Lever system, controls,descrete process control

1. Controlled systems
Lever system
Hydraulic servomotor
Temperature controller
Tank systems

2. Lever systems
b
a
b
x
e



 .
b
a
a
y
e




 .
)
,
( y
x
f
e 
y x

3. Lever systems
y
b
a
a
x
b
a
b
e




)
,
( y
x
f
e 
.....(1)
y
y
e
x
x
e
e
i
i 





The value of is obtained
by finding the ratio of change in e
for a change in x with all other
parameters(y) fixed.
i
x
e 

b
a
b
x
e
x
e
i
i 






lim
b
a
a
y
e
y
e
i
i








lim
Substituting values in equation (1)
y
as
e
because
arises
sign
Minus 

0

e
0

x
0

y
0

e

4. Lever systems
y
b
a
a
x
b
a
b
e




)
,
( y
x
f
e 
y
as
x
because
arises
sign
Minus 

e
+
-
b
a
b

x
b
a
a

y

5. Hydraulic valve and piston
Supply
Sink
Sink
q
q  f(x)
q = C1x
(linearize at Xi for small x)
x
x
Load
Spool
valve
Hydraulic
cylinder
and piston
x
q
x
C1
q
Piston
Xi
x=Xi

6. q = C1x
y= downward movement
of piston and Load
Then
A1Dy=C1x
y
A1
Here x and y are incremental quantities
Hydraulic valve and piston
q
x

7. x
y
e
e=0
Walking
beam
linkage
Hydraulic valve,
piston and lever
Hydraulic
servo
motor

8. Hydraulic servomotor
x
y
e
a b
y
b
a
a
x
b
a
b
e




For the case a=b
2
y
x
e



9. Hydraulic servomotor
x y
e
a a
2
y
x
e



10. 1/2
1/2
x y
e
a a
Block diagram representation
2
y
x
e


-
+ e

11. x y
e
a a
Hydraulic servomotor
System equation-
(1+tD)y(t)=x(t)
This is a 1st order differential Eqn.
Let’s solve this ODE for x = H, a constant.
H

12. x y
e
a a
Hydraulic servomotor
Solution of 1st order ODE
for x = H …..(a constant)
Y(t)
t
H
0.632H
t
0.368

13. Hydraulic servomotor
Y(t)
t
H
0.632H
Slope=H/t
t
t = t 1 - e-1 0.632
t = 4t 1 - e-4 0.982
t = 2t 1 - e-2 0.865
t = 8t 1 - e-8 0.9996
2t 4t
0.865H
0.982H

14. Temperature controller Example

15. Our first control system
• Lets use the concepts in studied in previous
slides and build a control system
• We shall take example of a fuel fired furnace
• In Fuel fired furnace temperature is controlled
by controlling fuel fed at the burner
• Temperature is measured in a closed loop
• Amount of fuel is controlled to maintain the
temperature at required value

16. Sensor: Temperature sensitive bellow
Length =L
Length of bellows is a function of its
temperature
Slope at Tb=Ti

17. Desired
temperature
200
500
800
Tin
Control arm
Pivot
z
Bellows
l
T0
Q0
x
y
e
Ta
Decrease heat
Increase heat Heat source
Temperature
control system
Qin
Fuel gas
valve

18. Desired
temperature
200
500
800
Tin Control arm
Pivot
Z
Bellows
T0
Q0
X
Y
Qin
Decrease heat
Increase heat Heat source
Temperature
control system
L
Tin = T0 = Ti
Ta
+tin
e
+q0
+ta
+qin
+t0
+x
+y
+z
+l
tin = Ti ± dTin
=0
Reference state

19. Desired
temperature
200
500
800
tin
Control arm
Pivot
z
Bellows
t0
q0
x
y
e
qin
ta
Decrease heat
Increase heat Heat source
Temperature
control system
l

20. z=C2tin
l=C3t0
x=z - l
y = f(x)
qin=C4y
Temperature control system
1/2
-
+ e
C2
tin z x
+
-
t0
C3
l
C4
qin
y
Desired
temperature
200
500
800
tin Control arm
Pivot
z
Bellows
l
t0
q0
x
y
e
Decrease heat
Increase heat
qin
ta
Heat source

21. Total heat accumulated
Now Q0= f(T)
Temperature control system
Desired
temperature
200
500
800
tin Control arm
Pivot
z
Bellows
l
t0
q0
x
y
e
Decrease heat
Increase heat
qin
ta
Heat source

22. Temperature control system
C4
y qin
C6
+
ta
+ t0
C6
-
t0
Desired
temperature
200
500
800
tin Control arm
Pivot
z
Bellows
l
t0
q0
x
y
e
Decrease heat
Increase heat
qin
ta
Heat source

23. C2
tin z +
-
t0
C3
l
1/2
-
+ e C4
qin
C4
y qin
C6
+
ta
+ t0
C6
-
t0
Temperature control system
Desired
temperature
200
500
800
tin Control arm
Pivot
z
Bellows
l
t0
q0
x
y
e
Decrease heat
Increase heat
qin
ta
Heat source

24. Desired
temperature
200
500
800
tin Control arm
Pivot
z
Bellows
l
t0
q0
x
y
e
Decrease heat
Increase heat
qin
ta
Heat source
C2
tin z
+
-
1/2
C4
qin
C6
+
ta
t0
C6
t0
t0
C3
l -
+
e
+
-
Temperature control system
z=C2tin
l=C3t0
x=z - l
e=x-y
qin=C4y
1/2
e
Total heat accumulated
Since
C6t0
-

25. Control systems

26. Drain Drain
Supply
Hydraulic servomotor

27. x y y
Drain Drain
Supply Area -A
Load
Passage-1
Passage-2
Hydraulic servomotor
1/2
-
+ e
C1
A1
e= x - y
q = C1e
q= A1Dy
y= (c1/A1D)e
q

28. Hydraulic power amplifier
x
y
Drain Drain
Supply
Piston
a
b
w
1/2
1/2
-
+ e
C1
A1
q
q=C1e
q=A1Dy
y
b
a
b
w



29. a
b
w
x e
B
K
y
To
load
Supply
w
b
a
a
x
b
a
b
e




)
( w
y
BD
Kw 

Hydraulic amplifier
C1
A1
y
BD
K
BD
w


BDy
BDw
Kw 

Write force balance at
q

30. q’
Single tank system
Qout
Area = A1
Qin
H1
Rf2
q’ = qin-qout
q’ = A1Dh1
qout = qin-q’ = qin- A1Dh1 = qin- A1Dqout.Rf2
+h1
+ qout
+ qin
qin qout
…..t = A1Rf2

31. Comparison with Thermal & Electrical systems
RT
CT
Q
Ti
T
C
I
Ei
E
R
Comparing with thermal and electrical
systems. (Solved for T & E0)
 D
C
R
T
DT
C
Q
R
T
T
Q
T
T
i
T
T
i
T





1
D
RC
E
E
CDE
I
dt
I
C
E
R
E
E
I
i
i
)
(
1
.
1







D
RC
E
E
i
)
(
1
1


 D
C
R
T
T
T
i
T 

1
1

32. Liquid level controller
Supply-Ps
Q
y
inc
dec
z e
h0
a
a
h0
Control
lever
H0
Q0
Qin
RF
A2
Kv
C1A1
q’

33. Supply-Ps
Q
y
inc
dec
z e
h0
a
a
h0
Control
lever
H0
Qin
RF
A2
Kv
C1A1
+
- h0
e y
Kv
+
Q0
C3
pS
+ q
+
h0
C2
z
qin
q0
Liquid level controller
-
q0
q’’
q’
z=C2qin
e=(z-h0)/2
y=(C1/A1D) . e
Q= f(Ps, Y)
Linearize at operating point
q=yKv+C3ps
q’=q-q0
q’=A2Dh0
q0=h0/Rf2

34. Flow and level control systems
QE
Area = A1 Qout
Qin
H1
q’
Rf2
In presence of an external flow qe
q’ = qin+ qe- qout
+h1
+ qout
+ qin
+ qe

35. Flow and level control systems
P0 = 0
Capacity = C1
Area = A1
qin, pin
h1
p1
Rf 2
Rf1
Qout
q’
qin = (pin-p1)/Rf1
q’=qin-q0
qout = h1/Rf2
p1=rgDh1

36. Flow and level control systems
Area = A2
Qout
h2
q2
Rf2
Area = A1
Q’
h1
q1
Rf1
q1=qin-q’
q1=A1Dh1
Rf
h
h
q
1
2
1
'


q2=q’- qout
q2=A2Dh2
Rf
h
qout
2
2

Qin

37. Flow and level control systems
h1
h1-h2
q’
qin
+
-
q’
q1
h2
q0
Area = A2
qo
h2
q2
Rf2
Area = A1
q’
qin
h1
q1
Rf1
+-
h2
+
-
q2
q0

38. Flapper nozzle
Nozzle
Fluid bleeds
through
auxiliary port
Output pressure
p0
pS
R
x
Flapper can be
moved axially
P0 < pS
Supply
fluid
p0 = Kx

39. K
A
C
R
qair
x
Kv
q0
Pneumatic control valve
pi
p0
pi q0

40. Liquid flow controller
Inc
dec
Qin
Flow control lever
K1 z
x
e=½ x
Diaphragm
p1
p2
Ad
y
Q0
Supply-Ps
A1 C1
Kv
fd
fd=K1(z-x)

41. Tension regulator
Tension
control
spring
Tension
control
lever
y
y
Fc Fc
Tm
Windup roll
Idler roll
R
x
Torque
control arm
Fr
Inc.
Dec.
Motor
K
a a
e
Tm=FCR
Fc

42. Electric speed control system
Motor
Tachometer
TL
JD2q
Field
Lf
Rf
If
Ef
Ampli
fier
Ka
Er=KrNin
EC=KCN0
q
Nin(RPM)
Input
potentiometer
N0
(RPM)
T=KmIf
The torque t produced by a motor
minus load torque tL is the net
torque available for acceleration
That is t-tL = Ja
JDN0
And the external load torque
Kr
nin +
-
Kc
n0
ef
Ka
Zf = Rf+LfD

43. Block diagram
ef
Ka
if
Km
t
tL
-
+ n0
t-tL =Ja = JDw
or (T-tL)/JD = w ….(in rad/s)
Convert to RPM
Kr
nin +
-
Kc
n0

44. b1
b2
y
z
u
a3
a1
a2
w
x
e
C1
A1
C2 A2
Pivots here
Pin
joints
hydraulic controller
y

45. Block diagram of
hydraulic controller
+
-u
y
+
+
w
x
e z
+
- z
e
z
w
a
a
a
a
a
a
a
a
2
1
3
2
1
3
2
1







46. q’
Single tank level control system
Qout
Area = A1
Qin
Rf2
H1
q’ = qin-qout
q’ = A1Dh1
h
+ qout
+ qin
qin
QE+ qe
+h1
kv
y
Required height=Hi
Actual height=H1
At steady state, H1=Hi
And other parameters are
Qin and Qout
Now we change Hi to Hi + hi
Corresponding changes are-
H1  H1+h1,
Qin  Qin+qin
Instantaneous error = hi - h1
h acts through valve to effect qin
= h
a
b
Now draw the block diagram
Let’s add a disturbance flow Qe
Qout  Qout+qout
Hi +hi

47. q’
Single tank level control with disturbance
Qout
Area = A1
Qin
Rf2
H1
q’ = A1Dh1
h
qin
QE+ qe
+h1
kv
y
a
b
Hi +hi
q’ =
Increased Qe should
result in increased Qout
qe -qin
-qout
and decreased Qin

48. pc
PS x
a
b
Pivot
Area=Ab
C
K K
R
Bellows and flapper nozzle
+
-
x
+
-
G
pc
-
+
pb
e
pb
qb
Supply pressure
Large resistance to flow is an
integral property of flapper nozzle
In flapper nozzle, supply pressure Ps
does not figure in flapper nozzle
equation
Bellows
Flapper nozzle
y
y
pc
Pc = G . x
pb
qb
pc
pc
For Bellows,
Pc.Ab = K.x Pc Pc
K
x
Ab

49. x = x1-x2
Main air
supply
qc
cam Guide roller
x0
Mass
M
Power
cylinder
Area, AC
Capacity, CC
Spring, K
Bellows
Area, Ab
Capacity, Cb
Pipeline
resistance, R qb Input pressure, Pi
Pressure
in bellows
pb
Bell
crank
ratio b/a
a
Force=PbAb
x2
x1
pC
Friction
force, ff Dx0
Pilot valve, Kv
Sf=MD2x0
Upper part of spring is
rigidly connected to the
bottom of bellows
assembly. Feedback
from bell crank
moves the entire spring
and bellows assembly
downwards by x2

50. To process
From
process
Set point
p0
G(D)
q p0
qi
q0
q
qi
q0
q
G(D)
p0

51. Summary: control systems
• Electric motor control (speed, torque, inertial
load, field resistance & inductance, tacho)
• Hydraulic positioning systems (spool valve,
cylinder, area, inertia)
• Pneumatic positioning systems (valve, cylinder,
pressure, capacity, area, flapper, friction, inertia)
• Single tank and dual tank level/flow control
systems (flapper, pneumatic valve, disturbance
flow)
• Heat flow (temperature, coupled conductors,
capacity, heat transfer coefficients, heat losses)

52. Block diagram reduction
To reduce a complex block diagram to a single
block representing the transfer function
G1 G2
X1
X1G1 X1G1G2
G1 G2
X1
X1G1G2

53. Next PPT 
5. Pneumatic controller
6. Block diagram reduction