Modeling and Analysis of Dynamic Systems

Transcription

1 Modeling and Analysis of Dynamic Systems by Dr. Guillaume Ducard Fall 2016 Institute for Dynamic Systems and Control ETH Zurich, Switzerland 1/22

2 Outline 1 Lecture 5: Hydraulic Systems Pelton Turbine: presentation Pelton Turbine: modeling 2 3 2/22

3 Outline Lecture 5: Hydraulic Systems Pelton Turbine: presentation Pelton Turbine: modeling 1 Lecture 5: Hydraulic Systems Pelton Turbine: presentation Pelton Turbine: modeling 2 3 3/22

4 Pelton Turbine: presentation Pelton Turbine: modeling Example: Pelton Turbine in a Hydro-electric Power Plant 4/22

5 Outline Lecture 5: Hydraulic Systems Pelton Turbine: presentation Pelton Turbine: modeling 1 Lecture 5: Hydraulic Systems Pelton Turbine: presentation Pelton Turbine: modeling 2 3 5/22

6 Pelton Turbine: presentation Pelton Turbine: modeling w v Rω ω R w Rω dm w Rω Figure: Pelton turbine, definition of system variables. Questions: 1 What is the mean force acting on the Pelton turbine? F T = f(w,ω, V)? 2 What is the resulting wheel torque: T T = g(w,ω, V)? 3 What is the power being transferred from the fluid to the turbine? 6/22

7 Outline Lecture 5: Hydraulic Systems 1 Lecture 5: Hydraulic Systems Pelton Turbine: presentation Pelton Turbine: modeling 2 3 7/22

8 Introduction Lecture 5: Hydraulic Systems Electromagnetic systems often can be formulated as RLC networks: resistances (R) inductances (L) capacitances (C) Two classes of reservoir elements are important: magnetic energy: stored in magnetic fields B; and electric energy: stored in electric fields E. 8/22

9 Mathematical modeling Element Capacitance Inductance Energy W E = 1 2 C U2 (t) W M = 1 2 L I2 (t) Level variable U(t) (voltage) I(t) (current) Conservation law C d d dtu(t) = I(t) L dti(t) = U(t) The electrical power is P(t) = U(t) I(t). 9/22

10 Working with RLC networks, two rules ( Kirchhoff s laws ) are useful: 1: The algebraic sum of all currents in each network node is zero. 2: The algebraic sum of all voltages following a closed network loop is zero. These rules are equivalent to the energy balance and are usually more convenient to use. 10/22

11 Outline Lecture 5: Hydraulic Systems 1 Lecture 5: Hydraulic Systems Pelton Turbine: presentation Pelton Turbine: modeling /22

12 I(t) L R u(t) U L (t) U R (t) U C (t) C y(t) Step 1: Inputs and Outputs 1 Input: u(t), input voltage 2 Output: y(t), output voltage Step 2: Energy reservoirs 1 Magnetic energy in L 2 Electric energy in C Step 3: Equivalent energy balance : Kirchhoff rule U L (t)+u R (t)+u C (t) = u(t) 12/22

13 Step 4: Differential equations C and L law : U L (t) = L d dt I(t), I(t) = C d dt U C(t) and Ohm s law: U R (t) = R I(t) Step 5: Reformulation and result Definitions: y(t) = U C (t), I(t) = d dt Q(t) Reformulation: I(t) = C d dt y(t), d d2 dti(t) = C y(t) dt 2 Result: L C d2 dt 2 y(t)+r C d dt y(t)+y(t) = u(t) 13/22

14 Outline Lecture 5: Hydraulic Systems 1 Lecture 5: Hydraulic Systems Pelton Turbine: presentation Pelton Turbine: modeling /22

15 Most electric motors used in control loops are rotational. Classification: according to the commutation mechanisms used: 1 Classical DC drives have a mechanical commutation of the current in the rotor coils and constant (permanent magnet) or time-varying stator fields (external excitation). 2 Modern brushless DC drives have an electronic commutation of the stator current and permanent magnet on the rotor (i.e., no brushes). 3 AC drives have an electronic commutation of the stator current and use self-inductance to build up the rotor fields. 15/22

16 DC motor Lecture 5: Hydraulic Systems Figure: Principle of a DC motor (picture from Wikipedia) 16/22

17 DC Motor Principle 17/22

18 Outline Lecture 5: Hydraulic Systems 1 Lecture 5: Hydraulic Systems Pelton Turbine: presentation Pelton Turbine: modeling /22

19 ω(t) T l (t) I(t) R L Θ u(t) U R (t) U L (t) T m (t) U ind (t) Step 1: Input & Output System input voltage u(t) (control input) and load torque T l (t) (disturbance). System output measurement of rotor speed ω(t). Remarks: The motor is permanently excited, parameters κ in the motor and generator laws are constant. The mechanical part has friction losses. 19/22

20 Step 2: relevant reservoirs the magnetic energy stored in the rotor coil, level variable I(t); the kinetic energy stored in the rotor, level variable ω(t). Step 3: two energy conservation laws L d dt I(t) = R I(t) U ind(t)+u(t) Θ d dt ω(t) = T m(t) T l (t) d ω(t) (1) 20/22

21 Step 4: generator and the motor laws P elec = P mech U ind (t) I(t) = κ ω(t) I(t) = T m (t) ω(t) (2) T m (t) = κ I(t) Inserting equation (2) into equation (1), L d dti(t) = R I(t) κ ω(t)+u(t) Θ d dt ω(t) = κ I(t) T l(t) d ω(t) (3) 21/22

22 Next lecture + Upcoming Exercise Next lecture Case study: Loudspeaker Thermodynamics systems Examples: Stirred Tank, Heat Exchanger, Gas Receiver Next exercises: Hydro Power Plant 22/22