ENV 6015 Solution to Mixing Problem Set

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Linda Hensley
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1 EN 65 Soluion o ixing Problem Se. A slug of dye ( ) is injeced ino a single ank wih coninuous mixing. The flow in and ou of he ank is.5 gpm. The ank volume is 5 gallons. When will he dye concenraion equal 6.8 ug/? Soluion: Q, o Q, Q, ass balance : in - ou acc Q m d ( ) m d where : in because here is no more inpu where m a a all poins and volume is consan. Q d m d ; and m Q (by definiion T is hydraulic residence ime, HRT) T d m d T m T ln[ ] m m T T ln ( ) ln ln Where 5 gallons

2 μg 6.8 g Q.5 min T 5 gal.5 gal/min min T ln 5 gal.5 gal/min ln (6.8 μg/) 5 gal , μg /gal 679 min..3. for 6.8µg/ m

3 . Is he differenial equaion shown below he correc differenial equaion for he sysem? dc d Τ Τ e T Q, o Q, Q, Soluion: For reacor ln () T / where T Q For reacor : ass Balance (No generaion) ; In Ou generaion accumulaion Q d[ ] Q bu consan d Q dc ; d Q T dc where T d T Q dc Rearranging : we ge () d T T 3

4 solving equaion () for we ge (3) /T e Subsiuing (3) in () dc d T T e /T Which is he same equaion, and i is he correc differenial equaion for he sysem. 4

5 3. The daa shown in he succeeding able represen concenraion of dye following ime of addiion. Assume he daa poins can be conneced by sraigh lines o provide a coninuous represenaion of mass flow ou of he ank. This daa is ypical of daa generaed for a residen ime disribuion sudy. How many pounds of dye were added o he ank if he ank volume was 53, gallons and he flow was 6,5 gallons/? Draw a concenraion verses ime curve, and a cumulaive mass verses ime curve for his daa. Repor he imes when %, 4 % and 9 % of he mass has exied he ank. Do you hink his sysem represens a plug flow, complee mix or combinaion plug flowcomplee mix sysem? Why? Time () Flow (gph) oncenraion (/) Soluion: The plo of dye concenraion wih ime in he sysem is shown in he nex figure: oncenraion vs Time Graph 8 oncenraion, / Time, s The following able shows he ime elapsed, he flow rae, effluen concenraion, cumulaive mass and percenage of he oal mass of dye ha have exied he sysem. 5

6 Time (s) Flow (gal/) oncenraion (/) umulaive mass (lb) % mass The cumulaive mass of dye is calculaed by: dye -ime Flow - ( ime - ime ) dye -ime3 dye -ime Flow 3 - ( ime 3 - ime ) as an example: a ime.5 hours dye -.5 6,5 gal gal lb 453, ( ) - dye lb a ime 4. hours: dye 4 6,5 (.5.75 ) gal gal ( 9 ) lb 453, (.5) ( 9 9) (.6 ) ( 9 5) ( 4.6 ) ( 5 ) Thus: dye lbs The cumulaive mass of dye versus ime is shown in he nex figure. The imes corresponding o, 4 & 9 % of he iniial mass exiing he ank can be obained by inerpolaing from he graph below. 6

7 umulaive ass vs Time Graph 4 umulaive ass, / 3 4% 9% % Time, s For % mass.344 lbs ou a.75 s, mass, a.5 s, mass.75 lbs (linear variaion) ass rae ou.75 lb lb.5.75 lbs/ Hence for.34 lbs: ime.34 s. Toal ime s For 4 % mass.38 lbs ou a.5 s, mass.75 lbs, a. s, mass.74 lbs (linear variaion) ass rae ou.74 lb.75 lb.5.98 lbs/ Hence for ( ).63 lbs, ime Toal ime s.63 lb.98 lb/.36 s. for 9 % mass 3. lbs ou a.6 s, mass.67 lbs, a 4. s, mass 3.44 lbs (linear variaion) ass rae ou 3.44 lb.67 lb lbs/ 7

8 Hence for ( ).43 lbs, ime Toal ime s.43 lb.55 lb/.77 s. The imes, percenage of dye ou and mass of dye ou are presened in he nex able: Time () % Dye Ou Dye ou (lb)

9 4. Prove ha he equaion for s order growh in a wo ank series (of equal size) is Q. o Q. Q. Soluion: ass balance for he firs ank ; In Ou gen Acc Q Q γ d d dc A seady sae d T k o Subsiuing: Q Q K or, Q (Q K ) Dividing boh sides by Q: K and T Q Q And T T KT ;T ; () KT () For second ank mass balance : 9

10 In Ou gen Acc ; a seady sae Acc ; Q Q K Q Q K K Q KT (3) KT Subsiuing from equaion () in equaion (3): KT KT KT which is he equaion for firs order growh in wo equal size reacors in series In general: n KT n n

11 5. An equalizaion basin receives ee flows (combined flow 5 /day). If l is added, he kill is approximaely a s order reacion wih a decay rae consan of. day -. The inpu from he ee pipes causes raher complee mixing. an we expec an 85% kill? How many comparmens would we have o divide he basin ino? (ixing provided in each). olume of basin, liers. o org/l. Soluion: ass balance for organisms: In-ou-decay acc Assuming seady sae condiions: acc Q o Q K d Dividing by Q we ge, K K d d ; Q Q K d ; Q K d Q Solving for K d. day - ; ; Q 5 day - (.day )( ) ( ) 5/day (.4) kill efficiency * η o η ( -.94)* 7.6% a) We will ge only a 7% kill using one comparmen. To achieve 85% kill we will have o divide he basin in several comparmens. b) Using he general formula: n K dt n Where n number of equal size comparmens, T Q and oal volume of basin

12 # basin comparmens % kill Need 5 comparmens of equal size o obain an 85% kill or more

13 6. A subsance is discharged a /; i can be made less oxic if neuralized wih NaOH (negligible flow). The reacion is no insananeous, bu firs order wih a rae consan of.4 -. Repor and. Does he second ank help much? The exising sysem serves as a mixing and reacion basin and has he following characerisics: olume () Q (/) q (/) Tank 5 Tank 5 Q, o q, Q, q, Soluion: ass balance for ank # In ou decay Acc Assuming seady-sae: Acc In ou decay q q K d K Solving for : d () q ass balance for ank # In ou decay Acc Assuming seady-sae: Acc Q q q Q K d o [ q Q K ] Q () q d subsiuing from equaion() in equaion () Q Kd q d q ( q Q K ) Kd Q d q q ( q Q K ) 3

14 K d q Q ( q Q K ) q d - (.4 )( ) - 5 (.4 )( ) 5, 3.5 ()( 3 5) 3. From equaion () 69.8 K d q 3. (.4)( ) b) If he firs reacor works alone: In ou decay Acc; (seady-sae) Q Q K d Q Q K d 5 5 (.4h )( ), 3 4

15 The change obained when adding a second ank is only of 7 / which is 9.3%. 5

Math 333 Problem Set #2 Solution 14 February 2003

Math 333 Problem Set #2 Solution 14 February 2003 Mah 333 Problem Se #2 Soluion 14 February 2003 A1. Solve he iniial value problem dy dx = x2 + e 3x ; 2y 4 y(0) = 1. Soluion: This is separable; we wrie 2y 4 dy = x 2 + e x dx and inegrae o ge The iniial