chapter 3 material balance- part 1

8/13/2019 Chapter 3 Material Balance- Part 1

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Material Balance-Part 1

Chapter 3

Material and Energy Balances


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This chapter will present the methods for

organizing known information about processvariables, setting up a material balance

equations for individual process units and

multiple-unit processes and solving the

equations for unknown variables.


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Material Balance

Processclassification

Balances

Material balancecalculations

Balances onMultiple-Unit

Process

Recycle and Bypass

Material BalancesProblems with

Chemical reaction

Material BalanceProblems withoutChemical reaction

What are in this chapter?


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Topic Outcomes

At the end of Chapter 3, you should:

Briefly and clearly describe batch process,semibatch, continuous, transient.

Briefly and cearly describe recycle, purge,degree-of-freedom, fractional conversion of alimiting reactant, percentage excess of reactant,yield and selectivitiy, dry and wet basiscompositions, theoretical air and also percent

excess air in a combustion reaction.


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At the end of Chapter 3, you should: Draw and fully label a flowchart representing a

chemical process (single unit or multiple unit),choose a convenient basis, perform a degree-of-

freedom analysis, write in order the equations tobe used to calculate specified process variablesand subsequently solving for material balances.

Calculate the feed rate of air from a given percent

excess of air given a combustion reactor andinformation about the fuel composition.

Topic Outcomes-cont.


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At the end of Chapter 3, you should: Calculate the flow rate and composition of the

poduct gas given information about the conversionof the fuel and the absence or presence of CO in

the product gas.

Topic Outcomes-cont.


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Process Classification

3 type of chemical processes;1. Batch process

Feed is charge to the process and product isremoved when the process is completed

No mass is fed or removed from the processduring the operation

Used for small scale production Operate in unsteady state


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2. Continuous process

Input and output is continuously red andremove from the process

Operate in steady state Used for large scale production3. Semibatch process

Neither batch nor continuous

During the process a part of reactant can befed or a part of product can be removed.


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2 type of process operations;

1. Steady state

All the variables (i.e. temperatures, pressure,volume, flow rate, etc) do not change withtime

Minor fluctuation can be acceptable2. Unsteady state or transient

Process variable change with time


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Define type and operation of processes given:

A balloon is filled with air at steady rate of 2g/min

Semibatch and unsteady state A bottle of milk is taken from the refrigerator

and left on the kitchen

Batch and unsteady state

Water is boiled in open flaskSemibatch and unsteady state

Try This


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Balances

General Balance Equation

A balanceon a conserved quantity(total mass,

mass of a particular species, energy, momentum)in a system( a single process unit, a collection ofunits, or an entire process) may be written as:

INPUT + GENERATION OUTPUT CONSUMPTION= ACCUMULATION


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IN GROUP DISCUSSIONEq. (4.2-1) from pg 85

[INPUT + GENERATION OUTPUT CONSUMPTION = ACCUMULATION]

See Example 4.2-1


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Differential vs. Integral Balances

2 types of balances:

1. Differential balances

balances that indicate what is happening in asystem at an instant time.

balance equation is arate(rate of input, rateof generation, etc.) and has units of thebalanced quantity unit divided by a time unit(people/yr, g SO2/s).

usually applied to a continuous process.


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2. Integral balances

Balances that describe what happens betweentwo instants of time.

balance equation is an amountof thebalanced quantity and has the correspondingunit (people, g SO2).

usually applied to a batch process, with thetwo instants of time being the moment afterthe input takes place and the moment beforethe product is withdrawn.


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Simplified Rules for Material BalanceEquation:

If the balanced quantity is TOTAL MASS, setgeneration = 0 and consumption = 0. Mass canneither be created nor destroy.

If the balanced substances is a NONREACTIVESPECIES(neither a reactant nor a product), setgeneration = 0 and consumption = 0.

If a system is at STEADY STATE, setaccumulation = 0, regardless of what is beingbalanced.


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Balances on Continuous Steady-StateProcess

Steady state; accumulation = 0

INPUT + GENERATION = OUTPUT + CONSUMPTION

If balance on nonreactive species or on totalmass; generation = 0, accumulation = 0

INPUT = OUTPUT


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Balances on Batch Process

Ammonia is produced from nitrogen and hydrogen

in a batch reactor. At time t = 0 there are n0mol ofNH3in the reactor, and at a later time tfthereaction terminates and the contents of the reactor,which include nf ammonia, are withdrawn. Between

t0and tfno ammonia enters or leaves through thereactor boundaries.

From GBE; input=0 and output=0

GENERATION CONSUMPTION = ACCUMULATIONFor batch reactor:ACCUMULATION= INITIAL INPUT FINAL OUTPUT


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Final General Balance Equation for batch processes:

INITIAL INPUT + GENERATION = FINAL OUTPUT

CONSUMPTION


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Material Balance Calculation

In material balances problems, equationsinvolving input and output stream variables arederived.

Solving the equations is simple, but derivingthem from a description of process is consideredtricky.

Hence, procedure to derive a set of equationsthat can be solved for unknown variables will be

outlined.


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Flowcharts

When you are given process information andasked to determine something about the process,it is essential to organize the information in away that is convenient for subsequent

calculations. The best way to do this is to draw a flowchart

using boxes or other symbols to representprocess units (reactors, mixers, separation

units, etc.)

lines with arrows to represent inputs andoutputs.


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Flowcharts-continue

The flowchart of a process can help get materialbalance calculations started and keep themmoving.

Flowchart must be fully labeled when it is firstdrawn, with values of known process variablesand symbols for unknown variables being writtenfor each input and output stream.

Flowchart will functions as a scoreboard for theproblem solution: as each unknown variable isdetermined its value is filled in, so that theflowchart provides a continuous record of wherethe solution stands and what must still be done.


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Labeling a flowchart

2 suggestions for labeling flowchart:

1. Write the values and unitsof all known streamvariables at the locations of the streamson theflowchart.

For example, a stream containing 21 mole% O2and 79% N2at 320C and 1.4 atm flowing at arate of 400 mol/h might be labeled as:

400 mol/h

0.21 mol O2/mol0.79 mol N2/molT = 320C, P = 1.4 atm


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Process stream can be given in two ways:

a) As the total amountor flow rate of the streamand the fractionsof each component

b) Or directly as the amount or flow rateof eachcomponent.

60 kmol N2/min40 kmol O2/min

0.6 kmol N2/kmol0.4 kmol O2/kmol

100 kmol/min

3.0 lbm CH44.0 lbm C2H43.0 lbm C2H6

0.3 lbm CH4/lbm0.4 lbm C2H4/lbm0.3 lbm C2H6/lbm

10 lbm

Labeling a flowchart-continue


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2. Assign algebraic symbols to unknown streamvariables [such as m(kg solution/min),x(lbmN2/lbm), and n(kmol C3H8)] and write thesevariable names and their unitson the flowchart.

mol/h

0.21 mol O2/mol0.79 mol N2/molT = 320C, P = 1.4 atm

n 400 mol/h

ymol O2/mol(1-y)mol N2/molT = 320C, P = 1.4 atm



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If that the mass of stream 1 is half that of stream2, label the masses of these streams as mand2mrather than m1and m2.

If you know that mass fraction of nitrogen is 3times than oxygen, label mass fractions as

y g O2/gand 3y g N2/grather than y1and y2.

When labeling component mass fraction or molefraction, the last one must be 1 minus the sum ofthe others.

If volumetric flow rateof a stream is given, it isgenerally useful to label the mass or molar flowrate of this streamor to calculate it directly, sincebalance are not written on volumetric qualities.



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Consistent on Notation

gasinfractionmolesy

liquidinmoles)or(massfractioncomponentx

rateflowvolumeV

volumeVrateflowmolarn

molesn

rateflowmassm

massm


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Try This..

Two methanol-water mixture are contained inseparate flask. The first mixture contains 40wt%methanol and the second flask contains 70wt%

methanol. If 200g of the first mixture are going tobe mixed with 150g of the second in a mixing unit,what are the mass and composition of the productof the mixing unit?


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Drawing the flowchart

Mixing

200 g

150 g

mg0.40 g MeOH/kg0.60 g H2O/kg

0.70 g MeOH/kg0.30 g H2O/kg

xg MeOH/kg(1-x)g H2O/kg


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Solution

Batch process- unit in amount

Balance Equation for batch processes:INITIAL INPUT + GENERATION = FINAL OUTPUTCONSUMPTION

But NONREACTICEprocess; generation=0,consumption=0

INPUT=OUTPUT


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Balance on TOTAL MASS (Input=Output)

200 g + 150 g = m

m = 350 g

Balance on MeOH (Input=Output)

x= 0.529 g MeOH/g

(1-0.529) g H2O/g = 0.471 g H2O/g

200 g 0.40 g MeOH+ 150 g 0.70 g MeOH = 350 g x g MeOH

g

g

g

Solution-continue


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An experiment on the growth rate of certainorganism requires an environment of humid airenriched in oxygen. Three input streams are fedinto an evaporation chamber to produce an outputstream with the desired composition.

A: Liquid water fed at rate of 20 cm3/min

B: Air (21% O2and 79% N2)

C: Pure O2with a molar flow rate one-fifth of themolar flow rate of stream B

The output gas is analyzed and is found to contain1.5 mole% water. Draw and label the flowchart ofthe process, and calculate all unknown streamvariables.

Try This..


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Solution

CLASS DISCUSSION


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Evaporation

20 cm3H2O (l)/min

mol H2O/min2n

mol O2/min1200.0 n

mol air/min1n

mol/min3n

0.21 mol O2/mol

0.79 mol N2/mol

0.015 mol H2O/mol

y mol O2/mol(0.985-y) mol N2/mol

Solution


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TEST YOURSELF

Page 93


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Flowchart Scaling & Basis of Calculation

Flowchart scalingprocedure of changing thevalues of all stream amounts or flow rates by aproportional amount while leaving the streamcompositions unchanged. The process would still

be balance.

Scaling-upif final stream quantities are largerthan the original quantities.

Scaling downif final stream quantities aresmallerthan the original quantities.


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1 kg C6H6

300 lbm/h

1 kg C7H8

300 kg C6H6

300 kg C7H8

300 lbm/h

2 kg

0.5 kg C6H6/kg

0.5 kg C7H8/kg

600 kg

0.5 kg C6H6/kg

0.5 kg C7H8/kg

600 lbm/h0.5 lbmC6H6/lbm0.5 lbmC7H8/lbm

x 300

kg kg/hReplace kg with lbm



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Suppose you have balanced a process and theamount or flow rateof one of the processstreams is n1.You can scale the flow chart tomake the amount or flow rate of this stream n2

by multiplying all stream amounts or flow rate bythe ratio n2/n1.

You cannot, however, scale masses or mass flow

rates to molar quantities or vice versa by simplemultiplication; conversions of this type must becarried out using the methods as discussed inmass fraction and mol fraction section.


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Basis of Calculation

A basisof calculation is an amount (mass ormoles) of flow rate (mass or molar) of onestream or stream component in a process. Allunknown variables are determined to beconsistent with the basis.

If a stream amount or flow rate is giveninproblem, choose this quantity as a basis

If no stream amount or flow rate are known,assume one streamwith known composition. If

mass fraction is known, choose total mass ormass flow rate as basis. If mole fraction isknown, choose a total moles or molar flow rateas basis.


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CLASS DISCUSSION

Example 4.3-2


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Balancing a Process3.0 kg/min of benzene and 1.0 kg/min of toluene are mixed

Two unknown quantitiesmandx, associated withprocess, so two equations are neededto calculatethem.

For NONREACTIVE process, so input = output.

3 possible balance can be written Balance on totalmass, benzene, and toluene any two of whichprovide the equations needed to determine mandx.

m(kg/min)

x(kg C6H6/kg)

(1-x) (kg C7H8/kg)

3 kg C6H6/min

1 kg C7H8/min


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Balancing a Process-continue

2 rules for balancing nonreactive processes:

1. The maximum number of independent equationsthat can be derived by writing balances on anonreactive system equals the number of

chemical speciesin the input and outputstreams. (balances for substance and totalmass)

2. Write balancesfirstthat involve the fewest

unknownvariables.


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Total Mass Balance:

3.0 kg/min + 1.0 kg/min = m kg/min = 4.0 kg/min

Benzene Balance:

3.0 kg C6H6/min = 4.0 kg/min (x kg C6H6/kg)

x = 0.75 kg C6H6/kg

Solution


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TEST YOURSELF

Page 98


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Degree-of-Freedom Analysis

Before do any material balance calculation, use aproperly drawn and labeled flowchart todetermine whether there is enough informationto solve a given problem.

The procedure for doing so is referred to asdegree-of-freedom analysis.


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4 stepsto perform a degree-of-freedom analysis:

1. Draw and completely label a flowchart

2. Count the unknown variables on the chart

(n unknowns)1. Count the independent equations (n indep. eq.)

2. Find number of degree-of-freedom (ndf)

ndf= n unknowns- n indep. eq.

Degree-of-Freedom Analysis-continue


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3 possibilitiesfrom degree-of-freedom analysis:

1. If ndf = 0

the problem can in principle be solved.2. If ndf > 0

there are more unknowns than independentequations relating to them

at least ndfadditional variable values must bespecified before remaining variable values can

be determined. Either relations have been overlooked or the

problem is underspecified.


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3. If ndf < 0

there are more independent equations thanunknowns.

Either the flowchart is incompletely labeled or

the problem is overspecified with redundantand possibly inconsistent relations.

There is little point wasting time trying to solvematerial balance for n df> 0 or n df


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Sources of equations of unknown process variables: Material balances.

An energy balance.

Process specifications

4. Physical properties and laws

5. Physical constraints

6. Stoichiometric relations


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General Procedure for Single Unit Process

Material Balance Calculation9 steps for single unit material balance calculation:

1. Choose as basis of calculation an amount or flowrate of one of the process streams.

2. Draw a flowchart and fill in all unknownvariables values, including the basis ofcalculation. Then label unknown streamvariables on the chart.

3. Express what the problem statement asks youto determine in terms of the labeled variables.


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4. If you are given mixed mass and mole units fora stream (such as a total mass flow rate andcomponent mole fractions or vice versa),convert all quantities to one basis.

5. Do the degree-of-freedom analysis.

6. If the number of unknowns equals the numberof equations relating them (i.e., if the systemhas zero degree of freedom), write theequations in an efficient order (minimizingsimultaneous equations) and circle the variablesfor which you will solve.

General Procedure for Single Unit ProcessMaterial Balance Calculation-continue


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7. Solve the equations.

8. Calculate the quantities requested in theproblem statement if they have not alreadybeen calculated.

9. If a stream quantity or flow rate ngwas given inthe problem statement and another value ncwas either chosen as a basis or calculated forthis stream, scale the balanced process by the

ratio ng/ncto obtain the final result.

General Procedure for Single Unit Process Material

Balance Calculation-continue


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Material Balance on a Mixing Unit

An aqueous solution of sodium hydroxide contains

20.0% NaOH by mass. It is desired to produce and8.0% NaOH solution by diluting a stream of the

20% solution with a stream of pure water.

Calculate the ratios (liters H2O/kg feed solution)

and (kg product solution/kg feed solution).

Material Balance Problems WithoutChemical Reactions


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9 steps for single unit material balance calculation

1. Choose as basis of calculation an amount or flowrate of one of the process streams.

2. Draw and label the flowchart

Choose a basis of 100kg of the 20% feedsolution.

Mixing100 kg m2kg

0.20 kg NaOH/kg0.80 kg H2O/kg

0.080 kg NaOH/kg0.920 kg H2O/kg

m1(kg H2O)V1 (liters H2O)

3 E h t th bl t t t k


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3. Express what the problem statement asks youto determine in terms of the labeled variables.

The desired quantities are V1/100 (liters H2O/kg

feed solution) and m2/100 (kg product/kg feedsolution). Our task is therefore to calculate thevariables V1and m2.

4. Convert unit into one basis, 5. Do degree-of-

freedom analysis. count unknown variables andequations relating them. 6. Write the balancesequations in an efficient order.

Unknowns:3 unknowns- m1, m2and V1.

Equations:nonreactive process, as there are 2species, can write 2 balances. As 3 balances areneeded, another solution for V1will be takenfrom the information of density of water.


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7. Outline the solution and 8. Calculate theproperties of unknown variables.

Balances for the system:

INPUT=OUTPUT

NaOH Balance:

(0.2 kgNaOH/kg)(100kg)=(0.008 kg NaOH/kg)m2m2= 250 kg NaOH

Total Mass Balance:

100kg+m1=m2(m2=250 kg NaOH)m1=150 kg NaOH


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7. Outline the solution and 8. Calculate theproperties of unknown variables.

Calculate the volume of water V1=

V1=

= 150 L

Ratios

V1/100 kg= 1.5 L H2O/ kg feed solutionm2/100 kg= 2.5 kg product /kg feed solution

150 kg 1 L

kg


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Material Balance on a Condenser

CLASS DISCUSSIONExample 4.3-4


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Material Balance on a Distillation Column

CLASS DISCUSSION

Example 4.3-5

chapter 3 material balance- part 1

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