17_SEM_KL_Ch7_W12 - Fitting First and Second-Order Models...

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Unformatted text preview: Fitting First and Second-Order Models Using Step Tests Development of Empirical Models From Process Data • Simple transfer function models can be obtained graphically from step response data. Chapter 7 • In some situations it is not feasible to develop a theoretical (physically-based model) due to: 1. Lack of information 2. Model complexity 3. Engineering effort required. • An attractive alternative: Develop an empirical dynamic model from input-output data. • Advantage: less effort is required • If the process of interest can be approximated by a first- or second-order linear model, the model parameters can be obtained by hand calculations using the response curve. • The response of a first-order model, Y(s)/U(s)=K/(s+1), to a step change of magnitude M is: / y t KM (1 e t ) (5-18) • Disadvantage: the model is only valid (at best) for the range of data used in its development. i.e., empirical models usually don’t extrapolate very well. • A plot of the output response of a process to a step change in input is sometimes referred to as a process reaction curve or bump test. 1 2 y (t ) 1/ t 1 e KM • The initial slope is given by: d y 1 dt KM t 0 τ (7-15) • The gain can be calculated from the steady-state changes in u and y: K y y u M Figure 7.3 Step response of a first-order system and graphical constructions used to estimate the time constant, τ. where Δy is the steady-state change in y. 3 4 First-Order Plus Time Delay Model G (s) K e -θs τs 1 For this FOPTD model, we note the following characteristics of its step response: 1. The response attains 63.2% of its final response at time, t = . 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 (99.3%) at t = 5 , the “settling” time. Figure 7.5 Graphical analysis of the process reaction curve to obtain parameters of a first-order plus time delay model. 5 6 Sundaresan and Krishnaswamy’s Method There are two generally accepted graphical techniques for determining model parameters , , and K. • They proposed that two times, t1 and t2 , be found from a step response curve, corresponding to the times for 35.3% and 85.3% of the final response, respectively. 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. • The time delay and time constant are then estimated from the values of t1 = t35.3 and t2 = t85.3 using the following equations: θ 1.3t1 0.29t2 τ 0.67 t2 t1 Alternatively, can be found from the time that the normalized response is 63.2% complete. (7-19) • These values of and approximately minimize the difference between the measured response and the model, based on a correlation for many data sets to yield the 5 “parameter numbers” of the method: 35.3, 85.3, 1.3, -0.29, 0.67. Method 2. Sundaresan and Krishnaswamy’s Method This method avoids use of the point of inflection construction entirely to estimate the time delay. 7 8 Estimating Second-order Model Parameters Using Graphical Analysis • “In general”, a better approximation to an experimental step response can be obtained by fitting a second-order model to the data (first-order models are a “subset” of 2nd-order ones). • Figure 7.6 shows the range of shapes that can occur for the 2nd-order overdamped step response model, K G s (5-39) τ1s 1 τ 2 s 1 • Figure 7.6 includes two limiting cases: τ 2 / τ1 0 , where the system becomes first order, and τ 2 / τ1 1 , the critically damped case. Note dimensionless time used. Figure 7.6 Step response for several overdamped secondorder systems. • The larger of the two time constants, τ1 , is called the dominant time constant. 9 10 11 12 Smith’s Method – not just overdamped • Assumed model: G s Ke θs τ 2 s 2 2ζτs 1 • Procedure: 1. Determine t20 and t60 from the step response and their ratio t20/t60 – after eliminating time delay. 2. Find ζ and t60/ from Fig. 7.7 (2 y-values). 3. Using t60/ from Fig. 7.7 and then calculate (since t60 is known) ...
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