CEE 370 Environmental Engineering Principles Lecture #11 - PDF document

CEE 370 Lecture #11 10/2/2019 Print version Updated: 2 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 Lecture #11 Dave Reckhow 1

CEE 370 Lecture #11 10/2/2019  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 Lecture #11 Dave Reckhow 2

CEE 370 Lecture #11 10/2/2019 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 = Q Q V C C r in out A dt  And because we’ve got a conservative substance, r A =0: dC  V A = 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 Lecture #11 Dave Reckhow 3

CEE 370 Lecture #11 10/2/2019 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:    C C Q    t  V in e    C C o in  Since we’re starting with clean water, C o =0    C C Q    t   Q  V    t in e   and    V C C C e    C in in in  And finally, C in     Q     t C    V C C 1 e     in   Decreasing Q/V t 8 CEE 370 L#11 David Reckhow Lecture #11 Dave Reckhow 4

CEE 370 Lecture #11 10/2/2019 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 = Q Q kVC 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, dC  Q  C ( t ) 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 Lecture #11 Dave Reckhow 5

CEE 370 Lecture #11 10/2/2019 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    A   Q  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 Lecture #11 Dave Reckhow 6

CEE 370 Lecture #11 10/2/2019  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 Lecture #11 Dave Reckhow 7

CEE 370 Lecture #11 10/2/2019 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 Lecture #11 Dave Reckhow 8

CEE 370 Lecture #11 10/2/2019 Retention Time V   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 Lecture #11 Dave Reckhow 9

CEE 370 Lecture #11 10/2/2019 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 Lecture #11 Dave Reckhow 10

CEE 370 Lecture #11 10/2/2019 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 Lecture #11 Dave Reckhow 11

CEE 370 Lecture #11 10/2/2019 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 a C X = 0.5 M (0.3 M x 0.75) B B o A o   1 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 Lecture #11 Dave Reckhow 12

CEE 370 Lecture #11 10/2/2019 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 Lecture #11 Dave Reckhow 13

CEE 370 Lecture #11 10/2/2019 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 Lecture #11 Dave Reckhow 14

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CEE 370 Lecture #11 10/2/2019 31 CEE 370 L#13 David Reckhow 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 Lecture #11 Dave Reckhow 16