Materials Balances with Multiple Materials

Only one equation can be written for each black box unless there is more than one material in the flowExample

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?

Draw a diagram of the system

Add all available informationIncluding the unknown variables

Draw the black box representation

Rate of Water

accumulated

Rate ofWater In

Rate ofWater Out

Rate ofWater

Generation= - +

0 = [340 + 460] – Q0 +0 Q0 = 800 cfs

Mass Balance for Water

Our book calls this a volume balance

Mass Balance for Silt

Rate of Silt

accumulated

Rate ofSilt In

Rate ofSilt Out

Rate ofSilt

Generation= - +

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.

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

Rate of Solids

accumulated

Rate ofSolids In

Rate ofSolids Out

Rate ofSolids

Generation= - +

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 Solids

Mass Balance on Water

0 = [QA + QB] - QC

0 = QA + 100 - QC Therefore: QA = QC – 100Substitute into the equation above

Rate of Water

accumulated

Rate ofWater In

Rate ofWater Out

Rate ofWater

Generation= - +

Now

50(QC – 100)) + 200,000(100) = 1000 QC

QC = 21,047 L/min

So:QA = 21,047 – 100 = 20,947 L/min

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

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)

Rate of SO2

accumulated

Rate ofSO2 In

Rate ofSO2 Out

Rate ofSO2

Generation

= - +

0 = 7.126 x 1013 – [CSO2 out (1.86 x 1012)]

CSO2 out = 38 g/m3

SO2

Where does SO2 come from?

Donora, PA - 1948

Geneva Steel?

SeparatorsThe 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.

Separators are not perfect. There are two measures used to describe how well a separator works: recovery and purity

x0, y0

x1, y1

x2, y2

Recovery

Rx1 = x1/x0 x 100

Ry2 = y2/y0 x 100

Purity

Px1 = x1/(x1 + y 1) x 100

ExampleConsider 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?

Rate of Water

accumulated

Rate ofWater In

Rate ofWater Out

Rate ofWater

Generation= - +

0 = 40 –(30 + QU) + 0

QU = 10 m3/hr

Rate of Solids

accumulated

Rate ofSolids In

Rate ofSolids Out

Rate ofSolids

Generation= - +

0 = (CiQi )– [CUQU + C0Q0] + 0

0 = (5000)(40) – [(CU(10) + (25)(30)]

CU = 19,900 mg/L

Recovery

RU = (CUQU)/(CiQi) x 100

RU = [(19,900)(10) x 100]/[(5000)(40)] x 100 = 99.5%

Other Effectiveness Measures

Worell-Stessel Effectiveness

EWS = [(x1/x0) x (y2/y0)] ½ x 100

Rietema Effectiveness

ER = [(x1/x0) – (y1/y0)] x 100

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

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

ExampleConsider 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.

Rate of Solids

accumulated

Rate ofSolids In

Rate ofSolids Out

Rate ofSolids

Generation= - +

0 = QA –[QC + QK] + 0

Rate of Volume

accumulated

Rate ofVolume In

Rate ofVolume Out

Rate ofVolume

Generation= - +

0 = (QACA) – [(QKCK) + (QCCC)] + 0

0 = QA(4) – QK(20) – QC(0.1)

Two equations, three unknowns

QA = QC + QK

By substitution: QC = 0.804 QA, and QK = 0.196 QA

Rate of Volume

accumulated

Rate ofVolume In

Rate ofVolume Out

Rate ofVolume

Generation= - +

0 = [QB + QK] – QE + 0

Rate of Solids

accumulated

Rate ofSolids In

Rate ofSolids Out

Rate ofSolids

Generation= - +

0 = [QBCB + QKCK] – (QECE) + 0

0 = QB(4) + QK(20) – QE(10) (2)

QE = QB + QK (1)

Substitute (1) into (2) QK = 0.6 QB Since, QK = 0.196 QA 0.6 QB = 0.196 QA

Or QB = 0.327 QA

Rate of Volume

accumulated

Rate ofVolume In

Rate ofVolume Out

Rate ofVolume

Generation= - +

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

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

Rate of Volume

accumulated

Rate ofVolume In

Rate ofVolume Out

Rate ofVolume

Generation= - +

0 = Q0 – QC - QE

0 = 1 – 0.605 –0.393 = 0.002 close enough