D
Don Kelly
- Jan 1, 1970
- 0
------------News said:Not so. The words "heat pump" was used. A heat pump moves heat. A
conventional heat pump with COP 5 produces 5kW for every 1 kW it uses.
Heat pumps are now in COP 6 and 7 I believe on air to air. Take a COP 7.
Instead of having an electric motor turning the refrigerant compressor
install a high efficient Stirling motor that produces 1kW. Now for every 1
kW of energy the Stirling uses the fridge compressor will produce 7 kW. The
heat and cooling side for Stirling to keep it running is provided by the
refrigerants condenser and evaporator. Heat from the air is extracted to
make the Stirling move. That should keep running.
Unfortunately this will not work. This idea has been proposed many times and
the net result is that the output from the Stirling engine will be less than
what it needs to drive the heat pump. This is, in effect, what you are
asking it to do (as well as producing excess energy for other things) This
is a fact of life and the second law of thermodynamics still holds.
Take your COP heat pump. The heat delivered to the hot reservoir is Qh=7 kWh
The input work is W=1 Kwh and the heat from the atmosphere is Qc =6 kWh (not
kW which is power not energy)
At the theoretical best, a "reverse" carnot cycle can be obtained where COP
=Th/(Th-Tc) where Th and Tc are the hot and cold reservoir temperatures
(degrees K) Real heat pumps don't reach this COP
Now put the heat back into a heat engine. Perfect heat pump and perfect heat
engine results in a 0 sum game.
Example:
Suppose that an ideal heat pump has a cold reservoir temperature Tc of 0C
or 273K and the hot reservoir is at 20C or 293K. The COP will be
293/(293-273) =14.65 Raising the Th or making the temperature differential
smaller will improve the COP (infinite if the cold and hot reservoirs are
the same temperature but then the heat pump will be doing nothing).
Now we use the extracted heat and use it in a Carnot engine (ideal heat
engine).\
The engine efficiency is (Th-Tc)/Th =20/293 =6.83 %
If 1kWh into the heat pump yields 14.65 kWh of heat , this 14.65kWh will
yield a useful engine output of 14.65*0.0683=1.00 kwh A real heat engine
over the same temperature range will have a lower efficiency. We have a zero
sum game.
Now we can increase the high temperature and reduce the low temperature so
that the heat engine efficiency will increase. However the COP of the heat
pump will decrease - still 0 sum. The heat pump works best over a narrow
temperature range where the heat engine is very inefficient. Now take a real
Stirling engine and a real heat pump and the COP will be lower and the
engine efficiency higher at any given temperature differential. Try it for
different temperature differences.
I suggest you read:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatpump.html#c1
Still a pipe dream- facts get in the way.