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1=CH4 2=H2 1 − εb ε 6.5 Case Study - Hydrogen PSA Table 6.2: Model equations of hydrogen PSA Component mass balance ∂2y 1∂y ∂T 1∂y ∂P −D 1−2 1 +2 1 + L∂x2 T∂x∂x P∂x∂x ∂y1 ∂y1 RT1−εb ∂q1 2 ∂qi ∂t +u∂x + P y2 =1−y1 Ergun equation ∂P 150μ (1 − εb)2 1.75ρg −∂x = 4R2 ε3 u+ 2R ε ρs ∂t −y1 ∂t =0 b i=1 (6.24) (6.25) (6.26) (6.27) (6.28) u 2 pbpb LDF equation ∂qi =ki(qi∗−qi) i=1,2 ∂t Energy balance (ερC +ρC)∂T+ρCεu∂T−K∂2T b g pg b ps ∂t g pg b ∂x L∂x2 2 ∂q4h −ρ ∆Hads i + w(T−T )=0 b i∂tDw i=1 P 2 y i M wi C = yCi ρ g = R T 2 i=1 pg ipg i=1 Ci ==ai+biT+ciT2+diT3 i=1,2 pg c c c c Langmuir isotherm ∗ aiyiP αT βT i=1,2 (6.29) (6.30) (6.31) qi = 1+2i=1biyiP ai =α1ie 2i Linear velocity profile bi =β1ie 2i u = uL(x/L) + u0(L − x)/L Cyclic steady state z(t0) = z(tcycle) z : yi, qi, T ∀i Chapter 6. Reduced-order Modeling for Optimization 114PDF Image | Design and Operation of Pressure Swing Adsorption Processes
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