Chapter 3. Material Balances and Separations (1)

7/23/2019 Chapter 3. Material Balances and Separations (1)

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Dr. BUNRITH SENG

Chapter 3

Material Balances and Separations

Mobile : +81 (0) 80 3259 9952

E-mail: [email protected]; [email protected]

Department of Civil Engineering, Zaman University

No. 8, St. 315, 12151 Phnom Penh, Cambodia

Zaman UniversityDepartment of Civil Engineering

No. 8, St. 315, 12151 Phnom Penh, Cambodia


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2

Material Balances and Separations

1. Material balances with a single material

2. Material balances with multiple materials

3. Material balances with reactor


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3

Material Balances with a Single Material

One Inflow and One Outflow

0X

The balance can be expressed in mass and volume.

Mass balance for Solids

Volume balance for Liquids

1X0 1

Influent Effluent

OUT

XoftimeunitperMass

IN

XoftimeunitperMass

10 XX


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Material Balances with a Single Material (Cont.)

One Inflow and Two Outflows

0X1X

0

1

2X2

OUT

Xoftime

unitperMass

IN

Xoftime

unitperMass

210 XX X

n

1i

i0 XX

The material X can be separated into more than two fractions, so

the material balance can be:

n : number of exit streamsof effluents


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Example:A city generates 102 tons/day of refuse, all of which goes

to a transfer station. At the transfer station the refuse is split intofour flow streams headed for three incinerators and one landfill. Ifthe capacity of the incinerators is 20, 50, and 22 tons/day, how muchrefuse must go to the landfill?

Solution:

OUTMassINMass

M225020102

tons/day10M



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Several Inflows and One Outflow

mX

1X

m

0eX

1

2X 2 eX

m

1i

iX



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Example: A trunk sewer, shown in bellowed figure has a flow

capacity of 4.0 m3

/s. If the flow to the sewer is exceeded, it will notbe able to transmit all the sewage through the pipe, and backups willoccur. Currently, three neighborhoods contribute to the sewer, andtheir maximum (peak) flows are 1.0, 0.5, and 2.7 m3/s. A builder wantsto construct a development that will contribute a maximum flow of 0.7m3/s to the trunk sewer. Would this cause the sewer to exceed itscapacity?

OUTeVolume/timINeVolume/tim eX 7.07.25.00.1

sXe /m9.43

Solution:



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Complex processes with a single material

inX ou tX

Reaction

The flow is not constant

Reaction taken place

- Some material produced

- Some material consumed

CONSUMED

Mass/time

PRODUCED

Mass/time

OUT

Mass/time

IN

Mass/time

SDACCUMULATE

Mass/time

CONSUMED

AofRate

PRODUCED

AofRate

OUT

ARate

IN

AofRate

SDACCUMULATE

AofRate



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At steady state condition (Constant flow) with reaction

CONSUMED

AofRate

PRODUCED

AofRate

OUT

ARate

IN

AofRate0

At steady state condition (Constant flow) with no reaction

00OUTARate

IN

AofRate

0

OUT

ARate

IN

AofRate



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Example: A sewer carrying stormwater to Manhole 1 has a constant

flow of 2000 L/min (QA). At Manhole1 it receives a constant lateral flowof 100 L/min (QB). What is the flowto Manhole 2 (QC)?

Solution:

CONSUMEDAofRate

PRODUCEDAofRate

OUTARate

INAofRate

SDACCUMULATEAofRate

00QQ0 A CBQ

L/min2100)1002000( CQ



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Make the process into the black box



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What is the black boxof material balance?

Qp = 82x108 L/yr



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Example:Suppose the rainfall is 102 cm/yr, of which 50% percolatesinto the ground. The farmer irrigates crops using well water. Of the

extracted well water, 80% is lost by evapotranspiration; the remainderpercolates back into the ground. How much groundwater could a farmeron a 8 km2 farm extract from the ground per year without depletingthe groundwater reservoir volume?

Check Your Answer

at Home!

Chapter 4, page 118



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Step by Step Material Balance Procedure

1. Draw the system as a diagram, including all flows (inputs and outputs) asarrows.2. Add the available information, such as flow rates and concentrations. Assign

symbols to unknown variables.3. Draw a continuous dashed line around the component or components that are

to be balanced. This could be a unit operation, a junction, or a combination of

these. Everything inside this dashed line becomes the black box.4. Decide what material is to be balanced. This could be a volumetric or massflow rate.

5. Write the material balance equation by starting with the basic equation:

CONSUMED

RateVolume

orMass

PRODUCED

RateVolume

orMass

OUT

RateVolume

orMass

IN

RateVolume

orMass

DACCUMULATE

RateVolume

orMass

6. If only one variable is unknown, solve for that variable.7. If more than one variable is unknown, repeat the procedure, using a

different black box or a different material for the same black box.


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Material balances with Multiple Materials

Mixing multiple material flow streams

CONSUMED

Rate

Volume

orMass

PRODUCED

Rate

Volume

orMass

OUT

Rate

Volume

orMass

IN

Rate

Volume

orMass

DACCUMULATE

Rate

Volume

orMass

Example 1: The Allegheny and Monongahela Rivers meet atPittsburgh to form the mighty Ohio. The Allegheny, flowing souththrough forests and small towns, runs at an average flow of 10 m3/s

and has a low silt load, 250 mg/L. The Monongahela, on the other hand,flows north at a flow of 13 m3/s through old steel towns and poor farmcountry, carrying a silt load of 1500 mg/L.

a. What is the average flow in the Ohio River?

b. What is the silt concentration in the Ohio?


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Material balances with Multiple Materials (Cont.)

Allegheny River

QA = 10 m3

/sCA= 250 mg/L

PITTSBURGH

Monongahela River

QM = 13 m3/s

CM= 1,500 mg/L

Ohio River

QO = ?CO= ?

Step 1.Draw the system. The figure shows

the confluence of the rivers with theflows identified.

Allegheny RiverQA = 10 m3/s

CA= 250 mg/s

Monongahela River

QM = 13 m3/s

CM= 1,500 mg/s

Ohio River

QO = ?

CO= ?

Step 2. All the available informationis added to the sketch,

including the known andunknown variables.

Step 3. The confluence of the riversis the black box, as shown by

the dashed line.

Step 4. Water flow is to be balancedfirst.


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Step 5.Write the balance equation:

CONSUMED

waterofRate

PRODUCED

waterofRate

OUT

waterofRate

IN

waterofRate

DACCUMULATE

waterofRate

Step 6.Solve for the unknown

00OUT

Water

IN

Water0

00Q13100 O

s/m23Q 3O

Step 6.Repeat the procedure for the Mass Balance of silt

L/mg969CO


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Example 2:Suppose the sewers shown in figure below have QB = 0and QAand QCare unknowns. By sampling the flow at the first manhole,it is found that the concentration of dissolved solids in the flow cominginto Manhole 1 is 50 mg/L. An additional flow, QB= 100 L/min, is addedto Manhole 1, and this flow contains 20% dissolved solids. (Recall that 1%= 10,000 mg/L.) The flow through Manhole 2 is sampled and found to contain1000 mg/L dissolved solids. What is the flow rate of wastewater in the sewer

(QA)?

(C )


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Solution:

Step 1.Draw the diagram.

Step 2.Add all information, includingconcentrations.

Step 3.Manhole 1 is the black box.

Step 4.What is to be balanced? If the flowsare balanced, there are two unknowns.Can something else be balanced?Suppose a balance is run in terms of thesolids?

Step 5.Write the material balance for solids.

CONSUMED

Solids

PRODUCED

Solids

OUT

Solids

IN

Solids

DACCUMULATE

Solids

00QQ0 CA CBBA CCQC

M i l b l i h M l i l M i l (C )


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)1000(Q)000,200)(

min

100()50(Q0 CA

L

mg

L

mgL

L

mg

Step 6.Solve for unknown. (Skip)

Step 7.Balance of the Volume Flow Rate

CONSUMED

Volume

PRODUCED

Volume

OUT

Volume

IN

Volume

DACCUMULATE

Volume

00QQ0 CA BQ

Q100Q0 CA

(a)

(b)

(a) & (b)QC = 21.05 L/minand QA = 20.95 L/min


M i l B l i h M l i l M i l (C )


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Example 3: Estimate the concentration of SO2 in the urban airabove the city of St. Louis if the mixing height above the city is 1210m, the width of the box perpendicular to the wind is 105m, the averageannual wind speed is 15,400 m/hr, and the amount of sulfur dioxidedischarged is 6242 108g/yr.

6242 x 108g/yr

Material Balances with Multiple Materials (Cont.)

C = 38g/m3

Answer



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Separating multiple-material flow streams



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Recovery

1000

1

1

x

xRx

1000

1

1

y

yR

y

Purity

10011

11 yx

xPx

100

22

2

2

yx

yPy


Efficiency

100

2/1

0

2

0

1

y

y

x

xEWS

1000

1

0

1

y

y

x

xER

EWS: Worrell-Stessel efficiency

ER: Rietema efficiency

Binary Separator

M t i l B l ith M lti l M t i l (C t )


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Example:Assume that an aluminum can separator in a local recyclingplant processes 400 cans/min. The two product streams consist of thefollowing:

Total in Feed Product Stream 1 Product Stream 2

Aluminum cans/min 300 270 30

Steel cans/min 100 0 100

Calculate the recovery of aluminum cans and the purity of the product.

Solution: )30270(300OUTIN0

00 OK

%90100300

270R

1cansAl

%100100

0270

270P

1cansAl

High recovery

Perfect purity



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Example:Suppose a thickener in a metal plating plant receives a feedof 40 m3/hr of precipitated metal plating waste with a suspendedsolids concentration of 5000 mg/L. If the thickener is operated in asteady state mode so that 30 m3/hr of flow exits as the overflow, andthis overflow has a solids concentration of 25 mg/L, what is theunderflow solids concentration, and what is the recovery of the solidsin the underflow?

Solution:

M t i l b l ith M lti l M t i l (C t )


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CONSUMEDVolume

PRODUCEDVolume

OUTVolume

INVolume

DACCUMULATEVolume

00)30(400 uQ

hrQu /m103

Volume balance

Mass balance (Solids)

00)(0 OOuuii QCQCQC

)/mmg/L)(3025(/m10.)/m40(/50000 333 hrhrChrLmg u

mg/L900,19uC




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Recovery of solids

100C u

ii

uu

QC

QR

%5.99/40m5,000mg/L

100/10/900,19

3

3

hr

hrmLmg

Ru

99.5% of solids are drained through the underflow




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Example: Calculate the efficiencies for the following data, using theWorrell-Stessel and Rietema formulas.

Feed

(tons/day)

Organics/Inorganics

Product Stream 1

(tons/day)

Organics/Inorganics

Product Stream 2

(tons/day)

Organics/Inorganics

Air Classifier 1 80/20 72/6 8/14

Air Classifier 2 80/20 76/8 4/12

Solution:

%7910020

14

80

722/1

WSE

%60100

20

6

80

72

RE

Air classifier 1 Air classifier 1

%7510020

12

80

762/1

WSE

%55100

20

8

80

76

RE



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Polynary Separator

10010

11

11

x

xRx

Recovery

Purity100

12111

11

11

n

xxxx

xP

Efficiency

100/1

030

33

20

22

10

11

n

n

nnWS

x

x

x

x

x

x

x

xE

1000

1

30

13

20

12

10

11

1

n

nR

x

x

x

x

x

x

x

xE

M t i l b l ith M lti l M t i l (C t )


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Complex processes with multiple materials

Q0=4 L/min

Material Balances ith Reactors


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Material Balances with Reactors

Example: An activated sludge system has an influent (feed) of440L/s at a suspended solids concentration of 50 mg/L. The wasteactivated sludge flow rate is 9 L/s at a solids concentration of 1.2%.The effluent (discharge) has a solids concentration of 20 mg/L. Whatis the yield of waste activated sludge in kg per day, or in other words,what is the rate of solids production in the system? Assume steadystate.

Q1=440 L/min

Material Balances with Reactors (Cont )


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Solution:

Q1=440 L/s

QW=9 L/s

Material Balances with Reactors (Cont.)

1st

Balance: Volume 0094400 EQL/s431EQ

2stBalance: Suspended solids

CONSUMED

Solids

PRODUCED

Solids

OUT

Solids

IN

Solids

DACCUMULATE

Solids

0XQQ0 E11 WWE CQCC X: Rate of solidsproduced in thereactor

X000,12920431504040

kg/d8,200mg/s94620X


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Chapter 3. Material Balances and Separations (1)

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