Presentation on theme: "PTT108/108 MATERIAL AND ENERGY BALANCE"— Presentation transcript:
1 PTT108/108 MATERIAL AND ENERGY BALANCE SEM 11 (2012/2013)Chapter 3MASS BALANCE
2 Process Classification SemibatchBatchFeed is charge to the process and product is removed when the process is completedNo mass is fed or removed from the process during the operationUsed for small scale production (pharmaceutical products)Neither batch nor continuousDuring the process a part of reactant can be fed or a part of product can be removed.ContinuousInput and output is continuously red and remove from the processUsed for large scale production
3 Operation of Continuous Process Steady stateUnsteady stateAll the variables (i.e. temperatures, pressure, volume, flow rate, etc) do not change with timeMinor fluctuation can be acceptableProcess variable change with time, in particular mass flow rate.
4 TEST yourself Define type and operation of process given below A balloon is filled with air at steady rate of 2 g/minA bottle of milk is taken from the refrigerator and left on the kitchenWater is boiled in open flaskAnswerSemibatch and unsteady stateBatch and unsteady state
5 General balances equation A balance on a conserved quantity (total mass, mass of a particular species, energy, momentum) in a system ( a single process unit, a collection of units, or an entire process) may be written in the following way:INPUT + GENERATION – OUTPUT – CONSUMPTION = ACCUMULATION
6 Types of balances Differential Integral balances that indicate what is happening in a system at an instant time.balance equation is a rate (rate of input, rate of generation, etc.) and has units of the balanced quantity unit divided by a time unit (people/yr, g SO2/s).usually applied to a CONTINUOUS process.Balances that describe what happens between two instants of time.balance equation is an amount of the balanced quantity and has the corresponding unit (people, g SO2).usually applied to a BATCH process, with the two instants of time being the moment after the input takes place and the moment before the product is withdrawn.
7 Rules to simplify the material balance equations If the balanced quantity is TOTAL MASS, set generation = 0 and consumption = 0. Mass can neither be created nor destroyed.If the balanced substances is a NONREACTIVE SPECIES (neither a reactant nor a product), set generation = 0 and consumption = 0.INPUT = OUTPUTGeneration and consumption = 0.
8 Rules to simplify the material balance equations If a system is at STEADY STATE, set accumulation = 0, regardless of what is being balanced.Steady state: accumulation = 0INPUT + GENERATION = OUTPUT + CONSUMPTION
9 (Reactor, mixer, separation units, etc) flowchartsWhen you are given process information and asked to determine something about the process, ORGANIZE the by drawing a flowchartRepresentINPUTSRepresentPROCESS UNIT(Reactor, mixer, separation units, etc)RepresentOUTPUTS
10 Flowcharts LABELLING Example Write the values and units of all known stream variables at the locations of the streams on the flowchart.ExampleA stream containing 21 mole% O2 and 79% N2 at 320˚C and 1.4 atm flowing at a rate of 400 mol/h might be labeled as:Usually we writeof the stream!400 mol/h0.21 mol O2/mol0.79 mol N2/molT = 320˚C, P = 1.4 atm
11 Flowcharts LABELLING Process stream can be given in two ways: As the total amount or flow rate of the stream and the fractions of each componentDirectly as the amount or flow rate of each component.0.6 kmol N2/kmol0.4 kmol O2/kmol100 kmol/min60 kmol N2/min40 kmol O2/min0.3 lbm CH4/lbm0.4 lbm C2H4/lbm0.3 lbm C2H6/lbm10 lbm3.0 lbm CH44.0 lbm C2H43.0 lbm C2H6
12 Flowcharts LABELLINGAssign algebraic symbols to unknown stream variables [such as m (kg solution/min), x (lbm N2/lbm), and n (kmol C3H8)] and write these variable names and their associated units on the flowchart.mol/h0.21 mol O2/mol0.79 mol N2/molT = 320˚C, P = 1.4 atm400 mol/hy mol O2/mol(1-y) mol N2/molT = 320˚C, P = 1.4 atm
13 Flowcharts LABELLINGIf that the mass of stream 1 is half that of stream 2, label the masses of these streams as m and 2m rather than m1 and m2.mm1m22m
14 Flowcharts LABELLINGIf you know that the mass fraction of nitrogen is 3 times than oxygen, label the mass fraction as y g O2/g and 3y g N2/g rather than y1 and y2.y1 O2/gy2 g N2/gy g O2/g3y g N2/gWhen labeling component mass fraction or mole fraction, the last one must be 1 minus the sum of the othersy mol O2/mol(1-y) mol N2/moly1 mol O2/moly2 mol N2/mol
16 exercise1000 kg/h of a mixture of benzene (B) and toluene (T) containing 50 % benzene by mass is separated by distillation into two fractions. The mass flow rate of benzene in the top stream is 450 kg B/h and that of toluene in the bottom stream is 475 kg T/h. the operation is at steady state. Write balances on benzene and toluene to calculate the unknown component flow rates in the output streams.450 kg B/h(kg T/h)500 kg B/h500 kg T/h(kg B/h)475 kg T/h
17 solution Steady state accumulation = 0 Since no chemical reactions occur generation & consumption = 0INPUT = OUTPUTBenzene Balance500 kg B/h = 450 kg B/h +450 kg B/h(kg T/h)= 50 kg B/h500 kg B/h500 kg T/hToluene Balance500 kg T/h = kg T/h(kg B/h)475 kg T/h= 25 kg T/h
18 solution Benzene Balance 500 kg B/h = 450 kg B/h + Toluene Balance (kg T/h)= 50 kg B/h500 kg B/h500 kg T/hToluene Balance500 kg T/h = kg T/h(kg B/h)475 kg T/h= 25 kg T/hCheck the calculation:Total Mass Balance1000 kg/h = (all kg/h)1000 kg/h = 1000 kg/h
19 exerciseAn experiment on the growth rate of certain organism requires an environment of humid air enriched in oxygen. Three input streams are fed into an evaporation chamber to produce an output stream with the desired composition.A: Liquid water fed at rate of 20 cm3/minB: Air (21% O2 and 79% N2)C: Pure O2 with a molar flow rate one-fifth of the molar flow rate of stream BThe output gas is analyzed and is found to contain 1.5 mole% water. Draw and label the flowchart of the process, and calculate all unknown stream variables.
20 solution 0.2 (mol O2/min) (mol/min) 0.015 mol H2O/mol y (mol O2/mol) An experiment on the growth rate of certain organism requires an environment of humid air enriched in oxygen. Three input streams are fed into an evaporation chamber to produce an output stream with the desired composition.A: Liquid water fed at rate of 20 cm3/minB: Air (21% O2 and 79% N2)C: Pure O2 with a molar flow rate one-fifth of the molar flow rate of stream BThe output gas is analyzed and is found to contain 1.5 mole% water. Draw and label the flowchart of the process, and calculate all unknown stream variables.(mol O2/min)(mol/min)0.015 mol H2O/moly (mol O2/mol)(0.985 – y)(mol N2/mol)(mol air/min)0.21 mol O2/mol0.79 mol N2/mol20.0 cm3 H2O (l)/min(mol H2O/min)
23 Scaling down What? Scaling up Flowchart scaling Procedure of changing the values of all stream amounts or flow rates by a proportional amount while leaving the stream compositions unchanged.if final stream quantities are smaller than the original quantities.If final stream quantities are larger than the original quantities.Scaling downWhat?Scaling upFlowchart scalingSuppose you have balanced a process and the amount or flow rate of one of the process streams is n1.You can scale the flow chart to make the amount or flow rate of this stream n2 by multiplying all stream amounts or flow rate by the ratio n2/n1.You cannot, however, scale masses or mass flow rates to molar quantities or vice versa by simple multiplication; conversions of this type must be carried out using the methods as discussed in mass fraction and mol fraction section.
24 Flowcharts scalING 1 kg C6H6 2 kg 0.5 kg C6H6/kg 0.5 kg C7H8/kg 300 lbm/h1 kg C7H8300 kg C6H6300 kg C7H82 kg0.5 kg C6H6/kg0.5 kg C7H8/kg600 kg600 lbm/h0.5 lbm C6H6/lbm0.5 lbm C7H8/lbmx 300kg kg/hReplace kg with lbm
25 3.0 kg/min of benzene and 1.0 kg/min of toluene are mixed Balancing a process3.0 kg/min of benzene and 1.0 kg/min of toluene are mixedTwo unknown quantities – m and x, associated with process, so two equations are needed to calculate them.For NONREACTIVE STEADY STATE process input = output.3 possible balance can be written – Balance on total mass, benzene, and toluene – any two of which provide the equations needed to determine m and x.For example,Total Mass Balance:3.0 kg/min kg/min = m kg/min = 4.0 kg/minBenzene Balance:3.0 kg C6H6/min = 4.0 kg/min (x kg C6H6/kg)x = 0.75 kg C6H6/kgm (kg/min)x (kg C6H6/kg)(1-x) (kg C7H8/kg)3 kg C6H6/min1 kg C7H8/min
26 Balancing a process Rules of thumb for NONREACTIVE process The maximum number of independent equations that can be derived by writing balances on a nonreactive system equals the number of chemical species in the input and output streams.Write balances first that involve the fewest unknown variables.
27 General Procedure for Single Unit Process Material Balance Calculation Choose as basis of calculation an amount or flow rate of one of the process streams.Draw a flowchart and fill in all unknown variables values, including the basis of calculation. Then label unknown stream variables on the chart.Express what the problem statement asks you to determine in terms of the labeled variables.If you are given mixed mass and mole units for a stream (such as a total mass flow rate and component mole fractions or vice versa), convert all quantities to one basis.Do the degree-of-freedom analysis.If the number of unknowns equals the number of equations relating them (i.e., if the system has zero degree of freedom), write the equations in an efficient order (minimizing simultaneous equations) and circle the variables for which you will solve.Solve the equations.Calculate the quantities requested in the problem statement if they have not already been calculated.If a stream quantity or flow rate ng was given in the problem statement and another value nc was either chosen as a basis or calculated for this stream, scale the balanced process by the ratio ng/nc to obtain the final result.
28 Basis of calculationA basis of calculation is an amount (mass or moles) of flow rate (mass or molar) of one stream or stream component in a process. All unknown variables are determined to be consistent with the basis.lightIf a stream amount or flow rate is given in problem, choose this quantity as a basisBasis of calculationIf no stream amount or flow rate are known, assume one stream with known composition. If mass fraction is known, choose total mass or mass flow rate as basis. If mole fraction is known, choose a total moles or molar flow rate as basis
29 exerciseAn aqueous solution of NaOH contains 20% NaOH by mass. It is desired to to produce an 8 % 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).
30 solutionAn aqueous solution of NaOH contains 20% NaOH by mass. It is desired to to produce an 8 % 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).(kg)100 (kg)0.2 kg NaOH/kg0.8 kg H2O/kg0.08 kg NaOH/kg0.920 kg H2O/kgkg H2Oliters H2O
31 solution NaOH Balance Total Mass Balance (kg) 100 (kg) 0.2 kg NaOH/kg 0.8 kg H2O/kg0.08 kg NaOH/kg0.920 kg H2O/kgkg H2Oliters H2ONonreactive steady-state process input = outputNaOH Balance(0.2 kg NaOH/kg)(100 kg)=(0.08 kg NaOH/kg)= 250 kg NaOHTotal Mass Balance100 kg == 150 kg NaOH
32 Diluted water volume150 kg1.00 literkg== 150 litersRatios requested in problem statement100 kg= 1.50 liters H2O/kg feed solution100 kg=2.50 kg product solution/kg feed solution
33 Degree of freedom Procedure to perform a degree-of-freedom analysis: A degree of freedom analysis (DFA) is simply an accounting of the number of unknowns in a problem and the number of independent equations that can be written.lightProcedure to perform a degree-of-freedom analysis:draw and completely label a flowchartcount the unknown variables on the chart (n unknowns)count the independent equations (n indep. eq.)find number of degree-of-freedom (ndf)ndf= n unknowns - n indep. eq.Degree of freedomIndependent equations Equations are independent if none of them can be derived from the others. For example, not one of the set of equations can be obtained by adding or subtracting multiples of the others.
34 Number of Degree of freedom light– n df = 0, there are n independent equations and n unknowns. The problem can be solved.– n df > 0, there are more unknowns that independent equations. The problem is underspecified. n df more independent equations or specifications are needed tosolve the problem.– n df < 0, there are more independent equations than unknowns. The problem is overspecified with redundant and possibly inconsistent relations.Possible outcomes of a DFA:
35 6 Sources of Equation for Balance Material balances.For a nonreactive process, number of independent equation can be written is not more than number of molecules species (n ms) of the processIf benzene and toluene is involve in stream, we can write balance on benzene, toluene, total mass, atomic carbon and etc., but only TWO INDEPENDENT balance equation existAn energy balance.If the amount of energy exchanged between the system and its surroundings is specified or if it is one of the unknown process variables, an energy balance provides a relationship between inlet and outlet material flows and temperatures.To be discussed in later chaptersProcess specificationsThe problem statement may specify how several process are related.i.e: Outlet flow rate is two times than flow rate stream 1 or etc.
36 6 Sources of Equation for Balance Physical properties and lawsTwo of the unknown variables may be the mass and volume of a stream material, in which case a tabulated specific gravity for liquids and solids or an equation of state for gases would provide an equation relating the variables.Physical constraintsFor example, if the mole fractions of the three components of a stream labeled xA, xB, and xC, then the relation among these variables is xA + xB + xC = 1.Instead label as xc, the las fraction should be 1-xA-xBStoichiometric relationsIf chemical reactions occur in a system, stoichiometric equation provide a relationship between the quantities of reactant and the productTo be discussed later
37 exerciseA stream of humid air enters a condenser in which 95 % of the water vapor in the air is condensed. The flow rate of the condensate (the liquid leaving the condenser) is measured and found to be 225 L/h. Dry air may be taken to contain 21 mole % oxygen, with the balance nitrogen. The entering air contains 10.0 mole % water. Calculate the flow rate of the gas stream leaving the condenser and the mole fractions of oxygen, nitrogen, and water in this stream.Degree of freedom analysis:5 unknownsmaterial balances ( since there are 3 molecular species in this nonreactive process)density relationship (relating the mole flow rate to the given volumetric flow rate ofthe condensatethe fractional condensation0 degrees of freedom
38 Balances on Multiple Unit Processes In real chemical industries, more than one unit processes exist such as a separation unit after reactor and so on.Need to know term called SYSTEM in order to solve material problemSYSTEM:Any portion of process that can be enclosed within a hypothetical box (or boundary)It can be the entire process, an interconnected of process unit, a single unit, a point which two or more stream come together into one stream or etc.The inputs and outputs to a system are the process streams that are intersect to the system boundary
39 System of Multiple Unit Processes FEED 2AECPRODUCT 3BDFEED 1UNIT 2UNIT 1PRODUCT 1PRODUCT 2FEED 3
40 Balances on Multiple Unit Processes Solving material balances in multiple unit process is basically the same as single unit processesIn multiple unit, must isolate and write balance on several subsystems to obtain enough equation to determine all unknowns stream variablesAlways perform degree-of-freedom analysis before solving a material balance of system.
41 Discussion in classA labeled flowchart of a continuous steady state two-unit process is shown below. Each stream contains two components, A and B in different proportions. 3 streams whose flow rates and/or compositions are not known are labeled 1, 2 and 3. Calculate the unknown flow rates and compositions of stream 1, 2 and 3.40.0 kg/h0.900 kg A/kg0.100 kg B/kg30.0 kg/h0.600 kg A/kg0.400 kg B/kg100.0 kg/h0.500 kg A/kg0.500 kg B/kg312Solution:X1=0.233 kg A/kgX2=0.255 kgA/kgm1= 60.0 kg/hm2= 90.0 kg/hm3= 60.0 kg/h30.0 kg/h0.300 kg A/kg0.700 kg B/kg
42 RecycleNormally in chemical reaction, some of unreacted reactant also found in the product.This unreacted reactant can be separated and recycle back to the reactorProductRecycle StreamFreshFeedReactorSeparator
43 Purpose of Recycle Recovery of catalyst – catalyst is very expensive Dilution of process stream – typically for slurry solutionControl of process variables – especially for the reaction that release heat, heat can be reduce by lowering the feed concentrationCirculation of working fluid such as in refrigerator system
45 Forty-five hundred kilograms per hour of a solution that is one-third K2CrO4 by mass is joined by a recycled stream containing 36.4% K2CrO4 , and the combined stream is fed into an evaporator. The concentrated stream leaving the evaporator contains 49.4% K2CrO4 , this stream is fed into a crystallizer in which is cooled (causing crystals K2CrO4 to come out solution) and then filtered. The filter cake consist of K2CrO4 crystals and a solution that contains 36.4% K2CrO4 by mass; the crystal account for 95% of the total mass of the filter cake. The solution that passes through the filter, also 36.4% K2CrO4, is the recycle stream1.Calculate the rate evaporation, the rate of production of crystalline K2CrO4, the feed rates that evaporator and the crystallizer must be designed to handle, and the recycle ratio (mass of recycle)/(mass of fresh feed)
46 BypassFraction of the feed to a process unit is diverted around the unit and combined with the output stream from the unitUsed to control the composition of a final exit stream from a unit by mixing the bypass stream & the unit exit stream in suitable proportions to obtain desired final composition.ProductBypass StreamFreshFeedProcess Unit
47 StoichiometryStoichiometry – theory of proportions in which chemical species combine with one another.Stoichiometric equation of chemical reaction – statement of the relative number of molecules or moles of reactants and products that participate in the reaction.2 SO2 + O2 ---> 2 SO3Stoichiometric ratioratio of species stoichiometry coefficients in the balanced reaction equationcan be used as a conversion factor to calculate the amount of particular reactant (or product) that was consumed (produced).2 mol SO3 generated 2 mol SO2 consumed2 mol SO2 consumed 1 mol O2 consumed
48 C4H8 + 6 O2 --------> 4 CO2 + 4 H2O Test YourselfC4H8 + 6 O > 4 CO2 + 4 H2OIs the stochiometric equation balance?YesWhat is stochiometric coefficient for CO24What is stochiometric ratio of H2O to O2 including it unit4 mol H2O generated/ 6 mol O2 consumedHow many lb-moles of O2 reacted to form 400lb-moles CO2600 lb-moles O2 reacted100 mol/min C4H8 fed into reactor and 50% is reacted. At what rate water is formed?200 mol/min water generated
49 Limiting Reactant & Excess Reactant The reactant that would run out if a reaction proceeded to completion is called the limiting reactant, and the other reactants are termed excess reactants.A reactant is limiting if it is present in less than its stoichiometric proportion relative to every other reactant.If all reactants are present in stoichiometric proportion, then no reactant is limiting.
50 Example C2H2 + 2H2 ------> C2H6 Inlet condition: 20 kmol/h C2H2 and 50 kmol/h H2What is limiting reactant and fractional excess?(H2:C2H2) o = 2.5 : 1(H2:C2H2) stoich = 2 : 1H2 is excess reactant and C2H2 is limiting reactantFractional excess of H2 = (50-40)/40 = 0.25
52 Extent of Reaction Extent of Reaction, ξ ξ = extent of reaction ni = moles of species i present in the system after the reaction occurrednio = moles of species i in the system when the reaction startsvi = stoichiometry coefficient for species i in the particular chemical reaction equation
54 Acrylonitrile is produced in the reaction of propylene, ammonia, and oxygen C3H6 + NH3 +3/2 O C3H3N + 3H2OThe feed contains 10.0 mole % propylene, 12.0% ammonia, and 78.0% air. A fractional conversion of 30.0 % of the limiting reactant is achieved. Taking 100 mol of feed as basis, determine which reactant is limiting, the percentage by which each of the other reactants is in excess, and the molar amounts of all product gas constituents for a 30% conversion of the limiting reactant (Assume basis 100 mol)
55 Chemical EquilibriumFor a given set reactive species and reaction condition, two fundamental question might be ask:What will be the final (equilibrium) composition of the reaction mixture? – chemical engineering thermodynamicsHow long will the system take to reach a specified state short of equilibrium? – chemical kineticsIrreversible reactionreaction proceeds only in a single direction (from reactants to products)the concentration of the limiting reactant eventually approaches zero.Reversible reactionreactants form products for forward reaction and products undergo the reverse reactions to reform the reactants.Equilibrium point is a rate of forward reaction and reverse reaction are equalHowever the discussion to get the chemical equilibrium point is not covered in this text- learn in chemical engineering thermodynamic
57 Multiples Reaction, Yield & Selectivity Some of the chemical reaction has a side reaction which is formed undesired product- multiple reaction occurred.Effects of this side reaction might be:Economic lossLess of desired product is obtained for a given quantity of raw materialsGreater quantity of raw materials must be fed to the reactor to obtain a specified product yield.selectivity=moles of desired productmoles of undesired product
58 Yield 3 definition of yield with different working definition Yield = Moles of desired product formedMoles that would have been formed if there were no side reaction and the limiting reactant had reacted completelyYield=Moles of desired product formedMoles of reactant fedYield=Moles of desired product formedMoles of reactant consumed
59 Extent of Reaction for Multiple Reaction Concept of extent of reaction can also be applied for multiple reactionOnly now each independent reaction has its own extent.
61 Balance of Reactive Processes Balance on reactive process can be solved based on three method:Atomic Species BalanceExtent of ReactionMolecular Species Balance
62 Atomic Species Balance No. of unknowns variables- No. of independent atomic species balance- No. of molecular balance on indep.nonreactive species- No. of other equation relating the variable=============================No. of degree of freedom
63 Extent of Reaction No. of unknowns variables + No. of independent chemical reaction- No. of independent reactive species- No. of independent nonreactive species- No. of other equation relating the variable===================================No. of degree of freedom
64 Molecular Species Balance No. of unknowns variables+ No. of independent chemical reaction- No. of independent molecular species balance- No. of other equation relating the variable======================================No. of degree of freedom
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