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mfb
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You do not have a cyclic transformation, you need work as input and get heat flow from cold to warm as output. In other words, just a regular heat pump.
Why is this a good idea?Thermolelctric said:It seems good idea to describe the pump as a system of 4 reservoirs: Input reservoir, pump cold reservoir, pump hot reservoir, and output reservoir. And try to calculate heat pumped from input reservoire to the output reservoire, and the work done to pump.
Andrew Mason said:Why is this a good idea?
The whole idea behind a heat pump is to move heat from the cold reservoir to the hot reservoir. By the second law, this can only occur by doing work on the system. The only reason it makes any sense at all is if you get more heat flow out at the hot reservoir than you put in as work. Otherwise, as DrClaude has pointed out, you might as well have the input energy source just provide heat flow (e.g. an electric heater).
AM
For refrigerators is COP = Qc/Qh-Qc. Is that the other possible choice you are referring to?mfb said:You could define COP with the other possible choice as well. This would give more pumped heat, as you had to add your work input to this quantity. The result would be the same, but the notation would differ a bit.
With the pump doing work and requiring energy input. The heat energy extracted from the hot reservoir does no work, so the process only continues as long as external work is supplied. Not perpetually.Thermolelctric said:Mechanical work is no extracted, heat energy is extracted from pump hot reservoire.
Agreed, but it doesn't apply to your scenario since there is another final result: work was done on the system requiring external input energy.Thermolelctric said:A cyclic transformation whose only final result is to transfer heat from a body at a given temperature to a body at a higher temperature is impossible. - German physicist Rudolf Clausius
Agree to that?
DrClaude said:No, you reach [itex]T_c = T_h[/itex] and you don't have a heat pump anymore! Or, see my previous post, you just continue heating the cold source as would a electrical heater.
No, it doesn't. I can't even figure out why you think it does prove that. There is not even a hint of perpetual motion involved.Thermolelctric said:My COP=5 example cycle proves that COP=Th/(Th−Tc) is not correct for that COP value
This appears to be the source of confusion here. First of all, you are using a poor translation. Clausius actually stated the second law as:Thermolelctric said:Mechanical work is no extracted, heat energy is extracted from pump hot reservoire.
A cyclic transformation whose only final result is to transfer heat from a body at a given temperature to a body at a higher temperature is impossible. - German physicist Rudolf Clausius
Agree to that?
DaleSpam said:With the pump doing work and requiring energy input. The heat energy extracted from the hot reservoir does no work, so the process only continues as long as external work is supplied. Not perpetually.
Agreed, but it doesn't apply to your scenario since there is another final result: work was done on the system requiring external input energy.
That formula still works for an ideal heat pump. In your scenario, all you have is that Tc is a function of time due to the leakage.Thermolelctric said:Yes indeed but the amount of external energy that is input to the system do not satisfy the formula Qh/W=Th/(Th-Tc)
since Th and Tc is starting conditions.
Such a device is called a heat engine. What do you think the maximum efficiency of a heat engine is?Thermolelctric said:And we could use the heat from the process output to do the work of the pump.
Thermolelctric said:OK let's ditch the 4 reservoir idea.
DaleSpam said:Such a device is called a heat engine. What do you think the maximum efficiency of a heat engine is?
See here for a discussion: https://www.physicsforums.com/showthread.php?t=667129
Yes, provided:Thermolelctric said:So when we run heat engine inside building and use the mechanical work produced to pump some heat energy from the outside(colder), total we get more heat energy, than when just burning the fuel inside the building with 100% efficiency?
Pick any major air source heat pump manufacturer and look at the ratings. I guarantee you won't find one rated below 3:1. In fact if I had to guess, the legally required minimum is probably 3.5 or 4.Thermolelctric said:According to my understanding COP >2 is not possible for air to air heat pump. Based on assumption that heat from the building is lost to the same air (cold reservoir) from where it is pumped to inside the building (hot reservoir).
The COP>2 is possible for land to air pump, when the cold reservoir is hotter than the air aruond the building to where the heat is lost. But not for the air to air.
Er, well, I was way off on that. The federal requirement is 2.3. Heat pumps are much less efficient than air conditioners, which surprises me. Must be an issue of wider D-T in heating mode. Could also be an issue of including the defrost cycle and electric heat backup.russ_watters said:Pick any major air source heat pump manufacturer and look at the ratings. I guarantee you won't find one rated below 3:1. In fact if I had to guess, the legally required minimum is probably 3.5 or 4.
Try the calculations. You'll be surprised at just how far "below about 70%" the heat engine efficiency is for a D-T that also gives an 8 COP.Thermolelctric said:I'm still trying to undrestand how we can have so high COP numbers without violating the thermodynamic laws.
The heat engine COP is below about 70% never more than 1, but the reverse process of heat pump can have COP=8 or more still we do not have the violation of thermodynamic laws.
Er, yes it is!Thermolelctric said:Want to calculate: how much less electrical energy will the house consume, when it is heated with air to air heat pump, versus when it is heated by direct electrical heating.
This is not the same value as COP. So there is the catch!
Your original point is well taken, though. It makes little practical sense because if the COP of the heat pump is less than about 3.5 there is not much point in having a heat pump powered by electricity produced from a thermal source.russ_watters said:Er, well, I was way off on that. The federal requirement is 2.3. Heat pumps are much less efficient than air conditioners, which surprises me. Must be an issue of wider D-T in heating mode. Could also be an issue of including the defrost cycle and electric heat backup.
COP is not the efficiency of producing thermal energy from a low entropy source such as a chemical or electrical source. That can never exceed 1.Thermolelctric said:I'm still trying to undrestand how we can have so high COP numbers without violating the thermodynamic laws.
So if the amount of heat is Q and the electrical heating energy to produce Q is E and the work to pump Q is W then if I am understanding correctly you want E-W? Or perhaps (E-W)/Q?Thermolelctric said:Want to calculate: how much less electrical energy will the house consume, when it is heated with air to air heat pump, versus when it is heated by direct electrical heating.
This is not the same value as COP.
Just to add to what Dalespam has said, the COP of the heat pump has to be ≥ 1 since COP = (Qc+W)/W. If Qc < 0 it is no longer a heat pump.Thermolelctric said:Thanks for the link.
So when we run heat engine inside building and use the mechanical work produced to pump some heat energy from the outside(colder), total we get more heat energy, than when just burning the fuel inside the building with 100% efficiency?
That is exactly how most heat pumps run. The input work is provided by electricity generated by heat engines operating in an outside power plant. As I explained in my earlier post #57, in that case you need a COP of about 3.5 just to break even (ie. to get the same amount of heat than you would from burning the fuel with 100% efficiency). If the COP>3.5 you get more heat than burning the fuel inside at 100% efficiency. And a COP > 3.5 is easily possible. If the temperatures are Th/Tc = 293K/273K, the maximum Carnot COP would be 293/20 ≈ 15Thermolelctric said:The other interesting question is: theoretically, can we run the heat engine outside the building and use the work from the heat engine to run the heat pump to heat the building from the outside air and get the same amount or more heat, than burning the fuel inside the building with 100% efficiency?
I answered this already in post 50.Thermolelctric said:The other interesting question is: theoretically, can we run the heat engine outside the building and use the work from the heat engine to run the heat pump to heat the building from the outside air and get the same amount or more heat, than burning the fuel inside the building with 100% efficiency?
DaleSpam said:I answered this already in post 50.
I was assuming that the heat engine and the heat pump were both using the outside air as the cold reservoir in both cases. The difference is that the heat engine uses a hot reservoir heated by the fuel and the heat pump uses the inside of the building as the hot reservoir. The equation I provided applies.Thermolelctric said:This was supposed to be the case when the heat engine is inside the house, you answered in post 50.
The usual case is the is that heat engine is outside the building (in power plant). I have feeling that when the heat engine in the power plant uses same temperature as outside for cooling then no way we can get more heat with the air to air heat pump to heat the building, than was used at the power plant.
I was out gathering some free energy (firewood) :)