thanks...
By the law of conservation of energy, the equation: Qh = Qc + W decribes the heat released from condenser is equal to the sum of heat absorbed by evaporator and the work by compressor.

However, when I investigate the COP = Qc/W. Though I accept COP can be greater than 1, but I still can figure out the depedent factors of COP in reality...
for example, what process involved in the compressor?? isothermal compression? adiabetic compression? etc... how would it affect the amount of work to be done by the compressor?
In the expansion valve, does it undergo adiabetic expansion? etc...

 

[rock.freak667]

However, when I investigate the COP = Qc/W. Though I accept COP can be greater than 1, but I still can figure out the depedent factors of COP in reality...
for example, what process involved in the compressor?? isothermal compression? adiabetic compression? etc... how would it affect the amount of work to be done by the compressor?
In the expansion valve, does it undergo adiabetic expansion? etc...

Ideally, I think it would undergo some sort of adiabatic compression and isentropically.

So the work done per unit time would be W = m_f(h₂-h₁)

Which would depend on the pressures and temperatures at the states of the compressor.