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  #1  
Old 11-17-2007, 05:23 PM
Whack-a-Mole Whack-a-Mole is offline
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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  
Old 11-17-2007, 05:40 PM
WarmNPrickly WarmNPrickly is offline
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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.
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  #3  
Old 11-17-2007, 05:44 PM
A.R. Cane A.R. Cane is offline
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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.
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  #4  
Old 11-17-2007, 05:50 PM
Jinx Jinx is offline
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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!
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  #5  
Old 11-17-2007, 05:58 PM
WarmNPrickly WarmNPrickly is offline
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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; 11-17-2007 at 06:00 PM..
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  #6  
Old 11-17-2007, 06:17 PM
Bill Door Bill Door is offline
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Here's what the steam table I use gives:

Pressure Temp Heat of liq Lat. heat Total Specific volume ft/lb
psi(a) psi(g) F 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; 11-17-2007 at 06:19 PM..
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  #7  
Old 11-17-2007, 06:21 PM
Rhubarb Rhubarb is offline
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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 0C. 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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  #8  
Old 11-17-2007, 06:36 PM
chorpler chorpler is online now
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Quote:
Originally Posted by Bill Door
All you need is the starting temperature and your wife's sister is married to Robert.
Bob's your brother-in-law?
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  #9  
Old 11-17-2007, 06:40 PM
Alex_Dubinsky Alex_Dubinsky is offline
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Quote:
Originally Posted by Whack-a-Mole
More specifically, how much energy does it take to boil 25 liters of water in a system to produce steam at 15 PSI?
Here's the main question: how big the system?

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; 11-17-2007 at 06:42 PM..
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  #10  
Old 11-17-2007, 06:50 PM
Alex_Dubinsky Alex_Dubinsky is offline
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Quote:
Originally Posted by Whack-a-Mole
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.
Ok, nevermind, THAT is a hard question. First, 15 PSI is not really enough. Second, you have this big, hairy consideration of power vs torque. E.g., if all you care about is static force, then it's simple, it's just pressure. If you care about power, which is the chamber expanding and contracting and all the other considerations of your engine being in motion, then it gets much more complicated.

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.
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  #11  
Old 11-17-2007, 06:58 PM
Alex_Dubinsky Alex_Dubinsky is offline
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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.
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  #12  
Old 11-17-2007, 06:58 PM
Whack-a-Mole Whack-a-Mole is offline
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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 Whack-a-Mole; 11-17-2007 at 06:58 PM..
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  #13  
Old 11-17-2007, 07:09 PM
Bryan Ekers Bryan Ekers is online now
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I just fill the pot, put it on the stove and watch TV. No sweat.
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  #14  
Old 11-17-2007, 07:36 PM
Aldizzy Aldizzy is offline
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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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  #15  
Old 11-17-2007, 08:20 PM
WarmNPrickly WarmNPrickly is offline
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Quote:
Originally Posted by Alex Dubinski
But here's the trick: 15 PSI is atmospheric pressure at sea level.
Aha! Of course I was assuming psig.

Quote:
Originally Posted by Aldizzy
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.
Fortunately, if there's one thing the straightdope does not have a shortage of, its engineers.
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  #16  
Old 11-17-2007, 08:57 PM
running coach running coach is online now
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Quote:
Originally Posted by chorpler
Bob's your brother-in-law?
Bob's your uncle.
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  #17  
Old 11-17-2007, 09:33 PM
treis treis is online now
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Quote:
Originally Posted by Aldizzy
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.
If you don't know how to do this you really need to refund this guys money. ME students spend most of a semester learning how to do these types of problems. There's a lot more that goes into something like this than a simple energy calculation like this.

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.
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  #18  
Old 11-17-2007, 09:46 PM
Alex_Dubinsky Alex_Dubinsky is offline
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Quote:
Originally Posted by Aldizzy
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.
What's a pebble? And contractee? Do we dopers get a comission too ?

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 high-pressure 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 kW-h). 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; 11-17-2007 at 09:49 PM..
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  #19  
Old 11-17-2007, 10:06 PM
Aldizzy Aldizzy is offline
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Quote:
Originally Posted by Alex_Dubinsky
What's a pebble? And contractee? Do we dopers get a comission too ?
"The Pebble" was the term the in-group used to refer to Calculus. The contractee is the anonymous and unnamed soul whose idle question prompted this arduous analysis.

And, thank you for your insight.
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  #20  
Old 11-17-2007, 10:10 PM
Alex_Dubinsky Alex_Dubinsky is offline
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ok, let me know how it goes. pics would be cool too.
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  #21  
Old 11-17-2007, 11:00 PM
treis treis is online now
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Quote:
Originally Posted by Alex_Dubinsky
What's a pebble? And contractee? Do we dopers get a comission too ?

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 high-pressure 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 kW-h). 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.
This analysis is incorrect. The heat vaporization of water is not a constant. It depends on temperature and pressure. The value you used is for water at 100 C and 1 atm. The Enthalpy of saturated steam at 5 bar is 2748.7 KJ/Kg. In other words, it takes 2748.7 KJ of energy to raise 1 KG of water from 0 C at 1 bar to saturated steam at 5 bar.
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  #22  
Old 11-17-2007, 11:58 PM
Alex_Dubinsky Alex_Dubinsky is offline
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Quote:
Originally Posted by treis
This analysis is incorrect. The heat vaporization of water is not a constant. It depends on temperature and pressure. The value you used is for water at 100 C and 1 atm. The Enthalpy of saturated steam at 5 bar is 2748.7 KJ/Kg. In other words, it takes 2748.7 KJ of energy to raise 1 KG of water from 0 C at 1 bar to saturated steam at 5 bar.
Your point is simply that I separated "warming up to 150" and "turning to steam." Enthalpy includes both those terms, but I was calculating start-up cost and marginal cost independently. Constants do shift, but that's a minor point (else they wouldn't be called constants).
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  #23  
Old 11-18-2007, 12:38 AM
treis treis is online now
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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 m3/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 m3/L)* (1 kg/.3749 m3)=.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.
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  #24  
Old 11-18-2007, 12:59 AM
Alex_Dubinsky Alex_Dubinsky is offline
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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 step-by-step what it is that one is trying to do.
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  #25  
Old 11-18-2007, 02:06 AM
Malacandra Malacandra is offline
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Quote:
Originally Posted by runner pat
Not if he's married to your wife's sister. Unless you're from Arkansas, in which case, who knows?
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  #26  
Old 11-18-2007, 02:58 AM
treis treis is online now
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Quote:
Originally Posted by Alex_Dubinsky
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 step-by-step what it is that one is trying to do.
I suppose. The change in the heat of vaporization with temperature and pressure is smaller than I thought (10% or so). Your method would give a rough approximation. Still, given how easy it is to do the more precise way there is little to no justification for calculating it the way you did.
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  #27  
Old 11-18-2007, 07:16 AM
Jinx Jinx is offline
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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...ated-water.asp

Last edited by Jinx; 11-18-2007 at 07:16 AM..
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  #28  
Old 01-24-2014, 08:57 PM
ozziemozzie@hotmail.co.uk ozziemozzie@hotmail.co.uk is offline
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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 2-3 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.
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Old 01-24-2014, 09:19 PM
randompattern randompattern is offline
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Only took 7 years.
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