Efficiency of Heat Pumps and Refrigerators

how to heat and cool efficiently

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Abstract


A new form of the 2nd law of thermodynamics is used to analyze the efficiency or performance of heat pumps and refrigerators.


How to Solve the Energy Crisis

Emission-free heat pumps can help to solve the energy crisis, and modern civilization without refrigerators for food preservation is unthinkable. Both heat pumps and refrigerators transport heat from a cold to a warm reservoir at the expense of driving a compressor. The cold reservoir can be the ground or a lake/the interior of a refrigerator, and the warm reservoir the interior of a house/exterior of a refrigerator. 

To reduce energy consumption and the threat of climate change, it is important to use as efficient heat pumps and refrigerators as possible.  A standard heat pump can have a  coefficient of performance = 3, which means that 3 units of heat energy is delivered for each unit of supplied electrical energy. If the performance coefficient could be raised to 10, heating houses and water would no longer contribute much to global warming, since there is an endless source of heat in the ground and the air.


                                                    Efficient heating by Al Gore.

How Does It Work?

A heat pump or refrigerator operates by a refrigerant circulating in a closed pipe in the following thermodynamic cycle: 
  • 1. expansion/evaporation with temperature drop 
  • 2. absorption of heat from cold reservoir
  • 3. compression/condensation with temperature increase
  • 4. delivery of heat to warm reservoir

as illustrated in the following figure:

where we here assume that the compressor is driven by an electrical engine.

Efficiency

We now analyze the performance or efficiency of the above cycle using the following form of the  2nd Law of Thermodynamics :


                      dK/dt = W − D,    dE/dt = −W + D + Q,

where 

  • K is Kinetic Energy 
  • E is Heat Energy
  • W = PdV is rate of Work with P pressure and dV volume change (dV > 0 in expansion)
  • D > 0 is rate of shock/turbulent Dissipation
  • Q is rate of Heat In or Heat Out or consumed electrical energy in the Compressor (Q-In, Q-Out, Q-Compressor).

Note that 1 is the key step with a temperature drop with dE/dt < 0 coming from 
  • expansion/evaporation with W > 0 
  • W > D > 0 with  -W + D < 0 and W – D > 0.


The performance coefficient of a heat pump equals

                      Q-Out / Q-Compressor

while that of a refrigerator is  

                      Q-In / Q-Compressor.


Here Q-In decreases with increasing shock/turbulent Dissipation D in step 1, which also decreases Q-Out and thus heat pump performance. Q-In and Q-Out are also reduced from incomplete temperature equilibration or heat exchange in steps 1 and 3, which reduces heat pump performance and refrigerator efficiency. 

Decreasing the temperature difference between the hot and cold recervoir (by using air below instead of the ground above the freezing point), decreases the performance coefficient of a heat pump.

The temperature drop in the expansion can be increased in combination with evaporation from fluid to gas, which can increase Q_In. The corresponding condensation from gas to fluid in compression increases the temperature rise and thus can increase Q_Out and the performance coefficient.




Improving Efficiency 

Altogether, the perfprmance is enhanced by

  • decreasing the turbulent dissipation D in steps 1 and 2
  • increasing the heat exchange i steps 2 and 4
  • expansion/evaporation and compression/condensation.


Turbulence enhances heat exchange and reduces/increases the temperature difference in step 2/3. Thus turbulence decreases efficiency in step 1 and 2 but not in step 3 and 4.



                                             Large scale ground source energy.



Carnot Cycle Efficiency

Text books views a one-phase heat pump without losses as an ideal Carnot cycle with performance coefficient 


1 / ( 1 -T-cold/T-hot)


where T-cold is the temperature of the cold reservoir and T-hot that of the hot reservoir. A small temperature differerence  would thus seem to be efficient, but this can be misleading because in this case also Q-out would be small.
The theoretical limits of performance coefficients set by ideal Carnot cycles with only one gas phase, can be suprpassed by using two-phase cycles with evaporation/condensaton, or combining two circuits without compressor.