CEE 370 Environmental Engineering Principles

Transcription

1 Updated: 7 October 2015 Print version EE 70 Environmental Engineering Principles Lecture #11 Ecosystems I: Water & Element ycling, Ecological Principles Reading: Mihelcic & Zimmerman, hapter 4 Davis & Masten, hapter 4 David Reckhow EE 70 L#11 1 MFR: non-ss omplicated when reaction is occurring Simpler for case without reaction See section in book on step load change Equations 4-29 to 4- Example 4-9 David Reckhow EE 70 L#11 2 Lecture #11 Dave Reckhow 1

2 Non-steady State MFR Problem: The MFR 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 /day. The volume of the reactor is 500 m. What is the concentration exiting the reactor as a function of time after it is started? A0 0 A A 0 Equalization tank David Reckhow EE 70 L#11 Non SS MFR (cont.) So the general reactor equation reduces to: dm dt A = in r And because we ve got a conservative substance, r A =0: da = in out dt da = in out dt Now let: y out in out A David Reckhow EE 70 L#11 4 Lecture #11 Dave Reckhow 2

3 Non SS MFR without reaction So that: dy dt = y Rearranging and integrating, y( t) t dy dt Which yields, y(0) y 0 y( t) ln t y(0) or y( t) e y(0) t David Reckhow EE 70 L#11 5 Non SS MFR w/o reaction (cont.) And substituting back in for y: o in in e t Since we re starting with clean water, o =0 in in e t and in t ine And finally, in in 1 e t Decreasing / David Reckhow EE 70 L#11 6 t Lecture #11 Dave Reckhow

4 Now add a reaction term Returning to the general reactor equation: dma = in out r A dt And now we ve got a 1 st order reaction, r A =k A =k out : da = in out kout dt da = in out kout dt This is difficult to solve, but there is a particular case with an easy solution: where in = 0 This is the case where there is a step decrease in the influent concentration to zero (see D&M, pg 155) David Reckhow EE 70 L#11 7 Non SS MFR; in =0 So that: da = out k dt Rearranging and integrating, Note missing negative sign in equation 4-9 in D&M Which yields, or d = k dt ( t) (0) e ( t) David Reckhow EE 70 L#11 8 (0) d ( t) ln k t (0) out k t t 0 k dt Lecture #11 Dave Reckhow 4

5 omparison of PFR and MFR MFR PFR A A Ao A Ao Ao 1 k 1 k k David Reckhow EE 70 L#11 9 A Ao 1 k A Ao e Example: =100L, =5.0 L/s, k=0.05 s -1 1 A e Ao e onclusion: PFR is more efficient for a 1 st order reaction MFR PFR Distance Across Reactor David Reckhow EE 70 L#11 10 Lecture #11 Dave Reckhow 5

6 Rate of reaction of A is given by k A David Reckhow EE 70 L#11 11 Response to Inlet Spikes David Reckhow EE 70 L#11 12 Lecture #11 Dave Reckhow 6

7 Selection of MFR or PFR PFR Requires smaller size for 1 st order process MFR Less impacted by spikes or toxic inputs David Reckhow EE 70 L#11 1 omparison Davis & Masten, Table 4-1, pg 157 David Reckhow EE 70 L#11 14 Lecture #11 Dave Reckhow 7

8 Retention Time David Reckhow EE 70 L#11 15 Analysis of Treatment Processes Basic Fluid Principles olumetric Flow Rate Hydraulic Retention Time onversion Mass Balances Reaction Kinetics and Reactor Design hemical Reaction Rates Reactor Design Sedimentation Principles David Reckhow EE 70 L#10 16 Lecture #11 Dave Reckhow 8

9 onversion or Efficiency aa + bb k pp + q And the onversion, X, is: X = ( Ao - A) Ao or A = Ao (1 - X) Davis & Masten use efficiency (ɳ) to indicate the same concept David Reckhow EE 70 L#10 17 onversion/efficiency (cont.) (N - N ) = b Bo B a ( N Ao - N 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 Ao. The expression then becomes, ( Bo - B) Ao = b ( Ao - A) a Ao = b a X David Reckhow EE 70 L#10 18 Lecture #11 Dave Reckhow 9

10 onversion/efficiency (cont.) This expression can then be solved for the concentration of B in terms of other known quantities: B = Bo - b Ao a X David Reckhow EE 70 L#10 19 onversion/efficiency Example The reactor shown in the Figure has an inflow of 750 L/hr. The concentration of A in the influent is 0. M and the concentration of B in the influent is 0.5 M. The conversion (of A) is The reaction is: A + 2B Products Find the conversion of B, X B, and the effluent concentration of A and B. Ao = 0.M Bo = 0.5M =750 L/hr A A =? B =? David Reckhow EE 70 L#10 20 Lecture #11 Dave Reckhow 10

11 Solution to onversion Ex. The first step in the solution is to determine the effluent concentration of A. This can be obtained as follows: A = Ao (1 - X) = 0.M x (1-0.75) A = M 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. M M) = 0.45 M David Reckhow EE 70 L#10 21 Solution to onversion Ex. (cont.) B = 0.5 M M = 0.05 M Alternatively, using Eqn: b B = B o A o a X = 0.5 M 2 1 (0. M x ) B = 0.05 M The conversion of B is then: Bo B X B = ( - ) 0.5 M M) = ( Bo) 0.5 M X B = 0.9 David Reckhow EE 70 L#10 22 Lecture #11 Dave Reckhow 11

12 Energy Balance First law of thermodynamics Energy can be neither created nor destroyed But the form can certainly change Thermal Energy haracterized by Temperature (T) and Specific heat capacity (c p ) H Mc p T David Reckhow EE 70 L#1 2 Heat Transfer onduction Transfer of energy without mass flux onvection Energy is carried by molecules in bulk motion David Reckhow EE 70 L#1 24 Lecture #11 Dave Reckhow 12

13 Lake olume Outflow (10 9 m ) (10 9 m y -1 ) Superior 12, Michigan 4,900 6 Huron, Erie Ontario 1, David Reckhow EE 70 L#1 25 Example: 90 Sr fallout in Great Lakes David Reckhow EE 70 L#1 26 Lecture #11 Dave Reckhow 1

14 David Reckhow EE 70 L#1 27 David Reckhow EE 70 L#1 28 Lecture #11 Dave Reckhow 14

15 David Reckhow EE 70 L#1 29 Loading Function lose to an impulse load, centered around 196 estimated value is: 70x10-9 i/m 2 same for all lakes David Reckhow EE 70 L#1 0 Lecture #11 Dave Reckhow 15

16 Non Steady State Solution Impulse Load 1 st lake c 1 2 nd lake rd lake c c o c 1o e 11 e t 1o c t 12 1t 2t c2 c2oe c1 o e e ( ) t c 2o 2 ( ) 1t 212 e e 2( 2 1 ) 1 David Reckhow EE 70 L#1 1 2 e t 2 t e e t 2t e k 2 2 t Hydrologic Parameters Parameter Units Superior Michigan Huron Erie Ontario Mean Depth m Surface 10 6 m 2 82,100 57,750 59,750 25,212 18,960 Area olume 10 9 m 12,000 4,900, ,64 Outflow 10 9 m /yr Michigan Superior Huron Erie Ontario m cm c11 s cs c11 h sh mh ch c21 c22 c22 e he she mhe ce c1 c2 c c o eo heo sheo mheo co c41 c42 c4 c44 c44 k David Reckhow EE 70 L#1 2 Lecture #11 Dave Reckhow 16

17 From hapra, 1997 David Reckhow EE 70 L#1 To next lecture David Reckhow EE 70 L#11 4 Lecture #11 Dave Reckhow 17