Current question and sources of information

**SunEagle** · 12-09-2016, 12:06 PM

Hello David and welcome to Solar Panel Talk

To maybe answer your first question I would say should not try to calculate the output of a panel based on its volts and amps since the volts will change a little and the amps will change a lot during the day. The number to use is the panels Imp (or max current rating). That is the most it will produce at any time during the day which will help determine how many you will need to charge the battery system.

Best to first determine your daily watt hour usage. That would require you to determine the wattage of each of those fans and how long you want to run them each day.

Once you get that watt hour number you can calculate your 12volt battery size in Ah. It is always a good idea not to use more than 25% of the battery each day and to have at least 3 to 4 days of no sun (no charging) built into the calculation.

So as an example lets say each fan is ~ 10 watts and you want to run them for 8 hours each day from the batteries. That will calculate to 4 x 10w x 8hr = 320watt hours.

Now take that 320wh / 25% = 1280 watt hours. Take the 1280wh and divide by the battery voltage you get 1280wh / 12v = 107Ah. So now you have an idea about the size of the battery.

From there you know you will be using 320watt hours a day so you need to put that back into the battery plus a little more due to losses. Which on a day that only has 3 hours of useful sunlight (depends on where you live and what part of the year it is) would require 320wh / 3hr x 1.25% = 133 watt panel or round up to a standard size of 150watts.

To go into more detail may confuse you so I would suggest your read the Sticky threads in the Off Grid Solar Panel System section. There are some pretty good directions to help you determine your battery and solar panel wattage along with the right charge controller and if needed inverter.

**Mike90250** · 12-10-2016, 04:04 AM

[QUOTE=David Maxwell;n338016]......Now my second question: are there any sources of information in which all this is laid out and explained? Could I have found out ahead of time that a 12 volt panel puts out wildly varying voltages, QUOTE]
The sticker on the back of most every panel will tell you this Sorry you fried fans.

**David Maxwell** · 12-10-2016, 10:39 AM

Thank you all ever so much. As SunEagle suggested, there is considerable useful information on the Off-Grid forum, (although I am not sure I would have known that a "12 Volt" panel puts out a lot more than 12 Volts, ( and , no, there was no indication on the back of the panel. I think what is at play here is a large divide between the community of solar enthusiasts dealing with large domestic systems, basically assembling parts from specialised sources, and tinkerers like me, putting things together from bits purchased on eBay from China for $15 or $20. My fans were actually designed for cooling computers. I suspect if I had purchased fans designed for solar applications the issue would never have arisen.)

I followed SunEagle's excellent calculations, and am satisfied that all bits are appropriately sized. (I have a load of ~30 Watts, which runs for up to 6 hrs/day, or 180 Watt Hrs/day. At 12 V, this works out to a 60 AmpHrs battery and a 75 Watt panel. My battery is 100 AHr and my panel is 90 Watts, so I shouild be fine.

But... I still am unclear about how a PV panel works - a battery, (which is doing essentially the same thing - pushing electrons around, (albeit with chemical reactions rather than photons)) maintains a fixed voltage, independent of load. It appears that a solar panel has markedly varying voltages, (mine ranges from 22 Volts down to 5 Volts with varying insolation). Is it maintaining a relatively steady current flow, and dropping its voltage to compensate?? (And, in fact, these voltage measurements were taken with the panel connected to a load. If I were to disconnect the load, would the open voltage in fact be back up around 22 V? In other words, to use a plumbing analogy, if the panel is viewed as a pump, will it pump to a given pressure as long as the taps are all closed, but this pressure drops considerably if the taps are opened to the point that the capacity of the pump is reached?

My sense is that this may be an explanation for your figure of 3 hours of "useful sunlight", which was floated in with no further explanation. The inference I draw from this is that, even if the sun is shining 7 or 8 hours a day, only a brief period around noon provides enough actual solar energy to drive useful work. Correct? (I am in Nova Scotia, at latitude 45 degrees, if that contributes anything useful to the discussion. And my panel is oriented at about 15 degrees west of due south, angled at 45 degrees.)

Finally, tell me whether I am asking my questions in the wrong forum - should I be somewhere else, (like the off-grid one, [which didn't seem to me quite right]). Who deals with physics and engineering like this? I don't want to burden you inappropriately.

**Sunking** · 12-10-2016, 12:11 PM

Originally posted by David Maxwell

But... I still am unclear about how a PV panel works - a battery, (which is doing essentially the same thing - pushing electrons around, (albeit with chemical reactions rather than photons)) maintains a fixed voltage, independent of load. It appears that a solar panel has markedly varying voltages, .

That is because you do not know the difference between a voltage source (battery), and current source (panel). They do not operate or behave the same way. Solar Panels are Current Sources, not voltage.

**bcroe** · 12-10-2016, 01:02 PM

Simple DC motors will take a large overload for a brief interval. But perhaps you have some of those fans used in electronics, which
have a solid state driver to run a brushless motor. The driver may not be so tolerant of over voltage.

The most efficient operating point for your panel is likely about 80% of the open circuit voltage. That is pretty constant but the current
available will be dependent on sun intensity. Your voltage regulator must deal with this, ideally in a very efficient way. Bruce Roe

**David Maxwell** · 12-10-2016, 02:42 PM

Sunking, your phrasing is unfortunate, and comes across as unnecessarily offensive. (I am not saying this was your intent). Saying that I "don't know the difference between a battery and a PV panel" is both accusatory, and in fact incorrect. Had I not recognised that they behave differently, I would not have asked the question in the first place. But I *think* you have actually answered my question - a PV panel seeks to deliver a fixed flow of electrons ( a current), and varies the pressure, (voltage) to compensate for variations in the "supply" (the number of electrons knocked loose by the incoming photons) Correct?
If so, this has a rather significant implication, (if I have analysed it correctly). The PV panel is trying to deliver a volume of electrons (current), but the battery will resist accepting these electrons unless they are "pushed" with sufficient force, (voltage). If my analysis is correct, there will be a threshold below which a solar panel simply will not do anything - essentially the insolation needed to generate a voltage of around 13.5 volts. Below this level the panel may well drive a load, which depends on current flow rather than voltage, (like my fans or little circulation pumps), and continue to function, (albeit more slowly) down to very low voltages. (That is the PV panel connected directly to the fans and pumps will work even in cloudy conditions, but a battery will not come even close to accepting a charge). An interesting dilemma. (And I have a notion that there is an additional factor here - the lower the demand placed by the load, the higher the voltage the panel will actually deliver. So, for example, if all one is trying to do is maintain a trickle charge in the battery, the voltage the panel will deliver will be high enough even in cloudy conditions. But if we are trying to charge a discharged battery, the extra demand will result in the voltage dropping below that needed to drive the process. But I am open to correction here)

Perhaps it would help if I told you what I am trying to do. I have a passive solar greenhouse, with two large tanks filled with water as heat storage. The heat transfer into the water is inefficient, (absorption through black front walls, plus conduction from the air) and so I have rigged up a system with two car radiators through which the water is pumped when the air temperature exceeds the temperature of the water in the tanks, plus the famous fans to blow the air through the radiators. At present this whole thing runs directly from the PV panel, with voltage regulators to drop the voltage to 12 volts, (which, interestingly simply pass voltages lower than 12 V on to the load, so my pumps and fans continue to operate down to about 5 V. Now, if I insert a battery and charge controller into the circuit, and draw my power from the battery, rather than the panel, the pumps and fans will continue to run even during periods of weak light, (the greenhouse actually captures a remarkable amount of heat even on heavily cloudy days), BUT the energy to recharge the battery will not be collected because it will charge only in full sun, (or almost so), for a very restricted portion of the day. (My current direct connection arrangement will run for 5 to 6 hours a day. If I interpose a battery, I will collect energy to drive it for no more than 2 or 3 hours a day, and only on sunny days.) Is there a flaw in this reasoning?

**Sunking** · 12-10-2016, 03:56 PM

A solar panel behaves like a Current Source from Isc to Vmp, and like a Voltage Source from Vmp to Voc. But the answer to your question is you get maximum power transfer when both Source and Load Impedance are EQUAL. This means your load must be Dynamic add can operate on variable limited power. In your application pumps made for solar panels. They have a Buck/Boost converter that does the Magic of balancing the Load demand to with what the Source is limited to. So if you have a 100 watt panel, the motor needs to be able to operate from say 2 watts up to 90 watts or however much the panel can supply at that moment in time. You do not need to reinvent the wheel as they make pumps to do what you want already using series wound DC motors with built in converter.

Now for a battery system the controller is simpler, just a Buck DC Converter is needed because you are always stepping the voltage DOWN to a higher current level. For a Series Wound DC Motor you need both Buck and Boost DC Converter which cost a few more coins due to added circuitry and complexity. The Buck Boost Converter will deliver the maximum available power that is possible at that moment in time depending on conditions. Thus the motor must be able to run on whatever power is available at the time and is Dynamic. A series DC wound Motor is controlled with Voltage and the Buck Boost Converter finds the motor Max Power point matching the impedance of the solar panel at that moment in time.

**J.P.M.** · 12-10-2016, 04:11 PM

Originally posted by David Maxwell

Sunking, your phrasing is unfortunate, and comes across as unnecessarily offensive. (I am not saying this was your intent). Saying that I "don't know the difference between a battery and a PV panel" is both accusatory, and in fact incorrect. Had I not recognised that they behave differently, I would not have asked the question in the first place. But I *think* you have actually answered my question - a PV panel seeks to deliver a fixed flow of electrons ( a current), and varies the pressure, (voltage) to compensate for variations in the "supply" (the number of electrons knocked loose by the incoming photons) Correct?
If so, this has a rather significant implication, (if I have analysed it correctly). The PV panel is trying to deliver a volume of electrons (current), but the battery will resist accepting these electrons unless they are "pushed" with sufficient force, (voltage). If my analysis is correct, there will be a threshold below which a solar panel simply will not do anything - essentially the insolation needed to generate a voltage of around 13.5 volts. Below this level the panel may well drive a load, which depends on current flow rather than voltage, (like my fans or little circulation pumps), and continue to function, (albeit more slowly) down to very low voltages. (That is the PV panel connected directly to the fans and pumps will work even in cloudy conditions, but a battery will not come even close to accepting a charge). An interesting dilemma. (And I have a notion that there is an additional factor here - the lower the demand placed by the load, the higher the voltage the panel will actually deliver. So, for example, if all one is trying to do is maintain a trickle charge in the battery, the voltage the panel will deliver will be high enough even in cloudy conditions. But if we are trying to charge a discharged battery, the extra demand will result in the voltage dropping below that needed to drive the process. But I am open to correction here)

Perhaps it would help if I told you what I am trying to do. I have a passive solar greenhouse, with two large tanks filled with water as heat storage. The heat transfer into the water is inefficient, (absorption through black front walls, plus conduction from the air) and so I have rigged up a system with two car radiators through which the water is pumped when the air temperature exceeds the temperature of the water in the tanks, plus the famous fans to blow the air through the radiators. At present this whole thing runs directly from the PV panel, with voltage regulators to drop the voltage to 12 volts, (which, interestingly simply pass voltages lower than 12 V on to the load, so my pumps and fans continue to operate down to about 5 V. Now, if I insert a battery and charge controller into the circuit, and draw my power from the battery, rather than the panel, the pumps and fans will continue to run even during periods of weak light, (the greenhouse actually captures a remarkable amount of heat even on heavily cloudy days), BUT the energy to recharge the battery will not be collected because it will charge only in full sun, (or almost so), for a very restricted portion of the day. (My current direct connection arrangement will run for 5 to 6 hours a day. If I interpose a battery, I will collect energy to drive it for no more than 2 or 3 hours a day, and only on sunny days.) Is there a flaw in this reasoning?

I designed and built a solar sunspace in Buffalo NY back in the '80's. Nice to hear from something of a kindred spirit.

I can confirm that the space did stay warmer than outside ambient during daylight hours. The space had about 4,000 lbm of water with black dye in it, contained in 6 translucent FRP cylinders oin the back wall, slightly more exposed to the afternoon sun than the morning sun. There was moveable thermal insulation over the vertical glazing operated by something of a differential thermostat I rigged that raised the curtain when irradiance was sufficient to increase the space temp. above the outside ambient temp. (Or lower the curtain during warm weather to help avoid overheating.)

If I correctly understand what you're doing with the auto radiators, from a heat transfer standpoint and from the net energy gained, I suspect, but don't know for certain, that you'd be further ahead doing nothing along those lines and instead trying to ensure the containers get max. solar exposure. Doing so, you may avoid a bit of overheating of the space. I don't have the flow parameters and setup your using for the radiator scheme, but, having spent most of an engineering career designing industrial heat transfer equipment and other energy use stuff, I suspect the radiator scheme may well cost more energy (and $$) than it saves.

As for the natural convection you speak of: For natural (gravity driven) convective (not conductive when talking about the air to container surface heat transfer BTW) heat transfer between the air and the container walls, will be something of the order of about 1 BTU/ft.^2*deg. F. delta T*hr. Not all that much.

As an example, the rate of convective heat transfer too/from my containers was about 170 BTU/(hr. deg.F.) or about 800 - 1,000 BTU/(day*deg.F. delta T.) for the whole assembly. On a sunny winter day, the radiative heat transfer due to solar insolation would pump about 40,000 BTU into the water. Thus, and this sounds contrary to what you may think, heat transfer into the water via Solar insolation/thermal radiation can be much more efficient than convection heat transfer if done correctly.

By way of comparison as to how ineffective natural convection is compared to solar radiation when transferring heat, if the water containers are reasonably dark, and exposed to the sun, each ft.^2 of container exposed to solar radiation may well (and somewhat conservatively) be expected to transfer something like 600 - 800 BTU of heat to the water over the course of a reasonably sunny day.

Most of the experienced logic around thermal mass for sunspaces and passive solar in general knows that the thermal storage is primarily for 2 purposes: To prevent overheating and to provide thermal storage, primarily at night or on sunless/overcast days. The trick to making it work most efficiently and effectively is to expose the thermal mass to as much solar insolation as possible and practical. Putting large thermal masses in the shade or with less than max. solar exposure, and thinking convective heat transfer, either natural or forced will get the job done is a big mistake.

Take what you want of the above. Scrap the rest.

**David Maxwell** · 12-11-2016, 04:11 PM

We are off on a bit of a tangent here - is it OK to be talking about heat transfer and storage in a PV forum? If nobody objects, (and I will happily go elsewhere if they do), I would be most grateful for your input.
This is an evolving process. The original heat storage consisted of racks of plastic 4 L water jugs with water dyed with black dye. These were stacked along the back (North) wall, and indeed were dependent on direct solar radiation for heat capture. The advantage was that, being small, they were not subject to stratification of the heat. But they were not very efficient at heat capture. I then added a 6 mil plastic sheet across the front plus a fan to blow the warm air from the greenhouse through the stack of jugs. This actually made a major difference in heat capture, basically adding conductive gain from the air to the direct radiation.) But the plastic jugs lasted only a year or so before breaking down and leaking. And getting at them to replace them was a pain. So this summer I replaced the whole array with two large fibreglass tanks. These are semi- opaque, so there is no benefit in dying the water, but the front (south) faces are painted with flat black paint. But the heat gain through the front faces is going to be pretty restricted. The tanks are 6 ft X 2 ft high, by 16 in. deep. (This works out to 16 cu ft (450 L) of water for each tank, but only 12 square ft ( just over 1 square meter)of surface through which to gain radiative heat) How to replicate my success with capturing heat from the air in the past? Set up a couple of air-water heat exchangers - two car radiators, with fans to blow the warm air through the radiators. And drive them with a solar panel so that they operate only when the sun shines. Good idea in principle, but turns out to have several flaws. 1) the greenhouse rapidly heats up above the temperature of the water in the tanks even on heavily cloudy days when the PV panel will not generate anywhere near enough power to drive my pumps and fans. (Needs 2.3 Amps). So I am collecting heat only half the time. 2) Just as it is difficult to get heat into the water, the heat does not come out readily - the radiative heat from the fronts of the tanks is insufficient to maintain the temperature in the green house at night. The temperatures recently (with ambient temperatures ranging from -3 to -12 C) fluctuate between +28C to -1.6C. (Fortunately the things growing in the greenhouse will tolerate temps a little below freezing like this without harm). (The temperature in the water has gradually fallen over the past several weeks, buit is still around 12 C). So now I am thinking maybe I should just run my pumps and fans continuously. When the greenhouse heats up, it will capture the heat in the water, and at night it will add it back into the air. (But I am given pause by your comments about the inefficiency of conductive heat transfer, (which, of course is what I am doing with my radiators.) Any wise advice?

**J.P.M.** · 12-11-2016, 06:45 PM

Originally posted by David Maxwell

We are off on a bit of a tangent here - is it OK to be talking about heat transfer and storage in a PV forum? If nobody objects, (and I will happily go elsewhere if they do), I would be most grateful for your input.
This is an evolving process. The original heat storage consisted of racks of plastic 4 L water jugs with water dyed with black dye. These were stacked along the back (North) wall, and indeed were dependent on direct solar radiation for heat capture. The advantage was that, being small, they were not subject to stratification of the heat. But they were not very efficient at heat capture. I then added a 6 mil plastic sheet across the front plus a fan to blow the warm air from the greenhouse through the stack of jugs. This actually made a major difference in heat capture, basically adding conductive gain from the air to the direct radiation.) But the plastic jugs lasted only a year or so before breaking down and leaking. And getting at them to replace them was a pain. So this summer I replaced the whole array with two large fibreglass tanks. These are semi- opaque, so there is no benefit in dying the water, but the front (south) faces are painted with flat black paint. But the heat gain through the front faces is going to be pretty restricted. The tanks are 6 ft X 2 ft high, by 16 in. deep. (This works out to 16 cu ft (450 L) of water for each tank, but only 12 square ft ( just over 1 square meter)of surface through which to gain radiative heat) How to replicate my success with capturing heat from the air in the past? Set up a couple of air-water heat exchangers - two car radiators, with fans to blow the warm air through the radiators. And drive them with a solar panel so that they operate only when the sun shines. Good idea in principle, but turns out to have several flaws. 1) the greenhouse rapidly heats up above the temperature of the water in the tanks even on heavily cloudy days when the PV panel will not generate anywhere near enough power to drive my pumps and fans. (Needs 2.3 Amps). So I am collecting heat only half the time. 2) Just as it is difficult to get heat into the water, the heat does not come out readily - the radiative heat from the fronts of the tanks is insufficient to maintain the temperature in the green house at night. The temperatures recently (with ambient temperatures ranging from -3 to -12 C) fluctuate between +28C to -1.6C. (Fortunately the things growing in the greenhouse will tolerate temps a little below freezing like this without harm). (The temperature in the water has gradually fallen over the past several weeks, buit is still around 12 C). So now I am thinking maybe I should just run my pumps and fans continuously. When the greenhouse heats up, it will capture the heat in the water, and at night it will add it back into the air. (But I am given pause by your comments about the inefficiency of conductive heat transfer, (which, of course is what I am doing with my radiators.) Any wise advice?

This forum is not restricted to PV. It does not however, have a subforum for passive solar. I suppose everyone may be somewhat tolerant of such things for a post or two.

As for any advice I may have, wise or otherwise, I'd respectfully suggest that if what you have done so far produces results, and you are happy with them, by all means continue. It doesn't sound (read) as if what you're doing is dangerous. Passive solar is pretty forgiving in many respects. The worst that usually happens is things get hotter or colder than expected (or not expected), and/or a few bucks get wasted.

Without being at your location to view what's actually there, both in material, dimensional and practical aspects, I'm reluctant to offer any further comment. Also, and without trying to do your thinking, with what seems to be your present state of knowledge in such matters, getting into details about what your doing and any further comments I'd make would probably involve a lot of frustration for both of us., with little to show for it.

I would respectfully suggest that as part of your evolving process, you find good source material on passive solar design and also familiarize yourself with the basic concepts of heat transfer as a guide for your future endeavors. Such efforts may save you some time, effort and money, and also make the effort more enriching intellectually - at least I found it that way.

BTW, and at the risk of some folks here accusing me of condescension, know that what you are calling conductive heat transfer is probably better described as convective heat transfer.

Good luck.

**David Maxwell** · 12-11-2016, 07:54 PM

Thank you most sincerely. I am intrigued that my terminology, ("conductive") is incorrect. The good folk who talk about heat losses in respect to insulation and energy conservation use the term "conductive loss" to refer to losses of heat horizontally through, for example, the studs, ("thermal bridging") or the wall cavity. I understood convective heat transfer as involving the movement of a warmed liquid or gas upwards, due to its decreased density as it increased in temperature. - the way the contents of a saucepan set on a hot stove heat through. I thank you for your gentle correction.

**J.P.M.** · 12-12-2016, 02:27 AM

Originally posted by David Maxwell

Thank you most sincerely. I am intrigued that my terminology, ("conductive") is incorrect. The good folk who talk about heat losses in respect to insulation and energy conservation use the term "conductive loss" to refer to losses of heat horizontally through, for example, the studs, ("thermal bridging") or the wall cavity. I understood convective heat transfer as involving the movement of a warmed liquid or gas upwards, due to its decreased density as it increased in temperature. - the way the contents of a saucepan set on a hot stove heat through. I thank you for your gentle correction.

You're welcome. Conduction involves molecular excitation through solids. There is no movement of material. Convection involves the movement of fluids as a way to transport heat from one place to another.

There are 3 modes of heat transfer: Conduction, convection and radiation.

The use of the term "conductive loss" or such like is technically a misuse of the term, but is usually acceptable when used to talk about insulation and its effect on heat loss. For most other applications involving heat transfer, convection is the common term used.

Here's why the term "conductive heat loss" may have gained traction: Most insulation for low temp. applications involves the use of air, or sometimes other "inert" gasses such as argon which have low thermal conductivity, and attempts to trap those gasses, keeping them "still", and in so doing avoid or reduce the tendency for fluids to respond to a gravity driven bouyancy difference that produces a convective driven heat loss. Example: a stud space in a 2 X 6 wall filled with insulation has almost as much air in it as the same space with no insulation. However, the heat transmission from the hot side of the wall to the cold side is a lot less for the insulated space because the fiberglass batting suppresses the tendency of the air in the space to set up a convective cell within the space.

Thermal bridging is another such term. It's commonly used, but in point of fact, the heat transfer of a stud and the air in the space are actually linked. The "thermal bridging" term is usually associated with insulated stud wall systems because the heat transfer across a stud in a wall is much greater than the heat transfer across an insulated stud space.

Heat loss through solid material such as wood is conductive heat transfer. Through a stud wall it's a combination of parallel mechanisms: Conduction through the studs and convection across the air space, that space loss mostly via natural convection across the air space. In the end, even though insulation in the space involves convection suppression, it's possible to argue that, in an insulated air space, since there is little to no movement of the air that is trapped by insulating material, most, or essentially all of the comparatively small, remaining heat loss across the space may well be due to the small thermal conductivity of the "still" air as well as some small, lateral heat transfer by the stud material to the air space. In such cases the heat transfer across the stud may well be an order of magnitude greater than across an adjacent stud space. 6"-8" wall spaces and alternating 2 X 4 studs on inside/outside walls or some other tricks work well for thermal breaks.

Another example of term misuse that's become acceptable: Most internal combustion engine driven vehicles have a "radiator to remove heat from the engine coolant. However, there is little, actual thermal radiation that takes place.

Two fluids come into close contact but are separated by a wall of conductive material. The change in temperature of both fluids is accomplished via forced convection of the fluids past a system of separating walls and usually extended surfaces (fins). There is thermal conduction across the separating wall, but movement of the working fluids is what accomplishes the heat transfer within each fluid, the warmer fluid giving up it's heat to cooler fluid on the other side of the wall. Without movement, both fluids will eventually reach the same temp. and heat transfer will cease. Without getting into areas such as boundary layer theory and the mechanisms of molecular momentum transfer where thermal conduction takes place on a molecular level in each fluid, on a macro scale it's all about forced convection, and thermal conduction across the solid boundaries. Radiation has little to do with it. But the term has survived the test of time.

My bride still calls a refrigerator an ice box.

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