INTRODUCTION TO MATERIAL BALANCES CHE 05 Chemical Process Principles Most problems you will have to solve in CHE 05 will look something like this: A liquid mixture containing 45.0 wt% benzene (B) and 55.0% toluene (T) is fed to a distillation column operating at steady state. Product streams emerge from the top and bottom of the column. The top product stream contains 95.0 wt% B, and the bottom product stream contains 8% of the benzene fed to the column. The volumetric flow rate of the feed stream is 000 liters/h and the specific gravity of the feed mixture is 0.87. Determine the mass flow rate of the overhead product stream and the mass flow rate and composition (mass fractions) of the bottom product stream. You can take two approaches to solving such problems: 1. Start writing and solving equations until all of the required quantities have been determined.. Draw and label a flow chart, do a degree of freedom analysis, lay out a solution strategy, and then write equations and solve for the required quantities. Students in CHE 05 almost always initially lean toward the first approach, which looks more efficient to them. That choice has several likely consequences that you should be aware of. Feature of Course Problem-solving time early in Chapter 4 Problem-solving time, most of Ch. 4 and Ch. 5 9 Course grade Writing and solving equations Slightly faster Slower, then much slower, then may never get the solution C-F (as opposed to A-C for the other approach) Suggestion: Start using the second approach immediately, even on the simple problems, so that you ll be used to it when the problems start getting harder (which they will very soon). This handout guides you through the basics of that approach. We also suggest that you read through pp. 4-1 to 4-6 of the Student Workbook. If you follow all the guidelines laid out in those pages, it may not guarantee a good grade in the course but it will increase the odds significantly. The labeled flow chart of the process described in the first paragraph might appear as follows: 000 L/h (S.G.=0.87) m (kg/h) 1 0.45 kg B/kg 0.55 kg T/kg Distillation Column m (kg/h) 0.95 kg B/kg 0.05 kg T/kg m m B3 T 3 (kg B/h) (8% of B in feed) (kg T/h) Figure 1. Flow Chart of a Distillation Process -1 Copyright
CHE 05 Chemical Process Principles We will refer back to Figure 1 throughout this handout. Nomenclature: Batch process: Add contents, then remove them later Continuous process: Continuously add then remove contents Semi-batch process: Neither batch nor continuous; e.g. allow contents of gas tank to escape; slowly blend liquids in a tank from which nothing is withdrawn Steady state process: Process variables (flowrates, T, P) are constant Transient process: Process variables change with time Balance equation We can write a balance on any quantity entering or leaving a system mass, moles, energy, people: (a) I + G O C = A (4.-1) Input + Generation Output Consumption = Accumulation What What is What What is Change in amount comes produced leaves consumed contained in the system in across within the across within the system system system system boundary (reaction) boundary (reaction) Simplifications Continuous process at steady state: A = 0 Nonreactive system, or balance on nonreactive species: G = 0, C = 0 Batch process: I = 0, O = 0, A = final value initial value -
CHE 05 Chemical Process Principles Example: Referring back to Figure 1, write balance equations on benzene, toluene, and total mass. Solution: The general balance equation is I + G O C = A In all balances on the process shown in Figure 1, we may drop the terms,, and. Why? The balance equation therefore reduces to Balance on benzene (B): Balance on toluene (T): Balance on total mass: Drawing and labeling flow charts The first step in approaching a typical CHE 05 problem is to draw and completely label the flow chart. What does it mean to be completely labeled? A stream on a flow chart is completely labeled when you can write an expression for the amount (batch) or flow rate (continuous) of each species in the stream in terms of numbers and variables written on the flow chart. A flow chart is completely labeled if every stream on it is completely labeled. For example, a continuous stream of natural gas containing 85 wt% methane and the balance ethane might be labeled as: 0.85 kg CH 4 /kg 0.15 kg C H 6 /kg Is this stream completely labeled? (Hint: No!) Why not? Suppose we let m (kg/s) denote the unknown mass flow rate of the stream. The labeling would then be m (kg/s) 0.85 kg CH 4 /kg 0.15 kg C H 6 /kg The stream is now completely labeled, since we can express the mass flow rates of methane and ethane in terms of what is written on the chart: Methane flow rate: m CH (kg CH 4 4/s) = Ethane flow rate: m CH (kg C 6 H 6/s) = -3
CHE 05 Chemical Process Principles Rules for labeling: You may label either a total flow rate (or amount) and component mass or mole fractions, or individual component flow rates (amounts). m(kg) x1(kg NaCl/kg) OR m1(kg NaCl) x(kg KCl/kg) m(kg KCl) (1 x x )(kg H O/kg) m (kg H O) 1 3 Once you have one, you can always calculate the other: In terms of m, x 1, and x, m 1 = In terms of m 1, m, and m 3, m = and x 1 = Label in the way that builds in given information about the stream. (For instance, if you know or can easily calculate either a total flow rate or values of component mass or mole fractions, label the total and fractions; otherwise label individual component flow rates or amounts (likely to get easier algebra that way). Suppose you know that a gas stream containing sulfur dioxide and air (1 mole% O, 79% N ) flows at a rate of 15 mol/s. The stream might be labeled 15 mol/s x (mol SO / mol) (1-x) (mol air/mol) 0.1 mol O /mol air 0.79 mol N /mol air Exercise: Prove that this stream is completely labeled. Solution: n (mol SO /s) SO If volumes or volumetric flow rates of streams are either given or required, include labels both for them and for mass or molar quantities. For practice, do Test Yourself on p. 93. -4
Exercise: Enriching Air with O. CHE 05 Chemical Process Principles A stream of air (1.0 mole% O, 79.0% N ), a stream of pure oxygen, and a stream of liquid water flowing at a rate of 0.0 cm 3 /min, are fed to a steady-state evaporation chamber in which all of the liquid evaporates. The flow rate of the pure oxygen is 0% of the flow rate of the air. The emerging gas stream contains 1.5 mole% water vapor. O (g) 0% of air flow rate Air 0.1 mol O /mol 0.79 mol N /mol Evaporator 0.0 cm 3 H O(l)/min 0.015 mol H O(v)/mol Completely label the flow chart (including the molar flow rate of the liquid water stream), following the rules given above. Evaporator 0.015 mol H O(v)/mol 0.1 mol O /mol 0.79 mol N /mol 0.0 cm 3 H O(l)/min Convince yourself that your labeling is complete. (For example, what is the flow rate of oxygen in the outlet stream in terms of labeled values and variable names.) Basis of Calculation and Flowchart Scaling. At least one stream quantity or flow rate (mass, moles, volume) should be specified before any calculations are done. If the problem statement doesn t do it, you should choose a convenient basis (an amount of a stream with known composition.) Having degrees of freedom = 1 is often a reminder that you must choose a basis. -5
CHE 05 Chemical Process Principles Example: A stream containing 30 wt% ethanol (E) and 70% water (W) is blended with a stream containing 60% E and 40% W. The product contains 35% E. What is the mass ratio of Stream 1 to Stream (= m 1 /m in the flow chart)? m 1 (g) 0.30 g E/g 0.70 g W/g m (g) 0.60 g E/g 0.40 g W/g Mixer m 3 (g) 0.35 g E/g 0.65 g W/g Solution: The problem statement does not specify a stream amount or flow rate. There is enough information to calculate the required stream ratio without specifying one, but arbitrarily choosing a basis of calculation (an amount or flow rate of one of the streams) makes the required math easier. We might proceed as follows: Basis of calculation: 100 g of Stream 1 (m 1 = 100 g): 100 g 0.30 g E/g 0.70 g W/g m (g) 0.60 g E/g 0.40 g W/g Mixer m 3 (g) 0.35 g E/g 0.65 g W/g We could then write balance equations on ethanol and water, solve for m 1 and m, and determine the requested ratio as 100/m. The same result would be obtained if we chose any other basis of calculation: the values of m 1 and m might be different, but the ratio would be the same. Read Section 4.3b to see how to scale a flow chart from one basis to another, and work through the Test Yourself on p. 95. Degree-of-Freedom Analysis and Solution Strategy Once a basis of calculation has been chosen and the flow chart is drawn and fully labeled, carry out a degree-of-freedom analysis: unknowns (the number of unknown variables on the flow chart) balances (minus the number of independent balance equations) other equations (minus the number of other equations relating the unknowns) = DF degrees of freedom DF = 0 number of equations equals number of unknowns can solve for all unknowns DF > 0 (more unknowns than equations) either you ve overlooked some equations, problem is underspecified, or you can solve for all requested quantities without knowing all the unknowns. DF < 0 (more equations than unknowns) either flowchart is not fully labeled or problem is overspecified. -6
Here are some simple examples to illustrate: CHE 05 Chemical Process Principles x + y = 10 unknowns - independent equations ------------------------------ DOF x + y = 10 x y = 0 unknowns - independent equations ------------------------------ DOF x + y = 10 x + y = 0 unknowns - independent equations ------------------------------ DOF x + y = 10 unknowns x y = 0 - independent equations xy = 3 ------------------------------ DOF -7
CHE 05 Chemical Process Principles In this course, when you have zero degrees of freedom, the problem can be solved. For a non-reactive process, the number of independent material balance equations equals the number of independent species involved in the process. Independent molecular species: We discussed this in the context of how many balances we can do around a process. If two molecular species are in the same ratio everywhere in the process and the ratio is incorporated in the flowchart labeling, the species are not independent, and neither are balances on them. 100 mol CH 4 /s n n (mol/s) Gas blender 1 (mol air/s) x1(mol CH 4/ mol) 0.1 mol O /mol (1 x1 )(mol air / mol) 0.79 mol N /mol 0.1 mol O / mol air 0.79 mol N / mol air Looks like 3 unknowns 3 equations (balances on CH 4, O, N ) = 0 D.F., but it won t work. Write balances on O & N & see what happens. O balance: N balance: -8
CHE 05 Chemical Process Principles Example. Reconsider the distillation process shown again below. The basis of calculation is 000 L/h of the feed stream (exact, not 1 significant figure). 000 L/h (S.G.=0.87) m (kg/h) 1 0.45 kg B/kg 0.55 kg T/kg Distillation Column m (kg/h) 0.95 kg B/kg 0.05 kg T/kg Degree-of-freedom analysis (count only variables labeled on the chart): m m B3 T 3 (kg B/h) (8% of B in feed) (kg T/h) unknowns balances (, ) 1 1 = 0 DOF All unknowns can be calculated Since the problem is solvable, we may proceed to write out the equations. All balances have the form ( ). 000 L kg Feed density: m1 m 1(kg/h) (000)(0.87) (1) h L 8%: m 3 () B Mass balance: m1 m mb3 m T3 (3) Benzene balance: (4) Either do the algebra and arithmetic manually (easy in this problem, not so easy in problems to come) or use Excel s APEx or Solver to calculate the unknown quantities. Equations: m1 = 000*0.87 mb3 = 0.08*m1*0.45 m1 = m + mb3 + mt3 m1*0.45 = m*0.95 + mb3 kg kg kg B kg T m1 1744, m 760, mb3=6.8, mt3=91 h h h h m kg B m kg overhead xb = 0.0638, R 0.436 m m kg m kg feed B3 OF B3 T3 1-9
CHE 05 Chemical Process Principles Scale the flow chart: Now suppose we are asked to calculate the volumetric feed rate needed to produce 500 lb m /day of overhead product. The scale factor (see Example 4.3-) is the desired value of the specified quantity (overhead product rate = 500 lb m /day) divided by the value of the same quantity corresponding to the original basis of calculation (= 760 kg/h). 500 lb /day lb m/day 760 kg/h kg/h kg lb m/day 3 lb m feed ( m1) new ( m1) old SF 1744 3.9 5.7410 h kg/h day m Scale Factor (SF) = 3.9 Work through Problem 4.11 in the Student Workbook. Follow every step. Fill in the blanks yourself, then check the solutions. -10
CHE 05 Chemical Process Principles Multiple-unit balances (Section 4.4) We ve looked at processes represented by single boxes mixer, distillation column. Think about plants you ve seen from the highway all those towers & stacks & pipes running all over the place. How do you analyze something like that? Overall system: large box around system enclosing all units, recycle & bypass streams. Subsystems: individual units or combinations of units, stream mixing points & stream splitting points. DOF analyses as with single units. Go through Ex. 4.4-1. Work through Prob. 4.9. Go through Ex. 4.4- on your own or in groups. If you just read it you think you get it but you don t. Ask questions about each step, don t go on until you can answer it. -11
Problem 4.4-1: A Two-Unit Process. CHE 05 Chemical Process Principles 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. Three streams whose flow rates and/or compositions are not known are labeled 1,, and 3. Calculate the unknown flow rates and compositions of streams 1, and 3. Solution. First, fully label the chart and mark systems about which balances might be written. -1
Degree-of-freedom analysis: CHE 05 Chemical Process Principles Overall system: Mixing point: Unknowns: Independent balances: Unknowns: Independent balances: = = Degrees of freedom Degrees of freedom Unit 1: Mixing point (after solving unit 1) Unknowns: Independent balances: Unknowns: Independent balances: = = Degrees of freedom Degrees of freedom Therefore, we will solve the problem by: -13
CHE 05 Chemical Process Principles Overall mass balance: Overall balance of A: Mass Balance on Unit 1: Balance of A on Unit 1: Mass balance on Mixing Point: Balance of A on Stream Mixing Point: -14
CHE 05 Chemical Process Principles Recycle & Bypass (Section 4.5). 100 kg B/min 10 kg A/min 00 kg A/min 100 kg A/min 100 kg B/min Dump Sell 90 kg A/min What s wrong with this picture? How would you redraw it to be more cost efficient and environmentally conscious? Notes: Real reaction and real separation are always less than perfect. There is always some unreacted raw material, and some unwanted material in the product. Recycling is not free (requires pumping, pipes, etc.) but is only a one-time cost and usually pays for itself quickly Reasons to recycle: o Recovering a catalyst o Diluting a process stream o Controlling a process variable o Circulating a working fluid o Increasing the overall percent conversion of a reaction Sometimes a stream might bypass a process unit, where a fraction of the stream is diverted around the unit and combined with the output stream. This allows us to vary the composition and properties of the product stream. A stream that bypasses a unit introduces a splitting point in which a stream is split into two streams. -15
Splitting point: A stream is split into two streams. CHE 05 Chemical Process Principles 100 mol/s 60 mol/s 0.70 mol A/mol x1 (mol A/mol) 0.30 mol B/mol (1 x )(mol B/mol) 1 m (mol/s) x (mol A/mol) (1 x )(mol B/mol) Do a degree-of-freedom analysis: Did you get 1 DOF? Wrong. If all we re doing is splitting a stream, the compositions of the outlet streams must be the same as that of the feed stream. OK, so here s the revised flow chart. 100 mol/s 60 mol/s 0.70 mol A/mol 0.70 mol A/mol 0.30 mol B/mol 0.30 mol B/mol m (mol/s) 0.70 mol A/mol 0.30 mol B/mol Before we do the formal D.F. analysis, how many degrees-of-freedom do we have? Q: We know there should be 0. What went wrong? A: We are only allowed 1 balance. Q: Why? A: Because # balances = # independent species, which A and B are not in this case. Try writing the possible balances Mass Balance: A Balance: B Balance: -16
CHE 05 Chemical Process Principles Go through Example 4.5-1 on p. 110. Note reasons for recycling on p. 11. Go through Ex. 4.5-. If you can do that without looking at solution, you can handle recycle problems. Further practice: Workbook problems 4.3, 4.36. Example 4.5-: An Evaporative Crystallization Process. The flowchart of a steady-state process to recover crystalline potassium chromate (K CrO 4 ) from an aqueous solution of this salt is shown below. Forty-five hundred kilograms per hour of a solution that is one-third K CrO 4 by mass is joined by a recycle stream containing 36.4% K CrO 4, and the combined stream is fed into an evaporator. The concentrated stream leaving the evaporator contains 49.4% K CrO 4 ; this stream is fed into a crystallizer in which it is cooled (causing crystals of K CrO 4 to come out of solution) and then filtered. The filter cake consists of K CrO 4 crystals and a solution that contains 36.4% K 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 36.4% K CrO 4, is the recycle stream. 1. Calculate the rate of evaporation, the rate of production of crystalline K CrO 4, the feed rates that the evaporator and the crystallizer must be designed to handle, and the recycle ratio (mass of recycle)/(mass of fresh feed). Suppose that the filtrate were discarded instead of being recycled. Calculate the production rate of crystals. What are the benefits and costs of the recycling? Solution. 1. First, fully label the chart and mark systems about which balances might be written. -17
CHE 05 Chemical Process Principles Degree-of-freedom analysis: Overall system: Recycle-fresh feed mixing point: Unknowns: Independent balances: Additional relations: Unknowns: Independent balances: Additional relations: = = Degrees of freedom Degrees of freedom Evaporator: Crystallizer/filter: Unknowns: Independent balances: Additional relations: Unknowns: Independent balances: Additional relations: = = Degrees of freedom Degrees of freedom Therefore, we will solve the problem by: -18
CHE 05 Chemical Process Principles Overall K CrO 4 balance: Overall total mass balance: 95% specification: Mass balance around crystallizer: Water balance around crystallizer: Mass balance around recycle-fresh feed mixing point: -19
CHE 05 Chemical Process Principles 4.6. Chemical Reaction Stoichiometry Stoichiometry (4.6a) the proportions in which chemicals combine in reaction. Stoichiometric equation: 1 mol O consumed mol SO 3 generated SO + O SO 3,, etc. mol SO consumed mol SO consumed The stoichiometric coefficients define conversion factors from moles of one species consumed or formed in the reaction to moles of another species consumed or formed. For instant, if 50 mol/s of SO 3 are produced in the reaction 50 mol SO 3 formed 1 mol O consumed mol O consumed oxygen consumed 5 s mol SO formed s 3 Work through Test Yourself on p. 117. Limiting and excess reactants (4.6b) Feed reactants in stoichiometric proportion (e.g., mol SO /1 mol O ) all reactants run out at the same time. What if they re not in stoichiometric proportion? SO + O SO 3 (Sulfuric acid production, acid rain) 100 mol SO /s n 1 (mol SO 3 /s) 90 mol O /s n (mol SO /s) n 3 (mol O /s) Limiting reactant: reactant that runs out first, fed in less than stoichiometric proportion to all other reactants. In example, SO limits. (Stoichiometric proportion = SO : 1 O, actual feed proportion = 10 SO : 9 O ) Excess reactants: all other reactants but limiting one. In example, O is in excess. Theoretical (stoichiometric) requirement of an excess reactant: How much would be required to react completely with the limiting reactant. 100 mol SO fed 1 mol O Theoretical oxygen = 50 mol O (theoretical) s mol SO Percent excess of a reactant: (Amount fed theoretical amount)/(theoretical amount) x 100% 90 mol O 50 mol O %XS O = 100% 80% excess O (=80%XS air if O comes from air) 50 mol O -0
CHE 05 Chemical Process Principles Suppose you are told that oxygen is now fed in 60% excess (instead of at a rate of 90 mol/s). Then ( no ) fed ( no ) theoretical ( no ) excess ( no ) theoretical 0.60( no ) theoretical ( n ) (1 0.60) 1.60 (50 mol O /s) 80 mol O /s O theoretical % XS or in general, if A is fed in excess, ( na) fed ( n A) theoretical 1 100 Note: The theoretical and excess quantities depend only on the feed amounts and the stoichiometric reaction, not on what actually happens in the reactor. Fractional conversion of a reactant: moles reacted moles in moles out mole fed moles in SO + O SO 3 SO : f 100 mol SO /s 80(mol SO 3 /s) 90 mol O /s 0 mol SO /s 50 mol O /s 100 mol in/s 0 mol out/s 0.80 mol SO react/mol fed 100 mol in/s SO (Percentage conversion of SO =100 f SO = 80%) 90 mol in/s 50 mol out/s O : fo 0.44 mol O react/mol fed 90 mol in/s (Percentage conversion of O =100 f O = 44%) -1
CHE 05 Chemical Process Principles Exercise N + 3H NH 3 00 mol N n 1 (mol N ) 300 mol H n (mol H ) n 3 (mol NH 3 ) (a) The limiting reactant is (b) The percentage excess of the other reactant is (c) The theoretical amount of NH 3 produced is Now suppose the percentage conversion of nitrogen is 5% (d) n 1 =, n =, n 3 = (e) The percentage conversion of hydrogen is -
CHE 05 Chemical Process Principles Balances on Reactive Processes (4.7). There are three approaches: 1. Molecular species balances. Look again at the flow chart: SO + O SO 3 100 mol SO /s n 1 (mol SO 3 /s) 90 mol O /s n (mol SO /s) n 3 (mol O /s) We have three unknowns and three species ( 3 balances), so it looks like we should have zero degrees of freedom. We don t, however no matter how hard you try, you can t calculate any of those unknown variables without getting more information. Try using 0 = In-Out+Gen-Con However, let s say we are now told that n = 0 mol SO /s. SO + O SO 3 100 mol SO /s n 1 (mol SO 3 /s) 90 mol O /s 0 mol SO /s n (mol O /s) If molecular species balances are used (there are alternatives), then DOF = #unknowns () #independent molecular balances (3) #other equations (0) + #independent reactions (1) = 0 3 Independent reactions (Section 4.7b). Chemical reactions are not independent if we can get one in terms of the other by adding, subtracting, and multiplying them. A B (1) A B () (1) and () are not independent [ () = x(1)] A B (1) B C () (1) and () are independent, but A C (3) (1), (), and (3) are not [(3) = (1) + x()] -3
CHE 05 Chemical Process Principles mol SO consumed SO balance : I GOC A C I O 80 s SO balance : I GOC A O G 3 consumed n 1 s mol SO consumed s 80 mol SO mol SO3 generated mol SO3 O balance : I GOC A O I C n mol O 80 mol SO consumed mol O consumed 3 90 50 mol O s s mol SO consumed The original system must therefore have had degree(s) of freedom. / s. Extent of reaction (p. 119) For SO + O SO 3, if 10 mol SO 3 /s are formed, 10 mol SO /s & 5 mol O /s must react. Moles formed & reacted are always proportional to stoichiometric coefficients. mol SO 3 generated mol SO consumed mol O consumed = (mol/s) (pronounced zi) 1 where (mol/s) is the extent of reaction (same for all species, must be positive). It follows that ( nso ) final ( nso ) initial ( no ) final ( no ) initial ( n ) ( n ) SO final SO initial 3 3 For any single reaction, ni nio i (Batch) n n (Continuous, steady - state) i io i (4.6-3) where i is the stoichiometric coefficient of species i (+ for products, for reactants). One of these equations may be written for every independent molecular species involved in the process. For nonreactive species ( = 0), the equation reduces to ni nio). Q: What is? A. It s a dummy variable that reflects how far each reaction proceeds. It reflects the stoichiometry of the reaction and helps us keep up with generation and consumption. SO + O SO 3 100 mol SO /s n 1 (mol SO 3 /s) 90 mol O /s 0 mol SO /s n (mol O /s) -4 3
CHE 05 Chemical Process Principles DOF = #unknowns () #extent of reaction equations (3) #other equations (0) + #independent reactions(1) = 0 Apply extent of reaction equation ( ni n io i ), starting with the species we know most about: SO : 0 mol/s = 100 mol/s = 40 mol/s O : n 3 = = 50 mol O /s SO 3 : n 1 = = 80 mol SO 3 /s 3. Atomic species balances. Same process once more. SO + O SO 3 100 mol SO /s n 1 (mol SO 3 /s) 90 mol O /s 0 mol SO /s n (mol O /s) Instead of writing balances on the molecular species involved in the process (SO, O, SO 3 ), let s do it on the atomic species (S, O). Since atomic species are neither generated or consumed (except in nuclear reactions, which we don t consider), all balances reduce to I = O. With atomic species balances, DOF = #unknowns () #independent atomic balances () #other equations (0) 100 mol SO mol S 0 mol SO mol S n (mol SO ) mol S 1 3 S balance : s 1 mol SO s 1 mol SO s 1 mol SO 3 n 80 mol SO / s 1 3 3 100 mol SO mol O 90 mol O mol O n (mol SO ) mol O 1 3 O balance : = s 1 mol SO s 1 mol O s 1 mol SO 3 + n 50 mol O / s 3 Independent atomic species: If two atomic species are in the same ratio everywhere in the process, they are not independent. Look at TY #1 on P. 18. Q: How many independent atomic balances are there? A: Two. Write balances on C and H and see what happens. For more practice, do the Test Yourself on p. 119. -5
CHE 05 Chemical Process Principles Summary: Balances on reactive systems Three ways to write balances on reactive processes: Form #DF Molecular (# unknowns on chart) species + (# independent reactions) balances (# independent molecular species balances a ) Atomic species balances Extents of reaction (# other equations) (# unknowns on chart) -- (# independent atomic species balances b ) (# other equations) (# unknowns on chart) + [# independent reactions: (1 per reaction)] (# independent species balances a ) (# other equations) a If two molecular species are in the same ratio everywhere in the process and the ratio is incorporated in the flowchart labeling, the species are not independent b If two atomic species are in the same ratio everywhere in the process, they are not independent Which approach to use? DOF: If you re confident about independence of atomic species, use the atomic species balance approach, & if you re more confident about independence of reactions, use the extent of reaction approach. Note that all DOF approaches should give the same answer, so if you re unsure, use more than one to confirm. Balances: Don t use molecular species balances for reactive species (especially for multiple reactions it gets hairy fast) Hand calculations atomic balances Solver or chemical equilibrium extents of reaction. (Be sure reactions are independent) Note: You can use one method for the DOF analysis & another for the calculations. Work through Test Yourself on p. 134 Work through Example 4.7-1 on p. 131. If you fully understand the solution, you can solve any single-unit reactive balance problem you are likely to encounter on homework and tests. -6
CHE 05 Chemical Process Principles Chemical Equilibrium Calculations (Section 4.6c). A + B C If you are given an initial composition of a mixture of A, B, and C and an expression for the equilibrium constant K(T), and you are told that the reaction proceeds to equilibrium at a final temperature T, you can calculate the final composition using extents of reaction. See Example 4.6- for an illustration. Multiple reactions (4.6d) C H 6 C H 4 + H (Ethylene is the desired product, hydrogen is a byproduct) C H 6 + H CH 4 (Methane is an undesired byproduct) C H 4 + C H 6 C 3 H 6 + CH 4 (Propylene is an undesired byproduct) The second and third reactions both consume ethylene and so are undesired. Sometimes byproducts can be sold, and at other times they are worthless and possibly hazardous and have to be disposed of (another cost). Yield = mole of desired product / [moles that would have been formed if there were no side reactions and the reaction went to completion (the limited reactant was completely consumed)] Selectivity = moles of desired product / moles of undesired product (note, selectivity always refers to selectivity of desired A with respect to undesired B) If yield and selectivity are high, then we have successfully suppressed the undesired reactions. Note: in the ethylene example, H is not an undesired byproduct. An undesired byproduct is the result of a competing reaction that results in less of the desired product. A B + C B D B: Desired product, D: Unwanted byproduct 0 mol A 100 mol A 10 mol B 80 mol C 40 mol D Maximum possible B produced = mol (All A fed reacts, no side reaction) Yield (Eq. 4.6-4): Y B = Selectivity B/D (Eq. 4.6-5): S BD = A process engineer might take two different approaches to this reaction system: Maximize yield: Get most B you can, even if it means producing more D. Maximize selectivity: Hold down production of D, even if it means producing less B. Q: Why would you want to suppress production of a byproduct if it means getting less of the product you re selling? (Think of several possible reasons.) A: -7
CHE 05 Chemical Process Principles Extents of reaction for systems with multiple reactions Batch: ni ni0 ij j [ j(mol), i=species, j=independent reactions] j Cont., St.State: n n [ (mol/s), i=species, j=independent reactions] i i0 ij j j j (4.6-6) Recall is positive for products, negative for reactants, and ZERO for inerts (N, etc.) that go through the process without reacting. Example: A B + C [Rxn 1] A1 = 1, B1 = +, C1 = +1 B D [Rxn ] B = 1, D = +1 (All others = 0) 100 mol A/s n A (mol A/s) 0 mol B/s n B (mol B/s) 10 mol I/s n C (mol C/s) n D (mol D/s) n I (mol I/s) n 100, n, n n A 1 B C D, n I DOF: 5 unknowns ( na, nb, nc, nd, n I ) 5 equations + reactions = DOF. Specify any of the 7 variable values, fractional conversion of A, yield of B, or selectivity of B/D, & calculate the others. (Write equations, solve with Excel Solver.) -8
CHE 05 Chemical Process Principles Product Separation and Recycle (Section 4.7f) Talk through flowchart on p. 135 Identify fresh feed, recycle, feed to the reactor What is the percentage conversion of A? Depends on the definition of conversion : Single pass conversion: based on what goes into and comes out of the reactor. In this case it is 75%. A fed to reactor A leaving reactor Single-pass conversion of A = A fed to reactor = Overall conversion: Based on what comes in and out of the overall process. Overall conversion of A = A fed to process A leaving process A fed to process = We must specify whether a given conversion is overall or single-pass. Do Test Yourself on p. 135. Work through Example 4.7-. Note: only 10% of the propane entering the reactor is converted to propylene in a single pass. 99% of the unconsumed propane is recovered in the separation unit and recycled back, where it gets another chance to react. Net result = 95% of propane entering the process is converted to propylene and 5% leaves with final product. To achieve high conversion: 1. Design the reactor to achieve a high single-pass conversion o Requires very large residence time in reactor, large reactor volume, expensive. Design the reactor to achieve a low single-pass conversion, follow with a separation unit to recover and recycle unconsumed reactant. Decreased reactor cost Incur cost of separation process and cost of recycle line -9
CHE 05 Chemical Process Principles (b) Purge Inert gases (e.g. nitrogen, argon) are used in processes because they undergo chemical reactions under given conditions Using an inert gas in a system prevents undesirable chemical reactions from occurring (e.g. oxidation, hydrolysis, combustion) Purging with nitrogen minimizes fire hazards of residual solvents or process fluids If an inert gas is introduced to the system and it doesn t react, it has to exit the system somewhere otherwise, it builds up and would shut down the process Look at Fig. 4.7- on p. 138 production of ethylene oxide from ethylene. N is an inert. How did it get in the process? Purge stream: note it also contains ethylene and O. Why are we throwing away product and reactant? Note there is a cost to discard. But there is also a cost to separate and recycle the last little bit. Gas separation is hard and expensive. Engineers must evaluate the options based on economics, environmental regulations, etc. Note that purge stream take-off is a splitter (we can only write one balance one equation in the DOF). Work through Example 4.7-3. If you understand Example 4.7-3, you know how to do recycle problems, which are the hardest problems to solve. The key is flowcharting & systematic DOF analysis. General Strategies and Tips for Problem Solving 1. Break the problem down into small sub-problems when possible.. Even if the DOF analysis is not asked for, do it. That s your roadmap and confirmation that the problem is solvable. (5 minutes on the DOF analysis can save hours of trying to solve an impossible problem.) 3. For multiple unit processes or system with split, mix, recycle, purge, etc., do DOF for each subprocess. 10 equations and 10 unknowns for the entire process is literally correct, but if it doesn t converge you don t have a clue as to where to start looking for the error. 4. Use standard nomenclature: m for mass, n for moles, and x or y for composition. Avoid A-F, all X i, all M i. Someone may have to decipher your solution later. You may have to decipher it later. Save your creativity for solving the problem, not labeling it. 5. Clearly label your equations and the basis for the DOF analysis. This helps remind you and anyone checking your work what you have done. 6. Reality check: negative flowrates, negative compositions are a signal. Very small or very large numbers relative to others are also suspicious. 7. Always show complete units on the flowsheet and in calculations. -30
CHE 05 Chemical Process Principles Combustion Processes (Section 4.8) Combustion: Rapid reaction of a fuel with oxygen (usually but not always in air) Complete combustion: All C in fuel forms CO, all H forms H O, all S forms SO Partial combustion: Some C forms CO Air: Actual composition given on p. 143. For most of our purposes, assume 1.0 mole% O, 79.0% N 3.76 mol N /mol O, 4.76 mol air/mol O. Percent excess oxygen (= percent excess air): Based on complete combustion of fuel, regardless of whether all fuel actually reacts and whether some CO is formed. Wet-basis and dry-basis product gas compositions: Mole fractions of components with water included and not included, respectively. Example 100 mol CH 4 /s 1190 mol air/s 0.1 mol O /mol 0.79 mol N /mol 0 mol CH 4 /s 95 mol O /s 940 mol N /s 70 mol CO /s 10 mol CO/s 160 mol H O/s Q: Is the flow chart balanced? A: Yes. (C: 100 = 0 + 70 + 10; H: 400 = 80 + 30; O: 500 = 190 + 140 + 10 + 160: N : 940 = 940) Q: What reaction(s) are taking place? A1: CH 4 + 7 O CO + CO + 4H O WRONG! If that really were the reaction, what would the ratio of CO to CO in the product be? A: CH 4 + O CO + H O CH 4 + 3 O CO + H O Q: What is the percent excess air fed to the reactor? A: CH 4 + O CO + H O 100 mol CH4 mol O 4.76 mol air mol air Theoretical air 95 s 1 mol CH 1 mol O s 4 1190 mol fed/s 95 mol required/s %XS air 100% 5% excess air 95 mol required/s -31
CHE 05 Chemical Process Principles Q: What about the fact that not all the methane reacted and some CO was formed? A: Doesn t matter by definition, the percent excess air is based on complete combustion. Q: What are the wet-basis and dry-basis compositions of the product gas? A: Wet: (0+95+940+70+10+160) = 195 mol/s, 0 mol CH 4 /s mol CH4 xch 0.0154, etc. 4 195 mol/s mol Dry: (0+95+940+70+10) = 1135 mol/s, 0 mol CH 4 /s mol CH4 xch 0.0176, etc. 4 1135 mol/s mol Do Test Yourself on p. 146 Most material balance problems on combustion reactors are no different from those on any other reactor. Go through Example 4.8-3 for an illustration. Sometimes composition of fuel may be unknown, but if you know the atomic constituents of the fuel you can determine their ratio. Example 4.8-4 illustrates such a computation. Read Section 4.9. It outlines the differences between this course and the real world. -3