What are they talking about?

Originally posted by wolfgang59
My heat pump works well enough. Lets say I get a newer better one.
It will give me 10kW of heat output and use 2kW of electricity.
That's just using heat from outside air (at say 5C) and warming my house
to 22C. That's what it does. (And it pumps very cold air back outside!)

Now my question is how can I get 10kW of heat to produce more than 2kW of electricity?

When you say '10kW of heat output' what exactly do you mean? Is 10kW the amount of energy that would be required to raise your rooms temperature by the amount it is doing? Is it the amount of energy above absolute zero? What is it a measure of?

Originally posted by wolfgang59
My heat pump works well enough. Lets say I get a newer better one.
It will give me 10kW of heat output and use 2kW of electricity.
That's just using heat from outside air (at say 5C) and warming my house
to 22C. That's what it does. (And it pumps very cold air back outside!)

Now my question is how can I get 10kW of heat to produce more than 2kW of electricity?

The efficiency of a heat engine is less than dT/T, where dT is the difference in temperature and T is the hotter temperature. From your figures dT = 17, and T = 300 K. This gives an efficiency of 5.7%. So with a power input of 10 kW, you would expect to get less than 0.57 kW back out again.

Originally posted by wolfgang59
Thanks.
http://en.wikipedia.org/wiki/Stirling_engine

Wiki says up to 30% efficiency on temperature differences as low as 0.5C.

So with current technology an efficient heat pump together with an efficient
Stirling engine would be very close to being a "perpetual motion" machine.

When those efficiencies increase it will be free energy!

Thoughts?

Wiki says up to 30% efficiency on temperature differences as low as 0.5C.

No Wikipedia didn't say that. Wikipedia said:

Very low-power engines have been built which will run on a temperature difference of as little as 0.5 K.

and later on in a different section said:

The thermal efficiency is also comparable (for small engines), ranging from 15% to 30%.

The engine that runs on a tiny temperature difference will have an efficiency of 0.5/T_h where T_h is the hotter of the two temperatures, at liquid nitrogen temperatures you get an efficiency of less than 0.65%, at room temperature that goes down to about 0.17%. The figure of 30%, ignoring losses, and assuming T_c ~ 300K implies a temperature difference of 120 Kelvin.

You cannot increase the efficiency of a thermodynamic engine above (T_h - T_c) / T_h. It is physically impossible.

Originally posted by twhitehead
When you say '10kW of heat output' what exactly do you mean?

10kJ per second of heat

Originally posted by DeepThought

Wiki says up to 30% efficiency on temperature differences as low as 0.5C.

No Wikipedia didn't say that. Wikipedia said:

Very low-power engines have been built which will run on a temperature difference of as little as 0.5 K.

and later on in a different section said:[quote]The thermal efficiency is also comparable (for smal ...[text shortened]... the efficiency of a thermodynamic engine above (T_h - T_c) / T_h. It is physically impossible.

Your right, I speed read the article on Stirling engines.

But what efficiency would you expect on say a 10C temp difference?

Basically when you have a Stirling Engine or similar with efficiency X
and a heat pump with COP Y then if X*Y > 100 you are effectively
getting "free" energy.

Originally posted by wolfgang59
Your right, I speed read the article on Stirling engines.

But what efficiency would you expect on say a 10C temp difference?

Basically when you have a Stirling Engine or similar with efficiency X
and a heat pump with COP Y then if X*Y > 100 you are effectively
getting "free" energy.

Your right, I speed read the article on Stirling engines.

I've done that before.

Regarding the rest, sorry it won't work. Suppose we have a house with perfect insulation so that, apart from via an ideal heat pump and an ideal heat engine, no heat leaks out of the house. Assume outside has temperature T_c, and inside the house it's T_h. We'll also assume that outside is a perfect heat reservoir (5 petatonnes of air should be). We select settings on the engine and pump so that the temperature of the house doesn't change.

Efficiency = Useful energy out / supplied energy.

The heat transferred from outside (indoors) in one cycle of the pump or engine is Q_c (Q_h). The work done by the engine or consumed by the pump = W = Q_h - Q_c

For the engine the heat supplied is Q_h, the useful energy out is W_e. By definition the entropy change when the engine dumps heat outside is Q_c = T_c*dS_e and it is the same as the entropy change when heat is supplied Q_h = T_h*dS_e so W = Q_h - Q_c = (T_h - T_c)*dS_e.

efficiency(engine) <= W/Q_h = (Q_h - Q_c)/Q_h = (T_h - T_c)/T_h

For a heat pump the useful energy out is the heat deposited indoors Q_h and the energy in is the work done W_p, we don't need to add on Q_c as we don't have to pay for it.

efficiency(heat pump) <= Q_h / W_p = T_h / (T_h - T_c) = 1/efficiency(ideal engine)

for a fridge we'd need Q_c on the top and get efficiency <= T_c/(T_h - T_c)

Because we are ensuring that the temperature of the room isn't changing we have
Overall efficiency = Work done by engine / Work done by pump = [efficiency(engine)*Q_h] / [Q_h/efficiency(pump)] = efficiency(engine)*efficiency(pump) = 1

So in the ideal case you have an extremal perpetual motion machine - it runs forever, but doesn't supply any extra energy, it doesn't break any physical laws, it's just not realistic.

Now take away the perfect insulation on the house, on each cycle the heat lost is Q_l (l = loss). The house would cool down, but we can adjust the settings on the pump and supply some work to compensate. The heat pump now supplies Q_h + Q_l heat per cycle.

efficiency = Work done by engine / Work done by pump = (efficiency(engine)*Q_h)/((Q_h + Q_l)/efficiency(pump)) = efficiency(engine)*efficiency(pump)*Q_h/(Q_h + Q_l) = Q_h/(Q_h + Q_l) < 1

So even for ideal engines and pumps there's no point.

I obviously have not explained this very well.

1. We know that we can get electricity from heat. (say geothermal)
The electrical energy produced is a fraction of the thermal energy.

2. We know that we can get heat from a heat pump.
The thermal energy produced is greater than the electrical energy used. (By the COP factor).
[b]This does not break any physical laws because we are extracting heat from ambient air[b]

Now providing that the efficiency of the generator (heat=> electricity) is
greater than 1/COP you are producing more electricity than you put in.

Originally posted by wolfgang59
10kJ per second of heat

I am still unclear about what you mean by 'heat'. Do you mean total heat above 0K? Or the change from the original temperature?

Originally posted by twhitehead
I am still unclear about what you mean by 'heat'. Do you mean total heat above 0K? Or the change from the original temperature?

heat from the heat pump. the figures are for comparison with other heating
devices.

same as you wold measure heat output from any heating device.

presumably it would equate to the heat required to raise the room temperature.

I am not talking about the total heat energy of the air from the heatpump (since a heatpump putting freezing air into a room would not be considered heating!!)

Originally posted by wolfgang59
I obviously have not explained this very well.

1. We know that we can get electricity from heat. (say geothermal)
The electrical energy produced is a fraction of the thermal energy.

2. We know that we can get heat from a heat pump.
The thermal energy produced is [b]greater than the electrical energy used. (By the COP factor).
This does not ...[text shortened]... (heat=> electricity) is
greater than 1/COP you are producing more electricity than you put in.

In my post I proved using physics everyone accepts that what you are saying won't work.

The maximum efficiency of a heat engine is (T_h - T_c)/T_h

The maximum efficiency of a heat pump is the COP is T_h/(T_h - T_c)

A heat pump is a heat engine working in reverse, their ideal efficiencies are reciprocal.

This gives a chained efficiency of 1. The best you can do is make a machine that exactly powers itself pumping heat back and forth, but does nothing useful.

Originally posted by twhitehead
I am still unclear about what you mean by 'heat'.

http://www.energywise.govt.nz/products/listing/100/heatpump?gclid=CKbzzLuj3bgCFQMwpAod-D4A5A

In this list the top performer has a COP of 5.4 at 7C (thats the outside
temp) and will deliver 2.5kW. The electricity used is less than 0.5kW

The outside part of these pumps (sits on external wall) chucks out very
[b]very[/b[ cold air. (I know because its right beside my woodpile!)

Originally posted by DeepThought
A heat pump is a heat engine working in reverse, their ideal efficiencies are reciprocal.

[/b]

Not exactly.

What would you consider the maximum efficiency of a heat engine?

What would you consider a maximum COP of a heat pump?

Now multiply those figures!

😀

Originally posted by wolfgang59
Not exactly.

What would you consider the maximum efficiency of a heat engine?

What would you consider a maximum COP of a heat pump?

Now multiply those figures!

😀

Ok. they give the outside temperature of 7 celcius. They do not state the inside temperature. I'll assume 22 celcius = 295 kelvin. The efficiency figures assume that the inside temperature is fixed, but in real life it isn't your house can lose heat and drop in temperature. These temperatures give theoretical maximum COP of 19.67, so the best heat pump misses perfection by a factor of 4. A perfect heat engine working at that temperature to generate the power for the heat pump has a maximum theoretical efficiency of 0.0508. Multiply them together and you get 1. Multiply the perfect generator with the realistic heat pump and you get 0.27. So even with a perfect generator and the best realistic pump you are going to have to provide 73% of the energy to keep the pump going if your house has perfect insulation; if it doesn't then you will use more power than you would if you had the generator switched off.

Originally posted by DeepThought
Ok. they give the outside temperature of 7 celcius. They do not state the inside temperature. I'll assume 22 celcius = 295 kelvin. The efficiency figures assume that the inside temperature is fixed, but in real life it isn't your house can lose heat and drop in temperature. These temperatures give theoretical maximum COP of 19.67, so the best heat pu ...[text shortened]... it doesn't then you will use more power than you would if you had the generator switched off.

Do you agree or not agree that I get more thermal energy out of my
heatpump than the electrical energy I put in?

Originally posted by wolfgang59
Do you agree or not agree that I get more thermal energy out of my
heatpump than the electrical energy I put in?

Do you agree that the amount of thermal energy you have to put into a generator is less than the amount of work you get out?

The pump has to supply as much heat as it takes to balance insulation losses as well as provide heat to run the engine. Since the engine generates less work than the extra heat you put in your electricity bill goes up and not down.