Heat pump applications

[Iseason]

Hi all

 

Just an interesting thought. Would it be viable to extract the energy in a dam as an aside to running the turbine with it. Just how efficient is it now ?

 

Since the water is already being dropped on the turbines , would it be efficient to use part of that process to run the same technology as we use in our homes to create heat for different generation at the same time. I would imagine the heat to be extracted from the dam body and the same water as runs the turbine to run the heat pump.

 

I'm not claiming to know a whole lot about heat pumps except they are extremely efficient.

 

Cheers iseason

[studiot]

It is a good question and you should keep asking such questions.

 

However I also suggest you consider the following comparison.

 

How much energy can you extract from 1kg of water falling 30 metres, assuming your hydrogenerator is 85% efficient in conversion to electricity?.

 

How much energy can you extract from 1kg of water as heat energy, by cooling it 3 degrees centirgrade and assuming you have obtain 3 times as much heat as you have to expend energy to run your heat pump (That is the coefficient of performance is 3 - a good figure for a heat pump)?

 

Furthermore, electricity is readily transmittable to be used elsewhere. What would you do with the heat when you had it?

Edited July 12, 2014 by studiot

[Iseason]

Hi studiott

 

You may have misunderstood. I was suggesting using the heat to generate electricity. As I said I'm not claiming to understand whether it could extract the sort of energy to run a steam turbine as an aside to running the turbines.

 

But removing it from commercial scale would a creek now be a viable source to heat your home . There are quite a few possible differences here . In the morning it would have more stored energy than air. But I assume that as the air heats up , it contains a more readily available source.

 

Or is straight turbine electricity still the most efficient.

 

Cheers

Iseeson

Edited July 12, 2014 by Iseason

[studiot]

The comparison I suggested is interesting because

 

1) 1kg of water falling 30m at 85% efficiency will net you 250 joules of electrical energy.

 

2) 1kg of water cooled 3 degrees centigrade will supply 12,600 joules of heat energy, but you would need to expend 4,200 in your heat pump leaving 8,400J.

 

These are extreme because the water head is maximised so the fall is usually greater than 30m and the cooling is normally measured in points of a degree.

 

But the principle is there to discuss.

Edited July 12, 2014 by studiot

[Iseason]

Thanks.

 

What I see as interesting in your answers is what might be possible in other varianions.

 

Say , if you multistage the system. Does the a return v work remain the same if the initial heat transfer is gathered to a different medium?

 

In systems less than 30 meters , does this give scope for more potential sites for gathering energy for power production.

 

By the way. Those figures look heavily in favour of the heat pump. I guess it comes down to a comparison of work required to cool the water and what that equals in litres of water dropped thirty meters.

 

Let me se if I can do this.

1 litre = 250 joules

So i need 16.8 litres to get my 4200 joules to run the pump.

If I can add (and it's not a given I can) each 16.8 litres of water is now giving me

 

16.8. Times 250 joules ( that hydro gave me) = 4200

16.8 times 250 joules. ( that ran my pump). = 4200

 

It seems , if I follow you correctly I have 4200 as extra joules. It couldn't seriously give me a 100% better return?

 

Thanks for you interest

Cheers

Iseason

Edited July 13, 2014 by Iseason

[John Cuthber]

Interesting thought.

Imagine I already plan to build a hydroelectric power plant.

If I arrange for that falling water to flow through a heat pump just after it has gone through the turbines I can grab a lot of fairly low grade heat.

If my dam was next to a town I could sell the heat for something like community heating in the same way that conventional CHP systems work.

As far as I can see, the real problem is that people don't generally build dams near towns.

[studiot]

Iseason

What I see as interesting in your answers is what might be possible in other varianions.

 

John Cuthber

Interesting thought.

Imagine I already plan to build a hydroelectric power plant.

If I arrange for that falling water to flow through a heat pump just after it has gone through the turbines I can grab a lot of fairly low grade heat.

If my dam was next to a town I could sell the heat for something like community heating in the same way that conventional CHP systems work.

As far as I can see, the real problem is that people don't generally build dams near towns.

 

 

Of course Nature has provided a catch, as JC notes.

 

On the face of it, there is an enormous amount of apparently available thermal energy compared to the mechanical/electric sort.

 

The problem comes in trying to use that heat. Iseason you have suggested that heat pumps are efficient.

They are not, they are hopelessly inefficient. It is just that there is a large amount of heat energy available so the inefficiency can be tolerated.

In fact, efficiency is an inappropriate term to use for heat pumps. This is why I talked about COP.

 

Efficiency in % is defined as output divided by input times 100.

But this is predicated upon the input being fixed or set (the independent variable) at a aprticular value and the output being the determined by the system.

So if I put 100 watts or Joules into an electric motor at 85% efficiency I will get 85 watts or Joules out.

So for efficiency calculations the input is fixed and the output varies.

 

For heat pumps it is the other way round. That is in order to extract a set or fixed quantity of energy you need to input a particular quantity of work energy.

The parameter that measures this is called the coefficient of performance (COP). The COP is normally reported as a fraction rather than a % however.

 

So the COP is defined as the output divided by the input work but not including the input heat from the heat source.

So, as I said earlier, you can get a good figure of receiving three times as much heat as you put work in. COP = 3.

 

Now you have to output the heat into something by raising its temperature. Ususally this is a working fluid, say for instance the hot water in your central heating radiators.

 

The catch is that the higher the temperature you raise the working fluid to the lower the COP.

So yes you can raise the temperature of your central heating water from say 20C to 25 C at a COP of 3.

But in order to do any useful heating your central heating water needs to be at least 55C and perhaps more.

At this temperature the COP falls to about 1.4 .

Worse if you actually want to boil water to create steam, the COP is actually less than 1 so you would have to spend more energy pumping than if you simply used that energy to boil the water.

 

 

That is why engineers distinguish between high grade energy such as electricity and low grade energy such as heat pumped heat

as John Cuthber has pointed out.

 

He also points out that for safety reasons planners try to keep areas downstream of dams (they do burst occasionally) free of housing.

 

I will stop here (for questions) because these are actually difficult concepts but a direct consequence of the laws of Thermodynamics.

 

Edited July 13, 2014 by studiot

[Iseason]

Hi guys

 

Thanks for posting.

 

I was sure there would be a catch. That's why I mentioned other methods and uses. So I have 100% extra joules....multi staging?

 

Also systems less than a commercial dam. Say where a river runs by my property. Would it be now more efficient to use a heat pump to warm my cows up in winter , keep a highway clear of snow , warm my two bedroom house. Things where a continuous output was more important than the high temperatures needed for electricity production.

 

Or is a turbine still going to convert better over a longer but slower system.

 

Cheers

Iseason