COP of a heat pump

[ThierrySan]

For my part, I think that the term COP to use in thermal does not correspond to much from the real point of view of the installation. It is for me, more misleading than anything else. We talk about COP as if it were a yield, but it is not.
In conclusion, it does not make much real physical sense since we are already drawing heat from the cold source ...

Comparison of definitions:
Quote http://www.ademe.fr/particuliers/Fiches/glossaire.htm

COP (Coefficient of performance): the energy performance of a heat pump is expressed by the ratio between the quantity of heat produced by it and the electrical energy consumed by the compressor. This ratio is the coefficient of performance (COP) of the heat pump.

or:
http://www.actu-environnement.com/ae/di ... e_cop.php4
Different from:
http://fr.wikipedia.org/wiki/Pompe_%C3%A0_chaleur


In conclusion, maybe you should try going through the yield study to show what you have to show ...
- mechanical study: electric motor efficiency, pump efficiency.
- calorific study according to the pressure restored by the heat pump AND the temperature of the cold source: adiabatic compression in carnot cycle (ideal, without energy losses with the outside), it seems to me.

[Capt_Maloche]

The COP is purely financial information: I pay 1 KW to produce 3.

[Christophe]

OH YEAH ?

these are the plans I'm starting to make

Well if you spend a day in the area we will show you around For the plans on the other hand, I have to start making them ... I know it's not logical but hey ...

[Christophe]

The COP is purely financial information: I pay 1 KW to produce 3.

Commercial you mean ... ???

Otherwise you know how to make a superunit heat generator?

Réponse: A generator set mounted in cogeneration (30 + 50 = 80% overall efficiency) which supplies a heat pump (COP = 3)

For each joule of fuel consumed by the group we have:
0.5 + 0,3 * 3 = 1,4 thermal joules ...Overall heating efficiency = 140% and I'm not even talking about condensation of fumes ...

By coupling the motor directly to the heat pump motor we would still gain a few tens of% (elimination of loss of generator and motor efficiency) ... but the regulation would be more difficult ... to see for industrial installations can be ?

[balou59]

in fact I don't have a lot of technical data because I want my program to be able to calculate the recovered heat from information that we find in the catalogs, I have information like the calorific power, the electric power and the COP (7 °, 35 °).

[Christophe]

in fact I don't have a lot of technical data because I want my program to be able to calculate the recovered heat from information that we find in the catalogs, I have information like the calorific power, the electric power and the COP (7 °, 35 °).

Have you watched this?

As for your question you will find EVERYTHING you want in this excellent summary document on heat pumps

[balou59]

Have you watched this?

my answer was addressed to ThierrySan


yes of course I read everything and I took the pumps which had the characteristics closest to the pumps for which I have to calculate the contribution, I quite roughly reduced the variation of the volume to a linear curve and I I applied a coefficient to have the cop (T ext, 50). the Dunkirk climate being rather particular, it is not hot all year round but the days with negative temperatures are rare and the use of pumps would be done in relay of a boiler.

[Other]

Hello

The COP is purely financial information: I pay 1 KW to produce 3.

To measure the performance of my heat pump I take a tank (60 gallon 200 liter water tank I measure the water temperature and I connect the 4500w electric elements I measure amperage and voltage and I measure the time
I take temperature readings every 2 c once 60 c
I stop and I empty the tank.
I fill it with cold water and I heat with the thermo pump
compressor and circulation pump amperage and voltage measurement
the efficiency exceeds 3 when the water in the balloon is at low temperature, but when the water in the balloon rises above 40c it starts to deteriorate towards 55c the efficiency becomes low.
The simplest way (for a water / water system is to measure the yield with a water meter the water flow in the condenser and the inlet temperature is out

The more heat is pumped, the better the efficiency, these conditions are at their best when operating with a freon overload.
the fault, the compressor arrives close to the limits of amperage, as it is a hermetic its internal heat is transmitted to the freon.
So several factors intervene to modify the cop, the temperature levels and the pressures of the freon.

Andre

[balou59]

Hello


here is my project progressing, I do not know where but it is progressing.

my program is finished (apart from some values ​​to modify) and I attacked my internship report (on lateX so it's long ).

all this to say that I have browsed a lot of sites and forum to find out and document myself and that in the end it is here that I found the most interesting opinions and I therefore tried to make my program take into account both ecological and economic interests.

[Christophe]

all this to say that I have browsed a lot of sites and forum to find out and document myself and that in the end it is here that I found the most interesting opinions and I therefore tried to make my program take into account both ecological and economic interests.

Well, we are the best !!!
hehehehehehe

Well without laughing, would it be possible to tell us about your work when it is finished in order to distribute it? What exactly is the program you're talking about ?? Did I miss an episode there?