Example 13-3 Heat Effects in a Semibatch Reactor |POLVER05_0 |1 d(Ca)/d(t) = ra-(v0*Ca)/V # d(Cb)/d(t) = rb+(v0*(Cb0-Cb)/V) # d(Cc)/d(t) = rc-(Cc*v0)/V # d(T)/d(t) = (Qr-Fb0*cp*(1+55)*(T-T0)+ra*V*dh)/NCp # d(Nw)/d(t) = v0*Cw0 # v0 = 0.004 # Cb0 = 1 # UA = 3000 # Ta = 290 # cp = 75240 # T0 = 300 # dh = -7.9076e7 # Cw0 = 55 # k = 0.39175*exp(5472.7*((1/273)-(1/T))) # Cd = Cc # Vi = 0.2 # Kc = 10^(3885.44/T) # cpa = 170700 # V = Vi+v0*t # Fb0 = Cb0*v0 # ra = -k*((Ca*Cb)-((Cc*Cd)/Kc)) # Na = V*Ca # Nb = V*Cb # Nc = V*Cc # rb = ra # rc = -ra # Nd = V*Cd # rate = -ra # NCp = cp*(Nb+Nc+Nd+Nw)+cpa*Na # Cpc = 18 # Ta1 = 285 # mc = 100 # Qr = mc*Cpc*(Ta1-T)*(1-exp(-UA/mc/Cpc)) # Ta2 = T-(T-Ta1)*exp(-UA/mc/Cpc) # t(0)=0 Ca(0)=5 Cb(0)=0 Cc(0)=0 T(0)=300 Nw(0)=6.14 t(f)=360