
#1
Jun1607, 10:46 PM

P: 1

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. 



#2
Jun1607, 10:52 PM

PF Gold
P: 8,961

Welcome to PF, Mflatford.
The approach to this is not to pass the current through the water itself, but through a resistance heater immersed in the water. 



#3
Jun1607, 11:22 PM

Sci Advisor
HW Helper
P: 1,572

Pure water is a good insulator.
It would be difficult to pass enough current to heat it to boiling. However, if you add some salt to the water, you could do it. Use the specific heat of water to find how much power is required. This calculation would be the same as for the immersion heater that Danger mentions. 



#4
Jun1607, 11:40 PM

Mentor
P: 21,999

Boiling water using electric current
The specific heat and heat of vaporization of water is readily available with a google...




#5
Jun1607, 11:56 PM

P: 2,050

I'm with Danger on this. Use a kettle (electrolysis isn't desired), and be aware that the turbine will "create" less energy than the boiler uses.




#6
Jun1707, 06:52 AM

Admin
P: 21,628

Power plants using fossil fuel, e.g. coal and oil, and nuclear plants use the thermal energy from the combustion of the fuel (stored chemical energy) or fission (stored nuclear energy) to produce steam to drive a turbine, which drives a generator, which produces electricity. Gas turbines (which could use light petroleum products like jet fuel/kerosine) drive generators directly, but the efficiency of generation can be increased by using the waste heat (from the exhaust) of a gas turbine to heat a steam cycle. Back to the original question, the rate of electricity consumption will depend upon the rate at which liquid water in converted to steam, E = [itex]\dot{m}(c_p\,\Delta{T}\,+\,h_f\,+\,\Delta{h}_s)[/itex], where [itex]\dot{m}[/itex] is the mass flow rate of water, [itex]c_p[/itex] mean specific heat over the temperature range of liquid water, [itex]\Delta{T}[/itex] is the change in temperature of the water, [itex]h_f[/itex] is the specific heat of vaporization and [itex]\Delta{h}_s[/itex] change in specific enthalpy of steam. For a steam turbine, it is best to use superheated steam, since moisture leads to higher rates of erosion. One should be familiar with saturated steam. Rather than use electricity to generate steam to drive a turbine to produce energy, it would seem best to simply use the electricity to power whatever device (motor or appliance) one wishes to use. First law of thermodynamics  http://hyperphysics.phyastr.gsu.edu...o/firlaw.html# Specific heat http://hyperphysics.phyastr.gsu.edu...ermo/spht.html Rankine thermodynamic cycle http://en.wikipedia.org/wiki/Rankine_cycle http://www.qrg.northwestern.edu/ther...gen/regen.html 



#7
Jun2007, 10:44 PM

P: 230

I hope you will not use the turbine for producing electric power




#8
Jun2007, 11:22 PM

PF Gold
P: 8,961





#9
Jun2107, 01:30 AM

Sci Advisor
PF Gold
P: 1,767

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. [tex] P_{watts}=I_{amps}\times V_{volts}[/tex] The current through an voltage across a resistor satisfies the equation: [tex] V = I\times R[/tex] where R is resistance in ohms. Putting them together you get: Power = [itex] P = I^2 R = V^2/R[/itex] 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 testtube 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 EasyBake 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 [itex]2500\Omega[/itex] for [itex]1cm^2[/itex] area electrodes separated by [itex]1cm[/itex] 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 [itex]2400\Omega[/itex] resistance. Then to generate 36 watts, you'll need a voltage of: [tex] V = \sqrt{P\times R} = \sqrt{36\times 2400}\simeq 294 (volts)[/tex] 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 [itex]50\Omega[/itex] 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. 



#10
Sep2608, 02:06 AM

P: 1

i have had a simmaler idea to what the first guy was asking but i dont know if i were to be running the electricity directly through the water and i had almost solid voltage with very little in the way of amps (for instance im using static electricity to pump through the water) would it still boil or do you have to have the current and amps in a clean electricity supply to boil the
water the reason i would run it through the water directly is i there is not enough of a current to run a burner off of with static electricity just wondering im just a highschool kid that hasnt taken physics since sophmore year thanks 



#11
Sep2708, 05:34 PM

Mentor
P: 11,985

Welcome to PF, kirkwillrule.
No, I'm afraid static electricity will not produce any appreciable power. 



#12
Sep2808, 12:09 AM

Mentor
P: 21,999

As I said (a year ago!), the heat of vaporization of water is something you can find with a quick google. It takes a certain number of joules or watthours to boil a certain mass of water. Getting that power/energy requires a certain voltage and amperage. These things are very easy to play with in a theoretical sense, to get the performance you want(please do!).
On the pratical side, since water's resistance varies greatly with what is dissolved in it, the best way to do this would be with an immersed resistance heater, as said above. Also, static electricity: static means not moving. If it isn't moving, there is no amperage. No amperage means no power. 


Register to reply 
Related Discussions  
Boiling Water ?  General Physics  8  
Boiling water by high frequency AC current  General Physics  5  
A pot of boiling water?  General Physics  3  
Boiling points of water  Chemistry  3  
Boiling water & salt  General Physics  41 