Using dark tiles to heat a pool via absorption of solar energy

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1. Apr 8, 2014

doughpat

Hi everyone--first post, thanks for any guidance if I've posted in the wrong place.

I've got a simple idea and I'm curious if it's worthwhile doing. How effective would putting a dark surface (perhaps black tiles, or black corrugated plastic) on the bottom of the pool, to warm the water by absorbing sunlight?

I'm not even remotely sure about how to go about doing the math to answer the question.

Details on the pool: Approximate dimensions: 6 meters long, 4 meters wide, average depth 1.5 meters. It receives about 3-4 hours of direct sunlight a day. I live very near the equator, with average air temperatures of about 30°C year-round. Currently the pool has white tiles.

Any thoughts on what kind of temperature I would expect? (even ballpark would be useful!)

2. Apr 8, 2014

Matterwave

Well, I can give you a ballpark estimate for the maximum (upper bound) possible difference in temperature of the water between a white surface pool and a black surface pool, but I probably can't give you even a ballpark estimate for the absolute temperature of the pool, that would require too many details of energy balance and energy transport.

Let's say that for 4 hours, the sun is roughly directly overhead the pool, so that the black surface at the bottom of the pool basically absorbs ~100% of the sunlight it receives (black asphalt for example would absorb about 90%, so we aren't making a big error here). Assume then that a white surface will reflect ~100% of the energy (a very white substance like snow would reflect ~90% so we aren't making a big error here either).

The flux of energy coming from the Sun is ~1400W/m^2, of which ~70% reaches the ground (the other 30% gets reflected by the atmosphere or clouds). Multiplying this by the surface area of your pool:

1400W/m^2*70%*(6m*4m+2*1.5m*4m+2*1.5m*6m)~53,000W of power that could be absorbed by your pool's surfaces. This is definitely an overestimate.

Now over 4 hours, that 53,000W corresponds to 220kWh of energy. If all of this energy was deposited into the water (assuming the pool is well insulated from the ground), then this energy would go into heating the water. How much heating is that? Well, the specific heat of water is ~4000J/kg/degree C, so we need to know how much water is in our pool. The density of water is 1000kg/m^3. Multiplying this by the volume of the pool, we have:

1000kg/m^3*(6m*4m*1.5m)~36000kg.

Now finally, we find that the temperature difference between a black pool and a white pool is:

$$\Delta T_{max}=\frac{220kWh}{4000\frac{J}{kg^\circ C}*36000kg}\approx 5.5^\circ C$$

So, roughly 5.5 degrees of difference at most, which is not bad if you ask me.

Last edited: Apr 8, 2014
3. Apr 8, 2014

doughpat

Fantastic! An approachable answer--and even better, my pool is going to warm up!

Now a more practical question....what to use on the bottom of the pool. I don't own it, so I need to think of a material that won't damage the pool (but *will* sink). I wonder how much more effective it would be to have the surfaces submerged vs. floating.

4. Apr 8, 2014

Bobbywhy

But WAIT! Why do you think all the sun's radiation, including the infrared wavelengths, will pass unattenuated through the water column to the bottom of the pool?

“Shortwave radiation emitted from the sun wavelengths in the visible spectrum of light that range from 360 nm (violet) to 750 nm (red). When the sun’s radiation reaches the sea-surface, the shortwave radiation is attenuated by the water, and the intensity of light decreases exponentially with water depth. The intensity of light at depth can be calculated using the Beer-Lambert Law.”
http://en.wikipedia.org/wiki/Attenuation

Also see these references:
http://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water
http://oceanworld.tamu.edu/resources/ocng_textbook/chapter06/Images/Fig6-17.htm

5. Apr 8, 2014

Staff: Mentor

It doesn't, but for typical swimming pool depths and remembering that swimming pools are generally maintained to keep the water as clear as possible (so minimizing attenuation) a significant fraction of the incoming light makes it to the bottom of the pool.

A black-bottomed pool will definitely end up being warmer than a light-bottomed one, although this effect is more often a nuisance (solar heating is most effective in the summer months, and at most latitudes you don't want your pool heated during the summer) than a desirable effect.

A digression, but interesting: An old backpacker's trick for melting snow into drinking water when hiking at altitude is to set out a pan of snow in the morning and put a black stone on top of the snow. Leave it in the sunlight all day and you have a pan of water that evening. Omit the stone, and you still have a pan of snow at the end of the day.

6. Apr 8, 2014

Matterwave

I was giving a very rough estimate for the upper bound of the effects this black pool could have. I would think that the largest unaccounted for effect would be due to cooling from the air and air currents around the water, but I'm not ready to go into that amount of detail...

7. Apr 8, 2014

doughpat

That may be true (even if the water is relatively clear), but isn't the attentuation, by definition, absorption of the radiant energy and, I would assume, transformation into thermal energy in the water?

In fact, maybe the really cheap n' easy solution is to just throw some fertilizer in there, shut off the pump and wait! A nice dark green soup would probably give a pretty high ΔT

8. Apr 9, 2014

Bobbywhy

Very true that the longest wavelengths of sunlight will be absorbed by the upper layers of the pool, and therefore, will heat that water near the surface somewhat.

Considering the attenuation graph in post #4 (oceanworld) and the stated depth of 1.5 meters I estimate only one thousandth of the incident infrared energy would reach the bottom so it appears that the calculation by matterwave in post # 2 is way too optimistic because very little, if any, infrared energy will ever reach some black absorbing material surface on the bottom of the pool.