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.
How Does It Work?
- 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
- 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).
- expansion/evaporation with W > 0
- W > D > 0 with -W + D < 0 and W – D > 0.
Q-Out / Q-Compressor
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.
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.
Carnot Cycle Efficiency
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.