Matlab Homework #2 solution

1. Consider the following parameters for the remainder of this assignment: k1 = 5/6, k2 = 5/3, k3

= 1/6 mol/L/min, and 𝐶𝐴𝑓 = 10 mol/L. Find the maximum achievable steady-state concentration of

B for the case where the dilution rate (F/V ) is the manipulated input.

Solution:

For the remainder of this assignment I will define the input u(t) as u(t)=F(t)/V. I will also define �� to be the value of x(t) at steady state.

At steady state, we know that

𝑑𝐶𝐴

𝑑𝑡= 0 = ��(𝐶𝐴𝑓 − 𝐶𝐴

) − 𝑘1𝐶𝐴 − 𝑘3𝐶𝐴

2

𝑑𝐶𝐵

𝑑𝑡= −��𝐶𝐵

+ 𝑘1𝐶𝐴 − 𝑘2𝐶𝐵

Solve the two equations, we get

𝐶𝐴 =

−(𝑘1 + ��) + √(𝑘1 + ��)2 + 4𝑘3𝐶𝐴𝑓��

2𝑘3

𝐶𝐵 =

−𝑘1(𝑘1 + ��) + 𝑘1√(𝑘1 + ��)2 + 4𝑘3𝐶𝐴𝑓��

2𝑘3(𝑘2 + ��)

So we can get the maximum achievable steady-state concentration of B in Matlab using max:

k1=5/6; k2=5/3; k3=1/6; Caf=10; u=[0:0.001:10]; Cb=(-k1*(k1+u)+k1*sqrt((k1+u).^2+4*Caf*k3.*u))./(2*k3*(k2+u)); [Cbmax,i]=max(Cb); u(i) Cb(i)

This returns, maximum 𝐶𝐵 = 1.266 for F/V=1.29.

2. Since we've already defined u and 𝐶𝐵 as arrays above, all we have to do is type plot(u,Cb) and we get the plot shown below.

3. Build an open-loop Simulink model with the dilution rate (F/V) as the input and concentration of B (𝐶𝐵) as the output. See the figure below:

4. The maximum achievable concentration of B corresponds to a singularity, and the system is uncontrollable at this point. Instead consider both of the steady states corresponding to 𝐶𝐵 = 1.117 mol/L (F/V= 0.5714 min−1 and F/V = 2.8744 min−1 ). Use the open-loop Simulink model to find the linear state-space model for both steady-state operating points. 1) For F/V= 0.5714, 𝐶𝐴=2.9999

M-file for S-function:

function [sys,x0] = HW5_2014(t,x,u,flag) % % Mike Henson, September 23, 2010 % %% Parameters

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

k1 = 5/6; k2 = 5/3; k3 = 1/6; Caf = 10;

switch flag,

%% Derivatives

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% case 1,

Ca=x(1); Cb=x(2);

dxdt = [u*(Caf-Ca) - k1*Ca - k3*Ca^2; -u*Cb+k1*Ca-k2*Cb];

sys = dxdt;

%% Outputs

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% case 3,

sys = x(2);

%% Initialization

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% case 0,

NumContStates = 2; NumOutputs = 1; NumInputs = 1;

sys = [NumContStates,0,NumOutputs,NumInputs,0,0];

x0 = [2.9999 1.117];

%% Unhandled flags

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% case { 2, 4, 9 }, sys = [];

%% Unexpected flags

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% otherwise error(['Unhandled flag = ',num2str(flag)]); end

>>sys = 'HW5_2014_openloop_model'; >>load_system(sys) >>[x1,u1,y1,dx1]=trim(sys,[1; 1],[]); >>x1 >>sys_io(1)=linio([sys,'/Input'],1,'in'); >>sys_io(2)=linio([sys,'/CSTR'],1,'out'); >>linsys = linearize(sys,sys_io)

x1 =

2.9999

1.1170

linsys =

a =

S-Function(1 S-Function(2

S-Function(1 -2.405 0

S-Function(2 0.8333 -2.238

b =

Input

S-Function(1 7

S-Function(2 -1.117

c =

S-Function(1 S-Function(2

CSTR 0 1

d =

Input

CSTR 0

Continuous-time state-space model.

>> A1=linsys.a; B1=linsys.b; C1=linsys.c; D1=linsys.d;

>> A1

A1 =

-2.4047 0

0.8333 -2.2381

>> B1

B1 =

7.0001

-1.1170

>> C1

C1 =

0 1.0000

>> D1

D1 =

0

2) For F/V= 2.8714, Ca=6.0828 M-file for S-function is the same but x0=[6.0828 1.117] The linear state-space model is A2 =

-5.7353 0

0.8333 -4.5411

B2 =

3.9172

-1.1170

C2 =

0 1.0000

D2=0

5. Compute the eigenvalues of each state-space model and determine the stability of each steady state.

>> eig(A1)

ans =

-2.2381

-2.4047

>> eig(A2)

ans =

-4.5411

-5.7353

Since both of these sets of characteristic values are real and negative, both steady states are stable. We would expect a perturbation from the steady state to decay without oscillation.

6. Find the transfer function G(s)=C B (s)/D(s) corresponding to each state-space model (D ≡F/V).

One can find it easily using MATLAB

>> [num1,den1] = ss2tf(A1,B1,C1,D1); G1=tf(num1,den1)

Transfer function:

G1 =

-1.117 s + 3.147

---------------------

s^2 + 4.643 s + 5.382

>> [num2,den2] = ss2tf(A2,B2,C2,D2); G2=tf(num2,den2)

Transfer function: G2 =

-1.117 s - 3.142

---------------------

s^2 + 10.28 s + 26.04

7. Find zeroes and poles

Matlab:

>> roots(num1) ans =

2.8177

>> roots(den1) ans =

-2.4047

-2.2381

>> roots(num2)

ans =

-2.8129

>> roots(den2)

ans =

-5.7353

-4.5411

8. Use the direct substitution method and the transfer function models to analytically determine the ultimate controller gain (K cu ) for each steady state.

The poles of the closed loop transfer function are the poles of the open loop transfer function, GOL (s) = Kc Gp (s) for a proportional controller, and the roots of 1 + GOL . Since Gp (s) is stable in this problem, we only need worry about 1+GOL . Substituting s=iω and thereby forcing the poles onto the imaginary axis, we get:

For steady state 1:

1 + 𝐾𝑐𝑢

−1.117𝑖𝜔 + 3.147

(𝑖𝜔)2 + 4.643𝑖𝜔 + 5.382= 0

(4.643𝜔 − 1.117𝐾𝑐𝑢𝜔)𝑖 + (−𝜔2 + 3.147𝐾𝑐𝑢 + 5.382) = 0

𝐾𝑐𝑢 = 4.157𝐿

𝑚𝑜𝑙. 𝑚𝑖𝑛

For steady state 2:

1 + 𝐾𝑐𝑢

−1.117𝑖𝜔 − 3.147

(𝑖𝜔)2 + 10.28𝑖𝜔 + 26.05= 0

(10.28ω − 1.117𝐾cu𝜔)𝑖 + (−𝜔2 + 3.147𝐾𝑐𝑢 + 26.05) = 0

𝐾𝑐𝑢 = 9.203𝐿

𝑚𝑜𝑙. 𝑚𝑖𝑛

9. Build a closed-loop simulink model for control of the concentration of B, 𝐶𝐵, by manipulation of the dilution rate, F/V . Use the closed-loop simulation to determine the ultimate controller

gain (Kcu) near the steady state corresponding to F/V = 0.5714. Make a small step change away from the steady state to slightly perturb the system.

M-file- function [sys,x0] = model(t,x,u,flag)

k1=5/6;

k2=5/3; k3=1/6; Caf=10;

switch flag,

case 1,

Ca=x(1);

Cb=x(2);

sys = [u*(Caf-Ca)-k1*Ca-k3*Ca^2, -u*Cb+k1*CA-k2*Cb];

case 3,

sys = x(2);

case 0,

NumContStates = 2;

NumOutputs = 1;

NumInputs = 1;

sys = [NumContStates,0,NumOutputs,NumInputs,0,0];

x0 = [2.9999 1.117];

case { 2, 4, 9 },

sys = [];

otherwise

error(['Unhandled flag = ',num2str(flag)]);

end

Changing the value of Gain, ultimate controller gain can be found out.

For Kc=4 (plot of 𝐶𝐵 vs time)

For Kc=4.25 (plot of 𝐶𝐵 vs time)

For Kc=4.16 (plot of 𝐶𝐵 vs time)

10. Use the ultimate gain found by simulation and the Ziegler-Nichols tuning method to find PI controller parameters. The Ziegler-Nichols parameters for PI controllers are Kc=0.45Kcu and τI =Pu/1.2. This means that Kc = 1.872 for our model. It's hard to find Pu from the plots I showed, so here's another that facilitates easier determination:

This indicates that Pu ~ 1.4 min, which gives us τI = 1.1667 min. 11. Evaluate the PI controller parameters. Consider step changes in the concentration setpoint 𝐶𝐵 (+/- 0.1 mol/L) and the feed concentration 𝐶𝐴𝑓 (+/- 1.0 mol/L) for a system initially at the

steady-state where F/V = 0.5714.

M-file for s-function function [sys,x0] = model2(t,x,u,flag)

k1=5/6;

k2=5/3;

k3=1/6;

switch flag,

case 1,

CA=x(1);

CB=x(2);

D=u(1);

CAf=u(2);

sys = [D*(CAf-CA)-k1*CA-k3*CA^2, -D*CB+k1*CA-k2*CB];

case 3,

sys = x(2);

case 0,

NumContStates = 2;

NumOutputs = 1;

NumInputs = 2;

sys = [NumContStates,0,NumOutputs,NumInputs,0,0];

x0 = [2.9999 1.117];

case { 2, 4, 9 },

sys = [];

otherwise

error(['Unhandled flag = ',num2str(flag)]);

end

For 𝐶𝐵 = 1.017

For 𝐶𝐵 = 1.217

For 𝐶𝐴𝑓=9

For 𝐶𝐴𝑓=11

12. Use Skogestad's method to find a first-order-plus-time-delay (FOPDT) model that approximates the higher order transfer function corresponding to the steady state at F/V=0.5714min-1.

Skogestad's “half rule" creates a first-order plus time-delay model by using the longest time constant plus half of the second-longest time constant as the model's time constant and the

other half of the second-longest time constant plus all other time constants as time delays. The

function G1(s) (Problem 7) has poles at s = -2.4047 min -1 and s =-2.2381 min -1 , and a zero at s = 2.8177 min -1 . Therefore the time constants are 1/2.4047 = 0.4158 min and 1/2.2381 = 0.4468 min in the denominator, and the numerator time constant is 1/2.8177 = 0.3549 min. The largest time constant, 0.4468 min, is added to one half of the next-largest time constant, 0.4158 min, to get the model time constant of = 0.6547 min. The transfer function has a steady-state gain of 0.5847 mol/L/min (Kss = G(s=0)), so that is the value of K we use.

𝜃 =1

2(0.4158) + 0.3549 = 0.5628

𝐺𝐹𝑂𝑃𝐷𝑇 =0.5847𝑒−0.5628𝑠

0.6547𝑠 + 1

13. Use your FOPDT model and the Tyreus-Luyben tuning relations to determine an additional set of PI controller parameters. From table, using Kcu, Pu from problem 10, Kc = 0.31 Kcu : Kc= 1.2896

τI =2.2 Pu : τI = 3.08

14. Consider step-changes in the concentration setpoint 𝐶𝐵 to compare the performance of the PI

parameters determined from the Tyreus-Luyben tuning relations with the ones previously determined by the Ziegler-Nichols tuning method.

It is clear that the Ziegler-Nichols tuning relations give a more conservative controller here.