I use a solar still that holds 182 cubic feet of water. Here is how it works:

There is an insulator on both sides of the metal bowl. On the outside it covers the whole thing. on the inside it covers the sides but not the bottom. This collects the rain and also heats up the rainwater during sunny days so that the evaporation rate is faster. I have the pipe nearest the metal bowl the largest it can be while the metal bowl still collects rain and gets heated up by the sun. The water and any gases pass through there and into the condenser. I have this cold enough to condense water into a liquid but not so cold that the water turns solid and the $\ce{CO2}$, $\ce{O2}$, $\ce{N2}$ etc. turn liquid. Last but not least there is a spigot at the very end and a bucket below it. This separates atmospheric gases from liquid water with $\ce{CO2}$ being the main concern.

There is also dust and pathogens from the air in rainwater.

The dust most likely will stay behind and just make the bowl dusty. The pathogens get killed by the heat that there still is when all the water has been evaporated.

I would say the maximum loss with this system is from 1 pint or so to 1 gallon or maybe more depending on the water level in the bowl.

However I think there is a way to calculate maximum and minimum loss(neither of these is 0).

How could I calculate maximum and minimum loss based on the depth of the water(because obviously at different depths there is a different amount of loss) and the size of everything(which by the way the pipe slowly tapers off to a certain size and so you have to take pressure into account as well as volume)?

Also how could I calculate the pressure in the pipe based on the fact that it tapers off and starts at the maximum size possible with the bowl still collecting rain and being heated by the sun?

Solar Still

  • 2
    $\begingroup$ I think we need more information. Does the water boil in the collector, or just evaporate? What are the possible loss points for water (is the container open, etc.?) A schematic would be very helpful. $\endgroup$
    – thomij
    Jul 1, 2014 at 21:30
  • $\begingroup$ well I think the majority just evaporates but some of the water might boil after a while. The possible loss points for water are the bowl and the pipes that come before the condenser $\endgroup$
    – Caters
    Jul 2, 2014 at 1:15
  • $\begingroup$ I really do not understand how the water gets into the bowl - is it open towards the sky? I might not understand your picture, but if your still is not closed over the evaporation container, then water just diffuses into thin air. A photograph would certainly help much more. $\endgroup$ Jul 2, 2014 at 2:50
  • $\begingroup$ The water gets into the bowl by rainfall and gets heated up by the sun. Also my pipe starts off as large as it can be without blocking sunlight and I am afraid that if the bowl is closed that 3 things would happen: 1) There would be an insulator up top making the water even hotter because of raised conduction and 2) the sunlight would be blocked by the insulator and lastly 3) Rain will not be able to fall into the bowl. $\endgroup$
    – Caters
    Jul 2, 2014 at 2:51
  • $\begingroup$ I still do not understand the setup - Is the rainwater collected from a roof and then passed into the bowl? I also think this question would fit much better in Physics. $\endgroup$ Jul 2, 2014 at 3:06

1 Answer 1


There is no way to answer your question accurately for this setup. It will depend on the temperature of the water in the bowl, the exposed surface area, the convective flow rate of air near the surface, and the relative humidity of the air - none of which are constant or even very predictable.

Your best bet is to set it up and measure the amount before and after distillation. The difference would be your loss.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.