LECTURE 4MASS BALANCE

The accounting of all mass in an industrial chemical process is referred to as a mass (or material) balance.

USES ‘day to day’ operation of process for

monitoring operating efficiency Making calculations for design and

development of a process i.e. quantities required, sizing equipment, number of items of equipment

SAMPLE PROBLEM – BATCH MIXING PROCESS

200 kg of a 40% w/w methanol/water solution is mixed with 100 kg of a 70% w/w methanol/water solution in a batch mixer unit.

What is the final quantity and composition?

SAMPLE PROBLEM:

Total initial mass = total final mass = 300 kg

Initial methanol mass = final methanol mass

80 + 70 = final methanol mass = 150 kgTherefore final composition of batch is

(150/300) x 100 = 50 % by wt.

SAMPLE PROBLEM 2

1000 kg of 8% by wt. sodium hydroxide (NaOH) solution is required. 20% sodium hydroxide solution in water and pure water are available. How much of each is required?

BATCH PROCESSES

Batch processes operate to a batch cycle and are non-steady state. Materials are added to a vessel in one operation and then process is carried out and batch cycle repeated. Integral balances are carried out on batch processes where balances are carried out on the initial and final states of the system.

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BATCH CYCLE

Sequence of operations/steps repeated according to a cycle

Batch cycle time Batch size

SIMPLE BATCH REACTION CYCLE

3 steps

Start cycle t=0 t, finish cycle

Add reactants etc reaction Empty reactor

Next cycle

CONTINUOUS PROCESSES

These processes are continuous in nature and operate in steady state and balances are carried out over a fixed period of time. Materials enter and leave process continuously.

LAW OF CONSERVATION OF MASS

When there is no net accumulation or depletion of mass in a system (steady state) then:

Total mass entering system = total mass leaving system

or total mass at start = total final mass

GENERAL MASS BALANCE EQUATION

Input + generation – output – consumption = accumulation

Notes: 1. generation and consumption terms refer only to generation of products and consumption of reactants as a result of chemical reaction. If there is no chemical reaction then these terms are zero.

2. Apply to a system3. Apply to total mass and component mass

DEFINITIONS System – arbritary part or whole of a

system Steady state/non-steady state Accumulation/depletion of mass in

system Basis for calculation of mass balance

(unit of time, batch etc) Component or substance

SAMPLE PROBLEM

1000 kg of a 10 % by wt. sodium chloride solution is concentrated to 50 % in a batch evaporator. Calculate the product mass and the mass of water evaporated from the evaporator.

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MIXING OF STREAMS

F1

F2

F3

F4

EXAMPLE

Calculate E and x

Fresh feed 1000kg, 15%by wt sodium hydrogen carbonate

Recycle stream 300 kg, 10% satd. soln.

evaporator feed E, composition x%

FLOWSHEETS Streams Operations/equipment sequence Standard symbols

PROCESS FLOW DIAGRAMS Process flow diagram

TYPICAL SIMPLE FLOWSHEET ARRANGEMENT

reactorSeparation & purificationFresh feed

(reactants, solvents,reagents, catalysts etc)

product

Recycle of unreacted material

Byproducts/coproductswaste

SAMPLE PROBLEM

A 1000 kg batch of a pharmaceutical powder containing 5 % by wt water is dried in a double cone drier. After drying 90 % of the water has been removed. Calculate the final batch composition and the weight of water removed.

SAMPLE PROBLEM – BATCH DISTILLATION

1000 kg of a 20% by wt mixture of acetone in water is separated by multistage batch distillation. The top product (distillate) contains 95% by wt. acetone and the residue still contains 2% acetone. Calculate the amount of distillate.

USE OF MOLAR QUANTITIES

It is often useful to calculate a mass balance using molar quantities of materials and to express composition as mole fractions or mole %.

Distillation is an example, where equilibrium data is often expressed in mole fractions.

MOLAR UNITS

A mole is the molecular weight of a substance expressed in grams

To get the molecular weight of a substance you need its molecular formula and you can then add up the atomic weights of all the atoms in the molecule

To convert from moles of a substance to grams multiply by the molecular weight

To convert from grams to moles divide by the molecular weight.

Mole fraction is moles divided by total moles Mole % is mole fraction multiplied by 100

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MOLAR UNITS

Benzene is C6H6. The molecular weight is (6x12) + (6x1) = 78

So 1 mole of benzene is 78 grams1 kmol is 78 kg

SAMPLE PROBLEM – BATCH DISTILLATION

1000 kmol of an equimolar mixture of benzene and toluene is distilled in a multistage batch distillation unit. 90 % of the benzene is in the top product (distillate). The top product has a benzene mole fraction of 0.95. Calculate the quantities of top and bottom products and the composition of the bottom product.

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STRATEGY FOR ANALYZING MATERIAL BALANCE PROBLEM

1. Read the problem and clarify what is to be accomplished.

A train is approaching the station at 105 cm/s. A man in one car is walking forward at 30 cm/s relative to the seats. He is eating a foot-long hot dog which is entering his mouth at the rate of 2 cm/s. An ant on the hot dog is running away from the man’s mouth at 1 cm/s. How fast is the ant approaching the station?

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2. Draw a sketch (flow diagram) 3. Label. Assign symbols to each variable.4. Put down the known values5. Select the basis.6. List the symbols7. Write down independent equations8. Count the numbers9. Solve the equations10. Check the answer.

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MASS BALANCE PROCEDURES

Process description Flowsheet Label Assign algebraic symbols to unknowns

(compositions, concentrations, quantities) Select basis Write mass balance equations (overall, total,

component, unit) Solve equations for unknowns

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MATERIAL BALANCE NOT INVOLVING REACTIONS

SAMPLE PROBLEM – BATCH DISTILLATION

1000 kg of a 20% by wt mixture of acetone in water is separated by multistage batch distillation. The top product (distillate) contains 95% by wt. acetone and the residue still contains 2% acetone. Calculate the amount of distillate.

MASS BALANCE –FILTRATION / CENTRIFUGATION

feed suspension

wash water/solvent

solid

waste water filtrate

Filtration

F1

5000 kg DM water

Impurity 55 kgWater 2600 kgAPI 450 kg Water 7300 kg

Impurity 50 kgAPI 2kg

Water 300 kgAPI 448 kgImpurity 5 kg

Mass balance - drier

feed product

water/evaporated solvent

MASS BALANCE – EXTRACTION/PHASE SEPARATION

A + B

S

A + B

S + B

A – feed solvent; B – solute; S – extracting solvent

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Example (single stage extraction; immiscible solvents)

E1

feed

solvent

raffinate

extract

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The next step in the synthesis of aspirin in water is to extract the aspirin with chloroform in a batch process. A 195-kg particular batch containing 0.11 kg aspirin/kg water is extracted with 596 kg chloroform. Extraction coefficient is give by y=1.72x where y = kg aspirin/kg chloroform and x = kg aspirin/kg water in raffinate. Calculate amount and composition of extract and raffinate

F = 195 kg; xf = 0.11 kg API/kgwaterS = 596 kg chloroformy = 1.72x, where y is kgAPI/kg chloroform in extract and x is kg

API/kg water in raffinate.

Total balance 195 + 596 = E + RAPI balance 19.5 = 175.5x1 + 596y1

19.5 = 175.5x1 + 596.1.72x1

x1 = 0.0162 and y1 = 0.029R is 175.5 kg water + 2.84 kg APIand E is 596 kg chloroform + 17.28 kg API

Note: chloroform and water are essentially immiscible

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MASS BALANCE– ABSORPTION UNIT

feed gas stream

feed solvent

waste solvent stream

exit gas stream

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SAMPLE PROBLEM- CRYSTALLIZER

A crystallizer contains 1000 kg of a saturated solution of potassium chloride at 80 deg cent. It is required to crystallise 100 kg KCl from this solution. To what temperature must the solution be cooled?

T deg cent Solubility gKCl/100 g water

80 51.1

70 48.3

60 45.5

50 42.6

40 40

30 37

20 34

10 31

0 27.6

At 80 deg cent satd soln contains (51.1/151.1)x100 % KCl i.e. 33.8% by wt

So in 1000 kg there is 338 kg KCl & 662 kg water.Crystallising 100 kg out of soln leaves a satd soln

containing 238 kg KCl and 662kg water i.e. 238/6.62 g KCl/100g water which is 36 g KCl/100g. So temperature required is approx 27 deg cent from table.

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MASS BALANCES– MULTIPLE UNITS

Overall balance Unit balances Component balances

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Multiple units

E – evaporator; C – crystalliser; F – filter unitF1 – fresh feed; W2 – evaporated water; P3 – solid product; R4 – recycle of

saturated solution from filter unit

R4

E C FF1

W2

P3

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MASS BALANCE PROCEDURES

Process description Flowsheet Label Assign algebraic symbols to unknowns

(compositions, concentrations, quantities) Select basis Write mass balance equations (overall, total,

component, unit) Solve equations for unknowns

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DEFINITIONS Stoichiometric quantities Limiting reactant Excess reactant Conversion Yield Selectivity Extent of reaction

Stoichiometry Refers to quantities of reactants and

products in a balanced chemical reaction.aA + bB cC + dDi.e. a moles of A react with b moles of B to

give c moles of C and d moles of D.a,b,c,d are stoichiometric quantities

Reactor mass balances

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Example – aspirin synthesis reaction

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Limiting reactant/excess reactant

In practice a reactant may be used in excess of the stoichiometric quantity for various reasons. In this case the other reactant is limiting i.e. it will limit the yield of product(s)

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continued

A reactant is in excess if it is present in a quantity greater than its stoichiometric proportion.

% excess = [(moles supplied – stoichiometric moles)/stoichiometric moles] x 100

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Example – aspirin synthesis

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Conversion Fractional conversion = amount reactant

consumed/amount reactant supplied % conversion = fractional conversion x 100

Note: conversion may apply to single pass reactor conversion or overall process conversion

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Yield

Yield = (moles product/moles limiting reactant supplied) x s.f. x 100

Where s.f. is the stoichiometric factor = stoichiometric moles reactant required per mole product

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Example – aspirin synthesis

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Selectivity

Selectivity = (moles product/moles reactant converted) x s.f. x100

ORSelectivity = moles desired product/moles

byproduct

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ExtentExtent of reaction = (moles of component leaving

reactor – moles of component entering reactor)/stoichiometric coefficient of component

Note: the stoichiometric coefficient of a component in a chemical reaction is the no. of moles in the balanced chemical equation ( -ve for reactants and +ve for products)

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ExamplesA Bi.e. stoichiometric coefficients a = 1; b =

1100 kmol fresh feed A; 90 % single pass

conversion in reactor; unreacted A is separated and recycled and therefore overall process conversion is 100%

reactor separationFR

P

Discussion - Synthesis of 3,3 dimethylindoline

Discussion - Aspirin synthesis

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References Elementary Principles of Chemical

Processes, R. M. Felder and R. W. Rousseau, 3rd edition, John Wiley, 2000