Control HW

8/13/2019 Control HW

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Homework 5CHE 4198

Group 11

Jerad Brockman (89-821-2827)

Richard Hembree (89-957-7936)

Robert Ingram (89-676-1905) - Leader

Kalen Minvielle (89-869-8579)


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PROBLEM NO. 1

The process given in the problem statement is shownin the figure to the right is anything that is red or thatis black.

The added feedback control in show in purple and ishighlighted in the yellow box. The feedback controllermeasures the output, the outlet water temperature,and controls the input, the amount of inlet heat

provided by the steam.

The block diagram is shown below. The values for gdand gpwere obtained in the problem statement. Sincegfand gmare not given, they are assumed to benegligible, and therefore 1. A proportional controllerwas added to this process.

[1.1] [1.2]

[1.3] [1.4] [1.5] [1.4]

[1.5] [1.6]Using a solver on Microsoft Excel, we obtain the value of =0.1178

[1.7] [1.8]

[1.9]kc I D Kc I D

P 0.5kcu --- ---

P 3.453 --- ---

PI 0.45kcu Pu/1.2 --- PI 3.108 44.435 ---

PID 0.6kcu Pu/2 Pu/6.3 PID 4.144 26.661 8.464

The following screenshots show the step-by-step process in which the value of kcuand Puare confirmed. Transfer

functions were added, values of kcwere guessed until a critically stable curve is drawn.

Tin

Steam

Tout

Temperature gauge

e(s) c(s) u(s) y(s)

ym(s)

d(s)


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kc=0.5 (too small) kc=10 (way too large) kc=7 (slightly too largebut almost there)

The plot at the left shows the output of the program when

the user inputs a value of 6.9 for the value of kc. Since the

graph oscillates, the function is critically stable and therefore,

this confirms that the value of the kcu.

The Open Loop Method starts from Equation 1.1

[1.10] [1.11] [1.12] [1.13] [1.14] [1.16] [1.17] [1.18]

[1.19] [1.19] [1.19] [1.20] [1.21] [1.22] [1.23] [1.24]

kc I D kc I D

P 0.5kcu --- --- P 3.8035 --- ---PI 0.45kcu Pu/1.2 --- PI 4.4232 41.13 ---

PID 0.6kcu Pu/2 Pu/6.3 PID 4.5642 24.68 7.834

The values of the Closed Loop Method start from Equation 1.11. This method yields similar results to those seen in the

Open Loop Method and graphically above.


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P

PI

PID

Using the setting from the ZN Controllers for the closed loop method, the following graphs are the performance for theset point (left) and disturbance (right) changes of different types of controllers. As expected, the PI is more oscillatory

than PID in both cases. However, both of these controllers have no offset unlike the P controller.


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PROBLEM NO. 2

[2.1]

[2.2]

[2.3]Given in the problem statement

[2.4] * +

[2.5]

*

+ *

+

[2.6]

[] [2.7] [2.8]Linearizing First DiffEQ

[2.9] * +

[2.10]

* + *

+

[2.11]

[ ] [2.12] [2.13]Linearizing Second DiffEQ

[2.14]

[2.15]Take Laplace Transform

[2.16] [2.17] [2.18] [2.19] [2.20]

[2.21]

[2.22]Solve for transfer function

e(s) c(s) u(s) y(s)

ym(s)


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The values of gf(s) and gm(s) are assumed to be 1 since there is no specific value given. The controller type isproportional so gc(s) is constant kc. gp(s) is obtained above. This is the Feedback Control Loop.

[2.23]

[2.24]

[2.25] [2.26] [2.27]The set point transfer function is defined in Quation 2.23. The characteristic equation is the denominator of the set pointtransfer function set to 0. The values of all three terms must be positive, which are reflected in the two Eqautions 2.28:

[2.28] [2.28]Solving the equations for the magnitude of the proportional controller, two maximum values are obtained, and the lowervalue is chosen as this is the limiting factor.

P controller with kc=1 P controller with kc=-1

The first set of controllers analyzed is the P controller. As expected, anything with a kc> 0.6655 yields positive poles and

therefore is unstable. An example, with kc=1 is shown in the above right figure. On the other hand, when the value of kcis less than 0.6655 as in the figure to the right, the system is stable as all the poles are negative. This also yields only rea

poles. Various values around 0.6655 were observed, with the 2 figures above just one setting observed.

kc= -10 ; I = 4 kc= -1 ; I = 4 kc= -1 ; I = 10

Unstable!


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The other set of controllers analyzed is the PI controller. When increasing the magnitude of the gain, the range of poles

increases, while when increasing the integral action reduces the range of the poles (and eliminates all the complex

roots). The size of the circle informs the user on the range of the poles. A pole at the origin that was not originally there

in P controllers is now available with the development of the PI controller.

The ZN tuning method incorporated a delay of 0.1 and was used in a different way than normal as this problem had a

upper limit, rather than a lower limit. Reducing the value of kcup to a point was the method performed in the class. A

sample of how the team did this tuning is shown below. Knowing the kcuand Pu, the values of the P, PI, and PID

controllers is just simply plugging these values into known equations.

kc= -1 (way to large) kc= -5 (still not small enough) kc= -8 (too small)

kcu= -7.555 -> Pu= 0.5

kc I D kc I D

P 0.5kcu --- ---

P -3.78 --- ---

PI 0.45kcu Pu/1.2 --- PI -3.4 .4167 ---

PID 0.6kcu Pu/2 Pu/6.3 PID -4.533 .25 0.08

The controller action with the PI controller is shown to the right.

This controller obviously performed well, even though it was a PID,

due to the little oscillatory action performed and the quickness of

convergence.

A screenshot of the closed loop can be seen in the figure in the next

page. The behavior of the controller performs well, as indicated in

the apcon2.


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PROBLEM NO. 3


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Characteristic equation:

Using Apcon2 in MATLAB

Using ZN method to get PID settings

Step Change using kc=-2.85, ti=1.5, td=0.375 Disturbance Change using kc=-2.85, ti=1.5, td=0.375


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Step change followed by disturbance change using ZN controller settings

Using CC method to get PID settings

(

)

Step Change using Kc=-3.4, Ti=1.62, Td=0.275 Disturbance Change using Kc=-3.4, Ti=1.62, Td=0.275


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PROBLEM NO. 4

From the text, the height dependence on flows is expressed as:

[4.1]This can now be rewritten as:

[4.2]

[4.3]

The values of gf(s) and gm(s) are assumed to be 1 since there is no specific value given. The controller type is assumedto be proportional so gc(s) is constant kc. gp(s) is obtained above from the textbook.

[4.4] *+ [4.5] [4.6] * + [4.7] [4.8] [4.9] [4.10]Equation 4.4 is the standard form of the characteristic equation. When applying the transfer function of each element,this can be simplified into Equation 4.5. A Pad Approximation of the time delay with Equation 4.6 can be substitutedinto 4.7 to obtain Equation 4.10, the characteristic equation.

All terms in the Routh Arrays first column must be positive:

s2 .25 s1 0s0

[4.11] [4.12] [4.13]Using the root locus technique, kC


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Using only a P-controller, the controller is critically stable at kC=101. The ultimate period is 0.4 min.

Using Table 12.3 from the text, we calculated the ZN parameters as kC=50.5, and I=0.33; the response to a

step change is shown below.

The gain and phase margins are 1.7 and 20.0547, respectively.


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We can achieve a gain margin of 2.5 by reducing the gain to 34. This is a much more stable system than the previous

one.

A step change now introduces a response as seen bellow.

These controller settings give a response that is less oscillatory (settles quicker), and has less overshoot.


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Brockman Hembree Ingram MinvielleTotal (%) 150 150 151 150

Overall Contribution (%) 150 150 151 150Did you have group

meetings to discuss andsolve this assignment?

Yes Yes Yes Yes

Did each member of thegroup collaborate to

solve this assignment?

Yes, didproblems and

reviewed others


reviewed others


reviewed others


reviewed othersDoes each member ofthe group understand

the assignment?Yes Yes Yes Yes

Any other comments tobe highlighted?

Good partner forhomework!

Good partner forhomework!

AmazingGood partner for

homework!

Name of assessing student: Robert Ingram

Signature of assessing student: Robert IngramDate: November 23, 2009

Control HW

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