Modeling and Analysis of Dynamic Systems - ETH Z · Modeling and Analysis of Dynamic Systems by Dr....

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....

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