mflatford said:
I am very new to physics. I am searching for a way to boil water to create steam and use that steam to power a turbine and therefore create energy.
I surmize that water can be boiled if enough electric current is passed through it.
My question is how do I determine how much current to pass through a quantity of water to make it boil?
Again.. I am new to this... Be as plain as you can tolerate.
Thanks.
Here are the essential equations you need.
1 calorie which is approximately 4.18 joules (watt seconds) of energy will raise the temperature of one cc (or 1 gram) of water one degree Celsius.
It takes 2260 joules (watt seconds) to boil one gram=1cc of water.
The power output in terms of heat of a resistor is the current time the voltage.
P_{watts}=I_{amps}\times V_{volts}
The current through an voltage across a resistor satisfies the equation:
V = I\times R
where R is resistance in ohms.
Putting them together you get:
Power = P = I^2 R = V^2/R
In such a problem you mainly want to figure how fast you boil the water as the time it takes to get it up to the boiling point will be much less. Say you want to boil 1 gram per minute (a pretty good roiling boil for a test-tube of water. In fact at this rate I think it would boil over. But a cup of water would be simmering at this rate.) This means you need on the order of 2260/60 watts or almost 38 watts.
Let's try it a little slower at say 36 watts (since this is 6x6). You'll then need for a 6volt battery the resistor to be 1 ohm. You will then draw 6 amps of current. (You'd need an old VW bug 6volt car battery)
For a 24 volt supply you'll use a 2 ohm resistor and will draw 1.5 amps.
For your 120volt AC outlet you'd use a 400 ohm heating element and draw only 0.3 amps. (On the order of a 40 watt light bulb i.e. an Easy-Bake Oven)
Now the other bit of math you need is the resistance of water. This will depend on how far apart the electrodes that you use are and how much surface area. The resistivity of pure water is about 2500\Omega for 1cm^2 area electrodes separated by 1cm distance.
The same if you say double the area and quadruple the separation. But let's use almost this cm, square cm case and say you get 2400\Omega resistance. Then to generate 36 watts, you'll need a voltage of:
V = \sqrt{P\times R} = \sqrt{36\times 2400}\simeq 294 (volts)
That's pretty dangerous. Of course with some salts desolved into the water this goes down but remember cutting the resistance by a factor of 1/4 this value only halves the voltage. Most any scenario which will boil the water using the water as your resistive heating element will surely also give you a nasty shock.
It will help if you use say stacked electrode plates separated by much smaller distances, i.e. say six 1cm^2 electrodes (with thus 5 gaps) separated by a 1mm = 1/10 cm will give you a resistance of about 50\Omega the way I figure it and that's still giving you a required voltage of about 42 volts.
This would be practical except for one thing. You will get electrolysis at much lower voltages and most of your energy will be used up separating the water into Hydrogen and Oxygen instead of generating heat.
You really are better off using a resistive heating element instead of trying to pass the current through the water directly. Mainly for safety reasons but even if you're careful the method will be inefficient due to the business about electrolysis.
That never crossed my mind, and delivers a disturbing picture. I figured that he meant to drive some little machine like a bench grinder or something.