modeling and analysis of dynamic systems · 2 lecture 5: electromagnetic systems recalls electric...

Modeling and Analysis of Dynamic Systems

by Dr. Guillaume Ducard

Fall 2016

Institute for Dynamic Systems and Control

ETH Zurich, Switzerland

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Outline

1 Lecture 5: Hydraulic SystemsPelton Turbine: presentationPelton Turbine: modeling

2 Lecture 5: Electromagnetic SystemsRecallsElectric Oscillator

3 Lecture 5: Electromechanical SystemsPermanently Excited DC Motor: introductionModeling of a DC Motor

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Lecture 5: Hydraulic SystemsLecture 5: Electromagnetic Systems

Lecture 5: Electromechanical Systems

Pelton Turbine: presentationPelton Turbine: modeling

Outline




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Example: Pelton Turbine in a Hydro-electric Power Plant

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Outline




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w

v

R

Rω

w −Rω

w −Rω

dmω

Figure: Pelton turbine, definition of system variables.

Questions:1 What is the mean force acting on the Pelton turbine?

FT = f(w,ω,∗

V )?

2 What is the resulting wheel torque: TT = g(w,ω,∗

V )?3 What is the power being transferred from the fluid to the

turbine?6 / 22



RecallsElectric Oscillator

Outline




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Introduction

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.

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Mathematical modeling

Element Capacitance Inductance

Energy WE = 1

2C · U2(t) WM = 1

2L · I2(t)

Level variable U(t) (voltage) I(t) (current)

Conservation law C ·d

dtU(t) = I(t) L ·

d

dtI(t) = U(t)

The electrical power is P (t) = U(t) · I(t).

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Working with RLC networks, two rules (“Kirchhoff’s laws”) areuseful:

1: The algebraic sum of all currents in each network node iszero.

2: The algebraic sum of all voltages following a closed networkloop is zero.

These rules are equivalent to the energy balance and are usuallymore convenient to use.

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Outline




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I(t) L R

Cy(t)u(t) UL(t) UR(t)

UC(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

UL(t) + UR(t) + UC(t) = u(t)12 / 22




Step 4: Differential equations

“C” and ”L law”:

UL(t) = L ·

d

dtI(t), I(t) = C ·

d

dtUC(t)

and Ohm’s law:UR(t) = R · I(t)

Step 5: Reformulation and result

Definitions: y(t) = UC(t), I(t) =d

dtQ(t)

Reformulation: I(t) = C ·d

dty(t), d

dtI(t) = C ·

d2

dt2y(t)

Result: L · C ·d2

dt2y(t) +R · C ·

d

dty(t) + y(t) = u(t)

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Permanently Excited DC Motor: introductionModeling of a DC Motor

Outline




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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 thecurrent in the rotor coils and constant (permanent magnet) ortime-varying stator fields (external excitation).

2 Modern brushless DC drives have an electronic commutationof the stator current and permanent magnet on the rotor (i.e.,no brushes).

3 AC drives have an electronic commutation of the statorcurrent and use self-inductance to build up the rotor fields.

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DC motor

Figure: Principle of a DC motor (picture from Wikipedia)

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DC Motor Principle

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Outline




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I(t) LR Θ

Tl(t)

Tm(t)u(t)

ω(t)

UL(t)UR(t)

Uind(t)

Step 1: Input & Output

System input voltage u(t) (control input) and load torqueTl(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.

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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

dtI(t) = −R · I(t)− Uind(t) + u(t)

Θ ·d

dtω(t) = Tm(t)− Tl(t)− d · ω(t)

(1)

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Step 4: generator and the motor laws

Pelec = Pmech

Uind(t) · I(t) = κ · ω(t) · I(t) = Tm(t) · ω(t)

→ Tm(t) = κ · I(t)

(2)

Inserting equation (2) into equation (1),

L ·d

dtI(t) = −R · I(t)− κ · ω(t) + u(t)

Θ ·d

dtω(t) = κ · I(t)− Tl(t)− d · ω(t)

(3)

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Next lecture + Upcoming Exercise

Next lecture

Case study: Loudspeaker

Thermodynamics systems

Examples: Stirred Tank, Heat Exchanger, Gas Receiver

Next exercises: Hydro Power Plant

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modeling and analysis of dynamic systems · 2 lecture 5: electromagnetic systems recalls electric...

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