Print version Updated: 1 October 2019 CEE 370 Environmental Engineering Principles Lecture #11 Ecosystems I: Water & Element Cycling, Ecological Principles Reading: Mihelcic & Zimmerman, Chapter 4 Davis & Masten, Chapter 4 David Reckhow CEE 370 L#11 1
Monday’s local paper 2 CEE 370 L#11 David Reckhow
dd Follow-up on Tuesday 3 CEE 370 L#11 David Reckhow
CMFR: non-SS Step loads are easier More complicated when reaction is occurring Simpler for case without reaction Example 4-5 (pg 128-129) With reaction Example 4-4 (pg 125-127) 4 CEE 370 L#11 David Reckhow
Non-steady State CMFR Problem: The CMFR is filled with clean water prior to being started. After start-up, a waste stream containing 100 mg/L of a conservative substance is added to the reactor at a flow rate of 50 m 3 /day. The volume of the reactor is 500 m 3 . What is the concentration exiting the reactor as a function of time after it is started? C A C A Q 0 C A0 Q 0 V Equalization tank 5 CEE 370 L#11 David Reckhow
Non SS CMFR (cont.) So the general reactor equation reduces to: dm − − A = V Q Q C C r in out A dt And because we’ve got a conservative substance, r A =0: dC − A V = Q Q C C in out dt dC Q ( ) − A = C C out in dt V Now let: = − y C C out in 6 CEE 370 L#11 David Reckhow
Non SS CMFR without reaction So that: dy Q − = y dt V Rearranging and integrating, y ( t ) t dy Q ∫ ∫ = − dt y V y ( 0 ) 0 Which yields, y ( t ) Q = − ln t y ( 0 ) V or Q y ( t ) − t = V e y ( 0 ) 7 CEE 370 L#11 David Reckhow
Non SS CMFR w/o reaction (cont.) And substituting back in for y: − Q C C − t = V in e − C C o in Since we’re starting with clean water, C o =0 − Q C C − t Q − = V t in e and − = − V C C C e − C in in in And finally, C in − Q − C t = V C C 1 e in Decreasing Q/V t 8 CEE 370 L#11 David Reckhow
Now add a reaction term Returning to the general reactor equation: dm − − A = Q Q V C C r in out A dt And now we’ve got a 1 st order reaction, r A =kC=kC out : dC − − V = kVC Q Q C C in out out dt dC Q ( ) − − = C kC C out in out dt V This is difficult to solve, but there is a particular case with an easy solution: where C in = 0 This is the case where there is a step decrease in the influent concentration to zero (M&Z, example 4.4) 9 CEE 370 L#11 David Reckhow
Non SS CMFR; C in =0 So that: dC Q ( ) − − = kC C out out dt V Rearranging, recognizing that in a CMFR, C=C out , and integrating, C ( t ) dC Q t dC Q ∫ ∫ − + = − + = k dt k dt C V C V C ( 0 ) 0 Which yields, C ( t ) Q = − + ln k t C ( 0 ) V or Q C ( t ) − + k t V = e C ( 0 ) 10 CEE 370 L#11 David Reckhow
SS Comparison of PFR & CMFR CMFR PFR C kV = − Ao C = Q C C e A V + 1 k A Ao 1 st order reaction Q C V C 1 − k = = Q A A e C V + C 1 k Ao Q Ao Example: V=100L, Q=5.0 L/s, k=0.05 s -1 ( ) C 1 C 100 − 0 . 05 = ( ) = A A e 5 100 C + 1 0 . 05 C Ao 5 Ao = = 0 . 37 0 . 50 11 CEE 370 L#11 David Reckhow
Conclusion: PFR is more efficient for a 1 st order reaction CMFR PFR Distance Across Reactor 12 CEE 370 L#11 David Reckhow
Rate of reaction of A is given by VkC A 13 CEE 370 L#11 David Reckhow
Response to Inlet Spikes 14 CEE 370 L#11 David Reckhow
Selection of CMFR or PFR PFR Requires smaller size for 1 st order process CMFR Less impacted by spikes or toxic inputs 15 CEE 370 L#11 David Reckhow
Comparison Davis & Masten, Table 4-1, pg 157 16 CEE 370 L#11 David Reckhow
Retention Time V θ = Q 17 CEE 370 L#11 David Reckhow
Analysis of Treatment Processes Basic Fluid Principles Volumetric Flow Rate Hydraulic Retention Time Conversion Mass Balances Reaction Kinetics and Reactor Design Chemical Reaction Rates Reactor Design Sedimentation Principles 18 CEE 370 L#10 David Reckhow
Conversion or Efficiency Stoichiometry k → aA + bB pP + qQ And the Conversion, X, is: X = (C - C ) Ao A = C (1 - X) C or A Ao C Ao Some use “efficiency” (ɳ) to indicate the same concept 19 CEE 370 L#10 David Reckhow
Conversion/efficiency (cont.) - N ) = b (N a(N - N ) Bo B Ao A where, N Ao = moles of A at t = 0, [moles] N A = moles of A at t = t, [moles] N Bo = moles of B at t = 0, [moles] N B = moles of B at t = t, [moles] If the volume of the reactor is assumed to remain constant, we can divide both sides of the expression by C Ao V. The expression then becomes, (C - C ) = b (C - C ) = b Bo B Ao A aX a C C Ao Ao 20 CEE 370 L#10 David Reckhow
Conversion/efficiency (cont.) This expression can then be solved for the concentration of B in terms of other known quantities: - b C = C X aC B Bo Ao 21 CEE 370 L#10 David Reckhow
Conversion/efficiency Example The reactor shown in the Figure has an inflow of 750 L/hr. The concentration of A in the influent is 0.3 M and the concentration of B in the influent is 0.5 M. The conversion (of A) is 0.75. The reaction is: A + 2B → Products Find the conversion of B, X B , and the effluent concentration of A and B. C Ao = 0.3M C A C Bo = 0.5M C A = ? Q=750 L/hr V C B = ? 22 CEE 370 L#10 David Reckhow
Solution to Conversion Ex. The first step in the solution is to determine the effluent concentration of A. This can be obtained as follows: = C (1 - X) = 0.3M x (1 - 0.75) C A Ao = 0.075 M C A For each mole of A converted to product, two moles of B are converted to product. Since we know the initial concentration of B we can calculate its final concentration: Moles/ L of B converted = 2 x (0.3 M - 0.075 M) = 0.45 M 23 CEE 370 L#10 David Reckhow
Solution to Conversion Ex. (cont.) C = 0.5 M - 0.45 M = 0.05 M B Alternatively, using Eqn: − b 2 − C = C X = 0.5 M 1 (0.3 M x 0.75) a C B Bo Ao C = 0.05 M B The conversion of B is then: X = (C - C ) = 0.5 M - 0.05 M) Bo B B (C ) 0.5 M Bo X = 0.9 B 24 CEE 370 L#10 David Reckhow
Reactors in Series w w w w Q w Q 25 CEE 370 L#11 David Reckhow
Reactors in Series 26 CEE 370 L#11 David Reckhow
Volume (10 9 m 3 ) Outflow (10 9 m 3 y -1 ) Lake Superior 12,000 67 Michigan 4,900 36 Huron 3,500 161 Erie 468 182 Ontario 1,634 211 27 CEE 370 L#13 David Reckhow
Example: 90 Sr fallout in Great Lakes 1 2 1 4 3 28 CEE 370 L#13 David Reckhow
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Loading Function Close to an impulse load, centered around 1963 estimated value is: 70x10 -9 Ci/m 2 same for all lakes 32 CEE 370 L#13 David Reckhow
Q + λ = k V Non Steady State Solution Impulse Load C 11 − λ = 1 st lake t c c o e 1 C 22 1 1 C 21 ( ) Q − λ − λ − λ = + − t t t 12 c c e c e e 2 nd lake 2 1 2 2 2 o 1 o λ − λ V ( ) 2 2 1 3 rd lake C 31 C 32 ( ) Q − λ − λ − λ = + − t t t 23 c c e c e e 3 2 3 o o 3 3 2 λ − λ V ( ) C 33 3 3 2 − λ − λ − λ − λ − t − t t t Q Q e e e e 1 3 2 3 + − 23 12 c 1 o λ − λ λ − λ λ − λ V V ( ) 3 2 2 1 3 1 3 2 33 CEE 370 L#13 David Reckhow
Hydrologic Parameters Parameter Units Superior Michigan Huron Erie Ontario Mean Depth m 146 85 59 19 86 10 6 m 2 Surface 82,100 57,750 59,750 25,212 18,960 Area 10 9 m 3 Volume 12,000 4,900 3,500 468 1,634 10 9 m 3 /yr Outflow 67 36 161 182 212 Q + Michigan = m c c λ = k m 11 V Superior = s c c s 11 Huron = + + h sh mh c c c c h 21 22 22 Erie = + + + e he she mhe c c c c c e 31 32 33 33 Ontario = + + + + o eo heo sheo mheo c c c c c c o 41 42 43 44 44 34 CEE 370 L#13 David Reckhow
From Chapra, 1997 35 CEE 370 L#13 David Reckhow
To next lecture 36 CEE 370 L#11 David Reckhow
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