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Materials Balances with Multiple Materials

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

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Materials Balances with Multiple Materials

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  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: • What is the average flow of the Ohio River at Pittsburgh? • What is the silt concentration in the Ohio River?

  2. Rate of Water Generation Rate of Water accumulated Rate of Water In Rate of Water Out = - + Draw a diagram of the system Add all available information Including the unknown variables Draw the black box representation Mass Balance for Water Our book calls this a volume balance 0 = [340 + 460] – Q0 +0 Q0 = 800 cfs

  3. Rate of Silt Generation Rate of Silt accumulated Rate of Silt In Rate of Silt Out = - + Mass Balance for Silt 0 = [CAQA +CMQM] – C0Q0 + 0 0 = [(250 x 340) + (1500 x 4600)] – [C0 x 800] C0 = 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 QB = 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 (QA)? Draw a picture and the black box

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

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

  7. Air Pollution – “Box or Bubble” Model Estimate the concentration of SO2 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 SO2 discharged within the city limits is 1375 x 106 Lbs/yr. 1375 x 106 Lbs/yr = 7.126 x 1013: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) Qair = (1210 x 100,000) x 15,400 = 1.86 x 1012 m3/hr

  8. Rate of SO2 Generation Rate of SO2 accumulated Rate of SO2 In Rate of SO2 Out = - + Obviously: Qair in = Qair out = 1.86 x 1012 m3/hr (volumetric flow) and, QSO2 in = QSO2 out = 7.126 x 1013:g/hr (mass flow) 0 = 7.126 x 1013 – [CSO2 out (1.86 x 1012)] CSO2 out = 38 :g/m3

  9. SO2 Where does SO2 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. x1, y1 x0, y0 x2, y2 Separators are not perfect. There are two measures used to describe how well a separator works: recovery and purity Recovery Rx1 = x1/x0 x 100 Ry2 = y2/y0 x 100 Purity Px1 = x1/(x1 + 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 m3/hr and has a suspended solids concentration of 5000 mg/L. If the thickener is operated at steady-state so that 30 m3/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 Solids Generation Rate of Water Generation Rate of Water accumulated Rate of Solids accumulated Rate of Water In Rate of Solids In Rate of Water Out Rate of Solids Out = = - - + + 0 = 40 –(30 + QU) + 0 QU = 10 m3/hr 0 = (CiQi )– [CUQU + C0Q0] + 0 0 = (5000)(40) – [(CU(10) + (25)(30)] CU = 19,900 mg/L

  14. Recovery RU = (CUQU)/(CiQi) x 100 RU = [(19,900)(10) x 100]/[(5000)(40)] x 100 = 99.5%

  15. Other Effectiveness Measures Worell-Stessel Effectiveness EWS = [(x1/x0) x (y2/y0)] ½ x 100 Rietema Effectiveness ER = [(x1/x0) – (y1/y0)]x 100

  16. Multiple Materials These methods can be extended to multiple material (polynary) systems as well as two Component (binary) systems Px11 = x11/(x11 + x21 + x 31 + … + xn1) x 100 Where xij = amount of material i in effluent stream j

  17. This can also be extended to the measures of effectiveness: EWS = [(x11/x10)(x22/x20)(x33/x30) … (xnn/xn0)] x 100 ER1 = [(x11/x10) – (x12/x20) – (x13/x30) - . . . – (x1n/xn0)] 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/cm3.) 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 Volume Generation Rate of Solids Generation Rate of Solids accumulated Rate of Volume accumulated Rate of Volume In Rate of Solids In Rate of Volume Out Rate of Solids Out = = - - + + 0 = QA –[QC + QK] + 0 QA = QC + QK 0 = (QACA) – [(QKCK) + (QCCC)] + 0 0 = QA(4) – QK(20) – QC(0.1) By substitution: QC = 0.804 QA, and QK = 0.196 QA Two equations, three unknowns

  20. Rate of Solids Generation Rate of Volume Generation Rate of Solids accumulated Rate of Volume accumulated Rate of Volume In Rate of Solids In Rate of Volume Out Rate of Solids Out = = - - + + 0 = [QB + QK] – QE + 0 QE = QB + QK (1) 0 = [QBCB + QKCK] – (QECE) + 0 0 = QB(4) + QK(20) – QE(10) (2) Substitute (1) into (2) QK = 0.6 QB Since, QK = 0.196 QA 0.6 QB = 0.196 QA Or QB = 0.327 QA

  21. Rate of Volume Generation Rate of Volume accumulated Rate of Volume In Rate of Volume Out = - + 0 = Q0 – [QB + QA] + 0 Substituting QB = 0.327 QA and Q0 = 1 gal/min 0 = 1 – 0.327 QA - QA QA = 0.753 gal/min QB = 0.327 (QA) = 0.327 ( 0.753) = 0.246 gal/min QC = 0.804(QA)= 0.804 (0.753) = 0.605 gal/min QK = 0.196 (QA) = 0.196 (0.753) = 0.147 gal/min

  22. Rate of Volume Generation Rate of Volume accumulated Rate of Volume In Rate of Volume Out = - + We also know from the blender box that: 0 = QK + QB – QE or QE = QB + QK QE = 0.246 + 0.147 = 0.393 gal/min We can check our answer using a volume balance around the entire system 0 = Q0 – QC - QE 0 = 1 – 0.605 –0.393 = 0.002 close enough

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