Chapter 4 Fundamentals of Material Balances Material Balance-Part 1
Process Classifications 3 type of chemical processes: - Concept of boundary of the process 1. Batch process 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 Operate in unsteady state n 3/29/15 n Prof. Shoukat Choudhury n 2
Process Classifications 2. Continuous process Input and output is continuously fed and remove from the process Operate in steady state Used for large scale production 3. Semibatch process Neither batch nor continuous During the process a part of reactant can be fed or a part of product can be removed. n 3/29/15 n Prof. Shoukat Choudhury n 3
2 type of process operations: 1. Steady state All the variables (i.e. temperatures, pressure, volume, flow rate, etc) do not change with time 2. Unsteady state or transient Process variable change with time n 3/29/15 n Prof. Shoukat Choudhury n 4
Try This Define type and operation of processes given: A balloon is being filled with air at steady rate of 2 g/min Semibatch and unsteady state A bottle of milk is taken from the refrigerator and left in the kitchen Batch and unsteady state Water is boiled in an open flask Semibatch and unsteady state n 3/29/15 n Prof. Shoukat Choudhury n 5
General Balance Equation: INPUT + GENERATION OUTPUT CONSUMPTION = ACCUMULATION Balances on Continuous Steady-State Process Steady state; accumulation = 0 INPUT + GENERATION = OUTPUT + CONSUMPTION If balance on steady-state non-reactive processes; generation = 0, consumption = 0 INPUT = OUTPUT n 3/29/15 n Prof. Shoukat Choudhury n 6
Example 4.2-2 One thousands kilogram per hour of a mixture of Benzene (B) and Toluene (T) containing 50% Benzene by mass is separated by a disellaeon into two fraceons. The mass flow rate of Benzene in the top stream is 450 kg/h and that of Toluene in the boiom stream is 475 kg/h. The operaeon is at steady state. Calculate the unknown components flow rates in the output streams. n 3/29/15 n Prof. Shoukat Choudhury n 7
Flowchart n 3/29/15 n Prof. Shoukat Choudhury n 8
Steps for Material Balance Calculations: 1. Learn how to organize information about process variables - Flow Chart drawing and Labeling 2. Choose a Basis 3. Set up material balance equations 4. solve the equations for unknown variables. n 3/29/15 n Prof. Shoukat Choudhury n 9
Flowcharts When you are given process information and asked to determine something about the process, it is essential to organize the information in a way that is convenient for subsequent calculations. The best way to do this is to draw a flowchart using boxes or other symbols to represent process units (reactors, mixers, separation units, etc.) lines with arrows to represent inputs and outputs. n 3/29/15 n Prof. Shoukat Choudhury n 10
Flowcharts The flowchart of a process can help get material balance calculations started and keep them moving. Flowchart must be fully labeled when it is first drawn, with values of known process variables and symbols for unknown variables being written for each input and output stream. Flowchart will functions as a scoreboard for the problem solution: as each unknown variable is determined its value is filled in, so that the flowchart provides a continuous record of where the solution stands and what must still be done. n 3/29/15 n Prof. Shoukat Choudhury n 11
Step 1.2: Labeling a flowchart 2 suggestions for labeling flowchart: 1. Write the values and units of all known stream variables at the locations of the streams on the flowchart. For 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 n 3/29/15 n Prof. Shoukat Choudhury n 12
Labeling a flowchart-continue Process stream can be given in two ways: a) As the total amount or flow rate of the stream and the fractions of each component b) Or directly as the amount or flow rate of each component. 10 lbm 3.0 lbm CH 4 4.0 lbm C 2 H 4 3.0 lbm C 2 H 6 0.3 lbm CH 4 /lbm 0.4 lbm C 2 H 4 /lbm 0.3 lbm C 2 H 6 /lbm 100 kmol/min 60 kmol N 2 /min 40 kmol O 2 /min 0.6 kmol N 2 /kmol 0.4 kmol O 2 /kmol n 3/29/15 n Prof. Shoukat Choudhury n 13
Labeling a flowchart(continued) 2. 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 units on the flowchart. n! mol/h 400 mol/h 0.21 mol O 2 /mol 0.79 mol N 2 /mol T = 320 C, P = 1.4 atm y mol O 2 /mol (1-y) mol N 2 /mol T = 320 C, P = 1.4 atm n 3/29/15 n Prof. Shoukat Choudhury n 14
Labeling a flowchart-continue 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. If you know that mass fraction of nitrogen is 3 times than oxygen, label mass fractions as y g O 2 /g and 3y g N 2 /g rather than y 1 and y 2. When labeling component mass fraction or mole fraction, the last one must be 1 minus the sum of the others. If volumetric flow rate of a stream is given, it is generally useful to label the mass or molar flow rate of this stream or to calculate it directly, since balance are not written on volumetric qualities. n 3/29/15 n Prof. Shoukat Choudhury n 15
Consistent on Notation m = m! = n = n! = V = V! = mass mass flow rate moles molar flow rate volume volume flow rate x = component fraction (mass y = moles fraction in gas or moles) in liquid n 3/29/15 n Prof. Shoukat Choudhury n 16
Example 4.2-3 Two methanol water mixtures are contained in separate flasks. The first mixture contains 40 wt% methanol, and the second contains 70% methanol. If 200 g of the first mixture is combined 150 g of the second, what are the mass and composieon of the product? n 3/29/15 n Prof. Shoukat Choudhury n 17
Try This.. n Example 4.2.3 Two methanol-water mixture are contained in separate flasks. The first mixture contains 40wt% methanol and the second flask contains 70wt% methanol. If 20 Kg of the first mixture are going to be mixed with 15000 g of the second in a mixing unit, what are the mass and composition of the product of the mixing unit? n 3/29/15 n Prof. Shoukat Choudhury n 18
Try This.. n Example 4.3.1 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. n 3/29/15 n Prof. Shoukat Choudhury n 19
0.200n! n! 1 1 mol O 2 /min mol air/min 0.21 mol O 2 /mol 0.79 mol N 2 /mol Solution Evaporation n! 3 mol/min 0.015 mol H 2 O /mol y mol O 2 /mol (0.985-y) mol N 2 /mol 20 cm 3 H 2 O (l)/min n! 2 mol H 2 O/min Ø convert water flow-rate to mol/min Ø Water balance Ø Total mol Ø Nitrogen / oxygen balance n 3/29/15 n Prof. Shoukat Choudhury n 20
Flowchart Scaling & Basis of Calculation Flowchart scaling procedure of changing the values of all stream amounts or flow rates by a proportional amount while leaving the stream compositions unchanged. The process would still be balance. Scaling-up if final stream quantities are larger than the original quantities. Scaling down if final stream quantities are smaller than the original quantities. n 3/29/15 n Prof. Shoukat Choudhury n 21
Flowchart Scaling & Basis of Calculation 1 kg C 6 H 6 1 kg C 7 H 8 2 kg 0.5 kg C 6 H 6 /kg 0.5 kg C 7 H 8 /kg x 300 300 kg C 6 H 6 300 kg C 7 H 8 300 lb m /h 300 lb m /h 600 kg 0.5 kg C 6 H 6 /kg 0.5 kg C 7 H 8 /kg kg kg/h Replace kg with lb m 600 lb m /h 0.5 lb m C 6 H 6 /lb m 0.5 lb m C 7 H 8 /lb m n 3/29/15 n Prof. Shoukat Choudhury n 22
Flowchart Scaling & Basis of Calculation Suppose you have balanced a process and the amount or flow rate of one of the process streams is n 1.You can scale the flow chart to make the amount or flow rate of this stream n 2 by multiplying all stream amounts or flow rate by the ratio n 2 /n 1. 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. n 3/29/15 n Prof. Shoukat Choudhury n 23
Basis of Calculation 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. n 3/29/15 n Prof. Shoukat Choudhury n 24
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Flowchart n 3/29/15 n Prof. Shoukat Choudhury n 38
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Product Separation Reactants Reactor Product Separation Unit Products n 3/29/15 n Prof. Shoukat Choudhury n 40
K-salt Recovery In a steady state process crystalline potassium chromate (K 2 CrO 4 ) is recovered from an aqueous solution of this salt. 10 ton per hour of a solution that is 30% K 2 CrO 4 by mass is fed into an evaporator. The concentrated stream leaving the evaporator contains 50% K 2 CrO 4 ; this stream is fed into a crystallizer in which it is cooled (causing crystals of K 2 CrO 4 to come out of solution) and then filtered. The filter cake consists of K 2 CrO 4 crystals and a solution that contain 35% K 2 CrO 4 by mass; the crystals account for 95% of the total mass of the filter cake. The filtrate drains out of the system also contains 35% K 2 CrO 4. n 3/29/15 n Prof. Shoukat Choudhury n 41
10 ton per hour of a solution that is 30% K 2 CrO 4 by mass is fed into an evaporator. The concentrated stream leaving the evaporator contains 50% K 2 CrO 4 ; this stream is fed into a crystallizer in which it is cooled (causing crystals of K 2 CrO 4 to come out of solution) and then filtered. The filter cake consists of K 2 CrO 4 crystals and a solution that contain 35% K 2 CrO 4 by mass; the crystals account for 95% of the total mass of the filter cake. The filtrate drains out of the system also contains 35% K 2 CrO 4.. Fresh Feed 10,000 kg/h 0.30 kg K/kg 0.70 kg W/kg m1 kg W/h Evaporator. m2 kg/h 0.50 kg K/kg 0.50 kg W/kg Crystallizer & Filter Filtrate. m5 kg/h Filter Cake. m3 kg/h. m4 kg/h 0.35 kg K/kg 0.65 kg W/kg 0.35 kg K/kg 0.65 kg W/kg n 3/29/15 n Prof. Shoukat Choudhury n 42
m 1 = 4000 kg W/h m 2 = 6000 kg K-soln/h m 3 = 1385 kg K(S)/h m 4 = 73 kg K-soln/h m 5 = 4542 kg K-soln/h Product rate, m 3 = 1385 Kg/h Filtrate soln, m 5 = 4542 Kg/h If filtrate is discarded, a huge loss in raw material and money Therefore, recycle is the solution n 3/29/15 n Prof. Shoukat Choudhury n 43
K-salt Recovery In a steady state process crystalline potassium chromate (K 2 CrO 4 ) is recovered from an aqueous solution of this salt. 10 ton per hour of a solution that is 30% K 2 CrO 4 by mass is joined by a recycle stream containing 35% K2CrO4, and combined stream is fed into an evaporator. The concentrated stream leaving the evaporator contains 50% K 2 CrO 4 ; this stream is fed into a crystallizer in which it is cooled (causing crystals of K 2 CrO 4 to come out of solution) and then filtered. The filter cake consists of K 2 CrO 4 crystals and a solution that contain 35% K 2 CrO 4 by mass; the crystals account for 95% of the total mass of the filter cake. The solution that passes through the filter also 35% K 2 CrO 4, is the recycle stream. Calculate the rate of evaporation, the rate of production of crystalline K 2 CrO 4, the feed rates that the evaporator and the crystallizer must be designed to handle and the recycle ratio n 3/29/15 n Prof. Shoukat Choudhury n 44
Product Separation & Recycling. m 2 kg W/h Fresh Feed 10,000 kg/h 0.30 kg K/kg 0.70 kg W/kg. m1 kg/h MP Evaporator. Recycle m3 kg/h 0.50 kg K/kg 0.50 kg W/kg Crystallizer & Filter Filter Cake. m4 kg/h. m5 kg/h 0.35 kg K/kg 0.65 kg W/kg. m6 kg/h 0.35 kg K/kg 0.65 kg W/kg n 3/29/15 n Prof. Shoukat Choudhury n 45
Product Separation & Recycling m1 = 19662 kg mixed feed /h m2 = 6900 kg Water evaporated/h m3 = 12762 kg fed to crystallizer/h m4 = 2945 kg crystals/h m5 = 155 kg K-soln/h m6 = 9662 kg recycle/h Recycle Ratio= 0.9662 Kg recycle/kg Fresh feed Comment: Crystal production increases 113% due to recycling n Without Recycle m2 = 4000 kg W/h m3 = 6000 kg K-soln/h m4 = 1385 kg K(S)/h m5 = 73 kg K-soln/h m6 = 4542 kg K-soln/h n 3/29/15 n Prof. Shoukat Choudhury n 46
Recycle and Bypass n What is wrong here? Reuse of reactants/feed Recovery of catalyst Dilution of process streams (improve filter operation) Control of a process variable (Reduce reactant concentration) Circulation of a working fluid (Refrigeration) n 3/29/15 n Prof. Shoukat Choudhury n 47
Product Separation and Recycle Normally, reactions are not complete Separation and recycle Improved yield, conversion, Reactants Reactor Product Separation Unit Products Recycle ü Overall conversion ü Once-through conversion ü Process feed ü Fresh feed ü Gross product ü Net product n 3/29/15 n Prof. Shoukat Choudhury n 48
Balances on Reactive Systems Stoichiometry The theory of proportions in which chemical species combine with one another. Example: 2 SO 2 + O 2 à 2 SO 3 Stoichiometric Ratio Ratio of stoichiometric coefficients Example 2 mol SO 3 produced 1 mol O 2 reacted 2 mol SO 2 reacted 2 mol SO 3 produced n 3/29/15 n Prof. Shoukat Choudhury n 49
Limiting reactants Terminology Exist less than stoichiometric proportion Excess reactants Exist more than stoichiometric proportion Example 2SO 2 + O 2 à 2 SO 3 (30 mol) (10 mol) Excess Limiting n 3/29/15 n Prof. Shoukat Choudhury n 50
Fractional excess Percent excess Example Terminology n n ns n n n s H 2 + Br 2 à HBr H 2 : 25 mol /hr Br 2 : 20 mol /hr Fractional Excess H 2 = (25 20 ) /20 = 0.25 Percent Excess = 25 % s s 100 n 3/29/15 n Prof. Shoukat Choudhury n 51
Terminology Fractional conversion Chemical reactions are not always completed. Factional conversion f = (moles reacted) / (moles fed) When fresh feed consists of more than one material the conversion must be stated for a single component, usually the limiting reactant. n 3/29/15 n Prof. Shoukat Choudhury n 52
Overall Conversions = Single-Pass Conversions = Terminology reactant input to process - reactant output from process reactant input to process reactant input to reactor - reactant output from reactor reactant input to reactor In general, high overall conversions can be achieved in two ways: Design the reactor to yield a high singlepass conversion, or Design the reactor to yield a low singlepass conversion and follow it with a separation unit to recover and recycle unconsumed reactant. n 3/29/15 n Prof. Shoukat Choudhury n 53
Problem Consider the reaction 6 NaClO 3 + 6 H 2 SO 4 + CH 3 OH 6 ClO 2 + 6 NaHSO 4 + CO 2 if the reactor feed has the composition (mol%) of 36% NaClO 3, 54% H 2 SO 4, and the rest CH 3 OH, which is the limiting reactant? Calculate the reactant flows required to n 3/29/15 produce 10 ton n Prof. per Shoukat hour Choudhury of ClO 2 assuming n 54
Balances of Atomic and Molecular Species Methods for solving mass balances with reactions Using balances on molecular species Using balances of atoms Using the extent of reaction For multiple reactions, sometimes it is more convenient to use atomic balances Atomic species balances generally lead to the most straightforward solution procedure, especially when more than one reaction is involved. Extents of reaction are convenient for chemical equilibrium problems and when equation solving software is to be used. Molecular species balances require more complex calculations than either of the other two approaches and should be used only for simple systems involving one reaction. n 3/29/15 n Prof. Shoukat Choudhury n 55
A simple Problem The oxidation of ethylene to produce ethylene oxide proceeds as: 2 C2H4 + O2 = 2 C2H4O The feed to the reactor contains 100 Kmol C2H4 and 200 Kmol O2. What is the limiting reactant? What is the percentage excess of the other reactant? If the reaction proceeds to a point where the fractional conversion of the limiting reactant is 50%, how much of each n 3/29/15 n Prof. Shoukat Choudhury n 56
Multiple Reaction, Yield, Selectivity Multiple reaction : one or more reaction Side Reaction : undesired reaction Example: Production of ethylene C 2 H 6 à C 2 H 4 + H 2 Side Reactions C 2 H 6 + H 2 à 2CH 4 C 2 H 4 + C 2 H 6 à C 3 H 6 + CH 4 Design Objective Maximize desired products (C 2 H 4 ) Minimize undesired products (CH 4, C 3 H 6 ) n 3/29/15 n Prof. Shoukat Choudhury n 57
YIELD Terminology moles of desired product formed = moles that would have been formed if there n Yield = n Yield = were no side reactions and the limiting reactant had reacted completely moles of desired product formed moles of reactant fed moles of desired product formed moles of reactant consumed SELECTIVITY = moles of desired product formed moles of undesired product formed Yield and Selectivity are used to describe the degree to which a desired reaction predominates over competing side reactions. n 3/29/15 n Prof. Shoukat Choudhury n 58
Determination of yield and Reaction: selectivity C 2 H 6 + 2.5O 2 2CO + 3H 2 O 100 mol C 2 H 6 500 mol O 2 C 2 H 6 + 3.5O 2 2CO 2 +3H 2 O Undesired product: CO n Products 80% reactor REACTOR conversion n 20 mol C 2 H 6 n 120 mol CO 2 n 40 mol CO n 240 mol O 2 n 240 mol H 2 O Yield = 0.6 or 1.2 or 1.5 Selectivity = 3.0 n 3/29/15 n Prof. Shoukat Choudhury n 59
Problem Ethane is burned with air in a continuous steady-state combustion reactor to yield a mixture of carbon monoxide, carbon dioxide, and water. The feed to the reactor contains 10% C 2 H 6. The percentage conversion of ethane is 80%, and gas leaving the reactor contains 8 mol CO 2 per mol of CO. n Reaction: Determine molar composition of product n Cgas. 2 H 6 + 5/2 O 2 2CO + 3H 2 O n C 2 H 6 + 7/2 O 2 2CO 2 +3H 2 O n 3/29/15 n Prof. Shoukat Choudhury n 60
Concept of Purge- why needed? Production of Ethylene Oxide Reaction: 2C 2 H 4 + O 2 2C 2 H 4 O Mixture of Ethylene and air stream is charged to the reactor Reactor effluent is charged to absorber and gas stream containing N 2, O 2 and unreacted ethylene is charged back to reactor Recycle Fresh Feed Reactor Absorber Solvent Products n 3/29/15 n Prof. Shoukat Choudhury n 61
Production of Ethylene Oxide Problem: accumulation of N 2 Solution: allow purging of inert species Recycle Purge stream 50 mol C 2 H 4 /s 10 mol C 2 H 4 /s 25 mol O 2 /s 5 mol O 2 /s Fresh Feed 60 mol C 2 H 4 /s 30 mol O 2 /s 113 mol N 2 /s 565 mol N 2 /s 100 mol C 2 H 4 /s 50 mol O 2 /s 565 mol N 2 /s Reactor 50 mol C 2 H 4 /s 50 mol C 2 H 4 O/s 25 mol O 2 /s 565 mol N 2 /s Absorber 113 mol N 2 /s Products Solvent 50 mol C 2 H 4 O/s solvent n 3/29/15 n Prof. Shoukat Choudhury n 62
Purging Getting rid of undesired materials in recycle stream. Reactants Reactor Product Separation Unit Products Recycle Purging n 3/29/15 n Prof. Shoukat Choudhury n 63
Definition A Recycle Stream is a term denoting a process stream that returns material from downstream of a process unit back to the process unit. A Bypass stream - one that skips one or more stages of the process and goes directly to another downstream stage. A Purge stream a stream bled off to remove an accumulation of inert or unwanted material that might otherwise n 3/29/15 build up in the recycle n Prof. Shoukat Choudhury stream. n 64
Example 4.7-3, page 139 Recycle and Purge in the Synthesis of Methanol Methanol is produced in the reaction of carbon dioxide and hydrogen: CO 2 + 3H 2 CH 3 OH + H 2 O The fresh feed to the process contains hydrogen, carbon dioxide and 0.40 mole percent inerts(i). The reactor effluent passes to a condenser that removes essentially all of the methanol and water formed and none of the reactants or inerts. The latter substances are recycled to the reactor. To avoid buildup of the inerts in the system, a purge stream is withdrawn from recycle. The feed to the reactor contains 28.0 mole% CO2, 70 mole% H2 and 2% inerts. The single pass conversion of H2 is 60%. Calculate the molar flow rates and molar compositions of the fresh feed, the total feed to the reactor, the recycle stream, and the purge stream for a methanol production rate of 155 kmol methnol/hr. n 3/29/15 n Prof. Shoukat Choudhury n 65
Reaction: CO 2 + 3H 2 CH 3 OH + H 2 O Basis: 155 kmol CH 3 OH/h Recycle N8 kmol/h x6c mol CO 2 /mol x6h mol H 2 /mol (1-x6C-x6H) mol I/mol Purge N7 kmol/h x6c mol CO 2 /mol x6h mol H 2 /mol (1-x6C-x6H) mol I/mol N6 kmol/h x6c mol CO 2 /mol x6h mol H 2 /mol (1-x6C-x6H)mol I/mol n1 kmol/h n2 kmol/h x1c mol CO 2 /mol 0.28 mol CO 2 /mol (0.996-x1C)mol H 2 /mol 0.70 mol H 2 /mol 0.004 mol I/mol 0.02 mol I/mol REACTOR n3 kmol CO 2 /mol n4 kmol H 2 /mol n5 kmol I/mol CONDENSER 155 kmol CH 3 OH/h 155 kmol H 2 O/h 155 kmol CH 3 OH/h 155 kmol H 2 O/h n 3/29/15 n Prof. Shoukat Choudhury n 66
Combustion Reaction Combustion A rapid reaction of a fuel with oxygen Fuels : coal, fuel oil, gas fuel, solid fuel, Complete combustion / incomplete combustion Wet basis composition / dry basis composition Remember: Orsat analysis yields dry basis composition n 3/29/15 n Prof. Shoukat Choudhury n 67
Terminology Theoretical oxygen : Amount of oxygen needed for complete combustion all carbon in the fuel is oxidized to CO 2 and all the hydrogen is oxidized to H 2 O Theoretical air : The quantity of air that contains theoretical oxygen Air (theo) = 4.76 x O 2(theo) Excess air : The amount by which the air fed to reactor exceeds the (molesair) fed (molesair) theroretical theoretical air 100% ( moles air) theoretical n 3/29/15 Percent excess n Prof. air Shoukat Choudhury n 68
Composition of Flue or stack gas Wet Basis = => Dry Basis - Basis: 1 mole wet gas - basic idea, subtract the water and express the rest in % Dry Basis = => Wet Basis - Basis: 1 mole wet gas - need one extra information: How much water is there in one mole of wet gas? (Say, y mole fraction out of 1 mole wet gas) - subtract water from one mole of wet gas and get the mole of dry gas. (dry gas=1-y) - Now, dry gas fraction is (1-y)*yi, where yi is mole fraction of ith dry gas components n 3/29/15 n Prof. Shoukat Choudhury n 69
A problem with purge and recycle The fresh feed to an ammonia production process contains nitrogen and hydrogen in stoichiometric proportion, along with 2 mole% inert gas. The feed is combined with a recycle stream containing the same three species and the combined stream is fed to a reactor in which a single pass conversion of 20% is achieved. The reactor effluent flows to a condenser. A liquid stream containing essentially all of the ammonia formed in the reactor and a gas stream containing all the inerts and the unreacted nitrogen and hydrogen leave the condenser. 10% of the gas stream leaving the condenser is removed as purge and the rest constitutes the recycle stream. a) Draw a complete flow chart of the process. b) Completely label the flow-chart c) Find the overall conversion of N2 d) Find the total feed flow rates to the reactor n 3/29/15 n Prof. Shoukat Choudhury n 70
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Chemical Reaction What is final composition? ˆ ν i Chemical equilibrium K = fi thermodynamics K = a ν i i d ln K = dt ΔH RT r How long it will take to reach equilibrium? m Chemical kinetics = kot r i = k( T) f ( composition ) e C C E / RT n l A b... n 3/29/15 n Prof. Shoukat Choudhury n 72
Sources of equations of unknown process variables: 1. Material balances 2. An energy balance 3. Process specifications 4. Physical properties and laws 5. Physical constraints 6. Stoichiometric relations n 3/29/15 n Prof. Shoukat Choudhury n 73
Variables in process design or analysis Temperature Pressure Flow rate Chemical composition Physical properties Physical properties Specific gravity Specific volume Density Specific heat Enthalpy Heat of reaction, etc. Chemical composition Mass fraction Mole fraction Mass and molar composition Concentration: mass conc., molar conc., molality n 3/29/15 n Prof. Shoukat Choudhury n 74
Topic Outcomes Make your conception clear about the following terms: Batch, semibatch, continuous, transient, and steady-state processes Recycle, bypass and purge Degrees-of-Freedom Fractional conversion of a limiting reactant Percentage excess of a reactant Yield and selectivity Dry-basis composition of a mixture containing water Theoretical air and percent excess air in a combustion reaction n 3/29/15 n Prof. Shoukat Choudhury n 75
Topic Outcomes Given a process description: Draw and fully label a flowchart Choose a convenient basis of calculation For a multiple-unit process, identify the subsystems for which balances might be written Perform DoF analysis for the overall system and each possible subsystems Write in order the equations you would use to calculate specified process variables Perform the calculations n 3/29/15 n Prof. Shoukat Choudhury n 76
Topic Outcomes Do these computations for single-unit and multiple-unit processes and for processes involving recycle, bypass, or purge streams If the system involves reactions, you should be able to use molecular species balances, atomic species balances, or extents of reaction for both the DoF analysis and the process calculations n 3/29/15 n Prof. Shoukat Choudhury n 77
Topic Outcomes Given a combustion reactor and information about the fuel composition calculate the feed rate of air from a given percent excess or vice versa Given additional information about the conversion of the fuel and the absence or presence of CO in the product gas calculate the flow rate and composition of the product gas n 3/29/15 n Prof. Shoukat Choudhury n 78