#1




How much energy does it take to boil water?
More specifically, how much energy does it take to boil 25 liters of water in a system to produce steam at 15 PSI?
I know it takes 1 calorie to raise 1 gram of water 1 degree so 25,000 calories to raise 25 liters of water 1 degree. If I assume 23C ambient temperature I need to raise the water another 77 degrees (Celsius) or 1,925,000 calories. Searching around I find 1 calorie = 0.003969 BTU So, you need 7,640 BTU = 2.24 Kilowatt hours of energy. However, some clever friend of mine said that you need more energy to boil water than you need to get it to boiling (they cited ~80 calories/gram to go from a liquid at 100C to a gas at 100C). Add to that I know raising the pressure raises the boiling point of water but I have no idea how much. Any ideas? (And no...this is not a homework question...a friend is looking for an alternate power source for a rotary steam engine and is trying to get a sense of how much energy is needed to make it run). 
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#2




Actually, this problem is considerably more complex than that. First of all, a calorie is the amount of heat required to raise the temperature of 1 gram of water at a starting temperature of 20˚C (At least that's the calorie we use in chemistry). So as the water gets warmer, the number of calories it takes to heat it changes. Next, you specify that you want your steam to be at 15 PSI. That means you nead to maintain an atmosphere at 15 PSI which means that the boiling point is higher than 100˚C. I don't know how much higher. Third, it is going to take a certain amount of energy to maintain that temperature while the steam is being released. That energy is going to depend on the rate at which the steam is being released. Of course the steam has to be released to prevent the pressure from getting higher than 15 PSI which will require a check valve.
We need a serious chemical engineer to solve this realistically. 
#3




You're refering to the latent heat of evaporation, te energy required to change the state from a liquid to a vapor. It doesn't change the temp. of the water/vapor. It's also present from a solid to a liquid.

#4




This is very simple. I have a steam table that will tell you how many BTUs are needed per lb of water for any Pressure. Naturally, one would have to interpolate between the increments of presure, and convert units (such as liters to pounds). I think it takes 220 BTUs/lb at 14.7 psi....don't have time to confirm...google on "steam tables". Will report back later, sorry!

#5




And since the pressure is specified to be 15 PSI, that figure changes (I didn't bother to confirm the 80 kcal/gram figure that was given). Without the rate of evaporation, you don't know the rate of energy required. If you have a known volume of liquid that needs to be evaporated at 15 PSI, the problem is simplified in that you can figure out the total energy required assuming you know all of the proper figures for the right boiling temperature and the right heat of vaporization.
ETA: Jinx may have something Last edited by Christopher; 11172007 at 06:00 PM.. 
#6




Here's what the steam table I use gives:
Pressure Temp Heat of liq Lat. heat Total Specific volume ft³/lb psi(a) psi(g) ºF BTU/lb BTU/lb BTU/lb Water Steam 29.7 15.0 250.34 218.33 945.6 1163.92 0.017004 13.88 So the total heat in the system would be 1163.9 BTU per pound. All you need is the starting temperature and your wife's sister is married to Robert. I copied and pasted this, and noticed on preview that the formatting makes it pretty complicated to read, but what the hell, I'm not charging for this. Edited to say: Jinx at 220 BTU/lb latent heat is pretty darn close to the actual 218.33 BTU/lb value, close enough for engineering, for sure. Last edited by Bill Door; 11172007 at 06:19 PM.. 
#7




BTW, the figure of 80 calories per gram is the heat required to change water from the solid state to a liquid one or the heat requires to melt ice at a constant temperature of 0°C. For the transition from liquid to gas, the number is 540 calories per gram. This relationship is what makes evaporation such an effective means of cooling. So good, that under the right conditions you can make ice just by evaporating enough of it away quickly enough.

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If you have 25 liters of water, and 25.01 liters volume in the whole system, then you don't have to think about water>steam at all. You have to find out the temperature at which the water's partial pressure equals 15 psi. Then figure out the energy to heat water to that temperature (using 1 calorie/gram). But here's the trick: 15 PSI is atmospheric pressure at sea level. Water boils when its partial pressure equals atmospheric pressure, so without looking anything up we know we need 100C for your 15PSI steam. And the answer is the one you already got, 2.24 kW. Good job. It's not so hard after all, as long as there's no space for steam in your system. Last edited by Alex_Dubinsky; 11172007 at 06:42 PM.. 
#10




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First, you have to think about the size of the chamber. Bigger the chamber, the more steam. But, continuing along this line, who says you need a big chamber? Length of the chamber is determined by the radius of the rotation of the wheel. A bigger radius gives you more torque. Then you consider the radius of the chamber. The bigger the radius, the more force on the piston, the more static force. And as you consider all the other details of your design, the headache only multiplies. In short, buddy, you asked the wrong question. The answer you're looking for is a book on steam engine design. 
#11




or, ok, fine, just find out the volume of your chamber, its pressure (including atmospheric pressure), and the frequency of the in and the out. You then figure out the mass of the water in its gaseous state that gets used every second, and that how much you've got to boil. The energy of water going 20>100 is actually irrelevant, since I'm assuming you want to know the energy to keep it running, not to start it up. Also, you're looking for power (energy/time) not energy itself.
And don't even think of asking "the least amount of input power to get it to move very slowly." That requires knowing all the heat losses, the steam losses, and the dirtiest bits of the engineering problem. 
#12




I looked at the following online steam tables but frankly they are beyond me to sort for a meaningful answer:
http://webbook.nist.gov/chemistry/fluid/ http://www.dofmaster.com/steam.html And thanks for the answers so far. Last edited by WhackaMole; 11172007 at 06:58 PM.. 
#13




I just fill the pot, put it on the stove and watch TV. No sweat.

#14




Hi, I'm the silly fellow that told the original poster that heat of vap for water was ~80 cal/g. Perhaps I was remembering in J/mol or something. Anyway, the important thing for this problem is that the contractee needs not just to boil his water, but he needs steam pressure above ambient. Well above ambient, if possible. Neither of us know exactly how to attack this problem. Especially if it needs the pebble, which I fear it might.

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Anyways, you know the volume of your tank, mass of the water in it, and the pressure you want. You need to either find some steam tables, a steam chart, or use an electronic calculator like this one. Calculate the specific volume of your system (volume of tank/mass of water), and decide what pressure you want. Plug those into the calculator. You want the specific Enthalpy of superheated steam. Multiply that number by the mass of the water, and there you have the energy stored in that system. 
#18




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Steam pressure builds up with temperature. How much pressure is given by a "vapor pressure" table, such as this one. As you can see, by 150C your pressure is 5 atm (70 PSI), so the water doesn't need to get very hot. The energy to get to that temperature is calculated (degrees) * 4.2 J/g/C("specific heat of water"). Or, 550 kJ/L. However, that's just the cost of starting the engine. The energy to get 1 liter of 5 atm steam from liquid (temperature doesn't matter) is figured out by first determining the mass of water in the steam. Water vapor density is 0.6 g/L at 1atm. At 5 atm, 3g/L. So to get 1 liter of your highpressure steam, you need 3 grams of water. 3g *2.2 kJ/g ("heat of vaporization of water") means 13 kJ per liter of 70 PSI steam. If your engine uses 2 liters of 70 PSI steam per second, you need 26 kJ/s or 26 kW of heat. 550 kJ is actually a small amount of energy (0.15 kWh). But 26 kW is a whole lot of juice. Anyway, I hope i've given you an outline of where to start. P.S. Of coures, all the constants change a bit depending on conditions, but with such a calculation that's the least of your worries. Last edited by Alex_Dubinsky; 11172007 at 09:49 PM.. 
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And, thank you for your insight. 
#20




ok, let me know how it goes. pics would be cool too.

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#23




No, I am saying that you are doing it wrong. The correct way to do this problem is to use steam tables. Saturated steam has the following properties at 5 bar (appx. 72.5 psi):
Specific volume: .3749 m^{3}/kg Enthalpy: 2748.7 kJ/kg If you want to know the energy for 1 liter you do the following: (1) Find out mass of 1 L 1 L *(.001 m^{3}/L)* (1 kg/.3749 m^{3})=.00267 kg (2) Calculate energy .00267 kg * (2748.7 kJ/kg)= 7.33 kJ That means it takes 7.33 kJ to get one liter of saturated steam at 5 bar (72.5 psi) from an appropriate amount of 0 C water at 1 atm. 
#24




er, I did make one mistake. "3g *2.2 kJ/g ("heat of vaporization of water")" should have multiplied to 6.6kJ not 13kJ. Aside from that, you used a more precise mass of 2.7g vs my 3g, and considered 0C water > 150C steam while i chose to calculate 150C water > 150C steam (hence 2.7kJ/g vs 2.2kJ/g). What I did wasn't wrong, and just using the steam tables isn't "right." Besides, they're impossible to use unless you already know what you're doing, and I tried to explain stepbystep what it is that one is trying to do.

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#27




Here's a link to online steam tables to find what you seek for any given input, in a large choice of units.
http://www.spiraxsarco.com/resources...atedwater.asp Last edited by Jinx; 11182007 at 07:16 AM.. 
#28




Conservation of energy and system losses
To heat water, produce steam and use the steam to generate electricity or do other forms of work you need to only apply a couple of simple theories.
A certain amount of energy is required to get your system up to pressure. This question has been approximately answered elsewhere but may have neglected heat losses from the boiler. As the temperature rises, so do the losses. As to how much energy in > out, I expect if you achieved better than 20% overall you would be doing well, so 10KW of heat in should get you around 23 kilowatts of energy out as electricity or other work. If you designed a closed loop steam system, you could expect less run up delay and slightly better efficiency. If you used a modern internal combustion engine, you would probably end up with slightly better fuel to electricity conversion, which is why we don't actually use steam engines in cars. 
#29




Only took 7 years.

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