# Figuring out power requirements of pool heater elements powered directly by solar panels (panel efficiency & sun,evaporation, insulation, ), etc.)

EDIT 1: firstly, thank you for the comments and the input

In this calculations the panels are already involved from the start as being the source (due to other connections to other systems when not heating the pool). But the most important factor is the rise of 17 Celsius in the range from 04 C to 21 C, taking into account very cold conditions - thus the comment that the sizeable amount of water would not cool down from 21 C to literally above freezing in a single night

EDIT 2:

The pool and temperature characteristics: 20ft x 10ft x5 foot deep water body (1000 cubic ft) with temperatures ranging from 21 Celsius in the last light of the day, so something like 25-26 Celsius during the peak of the heat cycle (sometime in the early afternoon) from whenceforward the heat is only maintained by a part of the panels with the other part providing power elsewhere; whereafter the temperature drops to the said 21 Celsius

I have a question regarding pool heating I was thinking about some time ago and if you feel like, share your thoughts.

For some 20ft x10ft and 5 foot deep pool in california in wintertime and outside, how much solar power converted directly into heat for the pool would you guess could heat the pool?

Heating it during the winter's day so that the water is still "mildly lukewarm" in the morning before re-heating (battery heating overnight would be a separate system and is ignored in the calculation).

I'm having an issue with calculating this because of including evaporation and heat loss during the night. But water is complicated since heat is also lost by evaporating since the change from liquid to gas requires latent heat. Anyway, at least for heating:

I was using the constant of 4.2 kW of energy for an change of 1K in 1L of water in 1 second. Meaning, a cubic metre (about 35 ft3) needs 1.1666 kW to heat up 1K/hour.

This is the tricky part: If there was no heat loss, a 1000 ft3 pool (28.3m3) would heat up 7°C in 7 hours and would need about 8.2kW for that. So, 20kW for a 16.8°C increase during 7 hours.

How much "rated" solar power do I need for the panels to have an average power (or, root mean sq power?) of at least 20kW, that is what I am calculating and into this formula should be included also the heat loss during the day.

Because, a rise of 17°C for this amount of water should surely be sufficient to be more than what the water would lose until the morning? At least this is what I assume at the moment. My guess is this outside pool in wintertime (but Californian winter) with a top protective cover/insulation shouldn't lose more than 10-15°C overnight.

In any case the point is how to incorporate evaporation and heat loss into this equation? What other necessary variables should be established? How much (roughly) is a minimum of 20kW average solar power power output to the heaters when taking into account sunlight variation and atmospherics, during an average "real" winter day (i.e. between 21 Nov and 21 Jan) in California?

The panels have an efficiency parameter, 19.78%. This is connected to some specified standard conditions for measuring the performance.

Standard Test Conditions (STC) of radiance of 1000 W/m2, spectrum AM 1.5 and cell temperature of 25°C with the normal operating solar-cell temperature is obtained under the Test Conditions: 800W/m2, 20°C ambient temperature, AM 1.5 Spectrum, 1m/s wind speed.

So, this efficiency is 20% of what? Of the energy in the sunlight, the total amount which could be possibly collected?

I am interested into how all of this actually and precisely is calculated, but primarily my requirement is to find a rough estimate - but, scientifically correct.

So, the zenith hour of the Californian wintertime sunshine could be converted into heat generated by electricity which is in turn converted with a factor of 0.2 from the sunlight on the panel in that hour. But how to incorporate the waxing and waning of sunshine during the day (cloud cover ignored for simplifying purposes!) into all of this? Is there any sort of data website from where it could be noted how much solar radiance changes in a day of a certain season (21 Nov - 21 Jan) for a certain spot on the planet (considering the different inclinations of the rays)?

Any thoughts much appreciated

Regards

• Panel efficiency? Just add a panel. Sep 8, 2021 at 5:36
• Why not consider solar thermal panels? Much more efficient. Sep 8, 2021 at 5:59
• There is too much information in this question, and still important bits are not there. Like, what do you want the set the temperature of your pool, or what is the temperature that you expect to have in the winter. in any case there are sizing calculator online that can help you with the heat losses like this at Engineering Toolbox.
– NMech
Sep 8, 2021 at 6:49
• There are black rubber pool heaters available which can just lay on the grass / edge of the pool. Area of panel needed is 50% to 100% of the surface area of the pool. Sep 8, 2021 at 8:55
• Hi, @SolarMike , thank you for suggesting the solar thermal heater elements (I am familiar and fond of these inventions - the rubber circuitry, or panels, or simply tanks; there are also some very good solutions made by a vlogger from russia on youtube. He uses plexiglass/metal sheet lenses to focus sunlight on the black rubber piping. It is very efficient and he orients the lenses north-south and they do not require movement; the focused horizontal ray just shifts to a different part of the heated black element (rubber or plastic). Very good. However in this situation the PV panels are used Sep 8, 2021 at 12:51

This is not an answer. I will just put a lot of comments about the question. You clearly have some physics background or at least understanding, so hopefully this following comments will be helpful to clarify and proceed.

## Heating up vs. heat losses

Heating up the pool and bringing it to a set temperature is the easy (and cheap) part. Maintaining the temperature is the difficult part, the evaporative losses are the most significant part.

In that respect its best to calculate the evaporative losses and calculate the average daily energy (not power) that you require. So measure things in kWh.

## power vs Energy

The following line drew my attention

I was using the constant of 4.2 kW of energy for an change of 1K in 1L of water in 1 second. Meaning, a cubic metre (about 35 ft3) needs 1.1666 kW to heat up 1K/hour.

IMHO, this approach will complicate maters. You calculate the average power required to heat one cubic meter in one hour. In that time the water will also have losses. And the thing is that you can't tell what those losses are because, you don't know if that water is going from 10$$^oC$$ to 11$$^oC$$ or 98$$^oC$$ to 99$$^oC$$. The losses in those two scenarios are totally different.

## rise of 17°C should be sufficient

Because, a rise of 17°C for this amount of water should surely be sufficient to be more than what the water would lose until the morning? At least this is what I assume at the moment. My guess is this outside pool in wintertime (but Californian winter) with a top protective cover/insulation shouldn't lose more than 10-15°C overnight.

If you are planning to heat up 17$$^oC$$ above the set temperature (e.g. 28$$^oC$$) then you'd have to bring the water to 45$$^oC$$. Then: a) it just won't be pleasant to use the pool at 45oC, and you will probably get that early in the afternoon, when ideally you'd like the pool to be cool.
b) the heat losses are proportional to the temperature difference. So it will very quickly loses the heat that it accumulated.

So the trick is to provide as much heat as the heat losses. Don't use the pool as a thermal battery or a boiler.

## Approach I would use

• find the heat requirements using a online calculator e.g.. Express that quantity ($$Q_{req}$$) in $$kWh$$.
• Go to PVGIS (or <!--globalsolaratlas) and find at your location and panel an estimate for the Specific photovoltaic power output (PVOUT) which is expressed in $$\frac{kWh}{ kWp}$$.

If you divide:

$$\frac{Q_{req}}{PVOUT}$$

you will get a rough estimation of the required photovoltaic installation kW.