P 9-4

UNIVERSIDAD DE CARTAGENA

FACULTAD DE INGENIERA

PROGRAMA DE INGENIERA QUMICA

ALVARADO, V. ; GALLEGO, H. ; MARRUGO, J . ; MARTNEZ, H. ; NIETO, Y. ; PALACIOS, J .

Process Control

Consider the jacketed continuous stirred tank reactor (CSTR) shown below. The following information is obtained from testing the reactor and its control system: The transfer function of the reactor temperature to the jacket temperature is a first-order lag with a gain of 0.6C/C and a time constant of 13 min. The transfer function of the jacket temperature to the coolant flow is a first-order lag with a gain of -2.0C/(kg/s) and a time constant of 2.5 min. The control valve is linear with constant pressure drop and is sized to pass 12 kg/s when fully opened. Its time constant is negligible. The reactor temperature transmitter is calibrated for a range of 50C to 100C and has a time constant of 1 min. The jacket temperature transmitter is calibrated for a range of 0 to 100C, and its time constant is negligible.

Jacketed Reactor

TC

101

TT

101

TT

102

Product

Reactant

Coolant

Coolant

a) Decide on the proper fail position of the control valve and the action of the controller for a simple feedback control loop with the reactor temperature controller manipulating the position of the coolant valve. Draw the block diagram showing all the transfer function and write the closed-loop transfer function of the reactor temperature to its set point. Pay particular attention to the signs wich must correspond to the fail position of the valve and the controller action.

b) Write the characteristic equation for the singe feedback loop and calculate its ultimate gain and period by direct substitution.

c) Design a cascade control system for the reactor temperature with the jacket temperature as the intermediate process variable, specifying the action of both controllers. Draw the complete block diagram for the cascade control system showing all the transfer functions and their signs.

d) Assuming a proportional slave controller with a gain of 2%CO/%TO, write the transfer function for the jacket temperature loop and redraw the block diagram with the jacket temperature loop as a single block.

e) Using the simplified block diagram from the previous part, write the characteristic equation of the reactor temperature loop in the cascade control system and calculate the ultimate gain and period by direct substitution.

In the table below are summarized the constants and parameters given by the excercise

Element Data Transfer function

Reactor

= 0.6

= 13 =

0.6

13 + 1

Jacket

= 2

= 2.5 =

2

2.5 + 1

Valve = 12 / =

Transmitter R

: 50 100

= 1 =

+ 1

Transmitter J : 0 100 =

The proper fail position of the control valve is fail-open (FO).If there is an electric fail in the control system, the refrigerant must continue flowing to the jacket in order to avoid any overheating in the CSTR.

The action of the controller for the valve is reverse, because when there is an increase in the reactor temperature, the refrigerant flow of the jacket must increase too, and the fail-open valve needs air to close so it needs less air.

The transfer function for the valve is determinated

Gain

The transfer function for the transmitter R is determinated

Gain

So, the transfer function is

Decide on the proper fail position of the control valve and the action of the controller.

Draw the block diagram showing all the transfer function and write the closed-loop transfer function of the reactor temperature to its set point.

= 100

=12 /

100 %= 0.12

=(100 0)%

100 50 = 2

%

=2

+ 1

%

()

()

()/

%

+%

0.12 2

2.5 + 1

0.6

13 + 1

2

+ 1

()

The block diagram that represents the system with all the transfer function is shown below

From the block diagram, the following equations were obtained

= ; = A

=

+ =

=

1 +

Replacing the transfer functions

=

2

2.5 + 1 0.120.6

13 + 1

1 +2

+ 1 2

2.5 + 1 0.120.6

13 + 1

=

0.144 + 1

2.5 + 1 13 + 1 + 1 + 0.288

Transfer Function of the characteristic polynomial is obtained

Considering an equation of 1st order

With negligible time constant, and

Developing

Write the characteristic equation for the singe feedback loop and calculate its ultimate gain and period by direct substitution.

2.5 + 1 13 + 1 + 1 + 0.288 = 0

=

+ 1

=

2.5 + 1 13 + 1 + 1 + 0.288 = 0

32.53 + 482 + 16.5 + 1 + 0.288 = 0

Using the direct substitution method, the values = y = are replaced in the characteristic equation

Then, solving for 2 = 1 and separating the real and imaginary parts

From this complex equation, we obtain the following two equations, because both the actual and the imaginary parts must be zero.

32.533 + 482

2 + 16.5 + 1 + 0.288 = 0

482 + 1 + 0.288 + 32.5

3 + 16.5 = 0 + 0

482 + 1 + 0.288 = 0

32.53 + 16.5 = 0

If = 0.7125 = 81.13

= 81.13

At this gain, the loop response oscillates with a frequency of 0.7125 or a period of

=2

=

2

0.7125= 8.81

Initially the transfer function for the transmitter J is determinated.

The gain of the transmitter is given by

Assuming a first order lag with a negligible time constant, the transfer function for the transmitter J is

Action of the controllers:

Jacket temperature controller (slave): The action of the controller for the valve is reverse, because when there is an increase in the jacket temperature, the coolant flow must increase too, and the fail-open valve needs air to close so it needs less air.

Reactor temperature controller (master): The action of the controller for the valve is reverse, because when there is an increase in the reactor temperature, the temperature fluid of the jacket must increase too, and the fail-open valve needs air to close so it needs less air.

Design a cascade control system for the reactor temperature with the jacket temperature as the intermediate process variable, specifying the action of both controllers.

Draw the complete block diagram for the cascade control system showing all the transfer functions and their signs.

=(100 0)%

100 0 = 1

%

= 1

TC

101

TC

102

TT

101

TT

102

Product

Reactant

Coolant

Coolant

Jacket CSTR with cascade control system

0.12

1

2

+ 1

2

2.5 + 1

0.6

13 + 1

% +

1()

% +

2()%

()

()

2()%

1()%

()/

()

The block diagram that represents the system with all the transfer function for a cascade control system is shown below

Assuming that the slave controller is proportional with a gain of 2%CO/%TO the block diagram can be simplified by writing a new function for the jacket temperature loop to a single block.

The new transfer function for the jacket temperature loop is given by

Given then, the proportional gain for the controller 2 = 2%/%

Dividing the last expression for giving the proper form we obtain the simplified transfer function

Assuming a proportional slave controller with a gain of 2%CO/%TO, write the transfer function for the jacket temperature loop and redraw the block diagram with the jacket temperature loop as a single block.

=2

2.5 + 10.12 2 1 1

=

22.5 + 1

0.12 2

22.5 + 1

0.12 2 1 + 11

1()=

22.5 + 1

0.12 2

22.5 + 1

0.12 2 1 + 1=

0.48

2.5 + 1.48

1()=

0.324

1.68 + 1

Redrawing the block diagram

%TO

C

0.324

1.68 + 1 1

0.6

13 + 1

2

+ 1

+

-

()

%TO

()

1()

1()

%/co

()

C

Transfer Function of the characteristic polynomial is obtained

Developing

Using the direct substitution method, the values = y = are replaced in the characteristic equation

Using the simplified block diagram from the previous part, write the characteristic equation of the reactor temperature loop in the cascade control system and calculate the ultimate gain and period by direct substitution.

1 +0.388

1.68 + 1 13 + 1 + 1= 0

21.84 3 + 36.522 + 15.68 + 1 + 0.388 = 0

21.843 3 + 36.6522

2 + 15.68 + 1 + 0.388 = 0

Then 2 = 1 separate the real and imaginary parts

From this complex equation, we obtain the following two equations, because both the actual and the imaginary parts must be zero.

36.52 2 + 1 + 0.388 + 15.68 21.84

3 = 0 + 0

36.52 2 + 1 + 0.388 = 0

21.84

3 + 15.68 = 0

If = 0.8473 = 64.995

= 64.995

At this gain, the loop response oscillates with a frequency of 0.846 or a period of

=2

=

2

0.8473= 7.415