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

 ‘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

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?

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.

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 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.

 Sequence of operations/steps repeated according to a cycle  Batch cycle time  Batch size Paul Ashall, 2008

3 steps Start cycle t=0t, finish cycle Add reactants etcreaction Empty reactor Next cycle

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.

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

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

 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

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.

Paul Ashall, 2008 F1 F2 F3 F4

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%

 Streams  Operations/equipment sequence  Standard symbols

 Process flow diagram

reactor Separation & purification Fresh feed (reactants, solvents, reagents, catalysts etc) product Recycle of unreacted material Byproducts/coproductswaste

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.

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.

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.

 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

Benzene is C 6 H 6. The molecular weight is (6x12) + (6x1) = 78 So 1 mole of benzene is 78 grams 1 kmol is 78 kg Paul Ashall, 2008

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.

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? Paul Ashall, 2008

2. Draw a sketch (flow diagram) 3. Label. Assign symbols to each variable. 4. Put down the known values 5. Select the basis. 6. List the symbols 7. Write down independent equations 8. Count the numbers 9. Solve the equations 10. Check the answer. Paul Ashall, 2008

 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 Paul Ashall, 2008

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.

feed suspension wash water/solvent solid waste water filtrate

F kg DM water Impurity 55 kg Water 2600 kg API 450 kg Water 7300 kg Impurity 50 kg API 2kg Water 300 kg API 448 kg Impurity 5 kg

feed product water/evaporated solvent

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

Paul Ashall, 2008 E1 feed solvent raffinate extract

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 Paul Ashall, 2008

F = 195 kg; x f = 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 y = 175.5x x 1 x 1 = and y 1 = R is kg water kg API and E is 596 kg chloroform kg API Note: chloroform and water are essentially immiscible

Paul Ashall, 2008 feed gas stream feed solvent waste solvent stream exit gas stream

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? Paul Ashall, 2008

T deg centSolubility gKCl/100 g 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.

 Overall balance  Unit balances  Component balances Paul Ashall, 2008

E – evaporator; C – crystalliser; F – filter unit F1 – fresh feed; W2 – evaporated water; P3 – solid product; R4 – recycle of saturated solution from filter unit Paul Ashall, 2008 R4 ECF F1 W2 P3

 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 Paul Ashall, 2008

 Stoichiometric quantities  Limiting reactant  Excess reactant  Conversion  Yield  Selectivity  Extent of reaction Paul Ashall, 2008

 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

Paul Ashall, 2008

 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) Paul Ashall, 2008

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 Paul Ashall, 2008

 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 Paul Ashall, 2008

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 Paul Ashall, 2008

Selectivity = (moles product/moles reactant converted) x s.f. x100 OR Selectivity = moles desired product/moles byproduct Paul Ashall, 2008

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) Paul Ashall, 2008

A B i.e. stoichiometric coefficients a = 1; b = kmol fresh feed A; 90 % single pass conversion in reactor; unreacted A is separated and recycled and therefore overall process conversion is 100% Paul Ashall, 2008 reactor separation F R P

 Elementary Principles of Chemical Processes, R. M. Felder and R. W. Rousseau, 3 rd edition, John Wiley, 2000 Paul Ashall, 2008