Hot water with electricity from solar panels

[mrm]

Originally Posted by BjarneK

[...]
The heater is 1500 W at 110 V which is a resistance of 110*110/1500 = 8.1 Ohm since P = U * I and U = R * I, so R = U*U / P.

When you connect the solar panels to the heater, the resistor is going to keep a constant ratio between voltage and current (U = R * I).

Ignoring the thermal coefficient of the heater element, that is correct. However..

Since you have two panels in parallel, each is going to see a resistance of 16 Ohm.

This, I think, requires some explanation. The resistance of a heater element will not magically change, it will still remain at 8 Ohms or thereabouts and the voltage across the resistance will be a function of a current flowing through it, so:

U(I) = I*R

What you are saying is that with two identical panels, each of them will supply half of current flowing through that heater element and that is correct, but I have doubts, whether the rest of matching analysis is correct?

[BjarneK]

Originally Posted by mrm

However..


Quote:

Since you have two panels in parallel, each is going to see a resistance of 16 Ohm.

Huh?

I admit that I skipped a few steps when I said that each panel "sees" 16 Ohm.

Think of it this way: The voltage is U (same for both panels, since they are parallel), the current from each panel is Ip, total current I = 2 * Ip, the total resistance is R (8 Ohm). Then we have: U = R * I = 2 * R * Ip, Then Ip = U / (2 * R) = U / 16 Ohm.

If you want to use the value 8 Ohm, just multiply all the currents on the Y axis of my graph by two since there are two panels and you will get the same result.

Edit: I see that you thought a little further after the initial "Huh?" response ;-)

[mrm]

Originally Posted by BjarneK

I admit that I skipped a few steps when I said that each panel "sees" 16 Ohm.

Think of it this way: The voltage is U (same for both panels, since they are parallel), the current from each panel is Ip, total current I = 2 * Ip, the total resistance is R (8 Ohm). Then we have: U = R * I = 2 * R * Ip, Then Ip = U / (2 * R) = U / 16 Ohm.

If you want to use the value 8 Ohm, just multiply all the currents on the Y axis of my graph by two since there are two panels and you will get the same result.

Edit: I see that you thought a little further after the initial "Huh?" response ;-)

Yes, I decided that more words are necessary
I still have doubts about your matching analysis. If you take performance graphs of a single panel, I think you still should draw an I/V relationship for R=8 on that graph, because that is what each individual panel will 'see'. But maybe I am wrong, need more coffee...

[senormechanico]

For you "Gloom and Doomers" who advise against using an inverter because of the inefficiencies, I have a question:

What difference does it make if you come out with hot water and full batteries at the end of the day having never started the engine or (in my case nonexistent) generator?

[mrm]

Originally Posted by BjarneK

I see that you thought a little further after the initial "Huh?" response ;-)

After more coffee I went even further and did some plots. From your I/V panel characteristics I calculated a number of panel power values for 800W/m2 insolation:

# V P
0 0
5 24.95
10 49.8
15 74.55
20 99.2
25 123.75
30 148.2
35 172.55
40 196
45 180
51 25
52 0

and then plotted the panel power across its voltage range as well as heater element power for same voltage range.

gnuplot> plot '/tmp/pvp.txt' using 1:2 with linespoints smooth csplines title'P_panel@800W/m2 smoothed' , \
'/tmp/pvp.txt' using 1:2 with linespoints title 'P_panel@800W/m2 raw data', \
(x*x)/8 title 'P_heater', \
0.5*(x*x)/8 title 'P_heater / 2'

See the resulting plots and tell me if you still think those panels are well matched to this use. Look at P_heater / 2 curve, as each panel has to supply half of the power.

[valhalla360]

Originally Posted by Dockhead

No, his analysis is sound. His approach is clever because using the very same heating element (expensive or impossible to mount dual heating elements in most calorifiers), he has a 4 times smaller one using the 110v inverter. This is small enough that it can be primarily driven by his solar panels without drawing power from the batteries. If he drove it at 230v, it would be sucking power from the batts like crazy.

Furthermore, it is simply not true, when using lead-acid batteries, that "To heat the same amount of water to the same temperature uses the same amount of power". Ever heard of Peukert?

His purpose is to divert excess solar power into the calorifier. His solution is elegant, avoiding a second immersion element, and avoiding power loads which would flatten bis batteries.

Why have a 110v water heater in the first place if you are running an off grid boat? Just go with the 12v heating element and be done.

Pukert effect doesn't come into play. Batteries are 12v. If you are going to draw X watts off them, you need to provide X/12 amps regardless of voltage...except when you figure efficiency losses, you really need to provide 10-20% more watts to run it all thru an inverter. If he's pulling off the batteries he's not diverting.

Diverting excess power is fine but if he was talking about pulling off the batteries this brings other problems. Better to get a DC water heater and use the battery charger to supply power when on shore power as shore power can run the charger indefinitely.

[valhalla360]

Originally Posted by senormechanico

For you "Gloom and Doomers" who advise against using an inverter because of the inefficiencies, I have a question:

What difference does it make if you come out with hot water and full batteries at the end of the day having never started the engine or (in my case nonexistent) generator?

You paid too much for an overbuilt system. I wouldn't rip it out if it's already there but I wouldn't purposely build it that way.

[Dockhead]

Originally Posted by valhalla360

Why have a 110v water heater in the first place if you are running an off grid boat? Just go with the 12v heating element and be done.

Pukert effect doesn't come into play. Batteries are 12v. If you are going to draw X watts off them, you need to provide X/12 amps regardless of voltage...except when you figure efficiency losses, you really need to provide 10-20% more watts to run it all thru an inverter. If he's pulling off the batteries he's not diverting.

Diverting excess power is fine but if he was talking about pulling off the batteries this brings other problems. Better to get a DC water heater and use the battery charger to supply power when on shore power as shore power can run the charger indefinitely.

You really didn't understand at all what he is doing.

First of all, it is rarely practical to put a second immersion element in a calorifier -- it needs a boss and it's a an expensive futz. That was written about. He can't put in a DC element because he doesn't want to give up the AC one.

He's got a variable amount of solar power. If he runs the AC immersion element at its normal voltage, then most of the power will come from batteries, and due to Peukert (which does indeed come into it) this will flatten his batteries, and plus there are further losses from the conversion into battery power and back. By reducing the voltage, he reduces the power of the immersion element 4x, and this makes it possible to cover all or most of it from his solar panels. What remainder comes from the batts won't flatten them because it's a much smaller amp draw. To occasionally draw part of the power from the batts does not defeat the purpose.



Is that more clear? It's good to think and understand, what someone proposes to do, before launching into a noisy critique of it. This is a clever idea. A second, DC immersion element would be better, but it's not practical in this particular case. A DC element could have a relay and voltage switch so that it only runs when there is excess power -- would be a good setup.

To the OP: you could do this with your setup as well. Just wire the remote switch of the Phoenix through a variable voltage sensing relay, so that it shuts down as soon as DC system voltage falls below 12.5v or whatever. Then you can prevent it from drawing anything at all from your batteries.


Even better would be a variable voltage inverter, which would adjust automatically to match the excess power from the panels and prevent the immersion element from drawing from the batteries at all. I bet it could be cobbled up.

[senormechanico]

Originally Posted by valhalla360

You paid too much for an overbuilt system. I wouldn't rip it out if it's already there but I wouldn't purposely build it that way.

The hot water tank is original with the original 120 volt, 1,000 watt tank element.
The 1500 watt sine wave inverter is used for our VitaMix, wife's hairdryer, and whatever other AC uses like power tools, soldering iron etc.
Cost me $199 and free shipping from ebay.

Tiger Claw 1500w/3000w Pure Sine WavePower Inverter DC-AC Power Converter | eBay

Oh, I almost forgot the cost of some wire and a big fuse with holder. $50 ?

All the above was already in place when I added a few bits to engage the hot water tank circuit.

I fail to see the validity of your point.

[botanybay]

20% is a bit large for modern high efficiency inverters sized to the load to hit their peak efficiency.

From memory my victron inverters are between 92% and 94% efficient at their peak efficiency.

Now I understand peak is only at one point so size the dedicated small inverter to match the load.

From memory this point is near 60 to 80 percent load. So if you want 200 watts of efficient conversion use a 350 watt inverter (or wherever the efficient point is)

Then again, the OP is talking about waste energy after the batteries are charged. So no battery inefficiency anyway. So pretty efficient conversion for waste power.

Sent from my SM-G935V using Cruisers Sailing Forum mobile app

[dlymn]

One of the ways you can connect your water heater to the batteries is through the inverter. This way the heater is being fed the voltage it was planned for. The loss through the inverter is not great. You could wire the inverter through a switch (DPDT) so that your mains voltage system is getting power either from the shore or the inverter but not both.

Now, if your solar panels are working, they feed into the batteries and either charge the batteries and heat water or heat water. Be warned that the water heater will take priority and the batteries will only charge when the thermostat cuts the current to the element. A 20 litre water heater will use 700w to raise the temp from 20c to 50c in 1 hour

If you can provide that power during the day and still charge your batteries then all is well.

You may not need that amount of electricity. Let's imagine you have a water heater that is also heated by the engine. Our experience is that 30+minutes of engine operation (ie in or out of a mooring) will heat our 20l hot water tank very well. The heater retains warmth at least enough for a shower for 2 days. Now if you are at anchor, you've got 20l of hot water on tap and after a shower the new mix of hot and clod water might be at say 30c and this will only take less than 500w to heat back up to 50c. A big solar system should easily cope with this and get a recharge in your batteries. A 200 amphour lead acid will need about 1000 ah of energy at 12v to fully recharge assuming discharge to 60% That charge should keep a small boat fridge going for a day or more. You can calculate the rest

But remember, this only works in high summer. Your generating power might reduce radically in winter depending on where you're based. You need to look at solar insolation charts to determine this.

[mitiempo]

Originally Posted by dlymn

A 200 amphour lead acid will need about 1000 ah of energy at 12v to fully recharge assuming discharge to 60%.

Possible typo. I think you mean 100, not 1000.

[dlymn]

Originally Posted by mitiempo

Possible typo. I think you mean 100, not 1000.

Yes quite right. I was thinking watts and ah and trying to keep it simple but then my brain clicked out. StuM will be caning me

[StuM]

Originally Posted by dlymn

Yes quite right. I was thinking watts and ah and trying to keep it simple but then my brain clicked out. StuM will be caning me

Nah, that's all right, you got the units correct, just a typo in the numbers.

Also, what people frequently forget is that the system is only nominally 12V.
It's not 100Ah @ 12V, it's more like 100Ah at somewhere between 13 and 14.4V to recharge the battery, so you need to add about another 10% or so to the Watt hours required from your panels)

[valhalla360]

Originally Posted by Dockhead

You really didn't understand at all what he is doing.

First of all, it is rarely practical to put a second immersion element in a calorifier -- it needs a boss and it's a an expensive futz. That was written about. He can't put in a DC element because he doesn't want to give up the AC one.

He's got a variable amount of solar power. If he runs the AC immersion element at its normal voltage, then most of the power will come from batteries, and due to Peukert (which does indeed come into it) this will flatten his batteries, and plus there are further losses from the conversion into battery power and back. By reducing the voltage, he reduces the power of the immersion element 4x, and this makes it possible to cover all or most of it from his solar panels. What remainder comes from the batts won't flatten them because it's a much smaller amp draw. To occasionally draw part of the power from the batts does not defeat the purpose.



Is that more clear? It's good to think and understand, what someone proposes to do, before launching into a noisy critique of it. This is a clever idea. A second, DC immersion element would be better, but it's not practical in this particular case. A DC element could have a relay and voltage switch so that it only runs when there is excess power -- would be a good setup.

To the OP: you could do this with your setup as well. Just wire the remote switch of the Phoenix through a variable voltage sensing relay, so that it shuts down as soon as DC system voltage falls below 12.5v or whatever. Then you can prevent it from drawing anything at all from your batteries.


Even better would be a variable voltage inverter, which would adjust automatically to match the excess power from the panels and prevent the immersion element from drawing from the batteries at all. I bet it could be cobbled up.

I suggest you take your own advice.i never suggested fitting two elements.
I also never suggested pulling off the batteries.

And the perkuet effect is not effected by the voltage selection. It is a function of the amperage at 12v. A 500w element at 120v, 240v or 200,000v will draw the same amps from the batteries (not sure what the efficiency of a 200,000v inverter is so that may not make as much sense)