Spring 2006 Process Dynamics, Operations, And Control 10 .

Spring 2006Process Dynamics, Operations, and Control10.450Lesson 4: Two Tanks in Series4.0context and directionIn Lesson 3 we performed a material balance on a mixing tank and deriveda first-order system model. We used that model to predict the open-loopprocess behavior and its closed-loop behavior, under feedback control. Inthis lesson, we complicate the process, and find that some additionalanalysis tools will be useful.DYNAMIC SYSTEM BEHAVIOR4.1math model of continuous blending tanksWe consider two tanks in series with single inlet and outlet streams.F, CAiF, CA1F, CA2volume V1volume V2Our component A mass balance is written over each tank.dV1C A1 FC Ai FC A1dtdV2 C A 2 FC A1 FC A 2dt(4.1-1)As in Lesson 3, we have recognized that each tank operates in overflow:the volume is constant, so that changes in the inlet flow are quicklyduplicated in the outlet flow. Hence all streams are written in terms of asingle volumetric flow F. Again, we will regard the flow as constant intime. Also, each tank is well mixed.Putting (4.1-1) into standard formdC A1 C A1 C AidtdC A 2τ2 C A 2 C A1dtτ1revised 2006 Mar 6(4.1-2)1

Spring 2006Process Dynamics, Operations, and Control10.450Lesson 4: Two Tanks in Serieswe identify two first-order dynamic systems coupled through thecomposition of the intermediate stream, CA1. If we view the tanks asseparate systems, we see that CA1 is the response variable of the first tankand the input to the second. If instead we view the pair of tanks as a singlesystem, CA1 becomes an intermediate variable. The speed of responsedepends on two time constants, which (as before) are equal to the ratio ofvolume for each tank and the common volumetric flow.We write (4.1-2) at a steady reference condition to findC A ,r C Ai ,r C A1,r C A 2,r(4.1-3)We subtract the reference condition from (4.1-2) and thus express thevariables in deviation form.dC 'A1τ1 C 'A1 C 'AidtdC 'A 2τ2 C 'A 2 C 'A1dt(4.1-4)4.2solving the coupled equations - a second-order systemAs usual, we will take the initial condition to be zero (response variablesat their reference conditions). We may solve (4.1-4) in two ways:Because the first equation contains only C′A1, we may integrate it directlyto find C′A1 as a function of the input C′Ai. This solution becomes theforcing function in the second equation, which may be integrated directlyto find C′A2. That istC'A1 1 t τ1 t τ1 'e e C Ai dtτ10(4.2-1)C 'A 2 tt1 t τ2 t τ 2 1 t τ1 t τ1 ' e e e e C Ai dt dtτ200 τ1 (4.2-2)On defining a specific disturbance C′Ai we can integrate (4.2-2) to asolution.Alternatively, we may eliminate the intermediate variable C′A1 betweenthe equations (4.1-4) and obtain a second-order equation for C′A2 as afunction of C′Ai. The steps are(1) differentiate the second equation(2) solve the first equation for the derivative of C′A1revised 2006 Mar 62

Spring 2006Process Dynamics, Operations, and Control10.450Lesson 4: Two Tanks in Series(3) solve the original second equation for C′A1(4) substitute in the equation of the first step.The result isτ1τ 2d 2 C 'A 2dC 'A 2() τ τ C 'A 2 C 'Ai12dt 2dt(4.2-3)Two mass storage elements led to two first-order equations, which havecombined to produce a single second-order equation. A homogeneoussolution to (4.2-3) can be found directly, but the particular solutiondepends on the nature of the disturbance:C'A 2 A1e tτ1 A 2e tτ2( ) C'A 2,part C 'Ai(4.2-4)where the constants A1 and A2 are found by invoking initial conditionsafter the particular solution is determined.4.3response of system to step disturbanceSuppose a step change ΔC occurs in the inlet concentration at time td.Either (4.2-2) or (4.2-4) yields (t t d ) (t t d ) τ1τ2τ1τ2 C'A2 U ( t t d ) ΔC 1 ee τ1 τ2 τ1 τ2 (4.3-1)Each tank contributes a first-order response based on its own timeconstant. However, these responses are weighted by factors that dependon both time constants.The result in Figure 4.3-1 looks somewhat different from the first-orderresponses we have seen. We have plotted the step response of a secondorder system with τ1 1 and τ1 1.5 in arbitrary units. At sufficientlylong time, the initial condition has no influence and the outletconcentration will become equal to the new inlet concentration; in thisrespect it looks like the first-order system response. However, the initialbehavior differs: the outlet concentration rises gradually instead ofabruptly. This S-shaped curve, often called “sigmoid”, is a feature ofsystems of order greater than one. Physically, we can understand this byrealizing that the change in inlet concentration must spread through twotanks, and it reaches the second tank only after being diluted in the first.revised 2006 Mar 63

Spring 2006Process Dynamics, Operations, and Control10.450Lesson 4: Two Tanks in SeriesC* Ai/ C10.50101234560123456C*A2/ C0.80.60.40.20tFigure 4.3-1: Response to step change in inlet composition4.4introducing the Laplace transformWe bother with the Laplace transform for two reasons: after the initial learning pains, it actually makes the math easier, so wewill use it in derivations some of the terminology in linear systems and process control is basedon formulating the equations with Laplace transforms.Definition: the Laplace transform turns a function of time y(t) into afunction of the complex variable s. Variable s has dimensions ofreciprocal time. All the information contained in the time-domainfunction is preserved in the Laplace domain. y(s) L{y( t )} y( t )e st dt(4.4-1)0(In these notes, we use the notation y(s) merely to indicate that y(t) hasbeen transformed; we do not mean that y(s) has the same functionaldependence on s that it does on t.)Functional transforms: textbooks (for example, Marlin, Sec. 4.2) usuallyinclude tables of transform pairs, so these derivations from definition(4.4-1) are primarily to demonstrate how the tables came to be.revised 2006 Mar 64

Spring 2006Process Dynamics, Operations, and Control10.450Lesson 4: Two Tanks in Series L{CU ( t t d )} CU ( t t d )e st dt0 C e st dt(4.4-2)td C stesCe st d s { td }L Ce at Ce at e st dt0 C (s a ) t e0s aC s a (4.4-3)Operational transforms: this allows us to transform entire equations, notjust particular functions. df ( t ) st df ( t ) L e dt dt 0 dt d sf ( t )e st (fe st ) dtdt 0 s f ( t )e st dt f ( t )e st0(4.4-4) 0 sL{f ( t )} f (0) t tL f (ξ )dξ f (ξ )dξ e st dt 0 0 0t 1 1 st f (t )e dt d f (ξ )dξ e st s 0 0s0 1 1 L{f (t )} f (ξ )dξ e sts 0s t (4.4-5)01 L{f (t )}srevised 2006 Mar 65

Spring 2006Process Dynamics, Operations, and Control10.450Lesson 4: Two Tanks in SeriesInverting transforms: use the tables to invert simple Laplace-domainfunctions to their time-domain equivalents. To simplify the polynomialfunctions often found in control engineering we may use partial fractionexpansion. The complicated ratio in (4.4-6) can be inverted if is expandedinto a series of simpler fractions.N (s)N (s)C1C2C3 . .n 1nnmD(s) (s α1 ) (s α 2 ) . (s α1 )(s α1 )(s α 2 ) m(4.4-6)In (4.4-6), α1 and α2 are repeated roots of the denominator. The inversetransform of each term will involve an exponential function of the root αi.f (s) C(s α) n f (t) Ct n 1e αt(n 1)!(4.4-7)Variety in the values of the coefficients Ci comes from the numeratorfunction N(s).how to write the expansionArrange the denominator so that the coefficient of each s is 1. If there areno repeated roots, each root appears in one term.C3C1C2N(s) (s α1 )(s α 2 )(s α 3 ) (s α1 ) (s α 2 ) (s α 3 )(4.4-8)If a root is repeated, it requires a term for each repetition.C3C1C2N(s) 22(s α1 ) (s α 2 )(s α 2 )(s α1 )(s α 2 )(4.4-9)Some roots may appear as complex conjugate pairs, so that, for exampleα1 a jbα 2 a jb(4.4-10)where j is the square root of -1.how to solve for the coefficients - it’s only algebra1) For each of the real, distinct roots, multiply the expansion by each RHdenominator and substitute the value of the root for s to isolate thecoefficient. This also works for the highest power of a repeated root.2) With some coefficients determined, it may be easiest to substitutearbitrary values for s to get equations in the unknown coefficients.revised 2006 Mar 66

Spring 2006Process Dynamics, Operations, and Control10.450Lesson 4: Two Tanks in Series3) For repeated roots, either(3a) multiply the expansion by s and take the limit as s .However, this will not isolate coefficients associated withrepeated complex roots.(3b) multiply the expansion by the RH denominator of highest power.Differentiate this equation with respect to s, and substitute thevalue of the root for s. Continue differentiating in this manner toisolate successive coefficients.4) For complex roots, solving for one coefficient is enough. The othercoefficient will be the complex conjugate.4.5solving linear ODEs with Laplace transformsWe return to (4.1-4), the two equations that describe concentration in thetanks.dC 'A1τ1 C 'A1 C 'AidtdC 'A 2τ2 C 'A 2 C 'A1dt(4.1-4)We perform the Laplace transform on the entire first equation. Itdistributes across addition, and constant τ1 may be factored out. dC ' L τ1 A1 C 'A1 L C 'Aidt dC ' τ1L A1 L C 'A1 L C 'Ai dt { }{ } { }(4.5-1)We next perform an operational transform on the derivative. Because thefunctional forms of the variables C′A1 and C′Ai are not yet known, wesimply indicate a variable in the Laplace domain.{}τ1 sC'A1 (s) C'A1 (0) C'A1 (s) C'Ai (s)τ1sC'A1 (s) C'A1 (s) C'Ai (s)(4.5-2)We can easily solve (4.5-2) for C′A1. If we similarly treat the secondequation in (4.1-4), we arrive at the equivalent formulation in the Laplacedomain.revised 2006 Mar 67

Spring 2006Process Dynamics, Operations, and Control10.450Lesson 4: Two Tanks in SeriesC 'A1 (s) 1C 'Ai (s)τ1s 11C (s) C 'A1 (s)τ 2s 1(4.5-3)'A2Equations (4.5-3) are not solutions - we have not solved anything! Wemerely have a new formulation of problem (4.1-4), a formulation that ismore abstract (what on earth is Laplace domain?) and yet simpler, byvirtue of being algebraic. It is important to remember that all theinformation held in the differential equations (4.1-4) is preserved in theLaplace domain formulation (4.5-3).We proceed toward solution by eliminating the intermediate variable in(4.5-3). We findC 'A 2 (s) 11C 'Ai (s)τ 2s 1 τ1s 1(4.5-4)With (4.5-4) we have gone as far as we can without knowing more aboutthe disturbance. That is, we cannot invert the right-hand side of (4.5-4)until we can actually substitute a functional transform for the variableC′Ai(s). In this sense, (4.5-4) resembles (4.2-2) and (4.2-4): a solutionneeding more specification.As in Section 4.3, suppose a step change ΔC occurs in the inletconcentration at time td.C 'Ai ( t ) U (t t d )ΔC(4.5-5)We must take the Laplace transform,C 'Ai (s) ΔC t d ses(4.5-6)which we may substitute into (4.5-4).C 'A 2 (s) 11 ΔC t dseτ 2s 1 τ1s 1 s(4.5-7)This IS the solution, the step response of the two tanks in series. Ofcourse, it really must be inverted to the time domain. We treat thepolynomial denominator from either the tables or partial fractionexpansion:revised 2006 Mar 68