dS is area, so it's differentiation over area dxH20 ds = dxN2 ds XH20 {KH20 =...
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dS is area, so it's differentiation over area dxH20 ds = dxN2 ds XH20 {KH20 = 2 (PHXH20 - PLуH20) + KN2 (PH*N2 - PLYN2)}] - KH2O (PHXH2O - PLYн20) q' [*N2(H20 (PHXH20-PLYH2O) + KN2 (PH*N2 - PLYN2)}] - KN2 (PH*N2 - PLYN2) q' == [H20 (PHXH20 - PLYн20) + da' ds δ KH20 is not constant and changes with XH20 KN2 δ (PHXN2 PLYN2) N2)}] da' can also be written as = JH20 +JN2 ds Kн20=C₁е₂PHXH20 YH20, YN2 are calculated over area using the equations below Унго JH20 JH20+JN2 and YN2 JN2 JH20+ JN2 where KH20 δ and JN2 = KN2 δ (PH*N2 - PLYN2) JH20= (PHXH20 PLYH2O) Also XH20 +XN2 = 1 The constants in this equations are: c1=0.00000000004: c2=0.9695; 6 =0.0023876; pb= 1; pl= 0.001; KN2 0.00000000000000104836; The initial conditions are: XH20 0.0231; XN2 =0.97692; q= 30 Solve until XH20 = 0.0001. Find area. Table of results for how XH20, XN2 change with area dS is area, so it's differentiation over area dxH20 ds = dxN2 ds XH20 {KH20 = 2 (PHXH20 - PLуH20) + KN2 (PH*N2 - PLYN2)}] - KH2O (PHXH2O - PLYн20) q' [*N2(H20 (PHXH20-PLYH2O) + KN2 (PH*N2 - PLYN2)}] - KN2 (PH*N2 - PLYN2) q' == [H20 (PHXH20 - PLYн20) + da' ds δ KH20 is not constant and changes with XH20 KN2 δ (PHXN2 PLYN2) N2)}] da' can also be written as = JH20 +JN2 ds Kн20=C₁е₂PHXH20 YH20, YN2 are calculated over area using the equations below Унго JH20 JH20+JN2 and YN2 JN2 JH20+ JN2 where KH20 δ and JN2 = KN2 δ (PH*N2 - PLYN2) JH20= (PHXH20 PLYH2O) Also XH20 +XN2 = 1 The constants in this equations are: c1=0.00000000004: c2=0.9695; 6 =0.0023876; pb= 1; pl= 0.001; KN2 0.00000000000000104836; The initial conditions are: XH20 0.0231; XN2 =0.97692; q= 30 Solve until XH20 = 0.0001. Find area. Table of results for how XH20, XN2 change with area
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