If I have pipe that is filled with water at 400 F and it is allowed to flash into a room of a certain area. The dimensions and the pressure in the pipe are known. How do I find the new pressure in the room caused by the flashing of some amount of the water as steam into the room?
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steam flashing
- Thread starter justout02
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justout - absent from your givens are flow rate and/or orifice diameter. Based on what you've posted, assuming no flow and a closed pipe and closed room, the pipe and room would reach a saturated steam mixture at a given pressure depending on the combined volume, with subcooling and condensation at a rate dependent on the net insulating value of the combined system. The pressures between the pipe and room would equalize (Pascal's Law). A key component to solving such a problem is determining and posting the right questions. The hardest part of engineering is determining and communicating the right questions and making the right assumptions.
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The main unknown is the heat loss to the room air as the water flashes. The flashing into steam and the heat exchange with room air would play a big role I think in the final outcome. If we can keep things simple by saying no heat loss during flashing and a room temperature of 400°F prior, I would
1. Figure the mass of water in the pipe. It's about 1,599 lbm based on saturated water at 400°F.
2. Figure what the mass would occupy under saturated vapor conditions in the combined volume (room + pipe), = 3,054 ft3.
3. Specific volume becomes 1.91 ft3/lbm (room volume divided by total water mass).
This is the saturation point for about 398°F. Assuming complete saturated vapor, pressure would be about 240 psi. Truth is, it would be a lot less because some of the steam would likely condense when it hits the air...
1. Figure the mass of water in the pipe. It's about 1,599 lbm based on saturated water at 400°F.
2. Figure what the mass would occupy under saturated vapor conditions in the combined volume (room + pipe), = 3,054 ft3.
3. Specific volume becomes 1.91 ft3/lbm (room volume divided by total water mass).
This is the saturation point for about 398°F. Assuming complete saturated vapor, pressure would be about 240 psi. Truth is, it would be a lot less because some of the steam would likely condense when it hits the air...
Is the "pipe" actually a fixed volume pressure vessel? Because water is so highly incompressible, very little water must expand from the vessel in order to fully vent the pipe. Does the rest of the water remain, unmixed, in the pipe/vessel?
ChasBean has done an accurate volumetric analysis based on complete mixing of the initial water mass, but has neglected conservation of energy.
Let us assume that all of the liquid leaves the pipe. Let us also assume that the final equilibrium pressure is 14.7 psia, but that the (presumed) air in the room does figure into the final state in any other way (first approximation).
(final enthalpy) = (orig. enthalpy) = 390 BTU/lbm
pressure = 14.7 psia
h(f) = 180.15 + X(970.4) [ X is thermodyn. quality ]
X ~ 21.6%
I am still bothered by your assumed "flashing" process. If there is a rigid fixed volume vessel, the above analysis is invalid. You will have approximately 5% of the liquid leaving the pipe before the pressure is completely vented (v = 0.01773 at 10000 psi, 400F) compare to (v = 0.01863 at 400F and low pressures; liquid spec.vol. is more sensitive to temperature than pressure). It is necessary to calculate another heat balance using 80 lbm of water, instead of 1600 lbm.
ChasBean has done an accurate volumetric analysis based on complete mixing of the initial water mass, but has neglected conservation of energy.
Let us assume that all of the liquid leaves the pipe. Let us also assume that the final equilibrium pressure is 14.7 psia, but that the (presumed) air in the room does figure into the final state in any other way (first approximation).
(final enthalpy) = (orig. enthalpy) = 390 BTU/lbm
pressure = 14.7 psia
h(f) = 180.15 + X(970.4) [ X is thermodyn. quality ]
X ~ 21.6%
I am still bothered by your assumed "flashing" process. If there is a rigid fixed volume vessel, the above analysis is invalid. You will have approximately 5% of the liquid leaving the pipe before the pressure is completely vented (v = 0.01773 at 10000 psi, 400F) compare to (v = 0.01863 at 400F and low pressures; liquid spec.vol. is more sensitive to temperature than pressure). It is necessary to calculate another heat balance using 80 lbm of water, instead of 1600 lbm.
poetix, I would think that if this pipe were suddenly opened to an unlimited volme (not a room) at atmospheric pressure that very nearly 100% of the water would flash because it's at 400°F, where the flash point is 212°F. My assumption is based on that - all the liquid flashing, even though in reality the pressure in the room would rise and the mixture's temperature would drop concurrently. Also in reality there would be energy transfer from the phase change.
If I were designing this room for a pressure rating, however, I would use my assumption and rate the room for 240 psig, plus about 15%.
You bring up a good point based on enthalpy - we can assume energy (initial) = energy (final) and cross-reference enthalpy on the tables. That I didn't do. I'm at home now and don't have the tables with me but will look into it... Thanks for your feedback. -CB
If I were designing this room for a pressure rating, however, I would use my assumption and rate the room for 240 psig, plus about 15%.
You bring up a good point based on enthalpy - we can assume energy (initial) = energy (final) and cross-reference enthalpy on the tables. That I didn't do. I'm at home now and don't have the tables with me but will look into it... Thanks for your feedback. -CB
ChasBean1:
poetix99 is correct that ~21.5% of the water in the pipe would flash (even if it opened to an unlimited volume). Even at high temperature and pressure, liquid has low enthalpy. At 400ºF and 10,000 psi, the enthalpy is ~388.9 BTU/lbm. Saturated liquid at atmospheric pressure has an enthalpy of ~180.1 BTU/lbm and Sat. Steam has an enthalpy of ~1150 BTU/lbm. Using the mixing rules (definition of quality):
388.9 = (1-x)(180.1) + (x)(1150)
solving for x gives us ~21.5%
The only way that all of the liquid in the pipe could flash into steam is if the saturation enthalpy of steam were equal to (or less than) 388.9 BTU/lbm.
Best Regards,
jproj
poetix99 is correct that ~21.5% of the water in the pipe would flash (even if it opened to an unlimited volume). Even at high temperature and pressure, liquid has low enthalpy. At 400ºF and 10,000 psi, the enthalpy is ~388.9 BTU/lbm. Saturated liquid at atmospheric pressure has an enthalpy of ~180.1 BTU/lbm and Sat. Steam has an enthalpy of ~1150 BTU/lbm. Using the mixing rules (definition of quality):
388.9 = (1-x)(180.1) + (x)(1150)
solving for x gives us ~21.5%
The only way that all of the liquid in the pipe could flash into steam is if the saturation enthalpy of steam were equal to (or less than) 388.9 BTU/lbm.
Best Regards,
jproj
...and, of course, properly accounting for the air in the room is going to introduce some more complexity in the energy balance calculations. The energy balance must consider the initial mass of air with its corresponding temperature rise (use constant pressure specific heat of 0.24 BTU/lbm-F).
Get your psychrometric charts out.
The energy balance depends a great deal upon the actual amount of water leaving this HP vessel. If, as I claim above, only about 80 lbm of liquid leaves before it is completely vented from 10000 psi down to atmospheric, most of the energy stays in the vessel!
I think that we're left with a foggy, humid room; maybe with 1.5 - 1.8 inches of (hot) standing water.
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- #9
Thanks to all for the advice and assitance. I just have to figure the pressure exerted on the walls of the room after 21.5% of the water flashes into the room. I know I set up an energy balance. Will the temperature of the steam in the room be 212F or something else. Any ideas?
The final equilibrium condition of the room must consider the air in the room. I was not joking when I mentioned psychrometric charts in an earlier post. The water discharge is not a lot of energy added to the room. The final conditions are calculable (with certain assumptions about heat transfer at the walls of the room). The final pressure is likely to be close the the initial pressure.
However, the PARTIAL pressure of the water vapor is certainly less than 14.7 psia and the final temperature is therefore not 212F. So, the earlier calculation of 21.5% vapor is also in error; it's a first guess.
In fact, I believe the solution must be iterated: To a very, very good approximation, the amount of liquid leaving the tank will not change as a function of whether the final room pressure is 10, 15 or 25 psia. This gives the energy flux into the room and leaves only the room conditions to determine.
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- #12
The value for quality solved above was obtained by using the enthalpy values. I was wandering why the quality obtained by the entropy values at the given temperature and pressure is different? Plus, which is the best approximation for the flashing situation?
The throttling process, which is the same as your flashing/splashing process, is isenthalpic, not isentropic.
You do not (cannot) know the final entropy of the flashed steam; you must use an energy balance (with the proper mixture of air and water vapor), and then calculate the entropy only after the final state is otherwise determined.
BTW, it still sounds to me as if you are calculating the flashed water in a vacuum - literally. The air/steam mixing must be accounted for.
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