Heat Pumps And The Peltier Effect

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SUMMARY

This discussion centers on the efficiency of heat pumps and the potential of Peltier junctions for home heating. Heat pumps utilizing R410 refrigerant can achieve a Coefficient of Performance (COP) of up to 600% under optimal conditions, significantly outperforming traditional electric heating elements. Peltier junctions, while capable of a temperature differential of up to 150 degrees Fahrenheit, face efficiency challenges due to the inherent limitations of semiconductor materials. The conversation highlights the importance of selecting appropriately sized units and understanding the efficiency equations that govern heat pump performance.

PREREQUISITES
  • Understanding of heat pump technology and refrigerants, specifically R410.
  • Knowledge of the Coefficient of Performance (COP) and its significance in heating systems.
  • Familiarity with Peltier junctions and their operational principles.
  • Basic thermodynamics related to heat transfer and efficiency calculations.
NEXT STEPS
  • Research the efficiency equations for heat pumps, focusing on COP calculations.
  • Explore the thermodynamic principles governing Peltier junctions and their applications.
  • Investigate the impact of outdoor temperature on heat pump performance and efficiency.
  • Learn about the design considerations for selecting heat pump systems, including sizing and zoning controls.
USEFUL FOR

HVAC professionals, engineers, and homeowners interested in energy-efficient heating solutions, particularly those considering heat pumps and Peltier technology for residential applications.

Bill McC
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I had seen a few topics from this forum while searching for information on the temperature differential possible in a reverse polarity thermocouple junction. I was reading some of the replies and I am wondering if the heat pump is understood.

It uses a refrigerant that can be anything that transfers heat. In actual use today the refrigerant is an HFC gas that is compressed and expanded, condensed, and evaporated respectfully to move heat in the system. Because there is heat available in the atmosphere down to more than minus 400 degrees Fahrenheit a system can absorb that heat and use it to warm a home. The higher the outdoor temperature the more heat can be absorbed.

R410 refrigerant is capable of remaining gaseous to a temperature of minus 51 degrees Fahrenheit which allows it to absorb many BTUs from the outside air and deliver them to the inside of the home. When temperatures are around 55 degrees Fahrenheit heat pumps can deliver six times the BTUs as would an electric heating element that is 100 percent efficient. The cost of electricity makes heating elements less attractive but with the use of a heat pump electricity becomes a viable alternative to oil heat and propane that is usually used in rural areas that do not have natural gas.

It is my understanding that Peltier junctions are capable of over 100 degrees Fahrenheit temperature differential perhaps up to 150 degrees which could prove to be an extremely effective way to heat a home if the price of manufacturing the thermocouples can be brought down. Just some food for thought.



William McCormick
 
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Bill McC said:
It is my understanding that Peltier junctions are capable of over 100 degrees Fahrenheit temperature differential perhaps up to 150 degrees which could prove to be an extremely effective way to heat a home if the price of manufacturing the thermocouples can be brought down.
Welcome to PF.

Nice idea, but delta-T works against efficiency overall (higher delta-T means lower efficiency), and Peltiers have low efficiency anyway. You can Google for data sheets with efficiency curves.
 
russ_watters said:
Welcome to PF.

Nice idea, but delta-T works against efficiency overall (higher delta-T means lower efficiency), and Peltiers have low efficiency anyway. You can Google for data sheets with efficiency curves.
Thank you for the welcome.

My thought is that like heat pumps that are often six times more efficient than resistive electric heat, which is 100 percent efficient, the thermocouple absorbing heat from outside, along with any resistive heat created should make it a rather efficient method for heating.



William McCormick
 
Bill McC said:
...the thermocouple absorbing heat from outside, along with any resistive heat created should make it a rather efficient method for heating.
It's not the thermodynamics per se that is the issue, it's the semiconductor that is inefficient.
 
Bill McC said:
My thought is that like heat pumps that are often six times more efficient than resistive electric heat, which is 100 percent efficient,
Can you please define the efficiency equations that get you to 600% efficiency for heat pumps? I see sources that say 300% depending on the heat sinks available...

https://brennanheating.com/how-energy-efficient-are-heat-pumps/
 
berkeman said:
Can you please define the efficiency equations that get you to 600% efficiency for heat pumps? I see sources that say 300% depending on the heat sinks available...

https://brennanheating.com/how-energy-efficient-are-heat-pumps/
Certain sized units in certain classes go up to 600 percent. If you pick a condenser or outdoor unit that has a very large coil for its rated tonnage they will perform in the 600 percent range. But if you chose the same company, the same model but a larger unit with the same square feet of condenser coil it will probably lower to high 300 percent range. You have to shop wisely. I have installed hundreds of systems and after noticing the difference in COP (Coefficient Of Performance) from differently sized units in the same category I started picking wisely.
Goodman two stage two ton.jpg

I put two of these systems in my son's house. With an EWC zone damper control system and a barametric bypass damper system with fresh air intake.

I will probably not be stopping back here they keep deleting my posts.



William McCormick
 
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berkeman said:
Can you please define the efficiency equations that get you to 600% efficiency for heat pumps? I see sources that say 300% depending on the heat sinks available...

https://brennanheating.com/how-energy-efficient-are-heat-pumps/
Every BTU requires 0.293 watts to create. So if a unit puts out 18,870 BTUs and uses 0.93 kilowatts to do that work at 55 degrees, 18,870 times 0.293 watts equals 5,528.91 watts which would be electric resistance heater efficiency. 5,528.91 watts divided by 930 watts of actual usage at 55 degrees outdoor temperature gives you 5.945 COP approximately.

But when you add in the zone dampers you actually get a little higher COP but that is speculative as there are no studies, just my keen observations.



William McCormick
 

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