Need a maths geek to answer a question about boiling water.

Hi all. I'm finally starting the prototype of my solar condenser this week, though this question doesn't relate to it directly, but to using it to power a steam distiller to purify water. How much water will ten kilowatt hours boil? ie, if I have one kilowatt of solar energy coming in over ten hours and 100% of it goes into heating the water. Also, for extra credit, assume the water is in a black plastic container also in the sun, so the water is preheated to whatever temperature that would be likely to heat it to. cheers!

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lemonie8 years ago
To use kelseymh's values, I calculate that you could vapourise (at boiling-point) ~16Kg of water. Or you could heat more than 100Kg to 100o.

Combining the two ~14Kg of water. (~30 pounds, 3.7 US Gal)

SolarFlower_org (author)  lemonie8 years ago
That's a little less than I was expecting but still a pretty good amount. What if the water was about 55 C, ie a hot sunny day in black plastic?
Like kelseymh said you ought to use this as a learning opportunity and work it out. Which is why I didn't post the equation.
But having the water pre-heated isn't going to make that much difference, as it's only 16Kg at 100oC

SolarFlower_org (author)  lemonie8 years ago
Learning? NNNNN()()()()()()()OOOOOO0000000ooooooo............ I know, I know. My father keeps telling me the same thing whenever I email him trying to avoid all the finer technical details. To be fair tho, I have been doing epic amount of research on all this, it's just when it comes to number crunching, it's so much easier to just ask you guys. you're, like, smart, and stuff. (and god, I used to be so good at maths in high school...) Anyway, thanks heaps for the answer.
If you want a solution to your question, not just an arbitrary number:

Energy required to boil water EV= Hv * m (specific heat of vaporisation * mass)
Energy required to heat water EH= SHC * dt * m (specific heat capacity of water * change in temperature * mass)

Total energy ET = EV + EH = (Hv + (SHC * dt)) * m

If you want energy to be a known and mass to be an unknown, rearrange for "m="

m = ET / (Hv + (SHC * dt))

The ever-useful Wolfram Alpha tells me Hv is 2.23*106 J/kg, and SHC is 4180 J/kg K. This means

m = ET / (2230000 + (4180 * (100 - starting temperature)))

For ten kilowatt hours (3600 * 1000 * 10 joules) this gives

m = 36000000 / (2230000 + (4180 * (100 - starting temperature)))

Calculating the value of m for starting temperatures between 0oC and 100oC shows the starting temperature doesn't make much difference, most of the energy goes into boiling the water. At 20oC you could boil 13.8L, at 80oC you could boil 15.6L.

Yay algebra :)
SolarFlower_org (author)  PKM8 years ago
Yay algebra. Ok, I'd hate to have to memorize all that, but it seems pretty straight forward. Next question: are there any ways other than lowering the pressure that makes water easier to boil? It seems like using solar to Pasteurise water would be very effective, but not so much to completely purify it through steam distillation. 14 L is enough for I guess two families of four or five if it's for drinking water only, which isn't bad at all, but if that could be increased then all the better. I'm thiniking of water solutions for the developing world. If they've got infected ground water then Pasteurisation should be enough, but if there's any chemical contaminants, or if they want to recycle grey, black, or salt water, then it's going to have to be steam. Also here in the west, especially in countries like Australia, recycling grey into potable water would be incredibly helpful. Unfortunately we use about 150 - 500 L of water per person per day (God only knows how), so having to use ten square meters of sunlight just to boil water seems a little excessive.
You don't have to memorise it, I derived it all from the definitions of specific heat capacity and specific heat of vaporisation. It's really just converting units- you know joules and joules per kilogram and want to work out kilograms, so multiply and divide the appropriate quantities together. Dimensional analysis is one of those things I always thought should be taught more, it's a very useful skill and prevents misunderstandings like treating amp-hours as amps and so on. (Sorry, pet peeve)

Here in the west we use maybe 6-10 litres of water flushing the toilet half a dozen times a day and 50 litres having a shower. Don't get me started on washing the car with a hose and lawn sprinklers...

I suspect that while you are thinking of concentrated solar for boiling water, you don't need to go that high tech- distilling wated with plain solar energy is a well established concept, which could always do with better engineering to make sturdier, cheaper and higher-yield designs but doesn't need "overengineering".
SolarFlower_org (author)  PKM8 years ago
Low-fi engineering is definitely where I'm aiming. My device is a bit more complicated than a hole in the ground, but not vastly more. And if it can Pasteurise over 100L per square meter, it'd be a lot more effective. Cheers guys, very helpful information.
Lowering pressure would require a vacuum-pump. You could ask pharmaceutical research-labs to donate old rotary-evaporators, but you'd need electricity to run them. Don't forget reverse osmosis and other membrane-filters L
kelseymh8 years ago
This isn't a math question, it's physics. You need to know the thermodynamic properties of water:
  • heat capacity of water (74.539 J/mol/K)
  • heat of vaporization of water (40.65 kJ/mol)
The total energy to boil water is just the sum of the energy to raise water from an initial temperature to 100 C, plus the heat of vaporization.

Conversion factors:
  • liters/mol (18 mL/mol)
  • watts = joules/s, so 1 kWh = 3.6 MJ
Initial conditions:
  • temperature of water (25 C ?)
  • quantity of water
Now you can do the arithmetic yourself. Go buy a calculator.
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