Heat from an A/C inverter calculation more than energy consumed?

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SUMMARY

The discussion centers on the efficiency of inverter air conditioners, specifically their ability to output more heat energy than the electrical energy consumed. Participants clarify that inverter units, functioning as heat pumps, can achieve a Coefficient of Performance (COP) greater than 1 by extracting low-temperature heat from the environment and converting it to high-temperature heat for indoor use. The COP can theoretically reach values of 4 or 5 under optimal conditions, although real-world performance may vary based on temperature differentials. The conversation also references the principles of thermodynamics, particularly the reversed Carnot cycle, to explain the mechanics behind heat pump efficiency.

PREREQUISITES
  • Understanding of heat pump technology and its applications
  • Familiarity with the Coefficient of Performance (COP) calculation
  • Basic principles of thermodynamics, particularly the Carnot cycle
  • Knowledge of temperature differentials and their impact on heating efficiency
NEXT STEPS
  • Research the Coefficient of Performance (COP) for various heat pump systems
  • Explore the thermodynamic principles behind the Carnot cycle and its applications in heat pumps
  • Investigate the efficiency of air-water and water-water heat pump systems
  • Examine advancements in heat pump technology that enhance performance in low-temperature environments
USEFUL FOR

Engineers, HVAC professionals, and homeowners interested in energy-efficient heating solutions will benefit from this discussion, particularly those looking to optimize the performance of inverter air conditioners and heat pumps.

pd2905
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Hi guys can anyone please tell me that it is possible with an air conditioner of type inverter to put less energy than what the piece is receiving as heat. So let's say that on the inverter it says output KWh/ input KWh =4.

Henthalpy= G+TS
4*Hinput= Houtput inside the house

So to heat my house I use 4 times less energy if I just used a normal electric heater or this is what they write in product description?
 
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If you use a standard electric heater, 100% of the electric energy consumed is converted in heat.

But, in if a 'heat pump', you use the electric energy to drive a compressor, and the machine converts low-temperature heat, available in the external surroundings, in the external air, mainly, to high temperature heat, that may be used to heat your room. The 'transportation cost' is lower than the energy costs of a resistance heater of the same heating power. How much lower depends on the difference of inside/outside temperatures. The larger the difference, the smaller the advantage with respect to a resistance heater.

A refrigerator (or heat pump, that is the same) is esentially a thermal engine working in reverse. Instead of taking heat from a high-temperature place, producing useful mechanical work, and discharging that low-temp heat in a low-temperature place, the refrigerator/heat pump takes low-temp heat from a low temperature place, injects mechanical power, and places the resulting high-temp heat in a high temperature environment...I believe that using '4 times less energy' may be too optimistic...
 
Last edited:
I guess that what you are describing is more commonly known as a heat pump. If that is the case, then indeed you get more heat out than the electricity input, as additional heat is extracted from the outside air. This is more efficient than direct electric heating.
 
This two very clear pictures are from 'The Second Law', by Henry Bent:

15599507266_916eb43343.jpg
15436560709_bc1ee81394.jpg
 
NTW said:
If you use a standard electric heater, 100% of the electric energy consumed is converted in heat.

But, in if a 'heat pump', you use the electric energy to drive a compressor, and the machine converts low-temperature heat, available in the external surroundings, in the external air, mainly, to high temperature heat, that may be used to heat your room. The 'transportation cost' is lower than the energy costs of a resistance heater of the same heating power. How much lower depends on the difference of inside/outside temperatures. The larger the difference, the smaller the advantage with respect to a resistance heater.

A refrigerator (or heat pump, that is the same) is esentially a thermal engine working in reverse. Instead of taking heat from a high-temperature place, producing useful mechanical work, and discharging that low-temp heat in a low-temperature place, the refrigerator/heat pump takes low-temp heat from a low temperature place, injects mechanical power, and places the resulting high-temp heat in a high temperature environment...I believe that using '4 times less energy' may be too optimistic...
Thanks for your input,
Just to add to this
For an ideal heat pump cycle:

COP = TH/(TL-TH)
source wikipedia.
 
DrClaude said:
I guess that what you are describing is more commonly known as a heat pump. If that is the case, then indeed you get more heat out than the electricity input, as additional heat is extracted from the outside air. This is more efficient than direct electric heating.

Thanks for the reply DrClaude, just to add to your thing there is heat pump systems that are air-water, water-water and air-air( air conditioner inverter).
For an ideal heat pump cycle:

COP = TH/(TL-TH)
source wikipedia.
 
NTW said:
This two very clear pictures are from 'The Second Law', by Henry Bent:

15599507266_916eb43343.jpg
15436560709_bc1ee81394.jpg
How does delta S stay zero in both directions? So the reversed Carnot is the refrigeration cycle, but do these heat pumps use only one cycle, and is it possible that they have some newer technologies so that the efficiency of an air-air heat pump reaches 4 or 5 up to -20 degrees C?
 
pd2905 said:
How does delta S stay zero in both directions? So the reversed Carnot is the refrigeration cycle, but do these heat pumps use only one cycle, and is it possible that they have some newer technologies so that the efficiency of an air-air heat pump reaches 4 or 5 up to -20 degrees C?

400K reservoir ---> ΔE (cal) = -1200 cal ΔS (cal/K) = -3 cal/K
300K reservoir ---> ΔE (cal) = +900 cal ΔS (cal/K) = +3 cal/K
Weight -------------> ΔE (cal) = +300 cal

Balance -----------> ΔE = 0 cal ΔS = 0 cal/k
 

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