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LECTURE 4 Mass Balance

LECTURE 4 Mass 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

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LECTURE 4 Mass Balance

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  1. LECTURE 4Mass Balance

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

  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

  4. 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?

  5. SAMPLE PROBLEM: Total initial mass = total final mass = 300 kg Initial methanol mass = final methanol mass 80 + 70 = final methanol mass = 150 kg Therefore final composition of batch is (150/300) x 100 = 50 % by wt.

  6. 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?

  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.

  8. BATCH CYCLE • Sequence of operations/steps repeated according to a cycle • Batch cycle time • Batch size Paul Ashall, 2008

  9. SIMPLE BATCH REACTION CYCLE 3 steps Add reactants etc reaction Empty reactor Next cycle Start cycle t=0 t, finish cycle

  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.

  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 ortotal mass at start = total final mass

  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

  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

  14. 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.

  15. MIXING OF STREAMS F2 F1 F4 F3 Paul Ashall, 2008

  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.

  17. FLOWSHEETS • Streams • Operations/equipment sequence • Standard symbols

  18. PROCESS FLOW DIAGRAMS • Process flow diagram

  19. TYPICAL SIMPLE FLOWSHEET ARRANGEMENT Recycle of unreacted material reactor Separation & purification product Fresh feed (reactants, solvents, reagents, catalysts etc) waste Byproducts/coproducts

  20. 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.

  21. 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.

  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.

  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

  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 Paul Ashall, 2008

  25. 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.

  26. STRATEGY FOR ANALYZING MATERIAL BALANCE PROBLEM • 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

  27. 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

  28. 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 Paul Ashall, 2008

  29. MATERIAL BALANCE NOT INVOLVING REACTIONS Paul Ashall, 2008

  30. 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.

  31. MASS BALANCE –FILTRATION / CENTRIFUGATION wash water/solvent solid feed suspension filtrate waste water

  32. Filtration 5000 kg DM water F1 Water 300 kg API 448 kg Impurity 5 kg Impurity 55 kg Water 2600 kg API 450 kg Water 7300 kg Impurity 50 kg API 2kg

  33. Mass balance - drier water/evaporated solvent product feed

  34. MASS BALANCE – EXTRACTION/PHASE SEPARATION A + B A + B S + B S A – feed solvent; B – solute; S – extracting solvent

  35. Example (single stage extraction; immiscible solvents) feed raffinate E1 solvent extract Paul Ashall, 2008

  36. 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

  37. 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 195 + 596 = E + R API balance 19.5 = 175.5x1 + 596y1 19.5 = 175.5x1 + 596.1.72x1 x1 = 0.0162 and y1 = 0.029 R is 175.5 kg water + 2.84 kg API and E is 596 kg chloroform + 17.28 kg API Note: chloroform and water are essentially immiscible

  38. MASS BALANCE– ABSORPTION UNIT exit gas stream feed solvent feed gas stream waste solvent stream Paul Ashall, 2008

  39. 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? Paul Ashall, 2008

  40. 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.

  41. MASS BALANCES– MULTIPLE UNITS • Overall balance • Unit balances • Component balances Paul Ashall, 2008

  42. Multiple units 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 E C F F1 P3 Paul Ashall, 2008

  43. 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 Paul Ashall, 2008

  44. DEFINITIONS • Stoichiometric quantities • Limiting reactant • Excess reactant • Conversion • Yield • Selectivity • Extent of reaction Paul Ashall, 2008

  45. Stoichiometry • Refers to quantities of reactants and products in a balanced chemical reaction. aA + bBcC + 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

  46. Reactor mass balances

  47. Example – aspirin synthesis reaction Paul Ashall, 2008

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