Question

2. (30 pts) CSTR Reactor Multiple Reactions: Consider the following set of series reactions which occurs in a CSTR reactor, with the given intrinsic reaction rate equations and kinetic constants Rxn 1: 2A → B; r1=kCA; ki = 0.5 (L/mole)-min Rxn 2: 2B → C; r = k Ce?; k2 = 0.2 (/mol)-min! a. Algebraic equation for each species: Write out the applicable algebraic equations for CA C using the space time parameter t, for these reactions occurring in a CSTR (according to approach from Chapter 8 for multiple reactions). The initial concentrations are CAD = 3.0 moles/L and Cno, and Cm=0. Write these equations using the parameters Cu ki, and t, ending up with three simultaneous equations involving for CA, Cs, and Cc. b. Product exit concentrations: Given a total volumetric flow rate vo of 10 L/minute and a CSTR volume of 100 liters, calculate the value of t. Then determine the concentration of Ca. Ce and Cc at the exit of this CSTR reactor by plugging in calculated T value, using the equations from part (a). e todos d

Answer 1

Ki=0.5 Girgen that 2A8 8,= KICA Rx DI (mal/min) K2=0.2 82= K2 CB Rxnz 1 mol/min) 23 CAO CBO cco general balance equation Now, woite the input - Cutput + disappearane by own + accumulation whese, accumulation - C at steady state) S input cut put + disappearance by rxn mass balance por component A input - (so te hun) (cao mos) input - CA. Oo mol mio output = at & Ca mol CA O und min at steady state 0 =( output = CAVO mol min disappearance by an env

prom eciation AYHATIAN CAGO = CALOotriv CAO GO - CABOTK, CAV CASO = 9 (CA+KI CAT CAO & CAT KI CAT K, CT + CA CAO - poo component B input = QCBO output = C disappearance by rx = (12-1) v Then equation will be ) v. OOCRO & Gols + (B Here, CBo=0 0 = 0oCB + (82-81) (82-7) (2,-2) v = - Gole DOCB Oo81-82 CB - T (8-8). CB = T (KICA - K2C2) CB = TKICA_TK2 GY 1 ↑ K2 CB + CB = 9k Cal por component C input = Gocco output = 0o cc

disappavance by Yxne -Yar negative C-ve) sign indicates cis porming & Cco o CC - 82v. these Coo o & Co - Ka Core golc - KCBV cc = K₂ CB VL cc = K2 CBT v=look vo= 10 h (6 we have, & -1004 Too Tolmin part I r- romin eruation & will be KIT CA +CACAO CAO - 3 m ( K,=0.5 80, 0.5x 10xCA + CA =3 SCA+GA - 320 slowing 0.68 -0.88 concentaation can not negative so Tex -0.68 mol foom equation , T K C + B = TK, CA 10x0.2 CB 7 cp = 10x0.5% (0.68)"

& CR + CB - 2.3/2 =0 80 wing we get CB - 0.85, -1.35 Then, TR=0.85 ma equatico from Cc = K₂ TCB -0.2x10x (0.85% Te = 1.445 mol