1. Materials Balances with Multiple Materials Only one equation can be written for each black box unless there is more than one material in the flow Example The Allegheny and Monongahela Rivers join in Pittsburgh to form the Ohio River. The Allegheny has an average flow of 340 cfs and a silt load of 250 mg/L. The Monongahela has a flow of 460 cfs and a silt load of 1500 mg/L: A.What is the average flow of the Ohio River at Pittsburgh? B.What is the silt concentration in the Ohio River?

2. Draw a diagram of the system Add all available information Including the unknown variables Draw the black box representation Rate of Water accumulated Rate of Water In Rate of Water Out Rate of Water Generation =-+ 0 = [340 + 460] – Q 0 +0 Q 0 = 800 cfs Mass Balance for Water Our book calls this a volume balance

3. Mass Balance for Silt Rate of Silt accumulated Rate of Silt In Rate of Silt Out Rate of Silt Generation =-+ 0 = [C A Q A + C M Q M ] – C 0 Q 0 + 0 0 = [(250 x 340) + (1500 x 4600)] – [C 0 x 800] C 0 = 969 mg/L In this example there was only one known, which makes this a trivial problem. Usually, this is not the case and it is a little harder to get the answer.

4. Example Consider a sewer system where flows A and B enter manhole 1 and leave after they are combined into flow C. Flow C then enters manhole C. Sampling has told us that Q B = 100 L/min and the dissolved solids Concentration in flows A, B, and C are 50 mg/L, 20%, and 1000 mg/L, Respectively. What is the flow coming into manhole 1 (Q A )? Draw a picture and the black box

5. Rate of Solids accumulated Rate of Solids In Rate of Solids Out Rate of Solids Generation =-+ 0 = [Q A C A + Q B C B ] – Q C C C + 0 0 = [Q A (50 mg/L) + (100 L/min)(200,000 mg/L)] – [Q C (1000 mg/L)] Two unknowns, so we need to do another balance Mass Balance on Solids Mass Balance on Water 0 = [Q A + Q B ] - Q C 0 = Q A + 100 - Q C Therefore: Q A = Q C – 100 Substitute into the equation above Rate of Water accumulated Rate of Water In Rate of Water Out Rate of Water Generation =-+

6. Now 50(Q C – 100)) + 200,000(100) = 1000 Q C Q C = 21,047 L/min So: Q A = 21,047 – 100 = 20,947 L/min

7. Air Pollution – “Box or Bubble” Model Estimate the concentration of SO 2 in the urban air above the city of St. Louis if the mixing zone is 1210 m deep, and the city is 100,000 m wide in a direction perpendicular to the wind direction. The average wind speed is 15,400 m/hr, and the amount of SO 2 discharged within the city limits is 1375 x 10 6 Lbs/yr. 1375 x 10 6 Lbs/yr = 7.126 x 10 13 : g/hr The volume of air moving into the box is equal to the wind velocity times the cross sectional area through which it flows (Q = Av) Q air = (1210 x 100,000) x 15,400 = 1.86 x 10 12 m 3 /hr

8. Obviously: Q air in = Q air out = 1.86 x 10 12 m 3 /hr (volumetric flow) and, Q SO2 in = Q SO2 out = 7.126 x 10 13 : g/hr (mass flow) Rate of SO 2 accumulated Rate of SO 2 In Rate of SO 2 Out Rate of SO 2 Generation =-+ 0 = 7.126 x 10 13 – [C SO2 out (1.86 x 10 12 )] C SO2 out = 38 : g/m 3

9. SO 2 Where does SO 2 come from? Donora, PA - 1948 Geneva Steel?

10. Separators The purpose is to separate components in a flow stream by using some property of the components. Consider the example of separating different colors or glass in a waste stream.

11. Separators are not perfect. There are two measures used to describe how well a separator works: recovery and purity x 0, y 0 x 1, y 1 x 2, y 2 Recovery R x1 = x 1 /x 0 x 100 R y2 = y 2 /y 0 x 100 Purity P x1 = x 1 /(x 1 + y 1 ) x 100

12. Example Consider the waste water thickener shown below. The flow into the thickener is called the influent. The low-solids effluent stream is called the overflow. The high-solids stream, which leaves the bottom of the thickener, is called the underflow. Suppose the influent flow to a thickener is 40 m 3 /hr and has a suspended solids concentration of 5000 mg/L. If the thickener is operated at steady-state so that 30 m 3 /hr exits the overflow with a solids concentration of 25 mg/L, what is the underflow solids concentration and what is the recovery of solids in the underflow?

13. Rate of Water accumulated Rate of Water In Rate of Water Out Rate of Water Generation =-+ 0 = 40 –(30 + Q U ) + 0 Q U = 10 m 3 /hr Rate of Solids accumulated Rate of Solids In Rate of Solids Out Rate of Solids Generation =-+ 0 = (C i Q i )– [C U Q U + C 0 Q 0 ] + 0 0 = (5000)(40) – [(C U (10) + (25)(30)] C U = 19,900 mg/L

14. Recovery R U = (C U Q U )/(C i Q i ) x 100 R U = [(19,900)(10) x 100]/[(5000)(40)] x 100 = 99.5%

15. Other Effectiveness Measures Worell-Stessel Effectiveness E WS = [(x 1 /x 0 ) x (y 2 /y 0 )] ½ x 100 Rietema Effectiveness E R = [(x 1 /x 0 ) – (y 1 /y 0 )] x 100

16. Multiple Materials These methods can be extended to multiple material (polynary) systems as well as two Component (binary) systems P x11 = x 11 /(x 11 + x 21 + x 31 + … + x n1 ) x 100 Where x ij = amount of material i in effluent stream j

17. This can also be extended to the measures of effectiveness: E WS = [(x 11 /x 10 )(x 22 /x 20 )(x 33 /x 30 ) … (x nn /x n0 )] x 100 E R1 = [(x 11 /x 10 ) – (x 12 /x 20 ) – (x 13 /x 30 ) -... – (x 1n /x n0 )] x 100

18. Example Consider a system in which sludge from a wastewater treatment plant is thickened by a centrifuge. A sludge with 4% solids is to be thickened to 10% solids. The centrifuge produces a sludge with 20% solids from the 4% feed. This means some of the sludge can by-pass the centrifuge and be mixed with the thickened sludge to produce the desired solids concentration. The question is: how much do you by-pass? The influent flow to the centrifuge is 1 gal/min. (We can assume the specific gravity of sludge solids is 1.0 g/cm 3.) The centrate from the centrifuge (the effluent stream with low solids concentration) is 0.1% solids. The cake (the high solids effluent) is 20% solids. Find the required flow rates.

19. Rate of Solids accumulated Rate of Solids In Rate of Solids Out Rate of Solids Generation =-+ 0 = Q A –[Q C + Q K ] + 0 Rate of Volume accumulated Rate of Volume In Rate of Volume Out Rate of Volume Generation =-+ 0 = (Q A C A ) – [(Q K C K ) + (Q C C C )] + 0 0 = Q A (4) – Q K (20) – Q C (0.1) Two equations, three unknowns Q A = Q C + Q K By substitution: Q C = 0.804 Q A, and Q K = 0.196 Q A

20. Rate of Volume accumulated Rate of Volume In Rate of Volume Out Rate of Volume Generation =-+ 0 = [Q B + Q K ] – Q E + 0 Rate of Solids accumulated Rate of Solids In Rate of Solids Out Rate of Solids Generation =-+ 0 = [Q B C B + Q K C K ] – (Q E C E ) + 0 0 = Q B (4) + Q K (20) – Q E (10) (2) Q E = Q B + Q K (1) Substitute (1) into (2)Q K = 0.6 Q B Since, Q K = 0.196 Q A 0.6 Q B = 0.196 Q A Or Q B = 0.327 Q A

21. Rate of Volume accumulated Rate of Volume In Rate of Volume Out Rate of Volume Generation =-+ 0 = Q 0 – [Q B + Q A ] + 0 Substituting Q B = 0.327 Q A and Q 0 = 1 gal/min 0 = 1 – 0.327 Q A - Q A Q A = 0.753 gal/min Q B = 0.327 (Q A ) = 0.327 ( 0.753) = 0.246 gal/min Q C = 0.804(Q A ) = 0.804 (0.753) = 0.605 gal/min Q K = 0.196 (Q A ) = 0.196 (0.753) = 0.147 gal/min

22. We also know from the blender box that: 0 = Q K + Q B – Q E or Q E = Q B + Q K Q E = 0.246 + 0.147 = 0.393 gal/min We can check our answer using a volume balance around the entire system Rate of Volume accumulated Rate of Volume In Rate of Volume Out Rate of Volume Generation =-+ 0 = Q 0 – Q C - Q E 0 = 1 – 0.605 –0.393 = 0.002 close enough