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Vapour Pressure Of Water At 500 Kpag And 5 Deg C
Started by SAprocessGD, Mar 25 2012 01:47 AM
vapour pressure water
3 replies to this topic
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#1
Posted 25 March 2012 - 01:47 AM
It's been a while since I've used vapour pressure and would like to get some clarification.
I'm dealing with emulsion entering a FWKO at 500 kPag and 5 deg C, with water, light HC and gas being separated. If I wanted to find the vapour pressure, in absolute kPa, of the produced water (at 500 kPag, 5 deg C) how is this calculated?
As all three phases are being separated and are in equilibrium, would the vapour pressure (kPaa) of the water be 500 kPag + atmospheric pressure?
OR,
Based on water vapour pressure tables at 5 degC, pressure of water is 0.872 kPa. Would the vapour pressure be 0.872 kPa + atmospheric pressure?
OR am I missing the point entirely.
All comments are welcome, and appreciate the help. Thanks!
I'm dealing with emulsion entering a FWKO at 500 kPag and 5 deg C, with water, light HC and gas being separated. If I wanted to find the vapour pressure, in absolute kPa, of the produced water (at 500 kPag, 5 deg C) how is this calculated?
As all three phases are being separated and are in equilibrium, would the vapour pressure (kPaa) of the water be 500 kPag + atmospheric pressure?
OR,
Based on water vapour pressure tables at 5 degC, pressure of water is 0.872 kPa. Would the vapour pressure be 0.872 kPa + atmospheric pressure?
OR am I missing the point entirely.
All comments are welcome, and appreciate the help. Thanks!
#2
Posted 25 March 2012 - 06:42 AM
I think water vapor pressure is 0.872 kPa a, determined by the temperature of 5 oC.
Total gas phase pressure of the emulsion (including water vapor's) is 500 kPa g = 601.33 kPa a.
Above under the assumption of ideal gas law and completely immiscible hydrocarbon / water system (additive vapor pressures), most probably representing the specific case.
Total gas phase pressure of the emulsion (including water vapor's) is 500 kPa g = 601.33 kPa a.
Above under the assumption of ideal gas law and completely immiscible hydrocarbon / water system (additive vapor pressures), most probably representing the specific case.
Edited by kkala, 25 March 2012 - 06:47 AM.
#3
Posted 25 March 2012 - 07:03 AM
SAProcessGD,
Going by the definition of vapor pressure
Vapor pressure of any mixture = Σxi*Pi
where:
xi = mole fraction of the ith component in the mixture
Pi = vapor pressure of the ith component @operating temperature in the mixture
Regards,
Ankur.
Going by the definition of vapor pressure
Vapor pressure of any mixture = Σxi*Pi
where:
xi = mole fraction of the ith component in the mixture
Pi = vapor pressure of the ith component @operating temperature in the mixture
Regards,
Ankur.
#4
Posted 25 March 2012 - 11:34 AM
I think above relation of vapor pressures is true for an ideal mixture of miscible liquids. But for a mixture of two immiscible liquids (in rather good dispersion, like an emulsion), its vapor pressure is the sum of vapor pressures of the two liquids, irrespectively of their proportion in the mixture. See http://www.separati.../DT_Chp01m.htm . Usual liquid hydrocarbons are practically immiscible with water (solubility in water and water in hydrocarbon is extremely low).Going by the definition of vapor pressure
Vapor pressure of any mixture = Σxi*Pi
where: xi = mole fraction of the ith component in the mixture
Pi = vapor pressure of the ith component @operating temperature in the mixture
Info on this matter can be also detected in http://www.cheresou...ntrifugal-pump .
Edited by kkala, 25 March 2012 - 11:44 AM.
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