Download IDEAL PLUG FLOW REACTOR and more Study Guides, Projects, Research Design in PDF only on Docsity! 1 IDEAL PLUG FLOW REACTOR Characteristics of ideal plug flow PERFECT MIXING IN THE RADIAL DIMENSION (UNIFORM CROSS SECTION CONCENTRATION) NO MIXING IN THE AXIAL DIRECTION, OR NO AXIAL DISPERSION (SEGREGATED FLOW) TRACER PULSE INPUT AT t = 0 TRANSLATED TO EQUAL PULSE OUTPUT AT t = = L/v (L = PFR length, v = average velocity) COMPARE WITH CSTR RESPONSE TO TRACER PULSE DISPERSION 0 Q CO Q CL L 0 time 0 time 2 In an ideal PFR, concentration is a function of both distance along the flow path, x, and time, t: C = C(x,t) For a mass balance on a reacting compound, take mass balance on differential axial element with uniform reaction potential (concentration), where dV = differential volume A = cross sectional area dx = differential distance and dV = Adx Mass balance over differential element on a reactant, C In = QCx Out = QCx+dx Generation = dVrC = AdxrC Accumulation = dV Cx t Adx Cx t QCx – QCx+dx + dVrC = dV Cx t Cx+dx = Cx + dCx Q(Cx – Cx – dCx) + dVrC = dV Cx t Q Cx V rC Cx t Cx V Q rC since Q is constant Q Cx Q Cx+dx x x+dx A 5 Where Q = 0.25 m 3 /s A = channel cross section between baffles = 18 m 2 rd = rate of microorganism kill in presence of chlorine = -kdX X = concentration of microorganisms at any point in contact reactor Xo = influent concentration of microorganisms = 10 6 E. coli/100 ml kd = 5 hr -1 rc = rate of chlorine decay (from microorganism Cl-demand) = -kcX kc = 10 -5 (mg-chlorine/L)(#/100mL) -1 hr -1 2 rate expressions, 2 constituents, 2 coupled mass balances find: 1. reactor volume and flow path length, L, such that XL < 10 3 cells/100 ml 2. chlorine concentration which must be added to insure that there is detectable chlorine at PFR exit (detection level = CL = 0.05 mg/L) 1. Steady-state mass balance on cells XL = Xoexp(-kd ) = (1/kd)ln(Xo/XL) = (1/5)(hr)ln(10 6 /10 3 ) = 1.4 hr V = Q = 0.25 m 3 /s*3600 s/hr*1.4 hr = 1,260 m 3 L = V/A = 1,260 m 3 /18 m 2 = 70 m 3. Steady state mass balance on chlorine 6 dCc d k cX k cXo exp( k dd ) dCc Cco CL k cXo exp( kd )d 0 CL Cco (k cXo) k d k cXo exp( kd ) kd CL Cco (k cXo) k d (1 exp( kd )) CCO = 0.05 + (10 -5 (10 6 )/5)(1-exp(-5(1.4)) = 2.05 mg/L Chlorine contact PFR 1.0E+02 1.0E+03 1.0E+04 1.0E+05 1.0E+06 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 = x/v (hr) E . c o li ( # / 1 0 0 m L ) 0.00 1.00 2.00 3.00 4.00 C c ( m g - c h lo r in e / L ) X Cc
