[QUOTE=Alex_Dubinsky]
What’s a pebble? And contractee? Do we dopers get a comission too :D?
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.
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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.
[QUOTE=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.
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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).
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[sup]3[/sup]/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[sup]3[/sup]/L) (1 kg/.3749 m[sup]3[/sup])=.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.
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.
[QUOTE=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.
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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.
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.
