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CHEE 221: Chemical Processes and Systems Module 1. Material Balances: Single Process Units without Reaction (Felder & Rousseau Ch 4.1‐4.3) General Material Balance Equation (“GMBE”) Accumulation = In – Out + Generation – Consumption system boundary Input streams to system System over which mass balance is made output streams from system Accumulation within the system (mass buildup) = Input through system boundary + CHEE 221 ‐ Generation within the system Output through system boundary ‐ Consumption within the system 2 What is the System? CHEE 221 3 Some Basic Process Unit Functions Splitter – divides a single input into two or more outputs of the same composition (no reaction) splitter Mixer – combines two or more inputs (usually of different compositions) into a single output) (no reaction) mixer Separator – separates a single input into two or more outputs of different composition (no reaction) separator CHEE 221 4 Basic Process Unit Functions…cont’d Reactor – carries out a chemical reaction that converts atomic or molecular species in the input to different atomic or molecular species in the output Heat exchanger – transfers heat from one input to a second input (no reaction) Pump – changes the pressure of an input to that of the corresponding output (no reaction) reactor heat exchanger pump Actual process units can combine these different functions into a single piece of hardware, and are given different names, e.g. a separator can be a distillation column, a filter press, a centrifuge, etc. F&R Encyclopedia of Chemical Engineering Equipment ( textbook website) CHEE 221 5 Steam Boiler Steam Boiler CHEE 221 Heat Exchanger (no reaction) + Reactor (reaction) 6 Distillation—A Very Common Separation Process CHEE 221 7 Material (Mass) Balances (“MBs”)‐No Reaction A material balance is simply an accounting of material. For a given system in which no reaction is occurring (you will not be told this, and will need to know this from the type of unit that is under consideration; crystallizer, evaporator, filter, furnace, etc.), a material balance can be written in terms of the following conserved quantities: 1. 2. 3. Total mass (or moles) Mass (or moles) of a chemical compound Mass (or moles) of an atomic species To apply a material balance, you need to define the system and the quantities of interest (e.g. mass of a component, total mass, moles of an atomic species). What is your system, and what are you keeping track of? System – a region of space defined by a real or imaginary closed envelope (envelope = system boundary) – may be a single process unit, collection of process units or an entire process CHEE 221 8 Process Classification Before writing a material balance (MB) you must first identify the type of process in question. Batch – no material (mass) is transferred into or out of the system over the time period of interest (e.g., heat a vessel of water) Continuous – material (mass) is transferred into and out of the system continuously (e.g., pump liquid into a distillation column and remove the product streams from top and bottom of column) Semibatch – any process that is neither batch nor continuous (e.g., slowly blend two liquids in a tank) Steady‐State – process variables (i.e., T, P, V, flow rates) do not change with time Transient – process variables change with time CHEE 221 F&R Ch 4.1 9 F&R Example 4.1 Classify the following processes as batch, continuous, or semibatch, and transient or steady‐state. 1. 2. 3. 4. A balloon is filled with air at a steady rate of 2 g/min. A bottle of milk is taken from the refrigerator and left on the kitchen table. Water is boiled in an open flask. Carbon monoxide and steam are fed into a tubular reactor at a steady rate and react to form carbon dioxide and hydrogen. Products and unused reactants are withdrawn at the other end. The reactor contains air when the process is started up. The temperature of the reactor is also constant, and the composition and flow rate of the entering reactant stream are also independent of time. Classify the process (a) initially and (b) after a long period of time has elapsed. CHEE 221 10 Material Balance Simplifications The following rules may be used to simplify the material balance equation: Accumulation = In – Out + Generation – Consumption If the system is at steady‐state, set accumulation = 0 In – Out + Generation – Consumption = 0 If the balanced quantity is total mass, set generation = 0 and consumption = 0 (law of conservation of mass) Accumulation = In – Out If the balanced substance is a nonreactive species, (neither a reactant nor a product) or for non‐reacting systems in general, set generation = 0 and consumption = 0 Accumulation = In – Out CHEE 221 11 Problems Involving Material Balances Initial procedures will be outlined for solving single unit processes – No reaction (consumption = generation = 0) – Continuous steady‐state (accumulation = 0) – And so the Conservation Equation becomes….. (what?) These procedures will form the foundation for more complex problems involving multiple units and processes with reaction Following a standard methodology to solve problems is the key to success. This standard methodology will be illustrated via many examples in class, and is the one used by F&R. CHEE 221 12 Example 1 450 kmol per hour of a mixture of n‐butanol and iso‐butanol containing 30 mole % n‐butanol is separated by distillation into two fractions. The flow rate of n‐butanol in the overhead stream is 120 kmol/h and that of iso‐butanol in the bottom stream is 300 kmol/h. The operation is at steady‐state. Calculate the unknown component flow rates in the output streams. What is the mole fraction of n‐butanol in the bottom stream? What is the mass fraction? CHEE 221 13 Fractionation of Oil Wk 2 pre‐tutorial exercise: Draw a schematic of a plate (or tray) distillation column (continuous operation), and briefly explain how separation occurs F&R Encyclopedia of Chemical Engineering Equipment (textbook website) CHEE 221 14 Material Balance Procedures All material balance calculations are variations on a single theme: Given values of some input and output stream variables (e.g. flowrates, compositions), derive and solve equations for the others Solving the equations is a matter of simple algebra (the math is easy!), however, you first need to: convert the problem statement into a process flow diagram; what are the streams in/out and what components are in each stream? label the PFD with the ‘knowns” (flows, compositions, etc.), assign variables to the unknowns (remaining flows, compositions), identify the system on which you are doing the MB, and decide on your basis (mass/moles/input/output….) derive the necessary equations from the component and/or overall MB equations, and process constraint (PC) equations follow the standard methodology to solve the problem CHEE 221 15 Process Flow Diagrams A flowchart, or process flow diagram (PFD), is a convenient (actually, necessary) way of organizing process information for subsequent calculations. To obtain maximum benefit from the PFD in material balance calculations, you must: 1. Write the values and units of all known stream variables (flows and compositions) at the locations of the streams on the chart. 2. Assign algebraic symbols to unknown stream variables (flows and compositions) and write these variable names and their associated units on the chart. Your PFD is an essential part of the problem solution, and will be assigned marks for completeness. CHEE 221 F&R Ch 4.3a, Example 4.3‐1 16 Note on Notation The use of consistent notation is generally advantageous. For the purposes of this course, the notation adopted in Felder and Rousseau will be followed. For example: n – moles m – mass n – molar flow rate m – mass flow rate V – volume V – volumetric flow rate x – component fractions (mass or mole) in liquid streams y – component fractions in gas streams CHEE 221 17 Example 2 A spent sulfuric acid solution is brought up to strength for a pickling process in a mixer. Spent solution at 3% sulfuric acid (by weight) is mixed with a 50% solution (by weight) to obtain the desired product concentration of 40% acid by weight. All are aqueous solutions. Determine all flowrates on the basis of 100 lbm/h of product. If the actual flow of the spent stream is 300 lbm/h, what must the flowrates of the streams be? CHEE 221 18 Basis of Calculation Basis of calculation – is an amount or flow rate of one of the process streams on a mass or mole basis If a stream amount or flow rate is given in the problem statement, use this as the basis of calculation (usually) If no stream amounts or flow rates are known, you can assume one, preferably a stream of known composition – if mass fractions are known, choose a total mass or mass flow rate of that stream (e.g., 100 kg or 100 kg/h) as a basis – if mole fractions are known, choose a total number of moles or a molar flow rate CHEE 221 19 Flowchart Scaling Scaling – the process of changing the values of all stream amounts or flow rates by a proportional amount while leaving the stream compositions and conditions unchanged. Scaling up – final stream quantities are larger than the original quantities Scaling down – final stream quantities are smaller than the original quantities 30 mol A/min 70 mol B/min 40C, 1 atm 100 mol/min 0.30 mol A/mol 0.70 mol B/mol 40C, 1 atm CHEE 221 60 mol A/min 140 mol B/min Scale up process by a factor of 2 40C, 1 atm 200 mol/min 0.30 mol A/mol 0.70 mol B/mol 40C, 1 atm 20 Methodology for Solving Material Balance Problems 1. Choose a basis of calculation (input, output, mass, moles) 2. Draw and fully label a flowchart with all the known and unknown process variables (flows, compositions) as well as the basis of calculation. Be sure to include units. 3. Write any Process Constraint (PC) equations that relate variables. 4. Determine the number of unknowns and the number of equations that can be written to relate them. That is, does the number of equations equal the number of unknowns? 5. Solve the equations 6. Check your solution – does it make sense? Calculate the quantities requested in the problem statement if not already calculated 7. Cleary present your solution with the proper units and the correct number of significant figures “Understanding the Concepts” is not good enough. You will not be tested on “Understanding the Concepts”. You will be tested on your ability to set up and solve problems, and to get the correct answer. CHEE 221 21 Example 3: Quiz 1 2007 A mixture containing 42 wt% benzene (B) and 58 wt% toluene (T) is fed to a distillation column at a flowrate of 100 kg/min. The product stream leaving the top of the column (the overhead product) contains 90 wt% benzene, and 85 wt% of the total benzene fed to the column exits in this overhead product stream. Calculate the mass flowrate and mass composition of the product stream leaving the bottom of the column. Calculate the volumetric flowrate of the overhead product, assuming that it exits the distillation column as a vapour stream at 82 ºC and 1 atm (abs). Physical Property Data (S.G.=specific gravity) from Table B1: • Benzene S.G.=0.879 MW=78.11 g/mol • Toluene S.G.=0.866 MW=92.13 g/mol • Water density = 1.00 kg/L MW=18.02 g/mol R = 0.08206 L∙atm/(mol∙K) CHEE 221 22 Degrees of Freedom Analysis: A Motivating Example A feed stream containing C8 and C10 hydrocarbons is split into 3 product streams: an overhead fraction, a middle cut and a bottom fraction, whose mole fraction compositions are shown below. Seventy per cent of the C8 entering the column in the feed is recovered in the overhead. On the basis of 100 lb‐moles/h of feed determine the molar flow rates of the 3 product streams. C8: 0.516 C10: 0.484 100 lb‐moles/h C8: 0.300 C10: 0.700 CHEE 221 C8: 0.352 C10: 0.648 C8: 0.146 C10: 0.854 23 Independent Equations A set of equations are independent if you cannot derive one by adding and subtracting combinations of the others. x 2y z 1 Is this set of equations independent? 2x y z 2 y 2z 5 1 2 1 x 1 2 1 1 y 2 0 1 2 z 5 1 0 0 x 6 0 1 0 y 5 0 0 1 z 5 row reduce Rank = 3. No non‐zero rows in reduced form CHEE 221 24 Independent Equations… cont’d Are these sets of equations independent? x 2y z 1 2 y 4 z 10 y 2z 5 x 2y z 1 2x y z 2 3x 3 y 3 CHEE 221 25 Degree‐of‐Freedom Analysis A degree‐of‐freedom analysis (DFA) is a determination of the number of unknowns in a problem, and the number of independent equations that can be written. The difference between the number of unknowns and the number of independent equations is the number of degrees‐of‐freedom, DF or ndf, of the process. ndf nunknowns nindependent equations Possible outcomes of a DFA: – ndf = 0, there are n independent equations and n unknowns. The problem can be solved. – ndf > 0, there are more unknowns that independent equations. The problem is underspecified. ndf more independent equations or specifications are needed to solve the problem. – ndf < 0, there are more independent equations than unknowns. The problem is overspecified with redundant and possibly inconsistent relations. CHEE 221 26 DFA: Sources of Equations Sources of equations that relate unknown process variables include: 1. Material balances – for a nonreactive process, usually but not always, the maximum number of independent equations that can be written equals the number of chemical species in the process 2. Process constraints– given in the problem statement 3. Physical constraints – e.g., mass or mole fractions must add to 1 (usually taken care of when setting up PFD) 4. Stoichiometric relations – systems with reaction (later) 5. Energy balances – 2nd half of course CHEE 221 27 Notes on DFA: Dependent Material Balances There are two common situations where you will find fewer independent equations than species, and they are: 1. Balance around a splitter – Single input – two or more outputs with same composition – Only 1 independent balance equation, since: m1 = m2 + m3 (Overall Balance) and x1m1 = x2m2 + x3 m3 (Balance on A) but since x1 = x2 = x3, these balances are not independent – Splitters are used for: • Purge streams (reactor systems with recycle) • Total condensers at the top of distillation columns m1 kg/h x1 kg A/kg (1‐x1) kg B/kg CHEE 221 m2 kg/h x2 kg A/kg (1‐x2) kg B/kg Splitter m3 kg/h x3 kg A/kg (1‐x3) kg B/kg 28 Notes on DFA: Dependent Material Balances… cont’d Distillation Column with Total Condenser CHEE 221 29 Notes on DFA: Dependent Material Balances… cont’d 2. If two species are in the same ratio to each other wherever they appear in a process and this ratio is incorporated in the flowchart labeling, balances on those species will not be independent equations. – Situation occurs frequently when air is present in a nonreactive process (21 mol% O2; 79 mol% N2) – E.g., vapourization of liquid carbon tetrachloride into an air stream CHEE 221 n1 mol O2/s 3.76n1 mol N2/s n3 mol O2/s 3.76n3 mol N2/s n4 mol CCl4(v)/s n2 mol CCl4(L)/s n5 mol CCl4(L)/s Best to treat air as a single species in this situation 30 Summary: MB applied to single process units without reaction • Standard procedures was developed for single‐unit processes (F&R 4.3) – No reaction (Consumption=Generation=0) – Continuous steady‐state (Accumulation=0) • Develop good habits now, and practice. Problems will get more complex as we extend the procedures to multiple‐unit processes (starting in ≈Week 3) and processes with reaction (starting in Week 4/5) • Standard procedures are summarized in F&R Section 4.3 and include: – drawing/labeling a process flow diagram (4.3a) These are critical sections – selecting a basis of calculation (4.3b) of the text and – setting up material balances (4.3c) form the basis for Quiz 1 – performing a degree of freedom analysis (4.3d) CHEE 221 31 Example 4 Hot soap is chilled on a roller and scraped continuously from the roller onto a moving conveyor belt which carries the soap into a dryer (see below). The entering soap contains 25% water by weight. It is desired to reduce the water content to 15% by weight and to produce 1200 lb/h of nearly dry soap chips. The entering air contains 0.3 mole % water vapour. The dryer manufacturer suggests that the dryer operates efficiently when the nearly dry air/wet soap flow ratio is 3.0. Calculate the unknown flowrates and compositions. Air is 21% oxygen and 79% nitrogen (mole basis) and has a molecular weight of 29.0 g/mol. moist air hot, nearly dry air Soap Dryer wet soap chips CHEE 221 dried soap chips 32 Condensers and Evaporators Wk 3 pre‐tutorial exercise: Dryers and dehumidifiers are two examples of a general class of separators known as condensers. Look into the difference between condensers and evaporators, and explain their industrial usage. Can you think of a household example of an evaporator? F&R Encyclopedia of Chemical Engineering Equipment (textbook website) CHEE 221 33 Example F&R 4.3‐1 An experiment on the growth rate of certain organisms 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 a rate of 20.0 cm3/min B: Air (21 mole% O2, the balance N2) C: Pure oxygen, 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 a PFD, and calculate all unknown stream variables (i.e. flows and compositions). Work through on your own, then check with solution in the textbook CHEE 221 34 Example: Quiz 1 2009 A continuous distillation column is to be used to separate a 3‐component mixture of acetic acid (AA), water (W) and Benzene (B), and a trial run gave the data below (mass basis). The data for the benzene in the feed (which consists of AA, W and B) was not taken because of an instrument malfunction. Use a degree of freedom analysis, and then calculate the benzene flow in the feed in kg/h. Waste 10.9% AA 21.7% W 67.4% B Aqueous Solution (containing 80% AA + 20% W) + B (data not available) Product 350 kg/h pure AA Answer: B in feed = 311 kg/h CHEE 221 35