Process Control, 3P4

Assignment 3Kevin Dunn, kevin.dunn@mcmaster.ca Due date: 03 February 2014

Assignment objectives: understand how to work with nonlinear systems; how to model time delays; furtherexperience with transfer functions and block diagrams. Getting some practice for the midterm.

Question 1 [10]

Consider the system from the previous assignment. It is a CSTR where the reaction A −→ B is occurring. You do nothave to derive the component material balance, i.e., the differential equation in the figure.

Using the above nonlinear model, create a linearized model and find the Laplace transform representation of it. Youwill need to create deviation variables 𝐶 ′

𝐴 for outlet concentration, and 𝐶 ′𝐴0 for inlet concentration. The transfer

function will relate the outlet concentration 𝐶 ′𝐴 to the incoming (input) concentration, 𝐶 ′

𝐴0.

Solution

The single nonlinear term requires linearization. After subbing in the linearization, the ODE will be:

𝑉𝑑𝐶𝐴

𝑑𝑡= 𝐹𝐶𝐴,0 − 𝐹𝐶𝐴 − 𝑘1𝐶𝐴,𝑠

1 + 𝑘2𝐶𝐴,𝑠− 𝑉 𝑘1

(1 + 𝑘2𝐶𝐴,𝑠)2(𝐶𝐴 − 𝐶𝐴,𝑠)

Writing the above equation at steady state, and subtracting, we see the 3rd term on the right side drop away, and wehave deviation variables:

𝑉𝑑𝐶 ′

𝐴

𝑑𝑡= 𝐹𝐶 ′

𝐴,0 − 𝐹𝐶 ′𝐴 − 𝑉 𝑘1

(1 + 𝑘2𝐶𝐴,𝑠)2𝐶 ′

𝐴

𝑑𝐶 ′𝐴

𝑑𝑡=

𝐹

𝑉𝐶 ′

𝐴,0 −𝐹

𝑉𝐶 ′

𝐴 − 𝛼𝐶 ′𝐴

𝛼 =𝑘1

(1 + 𝑘2𝐶𝐴,𝑠)2

allowing us to write:

𝐶 ′𝐴(𝑠)

𝐶 ′𝐴,0(𝑠)

=𝐹/𝑉

𝑠+ (𝐹/𝑉 + 𝛼)

Question: do you get the expected result when the reaction rate is zero, i.e. 𝛼 = 0?

Question 2 [8]

From a previous midterm

A thermometer having first-order dynamics, with a time constant of 0.2 minutes, is placed in a temperature bath, andafter the thermometer comes to equilibrium, with the bath, the temperature of the bath is increased linearly with time,at a rate of 0.5 degrees per minute.

1. Write down what the Laplace transfer expression would be for the thermometer temperature reading, 𝑇 . Usedeviation variables.

2. Find the analytical response of the thermometer temperature, 𝑇 ′(𝑡).

Solution

1. As described in tutorial, the dynamics refer to the ability of the thermometer to rapidly display the value of thewater bath, i.e.:

𝑇 ′(𝑠)

𝑇 ′𝑤(𝑠)

=𝐾

𝜏𝑠+ 1

must reflect the gain and time constant of the thermometer output, 𝑇 ′(𝑡) for a given input in the water bathtemperature, 𝑇 ′

𝑤(𝑡).

We know eventually (at final value state), the thermometer reading will match the bath reading, so the gain ofthe system is 1.0. The time constant, 𝜏 = 0.2 minutes was given.

So the transfer function is:

𝑇 ′(𝑠)

𝑇 ′𝑤(𝑠)

=1

0.2𝑠+ 1

for the input variable, bath temperature, 𝑇 ′𝑤(𝑠).

2. If the bath temperature is ramped up at 0.5 degrees per minute, we can write that as 𝑇 ′𝑤(𝑠) =

0.5

𝑠2.

So the output thermometer temperature, 𝑇 ′(𝑡) = ℒ−1

[︂0.5× 1

𝑠2 (0.2𝑠+ 1)

]︂.

Using partial fraction expansion,0.5

𝑠2 (0.2𝑠+ 1)= −0.1

𝑠+

0.5

𝑠2+

0.02

0.2𝑠+ 1

Inverting this to the time domain gives 𝑇 ′(𝑡) = −0.1 + 0.5𝑡+0.02

0.2

[︀1− 𝑒−𝑡/0.2

]︀= 0.5𝑡+ 0.1𝑒−5𝑡

Question 3 [12]

From a previous midterm

A second order reaction occurs in a well mixed tank (liquid phase reaction). The system is currently operating atsteady state, however the inlet feed flow and concentration are both known to vary (i.e. there are two inputs to thesystem)

The reaction is given by −𝑟A = 𝑘𝐶2A, and you may assume the tank volume is constant, and the reactor is isothermal.

The reaction rate constant has units of m3.s−1.mol−1.

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1. Derive a dynamic model that relates the outlet concentration of species A, denoted by 𝐶A, to the inlet flow rate,𝐹 , and inlet concentration, 𝐶A,in.

2. Use this model to derive the Laplace transform representation, linearizing where required.

Solution

This question was solved in detail in class on 03 February.

The ODE derived is:

𝑉𝑑𝐶𝐴

𝑑𝑡= 𝐹𝐶𝐴,0 − 𝐹𝐶𝐴 − 𝑘𝑉 𝐶2

𝐴

It requires linearized 3 nonlinear terms, as the question tells us that the inlet flow and inlet concentration vary withtime (the two inputs). The outlet concentration, 𝐶𝐴, is also time varying.

The solution is:

𝐶 ′𝐴(𝑠) =

𝐹𝑠

𝑉 𝑠+ 𝐹𝑠 + 2𝑘𝑉 𝐶𝐴,𝑠𝐶 ′

𝐴0(𝑠) +𝐶𝐴,0,𝑠 − 𝐶𝐴,𝑠

𝑉 𝑠+ 𝐹𝑠 + 2𝑘𝑉 𝐶𝐴,𝑠𝐹 ′(𝑠)

Note: even if you derive the solution as two separate transfer functions (one keeping 𝐹 constant while varying 𝐶𝐴,0,and the next one keeping 𝐶𝐴,0 constant while varying 𝐹 ), you still have to linearize the 𝐹𝐶𝐴 term, which requires aTaylor series expansion with two terms.

Question 4 [10]

From a previous midterm

Apply a step input of magnitude 3 units to the following system:

𝐺(𝑠) =𝑒−2𝑠

(𝑠2 + 2𝑠+ 5) (𝑠+ 2)

then, without explicitly inverting to the time domain, determine:

1. whether the response is stable

2. the steady (state) value of the response

3. whether the response will have oscillatory characteristics (justify your answer).

4. Give a rough sketch of the system’s output, starting from when the step response is applied.

Solution

There is no need to invert to the time-domain for these types of questions. However, recognize the input function is𝑢(𝑠) = 3

𝑠 , and it is the output response, 𝑦(𝑠) = 𝐺(𝑠)𝑢(𝑠) that we are interested in:

𝑦(𝑠) =3𝑒−2𝑠

𝑠 (𝑠2 + 2𝑠+ 5) (𝑠+ 2)

1. The response’s stability is judged from the final value theorem, given that we are putting a stable input into thesystem.

𝑦(𝑡) = lim𝑠→0

[𝑠𝑦(𝑠)] =3

10= 0.3

Since the final value is finite, the response is judged to be stable.

We’ve not yet studied in class about the roots of the denominator (these tell us about stability), however if youhave read ahead (p 109), you could answer that since the system has one root of zero, and that the other rootshave negative real part (𝑠 = −2; and 𝑠 = −1± 2𝑖), that the system is stable.

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2. The prior question showed the steady state value was 0.3

3. Again, you can observe that to find 𝑦(𝑡) would require a partial fraction expansion:

𝑦(𝑠) =𝛼1

𝑠+

𝛼2

(𝑠2 + 2𝑠+ 5)+

𝛼3

(𝑠+ 2)

that the second term would contribute oscillatory characteristics to the response; since from line 17, we wouldhave:

𝛼2

(𝑠2 + 2𝑠+ 5)= 𝛼′

2

2

(𝑠2 + 1)2+ 4

Note that the inversion would be of the form 𝑒−𝑡 sin(2𝑡), indicating the exponential gets damped down overtime (the 𝑒−𝑡 portion will force the sinusoid to have smaller and smaller amplitude).

We can also use this part of the question to judge stability, noting that all the terms in the inversion would steadyout to constant values as 𝑡 −→ ∞.

4. Putting all this information together, and noting that the response has a pure delay of 2 time units, the responsewould be expected to have the following characteristics:

Any plot showing (i) a 2 unit delay, (ii), steading out at 0.3 and (iii) oscillations of some sort, will be deemedcorrect.

END

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