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Unformatted text preview: More Gas Laws!
PTEC 3302 September 20th Homework Problem 26 (spreadsheet) uses the ClausiusClapeyron equation in linear form equation 25 in the text Problem 26 (spreadsheet) involves a densitytemperature diagram similar to figure 211and the property that the densities of the liquid and gas are identical at the critical point Gas Laws, Equations of State Dalton's Law of Partial Pressure:
The total pressure exerted by a mixture of gasses is equal to the sum of the pressures exerted by its components Law of Additive pressures Assume the following gas is at a pressure of 750 psia. What is the partial pressure exerted by the methane?
Component Methane Ethane Propane Mol fraction 0.85 0.10 0.05 Apparent Molecular Weight Multiply the molecular weight of each component by the mol fraction of that component in the mixture, then sum to get the apparent molecular weight of the mixture. To find specific gravity of a mixture, divide the apparent molecular weight by 29 (molecular weight of air) Zfactors and specific gravity Example 38 Example 39 (also find mass in lbs) When there is a heptanes+: Example 310 When the composition is unknown: Use specific gravity and Fig 31: Example 311 When H2S and CO2 are present: Example 312 Equation of van der Waals Considers the attractive force between molecules (which he defined as a/v2)
a ( p + 2 )(VM  b) = RT VM a,b are constants characteristic of the particular gas, Vm = v/n) is the molar volume, R is the universal gas constant Modifications and Improvements Clausius (1880) Berthelot (1889) Dieterici (1899) Wohl (1927) Lorentz (1881) a p+ 2 VM bVM VM  = RT VM + b In the spirit of van der Waals Beattie  Bridgeman EOS RT p= 2 VM c b 1  V  B0 1  3 M V T V M M A0 (1  a / VM )  2 VM "When two slowly moving molecules encounter one another, there is a tendency for them to move under the influence of each other for an appreciable length of time due to the intermolecular forces between them." B&B Current EOS SoaveRedlich and Kwong (SRK) Equation of State: aT temperature dependent term is the Pitzer acentric factor aT p+ (VM  b ) = RT VM (VM + b ) aT = ac RTc b = 0.08664 pc ac ( RTc ) 2 = 0.42747 = 1 + m(1  Tr ) ( pc ) 2 m = 0.48 + 1.574  0.176 2 PengRobinson EOS aT p+ (VM  b ) = RT VM (VM + b ) + b(VM  b ) aT = ac RTc b = 0.0778 pc ac ( RTc ) 2 = 0.45724 = 1 + m(1  Tr ) ( pc ) 2 m = 0.37464 +1.54226 0.269922 Mixing rules for SRK and PR ...and it gets worse! How do you deal with a j multiple component mixture? aT = yi y j aTij i j ij are binary interaction coefficients values aTij = 1  ij aTi aTj have to be given for each pair b = y jb j ( ) SRK example
yi
C1 0.65 VM=1.2 cu ft/lb mol Calculate pressure if T=709.6R
Tc
342.9 Pc
666.4 bj
0.4784 acj
8687 acentric factor
0.0104 mj
0.4964 j
0.612 aTj
5317 C2 0.25 549.5 706.5 0.7232 21042 0.0979 0.6324 0.8349 17569 nC4 0.10 765.2 550.6 1.2922 52358 0.1995 0.787 1.0591 55453 b=0.6210, (use .02 bic C1&C4, .01 C2 and C4, 0 C1 and C2) aT=10773 p=8223 psia ...
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This note was uploaded on 09/21/2008 for the course PTEC 502 taught by Professor Sue during the Spring '08 term at University of Texas Medical Branch at Galveston.
 Spring '08
 SUE

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