Process Control

Process Control Example Ruel G. Peneyra PROBLEM # 11.1. In the process shown in Fig. Pll. 1, the concentration of salt leaving the second tank is controlled using a proportional controller by adding concentrated solution through a control valve. The following data apply: 1. The controlled concentration is to be 0.1 lb salt/ft 3 solution. The inlet concentration ci is always less than 0.1 lb/ft3. 2. The concentration of concentrated salt solution is 30 lb salt/ft 3 solution. 3. Transducer: the output of the transducer varies linearly from 3 to 15 psig as the concentration varies from 0.05 to 0.15 lb/ft3. 4. Controller: the controller is a pneumatic, direct-acting, proportional controller. 5. Control valve: as valve-top pressure varies from 3 to 15 psig; the flow through the control valve varies linearly from 0 to 0.005 cfm. 6. It takes 30 set for the solution leaving the second tank to reach the transducer at the end of the pipe.

Tank A

Tank B

Draw a block diagram of the control system. Place in each block the appropriate transfer function. Calculate all the constants and give the units.

1

SOLUTION: Mass balance on Tank A at any time t V1

dc 1 dt

Fci

Fc1 V 1 c1

V 1 dc 1 F V 1 dt

F c F V1 i

c1

V1



let

1

dc 1 dt

Fc1 V 1 c1 Fc i

V1 F V1

1

dc 1 dt

1

c1

ci

V1 F

1

At steady state

using

1

dc 1 dt

1

c1

ci

V1 F

1

the steady state eqn

1

dc 1s dt

1

c 1s

c is

V1 F

1

Equating

d c 1 c 1s dt 1

dC 1 dt

C1

c1 1 1

1

c 1s Ci

V1 F

ci

V1 F

1

c is

using deviation C 1 c 1

transforming

1

sC 1 s

C1 0

c 1s andC i 1

C1 s

c is

Ci s

V1 F

1

ci

1

since C 1 0

0:

1

sC 1 s

since V 3 m and F 1 3

ft

1

C1 s

3

min

:

1 3 4

1

V1 F

then

Ci s

then

C1 s Ci s

1

C1 s Ci s

1 1 3 3 s 1 4

or

1

V1 F

s 1

C1 s Ci s

1 3s

4

At Tank B

V2

if

dc 2 dt

m A

Fc1

m

Fc2 V 2 c2

is far less than F

m A

m A

rearranging V 2

dc 2 dt

Fc2 V 2 c2

is negligible , dividing by F V 2 and setting

m A

V2 F V2

Fc1

m

2

2

2

dc 2 dt

F c F V2 1

c2

1

m

F V2

2

dc 2 dt

c2

1

1

V 2 c1 F

1

m

V2 F

F 1

At steady state 2

dc 2s dt

1

c 2s

1

V 2 c 1s

1

F 1

F

m s equating with the conditions at anytime t

V2 F

and using deviation variablesC 2 c 2 c 2s ; C 1 c 1 c 1s ; M m ms 2

dC 2 dt

1

C2

1

1

V2 C1 F

V2 F

F 1

M transforming

2

s

1

1 C2 s

1

V2 C1 s F

1

F 1

V2 F

M s

Finally 1

1

C2 s

V2 1 F 2

s 1

C1 s

F 1 2

V2 F

s 1

M s

sub the given C 2 S

1 5

C1 s 4 5

s

M s 1

Evaluating m (mass of concentrate) from steady state condition of Tank A

Fc is

Fc 1s V 1 c 1s

At tank B neglecting

Fc 1s if c 1s

ms

ms A

Fc 2s V 2 c 2s

1 c then 4 is

F c F V 1 1s

c 1s

ms

and setting c 2s

ms A

c 2s

0.5

ms

1

c 1s

1

3

c is

c 1s

1 c 4 is

0.1

Fc 2s V 2 c 2s

Fc 1s

ms

1 0.1

4 0.1

c 1s

0.25c is

CONTROL VALVE Sensitivity Kv

Kv

0.005 0 15 3

Kv

1 cfm 2400 psi

since

ms A

1 m s then p s 30

3

1 15 3 ms 30 0.005 0 3

ps

Hence the normal operating pressure is

m

1 ms 30

1 ms 30

K v ps

Valve equation is therefore

by steady state

ms

and by deviation variables M m ms where M m

1 m s and P p 30

taking the transform of M K v

A

3

80 m s

Kv p

3

3

80m s

and P p

80 m s

A

A

equating with the eqn at any time t

ps then

M Kv

A

P

80m s

3

M s P s

P

Kv

valve transfer function

A

TRANSDUCER Sensitivity Km

Km

15 3 0.15 0.05

Km

psi 120 lb3

ft

Normal signal = 0.1 0.05 15 0.15 0.05 Transducer equation 

3

3 9 psig

b 9 K m c2

By deviation variables  B

K mC2

0.1

where B b bs

By Laplace the transducer transfer function 

B s C2 s

and C 2

c 2 c 2s

Km

4

CONTROLLER Controller output pressure (p) and error relationship

p pS

K c cR

b

ps

Kc

Where

c RS

b 9 psig ; K c

controller sensitivity ;

error c R

b, psig and C R c r

c RS

Then

P Kc

and the transfer function of the controller by laplace

P s s

Kc

TRANSPORT LAG

d

30 min 0.5 min 60

Hence the transfer function is therefore

e

ds

e

0.5 s

CONSTANTS AT cis < 0.1

c is

ms

ps

1

ms

c 2s

c 1s

2

m at p=ps lb/min

A

lb/cu.ft 0.10 0.09 0.08 0.07 0.05 0.06 0.04 0.03 0.02 0.01 0.00

lb/min 0.48 0.48 0.48 0.48 0.49 0.49 0.49 0.49 0.50 0.50 0.50

psig 41.00 41.20 41.40 41.60 42.00 41.80 42.20 42.40 42.60 42.80 43.00

min 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75 0.75

min 0.80 0.80 0.80 0.80 0.80 0.80 0.80 0.80 0.80 0.80 0.80

lb/cu.ft

lb/cu.ft

0.03 0.02 0.02 0.02 0.01 0.02 0.01 0.01 0.01 0.00 0.00

0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10

cfm 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02

0.48 0.48 0.48 0.48 0.49 0.49 0.49 0.49 0.50 0.50 0.50

5

BLOCK DIAGRAM

1

C1

CR CR



Km

 psig

M

P +

 -

Kc



B

Kv

A

1



F

M F



3s

Ci

4 1

+

1



+

2

Km

e

s

v2 F

1

 C2



C2



0.5 s

psig OR

Ci

1



3s

4

C1

CR



+



KK c



1

+



+

2

-

e

1

s

v2 F

 1

0.5 s

6