GCH-2006: Calcul des réacteurs chimiques - Cours # 18 - Alain Garnier, génie chimique, H13
Plan Chapitre 9, régimes transitoires: Bilans molaires (chapitre 4) Bilans d énergie transitoires Histoire de cas ARSST Procédés semi-batch CSTR en régime transitoire Réactions multiples (Cliquez ici pour accéder au fichier Excel joint à ce document)
Bilan d énergie thermique transitoire Général: dt dt dt dt Q W S + Fi 0CPi ( Ti 0 T) + HirV i = NC i Pi i Pi Q W S + FA0CP 0( T0 T) + HRX rv A = NC En cuvée, pour une réaction simple: dt Q W S + HRX rv A = dt N C C X ( + ) A0 P0 P (réaction simple ou réversible) 0 dt Q WS + HRX ra V = dt N C A0 P0 (ΔCp=0)
Cuvée adiabatique dt HRX rv A = dt N C C X ( + ) A0 P0 P dt dt 0 ( Cp0 + CpX ) = ( H RX + Cp( T TR )) dx dt X C ( T T ) C ( T T ) = = H + C T T H P0 0 P0 0 0 RX P R RX ( ) ( ) H T H X T = T + = T 0 RX 0 RX 0 0 C C P0 CP0 P + X En présence d un échangeur de chaleur: dt Q W S + HRX rv A UAT ( a T) + HRX rv A = = dt N C C X N C C X ( + ) ( + ) A0 P0 P A0 P0 P...
Advanced Reactor System Screening Tool (ARSST) Principe de la bombe calorimétrique Utile pour déterminer: la chaleur de réaction Les paramètres de la loi d'arrhénius Le dimensionnement d'une soupape de sécurité
Température d'allumage («onset»)
Exemple 9-3 (p. 608) (voir feuille Excel pour calculs)
Dimensionnement d'une soupape de sécurité (PRS, chap9, DVD)
ARSST Source: ARST, Brown industries & U Michigan
Réactions emballées ( runaways ) For reactions taking place in the liquid (condensed) phase, this heat is absorbed by the liquid, increasing the vapor pressure and hence the pressure in the vessel. Such reactions are called tempered or vapor systems, since the heat of reaction increases the superheat and vapor pressure of the liquid. Another source of pressure from a reaction results when the products of the reaction are gases, and more moles of gas are generated than are consumed by the reaction. Such reactions are called gassy.
Calculs The required vent size is determined by an energy balance on the reactor, with the criterion that the size of the vent must be large enough to expel energy from the vessel (in the vented stream) at a rate equal to the maximum rate at which it is generated. For a tempered (vapor pressure) system, the energy balance on the vaporizing liquid is: The left side of the equation is the latent heat removed from the vessel by the fluid that is relieved at the rate ṁ through the vent, and the right side is the rate of heat generated by the reaction.
Calcul Assuming single-phase flow (i.e. complete disengagement of the vapor from the liquid before entering the vent) and choked vapor flow through the vent, the mass flux through the vent is given by application of Bernoulli s equation to an isentropic nozzle: Here k is the isentropic exponent for the gas/vapor (or the ratio cp/cv for an ideal gas) and Kd is the vent discharge coefficient. If m is eliminated from Eqns (1) and (2), the result can be solved for the ratio of the vent area A to the volume of the reacting mass, V: where V ρ is the test sample mass, V is the ARSST containment volume and ρ is the density of the reacting mixture. For a hybrid system, the required vent areas for both the vapor and gas release are added
Calcul
Gassy system, critical: Calcul Hybrid system:
Réacteur semi-batch avec transfert thermique Transfert thermique dans une cuve: UA Q = mc C Pc ( Ta 1 T) 1 exp mc C Pc éqn 8-49, p. 523 où: Lorsque le débit de réfrigérant est élevé et que la température Ta ne baisse pas significativement: ( ) Q = UA T T Bilan thermique: ( ) T = T T T e a2 a1 a UA m Cp C C dt dt Q + rv A HRX + Fi 0CPi ( Ti 0 T) = N C i Pi
Exemple 9-4: réaction semi-batch avec transfert thermique (p. 616) Saponification de l acétate d éthyle: Réaction de 2 nd ordre, en phase aqueuse Alimentation C B0 = 1000 moles/m 3 C W0 = 55000 moles/m 3 v 0 = 0,004 m 3 /s T 0 = 300 K Initialement V i = 0,2 m 3 C Ai = 5000 mole/m 3 C Wi = 30700 mole/m 3 C Bi = 0 T i =300 K C2H5( CH3COO) + NaOH Na( CH3COO) + C2H5OH A+ B C+ D UA = 3000 J/s/K ṁ c = 100 kg/s C Pc = 18 J/kg/K T a1 = 285 K k = 0,39175*exp(5472,7(1/273 1/T) m 3 /kmole/s K C = 10 3885,44/T ΔH RX o = 79076 J/mole C PA = 170,7 J/mole/K C PB = C PC = C PD = C PW = 75,24 J/mole/K Est-ce que l échangeur de chaleur permettra de maintenir la température du réacteur sous 315 K?
CSTR en régime transitoire (section 9.4) Exemple 9-5, démarrage d un CSTR Production de propylène glycol: A + B C Bilans molaire et thermique en régime transitoire Intérêt des diagrammes de phase C-T (voir PRS pour plus de détails sur l approche)
Système de réactions multiples non-isotherme dt dt = q j= 1 ( ) Q + rv H ( T) F C T T ij RXij i0 Pi 0 N C i Pi Exemple 9-7: réactions multiples en semi-batch k1a k2b 2 3 A B C 1 2 2 réactions de 1 er ordre UA = 35000 cal/h/k T a = 298 K y A0 = 1 C A0 = 4 M v 0 = 240 L/h T 0 = 305 K V i = 100 L C Ai = 1 M C Ii = 1M T i = 290 K k 1A = 1,25 h -1 @ 320K, E 1A = 9500 cal/mole k 2B = 0,08 h -1 @ 300 K, E 2B = 7000 cal/mole C PA = 30, C PB = 60, C PC = 20, C PI = 35 cal/mole/k ΔH RX1A = -6500 cal/mole de A ΔH RX2B = 8000 cal/mole de B