Modeling and Analysis of Dynamic Systems - ETH Z · Modeling and Analysis of Dynamic Systems by Dr. Guillaume Ducard ... Outline 1 Lecture 3: ... 07 AM

Modeling and Analysis of Dynamic Systems

by Dr. Guillaume Ducard

Fall 2016

Institute for Dynamic Systems and Control

ETH Zurich, Switzerlandbased on script from: Prof. Dr. Lino Guzzella

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Outline

1 Lecture 3: Modeling Tools for Mechanical SystemsSimplified Model of a Gas TurbineLagrange Formalism

2 Lecture 3: Examples with the Lagrange MethodNonlinear Pendulum on a Cart

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Lecture 3: Modeling Tools for Mechanical SystemsLecture 3: Examples with the Lagrange Method

Simplified Model of a Gas TurbineLagrange Formalism

Outline



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Model of Gas Turbines

yourdictionnary.com

http://www.aptech.ro

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Simplified model of a gas turbine

T1 T2

ω1 ω2

d1 d2

Θ1 Θ2ϕk

rotor 2: the turbine stagedriving torque T2, Moment of inertia: Θ2

rotor 1: the compressor stagebreaking torque T1, Moment of inertia: Θ1

shaft elasticity constant: k

friction losses: d1 and d2 [Nm.(rad/s)−1]

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Step 1: Inputs and Outputs

Inputs: Torques T1 and T2

Outputs: Rotor speed at the compressor stage: ω1

Step 2: Reservoirs and level variables

reservoir 2: kinetic energy of the turbine E2(t)a. Level: ω2

reservoir 1: kinetic energy of the compressor E1(t). Level: ω1

reservoir 3: potential energy stored in the elasticity of theshaft Ushaft(t). Level: ϕ

What is the energy associated with each reservoir?

E2(t) =

E1(t) =

Ushaft(t) =

a

the energies are noted E1 and E2 to avoid confusion with the torques T1 and T2 .6 / 17




Step 3: Dynamics equation - Mechanical power balance

dE2(t)

dt=

dE1(t)

dt=

dUshaft(t)

dt=

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Step 3: Dynamics equation - Mechanical power balance

Pmech,1 = compressor power = T1 · ω1

Pmech,2 = friction loss in bearing 1 = d1ω1 · ω1

Pmech,3 = power of the shaft elasticity at rotor 1 = kϕ · ω1

Pmech,4 = power of the shaft elasticity at rotor 2 = kϕ · ω2

Pmech,5 = friction loss in bearing 2 = d2ω2 · ω2

Pmech,6 = turbine power = T2 · ω2

d

dt

(

1

2Θ1ω

21(t)

)

= −Pmech,1(t)− Pmech,2(t) + Pmech,3(t)

d

dt

(

1

2Θ2ω

22(t)

)

= −Pmech,4(t)− Pmech,5(t) + Pmech,6(t)

d

dt

(

1

2kϕ2(t)

)

= −Pmech,3(t) + Pmech,4(t)

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Step 4: Dynamics equations of the level variables

Θ1

d

dtω1(t) = −T1(t)− d1 · ω1(t) + k · ϕ(t)

Θ2

d

dtω2(t) = T2(t)− d2 · ω2(t)− k · ϕ(t)

d

dtϕ(t) = ω2(t)− ω1(t)

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Outline



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Lagrange: 1736 -1813

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Lagrange Formalism: Recipe

1 Define inputs and outputs2 Define the generalized coordinates:

q(t) = [q1(t), q2(t), . . . , qn(t)] andq(t) = [q1(t), q2(t), . . . , qn(t)]

3 Build the Lagrange function:

L(q, q) = T (q, q)− U(q)

4 System dynamics equations:

d

dt

{

∂L

∂qk

}

−

∂L

∂qk= Qk, k = 1, . . . , n

Notes:

Qk represents the k-th “generalized force or torque” acting onthe k−th generalized coordinate variable qkn: number of degrees of freedom in the system

always n generalized variables12 / 17


Nonlinear Pendulum on a Cart

Outline



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replacements

y(t)

u(t)M = 1kg

ϕ(t)

l = 1m

mg

m = 1kg

Figure: Pendulum on a cart, u(t) is the force acting on the cart(“input”), y(t) the distance of the cart to an arbitrary but constantorigin, and ϕ(t) the angle of the pendulum.

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Step 1: Inputs & Outputs

Input: force acting on the cart: u(t)

Output: angle of the pendulum: ϕ(t)

Step 2: System’s coordinate variables

q1 = y, q1 = y

q2 = ϕ, q2 = ϕ

Step 3: Lagrange functions

L1(t) = T1(t)− U1(t)

L2(t) = T2(t)− U2(t)

L(t) = L1(t) + L2(t)

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Step 4: System’s dynamics equations

d

dt

{

∂L

∂q1

}

−

∂L

∂q1= Q1

d

dt

{

∂L

∂q2

}

−

∂L

∂q2= Q2

We are looking for dynamic equations of the form:

y(t) = f(ϕ(t), ϕ(t), u(t))

ϕ(t) = g(ϕ(t), ϕ(t), u(t))

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Next lecture + Upcoming Exercise

Next lecture

Ball on wheel example

Hydraulic systems

Next exercise: Online next Friday

Modeling of a clown balancing on a ladder

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Modeling and Analysis of Dynamic Systems - ETH Z · Modeling and Analysis of Dynamic Systems by Dr. Guillaume Ducard ... Outline 1 Lecture 3: ... 07 AM