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Let’s talk about saturated vapor pressure

2017-04-14 View Original

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The last edit to this post was made by Benpeng Guo Jibo45 on 2017-5-24 at 00:38. “Talking about saturated vapor pressure”: The definition of saturated vapor pressure is as follows: within a closed space. The gas pressure at which the liquid and gas phases of a substance coexist at a given temperature. At this point, the processes of evaporation and condensation reach dynamic equilibrium. The saturated vapor pressures of the main components of liquefied petroleum gas at various temperatures are shown in the table below. Temperature (°C), Substance: Propane (MPa), Propylene (MPa), n-Butane (MPa), Isobutane (MPa)
0: 0.457, 0.564, 0.100, 0.150
50: 0.533, 0.662, 0.121, 0.179
100: 0.617, 0.750, 0.143, 0.211
150: 0.711, 0.857, 0.171, 0.247
200: 0.817, 0.973, 0.201, 0.288
250: 0.933, 1.110, 0.235, 0.335
300: 1.061, 1.260, 0.275, 0.387
350: 1.201, 1.420, 0.318, 0.433

For some values that cannot be found in tables, the Anthony equation must be used. Here is the version of the Anthony equation applicable to liquefied petroleum gas: lgh = A – B/(t + C). In this formula, h represents the vapor pressure of the substance, in millimeters of mercury; t is the temperature, in °C. A, B, and C are constants as shown in the table below. The material constants ABC are as follows: propane – 6.82973813.2248.0; propylene – 6.81960785247; n-butane – 6.83029945.9240; isobutane – 6.74808882.8240. The conversion formula between pressure P (in Mpa) and height h (in mm) is: P = ρgh, where ρ is the density of mercury, 13.6 g/cm³, and g is the acceleration due to gravity, 9.8 N/Kg. As an example, let’s calculate the temperature of pure isobutane at a pressure of 1.8 MPa. The calculation process is as follows: 1800 = ρgh = 13.6 × 9.8 × h; therefore, h = 13.505 meters = 13,505 millimeters. Then, Lg13505 = 6.74808. Additionally, 882.8 / (t + 240) = 4.1305. Hence, 882.8 / (t + 240) = 6.74808 – 4.1305 = 2.61758. From this, 2.61758t = 882.8 – 2.61758 × 240 = 254.5808. Thus, t = 97.26. Therefore, at a pressure of 1.8 MPa, the temperature of pure isobutane is 97.26°C.
Reply #2 2017-04-14
Hehe, today I’m presenting this advanced topic; I hope it will be useful to everyone.
Reply #3 2017-04-14
This dish is too hard; the original poster is really thoughtful! “A, B, and C are all constants as shown in the table below; what if this table is applied to other gases? Like steam? Where can I look it up?
Reply #4 2017-04-14
This is only half of Anthony’s formula; you can look up these constants on Baidu. Writing a post isn’t the same as writing a thesis; I don’t need to be that detailed. The purpose is to let everyone know about this approach and method.
Reply #5 2017-04-14
I see, but the approach you suggested is truly excellent! I thought these three constants were commonly used values!
Reply #6 2017-04-14
Hehe, thank you; I hope it’s useful to everyone
Reply #7 2017-04-14
This post was last edited by LQ198619 on 2017-4-14 at 22:42. After seeing the Antoine formula, I came to promote my semi-original table: http://bbs.hcbbs.com/thread-1751494-1-1.html (Link to the Antoine constants database). The Antoine constants database contains information on 4,958 organic compounds; it’s not clear who created this database. . . It can no longer be verified. . . Its main function is to calculate the saturated vapor pressure of a substance. I added my own search and calculation functions, so it’s half repost and half original~~~

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