CEE 370 Environmental Engineering Principles

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1 Updated: 29 September 2015 Print version EE 370 Environmental Engineering Principles Lecture #9 Material Balances I Reading: Mihelcic & Zimmerman, hapter 4 Davis & Masten, hapter 4 David Reckhow EE 370 L#9 1

2 Basics Types of material balances Mass Balance Law of conservation of matter Energy Balance Law of conservation of energy General pproach Define a control volume Within that volume ccumulation Input Output Rate of ccumulation Rate of Input Rate of Output David Reckhow EE 370 L#9 2

3 Mass Flux Rate of input/output Flux In: Flow entering a control volume m mass time in Q in in volume time x mass volume David Reckhow EE 370 L#9 3

4 Mass input Q in, x in m ii Mass Balances onversion -r V m rrr System boundary or ontrol volume Mass output Q out, x out m ooo ccumulation rate Input rate - Output rate i i i i dm dt ( V) d dt d V dt David Reckhow EE 370 L#9 4 - onversion rate

5 Mass Balances (cont.) dm dt n ( Q ) in ( i1 n j1 Q ) out - r V where, M mass of a, [mass] in concentration of species entering the system, [mass/volume] out concentration of species leaving the system, [mass/volume] Q in volumetric flow rate bulk mass entering the system, [volume/time] Q out volumetric flow rate of bulk mass leaving the system, [volume/time] r reaction rate of species forming something else, [mass/volume-time] V volume of reactor nd for a reaction of order n, r k n David Reckhow EE 370 L#9 5

6 Mass Balances (cont.) For systems at steady state with no accumulation, the time dependent term goes to zero and the equation reduces to: dm dt 0 n ( Q ) in ( i1 n i1 Q ) out - r V r V n i1 ( Q ) in n i1 ( Q ) out Look at Example 4.1 David Reckhow EE 370 L#9 6

7 Mass Balances (cont.) onservative substances are those that do not react. For these the value of r is zero and the mass balance equation reduces to: dm dt n ( Q ) in ( i1 nd if the conservative substance is as steady state, the MB is even simpler: n n i1 Q ( Q ) in ( Q ) out i1 n i1 ) out David Reckhow EE 370 L#9 7

8 nalysis of Treatment Processes Basic Fluid Principles Volumetric Flow Rate Hydraulic Retention Time onversion Mass Balances Reaction Kinetics and Reactor Design hemical Reaction Rates Reactor Design Sedimentation Principles David Reckhow EE 370 L#9 8

9 Basic Fluid Principles Volumetric Flow Rate Q v where, Q volumetric flow rate, [m 3 /day, ft 3 /s] area across which the fluid passes, [m 2, ft 2 ] v fluid velocity, [m/d, ft/s] David Reckhow EE 370 L#9 9

10 Fluid Principles cont. Hydraulic Retention Time HRT θ V Q where, θ hydraulic retention time, [days] V volume, [m 3 ] Q volumetric flow rate, [m 3 /day] Work out Example 7.1 David Reckhow EE 370 L#9 10

11 Flux Density Flux is the movement of a mass past a surface, plane, or boundary. J i M i i t where, J i flux density crossing the boundary i, [Kg/m 2 -hr] M i mass crossing the boundary i in time t, [Kg] i area of boundary i, [m] t time for the mass to cross the boundary i, [hr] David Reckhow EE 370 L#9 11

12 Flux (cont.) Velocity, V i rea, i Length L If the right side of the above equation is multiplied by L/L, where L is the distance the approaching mass moves during time t, then the equation becomes: David Reckhow EE 370 L#9 12

13 Flux (cont.) J i M i i x t x L L M i x i L x L t J V i i where, i the concentration of the material crossing the boundary i, [Kg/m 3 ] V i the velocity of the material crossing the boundary i, [m/hr] i David Reckhow EE 370 L#9 13

14 larifier Example 25 m diameter secondary clarifier has an influent solids concentration of 2500 mg TSS/L. The flowrate into the clarifier is 17,500 m 3 /day. If the effluent solids are assumed to be zero, what return or recycle flow rate is required to attain a return solids concentration of 7500 mg TSS/L. lso, what is the solids flux across the boundary shown below. Q i 17,500 m 3 /d X i 2500 mg/l i Q e? X e 0 Q u? X u 7500 mg/l David Reckhow EE 370 L#9 14

15 larifier example (cont.) We can perform a mass balance to determine the underflow or recycle solids concentration, X u. ssuming no accumulation in the sedimentation tank, or Mass in Mass out Xi Q i XeQ e + Xu Qu Since X e is assumed to be zero, Q u Xi Q X u i (2500 mg TSS/L)(17,500 m 7500 mg TSS/L 3 /d) David Reckhow EE 370 L#9 15

16 larifier example (cont.) Q 5,800 m / day u To determine the flux across i, we need the mass moving across i per day, or, M i Xi V 3 where V is the volume applied per time. If we choose one day for t, then, V is 17,500 m 3. Thus, the mass is, M (2500 mg TSS/ L) x (17,500 m ) x i nd the flux is: J i 3 Kg mg x 10 L m 43,750 Kg ( π x (25 m/ 2) ) x (1day) J i 89 Kg/ m - day 3.7 Kg/ m 44,000 Kg David Reckhow EE 370 L# hr

17 River Example n industry is located adjacent to Spring reek. The industry uses copper cyanide for plating both copper and brass. Estimate the maximum concentration of copper that can be discharged in the effluent in order to meet the required maximum concentration, d, of mg u 2+ /L in the stream. The upstream copper concentration is below the detection limit, i.e. u 0 mg/l. ssume steady state conditions. Q u 0.25 m 3 /s u 0 Industry Q e 0.08 m 3 /s e? Spring reek Q d? d mg/l David Reckhow EE 370 L#9 17

18 Solution to River Ex. We first use a mass balance on the flow into and out of the system. The flow after discharge can be calculated by a mass balance on the water entering and leaving the system (the concentration of water in water is unity, and thus cancels): Q d Q u + Qe where the "e" subscript indicates effluent, the "u" subscript indicates up-stream, and the "d" subscript indicates downstream. d Q 0.25 m / sec m / sec 0.33 m / sec David Reckhow EE 370 L#9 18

19 Solution to River Ex. (cont.) The allowable concentration of copper in the effluent can then be determined by a mass balance on copper entering and leaving the system: Qu u + Qe e Qdd Solving for e (two equations and two unknowns): Qd d Qu u e Q m mg s L e m L e mg/ L e 0.25 m 3 s 0 mg L David Reckhow EE 370 L#9 19

20 Reactor Kinetics Batch Reactors Pg 129 ontinuous Flow ompletely Mixed (MFR) Pg Plug Flow (PFR) Pg Mixed Flow (non-ideal) David Reckhow EE 370 L#10 20

21 Plug Flow Reactors Like laminar flow through a pipe Examples Long, narrow Rivers Packed tower biofilters Drinking water distribution pipes David Reckhow EE 370 L#9 21

22 Plug Flow Reactors 0 Q 0 Q 0 Hydraulic Residence Time David Reckhow EE 370 L#10 22

23 PFR s (cont.) Minimal or no axial or longitudinal mixing s a slice of fluid progresses through the reactor, the reactants are converted to products. The reaction in the slice of fluid is analogous to the reaction in a batch reactor. The difference is that the fluid in this case is actually flowing through the reactor. The hydraulic residence time, θ, is the amount of time it takes the slice of fluid travel completely through the PFR. Thus, the mass balance equation for the PFR is: ccumulation rate dm dt Input rate - Output rate i i i i d V 0 0 rv dt k V O David Reckhow EE 370 L# onversion rate d dt e k kθ O e kt

24 Fluid Principles cont. Hydraulic Retention Time HRT θ V Q where, θ hydraulic retention time, [days] V volume, [m 3 ] of reactor or reactor segment Q volumetric flow rate, [m 3 /day] nd so it becomes: o e kv Q or: ln o k V Q k L u David Reckhow EE 370 L#10 24

25 PFR Example Disinfection (example) David Reckhow EE 370 L#10 25

26 PFR Example (cont.) David Reckhow EE 370 L#10 26

27 Batch Reactors General Reactor mass balance dm dt n n ( i Q ) i in ( j Q ) j out - r V i1 j1 Batch reactors are usually filled, allowed to react, then emptied for the next batch Fill & Draw Because there isn t any flow in a batch reactor: nd: d dt 1 V dm dt r - r Which for a 1 st order reaction is: d dt David Reckhow EE 370 L#10 27 k V

28 Batch Reactor Example wastewater contains contaminant "" with an initial concentration of 1200 mg/l. It is to be treated in a batch reactor. The reaction of to products is assumed to be first order. The rate constant, k, is 2.5/day. Determine the time required to convert 75 percent of to products. Plot the conversion of versus time for the first 10 days. r k So: d dt k t kt 0 ln ] ] o e o -kt David Reckhow EE 370 L#10 28

29 Batch Reactor Example (cont.) o t ln (1 - X) 1200 mg/ L x (1-0.75) 300 mg/ L t - k o 300 mg/ L ln 1200 mg/ L / day t 0.55 days David Reckhow EE 370 L#10 29

30 Batch Reactor Example (cont.) o -kt e (1 - X) X (1 - -kt e ) David Reckhow EE 370 L#10 30

31 General Reactor mass balance MFR (completely mixed) dm dt n i1 ( i Q i ) in n ( j1 j Q j ) out r V But with MFRs we have a single outlet concentration ( ) and usually a single inlet flow as well 0 Q 0 V Q 0 David Reckhow EE 370 L#10 31

32 MFR at SS nd at SS, dm /dt 0, so: where, Q o volumetric flow rate into and out of the reactor, [volume/time] o reactant concentration entering the reactor, [mass/volume] reactant concentration in the reactor and in the effluent, [mass/volume] V reactor volume, [volume] r V Q Q o o o nd if we define the hydraulic residence time: r ( o Q Q o o ) V θ V Q θ David Reckhow EE 370 L#10 32 r o

33 MFR with 1 st order reaction From the general equation We get: or k k o V Q o θ V Q r k o o θ o 1+ kθ V + k Q David Reckhow EE 370 L#10 33 o 1+ V o k Q

34 MFR SS Example #1 MFR is used to treat a wastewater that has a concentration of 800 mg/l. The hydraulic detention time of the wastewater in the reactor is 6 hours. The chemical reaction is an elementary irreversible second order reaction: 2 k Products The reaction constant, k, is 3.7 L/mg-day. Determine the conversion, X, for the process. David Reckhow EE 370 L#10 34

35 MFR SS Example #1 (cont.) The first step is to determine the reactor mass balance and the kinetic expressions. The reaction is second order irreversible so the kinetic expression is: r k 2 o The mass balance for a MFR is: nd: ( o - ) θ k David Reckhow EE 370 L# r ( - ) θ Rearranging, we obtain a quadratic equation with as the unknown: k θ o

36 MFR SS Example #1 (cont.) If the values for k, θ, and o are substituted into the relationship, the effluent concentration can be determined. It is 29 mg/l. The conversion can now be calculated: X ( o - ) o (800 mg/ L - 29 mg/ L) 800 mg/ L X 0.96 David Reckhow EE 370 L#10 36

37 To next lecture David Reckhow EE 370 L#9 37

CEE 370 Environmental Engineering Principles

CEE 370 Environmental Engineering Principles 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 &