/sci/ - Science & Math » Thread #13273515

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Anonymous Sun 13 Jun 15:34:52 2021 No.13273515 View Reply Original Report

Quoted By: >>13273673

If heat pumps are >100% efficient, why don't we generate "(almost) free energy" from them?
Say, a given heat pump is 400% efficient. If we could find a method to recover 25% of the energy generated, we could feed the pump that energy and keep it running on its own generated energy, and then we could use the cold/hot parts to generate some more energy (for example, the cold part could be used to amplify the difference in temperature needed to run stirling engines and peltier cells).

Pic unrelated.

Anonymous

Anonymous Sun 13 Jun 2021 15:53:59 No.13273564 Report

Quoted By: >>13273785

The efficiency of heat pumps >100% only because you count the result as "thermal energy added to whatever you're heating".
But you can't extract energy just from temperature (like taking 1L of water at 20°C, lowering its temperature to 10°C and pocketing the ~42 000J).
You can only extract energy from *differences* in temperature and, if you calculate the efficiency of heat pumps by taking the result as "recoverable thermal energy transferred", it's lower than 100%, so no perpetual motion.

Anonymous

Anonymous Sun 13 Jun 2021 15:54:11 No.13273566 Report

Quoted By: >>13273785 >>13275704

heat pumps only appear to be over-unity because we consider the energy input from the outside to be "free" - it's never included in our energy budget. in reality the energy pumped into the house is slightly less than the sum of the energy taken from outside and the energy used to run the whole system

Anonymous

Anonymous Sun 13 Jun 2021 16:39:07 No.13273673 Report

Quoted By: >>13273785

>>13273515
>If heat pumps are >100% efficient
They aren't. They just possibly have a coefficient of performance greater than 1. This is difference from thermal efficiency.

Anonymous

Anonymous Sun 13 Jun 2021 17:16:36 No.13273785 Report

Quoted By: >>13273793 >>13274901

>>13273564
well, the heat pump itself generates that difference in temp. I guess you could take advantage of that under certain conditions to amplify those differences and generate better conditions to extract energy.

>>13273566
>heat pumps only appear to be over-unity because we consider the energy input from the outside to be "free" - it's never included in our energy budget
as I said above, you could use that energy extracted from the outside, right?

>in reality the energy pumped into the house is slightly less than the sum of the energy taken from outside and the energy used to run the whole system
so in the end it is better to use an electric heater than a heat pump?

>>13273673
right. where can I learn more about this stuff?

Anonymous

Anonymous Sun 13 Jun 2021 17:18:23 No.13273793 Report

Quoted By:

>>13273785
>where can I learn more about this stuff?
in literally any introductory thermodynamics text
>well, the heat pump itself generates that difference in temp. I guess you could take advantage of that under certain conditions to amplify those differences and generate better conditions to extract energy
this is some nonsense

Anonymous

Anonymous Sun 13 Jun 2021 18:23:37 No.13273977 Report

Quoted By:

Coefficient of performance is not efficiency. If a heat pump has a COP of 4 (400%) then it moves four times as much energy as you put into it. In other words, if your heat pump is rated at 100w, then you move 400w of energy through the pump. That heat comes from, or moves to, a reservoir such as the air or ground so it's not free energy.

Anonymous

Anonymous Sun 13 Jun 2021 22:51:23 No.13274901 Report

Quoted By: >>13275704

>>13273785
>as I said above, you could use that energy extracted from the outside, right?
no, you'll get less than you originally spent moving the heat

>so in the end it is better to use an electric heater than a heat pump?
no because you'll spend more electricity running the heater than you get moving heat

Anonymous

Anonymous Mon 14 Jun 2021 03:23:02 No.13275704 Report

Quoted By: >>13275757

>>13274901
>no, you'll get less than you originally spent moving the heat
I realized after writing

>>13274901
>no because you'll spend more electricity running the heater than you get moving heat
hmm, isn't an electric heater 100% (or at least 95%+) efficient? you are converting all the electrical energy into heat in the heating elements. meanwhile, anon said here >>13273566 that you get less heat than the energy "used to run the whole system"

Anonymous

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Anonymous Mon 14 Jun 2021 03:52:34 No.13275757 Report

Quoted By: >>13275806

>>13275704
You're conflating multiple different meanings of "efficiency". CoP of a heat pump isn't the same as CoP for a A/C unit isn't the same as CoP for a refrigerator isn't the same as the thermal efficiency isn't the same as isentropic efficience...see what I mean? Apples and oranges, anon.
>anon said here that you get less heat than the energy "used to run the whole system"
Idk what that means but heat pumps do not generate heat from a heating element, they pull heat either from outside and put it into your home, or the opposite depending on if you want it warmer or cooler indoors. They can easily move this heat (in the form of refrigerant flowing through a circuit) faster than electrical/mechanical power is supplied at the compressor. Heat pumps are still limited by the second law of thermodynamics: they are less efficient than a Carnot heat pump operating under the same temperature diff. $\textrm{CoP}_{\textrm{heat pump}}=\frac{\textrm{rate of heat transfer}}{\textrm{mechanical power supplied at compressor}}=\frac{\dot mh}{\dot W}<\textrm{CoP}_{\textrm{Carnot}}=\frac{1}{1-T_L/T_H}$ Notice it is necessarily greater than 1.

Anonymous

Anonymous Mon 14 Jun 2021 04:15:10 No.13275806 Report

Quoted By: >>13275846

>>13275757
I kind of switched topics because I got interested in heat pumps when researching efficiencies, which apparently was the wrong thing to do. I'm in the southern hemisphere, here is (almost) winter right now.
I'm ignorant about the topic (should be obvious from this thread, I guess), but what I care about the most is the (electrical) input/output (in the form of heat inside my house) power ratio, that is obviously (IMO) the most practical way to compare these systems (because I can see it in my bills).

Anonymous

Anonymous Mon 14 Jun 2021 04:32:02 No.13275846 Report

Quoted By: >>13275897

>>13275806
Look into ground based heat pumps. They're expensive, but they use the heat from the summer in winter and the cold from winter in summer. Any heat pump will outperform a resistive heater if the COP is greater than 1 at the conditions you intend to use it. It's worth noting that any electrical divorce will dissipate heat equal to the power it consumes (basically it's power rating), so you can effectively use your energy twice if you do something like run a 3D printer or an electric washer, or mine for bitcoins in order to generate heat.

Anonymous

Anonymous Mon 14 Jun 2021 04:49:48 No.13275897 Report

Quoted By:

>>13275846
nah. those are for places where it snows. I live by the beach, here our lowest temps are like 5C and highest are 25C, so from my research, a shitty split system + inverter would work very well. I could have got one of those some months ago for cheap (a 24k btu unit for $400) but I fucked up and didn't buy it.
I get lots of sun here (even during winter) so I wanted to buy or make a solar water heater and install a bunch of radiators inside my house, but I realized I'd have to change a lot of things in my shitty house for that to work, so it gets complicated and expensive

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