33_BioNonLinearityW12

33_BioNonLinearityW12 - Page 1 of 10 Base Case G(s) =...

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Unformatted text preview: Page 1 of 10 Base Case G(s) = Bio2x2Glocal.p Numerical Linearization CHE 361 >> Bio2x2Glocal (after modifying and saving bioreac2x2.mdl Enter flow rate F = (1 for nominal ss) = 1 Enter feed nutrient conc. Ni = (5.2 for nominal ss) = 5.2 Enter estimate of SS Bugs (2 for Base Case) = 2 Enter estimate of SS Nutrient (1 for Base Case) = 1 SSbugs_B = 2 SSnutr_N = 1 Enter input number (F=1, Ni=2) = 1 Enter output number (B=1, N=2) = 1 model file) num/den = −0.85714 (1.1111s + 1) -0.95238 s - 0.85714 Glocal ( s ) = -------------------------0.47619 s 2 + 1.4286 s + 1 0.47619 s^2 + 1.4286 s + 1 G_gain = in standard form -0.8571 G_poles = -1.8873 -1.1127 Replace with your block. G_zero = -0.9000 Tau_2nd = bioreac2x2.mdl 0.6901 CHE 361 - OSU zeta = 1.0351 Chemical Engineering Tau_z = 1.1111 >> format long >> G_gain G_gain = 1 -0.857142857059012 Inport1_F Demux Mux >> G_poles G_poles = 2 -1.887298334834795 Bio361 Inport2_Ni -1.112701665362494 >> G_zero Mux Demux G_zero = -0.899999999999511 >> Tau_2nd Tau_2nd = 0.690065559308420 >> zeta zeta = 1.035098339030701 >> Tau_z Tau_z = 1.111111111112715 >> format short 1 Outport1_B 2 Outport2_N Page 2 of 10 >> Bio2x2Glocal Enter flow rate F = (1 for nominal ss) = 1.1 10% higher Enter feed nutrient conc. Ni = (5.2 for nominal ss) = 5.2 Enter estimate of SS Bugs (2 for Base Case) = 2 Enter estimate of SS Nutrient (1 for Base Case) = 1 SSbugs_B = 1.902415458937633 B goes down when F goes up. SSnutr_N = 1.222222222221314 Enter input number (F=1, Ni=2) = 1 Enter output number (B=1, N=2) = 1 ± 10% changes from F = 1.0 liter/hour num/den = -1.0735 s - 1.1057 ------------------------0.5643 s^2 + 1.4903 s + 1 G_gain = -1.1057 G_poles = -1.3205 + 0.1686i -1.3205 - 0.1686i G_zero = -1.0300 Tau_2nd = 0.7512 zeta = 0.9919 Tau_z = 0.9709 −1.1057 (0.97091s + 1) 0.56430 s 2 + 1.4903s + 1 in standard form Glocal+ ( s ) = −0.976 (1.43s + 1) 0.720 s 2 + 1.88s + 1 in standard form Gstep+ ( s ) = −0.976 0.482 s + 1 in standard form Gstep+,1st-order ( s ) = >> Bio2x2Glocal Enter flow rate F = (1 for nominal ss) = 0.9 10% lower Enter feed nutrient conc. Ni = (5.2 for nominal ss) = 5.2 Enter estimate of SS Bugs (2 for Base Case) = 2 Enter estimate of SS Nutrient (1 for Base Case) = 1 SSbugs_B = 2.075598086124823 B goes up when F goes down. SSnutr_N = 0.818181818180931 Enter input number (F=1, Ni=2) = 1 Enter output number (B=1, N=2) = 1 num/den = −0.66157 (1.3150 s + 1) -0.86994 s - 0.66157 Glocal− ( s ) = -------------------------0.41913s 2 + 1.4298s + 1 0.41913 s^2 + 1.4298 s + 1 G_gain = in standard form -0.6616 G_poles = -2.4294 −0.756 (1.13s + 1) -0.9821 Gstep − ( s ) = G_zero = 0.412 s 2 + 1.32 s + 1 -0.7605 Tau_2nd = in standard form 0.6474 zeta = −0.756 1.1043 Gstep-,1st-order ( s ) = but not "close" Tau_z = 0.303s + 1 1.3150 in standard form shape Page 3 of 10 Nonlinear Analysis in the Bioreactor Project CHE 361 (1) Calibration Curve (2) Nonlinearity Classification: Gain and “Dominant” Time Constant Changes (1) Calibration Curve You must plot your steady-state output value as a function of your steady-state input value. Use a circle to indicate the “nominal” steady state for operating your bioreactor. Bug Mass Concentration (g/L), Nominal SS = 2 K S F F B = Y N i − V µmax − F F + MYV N= KS F V µmax − F B/F Calibration Chart: Flow as Input, Bug Concentration as Output 2.5 2 1.5 1 0.5 0 0 0.5 1.5 1 Flow (L/h) , Nominal SS = 1 2 NOTE: The physical constraints that N < N i and B > 0 when a reactor contains bugs result in a limit on the steady-state value of the manipulated input variable for each transfer function as follows: F≤ N iV µmax An upper limit exists, i.e. F must be small enough that bugs are not flushed out. N i + KS KS F ≤ N i A lower limit exists, i.e. N i must be large enough that bugs are not starved for food. V µmax − F Page 4 of 10 (2) Nonlinearity Classification: Gain and “Dominant” Time Constant Changes. (2a) Gain Changes The basic idea is to calculate the change from the local gain and “dominant” time constant when a step change is made to a new steady state. You can complete a table similar to the one below by hand or spreadsheet to find the % change in gain when the manipulated variable is changed by +10%, +5%, +2%, -2%, -5% or –10%. The table would look as follows for the Base Case Bioreactor with the first line and last line as endpoints for the Gain curve in the “Classification of Bioreactor Nonlinearity” figure shown later in this handout. % F change F B DelF DelB 10 5 2 0 -2 -5 -10 1.10 1.05 1.02 1.00 0.98 0.95 0.90 1.9024 1.9543 1.9824 2.0000 2.0167 2.0402 2.0756 0.1000 0.0500 0.0200 NA -0.0200 -0.0500 -0.1000 -0.0976 -0.0457 -0.0176 NA 0.0167 0.0402 0.0756 DelB/DelF = NewK -0.9758 -0.9139 -0.8793 K= -0.8571 -0.8357 -0.8048 -0.7560 100* (NewK-K)/K 13.85 6.62 2.59 NA -2.50 -6.11 -11.80 Plot the first column on the x-axis and the last column on the y-axis of the “Classification of Bioreactor Nonlinearity” figure. Since the Gain change is less than 25% for steps of 10% in either direction, this process is classified as an “A” process based on the Gain analysis. But we must also analyze how much the dominant time constant changes as we change the input (manipulated) variable. The Base Case B/F Bioreactor Process turns out to be a “B” process based on dominant time constant changes, thus “overall” it is classified as “B” (worst case scenario approach). (2b)“Dominant” Time Constant Changes (a) If you have a second-order (no zero) transfer function bioreactor model that is overdamped, then your dominant time constant is τ 1 , the larger of your two first-order time constants. (b) If you have a second-order (no zero) transfer function that is underdamped, you can consider the second-order time constant value τ 2 nd-order as the dominant time constant. (c) If you have a first over second-order transfer function model (a LHP zero), then the value of a “dominant” time constant can be directly obtained by forcing step2g to fit a first-order step response. Do a step test and specify the “gain”=KM exactly - don’t let the computer algorithm fit the KM value. Specify KM as the true KM output deviation calculated from the nonlinear steady-state equations (pg 1 of this handout). Then specify a fixed time delay of 0 and let the step2g program find the best first-order time constant. That is your dominant time constant. Page 5 of 10 To find your “nominal” dominant time constant you can do a very small step up and down and average, or you can do a step test with your local transfer function G(s) as the process and force a fit to the correct local gain, but for the best first order instead of first/second form. After you have your local gain and dominant time constant values, you can do first-order fits to steps of +10 % and -10% of your input and find the corresponding % change for gain and dominant time constant. When you plot these on the Nonlinearity Classification figure, you can determine which classification of nonlinear behavior is reflected at the bioreactor’s nominal steady state. Classification by Most Severe Nonlinearity = Gain or Time Constant Changes A = nearly linear: the manipulated variable can be changed by 10% in either direction and neither gain nor dominant time constant changes by more than 25% B = slightly nonlinear: a change of between 5 and 10% in at least one direction for the manipulated variable results in a change of more than 25% in either gain or time constant C = nonlinear: a change of between 2 and 5% in at least one direction for the manipulated variable results in a change of more than 25% in either gain or time constant D = strongly nonlinear: a change of less than 2% in at least one direction for the manipulated variable results in a change of more than 25% in either gain or time constant Page 6 of 10 +10 % Step for Transfer Function model: Fnew = 1.10, F’new = 0.10 = M B/F “Local” Gain was –0.85714 = K in either F direction Thus KM for fitting should be fixed to be –0.085714 (g/L) / (L/h) Modified from HW5 file = HW5_1sim.mdl Enter choice 1,2,3 or 4 : 1 --------------------------------HOW to treat KM Enter 0 for computer adjusted Enter 1 for fixed KM : 1 For standard G(s) form Fixed KM = -0.085714 ------------------------------HOW to treat time delay Enter 0 for computer adjusted Enter 1 for fixed time delay : 1 For standard G(s) form Fixed time delay = 0 ------------------------------Enter estimate of tau : 0.5 End --------- Final Parameters ---Final KM = -0.0857 Final time delay = 0.0000 Final tau = 0.3800 -------- Poles and zeros -----One pole at s = -2.6318 ------------------------------- Page 7 of 10 For the Base Case, the first-order fit to the local G(s) step response results in a dominant time constant of τ dominant = 0.380 hours. A 10% flow step up to 1.1 L/h gave a value of 0.482 h for a +26.8 % increase. (Shown on next page.)This is just over the 25% limit, so in the positive direction the Base Case is class “B”. For a 10% decrease in flow to 0.9 L/h, the fitted dominant time constant was 0.303 h for a change of –20.3%, which is less than the –25% limit. So in the negative direction the class is only “A”. Overall then the Base Case is class “B” based on the effect of a positive step in F on the dominant time constant. The 3 step response experiments discussed above are shown on the following pages: step up, step down, and finally the “local” dominant time constant fit from step2g on the local transfer function model. Frequency Response Note: When you have a first/second process model, a pulse test of your bioreactor may yield a Bode plot that looks very much like a first-order Bode plot. In that case, the frequency where the phase angle is equal to -45 degrees can be considered as the inverse of the dominant time constant for comparisons to step test fits described above. However, to do the nonlinear classification test, use step response experiments. Page 8 of 10 For step response fits, consider +10% and –10% change in the input F, i.e. from 1.00 to 1.10 L/h and from 1.00 to 0.90 L/h. +10 % Step: Fnew = 1.10, F’new = 0.10 = M B/F Gain for +10% step was found from nonlinear steady-states to be –0.9758 = K Thus KM for fitting should be fixed to be –0.09758 (g/L) / (L/h) Enter choice 1,2,3 or 4 : 1 --------------------------------------HOW to treat KM Enter 0 for computer adjusted Enter 1 for fixed KM : >> 1 For standard G(s) form Fixed KM = -0.09758 --------------------------------------HOW to treat time delay Enter 0 for computer adjusted Enter 1 for fixed time delay : 1 For standard G(s) form Fixed time delay = 0 --------------------------------------Enter estimate of tau : 0.5 End --------- Final Parameters -----------Final KM = -0.0976 Final time delay = 0.0000 Final tau = 0.4819 --------- Poles and zeros ------------One pole at s = -2.0752 --------------------------------------- Page 9 of 10 -10 % Step: Fnew = 0.90, F’new = -0.10 = M B/F Gain for -10% step was found from nonlinear steady-states to be –0.7560 = K Thus KM for fitting should be fixed to be +0.07560 (g/L) / (L/h) Enter choice 1,2,3 or 4 : 1 --------------------------------------HOW to treat KM Enter 0 for computer adjusted Enter 1 for fixed KM : 1 For standard G(s) form Fixed KM = 0.07560 --------------------------------------HOW to treat time delay Enter 0 for computer adjusted Enter 1 for fixed time delay : 1 For standard G(s) form Fixed time delay = 0 --------------------------------------Enter estimate of tau : 0.5 End --------- Final Parameters -----------Final KM = 0.0756 Final time delay = 0.0000 Final tau = 0.3027 --------- Poles and zeros ------------One pole at s = -3.3031 Page 10 of 10 B A C Classification of Bioreactor Nonlinearity D B C A 30 Gain and Taudom Changes, % of nominal value 20 10 0 -10 -20 -30 -10 -8 -6 -4 -2 0 2 4 6 8 Input Step as % of Nominal Input Full Value: What is necessary to cause + or - 25% change in Gain or Taudom ? 10 ...
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This note was uploaded on 03/01/2012 for the course CHE 361 taught by Professor Staff during the Winter '08 term at Oregon State.

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