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PTT108/108 MATERIAL AND ENERGY BALANCE SEM 11 (2012/2013) MASS BALANCE Chapter 3.

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Presentation on theme: "PTT108/108 MATERIAL AND ENERGY BALANCE SEM 11 (2012/2013) MASS BALANCE Chapter 3."— Presentation transcript:

1 PTT108/108 MATERIAL AND ENERGY BALANCE SEM 11 (2012/2013) MASS BALANCE Chapter 3

2 Process Classification Semibatch Batch Continuous – Feed is charge to the process and product is removed when the process is completed – No mass is fed or removed from the process during the operation – Used for small scale production (pharmaceutical products) – Input and output is continuously red and remove from the process – Used for large scale production – Neither batch nor continuous – During the process a part of reactant can be fed or a part of product can be removed.

3 Operation of Continuous Process Unsteady stateSteady state – All the variables (i.e. temperatures, pressure, volume, flow rate, etc) do not change with time – Minor fluctuation can be acceptable – Process variable change with time, in particular mass flow rate.

4 Define type and operation of process given below A balloon is filled with air at steady rate of 2 g/min A bottle of milk is taken from the refrigerator and left on the kitchen Water is boiled in open flask Answer Semibatch and unsteady state Batch and unsteady state Semibatch and unsteady state

5 General Balance 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 IntegralDifferential – 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 SO 2 /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 SO 2 ). – 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with the two instants of time being the moment after the input takes place and the moment before the product is withdrawn.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 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 = OUTPUT Generation and consumption = 0.

8 If a system is at STEADY STATE, set accumulation = 0, regardless of what is being balanced. Steady state: accumulation = 0 INPUT + GENERATION = OUTPUT + CONSUMPTION

9 When you are given process information and asked to determine something about the process, ORGANIZE the by drawing a flowchart Represent INPUTS Represent OUTPUTS Represent PROCESS UNIT (Reactor, mixer, separation units, etc)

10 Write the values and units of all known stream variables at the locations of the streams on the flowchart. Example A stream containing 21 mole% O 2 and 79% N 2 at 320˚C and 1.4 atm flowing at a rate of 400 mol/h might be labeled as: 400 mol/h 0.21 mol O 2 /mol 0.79 mol N 2 /mol T = 320˚C, P = 1.4 atm Usually we write of the stream!

11 Process stream can be given in two ways: 60 kmol N 2 /min 40 kmol O 2 /min 0.6 kmol N 2 /kmol 0.4 kmol O 2 /kmol 100 kmol/min 3.0 lbm CH lbm C 2 H lbm C 2 H lbm CH 4 /lbm 0.4 lbm C 2 H 4 /lbm 0.3 lbm C 2 H 6 /lbm 10 lbm As the total amount or flow rate of the stream and the fractions of each component Directly as the amount or flow rate of each component.

12 Assign algebraic symbols to unknown stream variables [such as m (kg solution/min), x (lbm N 2 /lbm), and n (kmol C 3 H 8 )] and write these variable names and their associated units on the flowchart. mol/h 0.21 mol O 2 /mol 0.79 mol N 2 /mol T = 320˚C, P = 1.4 atm 400 mol/h y mol O 2 /mol (1-y) mol N 2 /mol T = 320˚C, P = 1.4 atm

13 If that the mass of stream 1 is half that of stream 2, label the masses of these streams as m and 2m rather than m 1 and m 2. m 2m m1m1 m2m2

14 If you know that the mass fraction of nitrogen is 3 times than oxygen, label the mass fraction as y g O 2 /g and 3y g N 2 /g rather than y 1 and y 2. y g O 2 /g 3y g N 2 /g y 1 O 2 /g y 2 g N 2 /g When labeling component mass fraction or mole fraction, the last one must be 1 minus the sum of the others y mol O 2 /mol (1-y) mol N 2 /mol y 1 mol O 2 /mol y 2 mol N 2 /mol

15

16 1000 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. 500 kg B/h 500 kg T/h 450 kg B/h (kg T/h) (kg B/h) 475 kg T/h

17 500 kg B/h 500 kg T/h 450 kg B/h (kg T/h) (kg B/h) 475 kg T/h Steady state accumulation = 0 Since no chemical reactions occur generation & consumption = 0 INPUT = OUTPUT Benzene Balance 500 kg B/h = 450 kg B/h + Toluene Balance 500 kg T/h = kg T/h = 50 kg B/h = 25 kg T/h

18 500 kg B/h 500 kg T/h 450 kg B/h (kg T/h) (kg B/h) 475 kg T/h Benzene Balance 500 kg B/h = 450 kg B/h + Toluene Balance 500 kg T/h = kg T/h = 50 kg B/h = 25 kg T/h Check the calculation: Total Mass Balance 1000 kg/h = (all kg/h) 1000 kg/h = 1000 kg/h

19 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 cm 3 /min B: Air (21% O 2 and 79% N 2 ) C: Pure O 2 with a molar flow rate one-fifth of the molar flow rate of stream B The 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 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 cm 3 /min B: Air (21% O 2 and 79% N 2 ) C: Pure O 2 with a molar flow rate one-fifth of the molar flow rate of stream B The 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 O 2 /mol 0.79 mol N 2 /mol mol H 2 O/mol y (mol O 2 /mol) (0.985 – y)(mol N 2 /mol) 20.0 cm 3 H 2 O (l)/min (mol H 2 O/min) 0.2 (mol O 2 /min) (mol air/min) (mol/min)

21 0.21 mol O 2 /mol 0.79 mol N 2 /mol mol H 2 O/mol y (mol O 2 /mol) (0.985 – y)(mol N 2 /mol) 20.0 cm 3 H 2 O (l)/min (mol H 2 O/min) 0.2 (mol O 2 /min) (mol air/min) (mol/min) 20.0 cm 3 H 2 O1.00 g H 2 O1 mol mincm g = H 2 O Balance = 1.11 mol/min Nonreactive steady-state process input = output (mol/min)= (mol)0.015 mol H 2 O (min)mol = 74.1 mol/min

22 0.21 mol O 2 /mol 0.79 mol N 2 /mol mol H 2 O/mol y (mol O 2 /mol) (0.985 – y)(mol N 2 /mol) 20.0 cm 3 H 2 O (l)/min (mol H 2 O/min) 0.2 (mol O 2 /min) (mol air/min) (mol/min) N 2 Balance (mol)0.79 mol N 2 = (mol)(0.985-y) (mol N 2 ) (min)mol (min)(mol) y = mol O 2 /mol Total Mole Balance = = 60.8 mol/min

23 Flowchart scaling If final stream quantities are larger than the original quantities. Scaling down What? Procedure of changing the values of all stream amounts or flow rates by a proportional amount while leaving the stream compositions unchanged. Scaling up if final stream quantities are smaller than the original quantities. Suppose 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 1 kg C 6 H lb m /h 1 kg C 7 H kg C 6 H kg C 7 H lb m /h 2 kg 0.5 kg C 6 H 6 /kg 0.5 kg C 7 H 8 /kg 600 kg 0.5 kg C 6 H 6 /kg 0.5 kg C 7 H 8 /kg 600 lb m /h 0.5 lb m C 6 H 6 /lb m 0.5 lb m C 7 H 8 /lb m x 300 kg kg/h Replace kg with lb m

25 3.0 kg/min of benzene and 1.0 kg/min of toluene are mixed Two 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/min Benzene Balance: 3.0 kg C 6 H 6 /min = 4.0 kg/min (x kg C 6 H 6 /kg) x = 0.75 kg C 6 H 6 /kg m (kg/min) x (kg C 6 H 6 /kg) (1-x) (kg C 7 H 8 /kg) 3 kg C 6 H 6 /min 1 kg C 7 H 8 /min

26 Rules of thumb for NONREACTIVE process 1.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. 2.Write balances first that involve the fewest unknown variables.

27 General Procedure for Single Unit Process Material Balance Calculation 1.Choose as basis of calculation an amount or flow rate of one of the process streams. 2.Draw a flowchart and fill in all unknown variables values, including the basis of calculation. Then label unknown stream variables on the chart. 3.Express what the problem statement asks you to determine in terms of the labeled variables. 4.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. 5.Do the degree-of-freedom analysis. 6.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. 7.Solve the equations. 8.Calculate the quantities requested in the problem statement if they have not already been calculated. 9.If a stream quantity or flow rate n g was given in the problem statement and another value n c was either chosen as a basis or calculated for this stream, scale the balanced process by the ratio n g /n c to obtain the final result.

28 light A 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. If a stream amount or flow rate is given in problem, choose this quantity as a basis If 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 Basis of calculation

29 An 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 0.2 kg NaOH/kg 0.8 kg H2O/kg 0.08 kg NaOH/kg kg H2O/kg 100 (kg) (kg) An 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 H2O liters H2O

31 Total Mass Balance Nonreactive steady-state process input = output = 250 kg NaOH 0.2 kg NaOH/kg 0.8 kg H2O/kg 0.08 kg NaOH/kg kg H2O/kg 100 (kg) (kg) kg H2O liters H2O NaOH Balance (0.2 kg NaOH/kg)(100 kg)=(0.08 kg NaOH/kg) = 150 kg NaOH100 kg + =

32 150 kg1.00 liter kg = = 150 liters 100 kg = 1.50 liters H2O/kg feed solution Diluted water volume Ratios requested in problem statement 100 kg =2.50 kg product solution/kg feed solution

33 light 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. Procedure to perform a degree-of-freedom analysis: 1.draw and completely label a flowchart 2.count the unknown variables on the chart (n unknowns ) 3.count the independent equations (n indep. eq.) 4.find number of degree-of-freedom (n df ) n df = n unknowns - n indep. eq. Degree of freedom Independent 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 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 to solve 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 1.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 process – If 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 exist 2.An 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 chapters 3.Process specifications – The 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 4.Physical properties and laws – Two 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. 5.Physical constraints – For example, if the mole fractions of the three components of a stream labeled x A, x B, and x C, then the relation among these variables is x A + x B + x C = 1. – Instead label as x c, the las fraction should be 1-x A -x B 6.Stoichiometric relations – If chemical reactions occur in a system, stoichiometric equation provide a relationship between the quantities of reactant and the product – To be discussed later

37 A 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 unknowns - 3 material balances ( since there are 3 molecular species in this nonreactive process) - 1 density relationship (relating the mole flow rate to the given volumetric flow rate of the condensate - 1 the fractional condensation 0 degrees of freedom

38 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 problem SYSTEM: – 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 FEED 1 FEED 2 PRODUCT 1PRODUCT 2FEED 3 PRODUCT 3 UNIT 2 UNIT 1 A B C D E

40 Solving material balances in multiple unit process is basically the same as single unit processes In multiple unit, must isolate and write balance on several subsystems to obtain enough equation to determine all unknowns stream variables Always perform degree-of-freedom analysis before solving a material balance of system.

41 100.0 kg/h kg A/kg kg B/kg 40.0 kg/h kg A/kg kg B/kg 30.0 kg/h kg A/kg kg B/kg 30.0 kg/h kg A/kg kg B/kg 12 3 A 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. Solution: X1=0.233 kg A/kg X2=0.255 kgA/kg m1= 60.0 kg/h m2= 90.0 kg/h m3= 60.0 kg/h

42 Recycle Normally in chemical reaction, some of unreacted reactant also found in the product. This unreacted reactant can be separated and recycle back to the reactor Product Recycle Stream Fresh Feed ReactorSeparator

43 Purpose of Recycle 1.Recovery of catalyst – catalyst is very expensive 2.Dilution of process stream – typically for slurry solution 3.Control of process variables – especially for the reaction that release heat, heat can be reduce by lowering the feed concentration 4.Circulation of working fluid such as in refrigerator system

44 EXERCISE

45 Forty-five hundred kilograms per hour of a solution that is one-third K 2 CrO 4 by mass is joined by a recycled stream containing 36.4% K 2 CrO 4, and the combined stream is fed into an evaporator. The concentrated stream leaving the evaporator contains 49.4% K 2 CrO 4, this stream is fed into a crystallizer in which is cooled (causing crystals K 2 CrO 4 to come out solution) and then filtered. The filter cake consist of K 2 CrO 4 crystals and a solution that contains 36.4% K 2 CrO 4 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% K 2 CrO 4, is the recycle stream 1.Calculate the rate evaporation, the rate of production of crystalline K 2 CrO 4, 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 Bypass Fraction of the feed to a process unit is diverted around the unit and combined with the output stream from the unit Used 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. Product Bypass Stream Fresh Feed Process Unit

47 Stoichiometry Stoichiometry – 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 SO 2 + O 2 ---> 2 SO 3 Stoichiometric ratio – ratio of species stoichiometry coefficients in the balanced reaction equation – can be used as a conversion factor to calculate the amount of particular reactant (or product) that was consumed (produced). 2 mol SO 3 generated2 mol SO 2 consumed 2 mol SO 2 consumed1 mol O 2 consumed

48 Test Yourself C 4 H O > 4 CO H 2 O 1.Is the stochiometric equation balance? – Yes 2.What is stochiometric coefficient for CO 2 –4–4 3.What is stochiometric ratio of H 2 O to O 2 including it unit – 4 mol H 2 O generated/ 6 mol O 2 consumed 4.How many lb-moles of O 2 reacted to form 400lb-moles CO 2 – 600 lb-moles O 2 reacted mol/min C 4 H 8 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 C 2 H 2 + 2H > C 2 H 6 Inlet condition: 20 kmol/h C 2 H 2 and 50 kmol/h H 2 What is limiting reactant and fractional excess? (H 2 :C 2 H 2 ) o = 2.5 : 1 (H 2 :C 2 H 2 ) stoich = 2 : 1 H 2 is excess reactant and C 2 H 2 is limiting reactant Fractional excess of H 2 = (50-40)/40 = 0.25

51 Fractional Conversion Fractional Conversion (f)

52 Extent of Reaction Extent of Reaction, ξ ξ = extent of reaction n i = moles of species i present in the system after the reaction occurred n io = moles of species i in the system when the reaction starts v i = stoichiometry coefficient for species i in the particular chemical reaction equation

53 EXERCISE

54 Acrylonitrile is produced in the reaction of propylene, ammonia, and oxygen C 3 H 6 + NH 3 +3/2 O 2 C 3 H 3 N + 3H 2 O The 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 Equilibrium For a given set reactive species and reaction condition, two fundamental question might be ask: 1.What will be the final (equilibrium) composition of the reaction mixture? – chemical engineering thermodynamics 2.How long will the system take to reach a specified state short of equilibrium? – chemical kinetics Irreversible reaction – reaction proceeds only in a single direction (from reactants to products) – the concentration of the limiting reactant eventually approaches zero. Reversible reaction – reactants 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 equal However the discussion to get the chemical equilibrium point is not covered in this text- learn in chemical engineering thermodynamic

56 Example 4.6-2

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: 1.Economic loss 2.Less of desired product is obtained for a given quantity of raw materials 3.Greater quantity of raw materials must be fed to the reactor to obtain a specified product yield. selectivity= moles of desired product moles of undesired product

58 Yield 3 definition of yield with different working definition Yield= Moles of desired product formed Moles that would have been formed if there were no side reaction and the limiting reactant had reacted completely Yield= Moles of desired product formed Moles of reactant fed Yield= Moles of desired product formed Moles of reactant consumed

59 Extent of Reaction for Multiple Reaction Concept of extent of reaction can also be applied for multiple reaction Only now each independent reaction has its own extent.

60 Example 4.6-3

61 Balance of Reactive Processes Balance on reactive process can be solved based on three method: 1.Atomic Species Balance 2.Extent of Reaction 3.Molecular 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 ======================================

65 Example 4.7-1

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