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Industrial Control Systems - Hydraulic Systems
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Transcript of Industrial Control Systems - Hydraulic Systems
Industrial Control
Behzad Samadi
Department of Electrical EngineeringAmirkabir University of Technology
Winter 2010Tehran, Iran
Behzad Samadi (Amirkabir University) Industrial Control 1 / 17
Hydraulic Systems
Electrical Analogy
Type of System Electrical HydraulicT-Variable i , current q, volumetric flow
A-Variable v , voltage p, pressure
Dissipator resistor orifice
Storage (A-Type) capacitor storage tank
Storage (T-Type) inductor long pipe
Unidirectional diode check valve
The fluid is assumed to be incompressible.
[Macia and Thaler, 2004, Ljung and Glad, 1994]
Behzad Samadi (Amirkabir University) Industrial Control 2 / 17
Hydraulic Systems
Electrical Analogy
Type of System Electrical HydraulicT-Variable i , current q, volumetric flow
A-Variable v , voltage p, pressure
Dissipator resistor orifice
Storage (A-Type) capacitor storage tank
Storage (T-Type) inductor long pipe
Unidirectional diode check valve
The fluid is assumed to be incompressible.
[Macia and Thaler, 2004, Ljung and Glad, 1994]
Behzad Samadi (Amirkabir University) Industrial Control 2 / 17
Hydraulic Dissipator
d’Arcy’s Law
For a thin tube:p = Rf q
For a sudden change in area, such as an orifice or valve:
p = ℋq2 sgn(q)
ℋ is a constant.
[Ljung and Glad, 1994]
Behzad Samadi (Amirkabir University) Industrial Control 3 / 17
Hydraulic Dissipator
d’Arcy’s Law
For a thin tube:p = Rf q
For a sudden change in area, such as an orifice or valve:
p = ℋq2 sgn(q)
ℋ is a constant.
[Ljung and Glad, 1994]
Behzad Samadi (Amirkabir University) Industrial Control 3 / 17
Hydraulic Capacitor
p =mg
A=
�g
AV =
�g
A
∫ t
0q(�)d�
p =pressure at the bottom of the tank
V =volume of the fluid in tank
A =cross section area of the tank
� =density of the fluid
Hydraulic Capacitor
Cf = A�g
[Ljung and Glad, 1994]
Behzad Samadi (Amirkabir University) Industrial Control 4 / 17
Hydraulic Capacitor
p =mg
A=
�g
AV =
�g
A
∫ t
0q(�)d�
p =pressure at the bottom of the tank
V =volume of the fluid in tank
A =cross section area of the tank
� =density of the fluid
Hydraulic Capacitor
Cf = A�g
[Ljung and Glad, 1994]
Behzad Samadi (Amirkabir University) Industrial Control 4 / 17
Hydraulic Capacitor
p =mg
A=
�g
AV =
�g
A
∫ t
0q(�)d�
p =pressure at the bottom of the tank
V =volume of the fluid in tank
A =cross section area of the tank
� =density of the fluid
Hydraulic Capacitor
Cf = A�g
[Ljung and Glad, 1994]
Behzad Samadi (Amirkabir University) Industrial Control 4 / 17
Hydraulic Inductor
F =ma = (�lA)dv
dt
p =F
A= �l
dv
dtq =Av
⇒ p =�l
A
dq
dt= Lf
dq
dt
Hydraulic Inductor
Lf = � lA
[Ljung and Glad, 1994]
Behzad Samadi (Amirkabir University) Industrial Control 5 / 17
Hydraulic Inductor
F =ma = (�lA)dv
dt
p =F
A= �l
dv
dtq =Av
⇒ p =�l
A
dq
dt= Lf
dq
dt
Hydraulic Inductor
Lf = � lA
[Ljung and Glad, 1994]
Behzad Samadi (Amirkabir University) Industrial Control 5 / 17
Hydraulic Inductor
F =ma = (�lA)dv
dt
p =F
A= �l
dv
dtq =Av
⇒ p =�l
A
dq
dt= Lf
dq
dt
Hydraulic Inductor
Lf = � lA
[Ljung and Glad, 1994]
Behzad Samadi (Amirkabir University) Industrial Control 5 / 17
Hydraulic Inductor
F =ma = (�lA)dv
dt
p =F
A= �l
dv
dtq =Av
⇒ p =�l
A
dq
dt= Lf
dq
dt
Hydraulic Inductor
Lf = � lA
[Ljung and Glad, 1994]
Behzad Samadi (Amirkabir University) Industrial Control 5 / 17
Hydraulic Junction
∑i Qi = 0 (KCL)
p4 = p1 + p2 + p3 (KVL)
[Ljung and Glad, 1994]
Behzad Samadi (Amirkabir University) Industrial Control 6 / 17
Hydraulic Junction
∑i Qi = 0 (KCL) p4 = p1 + p2 + p3 (KVL)
[Ljung and Glad, 1994]
Behzad Samadi (Amirkabir University) Industrial Control 6 / 17
Hydraulic Transformer
p1Q1 =p2Q2
p1 =�p2
Q1 =1
�Q2
� =A2
A1
[Ljung and Glad, 1994]
Behzad Samadi (Amirkabir University) Industrial Control 7 / 17
Hydraulic Transformer
p1Q1 =p2Q2
p1 =�p2
Q1 =1
�Q2
� =A2
A1
[Ljung and Glad, 1994]
Behzad Samadi (Amirkabir University) Industrial Control 7 / 17
Hydraulic Servomechanism
Behzad Samadi (Amirkabir University) Industrial Control 8 / 17
Hydraulic Servomechanism
xm: maximum displacement
x = xm ⇒{
Pa = Ps
Pb = P0 = 0
x = −xm ⇒{
Pa = P0 = 0Pb = Ps
Pa = 12 ( x
xm+ 1)Ps
Pb = 12 (− x
xm+ 1)Ps
Pa − P1 = Rqq
Pb − P2 = Rqq
q = Ay
Behzad Samadi (Amirkabir University) Industrial Control 9 / 17
Hydraulic Servomechanism
xm: maximum displacement
x = xm ⇒{
Pa = Ps
Pb = P0 = 0
x = −xm ⇒{
Pa = P0 = 0Pb = Ps
Pa = 12 ( x
xm+ 1)Ps
Pb = 12 (− x
xm+ 1)Ps
Pa − P1 = Rqq
Pb − P2 = Rqq
q = Ay
Behzad Samadi (Amirkabir University) Industrial Control 9 / 17
Hydraulic Servomechanism
xm: maximum displacement
x = xm ⇒{
Pa = Ps
Pb = P0 = 0
x = −xm ⇒{
Pa = P0 = 0Pb = Ps
Pa = 12 ( x
xm+ 1)Ps
Pb = 12 (− x
xm+ 1)Ps
Pa − P1 = Rqq
Pb − P2 = Rqq
q = Ay
Behzad Samadi (Amirkabir University) Industrial Control 9 / 17
Hydraulic Servomechanism
xm: maximum displacement
x = xm ⇒{
Pa = Ps
Pb = P0 = 0
x = −xm ⇒{
Pa = P0 = 0Pb = Ps
Pa = 12 ( x
xm+ 1)Ps
Pb = 12 (− x
xm+ 1)Ps
Pa − P1 = Rqq
Pb − P2 = Rqq
q = Ay
Behzad Samadi (Amirkabir University) Industrial Control 9 / 17
Hydraulic Servomechanism
xm: maximum displacement
x = xm ⇒{
Pa = Ps
Pb = P0 = 0
x = −xm ⇒{
Pa = P0 = 0Pb = Ps
Pa = 12 ( x
xm+ 1)Ps
Pb = 12 (− x
xm+ 1)Ps
Pa − P1 = Rqq
Pb − P2 = Rqq
q = Ay
Behzad Samadi (Amirkabir University) Industrial Control 9 / 17
Hydraulic Servomechanism
xm: maximum displacement
x = xm ⇒{
Pa = Ps
Pb = P0 = 0
x = −xm ⇒{
Pa = P0 = 0Pb = Ps
Pa = 12 ( x
xm+ 1)Ps
Pb = 12 (− x
xm+ 1)Ps
Pa − P1 = Rqq
Pb − P2 = Rqq
q = Ay
Behzad Samadi (Amirkabir University) Industrial Control 9 / 17
Hydraulic Servomechanism
xm: maximum displacement
x = xm ⇒{
Pa = Ps
Pb = P0 = 0
x = −xm ⇒{
Pa = P0 = 0Pb = Ps
Pa = 12 ( x
xm+ 1)Ps
Pb = 12 (− x
xm+ 1)Ps
Pa − P1 = Rqq
Pb − P2 = Rqq
q = Ay
Behzad Samadi (Amirkabir University) Industrial Control 9 / 17
Hydraulic Servomechanism
xm: maximum displacement
x = xm ⇒{
Pa = Ps
Pb = P0 = 0
x = −xm ⇒{
Pa = P0 = 0Pb = Ps
Pa = 12 ( x
xm+ 1)Ps
Pb = 12 (− x
xm+ 1)Ps
Pa − P1 = Rqq
Pb − P2 = Rqq
q = Ay
Behzad Samadi (Amirkabir University) Industrial Control 9 / 17
Hydraulic Servomechanism
My + By + Ky = A(P1 − P2)
P1 + P2 = Ps ⇒ P2 = Ps − P1
P1 = Pa − Rqq = 12 ( x
xm+ 1)Ps − RqAy
My + By + Ky =A(P1 − P2)
=A(2P1 − Ps)
=A((x
xm+ 1)Ps − 2RqAy − Ps)
My + (B + 2RqA2)y + Ky = APs
x
xm
Y (s)
X (s)=
APsxm
Ms2 + (B + 2RqA2)s + K
Behzad Samadi (Amirkabir University) Industrial Control 10 / 17
Hydraulic Servomechanism
My + By + Ky = A(P1 − P2)
P1 + P2 = Ps ⇒ P2 = Ps − P1
P1 = Pa − Rqq = 12 ( x
xm+ 1)Ps − RqAy
My + By + Ky =A(P1 − P2)
=A(2P1 − Ps)
=A((x
xm+ 1)Ps − 2RqAy − Ps)
My + (B + 2RqA2)y + Ky = APs
x
xm
Y (s)
X (s)=
APsxm
Ms2 + (B + 2RqA2)s + K
Behzad Samadi (Amirkabir University) Industrial Control 10 / 17
Hydraulic Servomechanism
My + By + Ky = A(P1 − P2)
P1 + P2 = Ps ⇒ P2 = Ps − P1
P1 = Pa − Rqq = 12 ( x
xm+ 1)Ps − RqAy
My + By + Ky =A(P1 − P2)
=A(2P1 − Ps)
=A((x
xm+ 1)Ps − 2RqAy − Ps)
My + (B + 2RqA2)y + Ky = APs
x
xm
Y (s)
X (s)=
APsxm
Ms2 + (B + 2RqA2)s + K
Behzad Samadi (Amirkabir University) Industrial Control 10 / 17
Hydraulic Servomechanism
My + By + Ky = A(P1 − P2)
P1 + P2 = Ps ⇒ P2 = Ps − P1
P1 = Pa − Rqq = 12 ( x
xm+ 1)Ps − RqAy
My + By + Ky =A(P1 − P2)
=A(2P1 − Ps)
=A((x
xm+ 1)Ps − 2RqAy − Ps)
My + (B + 2RqA2)y + Ky = APs
x
xm
Y (s)
X (s)=
APsxm
Ms2 + (B + 2RqA2)s + K
Behzad Samadi (Amirkabir University) Industrial Control 10 / 17
Hydraulic Servomechanism
My + By + Ky = A(P1 − P2)
P1 + P2 = Ps ⇒ P2 = Ps − P1
P1 = Pa − Rqq = 12 ( x
xm+ 1)Ps − RqAy
My + By + Ky =A(P1 − P2)
=A(2P1 − Ps)
=A((x
xm+ 1)Ps − 2RqAy − Ps)
My + (B + 2RqA2)y + Ky = APs
x
xm
Y (s)
X (s)=
APsxm
Ms2 + (B + 2RqA2)s + K
Behzad Samadi (Amirkabir University) Industrial Control 10 / 17
Hydraulic Servomechanism
My + By + Ky = A(P1 − P2)
P1 + P2 = Ps ⇒ P2 = Ps − P1
P1 = Pa − Rqq = 12 ( x
xm+ 1)Ps − RqAy
My + By + Ky =A(P1 − P2)
=A(2P1 − Ps)
=A((x
xm+ 1)Ps − 2RqAy − Ps)
My + (B + 2RqA2)y + Ky = APs
x
xm
Y (s)
X (s)=
APsxm
Ms2 + (B + 2RqA2)s + K
Behzad Samadi (Amirkabir University) Industrial Control 10 / 17
Hydraulic Servomechanism
My + By + Ky = A(P1 − P2)
P1 + P2 = Ps ⇒ P2 = Ps − P1
P1 = Pa − Rqq = 12 ( x
xm+ 1)Ps − RqAy
My + By + Ky =A(P1 − P2)
=A(2P1 − Ps)
=A((x
xm+ 1)Ps − 2RqAy − Ps)
My + (B + 2RqA2)y + Ky = APs
x
xm
Y (s)
X (s)=
APsxm
Ms2 + (B + 2RqA2)s + K
Behzad Samadi (Amirkabir University) Industrial Control 10 / 17
Hydraulic Integrator
If M = K = B = 0 then
Y (s)
X (s)=
APs
2xmRgA2
1
s
Hydraulic integrator
[Ogata, 1997]Behzad Samadi (Amirkabir University) Industrial Control 11 / 17
Hydraulic Integrator
If M = K = B = 0 then
Y (s)
X (s)=
APs
2xmRgA2
1
s
Hydraulic integrator
[Ogata, 1997]Behzad Samadi (Amirkabir University) Industrial Control 11 / 17
Hydraulic Proportional Controller
Y (s)
E (s)=
bK
(a + b)s + aK≃ b
a
[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 12 / 17
Hydraulic Proportional Controller
Y (s)
E (s)=
bK
(a + b)s + aK≃ b
a
[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 12 / 17
Hydraulic Damper
A(P1 − P2) = ky
q = P1−P2R
qdt = A�(dx − dy)
Y (s)
X (s)=
1
1 + 1Ts
[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 13 / 17
Hydraulic Damper
A(P1 − P2) = ky
q = P1−P2R
qdt = A�(dx − dy)
Y (s)
X (s)=
1
1 + 1Ts
[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 13 / 17
Hydraulic Damper
A(P1 − P2) = ky
q = P1−P2R
qdt = A�(dx − dy)
Y (s)
X (s)=
1
1 + 1Ts
[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 13 / 17
Hydraulic Damper
A(P1 − P2) = ky
q = P1−P2R
qdt = A�(dx − dy)
Y (s)
X (s)=
1
1 + 1Ts
[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 13 / 17
Hydraulic Damper
A(P1 − P2) = ky
q = P1−P2R
qdt = A�(dx − dy)
Y (s)
X (s)=
1
1 + 1Ts
[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 13 / 17
Hydraulic Damper
A(P1 − P2) = ky
q = P1−P2R
qdt = A�(dx − dy)
Y (s)
X (s)=
1
1 + 1Ts
[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 13 / 17
Hydraulic Proportional Integrator Controller
Y (s)X (s) = K
s
Z(s)Y (s) = 1
1+ 1Ts
Y (s)
E (s)≃ b
a
(1 +
1
Ts
)[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 14 / 17
Hydraulic Proportional Integrator Controller
Y (s)X (s) = K
s
Z(s)Y (s) = 1
1+ 1Ts Y (s)
E (s)≃ b
a
(1 +
1
Ts
)[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 14 / 17
Hydraulic Proportional Integrator Controller
Y (s)X (s) = K
s
Z(s)Y (s) = 1
1+ 1Ts
Y (s)
E (s)≃ b
a
(1 +
1
Ts
)[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 14 / 17
Hydraulic Proportional Integrator Controller
Y (s)X (s) = K
s
Z(s)Y (s) = 1
1+ 1Ts
Y (s)
E (s)≃ b
a
(1 +
1
Ts
)[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 14 / 17
Hydraulic Proportional Integrator Controller
Y (s)X (s) = K
s
Z(s)Y (s) = 1
1+ 1Ts Y (s)
E (s)≃ b
a
(1 +
1
Ts
)[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 14 / 17
Hydraulic Proportional Derivative Controller
Y (s)X (s) = K
s
k(y − z) = A(P1 − P2)Z(s)Y (s) = 1
Ts+1
Y (s)
E (s)≃ b
a(1 + Ts)
[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 15 / 17
Hydraulic Proportional Derivative Controller
Y (s)X (s) = K
s
k(y − z) = A(P1 − P2)Z(s)Y (s) = 1
Ts+1Y (s)
E (s)≃ b
a(1 + Ts)
[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 15 / 17
Hydraulic Proportional Derivative Controller
Y (s)X (s) = K
s
k(y − z) = A(P1 − P2)
Z(s)Y (s) = 1
Ts+1Y (s)
E (s)≃ b
a(1 + Ts)
[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 15 / 17
Hydraulic Proportional Derivative Controller
Y (s)X (s) = K
s
k(y − z) = A(P1 − P2)Z(s)Y (s) = 1
Ts+1
Y (s)
E (s)≃ b
a(1 + Ts)
[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 15 / 17
Hydraulic Proportional Derivative Controller
Y (s)X (s) = K
s
k(y − z) = A(P1 − P2)Z(s)Y (s) = 1
Ts+1
Y (s)
E (s)≃ b
a(1 + Ts)
[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 15 / 17
Hydraulic Proportional Derivative Controller
Y (s)X (s) = K
s
k(y − z) = A(P1 − P2)Z(s)Y (s) = 1
Ts+1
Y (s)
E (s)≃ b
a(1 + Ts)
[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 15 / 17
Hydraulic Proportional Derivative Controller
Y (s)X (s) = K
s
k(y − z) = A(P1 − P2)Z(s)Y (s) = 1
Ts+1Y (s)
E (s)≃ b
a(1 + Ts)
[Ogata, 1997]
Behzad Samadi (Amirkabir University) Industrial Control 15 / 17
Comparison
Electrical Hydraulic PneumaticEnergy source Usually from outside
supplierElectric motor or dieseldriven
Electric motor or dieseldriven
Energy storage Limited (batteries) Limited (accumulator) Good (reservoir)
Distribution system Excellent, with minimalloss
Limited, basically a lo-cal facility
Good, can be treated asa plantwide service
Energy cost Lowest Medium Highest
Rotary actuators AC and DC motors.Good control on DCmotors. AC motorscheap
Low speed. Good con-trol. Can be stalled.
Wide speed range. Ac-curate speed controldifficult
Linear actuators Short motion viasolenoid. Otherwise viamechanical conversion
Cylinders. Very highforce
Cylinders. Mediumforce
Points to note Danger from electricshock
Leakage dangerous andunsightly. Fire hazard
Noise
[Parr, 1999]
Behzad Samadi (Amirkabir University) Industrial Control 16 / 17
Analogy Summary
[Macia and Thaler, 2004]
Behzad Samadi (Amirkabir University) Industrial Control 17 / 17
Ljung, L. and Glad, T. (1994).Modeling of Dynamic Systems.Prentice Hall PTR, 1 edition.
Macia, N. F. and Thaler, G. J. (2004).Modeling and Control of Dynamic Systems.Delmar Learning.
Ogata, K. (1997).Modern Control Engineering.Prentice Hall, 3 edition.
Parr, A. (1999).Hydraulics and Pneumatics: A Technicians and Engineers Guide.Butterworth-Heinemann, 2 edition.
Behzad Samadi (Amirkabir University) Industrial Control 17 / 17