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Introduction to Chemical Engineering ProcessesPrint Version From Wikibooks the opencontent textbooks collection Contents E 0 1 Chapter 1 Prerequisites O O O O 11 Consistency of units 111 Units of Common thsical Properties 112 SI gkg m s System 1121 Derived units from the SI svstem 113 CGS gcm g s system 114 English system 12 How to convert between units 121 Finding equivalences 122 Using the equivalences 13 T analysis as a check on equations 14 Chapter 1 Practice Problems 2 Chapter 2 Elementam mass balances O O O O O 21 The quotBlack Boxquot approach to problem solving 211 Conservation equations 212 Common 0n the conservation equation 22 Conservation of mass 23 Converting Information into Mass Flows Y 24 Volumetric Flow rates 241 Why they39re useful 242 Limitations 243 How to convert volumetric ow rates to mass ow rates 25 Velocities 251 Why thev re useful 252 Limitations 253 How to convert velocitv into mass ow rate 26 Molar Flow Rates 261 Why they39re useful 262 Limitations 263 How to Change from Molar Flow Rate to Mass Flow Rate 27 A Typical Type of Problem 28 Single Component in Multiple Processes a Steam Process 281 Step 1 Draw aFlowchart 282 Step 2 Make sure vour units are J O 0 3 Chapter 3 Mass balances on mn n O O O O O O O O 283 Step 3 Relate your variables 284 So vou want to check vour guess Alright then read on 285 Step 4 Calculate vour unknowns 286 Step 5 Check your work 29 Chapter 2 Practice Problems svstems 31 Component Mass Balance 32 Concentration Measurements 321 Molarigy 322 Mole Fraction 323 Mass Fraction 33 C on Multi streams 331 Average Molecular Weight 332 Density of Liquid Mixtures 3321 First Eguation 3322 Second Equation 34 General Strategies for Multiple Component Operations 35 Multiple C in a Single Operatinn39 emm rm of Ethanol and Water 351 Step 1 Draw aFlowchart 352 Step 2 Convert Units 353 Step 3 Relate your Variables 36 Introduction to Problem Solving with Multiple Components and Processes 37 Degree of Freedom Analysis 371 Degrees of Freedom in Multiple Process Svstems 38 Using Degrees of Freedom to Make a Plan 39 Multiple Components and Multiple Processes Orange Juice Production 391 Step 1 Draw aFlowchart 392 Step 2 Degree of Freedom analvsis 393 So how to we solve it 394 Step 3 Convert Units 395 Step 4 Relate your variables 310 Chapter 3 Practice Problems 1 4 Chapter 4 Mass balances with recvcle O O O O 41 What Is Recycle 411 Uses and Benefit of Recycle 42 Differences between Recycle and non Recycle systems 421 Assumptions at the Splitting Point 422 A nmntinn at the Point 43 Degree of Freedom Analvsis of Recvcle Svstems 44 Suggested Solving Method 45 Example problem Improving a Separation Process 451 Recvcle on the Separation Process 4511 Step 1 Draw aFlowchart 4512 Step 2 Do a Degree of Freedom Analvsis 4513 Step 3 Devise a Plan and Carry it Out 46 Systems with Recvcle a Cleaning Process 461 Problem Statement 462 First Step Draw a Flowchart 463 Second Step Degree of Freedom Analvsis 464 Devising a Plan 465 Converting Units 466 Carrying Out the Plan 467 Check your work 0 5 Chapter 5 Massmole balances in reacting systems 0 51 Review of Reaction v 52 Molecular Mole Balances 53 Extent of Reaction 54 Mole Balances and Extents of Reaction 55 Degree of Freedom Analysis on Reacting Systems 56 Complications 561 I J J and Denendent 5611 Linearly Dependent Reactions 562 Extent of Reaction for Multiple Independent Reactions 563 Eguilibrium Reactions 5631 Liguid phase Analysis 5632 Gas phase Analysis 564 Special Notes about Gas Reactions 565 Inert Species 57 Example Reactor Solution using Extent of Reaction and the DOF 58 Example Reactor with Eguilibrium 59 I J to Reactions with Recvcle 510 Example Reactor with Recycle 5101 DOF Analysis 5102 Plan and Solution 5103 Reactor Analvsis 5104 Comparison to the situation without the separatorrecycle system 6 Chapter 6 Multiple phase systems introduction to phase equilibrium 7 Chapter 7 Energv balances on non reacting svstems 8 Chapter 8 Combining energy and mass balances in non reacting systems 9 Chapter 9 Introduction to energy balances on reacting systems 10 Appendix 1 Useful Mathematical Methods 0 101 Mean and Standard Deviation 1011 Mean 1012 Standard Deviation 1013 Putting it together 0 102 Linear Regression 1021 Example of linear regression 1022 How to tell how good your regression is o 103 Linearization 1031 In general 1032 Power Law 1033 Exponentials 00000 0000 o 104 Linear Integpolation 1041 General formula 1042I i i of Linear Intemolation 105 References 106 Basics of Rootfinding 107 Anal ical vs Numerical Solutions 108 Rootfinding Algorithms 1081 Iterative solution 1082 Iterative Solution with Weights 1083 Bisection Method 1084 Regula Falsi 1085 Secant Method 1086 Tangent Method Newton39s Method 0 109 Whatis a System of Equations 0 1010 Solvability 0 1011 Methods to Solve Systems 10111 Example of the Substitution Method for Nonlinear Systems 0 1012 Numerical Methods to Solve Svstems 10121 Shots in the Dark 10122 Fixed point iteration 10123 Looping method 101231 Looping Method with Spreadsheets 10124 Multivariable Newton Method 101241 Estimating Partial Derivatives 101242 Example of Use of Newton Method 0 11 Appendix 2 Problem Solving using Computers 0 111 Introduction to Spreadsheets o 112 Anatomy of a spreadsheet o 113 Inputting and Manipulating Data in Excel 1131 Using formulas 1132 Performing Operations on Groups of Cells 1133 Special Functions in Excel 11331 Mathematics Functions 11332 Statistics Functions 11333 Programming Functions 0 114 Solving Fauations in S J Goal Seek o 115 Graphing Data in Excel 1151 Scattegplots 1152 Performing of the Data from a Scattemlot o 116 Further resources for S J o 117 Introduction to MATLAB o 118 Inserting and Manipulating Data in MATLAB 1181 Importing Data from Excel 1182 Performing Operations on Entire Data Sets 0 119 Graphing Data in MATLAB 1191 Pol omial Regressions 1192 Nonlinear f 5 0 12 Amendix 3 Miscellaneous Useful Information 0 121 Whatis a quotUnit Operationquot 0 122 Separation Processes 1221 Distillation 1222 Gravitational Separation 1223 Extraction 1224 Membrane Filtration o 123 Purification Methods 1231 Adsoggtion 1232 Recrystallization o 124 Reaction Processes 1241 Plug ow reactors PFRs and Packed Bed Reactors PBRs 1242 Continuous Stirred Tank Reactors CSTRs and Fluidized Bed Reactors Bs 1243 Bioreactors o 125 Heat Exchangers 1251 Tubular Heat Exchangers 13 Agpendix 4 Notation o 131 A Note on Notation 132 Base Notation in quot order 133 Greek 134 Subscripts 135 Embellishments o 136 Units Sectionl Analeis 14 Amendix 5 Further Reading 0 15 Amendix 6 External Links 16 AQpendix 7 License 0 161 0 PREAMBLE 162 1 APPLICABILITY AND DEFINITIONS 163 2 VERBATIM COPYING 164 3 COPYING IN UANTITY 165 4 MODIFICATIONS 166 5 COMBINING DOCUMENTS 167 6 COI I FCTIONS OF DOCUMENTS 168 7 AGGREGATION WITH INDEPENDENT WORKS 169 8 TRANSLATION 1610 9 TERMINATION 1611 10 FUTURE REVISIONS OF THIS LICENSE O O O O O O O O O O O O O O git Chapter 1 Prerequisites git Consistency of units Any value that you39ll run across as an engineer will either be unitless or more commonly will have specific types of units attached to it In order to solve a problem effectively all the types of units should be consistent with each other or should be in the same system A system of units defines each of the basic unit types with respect to some measurement that can be easily duplicated so that for example 5 ft is the same length in Australia as it is in the United States There are five commonly used base unit types or dimensions that one might encounter shown with their abbreviated forms for the purpose of dimensional analysis Length L or the physical distance between two objects with respect to some standard distance Time t or how long something takes with respect to how long some natural phenomenon takes to occur Mass M a measure of the inertia of a material relative to that of a standard Temperature T a measure of the average kinetic energy of the molecules in a material relative to a standard Electric Current E a measure of the total charge that moves in a certain amount of time There are several different consistent systems of units one can choose from Which one should be used depends on the data available Units of Common Physical Properties Every system of units has a large number of derived units which are as the name implies derived from the base units The new units are based on the physical definitions of other quantities which involve the combination of different variables Below is a list of several common derived system properties and the corresponding dimensions 2 denotes unit equivalence If you don39t know what one of these properties is you will learn it eventually Mass 2 M Length 2 L Area 2 LAZ Volume 2 LA3 Velocity 2 Lt Acceleration 2 LtAZ Force 2 MLtA2 EnergyWorkHeat 2 MLA2tA2 Power 2 MLA2tA3 Pressure 2 M LtA2 Density 2 MLA3 Viscosity 2 M Lt Diffusiyity 2 LAZs Thermal conductivity 2 MLtA3T Specific Heat Capacity2 LAZTtA2 Specific Enthalpy Gibbs Energy 2 LA2tA2 Specific Entropy 2 LA2 tA2T Iedit SI kgms System This is the most commonly used system of units in the world and is based heavily on units of 10 It was originally based on the properties of water though currently there are more precise standards in place The major dimensions are L i meters m t seconds s M i kilograms kg T 5 degrees Celsius 0C E 39 Amperes A where idenotes unit equivalence The close relationship to water is that one mquot3 of water weighs approximately 1000 kg at 00C Each of these base units can be made smaller or larger in units of ten by adding the appropriate metric pre xes The specific meanings are from the g page on Wikipedia SI Prefixes Name yotta zetta exa peta tera giga mega kilo hecto deca SymbolYZEP TGMkhda Factor 1024 1021 1018 1015 1012 109 106 103 102 101 Name deci centi milli micro nano pico femto atto zepto yocto Symbol d c m u n p f a z y Factor 10 1 10 2 10 3 10 6 10 9 10121015 10481021 10 24 If you see a length of 1 km according to the chart the prefix quotkquot means there are 103 of something and the following quotInquot means that it is meters So 1 km 2 103 meters It is very important that you are familiar with this table or at least as large as mega M and as small as nano n The relationship between different sizes of metric units was deliberately made simple because you will have to do it all of the time You may feel uncomfortable with it at first if you re from the US but trust me after working with the English system you39ll learn to appreciate the simplicity of the Metric system Ldit Derived units from the SI system Imagine if every time you calculated a pressure you would have to write the units in kgmsquot2 This would become cumbersome quickly so the SI people set up derived units to use as shorthand for such combinations as these The most common ones used by chemical engineers are as follows Force 1 kgms02 1 Newton N Energy 1 Nm l J Power 1 Js 1 Watt W Pressure 1 NmAZ 1 Pa Volume 1 mA3 1000 Liters L Thermodynamic temperature 1 0C K e 27315 K is Kelvin Another derived unit is the mole A mole represents 6022quot 1023 molecules of any substance This number which is known as the Avogadro constant is used because it is the number of molecules that are found in 12 grams of the 12C isotope Whenever we have a reaction as you learned in chemistry you have to do stoichiometry calculations based on moles rather than on grams because the number of grams of a substance does not only depend on the number of molecules present but also on their size whereas the stoichiometry of a chemical reaction only depends on the number of molecules that react not on their size Converting units from grams to moles eliminates the size dependency CGS cmgs system The so called CGS system uses the same base units as the SI system but expresses masses and grams in terms of cm and g instead of kg and m The CGS system has its own set of derived units see wzcgs but commonly basic units are expressed in terms of cm and g and then the derived units from the SI system are used In order to use the SI units the masses must be in kilograms and the distances must be in meters This is a very important thing to remember especially when dealing with force energy and pressure equations edit English system The English system is fundamentally different from the Metric system in that the fundamental inertial quantity is a force not a mass In addition units of different sizes do not typically have prefixes and have more complex conversion factors than those of the metric system The base units of the English system are somewhat debatable but these are the ones I39ve seen most often r Length L i feet ft seconds s E i poundseforce lbf T 39 degrees Fahrenheit 0E The base unit of electric current remains the Ampere There are several derived units in the English system but unlike the Metric system the conversions are not neat at all so it is best to consult a conversion table or program for the necessary changes It is especially important to keep good track of the units in the English 111 11s us your calculations 1e for 1 force you ll end up with units like Btuin instead of Just pounds 111 1 W ufthestzme mt m mm mm same base 1mm edit How to convert between units 1 Finding equivalence 111 111 11111 u 1 u 111 order to yo want and the 11 its yo e To do this use a conversiontable See w enverslon ofunits for a fairly extensive But not exhaustive list of common units and their equivalences1 listed h 1 1nL 1 cm sometimes but not on Wtkipedia wntten 1n the form 1unitl numbcrunit2 numbsmumts For example you might recall the following conversion from chemistry class 1 atm 760 1man 1tl13106 Pa 1013 bar The table on W h 1 1 1 1s 1 11 m we have 1 of the middle is the definition of the uh1t on the left and on the far rightrhand foot International ft 13 yd 03048 m 1 Using the equivalence Once the equivalences ate detetmhed use the general form What vou want Vt lmr vou want What you have gtgt V1113 11 have The hattmh 0n the nght comes duectly from the tonvetsmh tables 16 t 4 t Example Convert 800 mmHg into bars Ami u u H HM d w dnectly 1013 bar bar 300 1an e 7 1066 hm g 760 mmHg nnedm quot then convert from the intermediate to the desired unit We would nd that 1 1man 2 Pamdl bar 10 pa Agam we have to set 1t up usmg the same general form Just we have to do 1t tw1ce 133322 Pa 11m bar 300 mmllg 7 s 1066 bar 1 mmHg 10 Pa r L L It s a very important sk111 for any engmeex keep from avmdmg domg 1t backwards 15 to wnte everythmg out and make suxe somethmg One way to your umts tahte1 out as they shau1dz f you hyta do 1t backwards you ll end up w1th hke this 760 mmHg 5mmHgZ bm s 7 sou mmHg W 7 60 e 10 V f you wnte everythmg even tahvetsmhs w1th1n the meme system out and make Sure that everything tahte1s you ll help m1tigate umtrchanging ems About 304070 of an mistakes Ve mbex them well edit Dimensional analysis as a check on equations Since we know what the units of velocity pressure energy force and so on should be in terms of the base units L M t T and E we can use this knowledge to check the feasibility of equations that involve these quantities a VAquot Example Analyze the following equation for dimensional consistency P 9 5 f1where g is the gravitational acceleration and h is the height of the uid SolutionWe could check this equation by plugging in our units FEML 2 t2 h39L gem g h39L2t2 7t ML a t Since gh doesn t have the same units as P the equation must be wrong regardless of the system of units we are using The correct equation in fact is P p nae g a h l l l MLI where p18 the dens1ty of the uid Density has base units of 4 so 2 2 1 2 p 9 1 sit Ail 1 i L t which are the units of pressure This does not tell us the equation is correct but it does tell us that the units are consistent which is necessary though not sufficient to obtain a correct equation This is a useful way to detect algebraic mistakes that would otherwise be hard to find The ability to do this with an algebraic equation is a good argument against plugging in numbers too soon edit Chapter 1 Practice Problems Problem 1 Perform L youimswei lb 153 i i a 5 hr 45 102 w Lquot 1 mm M a E umquot m J W 4187 4v 7 i1 9 00 7 0Fnote kWh means kilowattrhoux e 100 m3 g L 4 aim m m5 Problem 2 Perform a dimensional amiysis on the following equations is determine If they sie reasonable 1 zit WheIeV lsVelOclty d is distance andt istime m x y b r where F is force m is mass v is Velocity and i is miiiiis a distance r J V gwhexe is density v is Volume and g is gravitational acceleration V m 7 i1 Pwheie in is mass owmte Visvoiumemc flow rate and Pis density Problem 3 Recall that the uleal gas law IS PV nBTwhere P IS pressure v IS volume n IS number of moles R ls a constant and T IS th temperature a What are the unlts of R ln terms of the base unlt types length tlme mass and temperature these 831 b show L how two values of R are equwalent MM 4 R 008217 7 m0 gtk K mot gt2 Ii 5 fanldeal gas exrsts maclosedcontalnerwthamolaxdensltyof 39 L atapressureof 096 A 10 Pa what temperature lsthe contalner held at7 4139 na l Ln lu ran 4 F 10 Part the total pressure ln the tontalner ls 6 Blunt 2 At what temperatures and pressures IS a gas most and least llkely to be ldealt hlnt you can t use lt when you have a llquld d an a pressure of 500 Torr and contams 50 moles of water vapor and 30 moles ofwater at 700C Torr an 700C th that A h a o l l l You retomblne the gasses lnto a slngle tank the same slze as the rst two Assumng that the 1 atm pressure wlll lt blow upt Problem 1 H lt 7 gt H 7 4 Conslder the reat on 202 20 202 tr whlch IS tamed out by many organlsrns as a way to ehmlnate hydrogen peroxlde a nu r b What is the standard Gibbs energy change of this reaction Under what conditions does it hold In what direction is the reaction spontaneous at standard conditions 0 What is the Gibbs energy change at biological conditions 1 atm and 370C if the initial hydrogen peroxide concentration is 001M Assume oxygen is the only gas present in the cell d What is the equilibrium constant under the conditions in part c Under the conditions in part b What is the constant independent of 6 Repeat parts a through d for the alternative reaction 11202 I2 I 02 Why isn39t this reaction used instead FA Problem 5 Two ideal gasses A and B combine to form a third ideal gas C in the reaction A I B gt Suppose that the reaction is irreversible and occurs at a constant temperature of 250C in a 5L container If you start with 02 moles of A and 05 moles of B at a total pressure of 104 atm what will the pressure be when the reaction is completed 13 Problem 6 How much heat is released when 45 grams of methane are burned in excess air under standard conditions How about when the same mass of glucose is burned What is one possible reason why most heterotrophic organisms use glucose instead of methane as a fuel Assume that the combustion is complete ie no carbon monoxide is formed Problem 7 Suppose that you have carbon monoxide and water in a tank in a 1521 ratio a In the literature nd the reaction that these two compounds undergo hint look for the water gas shift reaction Why is it an important reaction 17 Using a table of Gibbs energies of formation calculate the equilibrium constant for the reaction c How much hydrogen can be produced from this initial mixture d What are some ways in which the yield of hydrogen can be increased hint recall Le Chatlier s principle for equilibrium 6 What factors do you think may in uence how long it takes for the reaction to reach equilibrium Chapter 2 Elementary mass balances editl The quotBlack Boxquot approach to problemsolving In this book all the problems you39ll solve will be quotblack boxquot problems This means that we take a look at a unit operation from the outside looking at what goes into the system and what leaves and extrapolating data about the properties of the entrance and exit streams from this This type of analysis is important because it does not depend on the specific type of unit operation that is performed When doing a blackbox analysis we don t care about how the unit operation is designed only what the net result is Let39s look at an example edit Conservation equations The formal mathematical way of describing the black box approach is with conservation equations which explicitly state that what goes into the system must either come out of the system somewhere else get used up or generated by the system or remain in the system and accumulate The relationship between these is simple 1 9 b The streams entering the system cause an increase of the substance mass energy momentum etc in the system The streams leaving the system decrease the amount of the substance in the system Generating or consuming mechanisms such as chemical reactions can either increase or decrease the stuff in the system What39s left over is the amount of stuff in the system With these four statements we can state the following very important general principle Acemnuiaiion In Out Generation Its so important in fact that you39ll see it a million times or so including a few in this book and it is used to derive a variety of forms of conservation equations Iedit Common assumptions on the conservation equation The conservation equation is very general and applies to any property a system can have However it can also lead to complicated equations and so in order to simplify calculations when appropriate it is useful to apply assumptions to the problem Closed system A closed system is one which does not have ows in or out of the substance Almost always when one refers to a close system it is implied to be closed to mass ow but not to other ows such as energy or momentum The equation for a closed system is Accumulation Generation The opposite of a closed system is an open system in which the substance is allowed to enter andor leave the system The funnel in the example was an open system because mass owed in and out of it No generation Certain quantities are always conserved in the strict sense that they are never created or destroyed These are the most useful quantities to do balances on because then the user does not need to worry about a generation term Accumulation I 71 O at The most commonly used conserved quantities are mass and energy It is important to note however that though the total mass and total energy in a system are conserved the mass of a single species is not since it may be changed into something else Neither is the quotheatquot in a system if a so called quotheat balancequot is performed Therefore one must be careful when deciding whether to discard the generation term Steady State A system which does not accumulate a substance is said to be at steady state Often times this allows the engineer to avoid having to solve differential equations and instead use algebra In Out Generation 0 All problems in this text assume steady state but it is not always a valid assumption It is mostly valid after a process has been running for long enough that all the ow rates temperatures pressures and other system parameters have reached equilibrium values It is not valid when a process is first warming up and the parameters wobble significantly How they wobble is a subject for another course edit Conservation of mass TOTAL mass ts a consewed quanttty except tn nuc1eat teacttons let s not go thete as ts the A A spectestth h th st Letus In Out 0 New thete ate two major ways tn whtch mass can entet or leave a system dtttuston and convectton However for tatgescate systems such as the ones constdeted hete tn whtch the ow Massm prA V into stgntty convecttve mass ow tate tn untts of m1155 time Stnce the total flow tn ts the sum of tndtvtduat ows and the same thh the ow out the followmg steady state mass balance ts ohtatned fo the ovemll mass tnthe system f tt ts a batch system 0 1fwe xe lookmg at how much has enteted and left tn a gtven pettod of tttne tathet than tnstantaneousty we can apply the same mass balance whom the tttne mn nPnV Yn tht h k 2mm Zm m D Thls is a fairly myial example but it gets the concepts of in out and accumulatlon on a ph ysical basis which is important for setting up problems Ln the next section it will be shown edit Converting Information into Mass Flows Introduction in any system there will be certain parameters that are often considerably easier to measure andor control than others When you are solying any problem and tryingto use a mass balance or any other equation X z mpumzvlt tn recagmze what piece of mfmrmmuw can be intercuvlwarted The purpose of this section is to show some of the more common altematiye for mnl w my velocity than it is to measure a mass flow rate directly edit Volumetric Flow rates A volumetric ow rate s a relation of how muda Volume of a gas or liquid solution passes oi i through a fixed p nt in a system typically the entrance or exit point of a process in a given amount oftlme it ls denoted as Vulume ttm in sneam n Volume in the metric system is typically expressed either in L dmAS mL cmAS or mAS Note that a cubic meter is very large a cubic meter otwater weighs about loookg 2200 pounds at room temperature Q Why they39re usetul rdpr A to usean mm l calcul Fm tin vthv one complesslblilty ls stnctly a lnctlon of temperature and pressure for any gas or gaseous mlxtuxe edit Limitations Volumetric Fiowrates are Not Conserved We can wrlte abalance on volume like anylhlng else ut the Volume generatlon term would be a complex lnctlon of system plopextles ll ldcl n rate before appiylng the balance equatlons Volumetnc flowmtes also do not lend themselves to spllttlng lnto components smce when we speak of V0 umes in practrcal terms we generally thlnk of the total soiutlon volume not the partral Volume of each component the latter ls a useful tool for thermodynamrcs but thats hm Hm l relatwelyuncommon edit How to uan volumetric ow ram to mass ow ram Volumetnc owrates are related to mass flow rates by a relatlvely easyrtormeasule physlcal property Slnce and 39 n d a pr p l mu r quotms1 01 um in order to convert them The density sewes thls purpose mcelyz W a pm mum streamn The l indicates that we re talking about one pamcuiar flow stream here smce each ow may have a dlffelent denslty mass ow rate or Volumemc flow rate edit Velocities The Veioclty of a bulk uld ls haw mm theml meme aluvlg the system mmlly e pipe p passe per pm me The Veioclty of a bulk uld llke any other has umts of listtmce U7 7 time in sneam n By detrmtron the bulk Veioclty of a uld ls related to the Voiumetllc ow rate by the uid at a la i ii the center of a pipe The bulk velocity is about the same as the instantaneous velocity for relatively fast flow or especially for ow of gasses For purposes of this class all velocities given Will be bulk velocities not instantaneous Velocities edit Why they39re useiui Bulk Velocities are useful because measure r constant pressure drop at steady state as they pass through the orifice or other similar merits Ln 1 i ii ike volumetric flow rates they are relatively easy to instru w Limitations 39 39 velocity changes With temperature and pressure of a gas though for a liquid velocity is generally constant along the length of a pipe Also Velocities can t be split into the ows of individual components since all of the components Will generally flow at the same speed They need to be convened into something m l v n v t be applied edit How to umvert Velocity inn mass ow rate in order to convert the velocity of a uid suearn into a mass ow rate you need two pieces of information i The cross sectional area of the pipe 2 The density ofthe uid Ln order to convert first use the definition of bulk velocity to convert it into a volumemc ow rate Vquot vn Aquot Then use the density to convert the volumemc ow rate into a mass flow rate 1m prn The cornbrnatron ofthese two equations is useful in Vquot In Arvin strearnn ed Molar Flow Rates of a solution or mixture that pass a fixed point per unit time males tuna in sueam n edit Why they39re usetul Molar ow rates are mostly useful because usmg mule Instead uf max alluw yuu tn wme r Tn ther words there are a lot less unknowns when you use a mole balance since the storchrornetry allows you to This will be discussed more in a latex chapter edit Limitations thk a reaction or not but the same is not true for the number of moles For example consider the reaction between hydrogen and oxygen gasses to form water i H 3 70 g H 20 This reaction consurnes 15 moles of reactants for every mole of products produced and However since neither rnass nor moles of individual components is consewed in a reacting systemr r A 7th M t v The molar ows are also somewhat less practical than mass ow rates since you can39t measure moles directly but you can measure the mass of something and then convert it to moles using the molar ow rate edit How to Change from Molar Flow Rate to Mass Flow Rate Molar ow rates and mass ow rates are related by the molecular weight also known as the molar mass of the solution In order to convert the mass and molar ow rates of the entire solution we need to know the average molecular weight of the solution This can be calculated from the molecular weights and mole fractions of the components using the formula JtflaVn HM W i yilln where i is an index of components and n is the stream number y39isignifies mole fraction of each component this will all be defined and derived later Once this is known it can be used as you would use a molar mass for a single component to find the total molar ow rate mu 7111 1 E39V nin stream 11 git A Typical Type of Problem Most problems you will face are significantly more complicated than the previous problem and the following one In the engineering world problems are presented as so called quotword problemsquot in which a system is described and the problem must be set up and solved if possible from the description This section will attempt to illustrate through example step by step some common techniques and pitfalls in setting up mass balances Some of the steps may seem somewhat excessive at this point but if you follow them carefully on this relatively simple problem you will certainly have an easier time following later steps Single Component in Multiple Processes a Steam Process d setuny and am V schatg39ea as wastethebuhet is fed mytifkgyheat exchanger whey it is cooled This sueam leaves the eat xmmger at a latent 15th pounds pethout calculate the owmte ofthe aischatge aha the efficiency otthe evaporatok V L 39 V t A tr A 39 rm gt ehtets a tutb ettwhele it loses enough energy to con eh leave 253 single sue f h li ams step 1 Draw a Flowchart The ptoblem as it stands contains an awful lot of text but it wont mean much until you draw what 1 gm tn yuu First ask youtself what ptocesses ate in use in this ptoblemv Make a list ofthe ptocesses m the ptoblem l Evapotatot A 2 HeatExchangelB 3 Tutbihe C ohceyou havea you heedto ml u winey you something use H theit connections and label the ptocesses it should look somethih like this Remember we don t cate what the actual ptocesses look like ot how they re deslgied At this in process theyte talking about ll all ul r A Tnthl case we have not drawn the feed stream mm the evaporator the waste sueam from the evaporator or the exlt streams from the turblne and heat exchanger The thlrd step ls to Label all your ows Label them Wlth any lnfolmatlon you are glven Any lnfolmatlon you are not glven and even lnfolmatlon you are glven should be glven a dlffelent vanallle r lfy0u calllt 39 ll 39 call sneam 1 units on the glven Values vapor mlo 5W5 feed m4 waste mo hquld m515001bhr Notlcethat for one of the streams a vulume ow rate ls glven rather than a max ow rate so lt ls labeled as such Tnls ls very lrnportant so that you avold uslng a value ln an equatlon that Isn t valld for example there s no such Lhng as consewatlon ofvolume for most cases tn ml m M ofthe lll ll my l stream ls glven so wrlte thls on the slde for future reference Carefully drawn owcharts and dlagxams are half of the key to solvlng any mass balance or unlts to gettlng the rlght answer edit step 2 Make sure your units are consistent ttn t that lt is n thls case since the pnnclple that we ll need to use to solve for the ow rate ofthe an also ln the same massr ow unlts tn t wl m bl m lt would save tlrne and rnlnlrnlze mlstakes to convert Va and msto kgs V pm1 Therefore 4 b a r ap r order l Slncethe ltwasntwe d 4 10quot L 1 In 1 min a 1min loooL 60 0004 1 1 kg IUUUg Note that slnce the denslty of a gas ls so small a huge volumemc ow rate ls necessary to achleve any slgamcant mass flow rate Thls IS talrly typlcal and ls a practlcal problem when deahng wlth gasrphase processes The mass ow rate Wscan be changed In a slrnllar manner but slnce lt IS already ln terms of mass or welght technlcally we don t need to apply a density 1kg 111 22 39 01893ka 39 7 1500amp m In 39 36005 Step 3 Relate your variables slnce we have the mass ow rate of the yapot sneam we can calculate the efflclency of the evaporator duectly m2 o f ficz39ency Flndlng quot14 A l g to wnte the mass balance on the eyapotatot slnce that wlll cenamly contaln the unknown we seek Assumlng that the process ls steady state we can wnte IniOu 0 7h1fng739n4mil0 Problem we don t know moso wlth only thls equatlon we cannot solve for me Have no fear m an H ttnttte ltouWTryto do so before you move on edit So you want to check your gum Alright then read on the heat exchanger ls slmply 7m7m3950 NOTE we need lnformatlon about Then we move to balances on other operatlons In order to gamer addltlonal lnformatlon about the unknowns tn the process Thls takes practice to c l t L et 4 A balances or look up lntotmatlon perform Thls IS not as much of an lssue wlth a Ielatlvely slmple problem llke thls but wlll become an lssue wlth more complex problems Therefore a steprbyrstep method exlsts to t you F 39 and therefore how many total lndependent equatlons you can use to help you solve bl s edit Step 4 Calculate your unknowns Canyng out the plan on this pmblem kg 7 I39m 7 01893 7 l s C m 11893 Hence from the mass balance on the evaporator k9 ma 66 62 mr 05 7 62666 e 61893 00441 So the flml answers are Evapornmr Ef ciency 533 k Vmsm scream mm 66447 Step 5 Check your work M l n l hl h 7 man l66 or other physically lmposslble happenlngs If something llke llals happens and It lll w created m the placesses It IS lmposslble for thls to mum Ln thls case Al Wm 4 back llke that Chapter 2 Practice Problems ra la Problem 1 a A salt solution is to be concentrated by evaporating the water in a salt pan with a condensing surface above it to gather the evaporated water Suppose 1200g of salt solution are emptied into the pan Once all the water is evaporated the salt is weighed and found to weigh 100g What percent of the original solution was water 17 Now suppose that 01 L of the evaporated water was added back to the salt to bring it to the desired concentration How much water remains to be used elsewhere 0 Do you think the salt solution would be safe to drink Why or why not 3L 1 J Problem 2 a In a stone quarry limestone is to be crushed and poured into molds for manufacture of oor tiles Suppose that a limestone company uses three trucks each of which is capable of carrying 3000 kg of limestone The quarry itself is 20 miles away from the processing plant and the trucks get there at an average speed of 30 mileshour Once at the plant the limestone is ground into fine powder and then melted and poured into the molds If each of the resulting slabs weighs 2 kg and the plant operates 24 hours a day how many slabs can the company make in a day 17 How could this plant become more efficient Plot the number of slabs the company can make as a function of distance from the quarry and capacity of the trucks What factors might keep the company from simply moving as close to the quarry as possible and using the largest trucks possible Ll ls Problem 3 What is the volumetric owrate of a solution with density 15 kgmquot3 owing at a velocity of 5 ms and a mass ow rate of 500 gmin What is the area of the pipe If it is circular what is the radius rl Lax 5 Problem 4 Suppose you have a pipe that constricts halfway through from a radius of 05 cm to a radius of 02 cm A liquid approaches the constriction at a velocity of 05 m s What is the velocity of the uid after the constriction Hint Apply conservation of mass on both sides of the constriction Challenge What kind of energy does the uid gain Energy is never created or destroyed so where does it come from Problem 5 Suppose that a river with a molar ow rate of 10000 mols meets another larger river owing at 500000 mquot3s at room temperature What is the mass ow rate of the river downstream of the intersection if you assume steady state 7 Evaluate the feasibility of the steady state assumption in this situation Also qualitatively evaluate the probability that the owrates are actually constant Ui L315 Problem 6 Suppose that the population of a certain country 11 years after year 2000 if there is no emigration can be modeled with the equation P 2 5 JH n Also suppose that in the country a net emigration of 100000 people per month actually occurs What is the total accumulation of people in this country from year 2000 to 2003 b What was the population of people in 2002 according to this model c What are some possible problems with this model For example what doesn39t it take into account What happens when n is 100 Where did those constants come from Would they be the same for every country or for the same country across generations git Chapter 3 Mass balances on multicomponent systems Component Mass Balance Most processes of course involve more than one input andor output and therefore it must be learned how to perform mass balances on this type of system The basic idea remains the same though We can write a mass balance in the same form as the overall balance for each component In Out Generation Accumulation For steady state processes this becomes I n Out Generation l The overall mass balance at steady state recall is 2mm Emuquot my J The mass of each component can be described by a similar balance S tAin 2361th mA gfn D The biggest difference between these two equations is that The total generation of mass 117quot is zero due to conservation of mass but since individual species can be consumed in a m quot 0 reaction A 5 for a reacting system edit Concentration Measurements You may recall from general chemistry that a concentration is a measure of the amount of some species in a mixture relative to the total amount of material or relative to the amount of another species Several different measurements of concentration come up over and over so they were given special names Molarity The first major concentration unit is the molarity which relates the moles of one particular species to the total volume of the solution 113 7 sm whele AlolurltyA A mat ViL A mole useful detlnltlon for flow systems that ls equally valld ls Molanty ls a useful measule of concenuatlon because lt takes lnto account the volumemc ang c occur when one creates a mlxtule from pule substances Thus lt ls a v ry plactlcal unlt of concenuatlon However slnce lt lnvolves volume lt can change wl mlxtule can also change wlth plessule so lt ls not usually used for gasses Mole Fraction detelmlne the molar ow late of any component from the total flowmte lt also convenlently ls that you can always use to help you solve problems The mole actlon of a component A ln a mlxtule ls de ned as nu 94 n l Vlkm vl39 lt you add up all you 1 wlthln calculatlon and measulement enol Note that each stream has its own independent set of concentrations Mass Fraction owmtes rather than mular Howrates when dus occurs t IS convement to express concentrattons In terms of mass fractions de ned stmtlatly to mole acttons ms 14 i 7m for batch systems where quotbus th e mass ofA It doesn t matter what the umts ofthe mass are as long as they are the same as the umts of the total mass of so1utton 211 edit Calculations 0n Multicomponent streams omponent sueams Thrs section shows some methods to combine the propertres of single component sueams mto somethmg usable for multiplercomponent streamsw1th some assumpttons w Average Molecular Weight at t t t the molecular wetght of a pure spectes It allows you to convert between the mass of a mtxtuxe and the number of moles whlch IS lmponant fox ieactlng systems especlally because balances must usually be clone 1 moles but measurements are generally in grams V 5111 A m 97 To find the value of mole sln we Split the solutlon up into its components as follows for k components g 5111 Elm male sin 7 nquot 7 nr m s n 2Mw m Therefore we have the followlng formula MW Ele r yr b A wnrcnlsrs a mixture edit Density of Liquid Mixtum Let us attempt to calculate the denslty of a liquid mixture from the denslty of its components slmllar to how we calculated the average molecular welght Thls tlme however we wlll notlce one cntlcal A F r make edit First Equation Byrl ttnttl n sin a solution IS F 7 V sln Followrng a Similar derivation to the above for average molecular weight ll X me mE V 5111 V VI rt m t The volume of 39 the mass but it is further Now h l on lianyr ms equation IS therefore useful for two substances with Similar pure densities If this IS true then V1 7 1m V mo edit Second Equation on is easier to derive if we assume the equation will have a form similar to that of a y a This equati mol mass Since density is given in terms of mass it makes sense to u using m ss 5 average fraction m m i 1m get this in tenns of only solution properties we need to get rid of ma We do this first by dwiding by the density ml mquot m Now ifwe add all ofthese up we obtain El a m Now we have to make an assumption and it s different from that in the first case This time we assume that the Volumes are additive This is true in two cases 1 In an ideal solution The idea of an ideal solution will be explained more later but for now you need to know that icleal solutions is so dilute that it doesnt effect the solution properties rnuch include ldeal Gas mlxtures at constant temperature and pressure In a Completely immiscible nomeacting mixture Ln other words if two substances dont mix at all like oil and water or if you throw a rock into a puddle the total volume will not change when you mlx them if the solution is ideal then we can write Sic Vquot 1 win 7 ma 7 p Hence for an ideal solution Note that this is Sigiiflcantly different from the previous equatlonl This equation is more accurate for most cases in all cases however it is most accurate to look up the value in a handbook such as Perry s Chemical Engineers Handbook If data is available on the solution of interest edit General Strategies for MultipleComponent Operations that far 1241 cumpuvlevlt yait mm with um im39epemz39evlt mass balance What do lrnean by independentv Well remember we can write the general overall mass balance for any steady state system 3mm 7 3mm D And we can write a similar mass balance for any cumpu e t of a stream E7hua m Emauut i may 0 Thls looks like we have three equatlons here but in reality only two of them are independent because l The sum of the masses of the components equals the total mass 2 The total mass generation due to reaction is always zero by the law of mass conservation Therefore if we add up all of the mass balances for the components we obtain the overall mass balance Therefore we can choose any set of n equations we want where n is the number of components but if we choose the overall mass balance as one of them we cannot use the mass balance on one of the components The choice of which balances to use depends on two particular criteria 1 Which components you have the most information on if you don39t have enough information you won39t be able to solve the equations you write 2 Which components you can make the most reasonable assumptions about For example if you have a process involving oxygen and water at low temperatures and pressures you may say that there is no oxygen dissolved in a liquid ow stream so it all leaves by another path This will simplify the algebra a good deal if you write the mass balance on that component Ldit Multiple Components in a Single Operation Separation of Ethanol and Water Following the stepby step method makes things easier ledit Step 1 Draw a Flowchart The first step as always is to draw the owchart as described previously If you do that for this system you may end up with something like this where x signifies mass fraction A signifies molarity of A and numbers signify stream numbers Vapor m2 rnllookgs KEMP x0120 Dlsullahon Column quuld m3 20 kys ELOH 4M edit Step 2 Coan Units Now we need to tum to convemng the concentratlons lnto approprlate unlts slnce the total the components would be most useful The vapor stream composltlons are glven as mass percents H l l l ln terms of a molanty ls not useful fox flndlng a mass flow rate of ethanol or of water Hence we must convert the concentmtlon to somethng more useful NOTE Convertlng between Concentratlon Measurements The easlest way to convert between concentratlons ls to take a careful look at the unit of tn na have molar mass denslty you Tn tht e amnle l lnto a mass fractlon We have from the detlnltlons that 7710 4 A i Sm 771 1 4 man To convert the numerators we need to convert moles of A to mass of A so we can se the molar mass for thls purpose slmrlarly to convert the enomrnators we need to change Lrters to Mass w rch means well use a densrty Hence the conversron from molanty to mass actlon ls MW 4 IA A PSLN Slnce we have ways to estlmate PSLNremember them we can lnterrconvert the conversrons ln order to convert the molanty rnto a mass hactlon then we need the molecular welght of ethanol an the densrty of a 4M ethanol solutron The former ls easy If you know the chemrcal formula of ethanol CHaCIIZOII Calculating the molecular weight as you dld ln chem 46 7 class you should come up wrth about mat urr nuuy rnr L Ju an an estrmate assumptrons llke these are all we have as long as we realrze the values wlll not be exact 1 7 g psw pt ln thrs case then 1790 175720 PSLN PEtrJH PHZU can look up the densrtres of pure water and pure ethanol they are as follows from lpedla s amcles thanol and wWater We Wlk i g 13th 7 07896quot 789L I Hzn 100 100 Thva 1 1 i may 1 7 Iaorr pm 89 100 MW 1 A 1 man EtOII gtr 7 Mia e g i 71005 Solvlng this equation yields Juana 0194unitless w Step 3 Relate your Variables mass ow rates Vanables Since with mass balances Remember that we can do a mass balance on any of overall mass but since e sum ofthe indiVidual masses equals the overall only N A 10fthese equations will e independent t is o en easiest mathematically to choose the overall mass balance and N 739 in it A A A A L for the overall measurements the N independent Specles and one on the Since our concentrations are now in appropriate units we can do any two mass balances we want Lets choose the overall first 1h1 m3912 739n30 Plugging in known values kg 9 in 100k 20 Now that we know muwe can do a mass balance on either ethanol or water to find the composition ofthe input stream Lets choose ethanol A 139Mi Ihaa l39 mar Written in terms of mass fractions this becomes 17 a m at arm2 lanai m Plugglng in what we know A mMIUO s 17 03152 k ls 08 gt1 20 lo194 gt1 80 9 Hence the feed is 32 Ethanol and 68 Water edit Introduction to Problem Solving with Multiple Components and Processes rhe desued resulr The calculauons for such processes as you can probably guess are consldembly more compl ed rhan rhose euher for only a slngle componenr or for a slngle operanon process Therefore several technlques have been developed to ard englneexs ln rherr analyses edit Degree of Freedom Analysis For r e l ll n nl vri uhavernr a m A u r w solvabrluy 1 de ned 2 The problem can be overdetermined also known as overspeci ed whlch means that ou have too much lnfoxmatlon and u is euher redundant or lnconslstent Thls could posslbly be xed by consolldatlng multlple data into a slngle functlon or ln extreme cases a slngle value such as a slope of a llnear correlanon or u could be flxed by t 0 5 ha a m remov 3 The problem can be underdetermined or underspeci ea whlch means that you don t have en ugh lnfoxmatlon to solve for all your u nowns ere are several ways of in s st obvlous is to gather addluonal lnfoxmatlon such as measunng addulonal temperatures ow rates and so on untll you have a wellrdeflned problem n ml out of a process such as how much converslon you obtaln in a reacuon how erhclenr a separauon process rs and so on Flmlly we can make assumptions in order to Simplify tlw rm H n E rs The method of analyzing systems to see whether they are over or under specified or if they are well defined is called a degree of freedom analysis It works as follows for mass balances on a single process 1 From your owchart determine the number of unknowns in the process What qualifies as an unknown depends on what you39re looking for but in a material balance calculation masses and concentrations are the most common In equilibrium and energy balance calculations temperature and pressure also become important unknowns In a reactor you should include the conversion as an unknown unless it is given OR you are doing an atom balance Subtract the number of Equations you can write on the process This can include mass balances energy balances equilibrium relationships relations between concentrations and any equations derived from additional information about the process 3 The number you are left with is the degrees of freedom of the process D If the degrees of freedom are negative that means the unit operation is overspeci ed If it is positive the operation is underspeci ed If it is zero then the unit operation is wellde ned meaning that it is theoretically possible to solve for the unknowns with a finite set of solutions Degrees of Freedom in MultipleProcess Systems Multiple process systems are tougher but not undoable Here is how to analyze them to see if a problem is uniquely solvable Label a owchart completely with all the relevant unknowns Perform a degree of freedom analysis on each unit operation as described above Add the degrees of freedom for each of the operations Subtract the number of variables in intermediate streams ie streams between two unit operations This is because each of these was counted twice once for the operation it leaves and once for the one it enters kaNt l The number you are left with is the process degrees of freedom and this is what will tell you if the process as a whole is overspecified underspecified or well defined NOTE If any single process is overspecified and is found to be inconsistent then the problem as a whole cannot be solved regardless of whether the process as a whole is well defined or not git Using Degrees of Freedom to Make a Plan Once you have determined that your problem is solvable you still need to figure out how you39ll solve for your variables This is the suggested method arezelo 2 calculate all of the unknowns lnvolved m thls comblnatlon 3 7 t quot4 A F m for h m Values as known rather than as Vanables NOTE N ha a to be aware of the sandwich effect in whlch calculatlons from one unlt operatlon can hlvlallze balall ll an uhex F 39 llke thls39 7 A 7gt B 7gt C 7 Su pose also that through mass balances on operatlons A and c you calculate the ext and the lnlet composition of c Once these are performed the man you can wn e an equation to solve an unknown write the equation an your degree of freedom analysls edit Multiple Components and Multiple Processes Orange Juice Production Step 1 Draw a Flowchart the pwblem m1n Product 80nH20 mS40 lmSl 20 s CRUSHER Sohd densxty 1 54 germ Mass ofan orange 0 4 kg Sohds m53 0 5mm N H e h l quot I quot sohds 1n the solrcl waste to the mass 1n the feed Thrs wrll be Important later because It l an additiuml piece ufmfuman uvt that i ecessary re eezte the prublem Also note that from here 1n solrds are referred to as S and water as W edit Step 2 Degree at Freedom analysis Recall that for each stream there are C mdependent unknowns where C is the number of 39 1 specres and the total mass ow rate smce wnh Cl concentrations we can nd the last one but we cannot obtmn the total mass ow rate from only concentratmn Let us apply the previously described algorithm to determining if the problem is well defined On the strainer There are 6 unknowns m2 XS3 m3 XS3 m4 and XS4 We can write 2 independent mass balances on the overall system one for each component 0 We are given a conversion and enough information to write the mass ow rate in the product in terms of only concentration of one component which eliminates one unknown Thus we have 2 additional pieces of information 0 Thus the degrees of freedom of the strainer are 6 2 2 2 2 DOF NOTE We are given the mass of an individual orange but since we cannot use that information alone to find a total mass ow rate of oranges in the feed and we already have used up our allotment of C 1 independent concentrations we cannot count this as quotgiven informationquot If however we were told the number of oranges produced per year then we could use the two pieces of information in tandem to eliminate a single unknown because then we can find the mass ow rate On the crusher There are 3 unknowns ml m2 and XS2 We can write 2 independent mass balances Thus the crusher has 3 2 1 DOF Therefore for the system as a whole 0 Sum of DOF for unit operations 2 2 1 3 DOF Number of intermediate variables 2 2 m2 and XS2 Total DOF 3 2 1 DOF Hence the problem is underspeci ed ledit So how to we solve it In order to solve an underspecified problem one way we can obtain an additional specification is to make an assumption What assumptions could we make that would reduce the number of unknowns or equivalently increase the number of variables we do know The most common type of assumption is to assume that something that is relatively insigni cant is zero In this case one could ask will the solid stream from the strainer contain any water It might of course but this amount is probably very small compared to both the amount of solids that are Ifwe th wateri t m stream is Therefore edit Step 3 Coan Units Thls step should be done after the degtee of heedom analysls because that analysls ls lndependent of yout unlt system and If you don t have enough lnfotmatlon to solve a ptohlem or wot you have too much you shouldn t waste tlme convemng unlts and should lnstea to the mrkgrs H u ul u u l system slnce most values ate alteady ln meme Here the lattet toute ls taken V 54cm 1m 7 71va m loocm702032m p5 154 1540 kg C77 m Now that everythlng ls ln the same system we can move on to the next step edit Step 4 Relate your variabla so that we can actually use that lntotmatlon ln a mass balance Ftom chapter 2 we can start wlth the quatlon pt uquot Aquot mquot 2 Smce the plpe ls sllsula and the aea of a clrcle ls 7r 7 we have 01297 uh2 Al 7r x 020322 So we have that 4 s 30 01297 38915 p4 m Now to flnd the density of st we assume that volumes are additive slnce the solids and eam 4 watet ate essentlally lmmlsclble does an orange dlssolve when you wash 1 Hence we can use the ldealr uld model tot denslty Isa 7 1 54 7 1540 1000 u 3 Q 7 EQUATION l 1940 1000 m4 New b 1 ln balances yet We of course have a cholce on whlch two to use ln this particular pmblem slnce we are directly given information concerning the amount of babe 1 sueam 4 the product stream It seems to make me sense to do the balance on hls component slnce we don t have lnfoxmatlon on sueam 2 and flndlng It would be polntless In 1 l a b s b l the sollds Emsm n E lsmm 0 NOTE Expandlng the mass balance In terms of mass actlons glves ml 051 ml 033 ml 054 b k values water 1 pl 39 and hence l sl EQUATION 2 02 ml 09 gtx 02 gtllt Flnally we can utlllze one further mass balance so let s use the easlest one the overall mass balance EQUATION We now have three equatlons In three unknowns m1 WM jvsako the problem IS solvable This ls Where all those systemrsolvlng skllls wlll come ln handy f you don t llke solvlng by hand nele are numerous computer programs out there to help you solve equations llke nus such as MATLAB POLYMATH and many others You ll probably Using either method the results are k ml 4786 ml 392507 orange 5 wk hr wk day 1 04 kg SGUDM 8 day was 124 mnm cmges 39 year Yearly Productlon Chapter 3 Practice Problems Fa la Problem 1 a Look up the composition of air Estimate its average molecular weight b Qualitatively describe whether the density of air should be large or small compared to the density of water 0 Qualitatively describe whether the mass density of air should be large or small compared to that of oxygen if the same number of moles of the two gasses are contained in identical containers d If the density of air under certain conditions is 106 gmquot3 how much does a gallon of air weigh Problem 2 a Using both of the formulas for average density calculate estimates for the density of a 50 by mass solution of toluene and benzene Comment on the results 17 Repeat this calculation for varying concentrations of toluene When does it make the most difference which formula you use When does it make the least Show the results graphically Would the trend be the same for any binary solution 0 Suppose that a 50 mixture of toluene and benzene is to be separated by crystallization The solution is cooled until one of the components completely freezes and only the other is left as a liquid The liquid is then removed What will the majority of the solid be What will the liquid be What temperature should be used to achieve this give an estimate d In the crystallization process in part c suppose that the after separation the solid crystals contained all of the benzene and 1 of the toluene from the original mixture Suppose also that after melting the solid the resulting liquid weighed 1435 g Calculate the mass of the original solution no u Problem 3 Consider a publishing company in which books are to be bound printed and shipped At 5 am every morning a shipment of 10000 reams of paper comes in as well as enough materials to make 150000 books and 30000 pounds of ink In this particular plant the average size of a book is 250 pages and each uses about 02 pounds of ink a How many books can be printed for each shipment Hint What is the limiting factor 17 Suppose that on average 4 of all books printed are misprints and must be destroyed The remaining books are to be distributed to each of 6 continents in the following proportions North America 15 South America 10 Europe 20 Africa 20 Asia 25 Australia 10 Each book that is printed including those that are destroyed costs the company US050 to print Those that are shipped cost the following prices to ship from the US North America 005 South America 008 Europe 010 Africa 020 Asia 012 Australia 015 If each book sells for an equivalent of US100 what is the maximum profit that the company can make per day 0 Challenge What is the minimum number of books that the company can sell from any continent in order to return a profit Hint what is the total cost of this scheme Does it matter where the books are sold once they are distributed d How many pounds of ink per day end up in each continent under the scheme in part b How many pages of paper 6 Can you think of any ways you can improve this process What may be some ways to improve the profit margin How can inventory be reduced What are some possible problems with your proposed solutions edit Chapter 4 Mass balances with recycle Q What Is Recycle Recycling is the act of taking one stream in a process and reusing it in an earlier part of the process rather than discarding it It is used in a wide variety of processes Uses and Benefit of Recycle The use of recycle makes a great deal of environmental and economic sense for the following reasons among others 0 Using recycle lets a company achieve a wider range of separations This will be demonstrated in the next section However there is a tradeoff the more dilute or concentrated you want your product to be the lower the owrate you can achieve in the concentrated or dilute stream 0 By using recycle in combination with some sort of separation process a company can increase the overall conversion of an equilibrium reaction You may recall from general chemistry that many reactions do not go to completion but only up to a certain point because they are reversible How far the reaction goes depends on the concentrations or partial pressures for a gas of the products and the reactants which are related by the reaction stoichiometry and the equilibrium constant K If we want to increase the amount of conversion one way we can do this is to separate out the products from the product mixture and re feed the purified reactants in to the reactor By Le Chatlier39s Principle this will cause the reaction to continue moving towards the products 0 By using recycle it is possible to recover expensive catalysts and reagents Catalysts aren39t cheap and if we don39t try to recycle them into the reactor they may be lost in the product stream This not only gives us a contaminated product but also wastes a lot of catalyst Because of the previous three uses recycle can decrease the amount of equipment needed to get a process meet specifications and consumer demand For example it may improve reaction conversion enough to eliminate the need for a second reactor to achieve an economical conversion Recycle reduces the amount of waste that a company generates Not only is this the most environmentally sounds way to go about it it also saves the company money in disposal costs Most tmportantly all of these thmgs can save a company money L e and probably gets the product faster too 1fthe proper analysts IS made edit Differences between Recycle and nonRecycle systems The btggeet dlffexence between recycle and nonexecycle system Is that the extra splitting amz39 a change occurs between two sneams peed egt Process egt Outlet Reecmbtnatzon Pomt Feed D Sphmng pomt ExitOutlet stream Recycle Stream me e us but VLutgu mm thepmcess uley the recumbmed stream dues Lnstead 39 one on the pllrtmg palm Iedit Assumptions at the Splitting Point The recombination point is relatively unpredictable because the composition of the stream leaving depends on both the composition of the feed and the composition of the recycle stream However the spliitng point is special because when a stream is split it generally is split into two streams with equal composition This is a piece of information that counts towards quotadditional informationquot when performing a degree of freedom analysis As an additional specification it is common to know the ratio of splitting ie how much of the exit stream from the process will be put into the outlet and how much will be recycled This also counts as quotadditional informationquot edit Assumptions at the Recombination Point The recombination point is generally not specified like the splitting point and also the recycle stream and feed stream are very likely to have different compositions The important thing to remember is that you can generally use the properties of the stream coming from the splitting point for the stream entering the recombination point unless it goes through another process in between which is entirely possible Degree of Freedom Analysis of Recycle Systems Degree of freedom analyses are similar for recycle systems to those for other systems but with a couple important points that the engineer must keep in mind 1 The recombination point and the splitting point must be counted in the degree of freedom analysis as quotprocessesquot since they can have unknowns that aren39t counted anywhere else 2 When doing the degree of freedom analysis on the splitting point you should not label the concentrations as the same but leave them as separate unknowns until after you complete the DOF analysis in order to avoid confusion since labeling the concentrations as identical quotuses upquot one of your pieces of information and then you can39t count it As an example let39s do a degree of freedom analysis on the hypothetical system above assuming that all streams have two components Recombination Point 6 variables 3 concentrations and 3 total ow rates 2 mass balances 4 DOF Process Assuming it39s not a reactor and there39s only 2 streams there39s 4 variables and 2 mass balances 2 DOF Splitting Point 6 variables 2 mass balances 1 knowing compositions are the same 1 splitting ratio 2 DOF So the total is 4 2 2 6 in between variables 2 2 DOF Therefore if the feed is specified then this entire system can be solved Of course the results will be different if the process has more than 2 streams if the splitting is 3 way if there are more than two components and so on git Suggested Solving Method The solving method for recycle systems is similar to those of other systems we have seen so far but as you ve likely noticed they are increasingly complicated Therefore the importance of making a plan becomes of the utmost importance The way to make a plan is generally as follows 5 6 1 Draw a completely labeled ow chart for the process 2 3 If it is solvable a lot of the time the best place to start with a recycle system is with a Do a DOF analysis to make sure the problem is solvable set of overall system balances sometimes in combination with balances on processes on the border The reason for this is that the overall system balance cuts out the recycle stream entirely since the recycle stream does not enter or leave the system as a whole but merely travels between two processes like any other intermediate stream Often the composition of the recycle stream is unknown so this simplifies the calculations a good deal Find a set of independent equations that will yield values for a certain set of unknowns this is often most difficult the first time sometimes one of the unit operations in the system will have 0 DOF so start with that one Otherwise it39ll take some searching Considering those variables as known do a new DOF balance until something has 0 DOF Calculate the variables on that process Repeat until all processes are specified completely git Example problem Improving a Separation Process It has been stated that recycle can help to This example helps to show that this is true and also show some limitations of the use of recycle on real processes Consider the following proposed system without recycle m2 m2 60 ofgntenng A ml 50 ofentenngB KA0 5 Separation E KB 0 5 Process m3 xA3 40 ofentenng A 50 ofentenng B A degree of freedom analysxs on ms process 4 unknowns t241xng and us 2 mass balances and 2 pieces of informauon A e 31 4 e an of B 60 of A and halfof B leaves m sueam 2 0 DOF Methods of prevlous chapters can be used to detexmme that g 9 m 557r n545 m 457 2 hr 2 3 hr and 5M3 0444 ms ls good pmmce for the interestedxeadex f we want to obtain a greater sep39amuonthan thxs one thmg that We can do 15 use a recycle NHL 4 4L4 44 4444 4 44 4 w Implauenting Recycle on the Separation Procas overcome A1115 IS a lather step at a me m Fwd W111 be W111 be le for the reader edit step 1 Draw a Flowchart A that enters u Into sueam 2 not 60 of the fresh feed stream Recomblnauon Pomt m2 xA2 sonn ofentenng A 50 ofentenng B Separaaon Process SphtterSU 50 m3 XAE m6 KAI 40 ofentenng A 50 ofentenng B m5 m5 edit step 2 Do a Degree of Freedom Analysis Recombination point 4 unknowns r 2 mass balances 2 degrees of freedom Separator a unknowns nothlng 1s speclfled r 2 lndependent places of lnfolmatlon r 2 mass balances 2 DOF Splmmg point 6 unknowns agaln nothng lS speclfled r 2 mass balances r 1 assumpuon that concentlauon xemalns constant 7 1 splmlng mtlo 2 DOF T0tal 2 2 2 r a 0 Thus the problem IS completely speclhed edit Step 3 Devise a Plan and Carry it Out Flrst kg U51007m x 39 m OverallmassbalanceonA h 2 A m A Thz i 1427 ha 1 145 5 Overall mass balance 011 B h A l n l m solvlng quulvP some lngenulty Fust lets see what happens when we comblne ths lnfoxmatlon wlth the spllttlng ratio and constant concentration at the sphtten m Wt Splitting Ratio 2 Comtantconcentration 145 TAB Plugglng these mm the overall balances we have 50 m 1 m3 1ltI 2 A2 7 A3 011A 2 50m2171l2 s171n I Total 2 eqmtzum tn the mlet cuereYLtm uVL m the epamtun Thls ls due to the convexslons we are glven 60 0fenteringA goesimo stream 2 means m2 a 7 06 i an m4 40 0fenteringA goes im0 stream 3 means We All 04 3044 I39m 50 of entering B goes into stream 2 means m4 gtr171742U517L A47h4 50 arentering B goes into stream 3 means m0 1 rm 05 1 144 I39m actlon and uanslatlng words lnto algebtalc equatlons r39nl 144 04 50 06m e 144 011A 2 0 r39 5005m417r ml 1 7 1 44 011 B Solvlng these equatlons glves t q 3 0484 NOTE Notlce that two thlngs happened as expected the concentzatlon of the stream enteran the evaporator went down because the feed ls mlxlng wrth a more drlute recycle stream and the total flowmte went up agaln due to contrrbutron from the recycle stream Thls ls always a good rough check to see lt your answer makes sense for example lt the flowmte was lower than the feed rate you d know somethng went wrong Once these values are known you can choose to do abalance ether on the separator or on the recombtnatton potnt smce oth now have 0 degrees of freedom We choose the separator because that leads dtrectly to what were looking for The mass balances on the separator can be solved uslng the same method as that wtthout a recycle system the results are k k 708379 05mmr 583314 0429 hr hr E 2l39 J 6151 1 1 0429 You should check to make sure that m2 and m6 add up to the total feed rate otherwise you made a mistake Now we can assess how effective the recycle is The concentration of A in the liquid stream was reduced by a small margin of 0015 mole fraction However this extra reduction came at a pair of costs the ow rate of dilute stream was significantly reduced from 45 to 29165 kghr This limitation is important to keep in mind and also explains why we bother trying to make very ef cient separation processes Systems with Recycle a Cleaning Process leditl Problem Statement clean wet ore W m Step Draw a Flowchart any of the desigq speci cauons the ef ciency of the settler the solubxhty of soil in water and the mass ow rates Therefore n z hzghly recummemz39ea39 x at W draw yum uva pmm even when am 1 pruvlded far we Make me you label all of the means and the unknown concentrations m5 H2O m4 3100 kyhr x04 Ore 17x04 H2O m9 P W lrxDZ Water 3 x1 Drrt 17x1 Water m8 x81 Drrt Addluonal lnfol39mahon l 7 x81 Water m7 0 gamma SE Solubrbty ofdm 0 4 germ m3 saturated wrth dln w Second Step Degree at Freedom Analysis er a lndependent unknowns In Around the wash mzt J7mt with Ina mm three rndependent mass balances ore drrt and water and one solubrlrty The washer has 2 Around the dryer 2 rndependent unknowns lady m5 and two rndependent equatrons 0 DOF NOTE shte the dryer has no degrees of freedom already we can say that the system variable behave as tf the stream gul g mm the met was at gul g ehywhete and therefore thrs stream should not be lncluded ln the lnrbetween Varlables talculatron Around the Seluerzs lndependent unknowns millIDJl h7l7n81x x two mass balances drrt and water the solubrlrty of saturated drrt and one addrtronal lnfoxmatlon 90 removal of drrt leavlng us wrth 1 no F 0 At the mixing point We need to include this in order to calculate the total degrees of freedom for the process since otherwise we39re not counting m9 anywhere 5 unknowns quot121 33 1 32 may Tny mg and 2 mass balances leaves us with 3 DOF Therefore Overall 2 321 6 intermediate variables not including X04 since that39s going to the dryer 2 0 The problem is well defined edit Devising a Plan Recall that the idea is to look for a unit operation or some combination of them with 0 Degrees of Freedom calculate those variables and then recalculate the degrees of freedom until everything is accounted for From our initial analysis the dryer had 0 DOF so we can calculate the two unknowns X04 and m5 Now we can consider X04 and m5 known and redo the degree of freedom analysis on the unit operations Around the washer We only have 5 unknowns now 33 1 quot12 17302 m 1 03 but still only three equations and the solubility 1 DOF Around the settler Nothing has changed here since X04 and m5 aren39t connected to this operation Overall System We have three unknowns 57701 quot17 mil since ml39xis already determined and we have three mass balances ore dirt and water Hence we have 0 DOF for the overall system Now we can say we know 3701 quot77 and mo Around the settler again since we know m7 the settler now has 0 DOF and we can solve for 7713 TD 7n85and Around the washer again Now we know m8 and XD8 How many balances can we write NOTE If we try to write a balance on the ore we will find that the are is already balanced because of the other balances we ve done If you try to write an ore balance you39ll see you already know the values of all the unknowns in the equations Hence we can39t count that balance as an equation we can use I39ll show you this when we work out the actual calculation The washer therefore has 2 unknowns m2 XD2 and 2 equations the dirt and water balances 0 DOF we have m2 25nd mm the system IS completely determlned edit Converting Units The only glven lnfoxmatlon m lnconslstent unlts IS the solulnllty whlch ls glven as g m d cm 1120 Hawever slnce we know the denslty of water or can look lt up we can q Art convert thlsto kg Hzoos follows cm 1120 g dirt kg dirt 717 7 47 0 91120 D 91120 0 1120 Carrying Out the Plan Flrst l L Remember that the thud balance ls not lndependent of the rst two Overall Balance ll m5 l ll Ore Balance 7m Jim m5 gtgt Ian 7m 1m Substltntlng the known Values Ovem 3100 ma l 2900 Ola7704 3100 1 2900 Solvlng glves m5 200 39 kg mm 7 09353 Now that we have flnlshed the dryer we do the next step In our plan whlch was the overall system balance Water Balance m9 inn Ore Balance 101 m1 m4 DirtBalance1 1m m1 m7 I A W my 0967 kg m7 100m an we move to the settler as planned thls one s 2 bn tnckxex smce the soluuons arent 1mmedmtely obvmus but a system must be solved Overall Balance m irtBalmcezm 7L7 1 137 ms t 17m Ef ciency of Removal m7 09 me mm 139n7 mx prupumumzl tn the mm Water and hence you can wnte that I mass dirt m sueam 3 04 mass ate 1n stream 3 Solubility me Im 04 m a 1 1m Pluggmg 1n known Values the following system of equauons ls obtamed mm 02863513 00385 quy we can go to the mmng pm and say S Overall mx 39 he Din quotL5 fus W 1132 From whxch the nal unknowns are obtmned 1139 L m 00229 Smce the problem was askxng for m2 we are now fxmshed edit Check your wcrk These Values should be checked by makxng a new owchart wnh the numencal Values and git Chapter 5 Massmole balances in reacting systems Review of Reaction Stoichiometry Up until now all of the balances we have done on systems have been in terms of mass However mass is inconvenient for a reacting system because it does not allow us to take advantage of the stoichiometry of the reaction in relating the relative amounts of reactants and of products Stoichiometry is the relationship between reactants and products in a balanced reaction as given by the ratio of their coefficients For example in the reaction I gt the reaction Stoichiometry would dictate that for every one molecule of 02112acetylene that reacts two molecules of II392hydrogen are consumed and one molecule of CEJU are formed However this does not hold for grams of products and reactants Even though the number of molecules in single substance is proportional to the mass of that substance the constant of proportionality the molecular mass is not the same for every molecule Hence it is necessary to use the molecular weight of each molecule to convert from grams to males in order to use the reaction39s coefficients Molecular Mole Balances We can write balances on moles like we can on anything else We39ll start with our ubiquitous general balance equation Input Output Accumulation Generation As usual we assume that accumulation 0 in this book so that Input Output Generation 0 Let us denote molar ow rates by ilto distinguish them from mass ow rates We then have a similar equation to the mass balance equation Eh Elia ugh 0 The same equatlon can be wrltten ln terms of each lndlyldual specles There ate a couple of lmportant thlngs to note about thls type ofbalance as opposed to a mass balance 1 Just llke Wlth the mass balance ln a mole balance a nonrreactlve system has quotm 0 for all Specles Unlike the mass balance the TOTAL genelatlon of moles Isn t necessanly 0 even for b To see thl s the above leactlon the flml number of moles will not equal the lnltlal number because 3 total moles ofmolecules ate leactlng to form 1 mole otploducts Why would We can nd 8 others from the reaction stoichiometry Thls ls a Very powerful tool because each reactluvl uley create um mzw mkvmwvl If you use thls method The followlng reactors edit Extent of Reaction ln Order to tolmallze the pleylous analysls of leactlons ln terms of a slngle yanable let us consldel the genenc leactlon allegtcf ldD The Molar Extent of Reaction X ls de ned as Ana 7 Ana 7 At Am a b c d slnce all of these ate equlyalent lt ls posslble to flnd the change ln moles of any specles ln a glven leactlon lfthe extent ofleactlon X ls known NOTE luunul quot 4 Hwy 7 ml can be detlned Some other de nitions ate dependent on the percent change of a pattlculat substlate and the stolchlometty ls used ln a dlfferent way to determlne the change ln the othels Thls detlnltlon makes x lndependent of the subsuate you choose The followlng example lllusuates the use of the extent of leactlon t a 4 Example Consider the reaction 11202 l 0a it 1120 39l 202 lfyou start wtth 50 g ot aOaand 25 grams of x and 25 ofthe moles otOxate consumed nd the molar extent ofxeacuon and the changes In the othet components Solution tenns of mass 1 01 m 50 g11202 a 34 g 1471 moles 1120 1 1 25g03gtr 0 05208 moles oJ 4 g Clearly ozone IS the hmtttng teactant hete Smce 2515 consumed we have that Amos 7025 t 05208 701302 males 03 2 X 730 11302 Hence by deflmuon 1 then we have 701302YAquot1I20 01302An02 2 01302 02601 ofthe apptopnate subsuate And A 1120 n 2 all 1 moles edit Mole Balances and Extents of Reaction w AMMe m de ned abuve l exactly the tdmeet uf mule efd gettetdted at cumumed by the edettmt NOTE This is only useful for imz39zvidwzl specie balances not the overall mole balance When to use an tnut utuat nt nan n at t k t A m all but one of the mdlvldual species This is because the total generation of moles m a teac genera ly not 0 so no algebraic advantage IS gamed by using the total matenal e system non IS balance on th Therefore we can write that 71AMquot AMA X 1 1 where X is the molar extent of reaction and a is the stoichiometric coefficient of A Plugging this into the mole balance derived earlier we arrive at the molecular mole balance equation ERAJIIM 271911511 A 9 a 0if A is consumed or Xa if it is generated in the reaction Degree of Freedom Analysis on Reacting Systems If we have N different molecules in a system we can write N mass balances or N mole balances whether a reaction occurs in the system or not The only difference is that in a reacting system we have one additional unknown the molar extent of reaction for each reaction taking place in the system Therefore each reaction taking place in a process will add one degree of freedom to the process NOTE This will be different from the atom balance which is discussed later Complications Unfortunately life is not ideal and even if we want a single reaction to occur to give us only the desired product this is either impossible or uneconomical compared to dealing with byproducts side reactions equilibrium limitations and other non idealities edit Independent and Dependent Reactions When you have more than one reaction in a system you need to make sure that they are independent The idea of independent reactions is similar to the idea of linear independence in mathematics Lets consider the following two general parallel competing reactions 39 LA I bB gtCCV dD lg41bgB reQE We can represent each of the reactions by a vector of the coefficients V A cooff B Geoff CT cooff D cooff E coeff 0 vi a39 b C d 1392 a21 b10103 2 This site NOTE The site above gives a nice tool to tell whether any number of vectors are linearly dependent or not Lacking such a tool it is necessary to assess by hand whether the equations are independent Only independent equations should be used in your analysis of multiple reactions so if you have dependent equations you can eliminate reactions from consideration until you39ve obtained an independent set By de nition a set of vectors is only linearly independent if the equation Kl 1391 If 7 I where K1 and K2 are constants only has one solution I A39L 112 0 Lets plug in our vectors Kl a b Cd0 Hg 1a ag 4120 092 0 Since all components must add up to 0 the following system follows K1ed0D a U1 g 20 Obviously the last three equations imply that unless c d 0 and e2 2 0 I A39l Iquot 0and thus the reactions are independent Lam Linearly Dependent Reactions Now let39s observe what happens if we add the reactions together a ailr l b 57ng 60 ED 83E We could solve the linear independence relation as we did in the previous case but let39s save ourselves the time it39s difficult to solve because there39s an infinite number of solutions and notice that a nontrivial solution to this equation is 1le 1I39g 1XV3 We can tell this beeeiese we notice that the second vector is simply the sum of the first two Therefore yield a third reaction then the three reactions are not independent All degree of freedom analyses in this book assume that the reactions are independent You shouldcl i edit Extent of Rmaion for Multiple Independent Reactims Whe you are setting up extents ofreaction in a molecular speciesbalance you mm make sure that you set up one for each reaction and include both in your mole balance So really your mole balance will look llke this 2mm 7 Em not l Eat Xi l for all In tl n due to the difficulty of solying such equations edit Equilibrium Reactions mm th t instead of reacting to completion it will stop at a certain point and not go any farther How far the reaction goes is dictated b the value of the eqttzlzbmtm ruef rle t Recall from general chemistry that the equilibrium coefficient for the reaction M l M A c l dDis defined ows 024 I i rq Cf thb B Iv1th concentration Ci expressed as molarity for liquid solutes or partial K pressure for gasses here A is the equilibrium concentration ofA usually expressed in molarity for an aqueous solution or partial pressure for a gas ms eqm u rim be remembered a 39pruduct erer Vedrttzvlts Usually solids and solvents are omitted by convention since their concentrations stay approximately constant throughout a reaction For example in an aqueous solution if water reacts it is left out ofthe equilibrium expression O en in ethbxium Ln order to do this rst recall that X iAnA a and smulat for the othet spectes w Liquidphase Analysis Rewntthg thxs m teth of molanty moles per Volume by dwxdmg by volume we have 5 A A z X 7 7 a or since the nal state wete htetested h is the equxhbxium state g Au 7 Aka V LL Solving for the desued equthbhum concentration we obtain the equauon for equthbhum concentration ofA in terms of conversion Similar equatth can be wntteh for B c and D usmg the defmmon of extent of teattmh Pluggmg h an the equatth into the expression for K we obtam 7 010 t rDJo t I 04W At equlhbnum for hquldrphase teattmhs only s g thxs equattoh knowing the value of K the teattmh stomhmmetry the mutual concentrations and the volume of the system the equlhbxium extent of reaction can be detetmthed NOTE If you know the reaction reaches equilibrium in the reactor this counts as an additional piece of information in the DOE analysis because it allows you to nd x If you to completion you can calculate the extent ofleactlon from that edit Gasphase Analysis By n entt n him for ldeal gasses are related to the mole actlon by the equatlon PA 1MP for ideal gasses only If A B C and D were all gases then the equlllbllum constant would look llke thls 1 ng PXszGasPhase Equilibrium Constant quot quot gases quotma VP 2 an ideal gas L 5160 a nonrldeal gas thls IS In general not one Slnce thls IS true we can saythat 71 P A 4 RT r above we obtain P ugglng this into the equation fol aX 7 7 7 7 PAN 7 a 1 Ala a RT Tr Therefore Slmllar equatlons can be wrltten for the other components Plugglng these lnto the equlllbrlum constant expresslon HPDDJr smarter MPH 7 RTb Gas Phase IdealGas Equilibrium Reaction at Equilibrium are at equlllbrlum and we ow the equlllbrlum coefflclent whllh can extent ofreactlon Agaln tr we know we kn o en be found in standard tables we can calculate the W Special Notes about Gas Reactions change a the mum gaze at unless the same number of moles are created as produced ln order to show that this ls true you nly need to wnte the lcleal gas law for the total amount of gas and reallze that the total number of moles ln the system changes Thls ls why we dont want to use tuttzlpressure ln the above equatlons for K we want to use puma pressures whlch we can convenlently wnte in terms ofextent orreattton edit Inert Species Notlce that all r r u t m one form or another Therefore tr there are specles present that don t react they may stlll ha Ve effect on the equillbrium because they wlll decrease the concentrations of the reactants and m tut t l l b e t Example Reactor Solution using Extent of Reaction and the DOE t alwayshelpsto draw a owchart rnl 100 kg m2 7 4PH3 802 egt P4H106H20 XQH3 0 5 moz 0 25 x02 0 5 There are four lndependent unknowns the total mass mole flowmte out of the reactor the r r t r e L L r r l and the extent ofxeactlon Mm H ll l m HPH P there are 0 DOF and tlns problem can be solved Let s lllustrote how to do lt fox tnls relotlvely slrnple system which lllustmtes some Very lmportantthlngs to keep ln mlnd Frrst recall that total mass is conserved even in a reacting system Therefore we can wnte at more m 100 kg ow lnce component masses aren t conserved we need to convert as much as we can lnto moles so we can apply the extent ofreootron 1 1 omm 05 s 100 kg 14705 moles P113 in 1 me I quotam 7 05 s 100 kg a m 7 15625 moles 02 All has m 025 z 100 kg x 78125 males 02 out 1 11101 0032 kg enters and how much leaves Recall that 2mm Ami a X U the teactton wntten above we have 1562 78125A8X 0 Solvmg gwes X 7 mol PH3 1470 6 hFszut 7 497136 0 2mm 10800 moles PH3 PAH 0 7 means 101766 0 01mm 97661noies p10 note 1t s hsteod of 7 because 1t s bemg genemted rather than consumed by the heathen H20 0 0mmquot k 697136 0 quotmm 5860 males 1120 Wm H obtam the mass percents As a samty check all of these plus 25 kg of oxygen should y1eld 100 kg total MassPIIvlout1080 moles0034kgmole 672kg Mass Homout 9766 moles 284 kgmo1e 74 kg Mass H20 out586 moles0018kgmole1055 kg Samty check 3672 2774 1055 25 oxygen 100 kg total so we re sun sane Hence we get 3 720513113 27 74P Hm 1 5H20 250Zby mass edit Example Reactor with Equilibrium mm Let s rst nd the number of male ofA and B we have murally 1 mol A mm 10 kg m W 400 mol A HE 100 kg solution w 16 kg B 12133 34043 mol B Now the volume conmbuted bythe 100kg of16 B soluuon rs r 100 k VTgl jmgu 57 22 lb 2 mL shoe addm the A conmbutes 5L to the Volume the Volume after the two ate mrxed IS 1093 L t 5 L 1143 L By de mtmn then the molantres ofA and B before the reactron occurs are 400 males A A0 W 35qu B 394 42 1 B 0 mi 297am Ln addrtmn there rs no C or D m the solutmn mrtmlly 010 01o 0 According to the storchrometry of the reaction a 1t b 2t c 3t d 1 Therefore we equrhbrrum equattoh rot hqurds the rouomhg equattoh IS obtamed Y 187 ii 35 7 mmms 7 Thts equottoh can be solved usmg Goalseek or one of the numerical methods h appeth 1 to we slnce we seek the amount of compound c that ls pmduced we have Aquot X 7539 gm C c 3mg o and X 14631 am no 3 14631 43893 moles c yields 43893 moles of c can be produced by thls lemon edit Introduction to Reactions with Recycle number used to improve the selectlvlty of multiple lemons push a Ieaction beyond its equlllbxlum reactor will generally contaln a sepamnon process m whlch unused lemme are partlally feed edit Example Reactor with Recycle is H2 02H 2Al B e C 0xqecLand7U 70xe ux qexte gofx V 0 j lsj slen7whalwouldua bf tlb ALheI e wa asepaxallanle ycle placess a erk assumelhauhemasspercen fhy mgenleavlngthe leader is the same whal h mllshoweffecllyexhls process can be7 Solution Let s rst draw our flowchart as usual 0 3 m5 xAS x135 FEED DOF Analysis mm A 1 On spllttex 3 unknowns r 0 equatlons we used all of them In labelng the than V 3 D Dupllcate vanables 8 quot1511451 Total DOF8 780 DOF edit Plan and Solution Genelally though not always u IS easlest to deal wuh the Ieactoz ltself 21 because u usually the st unknowns Lets begln by looklng at the overall system because we can o en get some valuable information from that E Overall System DOFoverall system 4 unknowns mmTAR X 3735 3 equations 2 1 DOF NOTE We CANNOT say that total mass of A and Bis conserved because we have a reaction here Therefore we must include the conversion X in our list of unknowns for both the reactor and the overall system However the total mass in the system is conserved so we can solve for quot15 Let39s go ahead and solve for m5 though because that39ll be useful later 784 100 03mm my 2280 kgh We can39t do anything else with the overall system without knowing the conversion so lets look elsewhere DOFseparator 4 unknowns mat 333 3345 3535 3 equations 2 1 DOF Let39s solve for those variables we can though We can solve for m3 because from the overall material balance on the separator 7771 7714 I Til5 m 100 7 2280 Th5 2380 kgh Then we can do a mass balance on A to solve for XA5 7i13324433rh4zr44 Iii533A 2380l15 100ll05 l 23803145 Since we don39t know 33B or rTBK we cannot use the mass balance on B or C for the separator so lets move on Let39s now turn to the reactor Ream Analysis ma1n 39r r 7 hi HL 4 7 DOF L t 1 V H e N we can solve for L eknow m mn by wr1t1ng a mule balance on A 1n the reactor n 39 AJrnr39tr 7 X gt1 a 7 Ma 1 may n14 Hm mol molecular we1ght 150002015 3 V kg 1 mnl rnol A M1 7 200 70002016 kg 7 992007 m5 e 115 07228001511 moi A nAmecyd 7 37 7 7 122000 h Thus the mad amount ofA entering the reactor is mol A 11A 99206 122000 221428 I The amount exrtrng 1s 80 0 mol A 7 7 1 0002010 77083 11 Therefore We have the followlng from the mole balance 221428 7 2X 7 177083 M males A 7 2213 h Now that We have thls We can calculate the mass of B and C generated my 7 7X MWB 7 22173 1 B 7 0026 kg 75705 kg B h mo B 11 mom l X MWp 7 22173 I C c 0030 kg I mo kg 0 h 06652 At thls polnt you balances on B and C 584 07 r35 2250 7 5705 7 tag s 2380 2 7 1 07 4 1 7 01544 7 135 e 2280 665 1 7 015 7 r53 2380 However these eqmnum are exactly the ame Therefore we have proven om assemon that there IS stlll l DOF In the leactol So we need to look elsewhere for somethng to calculate 2435 on B That place IS the sep arath balance m3 133 mu 134 m5 135 2 23801533 7 002000 2280135 I 0008564535 0008058 Note that thls means the pledomlnant specles m sueam 5 ls also C 50 the separatorrecycle setup does make a big dlffelence as we ll see next 0838 However 39 Now that we know now much ethane we can obtaln from the reactor after separatmg let s c ompare to what would happen wlthout any of the recycle systems In place wlth the same data as m the flrst part ofthls problem the new owchart looks llke thls m3 15 A m2584kyh3 ml 200 kyhA There are three unknowns Mullah X and three independent material balances so the 139n1m me la 7 9 m37 89h The result 1s that X 20250 moleng 077 77 was alloweu e n Mu n 4 mm m n W formed from the umeacted A and B edit Appendix 1 Useful Mathematical Methods edit Mean and Standard Deviation A lot of the time when youre conducting an experiment you will run it more than once especially if it is inexpensive Scientists run experiments more than once so that the mmz39um marks on a ruler cancel themselves out and leave them with a more precise measurement 4 l a u u a u a t more managable to usev edit Mean Suppose you have n data points taken under the same conditions and ou wish to consolidate them to as few as feasibly possible One thing which could help is is to use some centzallzed value which is in some way between all ofthe original data points This in tact is called the mean of the data set Ill mm A Other types of mean include the szeometrlc mean which should be used when the data are very widely distributed ex an exponential distribution and the lo grmean which occurs often in transport phenomena w Smndard Deviation iu ter tell you how spread out they are from the center A second statistical variable called the standard deviation is used for that The standard deviation is essentially the average imam2 to prevent negative deviations from lessonlng the effect ofposltlve deviations The mathematical formulation for the standard deviation ms The denominator is n71 instead of n because statlstlclans found that lt glves better results for w Putting it together The standard deviation of a data set measured under constant conditions is a measure of how precise the data set is Because thls is due the standaid deviation of a data set is often used in conjunctlon with the mean in oidei to iepoit expenmental iesults Typically iesults aie iepoited as i i 039 f a dlstnbutlon ls assumed kn win 0th the andard devlatlon can help us to estimate the piobability that the actual value of the vaiiable is within a ceitain range zf there l YLu systemotts bid in the data ftheie is such as use ofbioken equipment negllgence and so on then no statistics could predict the effects ofthat edit Linear Regression Suppose you have a set of data points it i iii taken under differing conditions which you suspect from a gmp can be ieasonably estimated by dmwlng a line through the polnts Any line that you could dmw will have or can be wntten In the following form y quotW t bwheie m is the slope of the line and b is the yrlntexcept Thls line will be most useful for both interpolating between points that we know and extmpolatlng to l A valuesr tb gow umdu which is called the residual Rt Wit W l Ln oidei to noiinalize the iesiduals so that they dont cancel when one s positive and one s Rtwhen dolng leastrsquares iegiession We use squared teiins and not absolute values so that the functlon is ditteientiable dont wony about this lfyou haven t taken calculus yet MA in i cm into squared residuals minimize 212i Using calculus we can take the derivative of this with respect to m and with respect to b and solve the equations to come up with the values of m and b that minimize the sum of squares hence the alternate name of this technique leastsquares regression The formulas are as follows where n is the total number of data points you are regressingm quot EU ilk Slit Elm quot EMA 20 b 7 Em m m 7 n W Example oi linear regmion The first thing we should do with any data like this is to graph it and see If a linear fit would be reasonable Plotting this data we can seeby inspection that a linear fit appears to be reasonable t m N0w we need 0 compute all 0f the Values 1m 0m Iegresslon formulas and 0 d0 mm by hand we set up a table t 2 1 11 559 5 313040 615 45 2 19 759 5 576340 1443 05 3 30 3 3 2 0676 2694 6 4 33 1116 1246126 4241 9 5 53 13037 1712695 693611 TOTAL 151 46422 4655464 1593115 N0w that we have um data we can plug 1t mt0 out meat Iegresslon equatmn n Em 91 7 2521 031 w 200 2a2 7 5 615931 13A 1513 46422 7 525039 5 1512 1779 yard mm 4 422A1 l 1 W 3912 yards Hence the equation for the line ofbest fit is D l779iltt39 2 The graph ofthis plotted against the data looks like this D yards tmin edit How to tell how good your regression is IS it stiii close enough to be useiuiv Though it will always come down to your own Judgment a ex seeing the fit iine graph against the data there is a mathematical tooi to help you called a com 39on coefficient I which can be defined in several different ways One of them is as follows 1 quot EU Qt Em Em tnmammtw quotexit942m it can be shown that this Value always iies between 71 and 1 The closer it is to 1 or 71 the more reasonable the linear fit Ln genetai the more data points you have the smaller I needs to be and then graph to see If the graph makes sense Sometimes it will sometimes it mon t the method is not fooiptoof Ln the above example we have 5 x1593113 1514 46422 ta 5635 1512 s 5 4 4655464 4 464222 7 1992 Hence the data conelates Very wellwuh a 11neat model edit Linearization edit In general Whenever you L r linearize the funcuon mst because lmear tegtessmn 1s much less mtenswe and more accurate Variable 1 Constant constant gtIlt Variable 2 The dxffexence between th1s and standard Imam tegtessmn 15 that Vanab1e 1 and Vanable 2 L you can t have quot339 yjas one vanable The techmque can be extended to more than two vanab1es usmg a method caued wzmuluple lmear Ie essmn but as thats more d1ff1cult to perform this secuon W111 focus on tword1mens1omlxegzess1on Power Law To see some of the power of lmeanzauon let s suppose that we have two vanab1es x and y related byapowex 1aw y A a atquot h 7 mi h 7 mm m f L wewould we can change 1t 1nto a lmearrtype funct1on1 quot L so 1n order log of both 5111155 in y n Axb y111 lt7gtlnyiln1bxrlnx The beauty ofthls equation is that it is in a sense hneat If we graph lny vs lnx obtain a straight line with slope h and yintercept lntA w Exponential Another common use of lineatiaation is with exponentials where x and y ate related by an expression of the form y A t 12 Again we take the natural 10g ofboth sides In older to get mi of the exponent yAbJ39lt gtlnyln1 l7lnb This time Graph in y vs x to obtain a line with slope 1111 and yintercept intA edit Linear Interpolation O en when you look up properties on a chart you will be looking them up at conditions in tow ti two charted conditions For example If you were looking up the Specific enthale of steam at 2 p 0 tips 1 tact u klkg 400 283241 2 4 450 Value First we set up a table including the unknown Value like this rtoct H klkg 1 400 28324 2 430 x 3 450 29434 milit im 7 with in ofalineisa I Therefore we can write that 2 r 7 450 e 430 7 450 e 400 Solving gives x 2899 kJkg for a given ll edit General formula To derive a more general formula though i always derive it from scratch anyways it s nice to have a f vmiila r yi 2 Xe ye ya Setting the siope between points 3 and 2 equal to that between 3 and 1 yields This equation can then be solved for 14 or 31 as appropriate edit Limitations ot Linea Interpolation exact How p two milon factors approximation 2 The difference between consecutive x Values on the table the smaller the distance the closer almost any function Wlll resemble a line Therefore it is not recommended to use iinear interpoiation 1f the spaces are Very widely separated Ho eve If no ther method of approximation is available iinear interpoiation is o en the onl option or other forms of interpoiation which may be just as inaccurate depending on what the actuai function is See also wzlntegolation References 1 Smith Karl J The Nature of Mathematics Pacific Grove California BrooksCole Publishing company 6e p 683 2 Sandler Stanley 1 Chemical Biochemical and Engineering Thermodynamics University of Deleware John Wiley and Sons inc 46 p 923 Q Basics of Rootfinding Rootfinding is the determination of solutions to single variable equations or to systems of n equations in n unknowns provided that such solutions exist The basics of the method revolve around the determination of roots A root of a function 13311 512 any number of variables is defined as the solution to the equation F0731 952 0 In order to use any of the numerical methods in this section the equation should be put in a speci c form and this is one of the more common ones used for all methods except the iterative method However it is easy to put a function into this form If you stait with an equation of the form F1L 1 2 F212 then subtracting szill yield the required form Do not forget to do this even if there is only a constant on one side Since any equation can be put into this form the methods can potentially be applied to any function though they work better for some functions than others git Analytical vs Numerical Solutions An analytical solution to an equation or system is a solution which can be arrived at exactly using some mathematical tools For example consider the function y m 3quot graphed below The root of this function is by convention when 5quot 0 or when this function crosses the X axis Hence the root will occur when 743 0 39T 1 The answer X1 is an analytical solution because through the use of algebra we were able to come up with an exact answer On the other hand attempting to solve an equation like 3 nut analytically is sure to lead to frustration because it is not possible with elementary methods In such a case it is necessary to seek a numerical solution in which guesses are made until the answer is quotclose enoughquot but you39ll never know what the exact answer is All that the numerical methods discussed below do is give you a systematic method of guessing solutions so that you39ll be likely and in some cases guaranteed to get closer and closer to the true answer The problem with numerical methods is that most are not guaranteed to work without a good enough initial guess Therefore it is valuable to try a few points until you get somewhere close and then start with the numerical algorithm to get a more accurate answer They are roughly in order from the easiest to use to the more difficult but faster converging algorithms Rootfinding Algorithms Iterative solution Iterative solutions in their purest form will solve the desired function so that it is in the form x x Then a value for X is guessed and fX is calculated The new value of X is then re inserted into fX and the process is repeated until the value of X changes very little The following example illustrates this procedure 39 I is Example Use an iterative solution to calculate the root of I l39 lnI 0 Solution Solve the equation for X First we need to guess an X to get it stalted Let39s try 17 05 Then we have 3 6 06065 mm awW 05453 323 63 05453 05796 34 2 6 0579 2 05601 ITS 8405601 3 6M7 05649 197 e4 wa 05584 Thus to two decimal places the root is 1 056 More iterations could be performed to get a more accurate answer if desired This method has some rather severe limitations as we ll see in this example 05798 0606 This example shows that the success of the iteration method strongly depends on the properties of the function on the righthand side In particular it has to do with how large the slope of the function is at the root If the slope is too large the method will not converge and even if it is small the method converges slowly Therefore it is generally undesirable to use this method though some more uSeful algorithms are based on it which is why it is presented here Q Iterative Solution with Weights Although the iterative solution method has its downfalls it can be drastically improved through the use of averaging In this method the function is still solved for X in the form From the initial guess 0 the function fx is used to generate the second guess 331 However rather than simply putting icjinto fX a weighted average of Jquotland inns made rimraw 11 0 1 a rlold0 S o S 1 The term Otis called the weight The most common value of the weight is one half in which case the next value to plug into fX is simply the average of 0and I10ld Old I Neu 71 I2 value ls then plugged rnto ttx averaged wrth the result and thrs ls repeated untrl convex nte Thls new Be an ru ruuru n r la Acmu t normal lteratlve solutlon Wu rd Example 1 a Flnd the root of7 l WI Ousng the lteratlve rnethod Wth a welght of 2 Solution Let s start wrth a guess of 05 lrkelast trrne and compare what happens thrs trrne from what happened wrth normal rteratron I 7 quot5 06065 05 06065 1171610 05533 2 quot35 05751 339 r 1 when 55542 m 0500 05688 Here a ex only three evaluatlons of the functlon Whlch usually takes the longest tlrne of all the steps 39 The method IS not only fastexrconvexglng but also more stable so that rt can actually be used solvlng the eguatlon the other way too t L 4 Example Stamnz wlth an lnltlal guess of z 05and using 30 5quotIand the welghted lteratlon 1 a method wlth 2 flnd the root ofthe eguatron Solution Stamng Wth 0 05we have 1 iln05 0693 1693 05 mneut 0597 52 440597 0517 I l jammy w 0557 33 450557 05856 05856 0557 I3nelu 0571 14 ln0571 0500 0560 U571 Moont f 0565 205 4110565 0570 Therefore we can slowly converge m thls case uslng the Welghted lteratlon method to the solutlon solutron but one way ls clearly faster than the other However welghtlng wlll accelerate the edit Bisection Method Let us consider an alternative approach to rootfinding Consider a function fx 0 which we desire to find the roots of If we let a second variable y f 5quot then y will almost always change sign between the left hand side of the root and the right hand side This can be seen in the above picture of y I ln which changes from negative to the left of the root x 1to positive to its right The bisection method works by taking the observation that a function changes sign between two points and narrowing the interval in which the sign change occurs until the root contained within is tightly enclosed This only works for a continuous function in which there are no jumps or holes in the graph but a large number of commonly used functions are like this including logarithms for positive numbers sine and cosine and polynomials As a more formalized explaination consider a function y fthat changes sign between 139 land 33 bWe can narrow the interval by 1 Evaluating the function at the midpoint 2 Determining whether the function changes signs or not in each sub interval 3 If the continuous function changes sign in a sub interval that means it contains a root so we keep the interval 4 If the function does not change sign we discard it This can potentially cause problems if there are two roots in the intervalso the bisection method is not guaranteed to find ALL of the roots Though the bisection method is not guaranteed to find all roots it is guaranteed to find at least one if the original endpoints had opposite signs The process above is repeated until you39re as close as you like to the root Note that convergence is slow but steady with this method It is useful for refining crude approximations to something close enough to use a faster but non guaranteed method such as weighted iteration Iedit Regula Falsi The Regula Falsi method is similar the bisection method You must again start with two x values between which the function fx you want to find the root of changes However this method attempts to find a better place than the midpoint of the interval to split itIt is based on the hypothesis that instead of arbitrarily using the midpoint of the interval as a guide we should do one extra calculation to try and take into account the shape of the curve This is done by finding the secant line between two endpoints and using the root of that line as the splitting point More formally Draw or calculate the equation for the line between the two endpoints afa and bfb Find where this line intersects the X axis or when y 0 giving you X c Use this X value to evaluate the function giving you fc The sub intervals are then treated as in the bisection method If the sign changes between fa and fc keep the inteval otherwise throw it away Do the same between fc and f Repeat until you re at a desired accuracy Use these two formulas to solve for the secant line y mx B m f0 7 M be a B N7 quotI b f m ayou canuseelthexj OI ma ll m l nan the blsectlon method dependlng on how long u takes to calculate the slope ofthe hue and the shape of the functlon 3 Example FmdlhemocofF 711Obutthstlmeusethexegulafalslmethod Solution lt s actual value For companson wuh lusecuah let s choosethe same lhulal guesses a 05m b 1 for whlchfll 0593ahdfb 1 nFlrstlntexVal 0 5 lt w lt1l0193 lt fxlt1l gem 1m 1 23861 1386 Root ofsecant llne T U 5 I 7 bl Functlonvalueatmotf 0581 39 MU581 0038 whlch would use an 075as the Spllttlng palm Secondlntewal05 lt I lt 0581 U lt x lt 0038 l Secant llhe 2 28521 7 1619 We come up wuh practlcally the exact root a ex only two llelalmhsz Ln some A n miln n the shape ofthe cuwe However u generally wonh uyhg for a couple of Iterations due to the dmstlc speed hcleases posslble edit Secant Method Tangent Method Newton39s Method In this method we attempt to find the root of a function y fx using the tangent lines to functions This is similar to the se cant method except it quotcuts loosequot from the old point and only concentrates on the new one thus hoping to avoid hang ups such as the one experienced in the example Since this class assumes students have not taken calculus the tangent will be approximated by finding the equation of a line between two very close points which are denoted x and 3quot 39l39 m The method works as follows 1 Choose one initial guess 331 2 Evaluate the function fx at 33 iI39I39Iand at x 1 quotI39 6where this a small number These yield two points on your approximate tangent line 3 Find the equation for the tangent line using the formulas given above 4 Find the root of this line This is 2 5 Repeat steps 2 4 until you39re as close as you like to the root This method is not guaranteed to converge unless you start off with a good enough first guess whichis why the guaranteed methods are useful for generating one However since this method when it converges is much faster than any of the others it is preferable to use if a suitable guess is available 01 6110 f0501 43190 853 161 git What is a System of Equations A system of equations is any number of equations with more than one total unknown such that the same unknown must have the same value in every equation You have probably dealt a great deal in the past with linear systems of equations for which many solution methods exist A linear system is a system of the form Linear Systems C1 1111 Help Cg bll l b25132 And so on where the a39s and b39s are constant Any system that is not linear is nonlinear Nonlinear equations are generally far more difficult to solve than linear equations but there are techniques by which some special cases can be solved for an exact answer For other cases there may not be any solutions which is even true about linear systems or those solutions may only be obtainable using a numerical method similar to those for single variable equations As you might imagine these will be considerably more complicated on a multiple variable system than on a single equation so it is recommended that you use a computer program if the equations get too nasty Solvability A system is solvable if and only if there are only a finite number of solutions This is of course what you usually want since you want the results to be somewhat predictable of whatever you39re designing Here is how you can tell if it will de nitely be impossible to solve a set of equations or if it merely may be impossible Solvability of systems 1 If a set of n independent equations has n unknowns then the system has a finite possibly 0 number of solutions 2 If a set of n independent equations has less than n unknowns then the system has an infinite number of solutions 3 If a set of n independent or dependent equations has more than n unknowns then the system has no solutions 4 Any dependent equations in a system do not count towards 11 Note that even if a system is solvable it doesn39t mean it has solutions it just means that there39s not an infinite number edit Methods to Solve Systems As you may recall there are many ways to solve systems of linear equations These include 0 Linear Combination Add multiples of one equation to the others in order to get rid of one variable This is the basis for Gaussian elimination which is one of the faster techniques to use with a computer Cramer39s rule which involves determinants of coefficient matrices Substitution Solve one equation for one variable and then substitute the resulting expression into all other equations thus eliminating the variable you solved for The last one substitution is most useful when you have to solve a set of nonlinear equations Linear combination can only be employed if the same type of term appears in all equations which is unlikely except for a linear system and no general analogue for Cramer39s rule exists for nonlinear systems However substitution is still equally valid Let39s look at a simple example Example of the Substitution Method for Nonlinear Systems Solution We want to employ substitution so we should ask which variable is easier to solve for In this case X in the top equation is easiest to solve for so we do that to obtain X 4 1 Substituting into the bottom equation gives 4 Y2 Y9 22 4 8Y2lY4 Y2 22 Y4 912 18 0 This can be solved by the method of substitution Let U Y2 Plugging this in UQQUlsr NOTE All Ys must be eliminated for this method to be Valid Do something like this when the 4 2 y form every time ie ifthe equation was somethinglike 1 7 9y 7 18 i c then the method would no simplify your calculations enough to make it worth doing Solving by factoring U 7 6U 7 3 o Uamp6 o Thus Since U Y we obtain four solutions for Y Yii3YiiE N ti h 7 make sense so think before you continue in we have 1 Again positive in t d 1 make sense then the ONLY soiution that is of any worth to us is the soiution X Y L kmce Y xesuits in anegative vaiue fotx Notice that even a small system like this has a large number of solutions and indeed some systems will have an infinite number such as ll y ed t Numerical Methods to Solve Systems There are numerical equivalents in multiple variables to some of the methods demonstrated in the previous section Many of them in their purest forms involve the use of calculus in fact the Taylor method does as well but as before they can be reduced to approximate algebraic forms at the expense of some accuracy Iedit Shots in the Dark If you can solve all of the equations explicitly for the same variable say y you can guess all but one and then compare how different the resulting values of y are in each equation This method is entirely brute force because if there are more than two equations it is necessary to guess all of the variables but one using this method and there is no way to tell what the next guess should be Trying to guess multiple variables at once from thin air gets to be a hastle even with a computer Since there are so many problems with this method it will not be discussed further edit Fixedpoint iteration Again the multivariate form of fixed point iteration is so unstable that it generally can be assumed that it will not work Weighted iteration is also significantly more difficult Looping method This is one method that does work and that is somewhat different from any single variable method In the looping method technique it is necessary to be able to solve each equation for a unique variable and then you39ll go around in a loop essentially starting with an initial guess on ideally a single variable say y and then evaluating all equations until you return to your original variable with a new value y39 If the result is not the same as the guesses you started with you need to make a new guess based on the trends in the results NOTE What kind of trends am I talking about If you have a well behaved system an increase in y will consistently lead to either an increase or a decrease in y39 so you can take advantage of this to see which way you need to adjust your original guess DO NOT attempt to use the value for yl as a new guess More specifically here is an algorithm you can use Solve all equations for a different variable Make a guess on one variable or as many as necessary to evaluate a second one if it39s more than one it gets harder though so it is recommended to use another method Go through all of the equations until you end up recalculating the variable or all of the variables which you had originally guessed Note whether the result is higher or lower than your guess Make another guess on the variables Go through the loop again D Nh l 4 LII After these two guesses we know whether increasing or guess will increase or decrease the recalculated value Therefore we can deduce whether we need to increase or decrease our guess to get a recalculated value equal to the guess 6 Keep guessing appropriately until the recalculated value equals the guess This technique is often necessary in engineering calculations because they are based on data not on explicit equations for quantities As we39ll see however it can be difficult to get it to converge and this method isn t that convenient to do by hand though it is the most reliable one to do realistically It is great however for inputting guesses into a computer or spreadsheet until it works First we need to solve one of them for x let39s choose the first one 1quot ny To stait off we make a guess y 01 Then from the first equation x 2303 Plug this back into the second equation and you39ll come out with y39 0834 The recalculated value is too high Now make a new guess on y say y 05 This results in x 06931 Plugging back into the second equation gives y39 03665 The recalculated value is too low NOTE Now we know that increasing the guess decreases the recalculated value yl and vice versa Since the second value of y is too low this means that we need the guess to be smaller than 05 likewise since the first yl was too high we need it to be greater than 01 Lets now try y 025 This results in X 1386 from the first equation and yI 0326 from the second Too high so we need to increase our guess Let s guess y 03 Thls ylelds x 1204 and thus 37 0185 whlch 15100 low Indicating the guessed value as too high Guess y 028 hence x 1273 and y 0241 The guess ls therefore stlll too hlgh Guess y 027 hence x1309 and y 169 Therefore we have now converged r 13094 027 edit Looping Method with Spreadsheets post t m three rows llke so c A B 1 y guess x y39 2 71nHX2 1ntgzt Ln r u L Hquot m llne A2 Canto cell D2 slnte we want y to equal y Just keep guesslng untll the value n m ls as close to zero as you hke As a more lnrdepth example whlch would be slmhtantly more dlmtult to do by hand consider the system ln order for thls to work we only need to solve each equatlon for a unlque vanable the and this wlll be evident shortly 2 P9 X273 clt39T T T 7 2 X T3 P P 01T NOW We need to am ln thls case the best answet ls T because from thls guess we can calculate P from equatlon 3 w n Fv m Manon 2 fr m POHZH n l thls new Value as a gauge afoul old guess NOTE Genetally you want to statt the loop Wlth a yatlable that allows you to calculate a second the case before solvlng because we want to ayold guessan on multlble yatlables If at all posslble Lets program thls lnto the spteadsheet a c D 1 r guess p x TV TV 7 r guess m2 sqrt123 7 gzt 72W322WCZ2 7 3sxpt7 2 e l CZAZHAz 7 2 7D2 7 Once all thls ls programmed ln you can just lnput guesses as before wlth the eventual tesult at P 02453X 38098 2453 Multivariable Newton Mahod NOTE You may want to sklp thls sectlon If you don t know how to lnyett maulces add them ot multlblythem Thete ls a multlyatlate extenslon to Newton s method whlch ls highly useful lt converges qulckly llke the slngleyatlable yetslon Wlth the downslde that at least by hand lt ls tedlous me to do thls Wlth llttle dlfflculty and the method ls not ping method ls ln add u ll ll ofval owevet a computer can e plogtam llmlted only to systems whlch can be expllcltly solved llke the loo ltlon a null l ues to use as a guess The method works as follows i rleletUFIr T l 2 Guess a Value for all Varlables and put them lnto a matilx X Calculate the Value of all functlons F at thls guess and put them lnto a culumvl mamx W 3 whlch ls descrllled later 4 Construct a matrix to become the Jacobian as follows make an empty mamx Wlth h rows and h colurnns where h ls the number of equatlons or the number of Vailables remember a Wlth the names of varlables and the rows Wlth the names of your tunctlons it should look somethng llke thls W 371 ya F1 F9 39 l 5 Put the approprlate pamal derlvatlve lnthe labeled spot For example put the partlal derlvatlve Wlth respect to 141 from functlon 1 ln the flrst spot 6 Once the Jacoblan mamx ls completely constructed flnd the lnverse of the matrix There are r thls one WARN use at your own rlskz Or you can do it byhand lfyou know how 7 Matilxrmultlply the inverse Jacoblan with the transposition of Fquot FT then add the transposition ofX again make lt a column maulx XL XnT a Jquot MAW Muitivariabie Newton Method Formula 8 The result ls your next guess Repeat untll convergence edit Estimating Partial Derivatives sure you carry out quite a few decimal places A ecause chan ing the variables by a very small amount may not change the function values too much but even small changes are important You MUST make when doing this b A Partial derivative is in its most basic sense the slope ofthe tangent line of a function with rethan n anal r tii c v t 1 Calculate one function F at your guess crease um variable x by a very small amount 6 Leave all other variables constant 3 Recalculate F at the modified guess to give you F F F The partial derivative ofthe function F with respect to x is then 6 edit Example of Use of Newton Method Let s go back to our archetypal example 116quot yln1 step 1 We needto solve each for zero step 2 h ofthe equations Then X 230301 The values of F at this guess are Fl 0r F 2 0734213 and hence by definition F 0 0724213 step 35 Now we need to stay organized so let s introduce some notation l the pamal denvauve of funcuon mm respect to vamble Lets choose 6 10LThen FM 7 r51 1 F1051 6 ngw E Fwym may 01 001 9 4701 7 1 101 7 7 The pamal denvauves of F2 can be sxmllarly calculated to be 65630 0433and Jngy 1 Therefore the Jacobxan ofthe system 15 I y F 701036 1 F2 0433 1 Step 6 r J y 1 636 18636 Jquot F F2 38069 411931 Step 7 The hansposmon of F 15 simply F 0 0734213 A 1quot FT 13582 01418 Therefore we should subtmct 13682 from x and 01418 from y to get the next guess 09373 2418 Notlce how much closer thls ls to the true answez than what we started wlth However thls Problem 1 TH n m Hm H n mm nfvm Whichls ofthe form Vm S 45 7 Km l S where Vm mand 1839 mare constants a Wnte thls equatloh lh a llneanzed form What should you plot to get a llne7 What will the slope be7 How about the yrmtercept 11 Gwen the followlng data and the llneanzed form of the equatloh pledltt the values of Vumahd Km ls n 5 Hs 0 02 0 oooe 00010 00014 0002a 030 00023 00030 0003a 00037 200 00033 Also calculate the R Value and comment on how good the flt ls u l lul xvhm mlgu Vm lxepxesent 1 Find the value of rrs when S is 10 M In three ways 1 Plug 10 into your expression for as with the bestrflt parameters 2 P rform a linear interpolation between the appropriate points nearby 3 Perform a linear extrapolatlon from the line between points 05 00030 and 08 6 Which is probably the most accurate whyv 4 Problem 2 Flnd the standard deviation ofthe following set of arbiuary data Write the data in i 0 form Are the data very preoisev 101 13900 03986 03993 03995 Problem a to the equation a 1 14139 i 15 02 solutions Use the quadratic formula to check your technque before moVing on to the next problems we A141 l 15 n 31 c e fr 1 solutlon it Appendix 2 Problem Solving using Computers git Introduction to Spreadsheets This tutorial probably works with other spreadsheets such as wzopen office with minor modifications A spreadsheet such as Excel is a program that lets you analyze moderately large amounts of data by placing each data point in a cell and then performing the same operation on groups of cells at once One of the nice things about spreadsheets is that data input and manipulation is relatively intuitive and hence easier than doing the same tasks in a programming language like MATLAB discussed next This section shows how to do some of these manipulations so that you don39t have to by hand git Anatomy of a spreadsheet A spreadsheet has a number of parts that you should be familiar with When you first open up the spreadsheet program you Will see something that looks like this the image is from the German version of open office ZJ39BBQQE watAWMq 39wg va 132131411 1E 7 lAnal Hl l lmi a 2 E l C quotquot39u lm lmZ First off notice that the entire page is split up into boxes and each one is labeled RUWS are labeled with numbers and columns with letters Also try typing something in and notice that the box above the spreadsheet to the right of Z will change automatically as you type When you39re just putting in numbers this info box will just have the same number in there But when you39re putting in formulas the cell will display the value calculated from the formula while the info box will display what the formula was Inputting and Manipulating Data in Excel The first step in any spreadsheet analysis is to input the raw data you want to analyze It is most effective if you put it in columns with one column for each variable It lets you see more data at once and it also is less limited because the maximum number of rows is much larger than the maximum number of columns It is good practice to use the first row for the names of the variables and the remaining for the data points Make sure you include units In this section the following data will be used as illustration A B 1 tmin D yards 2 11 5595 3 19 7595 4 30 8982 5 38 11163 6 53 13087 Using formulas In order to tell the spreadsheet that you want to use a formula rather than just enter a number you have to start the entry with an equal sign 2 You can then use combinations of decimal values and cell designations A cell designation is simply the column letter followed by the row number containing the value you wish to manipulate For example if you wanted to find the product of the distance traveled and the time spent traveling you could put in the formula A2 B2 into any empty cell and it would give you the answer From here out it will be assumed that this value is in cell C2 You should label the column with the type of calculation you39re performing ledit Performing Operations on Groups of Cells The question may arise why not just put in the numbers themselves instead of referencing the cell There are two major reasons for this 1 If you change the value in the referenced cell the value calculated in the formula will automatically change 2 The built in dragging capability of most spreadsheets The dragging capability is a simple concept If you have put a formula into a spreadsheet you can have it copied to any number of cells you want To do this select the cell with the formula and bring the mouse pointer to its lower right hand corner You should see a dark icon Info Bar A2 32 A B C l tmln D yards 2 11 5595 l 62545 I 3 19 7595 7777777 71 4 30 8982 5 38 11163 6 53 13087 Click on the and drag it down This will cause the formulas to change according to how you drag the box In this case if you drag it down to row 6 the spreadsheet will produce the following A B C 1 t min D yards tD 2 11 5595 61545 3 19 7595 144305 4 30 8982 26946 5 38 11163 424194 6 53 13087 693611 If you click on the last value in column C 693611 the info bar will display A6 B6 This is very useful for performing the same operations on multiple sets of data at once rather than having to do the multiplication 5 separate times here we just do it once and drag down the box Special Functions in Excel In order to do many mathematical operations in Excel or at least the easiest way it is necessary to use functions not to be confused with formulas A function is simply an implementation someone already wrote for the mathematical operation so all you have to do is know how to tell it to do the operation and where to put it when it39s done In excel you can call a function named quotfunctionquot by typing the following into a cell function inputs The function will then execute and the cell containing the call will display the answer The necessary inputs are sometimes numbers but are more often the cell addresses For example in the data above say you wanted to take the exponential 31 of all the time points in column A and place the result in column D The function for exponential is exp and it can only accept one input at a time but due to the dragging capability of Excel this will not matter much you can just call it once and then drag the cell as you would with any formula containing cell addresses So to do this you would type into cell D2 eXpA2 Hit enter then click the in the bottom right and drag the cell down You should end up with something like this after labeling the D column appropriately A B C D 1 t min Dyards tD eAt 2 1 5595 61545 3004166024 3 19 7595 144305 6685894442 4 3 8982 26946 2008553692 5 38 11163 424194 4470118449 6 53 13087 693611 20033681 All excel functions output only one value at a time though some can accept multiple cells at a time as input mostly statistical functions Following is a brief synopsis of the functions available For a complete list see the help files for your spreadsheet as the availability of each function may vary depending on which one you are using CELL signifies either the rowcolumn designation of the cell you want to pass to the function as input or some numerical value you enter manually edit Mathematics Functions Generally these only take one input at a time absCELL Absolute value of CELL sqrtCELL Square root of CELL to do nth roots use CELLA1n lnCELL Natural log of CELL log10CELL Log of CELL to base 10 logCELL NUM Log of CELL to the base NUM use for all bases except e and 1m eXpCELL EXponentialeAX of CELL Use since Excel doesn39t have a builtein constant quotequot sinCELL cosCELL tanCELL Trigonometric functions sine cosine and tangent of CELL CELL must be in radians asinCELL acosCELL atanCELL Inverse trigonometric functions returns values in radians sinhCELL coshCELL tanhCELL Hyperbolic functions asinhCELL acoshCELL atanhCELL Inverse hyperbolic functions Ldit Solving Equations in Spreadsheets Goal Seek Excel and possibly other spreadsheets have a very useful tool called goalseek which allows the user to solve single variable equations and can be used as an aid in guess and check for systems of algebraic equations Let39s suppose for the purposes of this tutorial that you wish to find a solution to the equation 0 l 2X2 X 41 In order to set up the problem in Goalseek it is necessary to define a cell for the variable you want to change X and a cell for the function you want to evaluate NOTE Goalseek will only work if you tell it to evaluate some function until it reaches a constant value You cannot tell it to equal something that can change so for example you cannot do something like this X X 2X2 11 because neither side is a constant The easiest way around this is generally to solve the function for zero and then use that as the evaluating function Here we could set up the cells as follows A B l X fX 2 fl A2A3 2A2A2 7 A2 1 To solve this one go to Tools gt Goalseek It39ll give you three boxes quotSet Cellquot quotTo Valuequot and quotBy Changing Cellquot Since we want the value in cell B2 to equal 0 enter B2 into the quotSet Cellquot box and 0 into the quotTo Valuequot box Since cell B2 depends on cell A2 we want to change A2 so that B2 equals 0 Hence the quotBy Changing Cellquot box should contain A2 Put that in and click quotOKquot and Goalseek will converge to an answer A B l X NM 2 7254683 7000013 Notice that the success of the goalseek depends on what your initial guess was If you try to put in an initial guess of 0 in this example instead of 1 goalseek will diverge It will tell you so saying quotGoal Seeking with Cell B2 May Not have Found a Solutionquot However the algorithm is generally fairly robust so it shouldn39t take too many guesses to obtain convergence NOTE You can only enter one cell into each of the quotSet Cellquot and quotBy Changing Cellquot boxes and the value in quotTo Valuequot must be a constant Graphing Data in Excel In Excel there are a variety of ways to graph the data you have inserted such as bar graphs pie charts and many others The most commonly used in my experience is the scatterplot which is the name Excel uses for the typical x y quotline graphquot plot that you probably think of first when you think of a graph Scatterplots Scatterplots can be made relating any one independent variable to any number of dependent variables though if you try to graph too many it will get crowded and hard to read Excel will automatically give each different dependent variable a different color and a different shape so that you can distinguish between them You can also name each quotseriesquot of data differently and Excel will automatically set up a legend for you This is how to make a scatterplot 1 Put the data into columns just like it was given in the problem statement 2 Now we need to set up the graph Go to 17186 Chart 3 Select quotXY scatterquot and click quotnextquot 4 Click the quotseriesquot tab on top If there are any series present remove them with the remove button since it usually guesses wrong what you want to graph Now we can add a series for each dependent variable we want to graph as follows 1 Click quotaddquot 2 Next to quotX valuesquot click the funky arrow symbol to the right of the text box A small box will pop up 3 Click on the first value for the independent variable and drag the mouse down to the last value Click the funky symbol again to bring you back to the main window 4 Do the same thing with the quotY valuesquot but this time you want to select the values of the dependent variable 5 Click next and give the graph a title and labels if you want Then click next and quotfinishquot to generate your graph edit Performing Regressions 0f the Data from a Scatterplot Once you have a scatterplot of your data you can do one of several types of regression logarithmic exponential polynomial up to 6th degree linear or moving average Excel will plot the regression curve against your data automatically and except for moving average you can tell it to give you an equation for the curve To do this 1 Right click on one of the data points it doesn39t matter which Click quotadd trendlinequot 2 A new window will come up asking you for the type of regression Choose the type of regression you want to use 3 Click on the quotoptionsquot tab and check the quotDisplay Equation on Chartquot box and if you want the quotDisplay R squared value on Cha quot box Click OK 9 If you chose a quotlinearquot regression with the sample data above the equation and R value appear quotr r Fr Iui 4 39 on the graph as 3 1 l Bquot J l 59125 R i 09545 Note Excel displays R39rather than R so that we don t need to worry about negative vs positive values if you want R just take the square root which is 09921 as we calculated in the section on linear regressions git Further resources for Spreadsheets Excel and other spreadsheets can do far more than what is described here For additional information see Excel szxcel or the help files for the program you are using Graphing Data in MATLAB Polynomial Regressions MATLAB is able to do regressions up to very large polynomial orders using the quotpolyfitquot function The syntax for this function is polyfitXDATA YDATA Order The X data and y data must be in the form of arrays which for the purposes of this application are simply comma separated lists separated by brackets For example suppose you want to perform the same linear regression that had been performed in the quotlinear regressionquot section The first step is to define the two variables gtgt XDATA ll19303853 gtgt YDATA 55957594898211163l3087 Then just call polyfit with order 39139 since we want a linear regression gtgt polyfitXDATA YDATA 1 ans l0e002 177876628209900 391232582806103 The way to interpret this answer is that the first number is the slope of the line 177810quot2 and the second is the y intercept 391210quot2 git Appendix 3 Miscellaneous Useful Information Ldit What is a quotUnit Operationquot A unit operation is any part of potentially multiple step process which can be considered to have a single function Examples of unit operations include 0 Separation Processes Purification Processes Mixing Processes Reaction Processes Power Generation Processes Heat Exchangers In general the ductwork between the processes is not explicitly included though a single pipe can be analyzed for purposes of determining friction loss heat losses pressure drop and so on Large processes are broken into unit operations in order to make them easier to analyze The key thing to remember about them is that the conservation laws apply not only to the process as a whole but also to each individual unit operation The purpose of this section is not to show how to design these operations that39s a whole other course but to give a general idea of how they work edit Separation Processes There are a large number of types of separation processes including distillation extraction absorption membrane filtration and so on Each of these can also be used for purification to varying degrees Distillation Distillation is a process which is generally used to separate a mixture of two or more liquids based on their boiling points The idea is that the mixture is fed into a column and is heated up until it starts to boil When a solution boils the resulting gas is still a mixture but the gaseous mixture will in general have more of the lower boiling compound than the higher boiling compound Therefore the higher boiling compound can be separated from the lower boiling compound Two examples of distillation processes are petroleum distillation and the production of alcoholic beverages In the first case oil is separated into its many components with the lightest on the bottom and the heaviest on top In the latter the gas is enriched in ethanol which is later recondensed Distillation has a limit however nonideal mixtures can form azeotropes An azeotrope is a point at which when the solution boils the vapor has the same composition as the liquid Therefore no further separation can be done without another method or without using some special tricks edit Gravitational Separation Gravitational separation takes advantage of the well known effect of density differences something that is less dense will oat on something that is more dense Therefore if two immiscible liquids have significantly different densities they can be separated by simply letting them settle then draining the denser liquid out the bottom Note that the key word here is immiscible if the liquids are soluble in each other then it is impossible to separate them by this method This method can also be used to separate out solids from a liquid mixture but again the solids must not be soluble in the liquid or must be less soluble than they are as present in the solution Iedit Extraction Extraction is the general practice of taking something dissolved in one liquid and forcing it to become dissolved in another liquid This is done by taking advantage of the relative solubility of a compound between two liquids For example caffeine must be extracted from coffee beans or tea leaves in order to be used in beverages such as coffee or soda The common method for doing this is to use supercritical carbon dioxide which is able to dissolve caffeine as if it were a liquid Then in order to take the caffeine out the temperature is lowered lowering the quotsolubilityquot in carbon dioxide and water is injected The system is then allowed to reach equilibrium Since caffeine is more soluble in water than it is in carbon dioxide the majority of it goes into the water Extraction is also used for purification if some solution is contaminated with a pollutant the pollutant can be extracted with another clean stream Even if it is not very soluble it will still extract some of the pollutant Another type of extraction is acidbase extraction which is useful for moving a basic or acidic compound from a polar solvent such as water to a nonpolar one Often the ionized form of the acid or base is soluble in a polar solvent but the non ionized form is not as soluble The reverse is true for the non ionized form Therefore in order to manipulate where the majority of the compound will end up we alter the pH of the solution by adding acid or base For example suppose you wanted to extract Fluoride F from water into benzene First you would add acid because when a strong acid is added to the solution it undergoes the following reaction with uoride which is practically irreversible F 1130 4 HF H30 The hydrogen uoride is more soluble in benzene than uoride itself so it would move into the benzene The benzene and water uoride solutions could then be separated by density since they39re immiscible The term absorption is a generalization of extraction that can involve different phases gas liquid instead of liquid liquid However the ideas are still the same Membrane Filtration A membrane is any barrier which allows one substance to pass through it more than another There are two general types of membrane separators those which separate based on the size of the molecules and those which separate based on diffusivity An example of the first type of membrane separator is your everyday vacuum cleaner Vacuum cleaners work by taking in air laden with dust from your carpet A filter inside the vacuum then traps the dust particles which are relatively large and allows the air to pass through it since air particles are relatively small A larger scale operation that works on the same principle is called a fabric filter or quotBaghousequot which is used in air pollution control or other applications where a solid must be removed from a gas Some fancy membranes exist which are able to separate hydrogen from a gaseous mixture by size These membranes have very small pores which allow hydrogen the smallest possible molecule by molecular weight to pass through by convection but other molecules cannot pass through the pores and must resort to diffusion which is comparatively slow Hence a purified hydrogen mixture results on the other side Membranes can separate substances by their diffusivity as well for example water may diffuse through a certain type of filter faster than ethanol so if such a filter existed it could be used to enrich the original solution with ethanol edit Reaction Processes Plug ow reactors PFRs and Packed Bed Reactors PBRs A plug ow reactor is a idealized reactor in which the reacting uid ows through a tube at a rapid pace but without the formation of eddies characteristic of rapid ow Plug ow reactors tend to be relatively easy to construct they39re essentially pipes but are problematic in reactions which work better when reactants or products are dilute Plug ow reactors can be combined with membrane separators in order to increase the yield of a reactor The products are selectively pulled out of the reactor as they are made so that the equilibrium in the reactor itself continues to shift towards making more product A packed bed reactor is essentially a plug ow reactor packed with catalyst beads They are used if like the majority of reactions in industry the reaction requires a catalyst to significantly progress at a reasonable temperature edit Continuous StirredTank Reactors CSTRs and Fluidized Bed Reactors FBs A continuous stirredtank reactor is an idealized reactor in which the reactants are dumped in one large tank allowed to react and then the products and unused reactants are released out of the bottom In this way the reactants are kept relatively dilute so the temperatures in the reactor are generally lower This also can have advantages or disadvantages for the selectivity of the reaction depending on whether the desired reaction is faster or slower than the undesired one CSTRs are generally more useful for liquid phase reactions than PFRs since less transport power is required However gas phase reactions are harder to control in a CSTR A uidized bed reactor is in essence a CSTR which has been filled with catalyst The same analogy holds between an FE and CSTR as does between a PFR and a PBR edit Bioreactors A bioreactor is a reactor that utilizes either a living organism or one or more enzymes from a living organism to accomplish a certain chemical transformation Bioreactors can be either CSTRs in which case they are known as chemostats or PFRs Certain characteristics of a bioreactor must be more tightly controlled than they must be in a normal CSTR or PFR because cellular enzymes are very complex and have relatively narrow ranges of optimum activity These include but are not limited to Choice of organism This is similar to the choice of catalyst for an inorganic reaction Strain of the organism Unlike normal catalysts organisms are very highly manipulable to produce more of what you39re after and less of other products However also unlike normal catalysts they generally require a lot of work to get any significant production at all Choice of substrate Many organisms can utilize many different carbon sources for example but may only produce what you want from one of them Concentration of substrate and aeration Two inhibitory effects exist which could prevent you from getting the product you39re after Too much substrate leads to the glucose effect in which an organism will ferment regardless of the air supply while too much air will lead to the pasteur effect and a lack of fermentation pH and temperature Bacterial enzymes tend to have a narrow range of optimal pH and temperatures so these must be carefully controlled op x U gt L1 However bioreactors have several distinct advantages One of them is that enzymes tend to be stereospecific so for example you don39t get useless D sorbose in the production of vitamin C but you get L sorbose which is the active form In addition very high production capacities are possible after enough mutations have been induced Finally substances which have not been made artificially or which would be very difficult to make artificially like most antibiotics can be made relatively easily by a living organism Heat Exchangers In general a heat exchanger is a device which is used to facilitate the exchange of heat between two mixtures from the hotter one to the cooler one Heat exchangers very often involve steam because steam is very good at carrying heat by convection and it also has a high heat capacity so it won39t change temperature as much as another working uid would In addition though steam can be expensive to produce it is likely to be less expensive than other working uids since it comes from water Tubular Heat Exchangers A tubular heat exchanger is essentially a jacket around a pipe The working uid often steam enters the jacket on one side of the heat exchanger and leaves on the other side Inside the pipe is the mixture which you want to heat or cool Heat is exchanged through the walls of the device in accordance to the second law of thermodynamics which requires that heat ow from higher to lower temperatures Therefore if it is desired to cool off the uid in the pipe the working uid must be cooler than the uid in the pipe Tubular heat exchangers can be set up in two ways c0 current or countercurrent In a co current setup the working uid and the uid in the pipe enter on the same side of the heat exchanger This setup is somewhat inefficient because as heat is exchanged the temperature of the working uid will approach that of the uid in the pipe The closer the two temperatures become the less heat can be exchanged Worse if the temperatures become equal somewhere in the middle of the heat exchanger the remaining length is wasted because the two uids are at thermal equilibrium no heat is released To help counteract these effects one can use a counter current setup in which the working uid enters the heat exchanger on one end and the uid in the pipe enters at the other end As an explanation for why this is more efficient suppose that the working uid is hotter than the uid in the pipe so that the uid in the pipe is heated up The uid in the pipe will be at its highest temperature when it exits the heat exchanger and at its coolest when it enters The working uid will follow the same trend because it cools off as it travels the length of the exchanger Because it39s counter current though the fact that the working uid cools off has less of an effect because it39s exchanging heat with cooler rather than warmer uids in the pipe git Appendix 4 Notation git A Note on Notation git Base Notation in alphabetical order plquot Molarity of species i in stream n A Area m mass MW Molecular Weight Molar Masw n moles N Number of components X Mass fraction y Mole fraction V velocity V Volume git Greek p Density Sum git Subscripts If a paiticular component rather than an arbitrary one is considered a specific letter is assigned to it 0 A is the molarity of A 0 173 Ais the mass fraction of A Similarly referring to a specific stream rather than any old stream you want each is given a different number 0 This the molar owrate in stream 1 nAjis the molar ow rate of component A in stream 1 Special subscripts If A is some value denoting a property of an arbitrary component stream the letter i signifies the arbitrary component and the letter 11 signifies an arbitrary stream ie Anis a property of stream 11 Note nnis the molar ow rate of stream 11 0 4 39iis a property of component i The subscript quotgenquot signifies generation of something inside the system The subscripts quotinquot and quotoutquot signify ows into and out of the system Q Embellishments If A is some value denoting a property then Andenotes the average property in stream 11 An denotes a total ow rate in steam n Aludenotes the ow rate of component i in stream 11 A indicates a data point in a set edit Units SectionDimensional Analysis In the units section the generic variables L t m s and A are used to demonstrate dimensional analysis In order to avoid confusing dimensions with units for example the unit In meters is a unit of length not mass if this notation is to be used use the unit equivalence character gather than a standard equal sign git Appendix 5 Further Reading Chapra S and Canale R 2002 Numerical Methods for Engineers 4th ed New York McGraw Hill Felder RM and Rousseau RW 2000 Elementary Principles of Chemical Processes 3rd ed New York John Wiley amp Sons Masterton W and Hurley C 2001 Chemistry Principles and Reactions 4th ed New York Harcourt Perry RH and Green D 1984 Perry s Chemical Engineers Handbook 6th ed New York McGraw Hill Windholz et al 1976 The Merck Index 9th ed New Jersey Merck General Chemistg For a more in depth analysis of general chemistry Matlab For more information on how to use MATLAB to solve problems Numerical Methods For more details on the rootfinding module and other fun math warning it39s written at a fairly advanced level git Appendix 6 External Links Data Tables Unit conversion table Wikipediag Enthalpies of Formation g ikipediaz Periodic Table Los Alamos National Laboratory Chemical Sciences Data Tables Has a fair amount of useful data including a fairly comprehensive List of Standard Entropies and Gibbs Energies at 250C also a list for ions a chart with molar masses of the elements acid equilibrium constants solubility products and electric potentials Definitely one to check out NIST properties You can look up properties of many common substances including water many light hydrocarbons and many gases Data available can include density enthalpy entropy Pitzer accentric factor surface tension Joule Thompson coefficients and several other variables depending on the substance and conditions selected To see the data in tabular form once you enter the temperature and pressure ranges you want click quotview tablequot and then select the property you want from the pull down menu It39ll tell you acceptable ranges edit Appendix 7 License Version 12 November 2002 Copyright C 200020012002 Free Software Foundation Inc 51 Franklin St Fifth Floor Boston MA 0211071301 USA Everyone is permitted to copy and distribute verbatim copies of this license document but changing it is not allowed 0 PREAMBLE The purpose of this License is to make a manual textbook or other functional and useful document quotfreequot in the sense of freedom to assure everyone the effective freedom to copy and redistribute it with or without modifying it either commercially or noncommercially Secondarily this License preserves for the author and publisher a way to get credit for their work while not being considered responsible for modifications made by others This License is a kind of quotcopylef quot which means that derivative works of the document must themselves be free in the same sense It complements the GNU General Public License which is a copyleft license designed for free software We have designed this License in order to use it for manuals for free software because free software needs free documentation a free program should come with manuals providing the same freedoms that the software does But this License is not limited to software manuals it can be used for any textual work regardless of subject matter or whether it is published as a printed book We recommend this License principally for works whose purpose is instruction or reference git 1 APPLICABILITY AND DEFINITIONS This License applies to any manual or other work in any medium that contains a notice placed by the copyright holder saying it can be distributed under the terms of this License Such a notice grants a world wide royalty free license unlimited in duration to use that work under the conditions stated herein The quotDocumentquot below refers to any such manual or work Any member of the public is a licensee and is addressed as quotyouquot You accept the license if you copy modify or distribute the work in a way requiring permission under copyright law A quotModified Versionquot of the Document means any work containing the Document or a portion of it either copied verbatim or with modifications andor translated into another language A quotSecondary Sectionquot is a named appendix or a front matter section of the Document that deals exclusively with the relationship of the publishers or authors of the Document to the Document39s overall subject or to related matters and contains nothing that could fall directly within that overall subject Thus if the Document is in part a textbook of mathematics a Secondary Section may not explain any mathematics The relationship could be a matter of historical connection with the subject or with related matters or of legal commercial philosophical ethical or political position regarding them The quotInvariant Sectionsquot are certain Secondary Sections whose titles are designated as being those of Invariant Sections in the notice that says that the Document is released under this License If a section does not fit the above definition of Secondary then it is not allowed to be designated as Invariant The Document may contain zero Invariant Sections If the Document does not identify any Invariant Sections then there are none The quotCover Textsquot are certain short 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either commercially or noncommercially provided that this License the copyright notices and the license notice saying this License applies to the Document are reproduced in all copies and that you add no other conditions whatsoever to those of this License You may not use technical measures to obstruct or control the reading or further copying of the copies you make or distribute However you may accept compensation in exchange for copies If you distribute a large enough number of copies you must also follow the conditions in section 3 You may also lend copies under the same conditions stated above and you may publicly display copies git 3 COPYING IN QUANTITY If you publish printed copies or copies in media that commonly have printed covers of the Document numbering more than 100 and the Document39s license notice requires Cover Texts you must enclose the copies in covers that carry clearly and legibly all these Cover Texts Front Cover Texts on the front cover and Back Cover Texts on 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permission to use the Modified Version under the terms of this License in the form shown in the Addendum below G Preserve in that license notice the full lists of Invariant Sections and required Cover Texts given in the Document39s license notice H Include an unaltered copy of this License 1 Preserve the section Entitled quotHistoryquot Preserve its Title and add to it an item stating at least the title year new authors and publisher of the Modified Version as given on the Title Page If there is no section Entitled quotHistoryquot in the Document create one stating the title year authors and publisher of the Document as given on its Title Page then add an item describing the Modified Version as stated in the previous sentence J Preserve the network location if any given in the Document for public access to a Transparent copy of the Document and likewise the network locations given in the Document for previous versions it was based on These may be placed in the quotHistoryquot section You may omit a network location for a work that was published at least four years before the Document itself or if the original publisher of the version it refers to gives permission K For any section Entitled quotAcknowledgementsquot or quotDedicationsquot Preserve the Title of the section and preserve in the section all the substance and tone of each of the contributor acknowledgements andor dedications given therein L Preserve all the Invariant Sections of the Document unaltered in their text and in their titles Section numbers or the equivalent are not considered part of the section titles M Delete any section Entitled quotEndorsementsquot Such a section may not be included in the Modified Version N Do not retitle any existing section to be Entitled quotEndorsementsquot or to con ict in title with any Invariant Section 0 Preserve any Warranty Disclaimers If the Modified Version includes new front matter sections or appendices that qualify as Secondary Sections and contain no 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distinguishing version number If the Document specifies that a particular numbered version of this License quotor any later versionquot applies to it you have the option of following the terms and conditions either of that specified version or of any later version that has been published not as a draft by the Free Software Foundation If the Document does not specify a version number of this License you may choose any version ever published not as a draft by the Free Software Foundation