30_BioreactorPulseTest_W12

30_BioreactorPulseTest_W12 - Bioreactor Pulse Testing = 3...

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Unformatted text preview: Bioreactor Pulse Testing = 3 “Hints” CHE 361 Hint #1. When doing a pulse test, the analysis is easiest to interpret when the the transfer function has a positive gain. In order to simplify the Bode plot that will be produced when your bioreactor G(s) naturally has a negative gain, you should reverse the signs in the Sum block used to define the third column, the deviation variable values (Store_tBout block example shown below). 1 Outport Clock1 1 in_1 Mux tBout t_B_B'_rec 2. B_initial_ss Sum Mux Store_tBout modified_Sum You double click on the Sum block and change +- to -+ to switch the signs on the block input signals, then hit “Apply”. When you close the Sum block, you should see the top input with a negative sign and the bottom input with a + sign. This chanes the sign of the deviation variable data so that it looks like your bioreactor has a positive gain and results in your phase angle starting at 0 degrees instead of -180. Of course you need to remember that your true gain is still negative. For example for the Base Case Bioreactor above: B ( s ) −0.95238s − 0.85714 G ( s) = = so the "real" gain is -0.85714 (negative) F ( s ) 0.4762 s 2 + 1.4286 s + 1 For a freqg.p analysis, we would analyze it as: 0.95238s + 0.85714 == gain is +0.85714, τ z 1.11, τ 1 0.90 = 0.53 and τ 2 0.4762 s 2 + 1.4286 s + 1 5 5 For a pulse width of 0.1 h, ωmax = = ≈ 50 / (CHE 361 factor 5)=10 rad/h tw 0.1 = G ( s) Amplitude Ratio For “large” pulse, Bout away from steady-state value, Pulse gain is not the same as Local gain - see Bode plots. 10 10 10 10 2 Double Bode Plot: chebode2i, solid=Biofreqg.dat dash=BioPulse0-1.dat 1 0 -1 Product Bug Concentration 2.05 -2 10 -2 10 2 10 -1 0 10 Frequency (rad/min) 10 1 10 2 90 1.9 0 1.85 Phase (deg) B out , [g / liter] 1.95 1.8 1.75 1.7 1.65 -90 -180 -270 0 1 2 3 4 Time , [hours] 5 6 -360 -2 10 10 -1 0 10 Frequency (rad/min) 10 1 10 2 5 5 = ≈ 500 / (CHE 361 factor 5)=100 tw 0.01 For a shorter pulse width of 0.01 h, ωmax = Amplitude Ratio 10 But standard “configuration” is for a max step size of 0.1 in the simulation which is too long for a very short pulse, so bad PA results above 10 rad/h. 10 Bode Plot: made by chebode.m using freqg.dat 0 -1 -2 10 -2 10 10 -1 0 10 Frequency (rad/min) 10 1 10 2 0 Phase (deg) -90 -180 -270 -360 -2 10 10 -1 0 10 Frequency (rad/min) 10 1 10 2 Hint #2. Use a max step size equal to or smaller than pulse width = duration. With a pulse width of 0.01 hours and n equal max step size of 0.01 h, we get a nice Bode Plot match to results from G(s) transfer function (Note frequencies in Bioreactor Bode plots should be rad/h) Product Bug Concentration Feed Volumetric Flow Rate 2.05 3 Bout , [g / liter] Fin , [liters / hour] 4 2 1 0 0 0.02 0.04 0.06 Time , [hours] 0.08 0.1 2 1.95 0 2 4 Time , [hours] 6 The result of this shorter pulse is that the ouput B stays closer to the nominal steady state and we get a very good match to the Bode plot of the local G(s) model from freqg.p. Amplitude Ratio 10 10 Double Bode Plot: chebode2i, solid=Biofreqg.dat dash=BioPulse0-01.dat 0 -1 -2 10 -2 10 10 -1 0 10 Frequency (rad/min) 10 1 10 2 Phase (deg) 0 -30 -60 -90 -2 10 10 0 -1 10 Frequency (rad/min) 10 1 10 2 Hint #3. Use of “grid” to remove grid lines to see differences above more easily. Amplitude Ratio 10 10 0 Double Bode Plot: chebode2i, solid=Biofreqg.dat dash=BioPulse0-01.dat -1 -2 10 -2 10 10 -1 0 10 Frequency (rad/min) 10 1 10 2 Phase (deg) 0 -30 -60 -90 -2 10 10 -1 0 10 Frequency (rad/min) 10 1 10 2 Also in simulation input/output time plots you can use: set(0,'ShowHiddenHandles','on') Then use xlim([xmin xmax]) or ylim([ymin ymax]) commands in Command Window to set plot limits. ...
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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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