Solar PV is now the most cost efective energy source.

[tszaboo]

If you can get "infinite amount" of not frozen water then a heat pump will be more effective than direct heating. Of course you need to step away from Peltier, that will be very expensive. Traditional systems use the same methods as refrigerators.
Also, I dont know how do you "feel" the cold or warmth in Canada. I have the issue, that I feel cold in a room if there isnt a warm radiator in the room, even if the room air is 20+ degrees warm. Its the infrared radiation of course.

Also, there are a lot of development for gas generators. They use the gas to generate electicity and heat, increasing the efficiency of the system. I believe it will be standard in the future, replacing every gas heating system.
https://en.wikipedia.org/wiki/Micro_combined_heat_and_power
Sure, it is still powered by dinosaurs fart, and not green.

Otherwise, I like the designs you are doing. There arent too many solar open source projects.

[splin]

Check last page and you can see when used at 3V will take about 1A so 3W of power and be able to pump about 12W when 10C delta is used. For high delta using two or 3 stages may be needed the COP will still remain the same.

No. For the second stage you need to provide another 3W of power to move the same 12W through another 10C delta. Plus an additional 3/12 * 3W to move the power added in the first stage, so the overall COP is now less than half that of the first stage. You end up with much the same COP as if you used a single stage. For large deltas, > 40C or so, you are forced to use multiple stages but the COP is terrible.

You do not move the same 12W you move additional 12W.
Second stage will of course need to be a bit more powerful since it will need to move the 12W + 3W from the hot side of the first stage.

No you're not moving an additional 12W - after all where is it supposed to have come from? The two stages are in series so the output from the second stage is the 12W transferred by the first stage from the object being cooled plus the I^2R losses of both stages.

Take a look at page 11 of http://www.pathways.cu.edu.eg/ec/text-pdf/part%20c-17.pdf to see the overall COP of single and multi-stage peltier coolers against temperature differences. Its not a very good quality diagram but its very clear that COP declines dramatically with increasing delta T no matter how many stages you use.

If you want the equations you can find them here: https://thermal.ferrotec.com/technology/thermoelectric-reference-guide/thermalref12/

Note in particular the equations for the amount of heat pumped for both a two stage and three stage cooler are identical - 12.2.1 b) and 12.2.2 c) being that pumped by the smallest module in both cases.

You forget that you get a 10C delta with single stage and 20C delta with two stage so is obviously more energy pumped.

No I didn't forget and its not obvious. A higher delta T does not mean that more energy is pumped. A single peltier can pump the same amount of heat, Qc (Watts), to different delta Ts but the COP reduces with increasing delta T.

[electrodacus]

If you can get "infinite amount" of not frozen water then a heat pump will be more effective than direct heating. Of course you need to step away from Peltier, that will be very expensive. Traditional systems use the same methods as refrigerators.
Also, I dont know how do you "feel" the cold or warmth in Canada. I have the issue, that I feel cold in a room if there isnt a warm radiator in the room, even if the room air is 20+ degrees warm. Its the infrared radiation of course.

Also, there are a lot of development for gas generators. They use the gas to generate electicity and heat, increasing the efficiency of the system. I believe it will be standard in the future, replacing every gas heating system.
https://en.wikipedia.org/wiki/Micro_combined_heat_and_power
Sure, it is still powered by dinosaurs fart, and not green.

Otherwise, I like the designs you are doing. There arent too many solar open source projects.

As you maybe seen in my presentation page 5 where I compare my solution with thermal solar and natural gas and both of those solutions are more expensive (and I was generous).
Geothermal and heat pump are also a way more expensive solution and I can get in to the detail if you want but basically the cost for unit of energy for direct PV heating (the solution I propose) is so low that you can not cost amortize the equipment needed even for an air heat pump much less a geothermal installation. Of course air heatpump will not work here since is way to cold.
That electricity generation and using the heat loss for heating was used for quite some time especially in East Europe from where I'm originally. The small electrical generators usually run on natural gas where also distributing the hot water and heating to the surrounding building.
There is no extra advantage there and that solution will be still much more expensive than PV solar direct heating plus electricity in most places around the world. The factor is just the amount of sun at that location and the cost of natural gas.
I do not insist much on the green part since my solution is cost effective compared to any other solution.
Traditionally natural gas was the cheapest form of energy that is why more than half of the buildings around the world are heated with natural gas but that is not the case anymore and PV solar direct heating is more cost effective.
   

[electrodacus]

Check last page and you can see when used at 3V will take about 1A so 3W of power and be able to pump about 12W when 10C delta is used. For high delta using two or 3 stages may be needed the COP will still remain the same.

No. For the second stage you need to provide another 3W of power to move the same 12W through another 10C delta. Plus an additional 3/12 * 3W to move the power added in the first stage, so the overall COP is now less than half that of the first stage. You end up with much the same COP as if you used a single stage. For large deltas, > 40C or so, you are forced to use multiple stages but the COP is terrible.

You do not move the same 12W you move additional 12W.
Second stage will of course need to be a bit more powerful since it will need to move the 12W + 3W from the hot side of the first stage.

No you're not moving an additional 12W - after all where is it supposed to have come from? The two stages are in series so the output from the second stage is the 12W transferred by the first stage from the object being cooled plus the I^2R losses of both stages.

Take a look at page 11 of http://www.pathways.cu.edu.eg/ec/text-pdf/part%20c-17.pdf to see the overall COP of single and multi-stage peltier coolers against temperature differences. Its not a very good quality diagram but its very clear that COP declines dramatically with increasing delta T no matter how many stages you use.

If you want the equations you can find them here: https://thermal.ferrotec.com/technology/thermoelectric-reference-guide/thermalref12/

Note in particular the equations for the amount of heat pumped for both a two stage and three stage cooler are identical - 12.2.1 b) and 12.2.2 c) being that pumped by the smallest module in both cases.

You forget that you get a 10C delta with single stage and 20C delta with two stage so is obviously more energy pumped.

No I didn't forget and its not obvious. A higher delta T does not mean that more energy is pumped. A single peltier can pump the same amount of heat, Qc (Watts), to different delta Ts but the COP reduces with increasing delta T.

Not sure what sort of analogy to make to explain this.
I will try with water pumps

Imagine you have two 100% efficient 12W water pumps and each can move X amount of water/minute at the height H
Then what you can do is to have them in parallel so they will move 2X amount of water at height H so total work done is two times higher.
Then imagine you have the first pump at first floor pumping X amount of water at first level ho height H and then second pump will be at the second level and move that to next level so total height is 2H but just X amount of water.

Now question is what is the equivalent efficiency of those pumps working together in first case with pumps in parallel and then in second case with pumps in series.
For me the answer is obvious in both cases there where 24W used and 24W of work done so in both cases the efficiency is 100%

Hope my analogy worked but yes the temperature (equivalent with height in the analogy) is important.

[splin]

Not sure what sort of analogy to make to explain this.
I will try with water pumps

Imagine you have two 100% efficient 12W water pumps and each can move X amount of water/minute at the height H
Then what you can do is to have them in parallel so they will move 2X amount of water at height H so total work done is two times higher.
Then imagine you have the first pump at first floor pumping X amount of water at first level ho height H and then second pump will be at the second level and move that to next level so total height is 2H but just X amount of water.

Now question is what is the equivalent efficiency of those pumps working together in first case with pumps in parallel and then in second case with pumps in series.
For me the answer is obvious in both cases there where 24W used and 24W of work done so in both cases the efficiency is 100%

Hope my analogy worked but yes the temperature (equivalent with height in the analogy) is important.

OK I think I understand where you are coming from but your analogy isn't useful - the work done by the 100% efficient pumps is equal to the electrical energy you provide to run them whereas the point of using a heat pump is to get more energy out than the electrical energy supplied - by transferring free low grade heat energy into more useful higher temperature energy.

What you are missing is that a peltier isn't doing any work - it merely moves energy from the low temperature side to the other, higher temperature side. The energy coming out the high side is equal to that moved from the low side plus the resistive heating losses incurred. Higher temperature out means the energy is more useful to you but it doesn't mean it provides more energy simply because it is hotter. A resistive heating element can create heat energy at any temperature you like (until it melts) but the heat energy won't exceed the electrical energy you input.

When you have a stack of peltiers the total heat output is the heat moved from the cold side of the first element plus the total losses in all the peltiers. So after a few stages, virtually all the heat being transferred is the heat created by the resistive losses in the preceding peltier stages, such that there is very little gain over using a resistive heater set to produce the desired temperature.

The conventional definition of overall COP for a multi-stage peltier is the amount of heat transferred from the cold side of the first stage divided by the total amount of power used. But in your case the total heat out is useful so a more relevant COP definition is the total amount of heat output divided by total electrical power used (assuming the heat transferred from the source by the first stage is 'free'). The attached shows a spreadsheet showing how the COP reduces as more stages are added, assuming each stage is scaled appropriately.

[Edit] Attached spreadsheet as well

[Kleinstein]

Besides the poor efficiency, TEC have another problem: the efficiency goes down even more if not used at the suitable current level for the temperature difference and there is quite some heat flow if turned off. This is especially a problem with volatile energy supply. So you would need a kind of thermal switch, like a fluid circuit with a pump - just the thing you wanted to avoid with the TEC.

[electrodacus]

OK I think I understand where you are coming from but your analogy isn't useful - the work done by the 100% efficient pumps is equal to the electrical energy you provide to run them whereas the point of using a heat pump is to get more energy out than the electrical energy supplied - by transferring free low grade heat energy into more useful higher temperature energy.

What you are missing is that a peltier isn't doing any work - it merely moves energy from the low temperature side to the other, higher temperature side. The energy coming out the high side is equal to that moved from the low side plus the resistive heating losses incurred. Higher temperature out means the energy is more useful to you but it doesn't mean it provides more energy simply because it is hotter. A resistive heating element can create heat energy at any temperature you like (until it melts) but the heat energy won't exceed the electrical energy you input.

When you have a stack of peltiers the total heat output is the heat moved from the cold side of the first element plus the total losses in all the peltiers. So after a few stages, virtually all the heat being transferred is the heat created by the resistive losses in the preceding peltier stages, such that there is very little gain over using a resistive heater set to produce the desired temperature.

The conventional definition of overall COP for a multi-stage peltier is the amount of heat transferred from the cold side of the first stage divided by the total amount of power used. But in your case the total heat out is useful so a more relevant COP definition is the total amount of heat output divided by total electrical power used (assuming the heat transferred from the source by the first stage is 'free'). The attached shows a spreadsheet showing how the COP reduces as more stages are added, assuming each stage is scaled appropriately.

I see so you think my analogy with water pumps is not applicable here.

I will try an example and please point out where you think I'm wrong.

I have one of those peltier elements mentioned before and it is on the cold side glued to a water cooling block while on the hot side is glued to say a huge copper cube (just for simplicity so that hot side temperature remains fixed and block will be at +27°C).
Now I have two buckets one is full of water at +27°C and the other is empty and I want to cool that water to +17°C an put that cold water in the empty bucket.
So there will be a variable speed pump pumping water from the full bucket to the empty bucket trough the water cooling block.
Since I will be supplying this peltier element with 3V and based on the graph it looks like it will be able to pump about 12W from the cold side and from the other graph COP is about 3.25 thus 12W/3.25 = 3.69W are used by peltier element so around 1.23A x 3V
Now we know that to change the temperature of one liter of water by one degree Celsius we need about 1.2Wh of energy and since our heat pump (peltier element) can move 12W and what we want is to reduce the water temperature by a 10 degree Celsius delta so we can calculate the appropriate flow rate required so that water is cooled to exactly +17°C
(1.2Wh per liter per degree Celsius x 10°C )/ 12W = 1 liter/hour and this will be the flow rate required.

Maybe you see where this is going
Adding a second peltier between the hot side of the first and the huge copper block you reduce the water temperature by 20°C thus twice the amount of energy is transferred to water.
So COP will be the average of the two stages and if both have a COP of 3.25 then total COP of two stages will still be the same 3.25

I know you will need to add the additional 3.69W to the cold side and the second stage will need to pump 12W + 3.69W = 15.69W so if same type peltier element will be used then it will have a slightly lower COP but you can also select a slightly larger peltier element so that you can maintain the same COP of 3.25

[splin]

I will try an example and please point out where you think I'm wrong.

I have one of those peltier elements mentioned before and it is on the cold side glued to a water cooling block while on the hot side is glued to say a huge copper cube (just for simplicity so that hot side temperature remains fixed and block will be at +27°C).
Now I have two buckets one is full of water at +27°C and the other is empty and I want to cool that water to +17°C an put that cold water in the empty bucket.
So there will be a variable speed pump pumping water from the full bucket to the empty bucket trough the water cooling block.
Since I will be supplying this peltier element with 3V and based on the graph it looks like it will be able to pump about 12W from the cold side and from the other graph COP is about 3.25 thus 12W/3.25 = 3.69W are used by peltier element so around 1.23A x 3V
Now we know that to change the temperature of one liter of water by one degree Celsius we need about 1.2Wh of energy and since our heat pump (peltier element) can move 12W and what we want is to reduce the water temperature by a 10 degree Celsius delta so we can calculate the appropriate flow rate required so that water is cooled to exactly +17°C
(1.2Wh per liter per degree Celsius x 10°C )/ 12W = 1 liter/hour and this will be the flow rate required.

Maybe you see where this is going

Adding a second peltier between the hot side of the first and the huge copper block you reduce the water temperature by 20°C thus twice the amount of energy is transferred to water.

No - if you don't increase the voltage to the first peltier it willl still only pump 12W with a 10C delta. That's where you have been going wrong. The water can now be cooled by 20C but the flow rate will have to be halved to .5l/hr. As you say, the second peltier will move 12W + 3.69W, requiring 4.83W (at a COP of 3.25) so outputing 20.52W to the copper block. Total electricity supplied is 3.69W + 4.83 = 8.52W

So COP will be the average of the two stages and if both have a COP of 3.25 then total COP of two stages will still be the same 3.25

No the overall COP will be 12W/8.52 = 1.41. Alternatively you get 20.52W of heat out for your 8.52W electrical input or 2.41x which isn't bad.

In reality other losses, such as heat leaking from the hot side of the peltier to the cold side through the insulation around them and the screws required to apply the necessary clamping forces between the heatsinks and peltiers will reduce these efficiencies.

Additional losses arise from the thermal resistances between the peltiers and the heatsinks, between adjacent peltiers and especially between the heatsinks and the source and destination mediums (eg. water or air) will also reduce the COP as the delta C seen by the peltiers will be higher than between the source and destination.

[electrodacus]

No - if you don't increase the voltage to the first peltier it willl still only pump 12W with a 10C delta. That's where you have been going wrong. The water can now be cooled by 20C but the flow rate will have to be halved to .5l/hr. As you say, the second peltier will move 12W + 3.69W, requiring 4.83W (at a COP of 3.25) so outputing 20.52W to the copper block. Total electricity supplied is 3.69W + 4.83 = 8.52W

I know I'm not great at explaining things but I was sure my example was quite clear.
When I added the second peltier element the hot side of the first peltier element has dropped from +27°C to +17°C provided by the cold side of the second peltier.
Each peltier element will pump around 12W (ignoring those IR losses) and those add up if the hot face of the first peltier is in contact with the cold face of the second one.
Same as if you inverse the current flow on one of the peltier you cancel the work of the first one.
So you either have the Qc1 + Qc2 or Qc1 - Qc2 and in case Qc1=Qc2 the you have either 2xQc or you have zero in the second case.
If you have two of those cheap peltier module you can do the experiment but is not necessary since it should be obvious.

I'm an EE so my knowledge of thermodynamics is limited but I'm smart enough so that if I'm wrong you will be able to explain where I'm wrong to me and so far I see no problems with my understanding of how a two stage heat pump will work.
I also have no problem admiring when I'm wrong since that is not something uncommon.

I also think my first analogy with water pumps was not that bad so you should look at that one again if you still do not get what I'm saying.
The height delta there is the equivalent of temperature delta here.
With the first stage you do half of the work by moving 12W of heat and with the second stage you take that work already done and move another 12W.
So in the example with one stage you will be able to cool 1liter of water with 10°C lower than hot side you can call that ambient (the huge copper cube so that the small energy transferred will not influence the cube temperature).
When you add the second stage that has sufficient power pumped to cool the hot side of the first stage 10°C lover than the huge copper cube it is almost like if there was a single stage but the copper cube had just +17°C instead of +27°C

[splin]

No - if you don't increase the voltage to the first peltier it willl still only pump 12W with a 10C delta. That's where you have been going wrong. The water can now be cooled by 20C but the flow rate will have to be halved to .5l/hr. As you say, the second peltier will move 12W + 3.69W, requiring 4.83W (at a COP of 3.25) so outputing 20.52W to the copper block. Total electricity supplied is 3.69W + 4.83 = 8.52W

I know I'm not great at explaining things but I was sure my example was quite clear.

Your example was clear enough, it was just that you made a mistake when you analysed the effect of the second peltier.

When I added the second peltier element the hot side of the first peltier element has dropped from +27°C to +17°C provided by the cold side of the second peltier.
Each peltier element will pump around 12W (ignoring those IR losses) and those add up if the hot face of the first peltier is in contact with the cold face of the second one.

Yes but the bit you keep missing is that they are the same 12Ws so they don't add up.

Same as if you inverse the current flow on one of the peltier you cancel the work of the first one.
So you either have the Qc1 + Qc2 or Qc1 - Qc2 and in case Qc1=Qc2 the you have either 2xQc or you have zero in the second case.
If you have two of those cheap peltier module you can do the experiment but is not necessary since it should be obvious.

The peltiers don't generate any power - ignoring their I^2R and thermal resistance losses, they simply transfer the power that's passed to their cold side to the hot side but increases the temperature. Higher temperature does not equate to more energy or power out than was put in. So if you have 5 in series you still only get 12W out of the hot side - each peltier passing on the 12W whilst increasing the temperature by 10C. The 12W that comes out of the last was the 12W that was passed into the first from the water. An analogy is a human chain passing along a bucket of water, where each person heats up the water by 10C but pours some away before passing on exactly the same thermal energy they received.  Alternatively the first person gets a load of paralleled batteries. They rearrange the batteries so some are in series and passes it on. The total energy in the batteries doesn't change but the output voltage increases at each step. So you don't get QC1 + QC2, you get QC1 only.

They don't remove or destroy energy or power either so QC1 - QC2 != 0. If you reverse the second, both will pump heat from their cold faces to the joined hot sides. That heat would have nowhere to go so the hot sides will get hotter and hotter until they destroy themselves. As the hot side gets hotter, and their cold sides remain at constant temperature, their COP reduce and the amount of heat they pump will reduce but not probably not enough to avoid damage. Consider this configuration as two peltiers in parallel with their hot sides connected to a common (very poor) heatsink but their cold sides connected to heatsinks at different temperatures and thus operating at different delta-T.

I'm an EE so my knowledge of thermodynamics is limited but I'm smart enough so that if I'm wrong you will be able to explain where I'm wrong to me and so far I see no problems with my understanding of how a two stage heat pump will work.
I also have no problem admiring when I'm wrong since that is not something uncommon.

I expect that's more likely to be one of D. Trump's many interesting characteristics 

I also think my first analogy with water pumps was not that bad so you should look at that one again if you still do not get what I'm saying.
The height delta there is the equivalent of temperature delta here.

In your analogy the water gains height and thus potential energy. Peltiers increase the temperature but don't increase the energy. Consider them more like transformers which change the voltage but not the power of a signal. 

With the first stage you do half of the work by moving 12W of heat and with the second stage you take that work already done and move another 12W.

Not another 12W, the same 12W. The second peltier can't pump another 12W as the only heat available for it to pump is that delivered by the first. It can't create energy, it can only move it from the cold side to the hot side whilst transforming the temperature.

So the total output is 12W plus the heat losses in the peltiers. It's really quite simple to calculate the COP - its the total heat moved from the input, cold side divided by the amount of electrical power used. The COP usually shown in peltier datasheets relevant for cooling performance where the heat out is a waste product and is better described as COPcooling. In cases like yours where the heat out is what you are after, COPheating is relevant and is the total heat delivered to the hot side dived by the electrical power used.

With one peltier  the output is 12W + 3.69W              = 15.69W; COPc = 12/3.69      = 3.25; COPh = 15.69/3.69 = 4.25
With two peltiers the output is 12W + 3.69W + 4.83W = 20.52W; COPc = 12W/(8.52) = 1.41; COPh = 20.52/8.52 = 2.41

The COP almost halve when delta T doubles from 10 to 20C

For an EE analogy consider a 12W, 12V 1A power supply driving a 12W constant power load. The power supply is the 12W that comes from your cold water, the voltage is analogous to the water temperature. Now add two 100% efficient 10V voltage boost convertors in series. The output voltage is now 32V but it still can only deliver 12W. The voltage/temperature has been increased by the boosters/peltiers but they don't add or subtract any energy/power.

To model the peltier I^2R losses, add constant power, power supplies in series with the output of each booster, the first set to 3.69W and the second to 4.83W. Reduce the boost voltages so that each stage still only increases the voltage by 10V (so the output is still 32V) and increase the load's power to 20.52W.[/quote]

So in the example with one stage you will be able to cool 1liter of water with 10°C lower than hot side you can call that ambient (the huge copper cube so that the small energy transferred will not influence the cube temperature).
When you add the second stage that has sufficient power pumped to cool the hot side of the first stage 10°C lover than the huge copper cube it is almost like if there was a single stage but the copper cube had just +17°C instead of +27°C

Yes the water will be 10C below ambient with one stage, 20C below with two stages. But the energy extracted from the water will be exactly the same in both cases - 12W - so you will have half the quantity of water in the second case after any given amount of time.

Hope that helps.

[electrodacus]

Yes the water will be 10C below ambient with one stage, 20C below with two stages. But the energy extracted from the water will be exactly the same in both cases - 12W - so you will have half the quantity of water in the second case after any given amount of time.

I can think on one more simplification for you to understand this.

Same example single stage on the hot side is the +27°C  huge copper cube and water is also +27°C  then I can cool water flowing at the rate of 1 liter/hour to +17°C  power transferred from water 12W
Now what you do is replace that huge copper cube with a similar one but this time the cube is at just +17°C so now you can cool water flowing at the same rate of 1liter/hour to +7°C so power transferred from water 24W

If you can agree with the above then second stage will do the exact same thing keeping the hot side of the first stage 10°C lower than the original copper cube so +17C.

[ez24]

How about add your youtube link to your profile like your web address (which seems also not to have a YT link)

Also I suggest you make a post with links to other YT reviewers on your project

Can I add you to this list:

https://www.eevblog.com/forum/other-blog-specific/dd/msg1093983/#msg1093983

[electrodacus]

How about add your youtube link to your profile like your web address (which seems also not to have a YT link)

Also I suggest you make a post with links to other YT reviewers on your project

Can I add you to this list:

https://www.eevblog.com/forum/other-blog-specific/dd/msg1093983/#msg1093983

Thanks for the suggestion, I just made the change to my profile.
Yes you can add me to that list, I had no idea that list existed.
My projects are fairly low volume so only Martin ( mjlorton ) and Julian Ilett made video reviews about my Solar BMS.
Maybe when I have the new SBMS and DMPPT ready will send a sample to Dave also.

[Someone]

Yes the water will be 10C below ambient with one stage, 20C below with two stages. But the energy extracted from the water will be exactly the same in both cases - 12W - so you will have half the quantity of water in the second case after any given amount of time.

I can think on one more simplification for you to understand this.

Same example single stage on the hot side is the +27°C  huge copper cube and water is also +27°C  then I can cool water flowing at the rate of 1 liter/hour to +17°C  power transferred from water 12W
Now what you do is replace that huge copper cube with a similar one but this time the cube is at just +17°C so now you can cool water flowing at the same rate of 1liter/hour to +7°C so power transferred from water 24W

If you can agree with the above then second stage will do the exact same thing keeping the hot side of the first stage 10°C lower than the original copper cube so +17C.

Or we can bring it back to electrical theory, first eliminating the concept of two bodies at constant temperature. Take some given temperature as your environment being a large thermal mass which you cannot materially effect. When you move thermal energy to or from this environmental temperature to something else the something else has an impedance (insulation from) to the environment so it cannot be an arbitrary temperature difference without a source of energy to keep it there.

Now the fun bit, when you model the peltier element thermal system with the electrical analogy it has the characteristics of a voltage source (the temperature difference) and then a current source pouring the thermal energy dissipated in the operation of the peltier back into the hot side:
http://www4.ee.bgu.ac.il/~pel/pdf-files/conf105.pdf
Its the addition of these dissipations at each stacked stage that cause the loss of performance, as each subsequent hotter peltier is supporting/carrying the thermal load of all of those above. Like supporting a pyramid you often see stacked peltier stages having successively smaller peltiers as the hottest end is pushing a lot more thermal power through it.

[splin]

Yes the water will be 10C below ambient with one stage, 20C below with two stages. But the energy extracted from the water will be exactly the same in both cases - 12W - so you will have half the quantity of water in the second case after any given amount of time.

I can think on one more simplification for you to understand this.

Same example single stage on the hot side is the +27°C  huge copper cube and water is also +27°C  then I can cool water flowing at the rate of 1 liter/hour to +17°C  power transferred from water 12W
Now what you do is replace that huge copper cube with a similar one but this time the cube is at just +17°C so now you can cool water flowing at the same rate of 1liter/hour to +7°C so power transferred from water 24W

If you can agree with the above then second stage will do the exact same thing keeping the hot side of the first stage 10°C lower than the original copper cube so +17C.

Sigh; one last try but I'm beginning to wonder if you actually read any of my previous posts.

Consider it slightly differently, again assuming 100% efficient peltiers. Take water from a lake containing water at 7C which is pumped through a 100% efficient heatsink which keeps the cold side of a peltier at a constant 7C. The hot side is attached to a large copper mass at a constant 17C. The peltier is driven at 3V and is sized to pump 12W from the lake water to the copper mass given the delta T of 10C. I'm sure we agree so far.

The 12W is determined *only* by the characteristics of the particular peltier, the driving voltage and delta T - it can be read from the graphs in the datasheet.

Now replace the 17C copper mass by a second peltier which has another copper mass at 27C. The second peltier will also have a delta T of 10C and because it is the same size and is also driven at 3V will also pump 12W from the cold to hot side. Which is convenient because that is exactly how much  power is being received from the first. The key fact is that the first peltier cannot tell the difference between the two scenarios and doesn't care. In both cases it is driven at 3V and sees a delta T of 10C so pumps 12W from the lake water to whatever is on the hot side. It won't start extracting 24W from the water as you keep suggesting as it doesn't know that anything changed between the first and scenario.

The only difference is the temperature of the final hot side but the rate that heat energy is transferred from the lake water, 12W, is the same in both cases.

[tszaboo]

The only difference is the temperature of the final hot side but the rate that heat energy is transferred from the lake water, 12W, is the same in both cases.

Energy is measured in Joules, or calories, not watts. Heating up 1 gram of water 10 degrees is 10 cal, heating it up 20 degrees is 20 cal. The first system only heats up 10 degrees, the second 20. Hence the second system delivers more energy.
The peltier is actually not going to be "100% efficient". It is not efficiency. The same way as a fuel tanker, moving a 10 tons of fuel, and just using 100Kg is not going to be 10000% "efficient". So stop thinking about it that way.

[electrodacus]

The only difference is the temperature of the final hot side but the rate that heat energy is transferred from the lake water, 12W, is the same in both cases.

Energy is measured in Joules, or calories, not watts. Heating up 1 gram of water 10 degrees is 10 cal, heating it up 20 degrees is 20 cal. The first system only heats up 10 degrees, the second 20. Hence the second system delivers more energy.
The peltier is actually not going to be "100% efficient". It is not efficiency. The same way as a fuel tanker, moving a 10 tons of fuel, and just using 100Kg is not going to be 10000% "efficient". So stop thinking about it that way.

Thanks for getting involved.
If you are able to convince him with this short explanation then you are an excellent communicator
While all you are saying is completely correct I will prefer if you where not using Joules or calories for energy and instead use Watt second or Watt hour 1Joule = 1Ws = 1/3600Wh
For all involved in electronics Wh is a more meaningful unit for energy. (at least that is my personal opinion)

[electrodacus]

Sigh; one last try but I'm beginning to wonder if you actually read any of my previous posts.

Consider it slightly differently, again assuming 100% efficient peltiers. Take water from a lake containing water at 7C which is pumped through a 100% efficient heatsink which keeps the cold side of a peltier at a constant 7C. The hot side is attached to a large copper mass at a constant 17C. The peltier is driven at 3V and is sized to pump 12W from the lake water to the copper mass given the delta T of 10C. I'm sure we agree so far.

The 12W is determined *only* by the characteristics of the particular peltier, the driving voltage and delta T - it can be read from the graphs in the datasheet.

Now replace the 17C copper mass by a second peltier which has another copper mass at 27C. The second peltier will also have a delta T of 10C and because it is the same size and is also driven at 3V will also pump 12W from the cold to hot side. Which is convenient because that is exactly how much  power is being received from the first. The key fact is that the first peltier cannot tell the difference between the two scenarios and doesn't care. In both cases it is driven at 3V and sees a delta T of 10C so pumps 12W from the lake water to whatever is on the hot side. It won't start extracting 24W from the water as you keep suggesting as it doesn't know that anything changed between the first and scenario.

The only difference is the temperature of the final hot side but the rate that heat energy is transferred from the lake water, 12W, is the same in both cases.

I do of course read all that you are writing since I try to understand how I can better explain to you where you are wrong.
But maybe you already realized where you made the mistake.

Original question was the COP for two staked (as in serial thermal connection) heat pumps.
If for simplification both stages operate at a COP of 3 the total COP of the two stage system will also be 3

And to answer your question. Yes in your example with two stage the system will move 24W from the lake water to the large copper mass (not sure why you got complicated with the lake when you could have used another large copper cube on the cold side also).

Still my example with the 1liter/hour flow was better since cooling that 1 liter of water by 10C will equate with 12Wh of energy while cooling the same 1 liter of water by 20C will equate with 24Wh of energy.
That 1 liter of water siting at 10C lower than ambient will have a 12Wh stored and if is 20C lower than ambient it has 24Wh energy stored. So while that water will get back to ambient  temperature will transfer that amount of energy to ambient.

This makes Peltier elements fairly efficient if properly used. I just think many people have misconception about peltier elements in therms of COP because most if not all designs optimize for cost and not COP and then COP is realy bad below 0.5 in most of this applications. 

[Kleinstein]

.....
This makes Peltier elements fairly efficient if properly used. I just think many people have misconception about peltier elements in therms of COP because most if not all designs optimize for cost and not COP and then COP is realy bad below 0.5 in most of this applications.

No: Peltier elements are rather low efficiency. The best (not the cheap Chinese ones) ones are something like 15% of the thermodynamic limit, while conventional ones get something like 50% of that limit, sometimes more.

A TEC can get a high COP - but this is only at very low temperature difference (e.g. < 10 K). To be useful for heating, one would need to move heat from near 0 C to something like 30 C. At a 30 C difference the COP of an TEC is more like 1.5 at best. Stacking several TECs is not helping much at this low temperature difference - in theory it might help a little (like a few percent), but in practice things likely get worse due to thermal resistance.

So a COP of 3 at 10 K might be really good for a TEC, but it is still not competitive, as at such a low temperature difference conventional heat pumps could give you a COP of maybe 10 or 15.

The TEC specs are sometimes a little misleading, as they are for the limiting case of zero or maximum temperature difference. For best efficiency one would have to use lower current and thus also reduce the power per unit and thus increase the specific costs.

@electrodacus: if you are looking for mistakes - read your own texts too.

[electrodacus]

No: Peltier elements are rather low efficiency. The best (not the cheap Chinese ones) ones are something like 15% of the thermodynamic limit, while conventional ones get something like 50% of that limit, sometimes more.

A TEC can get a high COP - but this is only at very low temperature difference (e.g. < 10 K). To be useful for heating, one would need to move heat from near 0 C to something like 30 C. At a 30 C difference the COP of an TEC is more like 1.5 at best. Stacking several TECs is not helping much at this low temperature difference - in theory it might help a little (like a few percent), but in practice things likely get worse due to thermal resistance.

So a COP of 3 at 10 K might be really good for a TEC, but it is still not competitive, as at such a low temperature difference conventional heat pumps could give you a COP of maybe 10 or 15.

The TEC specs are sometimes a little misleading, as they are for the limiting case of zero or maximum temperature difference. For best efficiency one would have to use lower current and thus also reduce the power per unit and thus increase the specific costs.

TEC are solid state so very reliable.
You can stack TEC modules to increase the temperature delta while maintaining high COP.
Yes cost will increase but you can get a TEC module now for $2 or $3 (that is what I paid a few years ago did not checked now).
And yes a compressor based heatpump can have a higher efficiency but as I mentioned I prefer the solid state reliability of TEC.
Based on the cost of raw materials TEC should be less expensive than compressor based so is maybe just about the production volume.

@electrodacus: if you are looking for mistakes - read your own texts too.

If you are referring to grammar then you are right there are sure plenty of mistakes and if you are referring to technical mistakes then please point out where.

[polowcz]

how about a system combined from PV doing electrolysis of a water down to oxygen and hydrogen during the day (sun presence) and for nights using a fuel cell? would this system be technically/economically possible for wide presence? Benefits are well known since Apollo programme - electricity, (hot) water and no pollution...

BR
Paul

[Seekonk]

Speaking of efficiency, I'd like to know how you PWM a battery off the same panels you power point the resistive heaters.  It seems you can be very selective when it comes to numbers.

[electrodacus]

how about a system combined from PV doing electrolysis of a water down to oxygen and hydrogen during the day (sun presence) and for nights using a fuel cell? would this system be technically/economically possible for wide presence? Benefits are well known since Apollo programme - electricity, (hot) water and no pollution...

BR
Paul

I think I seen somewhere a similar solution but was for long therm energy storage as in they used PV energy is summer to generate and store hydrogen and then used that hydrogen to heat the house in winter.
It will maybe be a solution in places with very bad solar energy in winter months but not sure if even there this will be cost effective.
PV energy is so inexpensive now that even installing oversized PV arrays may be more economical than generating and storing large quantities of hydrogen.
For short/medium therm energy storage as in days up to a week nothing can beat thermal storage as far as I know.

Speaking of efficiency, I'd like to know how you PWM a battery off the same panels you power point the resistive heaters.  It seems you can be very selective when it comes to numbers.

The pdf paper in the first post explains how everything is done (project will be open source).
I do not use any PWM and what I do is redirect some of the panels to battery charging.
In my example I have the large 10kW PV array split in 6 smaller array's of different sizes and I'm able to redirect any of the sub arrays to battery charging.
There are a total of 39 panels and I can redirect to battery charging one up to all 39 panels depending on the amount of energy available and the combination of those 6 inputs will give me 39 individual levels basically I can connect any number of panels between 1 and 39
All panels that are not redirected to battery charging can be used for heating and storing that heat in thermal mass.
On the heating side there is also no PWM and there are again 6 outputs that can give me a theoretical max 63 levels of power (I will only use 31 levels in my example) so that I can get a Digital maximum power point tracking.
While the idea seems simple I'm not aware of any such system.
Hope you got the idea but if you read the document all is explained in more details and also all the benefit of this are explained also.

So there is no PWM done in the Digital MPPT or even in the Solar BMS.  PWM is not useful in any of this cases.

PS: In the PDF presentation there is actually even a link to a simple online simulator and a file that you load there with an extremely simplified version of the DMPPT with just 3 inputs and 3 outputs and playing with that can give a better understanding of how the DMPPT works.

[Kleinstein]

....
TEC are solid state so very reliable.
You can stack TEC modules to increase the temperature delta while maintaining high COP.
Yes cost will increase but you can get a TEC module now for $2 or $3 (that is what I paid a few years ago did not checked now).
And yes a compressor based heatpump can have a higher efficiency but as I mentioned I prefer the solid state reliability of TEC.
Based on the cost of raw materials TEC should be less expensive than compressor based so is maybe just about the production volume.
...

Stacking TEC modules does not increase the temperature difference and still maintain the COP.  This just does not work - it is against the laws of thermodynamics. In a simplified picture two elements in series will also need twice the electrical energy to drive and thus this would cut the COP to half for two elements in series and twice the temperature difference.  In the complete picture things get a little more complicated as the heat flow will not be exactly the same for both elements as the electrical energy is converted to heat too - so this give some minor corrections.

If things would be so easy to improve efficiency you would see much more stacked modules. In the low temperature difference limit, stacking of modules is not much more than making the legs longer and thus use a lower power density. So no change in COP if used with the same total temperature difference - just twice the costs for half the power. This is why you don't see stacked elements for low temperature differences.

The usual TEC elements use rather expensive materials, like alloys with bismuth and tellurium. Alternative materials (e.g. germanium based) are often even more expensive. So it is not all about small quantity, but also about the material used. This gets even worse if TEC would be used in larger quantities, as the tellurium reserves are very limited - thus significant more demand would cause prices to increase. Chances are those $3 modules were a special limited offer (excess inventory or even factory rejects). For higher power (e.g. kW range) - I am not sure a TEC based version would be cheaper than a compressor based solution. You can't compare just a bare TEC element to a complete heat pump system. The TEC system would also need quite some extra parts (e.g. heat exchanger / fan or pumps).

[electrodacus]

Stacking TEC modules does not increase the temperature difference and still maintain the COP.  This just does not work - it is against the laws of thermodynamics. In a simplified picture two elements in series will also need twice the electrical energy to drive and thus this would cut the COP to half for two elements in series and twice the temperature difference.  In the complete picture things get a little more complicated as the heat flow will not be exactly the same for both elements as the electrical energy is converted to heat too - so this give some minor corrections.

Have you read my examples above ? I know this discussion is getting out of hand but at least a few people reading this may clear some misconceptions.
How will the laws of thermodynamics be against this ?
Yes two elements in series will need twice the amount of power (actually slightly more if you want each stage to have the same COP for 10C delta) but then the equivalent COP of this two modules in series will be the same since they also extract more power from the cold side (twice as much if delta temperature is 2x higer).
So two stacked TEC modules will require slightly more than 2x more power but also move two times more so equivalent COP will be almost the same slightly smaller.

I think I mentioned before here and maybe this help's you understand better. If you have two peltier elements and you connect the cold side together on both (use some ideal thermal insulation around that cold side so they can not pump energy from the ambient) and supply both with same power say each with a COP of 3.25 for cooling of course 3.25 + 1 = 4.25 for heating  then the resulting COP of this system will be zero.
Yes zero in therms of cooling and 1 in terms of heating since all the heat you will get will be from the I2R losses so basically the supplied power.
 

If things would be so easy to improve efficiency you would see much more stacked modules. In the low temperature difference limit, stacking of modules is not much more than making the legs longer and thus use a lower power density. So no change in COP if used with the same total temperature difference - just twice the costs for half the power. This is why you don't see stacked elements for low temperature differences.

No you do not improve the COP by staking modules (COP not the same thing with efficiency) The COP is improved by using the peltier module with lower power and temperature delta since the problem with peltier is the thermal bridging since materials used between cold and hot side are relatively good thermal conductors so the lower the power the better the efficient as you probably seen in the last page of the spec http://www.thermonamic.com/TEC1-12706-English.PDF
 

The usual TEC elements use rather expensive materials, like alloys with bismuth and tellurium. Alternative materials (e.g. germanium based) are often even more expensive. So it is not all about small quantity, but also about the material used. This gets even worse if TEC would be used in larger quantities, as the tellurium reserves are very limited - thus significant more demand would cause prices to increase. Chances are those $3 modules were a special limited offer (excess inventory or even factory rejects). For higher power (e.g. kW range) - I am not sure a TEC based version would be cheaper than a compressor based solution. You can't compare just a bare TEC element to a complete heat pump system. The TEC system would also need quite some extra parts (e.g. heat exchanger / fan or pumps).

The small TEC1-12706 that is one of the most popular can be had even in low volume at $1.66 in China right now (just checked the price). Try a search on Aliexpress for this model TEC1-12706
And while if used at full rated power or close to say 12V is typical in most application it will require 5A for a delta of 40C and move just 23W while using 60W so COP of 0.38
While same module used at just 3V will require 1.23A at a delta of 10C so will move 12W wile requiring just 3.69W so a COP of  3.25
If I where to stack another module I will probably be using the TEC1-12707 with the cold side in contact with the hot side of the TEC1-12707 thus at the same 3V I will be able to move the necessary 12W + 3.69W = 15.69W at 10C delta while consuming about 4.8W 3Vx1.6A and maintaining the same COP 3.25
This two TEC seen as a single unit will be capable of moving a total of 27.69W from the cold side while using a total of 3.69W + 4.8W = 8.49W.

Some numbers are rounded and some data are aprox read from the graphs in the spec but the idea remains the same.

If you disagree please provide the calculation for this same example above.