ENVIRONMENTAL POLLUTION ESGN 503
Colorado School of Mines
Popular in Course
PHYS 201 - A
verified elite notetaker
MTED 5322, Pedagogical Content Knowledge in Teach Math
verified elite notetaker
verified elite notetaker
Popular in Environmental Science & Engineering
This 15 page Class Notes was uploaded by Rubie Mertz on Monday October 5, 2015. The Class Notes belongs to ESGN 503 at Colorado School of Mines taught by Staff in Fall. Since its upload, it has received 62 views. For similar materials see /class/219637/esgn-503-colorado-school-of-mines in Environmental Science & Engineering at Colorado School of Mines.
Reviews for ENVIRONMENTAL POLLUTION
Report this Material
What is Karma?
Karma is the currency of StudySoup.
Date Created: 10/05/15
Review of a Chemical Reaction Problem Calculus amp ODES Developing Analysis Skills by Eric Vestal The objective of this monograph is to present and develop a method of problem solving and analysis The detail presented for this single problem is a result of writing down in detail what should become a matter of thought in your problem solving Of course it is speci c to this problem but the overall approach can be generally applied to rate expressions problems and ordinary differential equations Problem Statement Reaction 2A 9 C 1 Kinetics 3 14 2 kAC 2 dt Reaction Rate Constant k 1 m3molday Initial Conditions A 100 moim3 Co 10 molm3 and to O a Find general expressions for At amp Ct b Graph At and Ct versus t What does the kinetic expression Equation 2 tell us Inspection of the kinetic equation shows that the rate at which the concentration of A changes is dependent upon and proportional to the concentration of A and the concentration of C Hence it is a nonlinear relationship If for example the kinetic expression was dA kA 3 d l which we ve seen before then we would call it linear even though the solution At Age 4 is non linear ie not a line We say 3 is linear because the rate of change of A is linear in the variable A If on the other hand the righthand side was kA2 then this would be a non linear ODE because the rate of change dAdt is not linear in A So for the rate of change given by kAC in equation 2 the expression is nonlinear This is important because it means that we MUST either nd an explicit relationship between A amp C so that we can integrate or we must solve two equations simultaneously In either case we must use the Law of Mass Action to understand how the concentrations of A and C are related Step I MassAction Expression Via the Law of MassAction we need to express the relationship between the concentrations C and A using stoichiometry Equation 1 We note that for every two moles of A reacted we produce 1 mole of C the molezmole relationships is 21 We also note that the equation is irreversible Now we can ask the simple but important questions 0 What are the initial concentrations of A and C Ada A0 5 C10 C 6 This is always the case and cannot be violated our solution must satisfy these Is A increasing or decreasing or both Is it bounded A is the reactant so it could be decreasing Since it is not also a product then it cannot be increasing So we can say absolutely that A decreases with time theoretically there is nothing preventing it from remaining constant Assuming the reaction goes to completion all of A will be consumed And of course concentrations are never negative So we know the following A02 5 A07 for t gt t1 7 limAt 0 or ASS zAmrO 8 At 2 o 9 Is C increasing or decreasing or both Is it bounded C is the product so it could be increasing Since it is not also a reactant then it cannot be decreasing So we can say absolutely that C increases with time C017 2 C11 for t gt t1 10 What do we know about the nal amount of C assuming the reaction consumes all of A C is originally C0 and it is increasing or not decreasing so Ct 2 Co And for every two moles of A consumed 1 mole of C is produced So if all the original A A0 is reacted then Ao2 moles of C are produced So Ct is bounded by the amount of Vestal Review 2 A that is initially available and the reactantproduct relationship between them Hence we write limHmCtCoAo2 0r CsssCmCoAo2 11 0 What is the general MassAction expression All of our analysis above gives us the starting and end points of the relationship so it should be easy to write the overall expression Amount of C Original Amount of C C Produced by Reacting A 12 Which we write as Ct C0 C1 Amount of A reacted Ar Initia1Current A0 A 13 C Produced by reacting A C1 Ar 2 14 C C0 Arl2 C0 Ao A 2 or 15a Arle A0 AC Co 2 15b 0 Does this satisfy our understanding of Cto and Ct From 15 CtO C0 A0 A0 2 O matches Eqn 6 From 15 amp 8 Ct o C0 A Ass 2 C0 AoZ x matches Eqn 11 What is the resulting kinetic expression for A 94 kAC kAC0 L A A or dt 2 511 m22t A 16a dt where A C0 A0 2 which is C53 and K k2 16b Step II Integrating the kinetic expression to solve the ODE Integration of equation 16 is not difficult However we must make a few decisions before we begin First are we going to use integration tables or some other method such as partial fraction decomposition And should we use a de nite or inde nite integral No matter the method we will arrive at the same answer so use the method that is most comfortable to you We present them all below to show how they are related In any case we use separation of variables to arrive at the following integral expression Vestal Review dA fm del 17 I De nite Integral Partial Fraction Decomposition this would be my choice For a de nite integral we use the initial condition and some future state to set the limits of integration Ato0 A0 and At A so we have from 17 A l mhffd 8 As shown in Equation A10 from Appendix A we can write the lefthand side of 18 is as follows using the limits of integration A dA 1 Al quot ln 19 AZA A 2A lzit A A The righthand side of Equation 18 is easy to integrate so we arrive at 1 MI quot lenlyhAl Lb 211 What do we do with those annoying absolute value requirements Normally we pretend they are not there which is a potentially dangerous mistake We already know that A20 so we can safely ignore the absolute value in the numerator The denominator is similar I2A A gt 27L 2 A or from 16b we require 2C0 A0 2 A 22 Since A is decreasing A s Aquot and Ca 2 0 then 2A2C0A2A02A 23 So it turns out we can ignore the absolute value requirements and we proceed from 21 A A A 2A A In M A L ln 2L A A0 2th 24 and by standard manipulation with the transcendental functions In amp e Vestal Review 4 A 4 Exp 22m t zit A zit A0 25 We will stop here as we have basically arrived at the solution We only need to rearrange for an explicit expression in A We will do this at the very last Let s explore other methods we may have used to integrate II De nite Integral Integration Table Formula For the integral on the le hand side of Equation 17 we nd the following applicable form in the integral tables Ii 7iqlni Zil where Xax2 bxc and q4ac b2 lt0 Applying this to f from Equation 17 we see that a1b2 c0 q4 2 l q2tso dA 11n 2A2t 2t 11n A fA2L A 2 2A2L 22L A u WAIT What s going on Equations 19 and 27 give different answers Is this possible 1 A 1 19 dA 22 11in A Elmo 1140 39 A 1ln A Equation 27 2A A 2ti Let s check Recall the basic de nition of Calculus yffdx a J f Vestal Review 26 26b 27 28 So differentiating the solutions in Equations 19 and 27 should give the integrand 11 l A n12A A 2A AA 1 am nLZA Al2t A i2iA2 A2A V 29a 3 l A h1A2A A za A g 1 dxiZA nLA 2All2M A Apt22f A2A f 2913 OK we can show that both solutions are in fact correct by the Fundamental Theorem of Calculus So what s the difference The constant of integration This will become clearer The moral here is twofold 1 An integral may have different solutions and 2 It is possible to check if a solution is accurate by differentiating it This is the process of analysis Now continuing from Equations 27 and 18 1 r A 1 A a 2AlnlA2AJ Lo KI 30 or A A AazA Aa 2AExp 2111 31 Equation 31 has the same form as Equation 25 and we see that they are in fact exactly the same if we multiply 31 through by 1 we arrive at 25 Again we will stop here to further investigate methods of integration III Inde nite Integral The alternative to using de nite integrals as shown in the two examples above is to use the inde nite integral and the constant of integration We have two forms as shown in Expression 28 Solving Equation 17 using the Partial Fraction form of Equation A10 dA 1 A dt gt 1 32 fAZA A KI 2A n2A A quott 01 1 where i1 is the constant of integration Vestal Review 6 Rearranging and playing our usual game with the constant 01 Exp 2216 A ail A olExp 2 Kt 33 Now how do we nd a Well we know from Equation 5 thatA0 A0 so let s try substituting the initial condition into equation 33 i 0 1 34 2A Ao Substituting our expression for a 34 into Equation 33 we arrive at the explicit solution A L E ZA 35 zit A zit A M K Note that Equations 35 and 25 are identical Solving Equation 17 using the Table Form as given in Equation 27 dA 1 A A dt 9 l t 36 Lam 4 quotf ZARA 2 02 1 0139 A m ozExp 2111 Using the initial condition to solve for 02 70 02 such that 38 A A0 Ex 2 t AZA Ao ZA p K 39 And we note that Equation 39 is exactly Equation 31 As we have already shown 31 and 25 are equivalent and 25 and 3 5 are identical so all four solutions give the same answer as they should Now back to the issue of why the two integral forms are different as shown in Expression 28 Note the relationship between 01 and 02 a 01 It is only through the initial Vestal Review condition or the limits of integration that the solutions become identical without them we have a family of solutions Finally let s write an explicit expression for A by manipulating 35 First note that from 16b 2 m 2Co A0 and 2K k for the exponential argument Thus A 22V A 2quot A Exp 21Kt a 22 AExp Akt Azc0 AExp Zkt ZMOExpAkt A 2ampEx Akt 260 AOExpAkt and substituting for 2 gives 260 A AoExpt M A 39 2C0 AoExIX Ak 40 Which is our nal answer to the problem given in 16 What about Ct From Equation 15 A Am so 2 Knaco HW1 41 2 2c0 AOExp Akt j Ct a Co Equations 40 and 41 completely describe the chemical reaction given in the Problem Statement in terms of the starting concentrations and the reaction rate k Let s check our answers with our basic observations expressed in Equations 5 6 8 9 and 11 This is an important analysis step Do your equations match the expected behavior andor observations 2C AA Ata0 9 h 5 2C0A0 A0 Vmatc es At a co 32 0 matches 8 Vestal Review 8 If A0 and Co are 2 O which they must be then At 2 O matches 9 2C0Aoii CtOQ 2 l1 2QA0 J Co f matches6 2 Ct gt 0 C0 43 W1 C0 33 xmatches 11 2 2g A0 o j 2 Step III Graphical Analysis Graphing Equations 40 and 41 using the explicit initial conditions and reaction rate constant given in the problem statement is another important step in the analysis process Figure 1 Reaction 2Agt c Kinetics dAIdt kAC k 1 mslmolday A1 M A Concanttationmm camwhmmqmoo Time days Do the curves in Figure l behave as we would expect 2 At tends to 0 Good 39 Ct reaches a steadystate value of C9A 92 15 6 moin13 Is At decreasing Yes 0 Is Ct increasing Yes Do the curves satisfy the Law of Mass Action If so then for every 2 moles of A reacted 1 mole of C will be produced or AgA CCo 2 Equation 15b See Figure 2 it demonstrates that this is in fact satis ed everywhere Vestal Review Figure 2 Law of Mass Action ConcentrationChanges n A A O iNWhU39IOKImCOOA Time days Finally what do the reaction rates look like 2 39 dAdt kAC dCdt 05dAdt 05kAC Four relationships are graphed in Figure 3 as follows kAC amp AAAt These should overlay each other 05kAC amp ACAt These should overlay each other Figure 3 Rates Change inAamp c dAIdtkAc 8 dCIdtkACI2 9 E i Ev 5 t 3 a 000 025 050 075 100 125 150 Time days Vestal Review From Figure 3 we see that the rate of change of A is twice as fast as that of C Also we can see that C is a product as its rate of change is always positive while A s is negative and hence represents consumption Further we note that the reactions pick up speed as C is initially produced and then slow down as A is consumed and less reactant is available This shows the interdependence of A amp C If the rate was only a function of A or C then the concavity would not change in the curves of Figure 3 Step IV Thinking Ahead How would you answer the questions How long does it take Ct to reach 95 of its steadystate value Express your answer a graphically and b mathematically At what time does the concentration of C equal that of A Express your answer a graphically and b mathematically At what concentration of A is the rateofchange in A a maximum Express your answer a graphically and b mathematically Does the maximum rateofchange in A occur at the same time as the maximum rateofchange in C Vestal Review 11 Appendix Partial Fraction Decomposition Given the integral as in Equation 18 dx fx2t x A1 We will integrate it using partial fraction decomposition We want to simplify the algebraic expression which is a ratio of polynomials into easily integrable pieces 1 Px A2 x2 x Qx Since we have one polynomial P divided by another Q we can break the expression into a sum of simpler quotients where the denominator is a polynomial of the 1St degree The only requirement is that the degree of P be less than the degree of Q In A2 the degree of P is 0 ie A 1 and the degree of Q is 2 ie A2 so we can proceed When P and Q have the same degree we have to take some initial steps that are not shown below Because the order of Q is 2 there will be two pieces to the sum You ll see First we write the terms sum of unknown constants over a factor of Q 1 l x2tx x ZRx where a and b are constants A3 Multiplying through by Q 1 a2t x bx A4 Now choose values of x that eliminate one of the unknowns as follows x0 1 a2 9 a 12A A5a x 2 1 b2 b 12k ASb So from A3 and A5 we have 111ll11 A6 xm x 2M 2A2 x 2x 2A xJ Vestal Review 12 du These two pieces are each easy to mtegrate usmg the form f lnlul u Substituting Equation A6 into A1 gives 1 ii 1 A7 21fx2A xdx 1 or as two integrals 1f a dx 1 2f x f2t x A8 When integrated A8 is 1 Jam 14214 A9 or simply dx 1 XI fx2 x 571 3921 xl A10 Note that in the integration of A8 we have left out a constant of integration in either A9 or A10 This is commonly done when writing an integral form as it may be used with integration limits or as an inde nite integral so this is implied in the form Vestal Review 13 Integrating Factor A single energy or material balance on a wellmixed system results in a rst order ordinary differential equation Since this equation is o en linearized in dynamic analysis a linear rstorder di erentiai equation resuits The di erentiai equation is useful because it provides analytical relationships between the process equipment and operating parameters and key dynamic parameters such as time constants and gains Often an analytical solution is desired for the openloop system output in response to one or more relatively simple input forcing functions The integrating factor can be used to evaluate the analytical solution The general iinearized modei will be of the form dY 31 250 of 1 d The functions atbt and ct are known functions of time t When the mction at 2 0 during the time considered in the solution the equation I can be rearranged to give dY gtmeg0gt 2 where g0 is the forcing function This ordinary di 39erentiai equation is linear and rst order but not separable However it can be modi ed to be separable and directly solvable by multiplying by a term called the integrating factor The integrating factor is de ned as IFexpltiftodo 3 Now the standard equation 2 is multiplied by the integrating factor to give dY explt Mode N car new gm exp Much lt4 The leftuhand side of equation 4 can be recognized to be the expansion of the derivative of a product d d explt Ifltdcfltrexp much goon Mode 5 This can be substituted to yield a separable di erential equation r exp fmdt 7 where I is a constant of integration to be evaluated Earn the initial conditions This method is successful when the integral in Equation 7 can be evaluated analytically which is possible for some simple functions g0 such as impulse step and sine which are useful in understanding the dynamic behavior of process systems Reference quotProcess Control Designing Processes and Control Systems for Dynamic Performance Thomas E Marlin McGrawHill 1995 pg 905