Question

The production of methane gas, CH4, from a shale formation is being considered. Reservoir conditions have been estimated to be at T=212 °F and p=1000 psia. Using both Excel and Matlab to estimate the specific volume of methane gas at the given temperature and pressure using the Peng-Robinson equation of state. Use an ideal gas constant of R = 0.08206 [L*atm/gmmoles*K]. For methane, the following parameters are given: Tc = 190.6 K, pc = 34.99 bar, and ω = 0.012

Answer 1

P = RT/V_m - b - a alpha/V_m^2 + 2b V_m - b62 a = 0.45724 R^2 T_c^2/p_c b = 0.07780 RT_c/p_c alpha = (1 + k(1 - T_r^0.5))^2 k = 0.37464 + 1.54226 w - 0.26992 w^2 T_r = T/T_c Therefore, a = .45*.082^2*190.6^2/34.9 = 3.14 b =.077*.082*190.6/34.9 = 0.034 k = 0.37 + 1.54*.012-.26*.012^2 = 0.38 T = 212 F = 372K Tr = 372/190.6 = 1.95 alpha = (1+.38*(1-1.95^.5))^2 = 0.72 P = 1000 psia = 68.9 bar Putting in the equation, Therefore, 68.9 =.082*373/(V-.034) -.72*3.14/(V^2 + 2*.034*V -.034^2) Therefore,�Specific Volume = 0.41 Litre/mol

Therefore, a = .45*.082^2*190.6^2/34.9 = 3.14

b=.077*.082*190.6/34.9 = 0.034

k=0.37+1.54*.012-.26*.012^2 = 0.38

T=212 F = 372K

Tr = 372/190.6 = 1.95

α = (1+.38*(1-1.95^.5))^2 = 0.72

P=1000 psia = 68.9 bar

Putting in the equation,

Therefore,

68.9=.082*373/(V-.034)-.72*3.14/(V^2+2*.034*V-.034^2)

Therefore, Specific Volume = 0.41 LItre/mol