Process Dynamics and Control - Esercizi Svolti (Solved Exercises)

PROCESS DYNAMICS

and CONTROL

EXERCISES

1)​ Solve the following differential equations with Laplace transform.

●​ with

5 + 4 = 2 (0) = 1

2

●​ with and

+ 5 + 6() = 1 (0) = 1 = 0

2

First system:

Applying the Laplace transform to both sides:

2

→ →

⎡ ⎤

5 + 4 [ ] = [ 2 ] 5 [ () − (0) ] + 4() =

⎣ ⎦

2 2

→

5() − 5 + 4() = () ( 5 + 4 ) = + 5

+5 5+2 .

2

() = =

(

5+4

)

5+4 −1 −1 5+2

Now we apply the inverse transform: .

⎡ ⎤

() = [ () ] = ⎣ ⎦

(

5+4

)

By using the table in the formulary we obtain from row 11 that

+

−1 −1

⎡⎢ ⎤

5+2 .

3

⎡ ⎤

= ⎥

( )

( )

⎣ ⎦ +

(

5+4

) +

⎣ ⎦

2

1

In order to find the right coefficient we divide by 5 Y(s):

5+2 +0,4

⎡ ⎤

⎡ ⎤ = ⎣ ⎦

⎣ ⎦

(

5+4

)

(

+0,8

)

so ; and .

= 0, 4 = 0 = 0, 8

3 1 2

The solution is − −

− − −0,8

→

3 1 3 2

1 2

() = + () = 0, 5 + 0, 5

− −

2 1 1 2

Second system: 2

⎡⎢ ⎤⎥

Applying the Laplace transform to both sides: ⎡ ⎤

+ 5 + 6

[ () ] = [ 1

]

⎣ ⎦

2

⎣ ⎦

2 1

(0)

() − (0) − + 5() − 5(0) + 6() =

2 1

() − + 5() − 5 + 6() =

We need to isolate Y(s): 1 2

+5+ +5+1 .

() = =

2

(

+3

)

(

+2

)

+5+6

2 (+2)(+3)+()(+2)+()(+3)

+5+1 =

(

+3

)

(

+2

)

(

+3

)

(

+2

)

2

+ 5 + 1 = ( + 2)( + 3) + ()( + 2) + ()( + 3)

The three solution roots of the denominator are .

= 0, − 2, − 3

1

Considering the equation becomes → .

= 0 1 = · 6 = 6 5

Considering the equation becomes → .

=− 2 − 5 =− 2 = 2

5

Considering the equation becomes →

=− 3 − 5 = 3 =− 3 1

1/6 5/2 −5/3

The function is .

() = + +

+2 +3

From the table (funct. 2, funct. 5) now it’s possible to find the solution of the single fractions:

−2 −3

1 5 5 .

() = + −

6 2 3

2)​ A continuous perfectly-stirred tank blending system is initially full of water and is

3

fed with pure water at a constant volumetric flowrate q = 0.4 m /min. At a

particular time, an operator shuts off the pure water supply and replaces it with an

aqueous caustic solution at the same flowrate q, but with solute concentration c =

i

3 3

50 kg/m . The system volume V is maintained constant and equal to 2 m by an

overflow line, and there are no significant changes in the volume due to mixing.

The mixture density ρ does not depend on composition and is equal to 1000 kg/m3.

●​ What is the concentration response in the blender?

●​ How long does it take for the solute concentration in the exit stream to reach the

3

value of 40 kg/m ?

Data: 3

●​ = 2 =

3

●​ = 0, 4 =

●​ → is the disturbance

= 50 3

●​ ρ = 1000 3

●​ → is the response

() =?

* *

●​ given that

( ) = 40 3

It is like for t < 0 and for t > 1 that can be expressed through the step function

= 0 =

.

= · ()

()

(

ρ

)

Mass balance is → .

= ρ − ρ = ρ − ρ

2

Since density and volume is constant .

=

Considering that the tank is perfectly mixed .

() = ()

(

)

The species balance is → . The output is c

= () − () = () − ()

while c is the input. V and q instead are parameters.

i

+ () = ()

We divide by q to obtain the function with a “1” coefficient: →

+ () = ()

where is the time constant of the system.

τ + () = () τ = = 5

Applying the Laplace transform to both sides: →

τ [ () − (0) ] + () = ()

1

→ .

τ() + () = () = τ+1

( )

− −0,2

( )

From the table, using function 13, we obtain .

τ

() = 1 − = 50 1 −

−0,2

( )

The concentration reaches at a time

40 () = 40 = 50 1 −

3

* 1 .

=− 5 = 8, 05

5

We assume that the time to reach a steady state condition is , in fact

= 4 ÷ 5τ

−0,2·4·8,05

( ) .

() = 50 1 − = 49, 08 3

Dynamics (time to reach steady state) depends on design parameters and operating

parameters, while the final steady state does not depend on these parameters.

3)​ We consider a perfectly stirred tank (blender) that has two streams in input, w and

1

w with two different mass compositions, x and x correspondingly. The outlet

2 1 2

stream w is obtained from the overflow line and its composition is x, so the volume

remains constant. We assume that the flowrates do not change with time.

We assume that the density of the liquid is constant (does not depend on composition).

Write a transfer function G(s) for the system.

In a second case, consider that also x is a function of time.

2 3

(

ρ

)

The material balance of the tank is: →

= − = + −

1 2

(

ρ

)

The species balance over the tank is: = + −

1 1 2 2

Density is independent of composition, and since the composition depends on time, also

density does not depend on time, as well as the volume:

(

ρ

) →

= + − = 0 = +

1 2 1 2

(

ρ

) → (1)

= + − ρ = + −

1 1 2 2 1 1 2 2

For the last equation:

●​ the output is the outlet concentration x

●​ the input are the composition of the two inlets (x and x )

1 2

●​ density, volume, w , w and w are the parameters

1 2

Since we need to find the transfer function of the system, and therefore a function that links

the output with the inputs, we would need to retrieve two different TF, since there are two

inputs: ()

●​ () = ()

1 1

()

●​ () = ()

2 2

Let’s assume for a moment that x is constant (so it becomes a parameter).

2

()

We need to find only .

() = ()

1 1

We start writing the dynamic equation of the system (1) at steady state, so we need to

introduce the nominal values at steady state for those variables that can change with time:

→ (2)

ρ = + − 0 = + −

1 1 2 2 1 1 2 2

Now we subtract equation (2) from (1), obtaining:

→

ρ = () + () − () − − +

1 1 2 2 1 1 2 2

⎡⎢ ⎤⎥

ρ = () − + 0 −

[ () − ]

⎣ ⎦

1 1 1

We define and as the deviation variables

() − = '() () − = '()

1 1 1 4

At time 0:

●​ '(0) = (0) − = − = 0

1 1 1 1 1

●​ '(0) = (0) − = − = 0

'(0) (0) (0) '(0) (0)

●​ where is a constant thus and since at

1 1 1 1 1

= − =

1

(0) '(0)

time 0 we are at steady state, so

1 1

= 0 = 0

'(0)

●​ for the same reason = 0

for the definition of deviation variables.

The dynamic equation can now be expressed with the deviation variables:

'()

ρ = '() − '()

1 1

Isolating on the left the output: '()

ρ + '() = '()

1 1

Usually we try to have a coefficient “1” in front of the output variable so we divide by w:

ρ '() 1

+ '() = '()

1

Applying the Laplace transform on both sides:

⎡⎢ ⎤⎥

ρ '() 1

⎡ ⎤

+ '() = '()

⎣ ⎦

1

⎣ ⎦

ρ 1

[ '() − '(0) ] + '() = '()

1

by definition so

'(0) = 0

ρ 1

⎡ ⎤

'() + 1 = '()

⎣ ⎦

1

1

'()

= = ()

ρ

'() 1

+1

1

is the transfer function.

We define ρ

●​ is the time constant of the system

τ =

●​ as the steady state gain of the system

1

=

1

so the transfer function becomes

1

() = τ+1

1

which is an inherent property of the system (does not change as the inlet changes).

We can retrieve the order of the dynamic system by looking at the order of the denominator

of the transfer function: in this case, the highest power of s is 1 so it’s a first order system (it

can be also seen from the time domain point of view).

Now we consider that x is also a function of time.

2

Starting from equation (1) at steady state:

→ (3)

ρ = + − 0 = + −

1 1 2 2 1 1 2 2 5

Subtracting (1) with (3):

ρ = + − − − +

1 1 2 2 1 1 2 2

⎡⎢ ⎤⎥ ⎡⎢ ⎤⎥

ρ = () − + () − − [

() − ]

⎣ ⎦ ⎣ ⎦

1 1 1 2 2 2

'()

ρ = '() + '() − '()

1 1 2 2

Applying the Laplace transform to both sides:

ρ [ '() − '(0) ] = '() + '() − '()

1 1 2 2

'() [ ρ + ] = '() + '()

1 1 2 2

1 2

'() = '() + '()

ρ+ ρ+

1 2

DIviding by w on numerator and denominator of the left side we obtain:

1 2

'() = '() + '()

ρ ρ

1 2

+1 +1

1 2

'() = '() + '()

τ+1 τ+1

1 2

We therefore obtained

and .

1 2

() = () =

τ+1 τ+1

1 2

The dynamic system can be written as:

'() = () '() + () '()

1 1 2 2

4)​ Consider the blending system shown in the figure. Assume that the liquid density

does not change with the composition, that the tank volume remains constant (due

to an overflow line not shown), that stream 2 is a pure solute (x = 1), and that the

2

following variables can change with time: w , w , x . Derive the transfer functions

1 2 1

that relate the output x to the independent inputs.

w , w and x are inputs.

1 2 1

The material balance around the control volume of the tank is:

(

ρ

) → (1)

= + − + =

1 2 1 2

The species balance around the same volume is:

(

ρ

) → (2)

= + − ρ = + −

1 1 2 2 1 1 2 6

( )

We can substitute (2) in (1) obtaining (2)

ρ = − + ( 1 − )

1 1 2

Dividing by :

ρ () () ()() () ()() ( )

1 1 1 2 2

= − + − = , , ,

ρ ρ ρ ρ 1 2 1

The transfer function we need to express are 3, as the number of inputs:

'() '() '()

, and .

'() '() '()

1 2 1

By the way there is a problem: the derivative of the output is a function of the output itself,

but also of w , w and x so we need to linearize the function using Taylor expansion up to the

1 2 1

first order. ( )

( ) ( )

( )

() ≃ , , , + (

− ) + − +

1 2 1 1 1

1

( ) ( )

( ) ( )

+ − + −

2 2 1 1

2 1

( )

The function is evaluated at the steady state condition, but since it the

, , ,

1 2 1

function f is the derivative of the output, at steady state that derivative is equal to 0.

( )

()

()

( )

1 2 1 2

= − − =− − =−

ρ ρ ρ

ρ ρ

( ) ( ) −

() ()

1

1

= − =

ρ

ρ ρ

1

( ) ( )

1 () 1−

= − =

ρ

ρ ρ

2

( ) ( )

()

1

1

= =

ρ

ρ

1

Summing up all the contributions we obtain:

−

1−

() .

1 1

≃− ' + ' + ' + '

ρ ρ ρ ρ

1 2 1

We isolate the terms that contain the output x’ at the left hand side and we make the

coefficient of x’ to be 1: −

1−

() 1 1

+ ' ≃ ' + ' + '

ρ ρ ρ ρ

1 2 1

−

1−

ρ () 1 1

+ ' ≃ ' + ' + '

1 2 1

We can now apply the Laplace transform, obtaining:

1 2 3

'() = '() + '() + '()

τ+1 τ+1 τ+1

1 2 1

−

1− ρ

where , , and

1 1

= = = τ =

1 2 3

7

5)​ The process of drug ingestion, distribution and subsequent metabolism in an

individual may be represented by the simplified block diagram shown in the figure.

A simplified mathematical model for the process is made of the following differential

equations:

1 = −

1 1

2 = −

1 1 2 2

where all variables are in deviation form and:

●​ x = drug mass in the gastrointestinal tract (GIT)

1

●​ x = drug mass in the bloodstream (BS)

2

●​ u = drug ingestion rate.

k and k are positive constants representing physiological proper