1
in refrigeration CoP =Qo/w= Todelta S/(Tk-To)delta S= To/Tk-ToEc = 1/(Tk/To)-1in heat pump CoP = Qk/W= Tkdelta S/(Tk-To) delta S= Tk/Tk-ToEh = 1/1-(To/Tk)
Coefficiency of performance.
This looks like the beginning of a question - but it never gets around to the question. Without a question being asked, it's pretty hard to provide an answer. For the benefit of anyone who stumbles upon this non-question, I'm guessing that they might have been looking for the work required to supply 1 KW of heat to the reservoir of a system where the COP is 5. COP is the acronym for "coefficient of performance". It is defined as COP = Q/W where Q is the heat supplied to or removed from the reservoir. If Q is 1kw and the COP is 5, then it would take W = Q/COP = 1 kW/5 = 0.2 kW of work to supply that heat.
LTP is the power that cannot be used continuously, i.o.w. the peak current. COP means the continuous power, absorbed by -f.e.- a machine running on one level. André Hak - The Netherlands
In case of vapour compression cycle (VCC) the COP is given by (desired effect / work input). in the other words it can be defines as what we want and what we are paying for that... so in VCC the paying amount is very less as due to low temperature difference that why its value is more than 1. but in case of vapour absorption system the COP is given by (heat taken by evaporator/ heat given to generator). the heat input taken by evaporator is less as compared to heat given to generator.. that why its COP is less than 1......
Cop Out grossed $44,875,481 in the domestic market.
Kindergarten Cop grossed $91,457,688 in the domestic market.
Cop Land grossed $44,906,632 in the domestic market.
Beverly Hills Cop grossed $234,760,478 in the domestic market.
Beverly Hills Cop II grossed $153,665,036 in the domestic market.
Beverly Hills Cop III grossed $42,586,861 in the domestic market.
there are many factors affecting COP. maybe it ranges from 2.5-5
As of July 2014, the market cap for ConocoPhillips (COP) is $105,889,595,666.25.
COP of any refrigerating system mainly depends on the performance of the compressors used and the heat load at the evaporator. A similar or almost equal COP can be achieved from any system by varying its working parameters like load, kind of refrigerant used and the system working pressure. For more details contact maheshkannajpr@gmail.com
ONE that is Near the COP_R of the coeff for the carnot cycle
It is not."A typical 40°F(4°C) cooler's condensing unit might have a COP (Coefficient of Performance) of 2.5""A 0°F (-14°C) freezer's COP is more likely to be about 1.67"Source: http://www.freeaire.com/freezers.html
It is highly unlikely.