Process Dynamics Operation and Control (MIT Course) Lessons

8/12/2019 Process Dynamics Operation and Control (MIT Course) Lessons

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10.450 Process Dynamics, Operations, and ControlLecture Notes - 1

Lesson 1. Design of a surge tank to smooth out fluctuations in flow. Definition ofimportant process control terms.

1.0 ContextMuch of the chemical engineering curriculum concerns continuous

processes operating at steady state. Well and good, but there's more to it:continuous processes may be disturbed in a variety of ways, and theeffects propagate through the process as a function of time throughoutthe process, temperature, pressure, flow, and composition may rise or fall.Process Control is about managing disturbances, for product quality, foreconomics, for safety. We begin with a simple example:

1.1 Surge tankEnvision two continuous processes operating in series. Process 1 feeds astream w i to process 2.

from other processes

process 1 process 2w i

to other processes

Stream w i has some steady design flowrate, but in practice it varies,causing problems in process 2. We might attack the problem by reducingthe cause of variation in process 1; we might also attempt to mitigate theeffect of the variation on process 2. Propagation of disturbances between

processes is a common problem, and a common solution is the surge tank ;its job is to damp out changes in w i from the upstream process and thusdeliver a steadier w o to the downstream process.

from process 1w i

h

to process 2

wo

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This tank has a free inlet and a pumped outlet; intuitively it seems possiblethat the tank may overflow or run dry. We can confirm this soberingthought by applying our tank model (1.2.2) to a persistent imbalance

between w i and w o. Suppose the simple case of

wi wo = C (1.2.4)

Substituting (1.2.4) into (1.2.2), we find

C h(t ) = ho +

At (1.2.5)

Therefore, we should protect our tank with some automatic processcontrol something that will measure the level and take corrective actionshould the level become too high or low.

1.3 Definitions to get us started process : the equipment within some boundary, along with the streams of

matter and energy that cross that boundary -- what we usuallymean when we think of 'chemical process'. In this example, it's thetank, pump, piping, and fluid.

disturbance : a change imposed on the process. In this example, the inputstream w i varies with time.

controlled variable : some feature of the process that we would like tocontrol. It may be a stream crossing the boundary or somequantity within. We want to control it because the disturbancemakes it change with time, in a way that we don't like. In thisexample, the controlled variable is the liquid level.

set point : the desired value of the controlled variable. In this example, wehave no set point, but we do want to confine the controlled variable

between high and low limits.manipulated variable : some feature of the process that we adjust so that

we can exert influence on the controlled variable. In mostchemical processes, the manipulated variable will be the flowrateof a stream. In this example, it seems reasonable to manipulate theoutlet flow.

final control element : a device that adjusts the manipulated variable. If themanipulated variable is usually a stream, the final control elementis usually a valve, referred to as a control valve .

measured variable : most often, synonymous with the controlled variable we measure it so that we can tell how well our control scheme isworking. Of course, we may also measure the manipulatedvariable and other variables, as well.

sensor : a measuring instrument. For chemical processes, the mostcommon measurements are of flow (F), temperature (T), pressure

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(P), level (L), and composition (A, for analyzer). The sensor willdetect the value of the measured variable as a function of time.

controller : the device that detects the output of the sensor, decides howseriously the controlled variable deviates from the set point, anddirects the final control element in response. The controller

performs calculations based on its control algorithm .transducer : this may be more than you wanted to know. The controllermust be able to communicate with sensor and final element.Transducers convert and transmit signals to make this possible.

There are more details, of course, but they can wait. We install a levelsensor on the tank, put a control valve at the pump discharge, and connectthe two with a controller. Notice the symbols: the circle containing Lrepresents the sensor, and LC represents the controller. The control valvehas a mushroom on it for reasons we'll cover later. In the schematic, thesensor communicates with the process by a solid line, and with the

controller and valve by a dashed line. We call this control structure feedback control - the value of the controlled variable is fed back to acontroller, which adjusts the manipulated variable in response.

w i

final control element(control valve)

wo

hL

L = level sensorC = calculation or controller

LC

When the level sensor indicates approach to high or low limits, thecontroller computes a response by its algorithm and directs the controlvalve to open or close appropriately. The outlet flow w o may not beconstant, as we wanted, but by suitable choice of tank size and controlalgorithm we can significantly reduce its variability, and hence the effectson downstream processes.

1.4 Thats process control?Enough for now. We've modeled a simple process, defined our terms, andsketched out a control scheme. Before we attempt to specify more about acontroller, however, we must learn more about the ways in which

processes might be disturbed.

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10.450 Process Dynamics, Operations, and ControlLecture Notes 2

Chapter 2. Dynamic system

2.0 ContextIn this chapter, we define the term 'system' and how it relates to 'process'and 'control'. We will also show how a simple dynamic system responds

to several disturbances.

2.1 SystemIn Chapter 1, we introduced a process - a surge tank with pumped outlet -that was subject to disturbances in time. We thought of the process as acollection of equipment and other material, marked off by a boundary inspace, communicating with its environment by energy and materialstreams.

w i

h

wo

'Process' is a good notion; another useful notion is that of 'system'. A

system is some collection of equipment and operations, usually with a boundary, communicating with its environment by a set of inputs andoutputs. By these definitions, a process is a type of system, but system ismore abstract and general. For example, the system boundary is oftentenuous: suppose that our system comprises the equipment in the plant andthe controller in the central control room, with radio communication

between the two. A physical boundary would be in two pieces, at least; perhaps we should regard this system boundary as partly physical (aroundthe chemical process) and partly conceptual (around the controller).

Furthermore, the inputs and outputs of a system need not be material and

energy streams, as they are for a process. System inputs are "things thatcause" and outputs are "things that respond".

inputs system outputs

(causes) (responses)

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To approach the problem of controlling our surge tank process, lets thinkof it in system terms: the primary output is the liquid level h -- not astream, certainly, but an important response variable of the system.Disturbances are of course inputs, and so the stream w i is an input. And

peculiar as it first seems, the outlet flow w o is also an input, because it

influences the liquid level, just as does w i.

The point of all this is to look at a single schematic and know how to viewit as a process, and as a system. View it as a process (w o as an output) towrite the material balance and make fluid mechanics calculations. View itas a system (w o as an input) to analyze the dynamic behavior implied bythat material balance and make control calculations.

2.2 Systems within systemsWe call something a system and identify its inputs and outputs as a firststep toward understanding, predicting, and influencing its behavior. We

recognize that our understanding may improve if we determine some ofthe structure within the system boundaries; that is, if we identify somecomponent systems . Each of these, of course, would have inputs andoutputs, too.

inputs

system

12

outputs

Considering the relationship of these component systems, we recognizethe existence of intermediate variables within a system. Neither inputsnor outputs of the main system, they connect the component systems.Intermediate variables may be useful in understanding and influencingoverall system behavior.

When we add a controller to a process, we create a single time-varyingsystem; however, it is useful to keep process and controller conceptuallydistinct as component systems. This is because relatively few controlschemes (relationships between process and controller) suffice for myriad

process applications. Using the terms we defined in Chapter 1, werepresent a control scheme called single-loop feedback control in thisfashion:

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system

other inputs process

final element

sensor

controller

manipulated variable (intermediate processvariables)

other outputs

set point controlledvariable

Inside the block called "process" is the physical process, whatever it might be, and the block is the boundary we would draw if we were doing anoverall material or energy balance. HOWEVER, we remember that theinputs and outputs for the block are NOT necessarily the same as thematerial and energy streams that cross the process boundary. From amongthe outputs, we may select a controlled variable (V C), and provide asuitable sensor to measure it. From the inputs, we choose a manipulatedvariable (V M) and install an appropriate final control element. Themeasurement is fed to the controller, which decides how to adjust V M tokeep V C at the set point. Other inputs are disturbances that affect V C, andso require action by the controller.

We keep in mind this feedback control scheme, and how it relates thecontroller to the process, when we represent the equipment in schematicform, as with the surge tank of Chapter 1.

w i

final control element(control valve)

wo

hL

L = level sensorC = calculation or controller

LC

We'll have much more to say about feedback control later. For now, it'simportant to think of a chemical process as a dynamic system thatresponds in particular ways to its inputs. We attach other dynamicsystems (sensor, controller, etc.) to that process in single-loop feedbackscheme and arrive at a new dynamic system that responds in different

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ways to the inputs. If we do our job well, it responds in better ways, so to justify all the trouble.

To do our job well, we must understand more about system dynamics --how systems behave in time. That is, we must be able to describe how

important output variables react to arbitrary disturbances.

2.3 Dynamics of a tank, without any controlFrom Chapter 1, our process model was

dh A

dt= wi wo h(0) = ho

1 t(2.3.1)

h(t ) = ho + A (wi (t ) wo (t ))dt

0

Now mindful of our system concepts, we recognize h(t) as the output andw i(t)-w o(t) as the input. Indeed the flow rates are separate inputs, but ourmodel of the process indicates that they always influence the output liquidlevel by their difference. For convenience, let us represent this differenceas x(t). Our model (2.3.1) captures the system dynamics; it tells us howthe output h(t) responds in time to input disturbances x(t). We nowintegrate (2.3.1) for several specific cases of x(t).

2.4 Response to rectangular pulse at time t dLet the tank be operating at steady state, so that the flows are initially

balanced, and x is zero. Suppose that at time t d, extra liquid is injected

into the feed stream: mass M is added over time interval t before the inletflow returns to normal. We can idealize this as a rectangular pulse.

x(t ) = 0, 0 t < t d M t

, t d t t d + t (2.4.1)

0, t d + t < t

Inserting the disturbance (2.4.1) into process model (2.3.1), we computethe response.

h(t ) = ho , 0 t < t d M

ho + At(t t d ), t d t t d + t (2.4.2)

M ho + A

, t d + t < t

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0

1

2

3

0 3

t/t d

n o r m a

l i z e

d i n p u

t a n

d o u

t p u t

input

output

w x

( wt

Ahh

d

o

2 1 4

)

Figure 2-3 . Of course, the model ceases to be applicable when the liquid level reachesthe top of the tank.

2.7 SineAt time t d, the inlet flow begins to oscillate with radian frequency .

x(t ) = wsin( t t d )w

h(t ) = ho + A (1 cos( t t d ))

(2.7.1)

The liquid level oscillates at the frequency of the disturbance. Its responseis delayed, in that the level reaches its peak some time after the inlet flowhas peaked. Notice that the amplitude of oscillation decreases as thefrequency increases. This indicates that the tank cannot follow fastchanges.

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-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

0 4 6 8 10 12 14

t

n o r m a

l i z e

d i n p u

t a n

d o u

t p u t

inlet flow

liquid level

w

x

(w

Ahh o

2

)

Figure 2-4 . The onset of disturbance td has arbitrarily been set to 1.

More detailResponse to a sine disturbance has two parts the initial transient,and a recurring oscillation. We can recast the response in this form

by using the sum-of-angles formula to write the cosine as a sinethat includes a phase angle.

w w h(t ) = ho + A

+ A

sin (t t d ) 2

(2.7.2)

Equation (2.7.2) shows that the output lags the input by /2radians, or 90 . The liquid level differs at most from its initialvalue by twice the amplitude. It either exceeds or stays below the

initial level according to the sign of w; that is, whether the flowinitially increased or decreased.

2.8 Typical disturbancesKnowing how a system responds to disturbances is a prerequisite forcontroller design. We will use the impulse, step, and sine disturbancesrepeatedly to test various dynamic systems. While we can never test ourcontrol designs against every conceivable disturbance, testing against

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these standard ideal disturbances will usually tell us what we need toknow.

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Lesson 3. Math review.

3.0 ContextIn the previous chapters, we solved a differential equation for differentforcing functions. Here we will review this and other mathematical topics

that we will need.

3.1 Quadratic equationThe roots of

s 2 + s +1 = 0 (3.1.1)

are

s = 42

2 (3.1.2)

If is less than zero, the roots s will be real, of opposite sign, and ofunequal magnitude. For greater than zero, the roots may be real orcomplex; the real parts will have the same sign. The term under theradical, aptly called the discriminant, determines whether the roots arecomplex.

3.2 complex numbersConsider the complex number

z = a + jb, where j =

1

, Re(z) = a, Im(z) = b (3.2.1)

The same number may be written in polar form

z = z e j , where 22 ba z + = and = tan 1 b (3.2.2)a

The diagram shows the complex plane, in which real numbers areconfined to the horizontal axis. A complex number appears as a vectorfrom the origin. The diagram relates the Cartesian and polar forms of thecomplex number. The phase angle is measured counterclockwise fromthe positive real axis.

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math function that doesn't care what our time units are. View = 2 /p =2f as the conversion factor between time units and radians.

The phase angle represents an advance in the signal y(t) with respect tosome other signal. That is, if

x(t ) = B sin ( ) and y(t ) = C sin ( t + ) (3.3.2) t

the oscillation y(t) is ahead of that of x(t) at any time t. However, we willmost often encounter phase lags , so that the phase angle will have anegative value. If = 0, x(t) and y(t) are said to be in phase .

Representing an oscillation with a phase angle is quite useful, but onoccasion it is helpful express the same signal in another fashion. This

phase angle identity, derived from the trigonometric sum-of-anglesformula, shows that the signal can be expressed as a combination of sineand cosine functions:

CS 22 + sin t + tan 1 C = S sin t + C cos t (3.3.3) S

3.4 ArctangentThe arctangent is a multi-valued function, and thus must be treatedcarefully in calculations. For example, suppose we wish to express thecomplex number -1-j in polar form. From the figure we see that the phaseangle should be designated as either 225 (3.927 radians) or -135 (-2.356 radians).

real

imaginary

However, most calculators and spreadsheets will process (3.2.2) to give45 (0.785 radians).

calculator = tan 1 1 = tan 1(1) = 45 o (3.4.1) 1

Calculators and spreadsheets tend to work between -90 and +90 .Because we will be considering phase lags, we will instead tend to apply

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the arctangent between -180 and 0 . Hence angles in the upper rightquadrant should be corrected.

o o

= calculator 180 , 0 calculator 90 (3.4.2)

Arctangent Function

-5

-4

-3

-2

-1

0

1

2

-30 -20 -10 0 10 20 30

argument

a r c t a n

( r a

d i a n s

)

as calculated

180 deg offset

By the way, the conversion factor between degrees and radians is

o

1801 =

radians(3.4.3)

3.5 First-order, linear, variable-coefficient ODEWe are addressing systems that vary in time, so our independent variableis always t.

a (t )dy + y(t ) = Kx(t ) y(t 0 ) = known (3.5.1)dt

In writing (3.5.1) we have arranged the coefficient functions to isolate thedependent variable y(t). By this means, a(t) must have dimensions oftime t, and K has dimensions of y/x. We solve this equation by definingthe integrating factor p(t)

dt p(t ) = exp a (t ) (3.5.2)

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Kx(t )dt dy =

a (t )t (3.8.1)

y = y(0) + K x(t ) dt0 a (t )

Separable equations are convenient to solve by straightforward functionintegration. The surge tank of Chapters 1 and 2 was described by aseparable equation of form (3.8.1).

3.9 First-order ODE, delayed disturbanceLet the forcing function be delayed; suppose x(t) is a unit step at time t d >0. Then from (3.5.3)

Kt d p(t )(0)

dt + t p(t )(1)

dt + p(0) y(0) y(t ) =

p(t ) 0

a (t )t d

a (t )

p(t ) (3.9.1)

t

y(t ) = p K(t ) t

d

a p

((t t))

dt + p(0 p

)( yt )

(0)

3.10 Second-order, linear, constant-coefficient ODE

d 2 y + dy + y(t ) = Kx(t ) y(0), dy = known (3.10.1)dt 2 dt dt 0

The coefficient a 2 has the dimension of time squared. The solution to(3.10.1) is the sum of two terms:

y(t ) = yh (t ) + y p (t ) (3.10.2)

The homogeneous solution y h, which depends only on the left-hand-sideof (3.10.1), is itself the sum of two linearly independent exponentialfunctions

yh (t ) = C 1e r 1t + C 2e r 2t (3.10.3)

where r 1 and r 2 are the roots of the characteristic equation of (3.10.1).

r 1,2 = 42

2 (3.10.4)

The value of the discriminant determines three distinct forms of thesolution.

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real, unequal roots for r

2t

242 t

242 t

yh (t ) = e C 1e + C 2e (3.10.5)

In process control, we prefer stable systems, those in which disturbancesdo not grow with time. We observe that (3.10.5) decays if and havethe same sign.

real, equal roots for r

yh (t ) = et

2

(C 1 + C 2t ) (3.10.6)

The solution will decay if and have the same sign.

complex roots for r t

2

+

Ct 4sin422

yh (t ) = e C 1 cos 2 2 2 t (3.10.7)

Once again, the solution will decay if and have the same sign. Thecoefficient of t in the trigonometric functions is the radian frequency of theoscillation.

The particular solution y p for any disturbance x(t) may be determined bythe 'method of undetermined coefficients', or the 'method of variation of

parameters'. The initial conditions are then applied to the solution y(t) todetermine coefficients C 1 and C 2.

The response of the system (3.10.1) then depends on the character of the system itself (through the left-hand-side coefficients, affecting the

exponential and trigonometric terms in the homogeneous solution) the initial conditions (affecting coefficients C 1 and C 2 in the homogeneous solution) the nature of the disturbance (through the particular solution, as well as C 1 and C 2, if the

disturbance is initially non-zero)

In a later lesson, we will introduce Laplace transforms as an alternativemethod for determining the solution.

3.11 Representing functions by Taylor seriesWe specify some reference value of the independent variable, andrepresent the function in the neighborhood of that reference as a series ofterms. For a function of one variable:

f ( x) = f ( x s ) + df ( x x s ) + O(( x x s )2 ) (3.11.1)dx x s

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For a function of more than one variable:

f ( x, y,...) = f ( x s , y s ,...) + f ( x x s ) x x s , y s ,...

(3.11.2)+ f ( y y s ) + ... + O(( x x s )2 , ( y y s )2 ,...) y

x s , y s ,...

By retaining only linear terms, we obtain a linear approximation. Thederivatives are evaluated at the reference point. Of course, theapproximation is exact at the reference, and it is often satisfactory in someregion about the reference value. As the figure indicates, however,extrapolation to x = 0 would be erroneous.

3.12 Chain rule for differentiation

dds g ( f ( s)) = dg df (3.12.1)df ds

The functions g and f are said to be nested. For example, let g be theexponential and f the square root.

Taylor Series Linearization about x = 2

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

0 0.5 1 1.5 2 2.5 3

x

ynonlinear

linear

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d e (= e s s ) 1 s 21

ds 2

The chain rule applies to the special cases of a product

d dfds

f ( s) g ( s) = f ( s) dg + ds

g ( s )ds

or a quotient

dN dDd N ( s) D( s) ds

N ( s )ds=

ds D ( s) D ( s)2

3.13 Must we?

The math topics collected here will be used during the course.review any that seem unfamiliar.

(3.12.2)

(3.12.3)

(3.12.4)

Please

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solve for the output variables as functions of the inputs introduce particular disturbances and calculate the responses

First, we write a component material balance on the solute.

ddt

VC o = FC i (t ) FC o (t ) C o (0) = C s (4.1.1)

Because the flow F and volume V are constant, there are no nonlinearterms in the equation. We write (4.1.1) at steady state with reference inletconcentration C s.

dVC odt

= 0 = FC s FC os (4.1.2) s

From (4.1.2) we see that the outlet concentration at the reference conditionis also C s. Subtracting (4.1.2) from (4.1.1), we obtain the process modelin terms of deviation variables, indicated by an asterisk superscript. Thesevariables are zero when the process is at the reference condition; nonzerovalues indicate deviation from the reference.

ddt

V(C o (t ) C s )= F (C i (t ) C s ) F (C o (t ) C s ) (4.1.3)

d * * * *dt

VC o = FC i (t ) FC o (t ) C o (0) = 0

This is a first-order ODE with constant coefficients. We rearrange it tostandard form.

*V dC o * * F dt

+ C o (t ) = C i (t ) (4.1.4)

In mathematical nomenclature, C i* is the forcing function and C o* the*dependent variable. In our system nomenclature, C i* is the input and C o

the output. In standard form, the ratio of tank volume to flow rate clearlytakes on the significance of a characteristic time, the time constant .

*dC o * * * dt

+ C o = C i (t ) C o (0) = 0 (4.1.5)

The solution (by (3.6.1)) is

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t* e

t t

*C o (t ) = e C i (t )dt (4.1.6)0Equation (4.1.6) is the process model for the mixing tank, showing how

the outlet concentration behaves for arbitrary disturbances in the inlet. Forexample, if inlet concentration undergoes a step change C at t d,

C o* = C

1 e

( t t d

)

(4.1.7)

As the disturbance is introduced, the outlet concentration begins tochange; it gradually becomes equal to the inlet concentration. Notice thatthe tangent to the initial response reaches the final value in one timeconstant.

Fig 4-1. The ordinate has been normalized by the magnitude of the stepchange, and the abscissa by the time constant. Thus this non-dimensional

plot is characteristic of all first order lag step responses. The time t d atwhich the input occurs has been set for convenience to equal the timeconstant.

0

0.2

0.4

0.6

0.8

1

1.2

0 2 4 6 7

(t - )/

n o r m a

l i z e

d i n p u

t a n

d o u

t p u

t

inlet concentration

outlet concentration

1 3 5

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Systems described by (4.1.5), with solution (4.1.6), are called first-orderlags. "First order" refers to the order of the governing differentialequation (4.1.5). "Lag" refers to the way in which the output lags behindthe input. The lag occurs because the system has storage capacity, and

that capacity takes time to fill or deplete when conditions change. In this problem, the system stores the dissolved component.

First order lags always feature a time constant that indicates the speed ofresponse, because time is normalized by, or scaled to, the time constant.From the properties of the exponential function, we see that the step is95% complete when time equal to three time constants has elapsed. If thetank time constant is large (large volume, low flow) this time will be large.If the time constant is smaller (small volume, large flow) the outletconcentration will respond more quickly. This is consistent with intuitionand experience.

In addition to speed of response, we are also interested in the degree towhich a dynamic system amplifies or attenuates the input signal. This isoften expressed by the steady-state gain, which is the ratio of steadyoutput change to input change following a step disturbance.

* *

gain = C o*() C o

*

(0) = C = 1 (4.1.8)C i () C i (0) C

For the mixing tank, the gain is 1, showing that permanent disturbancesare merely passed through the system. Both time constant and gain areindependent of the size of the disturbance C.

4.2 Integrator: pumped outlet tankThe pumped outlet tank of Lessons 1 and 2 is an example of a first orderintegrator.

dh A

dt= wi wo h(0) = h s (4.2.1)

All terms in the equation are linear. We define a steady state referencecondition in which the liquid level is h s, and the inlet and outlet flows areequal to w s. In deviation variables, (4.2.1) becomes

*dh * * * Adt

= wi wo h (0) = 0 (4.2.2)

To be strict about placing (4.2.2) in standard form, we should define a gainand a time constant. Gain always has dimensions of output/input, or in

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this case, length divided by mass flow. Hence we multiply (4.2.2) by theheight of the tank, and divide by w s.

*

AhT dh = hT (wi

* wo* ) h*(0) = 0 (4.2.3)

w dt w s s

Collecting terms, we find

*dh * * * dt

= K (wi wo ) h (0) = 0 (4.2.4)

Equation (4.2.4) is separable; its solution is

* * * t h = K (wi wi ) (4.2.5)

and the response to a step change of magnitude w in inlet flow at time t dis

h* = K w (t t d ) (4.2.6)

The integrator has no steady state response to a step disturbance, so Kcannot be viewed as a steady-state gain. The time constant is theresidence time for a full tank.

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0

1

2

3

4

0 2 4

(t - )/

n o r m a

l i z e

d i n p u

t a n

d o u

t p u t

change in inlet flow

liquid level response

1 3

Fig 4-2. The ordinate has been normalized by the product of the gain andstep disturbance, and the abscissa by the time constant. The time t d atwhich the input occurs has been set for convenience to equal the timeconstant.

4.3 Summary of disturbance responsesInitial condition is zero; disturbance is introduced at time t d.

system first order lag first order integratormodel

)()( t Kxt ydtdy =+ )(t Kx

dtdy =

step:

x = Au(t - t d)

( )

d t t

e AK 1

steady-state gain = K

( )

d t t AK

impulse:

x = A (t - t d)

( )

d t t

e AK

AK

sine:

x = Asin( t - td)

( ) ( )( )

)(tan

1

sin1

1

2222

= +

++ +

d

t t

t t AK e AKd

( )

+

2sin1

d t t

AK

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4.4 Multiple system inputs from multiple inlet streamsThe first order lag equation in Section 4.3 is the mathematical form thatresults from applying material and energy balances to well-mixed

volumes, as illustrated by the mixing tank in Section 4.1. Of course, atank may have more than one inlet stream. If so, it will usually be addedto the right-hand side of the equation. In illustration, consider a well-stirred tank heated by an electric resistance element of output power Q.

Our previous first-order systems have stored mass; this one stores energy.The energy balance is

ddt

( C pV(T o T ref ))= C p F (T i T ref ) C p F (T o T ref )+ Q (4.4.1)

where T ref is a reference temperature for computing the enthalpy of theflowing stream. Presuming that flow F is constant and the physical

properties are not a function of temperature, we see that (4.4.1) is a first-order linear ODE with constant coefficients. The energy balance at steadyconditions is

ddt

( C pV(T os T ref ))= 0 = C p F (T is T ref ) C p F (T os T ref )+ Q s (4.4.2)

Subtracting (4.4.2) from (4.4.1) and introducing deviation variables

ddt

( C pV(T o T os ))= C p F (T i T is ) C p F (T o T os )+ Q Q sd

(4.4.3)

dt( C pVT o*)= C p FT i* C p FT o* + Q *

We rearrange (4.4.3) to standard form and consider the case of initialsteady state.

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*V dT o * * 1 * *+ T o = T i + C F Q T o (0) = 0 (4.4.4) F dt p

The time constant for temperature change is the tank residence time, equalto the tank volume divided by the volumetric flow rate. We find that theoutlet temperature response depends on two inputs: the inlet temperatureand the heater power; either can act as a disturbance to the first-ordersystem. The gain for inlet temperature disturbances is unity; thus a stepchange in temperature would ultimately propagate through the tank. Thegain for power disturbances converts dimensions of power to dimensionsof temperature. This gain is a function of the flow rate, so that, forexample, power disturbances have less effect on T o when the flow F islarge.

Equation (4.4.4) is linear first-order, and can be solved by (3.6.1), just aswe did in Section 4.1.

et

t t

* Q * *T o (t ) = 0

e T i + C p F

dt (4.4.5)

In a linear model, the effect of the disturbances is additive. That is, eachaffects the response independently of the other, and the effects are simplyadded. Consider a step increase in inlet temperature at time /2, followedat 2 by a compensating step decrease in heater power. The outlettemperature first rises and then falls in response.

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Figure 4-3 The disturbances to linear equations produce additive responses.

4.5 Multiple Outlet StreamsA tank may have multiple outlet streams, as well. In modeling first ordersystems, we often find that these streams depend on the response variable.In this case, the effect of additional outlet streams is to alter the timeconstant and gain of the system. For example, suppose that a mixedoverflow tank is cooled by convective heat transfer to a condensing vapor.Thus there are two outlet streams: the enthalpy carried out with the outletflow, and the heat transferred to the cooling coil. The energy balance is

ddt

( C pV(T o T ref ))= C p F (T i T ref ) C p F (T o T ref ) UA(T o T c ) (4.5.1)

where the overall heat transfer coefficient is U and the coolant condensesat temperature T c. Writing (4.5.1) at steady state and subtracting thisresult from (4.5.1) gives

ddt

( C pVT o*)= C p FT i* C p FT o* UAT o* + UAT c* (4.5.2)

-1.5

-1

-0.5

0

0.5

1

1.5

0 3 4 5

(t - )/

n o r m a

l i z e

d i n p u

t a n

d o u

t p u t

outlet temperature response

change in heater power

change in inlet temperature

21 6

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*where we have two terms that depend on the response variable T o .Writing (4.5.2) in standard form gives

* C V dT * C F * UA * p o p C F + UA dt

+ T o = C F + UAT i + C F + UA

T c (4.5.3) p p p

Because there are two paths out for enthalpy - flow and heat transfer - thetime constant for temperature change is less than the tank residence time,in contrast to (4.4.4). Equation (4.5.3) also features two enthalpy inputs:inlet flow and the coolant. In fact, if the coolant temperature T c exceedsT i, (4.5.3) will describe heating of the tank. Notice that the gain for aninlet temperature disturbance is less than unity: because there are two

paths out, the outlet response will not grow to equal a permanent inletdisturbance. Equation (4.5.3) is solved as before, and the response isshown in the figure for a gain of 0.5 (that is, UA = C pF) and a stepdecrease in T i at time .

-12

-10

-8

-6

-4

-2

0

0 2 3 4

(t - )/

i n p u

t a n

d o u t p

u t ( d e g

C )

change in inlet temperature

outlet temperature response

1 5

4.6 Summing upFirst order systems arise from material and energy balances on perfectlymixed volumes. The system output or response variable is a measure ofthe storage in the system - for example, liquid level or concentration for

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mass and temperature for energy. Two parameters, the time constant andthe gain, characterize the response of the output variable to disturbances.

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10.450 Process Dynamics, Operations, and ControlLecture Notes Lesson 5

Laplace transforms

5.0 ContextRemember how one multiplies with logarithms:

Given 3.5 4.3- transform the problem into "log space"log (3.5 4.3) = log 3.5 + log 4.3

- perform the logarithm additionlog (3.5 4.3) = 0.5441 + 0.6335 = 1.1775

- transform the problem back into original terms by finding the antilog of the sum3.54.3 = log -1(1.1775) = 15.05

We exchange one multiplication for two transformations and an addition.In an analogous way, the Laplace transform can be used to solve linearordinary differential equations with constant coefficients.

5.1 The Laplace transformThe Laplace transform changes a function of time t into a function of anew independent variable s. This new variable may take complex values.We will go through the trouble for the same reason we use logarithms itmay make the original problem easier to solve.

y( s) = L[ y(t )]= y(t )e st dt (5.1.1)0

We will indicate a Laplace transform of y(t) or y by writing y(s). This isnot to be interpreted as a simple variable substitution of s for t in thefunction y.

5.2 IllustrationConsider the unit step function u(t). From (5.1.1) we find the Laplacetransform of u(t) as

u( s ) = L[u(t )]= u(t )e st dt (5.2.1)0

Before integrating, however, lets get a sense of what this looks like: plotthe product u(t)e -st versus t for different values of s.

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0

0.2

0.4

0.6

0.8

1

1.2

0 .5 1 .5 2 .5 3 .5

t

u ( t ) e x p

( - s

t ) 0.1

0.5

1

5

s =

0 1 2 3

Integrating under each of these curves gives an area associated with avalue of s. Plotting these versus s, we see the behavior of the Laplacetransform of u(t), at least along the real axis.

0

2

4

6

8

10

12

0 1 2 4 6

s

i n t e g r a

l o

f u

( t ) e x p

( - s

t )

3 5

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td


At each value of s, y(s) is based on the entire t-dependence of y. Themagnitude of s acts as a weighting parameter to emphasize differentregions of the t domain. At large values of s, the exponential function will

be appreciably different from zero only for small values of t; hence large semphasizes the initial region of y. By contrast, small values of s allow the

influence of longer times on the value of y(s). The plot of y(s) looks likes-1; performing the integration of (5.1.1) verifies this guess.

5.3 Functional and operational transformsThere are two categories of Laplace transforms we'll consider: transformsof particular functions and transforms of general operations.

Example functional transforms:unit step function u(t - t d).

L(Cu (t t d ))= Cu (t t d )e st dt

0 td

11

= C e st dtt d (5.3.1)

C st = e s t d

Ce st d= s

exponential decay

L(Ce at )= Ce at e st dt0

C e( s +a )t (5.3.2)=

s + a 0C=

s + a

Example operational transforms:the first derivative.Using integration by parts,

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L df (t ) = df (t )e st dt dt 0 dt

= s f (t )e st dt + f (t )e st 0 (5.3.3)0

= sL( f (t )) f (0)

Further detail

L df (t ) = df (t )e st dt dt 0 dt

uv = udv + vdudf

u = f du = dt dt

st v = e dv = se st dt

f (t )e st = s f (t )e st dt + df (t )e st dt0

0 0 dt

L df

dt(t )

= s

0

f (t )e st dt + f (t )e st0

= sL( f (t )) f (0)

the second derivativeUsing (5.3.3),

L d 2 f (t )

d 2 f (t )

e st dt dt 2

= 0 dt

2

df df= s

0 dte st dt

dt 0 (5.3.4) df= s s f (t )e st dt f (0)

dt 0 0

= s 2 L( f (t )) sf (0) dfdt 0

the integralUsing integration by parts,

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t t L f ( )d = f ( )d e st dt

0 0 0

t= f (t ) 1 se

st dt 1( f ( )d )e st (5.3.5)0

s0 0

= 1 L( f (t )) s

Further detail t t

L f ( )d = f ( )d e st dt 0 0 0

uv = udv + vduu = f ( )d du = fdt

st v = e dv = se st dtt

f ( )d ( )e st dt + f (t )e st dt ( f ( )d )e st

=

t s

0 0 0 0 0

L t

f ( )d

= 1 s

0

f (t )e st dt 1( t f ( )d )e st 0 s 0 0= 1 L( f (t ))

s

the time delay Suppose that some function f(t) is delayed by an interval of time . Thusthe delayed function g(t) may be written

1f(t) g(t)

1f(t) g(t)

0 0 < t


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L( g (t ))= f ( )e s ( + )d 0

= f ( )e s e s d (5.3.8)0

= e s f ( )e s d

0

= e s L( f (t ))

The last step happens because there is no reason not to rename theindependent variable as t. Equation (5.3.1) is an example of the timedelay: the transform of the unit step at t d may be viewed as the transformof the unit step at time 0 multiplied by the exponential in t d.

5.4 Solving the first-order lag with Laplace transformsRecall the mixing tank of Section 4.1.

*dC o * * * dt

+ C o (t ) = C i (t ) C o (0) = 0 (5.4.1)

Take Laplace transforms of (5.4.1)

[ sC o* ( s) C o* (0)]+ C o* ( s) = C i* ( s) (5.4.2)

Solve for the outlet concentration transform

* 1 *C o ( s ) = s + 1C i ( s) (5.4.3)

As before, the inlet concentration undergoes a step change C at t d.Transforming the step change by (5.3.1) and inserting, we get

* 1 Cet d s

C o ( s ) = ( s + 1) s(5.4.4)

To perform the reverse transform, it is convenient to use a table oftransform pairs, such as that provided by Marlin (2000). The reversetransform of everything except the exponential factor is

L1 C t (5.4.5)

s( s +1) = C 1

e

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The effect of the exponential factor is to introduce a time delay into thisfunction.

t t eC

t t t C

d

t t

d o

d

=


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1 y(t ) = A

e

t(5.6.4)

The solution decays from the impulse to a steady value of zero. Certainly

(5.6.4) has the same form as (5.5.4), but one should not confuse the initialcondition of (5.5.1) with the delta function disturbance in (5.6.1).

5.7 Example - first-order lag driven by impulse disturbance at time t d

dy + y = A (t t d ) y(0) = 0 (5.7.1)dt

Transforming,

[ sy( s) y(0)]+ y( s ) = Aet d s (5.7.2)

Rearranging and substituting,

y( s) = s

A+1

et d s (5.7.3)

The exponential factor contributes a time delay in the inverse transform.

y(t ) = 0 0 t < t d1 (t t d

) (5.7.4)= A

e t d t

Solution (5.7.4) is just (5.6.4) shifted to a later time of occurrence.

5.8 Example - first-order lag driven by step disturbance at time t d

dy + y = Au(t t d ) y(0) = 0 (5.8.1)dt

Transforming, substituting, and rearranging,

1 Aet d s (5.8.2) y( s) =

s+

1 sInverting with the aid of a transform pair table,

y(t ) = 0 0 t < t d

= A 1 e

( tt d

)

t d t (5.8.3)

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Solution (5.8.3) is just the general form of (5.4.6). Under a continuingsteady disturbance, a first-order lag approaches a new steady value.

5.9 Example - first-order lag driven by initial conditions and step disturbance at time t d

dy + y = Au(t t d ) y(0) = B (5.9.1)dt

Transforming

( sy( s) y(0) )+ y( s) = A e t d s (5.9.2) s

Substituting and rearranging

B + A e t d s (5.9.3) y( s) = s + 1 s( s + 1)

The two terms on the RHS are inverted separately; the time delay appliesonly to the second term.

y(t ) = Be

t

+ A 1 e

( t t d

)

(5.9.4)

The solution starts at the initial condition B and decays; at time t d, it begins to respond to the step input. At long times, the influence of theinitial conditions is negligible, and the solution approaches the value of the

step, A, as in (5.8.3). In interpreting (5.9.4), the second term is understoodto be zero at times less than t d.

5.10 Doing it wrong: forgetting coefficients when substituting the derivative transformsLets return to the problem of (5.5.1):

dy + y = 0 y(0) = A (5.10.1)dt

The constant A has the dimensions of variable y. Take the Laplacetransform of the equation

[ sy( s) y(0)]+ y( s ) = 0 (5.10.2)

Its simple algebra, but its possible to neglect to distribute the timeconstant across the transform

[ sy( s )] y(0) + y( s ) 0 (5.10.3)

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Solve for y(s), substituting the known initial condition.

A y( s)

s +1 (5.10.4)

Now use a table of transform pairs to invert this function

1 y(t ) A

e

t(5.10.5)

Compared to (5.5.4), the solution has the right form, but the dimensionsare wrong: y is set equal to a quantity with dimensions of y/time.

5.11 Doing it wrong: inverting before applying disturbanceThe Laplace transform gives a solution in the Laplace domain; apply thedisturbance before inverting to find the time-domain solution. Heres thewrong way: consider a first-order lag.

dy + y = x(t ) y(0) = 0 (5.11.1)dt

Transforming and rearranging,

1 y( s) =

s +1 x( s) (5.11.2)

If, for example, x is a step disturbance Au(t), the correct procedure is tomake the functional transform of the step, substitute that into (5.11.2) forx(s), and then invert the right-hand-side. Dont invert as if x(s) is aconstant, and then apply the disturbance!

y(t ) et

Au(t ) (5.11.3)

This solution will not satisfy the initial conditions on y; neither will itshow a long-term change from the step input.

5.12 Whats missingLots. We are skipping the mathematical questions of existence anduniqueness, extensions to the related Fourier transforms, and the wholequestion of how to invert an arbitrary Laplace transform.

5.13 References Marlin. Process Control . 2 nd ed. McGraw-Hill, 2000, p.100.

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Lesson 6. Further topics on Laplace transforms and system models

6.0 ContextIn Lesson 4 we saw how first-order models could describe several

physical systems of interest. We derived the models from conservation

equations, and solved them by the integrating factor method. In Lesson 5we examined the Laplace transform as an alternative method of solvingthe models. Here we emphasize several key topics on which the makingand solving of models depends: making equations linear, using deviationvariables, and inverting Laplace transforms.

6.1 LinearizationLaplace transforms are particularly useful for linear equations withconstant coefficients. If the process models include products or non-linearfunctions of any of the dependent variables, it is common to make linearapproximations by Taylor series so that they can be more easily solved.

Because control is intended to keep a process near a set point, and becausea linear approximation is usually accurate around a fixed point, the linearmodel is often perfectly satisfactory for control design.

6.2 Approximating a nonlinear function of the dependent variableFor turbulent flow through an obstruction, the volumetric flow is

proportional to the square root of the head loss.

F o = kh0.5 (6.2.1)

Fo is a function of one variable, h; it may be approximated by Taylor

series.

F o = F o (h) F o kh s0.5 + 0.5kh s0.5 (h h s )

(6.2.2)

where h s is the reference head loss. The approximation is exact at h = h s.

6.3 Approximating a product of variablesEven if the governing equation is linear, it may have variable coefficientsthat include products of variables. For example, the flow of solute Adepends on its concentration and the total volumetric flow rate.

w A = FC A (6.3.1)

A two-variable Taylor expansion is required.

w A = w A ( F ,C A )w A F sC As + F s (C A C As ) + C As ( F F s )

(6.3.2)

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where the subscript s denotes reference values of flow and concentration.

6.4 Approximating a product including a nonlinear functionAs a further example, consider a reaction rate term that might appear in a

material or energy balance:

k o e RT E

C A = f (T ,C A ) (6.4.1)

Both temperature T and concentration C A might be variables in a dynamicsystem; a two-variable Taylor expansion is required.

f E = k dT o RT 2

e RT E

C A

f

dC A RT

E RT

E

RT E

o

ekC

ek

=

E RT E k oe A o s C As + k o RT 2e sC As (T T s )+ k oe RT

E s (C A C As )

s

(6.4.2)

6.5 Deviation variablesIn process control, we try to keep a controlled variable at a set point; thus,we are concerned to reduce deviation from the set point. This suggeststhat we express our models in deviation variables, so that any non-zerovalue of the deviation indicates a problem.

We used deviation variables when we analyzed the mixing tank in Lesson4. Let's examine the procedure for the first-order lag in general.

dy + y = Kx(t ) y(0) = A (6.5.1)dt

where y(t) is the system output, x(t) the input, a characteristic timeconstant, K the steady state gain, and A the initial value of y. Let'sconsider the deviation of the input and output variables from a constantreference state x ref , y ref .

* y (t ) = y(t ) y ref(6.5.2)

* x (t ) = x(t ) xref

Substituting into (6.5.1), we obtain

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*

d ( y + y ref ) + ( y* + yref ) = K ( x* + xref ) y* (0) + y ref = Adt

(6.5.3)

and rearranging,

*

dy + y* = Kx* + ( Kxref y ref ) y* (0) = A yref (6.5.4)dt

Equation (6.5.4) is a first order lag with output y* determined by inputKx * + (Kx ref y ref ) and initial condition A y ref . The fundamental

behavior of the solution will not differ from that of (6.5.1); we merely maychoose the reference condition for our convenience. We consider severalcases:

reference at initial conditions: x ref = x(0); y ref = ASubstituting into (6.5.4)

*

dy + y* = Kx* + ( Kx(0) A) y* (0) = 0 (6.5.5)dt

The deviation variable y * will start at zero and respond to the right-handforcing function. The deviation input Kx * may begin at some arbitrarytime, but the second term affects the solution from time zero. At longtime,

* *

y () = Kx () + Kx(0) A (6.5.6)

which may or may not be zero.

reference at initial steady state: the classic process control problemFor much of our development of process control theory, we will conceiveof a system operating at steady state that is then disturbed at somearbitrary time. In (6.5.4), then,

yref = A = Kxref (6.5.7)

so that (6.5.4) becomes

*

dy + y* = Kx* y* (0) = 0 (6.5.8)dt

The deviation variable y* will have no contribution from the initialcondition and will become nonzero only when disturbance x* becomes

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nonzero. The difference between (6.5.5) and (6.5.8) is that the initialcondition is a steady state, as specified in (6.5.7). At long time,

* * y () = Kx () (6.5.9)

If the disturbance returns to zero, as for an impulse, so will the response.A persistent disturbance, such as a step, gives a persistent response.

reference at long-term conditions: x ref = x( ); y ref = Kx refOther reference conditions are possible. If a long-term condition isidentified, (6.5.4) becomes

*

dy + y* = Kx* y*(0) = A Kx() (6.5.10)dt

The deviation variable y* will start at A - Kx s, which may or may not bezero, and respond to the right-hand forcing function. Because thereference has been selected as the long-term state, y* will necessarily tendto zero.

reference at some arbitrary condition: x ref , y refThe deviation variable will in general not start at zero, nor ultimately tendto zero. The solution will mark the deviation from the reference state, asdriven by the initial conditions and forcing function.

6.6 An example with linearization and deviation variablesConsider a tank in which the outlet flow occurs by gravity.

h

ho Fo

Fi

h

ho Fo

Fi

Following the modeling procedure in 4.1, we begin with a material balance.

ddt

Ah = F i F o (6.6.1)

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A mechanical energy balance relates the outlet flow to the liquid level inthe tank through the frictional loss in the exit valve. For the normalcondition of turbulent flow through the valve,

Adh = F i (h + ho )0.5 (6.6.2)dt

where h o is the vertical distance between the tank bottom and the exitnozzle.

Further detailWe write the mechanical energy balance between the liquidsurface (point 1) and the outlet nozzle (point 2).

2 2

z 1 z 2 = v2

2

g+ K L

v2

2

g(6.6.3)

where K L is the loss coefficient. We represent the distance z 1 z 2 by the sum of the liquid level h and the distance between tank bottom and outlet nozzle, h o. In addition, we express the outletvelocity in terms of the volumetric flow and the cross-sectionalarea A o of the exit pipe.

21+ K L F o h + ho = 2 g

Ao

(6.6.4)

Solving for flow

0.5

F o = 2 gAo

2

(h + ho ) = (h + ho )0.5 (6.6.5)

1+ K L

and substituting (6.6.5) into (6.6.1), we obtain (6.6.2).

The nonlinear term may be approximated by Taylor series, referred to thesteady level h s.

1(h + ho )0.5 (h s + ho )0.5 + 2(h s + ho )0.5

(h h s ) (6.6.6)

On substituting into (6.6.2), we obtain the linearized model

A

dh = F i (h s + ho )0.5 2(h s + ho )0.5(h h s ) (6.6.7)dt

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Writing (6.6.7) at the steady reference condition,

A

dh s = 0 = F is (h s + ho )0.5

2(h s + ho )0.5(h s h s ) (6.6.8)dt

From (6.6.8) we relate the steady inflow and liquid level.

2

h s = F is ho (6.6.9)

Subtracting reference (6.6.8) from model (6.6.7), we obtain deviationvariables.

A d (hdt h s ) = ( F i F is ) 2(h s + ho )0.5

(h h s ) (6.6.10)

We indicate the deviation variables by * and rearrange into standard form,thus identifying the characteristic time constant and steady-state gain.

2 A(h s + ho )0.5 dh * + h* = 2(h s + ho )0.5 F i* (6.6.11) dt

The time constant can be rewritten using (6.6.9) to show that it is more

than twice the residence time of the volume of liquid at the steadyreference condition.

= 2 A(h s + ho ) (6.6.12)

F is

Equation (6.6.11) may be solved by Laplace transforms for particulardisturbances in the inlet flow rate F i.

6.7 Deviation variables are best in linearized equationsConsider a first-order lag in y disturbed by the ratio of independent inputs

x1 and x 2.

dy + y = K x1(t ) y(0) = y s (6.7.1)dt x 2 (t )

We write (6.7.1) at steady state and subtract the result from (6.7.1).

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N ( s) C 1 Bs + C= ( s 1 )( s 2 + bs + c) ( s 1 )

+ ( s 2 + bs + c)2

(6.8.5)

how to solve for the coefficients(1) For each of the real, distinct roots, multiply the expansion by each RH

denominator 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 roots determined, it may be easiest to substitute arbitraryvalues for s to get equations in the unknown coefficients.

(3) 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 written as separate roots in (6.8.3), solving for onecoefficient is enough. The other will be the complex conjugate.

6.9 Partial fraction examples to make sense of those rulesthe first order lag step responseFrom (5.4.4),

* Ce t d s Ce t d s 1C o ( s) = s( s +1)=

1 (6.9.1)

s s +

Partial fraction expansion is applied only to the polynomial part of thisfunction. The denominator has already been factored, and we havedivided top and bottom by to put it in the form of (6.8.1). We write theexpansion as in (6.8.3); there are no repeated factors to consider.

1

1 = C

s1 +

C 2

1 (6.9.2)

s s + s +

Using step (1), we solve (6.9.2) first for C 1, and evaluate the equation at s= 0.

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10.450Lecture Notes - 6

11

+

+

+

=

+

1010

1

10

01 22

3

22

C (6.9.12)

which gives C 2 = -2.

Alternatively, we proceed by step (3a).

+

+

+

=

+

1111 2

2

3

2

2

2

s

sC

s

s

s

s (6.9.13)

Taking the limit as s , we find, as before,

22 01

C +=

(6.9.14)

By step (3b), it is necessary to differentiate (6.9.11) with respect to s.leads quickly to the same result for C 2. ting into (6.9.10),

+

+

+

=

+

1

1

1

1

)1(

2

2

3

2

s s s

s (6.9.15)

and the inverse transform is

tt t

etete

=

+

)(

1)!11(

1

)!12(

1

3

23

(6.9.16)

repeated complex roots

22 )22(8

++ s s s (6.9.17)

Write the partial fraction expansion. Each repeated root accounts for 2terms.

)1()1()1()1()1()1(8 5

243

221

22 j sC

j sC

j sC

j sC

sC

j s j s s ++ +

++ +

+ +

+ +=

+++ (6.9.18)

Process Dynamics, Operations, and Control

ThisThus, substitu


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Lesson 7. Transfer functions and block diagrams

7.0 ContextIn Lesson 5, we introduced Laplace transforms as a method of solving thelinearized equations of system dynamics. However, a more important

reason is that so many of the concepts of dynamics and control theory areexpressed in the LT language, even if the methods are not required forsolution. In this lesson we present the transfer function, a concisedescription of a dynamic system that is based on Laplace transforms. Wealso present block diagrams, a convenient way to represent the structure ofdynamic systems.

7.1 Dynamics of systemsProcess control deals with systems that change in time. In Lesson 2, weasserted that systems are characterized by input disturbances (causes), andoutput responses (effects).

inputs system outputs

We have claimed that a variety of physical systems can be satisfactorilydescribed by relatively few mathematical models. We have dwelt on thefirst-order lag as a prime example:

*

dy

+ y

*

= Kx

*

(t ) y*

(0) = 0 (7.1.1)dt

The system model is the ordinary differential equation (7.1.1), relatinginput x * and output y *as they vary in time. In (7.1.1), x *(t) is themathematical forcing function, and y *(t) the dependent variable. Aftertaking Laplace transforms, we can relate input and output by an algebraicequation:

* K * y ( s ) = s + 1

x ( s ) (7.1.2)

The ratio in (7.1.2) contains all the information about the ODE (7.1.1).When multiplying x *(s), the transform of disturbance x *(t), the ratioconverts it into y *(s), the transform of the response y *(t). We call this ratiothe transfer function .

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7.2 Transfer functionsLet's take a larger view of getting transfer functions from differentialequations: we have a lot more to learn about dynamic systems, but its nottoo early to speculate a bit from what we already know. Perhaps systemsmore complicated than our first-order lag may be described by higher-

order equations. If we linearize such equations, and express them indeviation variables, they must look like

n * n1 * *

a nd dt

yn

+ a n1d dt n

y

1+ K + a1

dy + y* = f ( x* ) (7.2.1)dt

Its not outlandish to speculate that a complicated dynamic system mightdepend not only on the disturbance x *, but its rate of change, as well. Forthat matter, it may depend on higher derivatives of x *, leading us to write(7.2.1) as

n * d n1 y* * d l x* dx*a nd dt

yn

+ a n1 dt n1+ K + a1

dy + y* = bl dt l+ K + b1 dt

+ b0 x* (t )

dt(7.2.2)

We have already encountered systems with multiple disturbances inSection 4.4. Hence we may expand our speculative model further.

n * n 1 * * * *

a nd dt

yn

+ a n 1d dt n

y

1+ K + a1

dy + y* = bld l x

l1 + K + b1

dx1 + b0 x1* (t )

dt dt dtm * *

+ cm

d xm

2 + K + c1

dx2 + c0 x2

*

(t )dt dt+ K

(7.2.3)

As is usual in process control, we presume all initial conditions are zero,which describes a system expressed in deviation variables initially atsteady state. Taking Laplace transforms of (7.2.3) leads to

y* ( s ) = (bl sl + bl 1 s

l 1 + K + b1 s + b0 ) x1* ( s )(a n s n + a n 1 s n 1 + K + a1 s + 1)

+ (cm s m + cm1 s m1 + K + c1 s + c0 ) x2* ( s ) (7.2.4)(a n s n + a n 1 s n 1 + K + a1 s + 1) + K

If we set all the (7.2.4) coefficients except b 0 and a 1 to zero, we recoverthe particular example of (7.1.2). The ratios of polynomials in (7.2.4), like

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the ratio in (7.1.2), are transfer functions. We will represent a generaltransfer function by G(s). Thus (7.2.4) becomes

* * * y ( s ) = G1( s ) x1 ( s ) + G2 ( s ) x2 ( s ) + K (7.2.5)

G1(s) is the transfer function that relates y *(s) to x 1*(s). G 2(s) similarlyrelates y *(s) to x 2*(s). Notice that

The Laplace transforms of the disturbances, when substituted for thexi(s) variables, will not change the polynomial nature of the G i(s)terms in (7.2.4). Thus polynomial ratios in the Laplace variable s willalways result from the linear, constant-coefficient ordinary differentialequations of process control.

It is the nature of the linear ODE that the effects of the inputs areadditive. Each disturbance x i*(s), when processed through its

particular transfer function, contributes to the overall response of y *(s).

As we learned from the partial fraction expansion of Lesson 6, thetime-domain response will finally be a sum of exponential andtrigonometric terms. The various time constants, frequencies, and

phase lags in these terms are determined by the coefficients in thetransfer functions, and thus the original differential equation.

The dynamic response calculated from Equation (7.2.5) may becomplicated indeed, but the essential concept - a dynamic systemapproximated by a linear equation and expressed in terms of transferfunctions - is no different from what we have already studied in thesimpler first order system of (7.1.2).

7.3 Using the transfer functionA transfer function represents a differential equation. Just as we classifydifferential equations into recognizable types, we will classify transferfunctions, learn their characteristics, and use them as a conciserepresentation of particular behaviors. For example, the transfer functionin (7.1.2) represents a first-order lag; it contains the same information asthe ODE of (7.1.1).

Given the transfer function for a system, therefore, we can predict somefeatures of its behavior without actually calculating its response to

particular disturbances. Consider these terms:

order the highest power of s in the denominator. Equivalent to the orderof the differential equation describing the system.The first-order lag is described by a first-order differentialequation; its transfer function has a single s in the denominator.

pole root of the denominator. In later lessons, we will learn that poleswith negative real parts result in output signals that decay in time,

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so that the system will be stable. If there exist poles withimaginary parts, the system may oscillate, even without oscillatorydisturbances.The first order system has a pole at - -1; this negative, real valueindicates a stable response with no oscillation.

zero root of the numerator. These generally have no influence onstability, but can influence the rate and character of the dynamicresponse.The first-order system has no zeroes.

steady state gain the ratio of long-term output change to input stepchange. The gain is a measure of how sensitive the system is todisturbances. If the system is a chemical process, we would like alow value of gain, so that disturbances would have little effect onthe output variable. In a sound system, we would like a large gain,so that tiny input signals from the source (tape, vinyl, CD) areamplified to audibility. The gain is found by setting s = 0 in the

transfer function.

Here we summarize the first-order lag and integrator with respect to these properties

type equation transferfunction

poles steady state gain

lag)()( t Kxt y

dtdy =+

1+ s K

--1 K

integrator)(t Kx

dtdy =

s K

0 none; increaseswithout bound

7.4 Transfer function for the stirred reactorLets combine our knowledge of modeling first-order systems fromLesson 4, Laplace transforms from Lessons 5 and 6, and notions of thetransfer function. In Section 4.1 we modeled a stirred overflow tankcontaining a dissolved substance A. Lets now assume that A disappears

by first order chemical reaction.

r A 1 dN A = kC A (7.4.1)V dt

where the negative sign shows that A is consumed in the reaction. Thecomponent material balance is written assuming that the volumetric flowrate F is constant.

VdC Ao = F (C Ai C Ao ) VkC Ao C Ao (0) = C As (7.4.2)dt

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There are no nonlinear terms; subtracting the steady state from (7.4.2)leaves deviation variables.

*

VdC Ao = F (C

Ai

* C Ao

*) VkC Ao

* C Ao

*(0) = 0 (7.4.3)dt

Equation (7.4.3) is then put into standard form:

*V F + Vk

dC dt

Ao + C Ao* =

F + F

VkC Ai

*

(7.4.4)*

dC Ao + C Ao

* = KC Ai* C Ao

* (0) = 0dt

The time constant is smaller than that for the mixing tank in Equation

(4.1.4) -- in a manner analogous to the multiple outlet streams in Section4.5, the combination of outflow and chemical consumption in (7.4.4)reduces the time response of the outlet concentration. Similarly the gain isless than unity -- a disturbance in inlet concentration is only partlytransmitted to the outlet stream

Taking Laplace transforms of (7.4.4) gives

* K *C Ao ( s ) = s + 1

C Ai ( s ) (7.4.5)

Guided by (7.1.2), we identify the transfer function of the mixing tank.

* K C Ao ( s) (7.4.6)G( s) = s +1

= * C Ai ( s )

As we remarked in Section 7.2, the transfer function depends only on thegeometry and operating conditions of the tank itself, not on thedisturbance. The particular nature of the inlet disturbance C Ai*(s), whenworked through the transfer function G(s), gives the particular nature ofthe output response C Ao*(s).

DetailIn process control, we think of the gain K as a measure of how a

permanent change in input C Ai* affects the output C Ao* in the longterm. However, the steady-state performance of the reactor -- thatis, how well it converts C Ais to C Aos -- is also indicated by the gain.From material balance (7.4.2), written at steady state, we find that

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the gain depends on the reaction rate constant k and the reactorresidence time R .

C Aos F 1= C Ais

= K = F + Vk 1+ k R

(7.4.7)

Low gain means good conversion of reactant A.

By placing the equations in standard form, we found that the timeconstant depends on the tank volume, the flow, and the reactionrate constant.

V =

F + Vk=

1+ k R

R

(7.4.8)

These parameters are important individually, of course, but whenwe are concerned with the dynamic response, it is important toidentify how they interact to affect the time constant. We see thatthe time constant is related to the reactor residence time R , whichwe use in designing a stirred reactor to produce a desired outletconcentration of reactant. For understanding how that outletconcentration varies in time, however, the time constant is moresignificant. Notice that for no reaction, the time constant reducesto the residence time. As the reaction rate increases, the timeconstant decreases, indicating that the outlet concentrationresponds more quickly to disturbances

7.5 Block diagramsThe block diagram is a graphical display of the process model in theLaplace domain. It comprises blocks and arrows, and thus resemblesmany other types of flow diagram. In our use with control systems,however, the arrows represent signals, variables that change in time, whichare not necessarily actual flow streams . The block contains the transferfunction, which may be as simple as a units conversion between x and y,or as complicated as a full chemical process. Remember that the transferfunction incorporates all the dynamic information in the process model.

x(s)G(s)

y(s)

This diagram means

y( s) = G( s) x( s ) (7.5.1)

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inputs outputs


Returning to the mixed reactor in Section 7.4, we can represent thedynamic behavior of the reactor by a block diagram that is equivalent toEquation (7.4.5):

CAi*(s) C Ao*(s)1+ s

K

CAi*(s) C Ao*(s)1+ s

K

7.6 Block diagram structuresThe real value of block diagrams is to represent the flow of signals amongmultiple blocks.

The Block Diagram Rules: only one input and output to a block. The figure in Section 7.1 was fine for its purpose, but

does not qualify as a process control block diagram.

inputs systemsystem outputs

two signals may be summed at an explicit summing junction. Thealgebraic sign is indicated at the junction.

x1(s) G1(s)+

G2(s)-

y1(s)

y2(s)

x3(s)G3(s)

y3(s)

x2(s)

a single signal may feed its value to multiple blocks. This does NOTindicate that the signal is split among the blocks.

x1(s)G2(s)

G3(s)

x2(s)

x2(s)

y1(s)

y2(s)

G1(s)

Block diagrams may be turned into equations by simple algebra. It isusually most convenient to start with an output and work backwards bysubstitution. In the summing diagram

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wo1

wo2

w i


For example, suppose that an inlet flow always divides so that one branchreceives fraction of the flow and a second receives the remainder. Thenthe material balance is

wi = wo1 + wo 2

= wi + (1 )wi(7.7.1)

and the block diagram is drawn

w i

1-

1-

wo1

wo2

The gain of each transfer function is less than unity, showing that eachoutput signal is diminished from the input value.

DetailIn fact, the transfer functions are nothing but gain, which impliesthat the response of the outlet to inlet disturbances is instantaneous.Of course, we assumed this in writing the material balance (7.7.1).For an incompressible flow, this is a good description.

The flow junction might behave so that the flow fraction dependson the magnitude of the inlet flow w i. For a compressible flow, thedynamic response of the branches might differ, so that the transferfunctions would depend on Laplace variable s. The complexity ofthe transfer functions depends on the detail of our modeling;however, the principle of splitting a signal on a block diagram isthe same as in our simple incompressible case (7.7.1).

7.8 Describing systems

We began our study of dynamic systems by writing differential equations.Then we adopted the Laplace transformation of these equations as anequivalent description. Now we have introduced block diagrams as yetanother description. That does it for descriptions -- we will now applythem to increasingly complicated systems. Whatever the means, our

purpose is to calculate an output response for an input disturbance, as wehave in many examples of first-order systems.

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Lesson 8. Arrangements of first-order systems -- series, parallel, and recycle

8.0 ContextWe have studied varieties of the first-order lag in Lessons 4 - 7. Here wetake two stirred reactors and arrange them in different ways. This will

require us to use more than one differential equation in modeling, and thuslead us to higher-order systems. We shall also gain more experience inusing Laplace transform and block diagram methods.

8.1 Reactors in seriesCombine identical well-mixed overflow reactors in series.

F, C A0

F, C A1

F, C A2V

V

Recalling Section 7.5, the component material balance is written for eachtank, assuming that the volumetric flow rate F is constant.

VdC A1 = F (C A0 C A1 ) VkC A1dt (8.1.1)

VdC A2 = F (C A1 C A2 ) VkC A2

dtThe balances are evaluated at a steady state reference condition.

0 = F (C A0 s C A1 s ) VkC A1 s0 = F (C A1 s C A2 s ) VkC A2 s

(8.1.2)

The steady state balances are subtracted from the transient balances so thatthe concentrations may be expressed as deviation variables. We choosethe initial condition as the reference steady state, so that we may examinedisturbances to that condition. Rearranging to standard form, werecognize first-order systems in which we can group the physical

parameters of flow, volume, and rate constant to identify a time constantand gain.

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10.450Lecture Notes - 8

2

0)0(

0)0(

*2

*1

*2

*2

*1

*0

*1

*1

= +

=+ +

= +

=+ +

A A A A

A A A A

C C Vk F

F C

dtdC

Vk FV

C C Vk F

F C

dtdC

Vk FV

(8.1.3)

Taking the two equations together to represent a single system, we identifythe input disturbance as C A0 *, the output response as C A2*, and anintermediate variable as C A1*.

8.2 In series: time domain solution by successive integrationThese first-order equations are sequential: we can integrate the firstequation to determine C A1* for C A0* as a forcing function.substitute C A1* as the forcing function in the second equation.equations, and two integrals in the solution, indicate a 2 nd order system.

dt dt t C ee

K C

dt t C ee K C

t

A

tt

A

t

At

t

A

=

=

0

*02

2*2

0

*0

*1

)(

)(

(8.2.1)

where

Vk F F

K Vk F

V+

= +

= (8.2.2)

Upon substituting a particular disturbance for the input C A0*, we can perform the double integration to obtain the response C A2*.

8.3 In series: Laplace transform solutionTransforming (8.1.3),

( )( )()()0()(

)()()0()(*

1*

2*

2*

2

*0

*1

*1

*1

s KC sC C s sC

s KC sC C s sC

A A A A

A A A A

=+ =+

(8.3.1)

Eliminating intermediate variable C A1*, we relate C A2* to disturbance C A0*

through the transfer function for the system.

)()1(

)( *022

*2 sC s

K sC A A +

=

(8.3.2)

Upon substituting a particular disturbance for the input C A0*, we can perform the inverse transform to obtain the response C A2*(t).

Process Dynamics, Operations, and Control

Then weTwo

t

0

)

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without doing any calculations, we notice that the second order transferfunction has a repeated pole at -1. Thus all poles are real and negative,indicating a stable, non-oscillating response to step and pulse disturbances.

8.4 In series: block diagram solution

It is instructive to express Equations (8.3.1) as transfer functions, as shownin the block diagram.

CA0*(s)

1+ s K

CA1*(s) C A2

*(s)

1+ s K

*By the rules of block diagrams, we relate output C A2* to input C A0 :

* K *C A2 ( s) = ( s +1)C A1 ( s )

(8.4.1) K K * =

( s +1) ( s +1)

C A0 ( s )

which leads again to (8.3.2).

8.5 In series: step responseIf C A0* is a step disturbance C at time t d, the response from (8.3.2) is

* K 2 Ce t d sC A2 ( s) = ( s +1)

2

s(8.5.1)

Inverting via partial fractions or a table of transform pairs,

C A2* (t ) = CK 2

1

1+ (t

t d ) e( t t d

) (8.5.2)

where the response is understood to be zero before time t d. Equation(8.5.2) could also have been obtained by integrating the time domainsolution (8.2.1) for a step disturbance.

step response exampleFor example, the step in input concentration is 1 mol m -3, the timeconstant is 8 minutes, and the process gain is 0.5, where both gain andtime constant depend on flow and kinetics by (8.2.2). If the stepdisturbance occurs at t d = 5 minutes, the outlet concentration responds asshown in Figure 8-1.

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C A0


0

0.2

0.4

0.6

0.8

1

1.2

0 5 10 15 20 25 30 35

time (min)

d i s t u r b a n c e a n

d r e s p o n s e

( m o

l m

i n - 1 )

disturbance

intermediate C 1

equivalent single tank

output C 2

Figure 8-1. Series response compared to first order

Notice that outlet concentration C 2 reacts immediately to the disturbance(a consequence of the complete mixing assumption). However, the rise isgradual and not sharp as for a first-order system (such as intermediateconcentration C 1 leaving the first reactor). Also shown is the response of asingle tank. Its volume is equal to the combined volume of the tanks inseries, so that the reactor residence times are identical. At 32 minutes, theequivalent tank has reached 92% of its ul

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Process Dynamics Operation and Control (MIT Course) Lessons