Calculation Problem

[pottersj]

Hello, I am having difficulty in calculating F and was wondering if anyone could help.

F is the heat flux in W/m2 between the Heat Pump-House system from the Atmosphere.

Now I have the Energy 'lost' from the house walls into the atmosphere. If we don't consider any other parameters, this energy would be the same exact amount of energy that I need to add into the House to maintain a constant temperature.

By looking at some of the Heat Pump specs and by knowing that the Coefficient of performance is 2.5 I should be able to calculate the energy heat has to transfer from the outside into the inside.

The difference between the heat flux lost from the House into the atmosphere and what the heat pump extract from the atmosphere would give me the NETT COOLING of the atmosphere due to the Heat Pump. Which is my F in W/m2.

Now I'm a bit confused on how I go and do that really... Im fine with the sentence before the bold text but then I just get lost...

Thanks
John

 

[IRstuff]

You cannot cool the "atmosphere" in steady-state; that's like trying to use your refrigerator as an air conditioner. While the front of the refrigerator is nice and cool, the back of the refrigerator is dumping truckloads of heat.

Moreover, you describe a condition where the atmosphere is colder than the house, so there is no heat from the atmosphere to pump into the house.

Your definition of COP is incorrect and is why you're not getting plausible answers. The COP is the ratio of heating or cooling to plug power, not atmosphere.

Seems like you need to get an HVAC professional to do whatever it is that you are trying to do. I don't know what you are trying to do.

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[pottersj]

of course you can! Try and solve it analytically and you might realize that it is possible. The point is not if nor how, as I'm actually telling you how to do it, is just the solution I can't get . But I'm afraid, I'm just not clever enough hence my question. Try and use google scholar instead of plain google you'll find tons of papers written on that subject.

 

[DRWeig]

Now I have the Energy 'lost' from the house walls into the atmosphere. If we don't consider any other parameters, this energy would be the same exact amount of energy that I need to add into the House to maintain a constant temperature.

Correct.

Break down the sources of heat that you add to the house:

1. Heat extracted from the atmosphere by the outdoor coil and transferred to the refrigerant.
2. Heat transferred to the refrigerant due to the compressor's work and friction.
3. Heat rejected by the condenser into the house.

Since the compressor adds heat, you actually will extract less heat from the atmosphere than you will reject into the house. The house gets atmospheric heat plus compressor heat.

So how much heat does the compressor add? Look at the definition of COP. It's Btuh transfered to the conditioned space per Btuh input power in terms of electricity.

Assuming (wrongly but good enough for this discussion) that all compressor heat goes into the refrigerant, a COP of 2.5 means that for every 2.5 Btuh the house needs, 1.0 will come from the compressor and 1.5 will come from the atmosphere.

So if my house needs 10,000 Btuh to maintain constant temperature, it must be losing all of that heat through its envelope and by convection and radiation.

To replace the 10,000 Btuh with a heat pump running at 2.5 COP, I would need 10,000 / 2.5 or 4000 Btuh of electrical power input (at 3.1 Btuh per Watt, that's about 1.3 kW). Thus, my heat pump grabs about 6000 Btuh from the outdoor air, and another 4000 Btuh from the machine, and puts them both in my house.

Ignored by me in the interest of clarity: indoor fan, outdoor fan.

Best to you,

Goober Dave

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[zekeman]

Heat pump equations
In a cycle
Qr-Qin=W
Qr=heat removed from refrigerant and added to room
Qin= heat absorbed by refrigerant removed from atmosphere
W= work done on refrigerant
Qr/W= COP
So it is correct to say that the atmosphere is being "cooled"
albeit infinitesimally.
Qr is heating of the room

 

[pottersj]

Thank you both DRWeig and zekeman!

Now the problem that I have is that I have that I have two different units: one is power and the other is energy. BTW I KEEP EVERYTHING IN SI UNITS TO KEEP THE MATHS SIMPLE BUT WE CAN USE USCS or the OLD GOOD IMPERIAL SYSTEM)

So I know the Heat dissipation of the wall in W/m2 which let's say it's 2W/m2 (figurative)

So if the house has 4 identical walls and it's a perfect cube and it's only dissipating energy from 5 of its sides uniformly (we wont count its base - earth) and the wall was 10x10m then we would have 10*10*5*2=200W
than this is the amount of energy the house is dissipating right?

then I have the Heat pump systm which is expressed in KW/h and it has to compensate for that energy lost to mantain equilibrioum...

how do I go about and do that transformation now?
Thanks.
John

 

[DRWeig]

Let's leave the K out of it. That's just 1000.

First, both Btu/h and Watts are power or rate of energy flow per unit of time. Btu and Wh are energy.

Convert btu/h to watts thus: 1 watt = 3.413 Btu/h
Convert Btu to wh thus: 1 wh = 3.413 Btu

COP is unitless. If COP is 2.5, then it equals 2.5 Btu/h out : 1.0 Btu/h in or 2.5W out : 1.0W in.

In your 200W case, you get 120W from the atmosphere and 80W from the electric wires powering the compressor.

Best to you,

Goober Dave

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[pottersj]

hi DRWeig, sorry but how can it be?

In my case I have 2000W which is just Watt it's not watt per second therefore I how can I divide that for the COP even if it's unitless. if I want to turn my 2000watts into power than I would have to do 2000* amount of seconds. So if i do it with 1 second I would have 2000W*1s = 2000Joules which is the amount of power required to keep my cubic house balanced.

now as you can see that is also nonsense at it does not take into consideration the temperature difference between the inside and outside.

Now any ideas how I do that?

Oh I need to set the conditions so let's say outside air temp 270K and inside 300K

Thanks
John

 

[IRstuff]

You seem to be confused. A COP of 2.5 means that your heat pump consumes 5kW of electrical power to provide 2kW of cooling. It's that simple.

"now as you can see that is also nonsense at it does not take into consideration the temperature difference between the inside and outside."

No, it's not nonsense. Again, you are confused. COP is defined for a very specific temperature condition, such as 0ºC and 35ºC. You would need to scale the power according to your actual temperature condition.



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[DRWeig]

pottersj, You simply stated that your COP was 2.5, and I gave you an example to follow. You'll have to work out the units for yourself. If you want to do an energy analysis that includes varying temperatures, you'll need to hire an engineer or make an approximation like IRstuff mentioned.

In my case I have 2000W which is just Watt it's not watt per second therefore I how can I divide that for the COP even if it's unitless

I don't recall anyone mentioning "watt per second" in this thread... COP is watts out / watts in. That shouldn't be a problem.

IRstuff, you might want to check your math on that COP thing, I think you got it upside-down. If it's 2.5, you get 5.0 kW of cooling for 2.0 kW input electric energy. For a heat pump, just change "cooling" to "heating."


Best to you,

Goober Dave

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[DRWeig]

OK, I'll go one more step. I have to prepare a trainer on some basic HVAC concepts for later this year, so I made a quick visual. It's attached to this post.

Tell me what it is about the attached picture that you don't understand?

The heat pump performance vs temperature lift stuff is on the second page.


Best to you,

Goober Dave

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[DRWeig]

I forgot to cite the source for the chart:

http://www.heatpumpcentre.org/en/aboutheatpumps/heatpumpperformance/Sidor/default.aspx

There are more details there too.

Best to you,

Goober Dave

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[pottersj]

Thank you.

 

[chicopee]

The question should be where is the heat pump evaporator located? If it is outside and the ambient temp is capable of changing the state of the refrigerant from liquid to vapor then you you can heat the heat- pump house when the hot refrigerant is coolded thru the condenser located inside that pump house. Now if the evaporator is inside the heat-pump house, you'll probably blown up the refrigerant line from excessive pressure in the condenser; however, in theory it could work. I got a better idea, use a dehumidifier which would be a smaller system to heat up the pump house since that object acts as a heat pump. In the case of a refrigerator as mentioned above, if the unit door was left opened, essentially in due time you would heat up the room which incidentally was a quizz question when I took thermo back in the early 70's.

 

[racookpe1978]

Refrigerator cooling "could" work ... Just put the fridge halfway through the kitchen to the porch: Then, in summertime, you leave the door open on the kitchen side and the compressor is heating the porch and the great outdoor and cooling the kitchen.

Turn the fridge around in winter, and leave the door open to the outside. In cold weather, you don't need any power. In slightly warmer weather, you cool the food and heat the kitchen. 8<)

 

[chicopee]

"....In cold weather, you don't need any power...." yes you will to get the refrigerant moving in the cooling coil so as to change the state of the refrigerant from liquid to vapor before compression. The kitchen would then be heated from the heat of compression and hermatic motor heat load absorbed by the refrigerant.