Waste water treatment: Reactors

Mass Balance and Reactors Objectives • Introduce the concept of mass balance • Identify the different reactor types Conservation of Mass • • Mass cannot be created nor destroyed Atoms are conserved but molecules may change to other forms Mass Balance- General Equation Accumulation Input, Co System boundary Input rate = Ouput rate Output, C Decay + Decay + rate Steady state Accumulation = Conservative Decay rate = } Accumulation rate Input rate = Ouput rate Reactions Transform Chemical Transfer Biological Volatilization Hydrolysis Aerobic degradation Radioactive decay Anaerobic degradation Sorption Photodegradation Combustion Oxidation-reduction Ion exchange Bio uptake Sedimentation Filtration Types of Reactors • Batch Reactors V • Completely Stirred Tank Reactors (CSTR) Q, Co Q, C V • Plug-Flow Reactors Q Q Q Q to t1 t2 Q Q • Dispersed Plug-Flow Reactors Q Q to Q Q t1 Q Q t2 • Packed bed Reactors • Fluidized Packed bed Reactors Q, Co Q, C Q, C Q, Co Batch Reactor V,k,Co V,k,C time=0 time=t dC V = −VkC dt C − kt =e Co Example An industrial facility generates 1.2 m of waste with a toxic chemical at a level of 25 ppm Regulations allow the disposal of the waste into the marine environment only if the chemical at a level that does not exceed 0.5 ppm The industry decided to employ a chemical reaction (k = 0.45 day-1) using a batch reactor with a detention time of days Is the reaction time sufficient to meet the regulatory limits? Solution Using the equation for batch reactors with C = 0.5, Co =25 and k =0.45, the value of t would be 8.7 days Therefore, days will not be sufficient to reduce the concentration of the chemical to 0.5 ppm (Completely-Mixed Tank Reactor (CSTR CSTR Qin, Cin V Qout, Cout k Under steady-state conditions Input Ouput = + rate rate Decay rate Assume the decay rate is first-order = kCV The concentration inside the reactor is the same as the effluent concentration because of complete mixing QinCin = QoutCout Cout = Cin Qin Qout + kV + kCoutV Example Design a CSTR to treat the industrial wastewater described in the sketch Q= 0.05 m3/hr Co= 25ppm CSTR k= 0.45 d-1 V=? Q= 0.05 m3/hr C= 0.5 ppm Solution Applying the steady-state equation for CSTR CoQ = CQ + kCV CoQ − CQ (0.05 × 25) − (0.05 × 0.5) V= = = 130.7 m3 0.45 kC ( × 0.5) 24 Plug-Flow Reactor Q, Co Q, C dVd[ dV At steady-state C + C + (∂C / ∂x )dx ] ∂C = Q C − Q (C + dx ) − dVkC dt ∂x ∂C ∂C = −Q dx − dVkC ∂t ∂x dV = − Q dC k C Q C dC A ∫ dx = − ∫ k Co C o L C = e −k ( V / Q ) Co Example Design a plug flow reactor to treat the industrial wastewater described in the sketch Q= 0.05 m3/hr Co= 25ppm Plug flow k= 0.45 d-1 V=? Q= 0.05 m3/hr C= 0.5 ppm Solution Applying the steady-state equation for plug reactors 0.5 =e 25 − 0.45 (V / 0.05 ) 24 The required volume of the reactor is 10.43 m This volume is about 12 times smaller than that needed for the case of a CSTR Plug-Flow with dispersion L Q, Co At steady-state with a continuous mass inflow, Wehner and Wilhelm found: C 4ae ( Pe / ) = ( aPe / ) ( − aPe / ) Co (1 + a ) e − (1 − a) e 0.5 4k t R   a = 1 +  Pe   Pe= Peclet number=vxL/D D=dispersion coefficient vx=axial velocity= Q/A L=length of the reactor tR= retention time= V/Q Q, C Example Determine the effluent concentration for the PFD reactor shown below if the reactor has a length of m, width 1.4 m and depth 1.49 m Assume the dispersion coefficient is m 2/hr Q= 0.05 m3/hr Co= 25ppm Solution Plug flow k= 0.45 d-1 V=10.43 m3 vx = Q 0.05 = = 0.024m / hr Ax 1.4 ×1.49 tR = V 10.43 = = 208.6 hr Q 0.05 a = (1 + Q= 0.05 m3/hr C= ? v x L 0.024 × Pe = = = 0.12 D 4k t R 0.5 × 0.019 × 208.6 0.5 ) = (1 + ) = 11.5 Pe 0.12 C 4ae ( Pe / ) = = 0.19 Thus, C = 4.76 ppm ( aPe / ) ( − aPe / ) Co (1 + a ) e − (1 − a ) e Experimental determination of dispersion coefficient For a pulse injection of ideal tracer, moment analysis can be used to determine the dispersion coefficient following the steps below The centroid (actual retention time) The variance σ2 t C ∑ = ∑C i tC ∑ t= ∑C i i i i −t2 i The normalized variance σ σ θ2 = t The normalized variance is related to the dispersion number by D D σ =2 − 2( ) (1 − e −v x L / D ) vx L vx L θ Example Dispersed plug flow through a compartmented aeration tank was analyzed by injecting a pulse of lithium chloride tracer in the influent From the time and output concentration data listed, plot C (kg/m 3) versus time (min)-response curve Calculate the location of the centroid of the distribution, variance of the curve , normalized variance, and the reactor dispersion number D/vxL t C t C t C t C 0 105 89 210 33.5 315 15 120 95 225 25.8 330 4.6 30 135 88 240 20 345 3.5 45 3.5 150 78.2 255 15.4 360 2.6 60 16.5 165 65 270 12.1 375 1.7 75 46.5 180 55.2 285 9.5 390 0.7 90 72 195 43 300 7.5 405 Solution The response curve is shown in the figure below 100 80 C 60 40 20 0 100 200 300 time, 400 500 The centroid (actual retention time) The variance σ2 t C ∑ = ∑C i t = σ2 = 120770 = 152 795 21277913 − (152) = 3682.7 795 3682.7 σ = = 0.16 (152) θ i i i −t2 i The normalized variance tC ∑ t= ∑C σ σ θ2 = t i Using Excel a table similar to the one shown is generated The normalized variance is related to the dispersion number by σ θ2 = D D − 2( ) (1 − e −v x L / D ) vx L vx L The dispersion number can be found by trial and error or using the figure shown in the slide For σ θ2 = 0.16 , the dispersion number is 0.0875 Knowing the length and axial velocity one can determine the dispersion coefficient of the reactor D/vxL σ θ2 0.1 0.180 0.2 0.321 0.09 0.164 0.085 0.156 0.088 0.161 0.087 0.159 0.0875 0.160 [...]... dC k C Q C dC A ∫ dx = − ∫ k Co C o L C = e −k ( V / Q ) Co Example Design a plug flow reactor to treat the industrial wastewater described in the sketch Q= 0.05 m3/hr Co= 25ppm Plug flow k= 0.45 d-1 V=? Q= 0.05 m3/hr C= 0.5 ppm Solution Applying the steady-state equation for plug reactors 0.5 =e 25 − 0.45 (V / 0.05 ) 24 The required volume of the reactor is 10.43 m 3 This volume is about 12 times