Hello everyone,
i am doing a project
to build a solar water treatment (to provide drinkable water) and also a rechargeable battery which will be automatically switched off when it is fully charged. this battery will heat water at night. we have to build a working prototype and were asked to boil at least 5 litre of water. Our solar water treatment is rectangular shape and it is made of polystyrene. is it okay or is there any other good alternative? look, we just started the project and got some serious problems.
1) how do we calculate how much power we need to boil 5 litre of water?
2) what type of solar panel we should use (we are first time using solar panel and choose 200 Watt solar panel)?
3) is it good to use copper coil or copper plate or other material as hot plate:?:?
4) how we calculate the heat loss?
5) what type of battery should we use (Lithium-Ion battery:??
6) Most importantly are we ignoring any other obvious constraint?
N.B. the project's budget must not exceed $600.
any advice will be much appreciated.
At first blush, you're gonna need a big battery.
Here's how to calculate some of the power requirements. To boil water, you need to raise the volume of water to 100°C. By knowing the specific heat of the fluid (water), you can calculate how much energy is needed to raise it's temperature. The spec. heat of water is 4.186 Joule/gram*°C.
To raise 5L of water from room temp (20C) to 100C, you'd need:
5 L * 1 kg/L * 1000 g/kg * (100-20)°C * 4.186 J/g°C = 1,674 kJ
Once the water is at 100C, you need to add more energy to get it to boil. That requires knowing the heat of vaporization, 40.7 kJ/mol or 2261 kJ/kg.
To boil the now 100C water, you'll need to add an additional:
5 L * 1 kg/L * 2261 kJ/kg = 11,305 kJ
All told, to bring water from 20C to boiling will take 11305+1674 ~ 13,000 kJoules.
1 Joule = 1 Watt*second. Depending on how fast you want to boil off the water, you can calculate how many watts it will take. For a quick example, lets say you want to accomplish 5L boil-off in 60 minutes.
13,000 kJ = 13 MJ = 13 MW*seconds
13 MW*s / (60 mins * 60 sec/min) = 13 MW*s / 3600 s = 3611 Watts for 60 minutes.
If you relate that amount of power to a regular stove, running on 220Vac, then 3611W / 220V = 16 Amps. That would be reasonable for an electric stove to pull for current.
So, to calculate the capacity of your battery, you'll need to figure out the capacity in Amp-hours (Ah). So, let's assume you use a 12V battery pack.
3611 W / 12V = 300.9 Amps for 60 minutes, or ~300 A for 1 hour = ~300 Amp-hours @ 12V. To pull 300 amps from a battery is REALLY going to tax the cells, and you probably won't get the rated capacity from a few cells, since most batteries have less capacity, the faster the current drain is.
A standard car battery should get you around 70-100 Ah @ 12V. If you used 3 or 4 deep-cycle batteries in parallel, then you might be able to pull it off. This is still not counting any additional losses, such as inefficient conversion of battery current into heat, and the transfer of that heat into the water.
Next, you'll need to figure out how to recharge those batteries. If you use lead-acid batteries (cheap, easy to work with), then you'll need to take about 10 hours to give them a full charge. For quick math, that's four 100 Ah batteries. 400 Ah / 10 hours = 40 Amps (10A / battery). So you'd need an array that can put out somewhere around 40A @ 14.4V (2.40V/cell, 6 cells per 12V battery). So you'd need at least 14.4V * 40A = 576 W supply. Once again, not counting conversion inefficiencies and charge profiles for the batteries.
For the heating element, I'd go find a hot-coil from an old stove, or hot-plate... something like this:
This may or may not be the best option, as they are probably meant to run off of 110V/220V, so you'd either want a power inverter (BIIIG, and horribly inefficient to get 15+ amps), or you'll need to find another metal that can withstand high temps without melting. Maybe a loop of nichrome (hot-wire for cutting sytrofoam), or some alloy of tungsten (think lightbulb filament).
That's all I have for ideas at the moment. Hopefully this will get you moving in the right direction to calculate some of the rough values you'll need to pull this off. Again, none of those equations took losses or conversion efficiencies into account... which could be appreciable.