1. Module 9001 Mass Balance
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2. The accounting of all mass in a chemical/pharmaceutical process is referred to as a mass (or material) balance.
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3. 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
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4. Simple example – 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?
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5. continued Total initial mass = total final mass = 300 kg
Initial methanol mass = final methanol mass
= final methanol mass = 150 kg
Therefore final composition of batch is (150/300) x 100 = 50 % by wt.
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6. Exercise
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?
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7. 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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8. Batch cycle Sequence of operations/steps repeated according to a cycle
Batch cycle time
Batch size
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9. Simple batch reaction cycle
3 steps
Add reactants etc
reaction
Empty reactor
Next cycle
Start cycle t=0
t, finish cycle
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10. 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.
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11. 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
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12. 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 system
3. Apply to total mass and component mass
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13. 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
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14. Exercise
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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15. Mixing of streams F2 F1 F4 F3
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16. Example Calculate E and x evaporator feed E, composition x%
Fresh feed 1000kg, 15%
by wt sodium hydrogen carbonate
Recycle stream 300 kg, 10% satd. soln.
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17. Flowsheets Streams Operations/equipment sequence Standard symbols
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18. Flowsheets
Process flow diagram
PID
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19. Typical simple flowsheet arrangement
Recycle of unreacted material
reactor
Separation &
purification
product
Fresh feed
(reactants, solvents,
reagents, catalysts etc)
waste
Byproducts/coproducts
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20. Exercise
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.
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21. Exercise – 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 still contains 2% acetone. Calculate the amount of distillate.
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22. 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.
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23. 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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24. Molar units
Benzene is C6H6. The molecular weight is (6x12) + (6x1) = 78
So 1 mole of benzene is 78 grams
1 kmol is 78 kg
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25. Exercise – 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 Calculate the quantities of top and bottom products and the composition of the bottom product.
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26. Mass balance - crystalliser
A crystalliser 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?
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27. Solubility gKCl/100 g water 80 51.1 70 48.3 60 45.5 50 42.6 40 30 37
T deg cent
Solubility gKCl/100 g water 80
51.1 70
48.3 60
45.5 50
42.6 40 30 37 20 34 10 31
27.6
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28. So in 1000 kg there is 338 kg KCl & 662 kg water.
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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29. Mass balance filtration/centrifuge
wash water/solvent
solid
feed suspension
filtrate
waste water
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30. Filtration F1 5000 kg DM water Water 300 kg Impurity 55 kg API 448 kg
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31. Mass balance - drier water/evaporated solvent product feed
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32. Mass balance – extraction/phase split
A + B
A + B
S + B S
A – feed solvent; B – solute; S – extracting solvent
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33. Example (single stage extraction; immiscible solvents)
feed
raffinate E1
solvent
extract
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34. F = 195 kg; xf = 0.11 kg API/kgwater S = 596 kg chloroform
y = 1.72x, where y is kgAPI/kg chloroform in extract and x is kg API/kg water in raffinate.
Total balance = E + R
API balance 19.5 = 175.5x y1
19.5 = 175.5x x1
x1 = and y1 = 0.029
R is kg water kg API
and E is 596 kg chloroform kg API
Note: chloroform and water are essentially immiscible
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35. Mass balance – absorption unit
exit gas stream
feed solvent
feed gas stream
waste solvent stream
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36. Mass balances – multiple units
Overall balance
Unit balances
Component balances
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37. Multiple units W2 R4 F1 E C F P3
E – evaporator; C – crystalliser; F – filter unit
F1 – fresh feed; W2 – evaporated water; P3 – solid product; R4 – recycle of saturated solution from filter unit W2 R4 F1 E C F P3
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38. 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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39. Exercise
A mass balance and tracking of usage of a solvent used in an API production process is required for a Pollution Emission Register (PER).
Discuss and outline in general terms how you would do this.
Ref.
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40. Definitions Stoichiometric quantities Limiting reactant
Excess reactant
Conversion
Yield
Selectivity
Extent of reaction
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41. Stoichiometry
Refers to quantities of reactants and products in a balanced chemical reaction.
aA + bB cC + dD
i.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
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42. Reactor mass balances
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43. Example – aspirin synthesis reaction
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44. 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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45. 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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46. Example – aspirin synthesis
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47. 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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48. 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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49. Example – aspirin synthesis
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50. Selectivity
Selectivity = (moles product/moles reactant converted) x s.f. x100 OR
Selectivity = moles desired product/moles byproduct
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51. Extent
Extent 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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52. Examples A B i.e. stoichiometric coefficients a = 1; b = 1
100 kmol fresh feed A; 90 % single pass conversion in reactor; unreacted A is separated and recycled and therefore overall process conversion is 100% R F
reactor
separation P
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53. Discussion - Synthesis of 3,3 dimethylindoline
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54. Discussion - Aspirin synthesis
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55. References
Elementary Principles of Chemical Processes, R. M. Felder and R. W. Rousseau, 3rd edition, John Wiley, 2000
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