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...
I must admit that there was logical mistake in my COP=5 cycle. The heat pump will not pump any more when the cold reservoir is heated(above the ambient) back from hot reservoir. Thanks everybody for explaining patiently.
So there is no other limitation to the COP than COP≤Th/(Th−Tc).
The COP...
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!
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.
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?
And we could use the heat from the process output to do the work of the pump. It is possible to convert heat difference back to the work, with known coefficent of efficiency. Then it is possible to use iterative method to calculate the maximum COP before perpetum mobile kicks in. There must be...
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.
So now conclusion for the development of the thread:
DrClaude got the point there: in the limiting conditions of Tc=Th the formula COP=Th/(Th−Tc) does not hold. When Tc=Th there is no more heat pump. So the formula COP=Th/(Th−Tc) can not be used to determine the limiting values of the heat...
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?
Problem is that I tried to do that, but it does not work. I can not get the heat energy promised by the heat pump sales man. And instead there are some fundamental laws of nature that prohibit from getting such a good results from heat pumps.
That would be no problem if the pumps really could...
This is good observation. As long as the cool reservoir is below the ambient(outside temperature) it can absorb the heat from the ambient. So Tc < ambient temp. This brings in new reservoire.
It seems good idea to describe the pump as a system of 4 reservoirs: Input reservoir, pump cold...
Both will slowly come to the temperature Tf, near the Tf the COP will be very high. But this do not reflect the real Coefficent of Performance, because the heat goes to heating the low temp resevoire, we need to subtract the heat that goes to heating the cold side from Qh. Could this be the...
How would be the good way to describe the point when the pump begins to heat the cold source instead of hot reservoire as intended? This is the point from where where Qh/W≠Th/(Th−Tc). Simply saying the heat is not pumped any more to the hot reservoire but instead goes to the cold reservoire ...