analysis of water injection into high-temperature mixture of ...

20∂f∂y43∂f∂y44∂f∂y45∂f∂y46= 2 + D + D= 0103 113= D −d D104 3 14= dD −d3 15 3Eq.3-65Eq.3-59 may be solved using by Gauss elimination. The second approximationy = y + δk+ 1 k ky = 3, 4,5, 6 . The process of forming the Jacobian,y is repeated until a stop criterion isis then{ } { } { }solving Eq.3-59 and calculating new values for { }met.The above results in the molar concentrations of the 11 product species andwater injection, but the two model for mixing rate prediction requires the rates ofchange of certain gas properties with respect to temperature, pressure. This requiresthe rates of change of species concentration with respect to these same variables.Differentiating the four non-linear equations given in Eq.3-37 through Eq.3-40 withrespect to temperature and pressure given as,∂f j∂f j ∂y ∂f 3 j ∂y∂f 4 j ∂y ∂f5 j ∂y6+ + + + ≈ 0, j = 1, 2,3, 4∂T ∂y ∂T ∂y ∂T ∂y ∂T ∂y ∂T3 4 5 6Eq.3-66∂f j∂f j ∂y ∂f 3 j ∂y∂f 4 j ∂y ∂f5 j ∂y6+ + + + ≈ 0, j = 1, 2,3, 4∂P ∂y ∂P ∂y ∂P ∂y ∂P ∂y ∂P3 4 5 6Eq.3-67Evaluating the equilibrium specific heat requires the change in mole fraction due tochange in temperature ∂yi/ ∂ T, i = 3,4,5,6 since the remaining derivative can beexpressed in term independent set.( cyy1/2)∂y∂∂ ∂∂T ∂T ∂T ∂T∂T1 1 5 4 1/2 c1y4 ∂y5= = y4 y5 + D14 + D151/2( cy y)∂y∂∂ ∂∂T ∂T ∂T ∂T∂T2 2 4 6 1/2 c2y4 ∂y6= = y4 y6 + D24 + D261/2( cy )∂y ∂7 4 6 1/2 ∂c4∂y = = y66+ D76∂T ∂T ∂T∂TEq.3-68Eq.3-69Eq.3-70