Heat Pumps And The Peltier Effect

In summary: 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.
  • #1
Bill McC
4
1
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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  • #2
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.
 
  • #3
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
 
  • #4
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.
 
  • #5
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/
 
  • #6
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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  • #7
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
 

1. What is a heat pump and how does it work?

A heat pump is a device that transfers heat from one location to another using the Peltier effect. This effect is based on the principle that when an electric current flows through two different conductors, heat is either absorbed or released at the junction between them. The heat pump uses this effect to transfer heat from a cooler location to a warmer one, effectively "pumping" heat against its natural flow.

2. How is the Peltier effect related to heat pumps?

The Peltier effect is the underlying principle behind the operation of heat pumps. It is used to transfer heat from one location to another, allowing the heat pump to cool one area while simultaneously heating another. This effect is made possible by the use of two different conductors, typically made of different materials, which are connected to a power source.

3. What are the advantages of using a heat pump?

One major advantage of using a heat pump is its energy efficiency. Unlike traditional heating and cooling systems, which use energy to create heat or cool air, heat pumps simply transfer heat from one location to another. This means they require less energy to operate, resulting in lower utility bills. Additionally, heat pumps can be used for both heating and cooling, making them a versatile option for maintaining comfortable temperatures in a home or building.

4. Are there any limitations to using a heat pump?

One limitation of heat pumps is that they are most effective in moderate climates. In extreme temperatures, such as very cold or very hot weather, the heat pump may struggle to transfer heat effectively. Additionally, the initial cost of installing a heat pump can be higher than traditional heating and cooling systems. However, the long-term energy savings can offset this cost over time.

5. Can heat pumps be used in all types of buildings?

Yes, heat pumps can be used in a variety of buildings, including residential homes, commercial buildings, and industrial facilities. They are available in different sizes and configurations to suit different needs and can be used for both heating and cooling. However, it is important to consult with a professional to determine the best type and size of heat pump for a specific building and its heating and cooling needs.

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