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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+,1storder ( 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,1storder ( 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 steadystate output value as a function of your steadystate 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
steadystate 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*
(NewKK)/K
13.85
6.62
2.59
NA
2.50
6.11
11.80 Plot the first column on the xaxis and the last column on the yaxis 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 secondorder (no zero) transfer function bioreactor model that is overdamped, then
your dominant time constant is τ 1 , the larger of your two firstorder time constants. (b) If you have a secondorder (no zero) transfer function that is underdamped, you can consider the
secondorder time constant value τ 2 ndorder as the dominant time constant. (c) If you have a first over secondorder transfer function model (a LHP zero), then the value of a
“dominant” time constant can be directly obtained by forcing step2g to fit a firstorder 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 steadystate equations
(pg 1 of this handout). Then specify a fixed time delay of 0 and let the step2g program find the best
firstorder 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 firstorder 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 firstorder 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 firstorder 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 steadystates 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 steadystates 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.
 Winter '08
 Staff

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