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DYNAMIC BEHAVIOR DYNAMIC BEHAVIOR OF PROCESSES :OF PROCESSES :

Development of Empirical Development of Empirical Models from Process DataModels from Process Data

ERT 210ERT 210Process Control & Process Control &

DynamicsDynamics

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COURSE OUTCOME 1 CO1)

1. Theoretical Models of Chemical Processes2. Laplace Transform 3. Transfer Function Models 4. Dynamic Behavior of First-order and Second-order Processes5. Dynamic Response Characteristics of More Complicated Processes

6. Development of Empirical Models from Process Data - DEFINE, REPEAT and DEVELOP empirical steady-state and dynamic models for process. DEFINE and IDENTIFY the concept of discrete-time model, APPLY steady-state and dynamic empirical models for both continuous and discrete- time model, ANALYZE first-order and second-order dynamic model from step response data using analytical solutions.

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Development of Empirical Models from Process Data

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• Is the statistical analysis, that can be used to estimate unknown model parameters and to specify the uncertainty associated with the empirical model.

Figure 7.2 Three models for

scattered data

• In Fig. 7.2, straight-line model (Model 1) provides a reasonable fit, higher-order polynomial relations (Model 2) result in smaller errors between the data and the curve representing the empirical model (Model 3).

• For linear model, the least-squares approach is widely used to estimate model parameters.

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• As an illustrative example, consider a simple linear model

between an output variable y and input variable u,

where and are the unknown model parameters to be

estimated and is a random error.

• Predictions of y can be made from the regression model,

y

1 2β β εy u

1β 2β

1 2ˆ ˆˆ β β (7-3)y u

where and denote the estimated values of 1 and 2, and denotes the predicted value of y.

1 2

1 2β β ε (7-1)i i iY u

• Let Y denote the measured value of y. Each pair of (ui , Yi)

observations satisfies:

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The Least Squares Approach

221 1 2

1 1

ε β β (7-2)N N

i ii i

S Y u

• The least squares method is widely used to calculate the values of 1 and 2 that minimize the sum of the squares of the errors S for an arbitrary number of data points, N:

• Replace the unknown values of 1 and 2 in (7-2) by their estimates. Then using (7-3), S can be written as:

2

1

where the -th residual, , is defined as,

ˆ (7 4)

N

ii

i

i i i

S e

i e

e Y y

=

•This approach can be extended to more general models, with; 1. More than one input or output variable. 2. Functions of the input variables u, such as polynomials and exponentials, providing that the unknown parameters appear linearly.

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• The least squares solution that minimizes the sum of squared errors, S, is given by:

1 2β (7-5)uu y uy u

uu u

S S S S

NS S

2 2β (7-6)uy u y

uu u

NS S S

NS S

where:

2

1 1

N N

u i uu ii i

S u S u

1 1

N N

y i uy i ii i

S Y S u Y

The Least Squares Approach (continued)

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• A general nonlinear steady-state model which is linear in the parameters has the form,

1

β ε (7-7)p

j jj

y X

Extensions of the Least Squares Approach

Then p unknown parameters (j) are to be estimated, and the Xj are the unknown parameters j appear linearly in the model, and a constant term can be included by setting X1 = 1.

The sum of the squares function analogous to (7-2) is2

1 1

β (7-8)pN

i j iji j

S Y X

For Xij the first subscript corresponds to the ith data point, and the second index refers to the jth function of u.

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(7-9)T

S β βY - X Y X

where the superscript T denotes the matrix transpose and:

1 1β

βn p

Y

Y

β Y

This expression can be written in matrix notation as

11 12 1

21 22 2

1 2

p

p

n n np

X X X

X X X

X X X

X

The least squares estimates is given by,β

1ˆ (7-10)

β T TX X X Y

providing that matrix XTX is nonsingular so that its inverse exists. Note that the matrix X is comprised of functions of uj; for example, if: , then X1 = 1, X2 = u, and X3 = u2.

21 2 3β β β εy u u

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Conclusions:

1. If the number of data points is equal to the number of model parameters, (N = p), Eq. 7-10 provides a unique solution to the parameter estimation problem, one that provides perfect agreement with the data points, as long as XTX is nonsingular.

2. For N > p a least-squares solution results than minimizes the sum of the squared deviations between each of the data points and the model predictions.

3. The least-squares solution in Eq.7-10 provides a point estimate for the unknown model parameters i but it does indicate how accurate the estimate are.

4. The degree of accuracy is expressed by confidence intervals that have the form, . The 1 are calculated from the (u,y) data for a specified confidence level.

ii ˆ

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Example 7.1An experiment have been performed to determine the steady-state power delivered by a gas turbine-driven generator as a function of fuel flow rate. The following normalized data were obtained:

Fuel Flow Rate, ui Power Generated, Yi1.0 2.02.3 4.42.9 5.44.0 7.54.9 9.15.8 10.86.5 12.37.7 14.38.4 15.89.0 16.8

The linear model in (7-1) should be satisfactory because a plot of the data reveals a generally monotonic pattern. Find the best linear model and compare it with the best quadratic model.

1 2β β ε (7-1)i i iY u

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SOLUTIONTo solve for the linear model, Eqs. 7-5 and 7-6 could be applied directly. However to illustrate the use of Eq. 7-7, first define the terms in linear model: X1 = 1, and X2 = u. The following matrices are then obtained:

8.16 8.15 3.14 3.12 8.10 19. 5.7 .45 .44 2.0

9.0 8.4 7.7 6.5 5.8 4.9 4.0 2.9 2.3 1.0

1 1 1 1 1 1 1 1 1 1

T

T

Y

X

10

1ii

10

1ii

10

1i

21

10

1ii

10

1ii

10

1i

2

u

Y

u u

u (1)

i

T

T

YYX

XX

21)(

ˆuuu

uuyyuu

SNS

SSSS

22)(

ˆuuu

yuuy

SNS

SSNS

(7-5)

(7-6)

YXXXβ TT 1)(ˆ (7-10)

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The predicted values of are compared with the measured values (actual data) in Table 7.1 for both the linear and quadratic models. It is evident from this comparison that the linear model is adequate and that little improvement results from the more complicated quadratic model.

Solving for and using Eq. 7-10 yields the same results given in Eqs. 7-5 and 7-6, with and (95% confidence limits are shown).

1 20039.00785.01 093.0859.1ˆ

2

To determine how much the model accuracy can be improved by using quadratic model, Eq. 7-10 is again applied, this time with X1 = 1, X2 = u, and X3 = u2. The estimated parameters and 95 % confidence limits for this quadratic model are:

,0085.01707.01 096.0811.1ˆ2 0002.00047.0ˆ

3 and

)ˆ(yy

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ui yi Linear Model Prediction Quadratic Model Prediction

1.0 2.0 1.94 1.992.3 4.4 4.36 4.362.9 5.4 5.47 5.464.0 7.5 7.52 7.494.9 9.1 9.19 9.165.8 10.8 10.86 10.836.5 12.3 12.16 12.147.7 14.3 14.40 14.408.4 15.8 15.70 15.729.0 16.8 16.81 16.85

(S = 0.0613) (S= 0.0540)

ii uy 211ˆˆˆ 2

3212ˆˆˆˆ iii uuy

Table 7.1 A Comparison of Model Predictions from Example 7.1

Sum of the squares of the errors.

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• If the empirical model is nonlinear, then the nonlinear regression must be used.• Why nonlinear: - For example, suppose that a the reaction rate expression of the form, rA = kcA

n, is to be fit to experimental data, where rA is the reaction rate of component A, cA is the reaction concentration, and k and n are model parameters. - This model is linear with respect to rate constant k but is nonlinear with respect to reaction order n.

•A general nonlinear model can be written as; y = f(u1, u2, u3, …, 1, 2, 3…)

where y is the model output, uj are inputs, and j are the model parameters to be estimated.

(7-11)

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• In this case, the j do not appear linearly in the model.

• However, we can define a sum of squares error criterion to be minimized by selecting the parameter set j

N

iiyiY

jS

1

2)ˆ(min

(7-12)

where Yi and denote the ith output measurement and model prediction corresponding to the ith data point, respectively.

iy

• The least-squares estimates are again denote by j .

• For step response of higher-order model, nonlinear regression cannot be used.

• As alternative, a number of graphical correlation can be applied quickly to find the approximate values of 1 and 2 in second-order model (refer Eq. 5-48, page 117).

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• Simple transfer function models can be obtained graphically from step response data.

• A plot of the output response of a process to a step change in input is sometimes referred to as a process reaction curve.

• If the process of interest can be approximated by a first- or second-order linear model, the model parameters can be obtained by inspection of the process reaction curve.

• The response of a first-order model, Y(s)/U(s)=K/(s+1), to

a step change of magnitude M is: /(1 ) (5-18)ty t KM e

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0

1(7-15)

τt

d y

dt KM

• The initial slope is given by:

where Δ is the steady-state change in .

y yK

u M

y y

• The gain can be calculated from the steady-state changes in u and y:

with y = KM

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Figure 7.3 Step response of a first-order system and graphical constructions used to estimate the time constant, τ.

Conclusion:

1.The intercept of the tangent at t = 0 with the horizontal line y/KM = 1, occurs at = t.

2.As an alternative, can be estimated from a step response plot using the value of t at which the response is 63. 2% complete.

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Example 7.2

Figure 7.4 shows the response of the temperature T in a continuous stirred-tank reactor to a step change in feed flow rate w from 120 to 125 kg/min. Find an approximate first-order model for the process and these operating conditions.

Figure 7.4 Temperature response of a stirred-tank reactor for a step change in feed flow rate.

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Example 7.2 (cont’)

SOLUTION

First note that w = M = 125-120 = 5 kg/min. Because T = T()- T(0) = 160 – 140 = 20 oC, the steady-state gain is

kg/min

C4

kg/min 5

C20 oo

W

TK

The time constant obtained from the graphical construction shown in Fig. 7.4 is = 5 min. Note that this result is consistent with the “63.2% method” because

C 6.152)20(632.0140 oT

Consequently, the resulting process model is

15

4

)('

)('

ssW

sT

where the steady-state gain is 4 oC/kg/min.

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Conclusions:Very few experimental step responses exhibit exactly first-order behavior because:

1. The true process model is usually neither first order nor linear. Only the simplest processes exhibit such ideal dynamics.

2. The output data are usually corrupted with noise; that is; the measurements contain a random component.

3. Another process input (disturbance) may change in an unknown manner during the duration of the step test. In the CSTR example, undetected changes in inlet composition or temperature are examples of such disturbances.

4. It can be difficult to generate a perfect step input. Process equipment, such as the pumps and control valves discussed, cannot be changed directly from one setting to another but must be ramped over a period of time.

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• In order to account for higher-order dynamics that are neglected in a first-order model, a time-delay term can be included.• This modification can improve the agreement between model and experimental response.• The fitting of a first-order plus time-delay model (FOPTD) to the actual step response can be used for this propose, as shown in Figure 7.5.

Figure 7.5 Graphical analysis of the process reaction curve to obtain parameters of a first-order plus time delay model.

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2. The line drawn tangent to the response at maximum slope (t = ) intersects the y/KM=1 line at (t = ).

3. The step response is essentially complete at t=5. In other words, the settling time is ts=5.

1. The response attains 63.2% of its final response at time, t = .

-θ

( )τ 1

Ke sG s

s

First-Order Plus Time Delay (FOPTD) Model

For this FOPTD model, we note the following charac-teristics of its step response:

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Figure 7.5 Graphical analysis of the process reaction curve to obtain parameters of a first-order plus time delay model.

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There are two generally accepted graphical techniques for determining model parameters , , and K.

Method 1: Slope-intercept method

First, a slope is drawn through the inflection point of the process reaction curve in Fig. 7.5. Then and are determined by inspection.

Alternatively, can be found from the time that the normalized response is 63.2% complete or from determination of the settling time, ts. Then set =ts/5.

Method 2. Sundaresan and Krishnaswamy’s Method

This method avoids use of the point of inflection (as a result of measurement noise and small-scale recorder charts or computer display), construction entirely to estimate the time delay.

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• They proposed that two times, t1 and t2, be estimated from a step response curve, corresponding to the 35.3% and 85.3% response times, respectively.

• The time delay and time constant are then estimated from the following equations:

1 2

2 1

θ 1.3 0.29(7-19)

τ 0.67

t t

t t

• These values of and approximately minimize the difference between the measured response and the model, based on a correlation for many data sets.

Sundaresan and Krishnaswamy’s Method

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• In general, a better approximation to an experimental step response can be obtained by fitting a second-order model to the data.

• Figure 7.6 shows the range of shapes that can occur for the step response model,

1 2

(5-39)τ 1 τ 1

KG s

s s

• Figure 7.6 includes two limiting cases: , where the system becomes first order, and , the critically damped case.

• The larger of the two time constants, , is called the dominant time constant.

2 1τ / τ 02 1τ / τ 1

1τ

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Figure 7.6 Step response for several overdamped second-order systems.

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Smith’s Method

1. Determine t20 and t60 from the step response.

2. Find ζ and t60/ from Fig. 7.7.

3. Find t60/ from Fig. 7.7 and then calculate (since

t60 is known)

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• Assumed model:

θ

2 2τ 2ζτ 1

sKeG s

s s

• Procedure:

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In Chapter 5 we considered the response of a first-order process to a step change in input of magnitude M:

/ τ1 M 1 (5-18)ty t K e

For short times, t < , the exponential term can be approximated by

/ τ 1τ

t te

so that the approximate response is:

1M

M 1 1 (7-22)τ τ

t Ky t K t

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is virtually indistinguishable from the step response of the integrating element

22 (7-23)

KG s

s

In the time domain, the step response of an integrator is

2 2 (7-24)y t K Mt

Hence an approximate way of modeling a first-order process is to find the single parameter

2 (7-25)τ

KK

that matches the early ramp-like response to a step change in input.

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If the original process transfer function contains a time delay (cf. Eq. 7-16), the approximate short-term response to a step input of magnitude M would be

θ θKM

y t t S tt

where S(t-) denotes a delayed unit step function that starts at t=.

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Figure 7.10. Comparison of step responses for a FOPTD model (solid line) and the approximate integrator plus time delay model (dashed line).

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• Will not be discussed further in this chapter.

• Neural networks (NN) or artificial neural networks are an important class of empirical nonlinear models.

• It have been used extensively to model a wide range of physical and chemical phenomena and to model other nonengineering situations such as stock market analysis, speech recognition and medical diagnose.

Figure 7.12: Multilayer neural

network with three layers

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• A digital computer by its very nature deals internally with discrete-time data or numerical values of functions at equally spaced intervals determined by the sampling period.

• Thus, discrete-time models such as difference equations are widely used in computer control applications.

• One way a continuous-time dynamic model can be converted to discrete-time form is by employing a finite difference approximation.

• Consider a nonlinear differential equation,

, (7-26)dy t

f y udt

where y is the output variable and u is the input variable.

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• This equation can be numerically integrated (though with some error) by introducing a finite difference approximation for the derivative.

• For example, the first-order, backward difference approximation to the derivative at is

where is the integration interval specified by the user and y(k) denotes the value of y(t) at . Substituting Eq. 7-26 into (7-27) and evaluating f (y, u) at the previous values of y and u (i.e., y(k – 1) and u(k – 1)) gives:

1(7-27)

y k y kdy

dt t

t k t

tt k t

11 , 1 (7-28)

1 1 , 1 (7-29)

y k y kf y k u k

t

y k y k tf y k u k

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Example 7.4

For the first-order differential equation,

)()()(

tKutydt

tdy

Derive a recursive relation for y(k) using first-order backwards difference for dy/dt.

SOLUTION

)1()1()1()((

kKuky

t

kyky

)1()1(1)(

kut

Kkyt

ky

(7-31)

(7-32)

(7-30)

The corresponding differences equation after approximating the first derivative is;

Rearranging gives

The new value of y(k) is a weighted sum of the previous value y(k-1) and the previous input u(k-1).

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Second-Order Difference Equation Models

1 2 1 21 2 1 2 (7-36)y k a y k a y k b u k b u k

• Parameters in a discrete-time model can be estimated directly from input-output data based on linear regression.

• This approach is an example of system identification (Ljung, 1999).

• As a specific example, consider the second-order difference equation in (7-36). It can be used to predict y(k) from data available at time (k – 1) and (k – 2) .

• In developing a discrete-time model, model parameters a1, a2, b1, and b2 are considered to be unknown.

t t

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• For more details:

-Try analyze EXAMPLE 7.5 AND EXAMPLE 7.6

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