Hot and Cold Generation: A Physics Question

In a current G.Q. thread, this quote is provided by a Doper

Is this true?



How are air conditioner star ratings calculated?

The U.S. uses a different rating system (non metric), but the end result is the same. If the EER is 2.0, one calorie of heat produces 2 calories of cooling.

Thermodynamically speaking, there is a theoretical limit to how high the COP can be. It’s equal to:
[ul]1 / (1 - T[sub]COLD[/sub] / T[sub]HOT[/sub])[/ul]If it’s 310K outside and 300K inside, the COP has a theoretical maximum of 31.

The way the original quote is worded, I think you need to factor in the efficiency of the power generator (power plant) and the distribution system. If the EER is 2.0 and the power plant efficiency is 33%, you need to burn 3 calories to produce 1 calory of electricity which then produces 2 calories of cooling.

I only just noticed this thread, but as I said in the other thread, due to the laws of thermodynamics you must always generate more heat than ‘cold’, the exact ratio is dependant on the efficency of the system.

Expanding slightly on the example given in the other thread and considering a fridge: taking a room containing a fridge to be a closed system, though inside the fridge the temp. will drop, the overall temp. of the room will rise due to the fridge, even taking into account the cooled region.

Government ignorance exposed! Let’s hope Australians are not trying to counteract global warming by turning on their air conditioners full-blast and opening all their doors and windows!

(From the site linked by Desmostylus .)

Huge Huge Misconcept here. EER (Energy Efficiency Rating) is a bastardized measure, conceived by government lackeys intent on confusing the public. EER is the cooling capacity (in Btu/hr) divided by the electrical input (watts). Therefore, EER = COP x 3.412.

Air conditioners might produce a COP of 3 and an EER of over 10. In this case, the Air Conditioner removes three units of heat from the room and exhausts 4 units of heat outside (the three from the room and one from the work intput to the refrigeration cycle.)

In any event, no the French guy is way wrong. (A French environmentalist? One more strike and he’s out.)

A quick back of the envelope computation tells me that the overall efficiency from the heat produced by burning a lump of coal at the generating plant to the actual heat removed from a building into the atmosphere is between 30% and 40%. So the Minister doesn’t seem to me to be that far off.

I don’t think the thermodynamic efficiency of 1-T[sub]out[/sub]/T[sub]in[/sub] is applicable in this case. The total energy contained in the lump of coal that doesn’t get transformed to heat on burning doesn’t show up as heat does it?

To answer the question I think you’re asking:

The total energy in the lump of coal (meaning the chemical energy produced by reacting all the carbon, hydrogen and sulphur in the coal with oxygen) all leaves the power station as either heat or electricity, except for a small amount of unburned carbon in the ash.

To address the OP:
If you choose to consider the fuel into the power station as your energy in, and the cooling effect of the air conditioner as your energy out, you could figure something like:
GCV exported efficiency of power plant: 33%
Efficiency of transmission and distribution: 90%
COP of air conditioner: 3
Giving a cooling effect of approximately 0.9 joules for every 1 joule of coal burned.

The heat delivered to the environment will be 1.9 joules, of which:
0.67 joules at the power station (to cooling water and flue gases)
0.033 joules in the transmission and distribution system (wires and transformers)
0.9 + 0.297 = 1.197 joules transferred to the atmosphere from the condenser or cooling tower of the air conditioner.

So for each calorie of cooling you produce a little over 2 calories of heat, using these assumptions.

Does anyone disagree that the heat transferred by the air conditioner should be included in reckoning the amount of “heat produced”?

Not on my part. Our EER is in the same units as the COP. Watts/Watt. Or calories/calorie. With the U.S. ratings, you need to do a unit conversion, from Btu/Watt to calories/calorie, but you’ll still get the same answer in the end.

I guess someone’s ignorance has been exposed. You need to understand how the terms “input” and “output” are defined. “Input” only includes electrical power input. “Output” is the amount of useful heat transferred, and because that includes heat shifted from point A to point B, you can easily get efficiencies greater than 100%.

In the context of the original quote, which refers to “generating heat”, it probably shouldn’t be included. That heat existed independently of the air conditioner, i.e. the use of the air conditioned did not “generate” that heat.

Sorry, I should have read a little more carefully. I didn’t realize Australia had shifted to consistent units.

However, on the issue of efficiency, the original statement is still wrong. At worst it takes about three units of heat to generate one unit of electricity, which then transfers three units of heat from the conditioned space to the environment. 6 units total are rejected to the environment. However, only 3 of these were “generated” in that fuel was burned. So the original statement is still wrong by a factor of three. (Which isn’t close enough to count in horseshoes and maybe not even in hand grenades.)