# 03-17 - To model pollutant fate we need to: 1. Describe...

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Unformatted text preview: To model pollutant fate we need to: 1. Describe pollutant characteristics [Ex. - need a rate equation for non-conservative materials]. 2. Describe the system [model as appropriate for the degree of exactness desired - CSTR vs. PFR vs. AFR] EXPECTATIONS: consider the response of various reactor types to a continuous input of a conservative substance [Ex. - non reactive tracer] CO C↑ INPUT: Continuous Input (starts @ time = 0) t=0 time Reactor Response CO C↑ t=θ t = 0 time PFR CO C↑ CSTR t=θ CO C↑ AFR t=θ t = 0 time t = 0 time define θ =V/Q = average hydraulic retention time [i.e., average amount of time water spends in the reactor] Note: for a PFR: θ =LxA / uxA = L/u Consider the response of various reactor types to an input of a conservative tracer INPUT: C Mass added = W @ t = tO = 0 Pulse Input time Response: t=0 PFR Cmax C t=θ C CSTR Cmax = W/V t=θ time C AFR “washout” (input flow t=θ Area = W/Q time has zero concentration) t=0 time t=0 t=0 Area = W/Q CSTR RESPONSE EQUATIONS I. Continuous input (to reactor with Linitial = 0) We have already REVIEW Q LO - Q L - k1L V V || || || input output reaction The resulting integrated equation was: shown (For BODL ) that dL = dt L= Lo Q 1 − exp − t k1 + k1V V 1+ Q For Li = 0 OR, in general for any pollutant C with input concentration CO and initial concentration Ci=0 Co Q 1 − exp − t k1 + k1V V 1+ Q Q For a conservative pollutant: k1= 0 and (dC)/(dt)= (Co − C ) V separate variables and integrate: C= C dC ∫ CO − C 0 Q = ∫ dt V0 Q −t =e V t to get: Q − ln ( CO − C ) + ln CO = t V or CO − C CO and C =C0(1-e-(Qt/V)) Note that we arrive at the last equation on the previous page if we just substitute k1=0 into the general integrated equation. Co Q C= 1 − exp − t k 1 + kV 0 V 1+ 1 Q0 to a CSTR II. Pulse input. This may be visualized as a continuous input of CO= 0 into a reactor with initial concentration C=Ci For the BOD problem when Linitial≠0 we had C = W/V i Lo Q L= 1 − exp − t k1 + + Li k1V V 1+ Q Q − t k1 + e V W t=0 ∴ If LO= 0 and k1= 0 this simplifies to L= L i e −t (Q / V ) Q≠ 0 Q≠ 0 CO=0 Ci=W/V t= 0 +1 nanosecond or for the general case: C = Ci e −t ( Q / V ) Alternatively, derive using the general form: 0 dC dt = Q Cinput -Q Coutput - kC V V 0 for a conservative substance C= C i @ t=0 where Cinput=0 and boundary conditions are: Separate variables to get: C ∫ C Q =− t Integrating gives: ln Ci V or C = Ci e − t(Q / V) Ci dC Q = − ∫ dt C Vo t and Ci = W/V Note: For a CSTR fluid elements have a distribution of residence times since the reactor is completely mixed. Some fluid elements of the input flow leave the reactor immediately while others stay around for infinity. V/Q for an ideal PFR = LA/uA = L/u In general θ may be defined as a concentration weighted average time (for a pulse input [see below]). n θ= t = alternate symbol ∞ ∫0 tC( t )dt ∞ ∫0 C( t )dt j=1 ∑ C ( j) ∆ t j=1 n ∑ t jC( j) ∆t for discrete samples of effluent over time For a CSTR it may be shown that θ = V/Q as follows: ∞ ∫ tCdt 0 ∞ 0 For wash-out of a pulse of a conservative material − t (V / Q) θ= and C = C e i ∫ Cdt ∞ ∴ θ= tCi e − t /(V / Q)dt ∫ 0 ∞ 0 = t ( − Qt − 1) − Q / V V e Ci (Q / V)2 0 Ci e t − Q/V ∞ Ci e − t /(V / Q)dt ∫ θ 1 (Q / V)2 = = 1 Q/V 1 Q / V 0 ∞ V/Q A brief comment on how to compute your residence time. Consider the summation: ∑ C( j) ∆t j=1 n This is just the area under the tracer response curve (i.e., the plot of C vs. t) C↑ C(j) = (C4 + C3)/2 Time → 1 2 3 45 6 (C 4 + C 3 ) (t 4 − t 3 ) Area = 2 t(j) In the summation: ∑ t jC( j) ∆t j=1 n tj = (t4 + t3)/2 ...
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## This note was uploaded on 04/13/2010 for the course CEE 3510 taught by Professor Lion during the Spring '10 term at Cornell.

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