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# Industrial Control Systems - Hydraulic Systems

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### Industrial Control Systems - Hydraulic Systems

1. 1. Industrial Control Behzad Samadi Department of Electrical Engineering Amirkabir University of Technology Winter 2010 Tehran, Iran Behzad Samadi (Amirkabir University) Industrial Control 1 / 17
2. 2. Hydraulic Systems Electrical Analogy Type of System Electrical Hydraulic T-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 Behzad Samadi (Amirkabir University) Industrial Control 2 / 17
3. 3. Hydraulic Systems Electrical Analogy Type of System Electrical Hydraulic T-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
4. 4. Hydraulic Dissipator d’Arcy’s Law For a thin tube: p = Rf q Behzad Samadi (Amirkabir University) Industrial Control 3 / 17
5. 5. 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
6. 6. Hydraulic Capacitor Behzad Samadi (Amirkabir University) Industrial Control 4 / 17
7. 7. Hydraulic Capacitor p = mg A = 𝜌g A V = 𝜌g A ∫ t 0 q(𝜏)d𝜏 p =pressure at the bottom of the tank V =volume of the ﬂuid in tank A =cross section area of the tank 𝜌 =density of the ﬂuid Behzad Samadi (Amirkabir University) Industrial Control 4 / 17
8. 8. Hydraulic Capacitor p = mg A = 𝜌g A V = 𝜌g A ∫ t 0 q(𝜏)d𝜏 p =pressure at the bottom of the tank V =volume of the ﬂuid in tank A =cross section area of the tank 𝜌 =density of the ﬂuid Hydraulic Capacitor Cf = A 𝜌g [Ljung and Glad, 1994] Behzad Samadi (Amirkabir University) Industrial Control 4 / 17
9. 9. Hydraulic Inductor Behzad Samadi (Amirkabir University) Industrial Control 5 / 17
10. 10. Hydraulic Inductor F =ma = (𝜌lA) dv dt p = F A = 𝜌l dv dt q =Av Behzad Samadi (Amirkabir University) Industrial Control 5 / 17
11. 11. Hydraulic Inductor F =ma = (𝜌lA) dv dt p = F A = 𝜌l dv dt q =Av ⇒ p =𝜌 l A dq dt = Lf dq dt Behzad Samadi (Amirkabir University) Industrial Control 5 / 17
12. 12. Hydraulic Inductor F =ma = (𝜌lA) dv dt p = F A = 𝜌l dv dt q =Av ⇒ p =𝜌 l A dq dt = Lf dq dt Hydraulic Inductor Lf = 𝜌 l A [Ljung and Glad, 1994] Behzad Samadi (Amirkabir University) Industrial Control 5 / 17
13. 13. Hydraulic Junction ∑ i Qi = 0 (KCL) Behzad Samadi (Amirkabir University) Industrial Control 6 / 17
14. 14. Hydraulic Junction ∑ i Qi = 0 (KCL) p4 = p1 + p2 + p3 (KVL) [Ljung and Glad, 1994] Behzad Samadi (Amirkabir University) Industrial Control 6 / 17
15. 15. Hydraulic Transformer Behzad Samadi (Amirkabir University) Industrial Control 7 / 17
16. 16. Hydraulic Transformer p1Q1 =p2Q2 p1 =𝛼p2 Q1 = 1 𝛼 Q2 𝛼 = A2 A1 [Ljung and Glad, 1994] Behzad Samadi (Amirkabir University) Industrial Control 7 / 17
17. 17. Hydraulic Servomechanism Behzad Samadi (Amirkabir University) Industrial Control 8 / 17
18. 18. Hydraulic Servomechanism xm: maximum displacement Behzad Samadi (Amirkabir University) Industrial Control 9 / 17
19. 19. Hydraulic Servomechanism xm: maximum displacement x = xm ⇒ { Pa = Ps Pb = P0 = 0 Behzad Samadi (Amirkabir University) Industrial Control 9 / 17
20. 20. Hydraulic Servomechanism xm: maximum displacement x = xm ⇒ { Pa = Ps Pb = P0 = 0 x = −xm ⇒ { Pa = P0 = 0 Pb = Ps Behzad Samadi (Amirkabir University) Industrial Control 9 / 17
21. 21. Hydraulic Servomechanism xm: maximum displacement x = xm ⇒ { Pa = Ps Pb = P0 = 0 x = −xm ⇒ { Pa = P0 = 0 Pb = Ps Pa = 1 2( x xm + 1)Ps Behzad Samadi (Amirkabir University) Industrial Control 9 / 17
22. 22. Hydraulic Servomechanism xm: maximum displacement x = xm ⇒ { Pa = Ps Pb = P0 = 0 x = −xm ⇒ { Pa = P0 = 0 Pb = Ps Pa = 1 2( x xm + 1)Ps Pb = 1 2(− x xm + 1)Ps Behzad Samadi (Amirkabir University) Industrial Control 9 / 17
23. 23. Hydraulic Servomechanism xm: maximum displacement x = xm ⇒ { Pa = Ps Pb = P0 = 0 x = −xm ⇒ { Pa = P0 = 0 Pb = Ps Pa = 1 2( x xm + 1)Ps Pb = 1 2(− x xm + 1)Ps Pa − P1 = Rqq Behzad Samadi (Amirkabir University) Industrial Control 9 / 17
24. 24. Hydraulic Servomechanism xm: maximum displacement x = xm ⇒ { Pa = Ps Pb = P0 = 0 x = −xm ⇒ { Pa = P0 = 0 Pb = Ps Pa = 1 2( x xm + 1)Ps Pb = 1 2(− x xm + 1)Ps Pa − P1 = Rqq Pb − P2 = Rqq Behzad Samadi (Amirkabir University) Industrial Control 9 / 17
25. 25. Hydraulic Servomechanism xm: maximum displacement x = xm ⇒ { Pa = Ps Pb = P0 = 0 x = −xm ⇒ { Pa = P0 = 0 Pb = Ps Pa = 1 2( x xm + 1)Ps Pb = 1 2(− x xm + 1)Ps Pa − P1 = Rqq Pb − P2 = Rqq q = A˙y Behzad Samadi (Amirkabir University) Industrial Control 9 / 17
26. 26. Hydraulic Servomechanism M¨y + B ˙y + Ky = A(P1 − P2) Behzad Samadi (Amirkabir University) Industrial Control 10 / 17
27. 27. Hydraulic Servomechanism M¨y + B ˙y + Ky = A(P1 − P2) P1 + P2 = Ps ⇒ P2 = Ps − P1 Behzad Samadi (Amirkabir University) Industrial Control 10 / 17
28. 28. Hydraulic Servomechanism M¨y + B ˙y + Ky = A(P1 − P2) P1 + P2 = Ps ⇒ P2 = Ps − P1 P1 = Pa − Rqq = 1 2( x xm + 1)Ps − RqA˙y Behzad Samadi (Amirkabir University) Industrial Control 10 / 17
29. 29. Hydraulic Servomechanism M¨y + B ˙y + Ky = A(P1 − P2) P1 + P2 = Ps ⇒ P2 = Ps − P1 P1 = Pa − Rqq = 1 2( x xm + 1)Ps − RqA˙y Behzad Samadi (Amirkabir University) Industrial Control 10 / 17
30. 30. Hydraulic Servomechanism M¨y + B ˙y + Ky = A(P1 − P2) P1 + P2 = Ps ⇒ P2 = Ps − P1 P1 = Pa − Rqq = 1 2( x xm + 1)Ps − RqA˙y M¨y + B ˙y + Ky =A(P1 − P2) =A(2P1 − Ps) =A(( x xm + 1)Ps − 2RqA˙y − Ps) Behzad Samadi (Amirkabir University) Industrial Control 10 / 17
31. 31. Hydraulic Servomechanism M¨y + B ˙y + Ky = A(P1 − P2) P1 + P2 = Ps ⇒ P2 = Ps − P1 P1 = Pa − Rqq = 1 2( x xm + 1)Ps − RqA˙y M¨y + B ˙y + Ky =A(P1 − P2) =A(2P1 − Ps) =A(( x xm + 1)Ps − 2RqA˙y − Ps) M¨y + (B + 2RqA2 )˙y + Ky = APs x xm Behzad Samadi (Amirkabir University) Industrial Control 10 / 17
32. 32. Hydraulic Servomechanism M¨y + B ˙y + Ky = A(P1 − P2) P1 + P2 = Ps ⇒ P2 = Ps − P1 P1 = Pa − Rqq = 1 2( x xm + 1)Ps − RqA˙y M¨y + B ˙y + Ky =A(P1 − P2) =A(2P1 − Ps) =A(( x xm + 1)Ps − 2RqA˙y − Ps) M¨y + (B + 2RqA2 )˙y + Ky = APs x xm Y (s) X(s) = APs xm Ms2 + (B + 2RqA2)s + K Behzad Samadi (Amirkabir University) Industrial Control 10 / 17
33. 33. Hydraulic Integrator If M = K = B = 0 then Y (s) X(s) = APs 2xmRg A2 1 s [Ogata, 1997] Behzad Samadi (Amirkabir University) Industrial Control 11 / 17
34. 34. Hydraulic Integrator If M = K = B = 0 then Y (s) X(s) = APs 2xmRg A2 1 s Hydraulic integrator [Ogata, 1997] Behzad Samadi (Amirkabir University) Industrial Control 11 / 17
35. 35. Hydraulic Proportional Controller Behzad Samadi (Amirkabir University) Industrial Control 12 / 17
36. 36. Hydraulic Proportional Controller Y (s) E(s) = bK (a + b)s + aK ≃ b a [Ogata, 1997] Behzad Samadi (Amirkabir University) Industrial Control 12 / 17
37. 37. Hydraulic Damper Behzad Samadi (Amirkabir University) Industrial Control 13 / 17
38. 38. Hydraulic Damper A(P1 − P2) = ky Behzad Samadi (Amirkabir University) Industrial Control 13 / 17
39. 39. Hydraulic Damper A(P1 − P2) = ky q = P1−P2 R Behzad Samadi (Amirkabir University) Industrial Control 13 / 17
40. 40. Hydraulic Damper A(P1 − P2) = ky q = P1−P2 R qdt = A𝜌(dx − dy) Behzad Samadi (Amirkabir University) Industrial Control 13 / 17
41. 41. Hydraulic Damper A(P1 − P2) = ky q = P1−P2 R qdt = A𝜌(dx − dy) Behzad Samadi (Amirkabir University) Industrial Control 13 / 17
42. 42. Hydraulic Damper A(P1 − P2) = ky q = P1−P2 R qdt = A𝜌(dx − dy) Y (s) X(s) = 1 1 + 1 Ts [Ogata, 1997] Behzad Samadi (Amirkabir University) Industrial Control 13 / 17
43. 43. Hydraulic Proportional Integrator Controller Behzad Samadi (Amirkabir University) Industrial Control 14 / 17
44. 44. Hydraulic Proportional Integrator Controller Y (s) X(s) = K s Behzad Samadi (Amirkabir University) Industrial Control 14 / 17
45. 45. Hydraulic Proportional Integrator Controller Y (s) X(s) = K s Z(s) Y (s) = 1 1+ 1 Ts Behzad Samadi (Amirkabir University) Industrial Control 14 / 17
46. 46. Hydraulic Proportional Integrator Controller Y (s) X(s) = K s Z(s) Y (s) = 1 1+ 1 Ts Behzad Samadi (Amirkabir University) Industrial Control 14 / 17
47. 47. Hydraulic Proportional Integrator Controller Y (s) X(s) = K s Z(s) Y (s) = 1 1+ 1 Ts Y (s) E(s) ≃ b a ( 1 + 1 Ts ) [Ogata, 1997] Behzad Samadi (Amirkabir University) Industrial Control 14 / 17
48. 48. Hydraulic Proportional Derivative Controller Behzad Samadi (Amirkabir University) Industrial Control 15 / 17
49. 49. Hydraulic Proportional Derivative Controller Y (s) X(s) = K s Behzad Samadi (Amirkabir University) Industrial Control 15 / 17
50. 50. Hydraulic Proportional Derivative Controller Y (s) X(s) = K s k(y − z) = A(P1 − P2) Behzad Samadi (Amirkabir University) Industrial Control 15 / 17
51. 51. Hydraulic Proportional Derivative Controller Y (s) X(s) = K s k(y − z) = A(P1 − P2) Z(s) Y (s) = 1 Ts+1 Behzad Samadi (Amirkabir University) Industrial Control 15 / 17
52. 52. Hydraulic Proportional Derivative Controller Y (s) X(s) = K s k(y − z) = A(P1 − P2) Z(s) Y (s) = 1 Ts+1 Behzad Samadi (Amirkabir University) Industrial Control 15 / 17
53. 53. Hydraulic Proportional Derivative Controller Y (s) X(s) = K s k(y − z) = A(P1 − P2) Z(s) Y (s) = 1 Ts+1 Behzad Samadi (Amirkabir University) Industrial Control 15 / 17
54. 54. 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
55. 55. Comparison Electrical Hydraulic Pneumatic Energy source Usually from outside supplier Electric motor or diesel driven Electric motor or diesel driven Energy storage Limited (batteries) Limited (accumulator) Good (reservoir) Distribution system Excellent, with minimal loss Limited, basically a lo- cal facility Good, can be treated as a plantwide service Energy cost Lowest Medium Highest Rotary actuators AC and DC motors. Good control on DC motors. AC motors cheap Low speed. Good con- trol. Can be stalled. Wide speed range. Ac- curate speed control diﬃcult Linear actuators Short motion via solenoid. Otherwise via mechanical conversion Cylinders. Very high force Cylinders. Medium force Points to note Danger from electric shock Leakage dangerous and unsightly. Fire hazard Noise [Parr, 1999] Behzad Samadi (Amirkabir University) Industrial Control 16 / 17
56. 56. Analogy Summary [Macia and Thaler, 2004] Behzad Samadi (Amirkabir University) Industrial Control 17 / 17
57. 57. 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