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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
1 / 17
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
2 / 17
Lecture 3: Modeling Tools for Mechanical SystemsLecture 3: Examples with the Lagrange Method
Simplified Model of a Gas TurbineLagrange Formalism
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
3 / 17
Lecture 3: Modeling Tools for Mechanical SystemsLecture 3: Examples with the Lagrange Method
Simplified Model of a Gas TurbineLagrange Formalism
Model of Gas Turbines
yourdictionnary.com
http://www.aptech.ro
4 / 17
Lecture 3: Modeling Tools for Mechanical SystemsLecture 3: Examples with the Lagrange Method
Simplified Model of a Gas TurbineLagrange Formalism
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]
5 / 17
Lecture 3: Modeling Tools for Mechanical SystemsLecture 3: Examples with the Lagrange Method
Simplified Model of a Gas TurbineLagrange Formalism
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
Lecture 3: Modeling Tools for Mechanical SystemsLecture 3: Examples with the Lagrange Method
Simplified Model of a Gas TurbineLagrange Formalism
Simplified model of a gas turbine
Step 3: Dynamics equation - Mechanical power balance
dE2(t)
dt=
dE1(t)
dt=
dUshaft(t)
dt=
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Lecture 3: Modeling Tools for Mechanical SystemsLecture 3: Examples with the Lagrange Method
Simplified Model of a Gas TurbineLagrange Formalism
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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Lecture 3: Modeling Tools for Mechanical SystemsLecture 3: Examples with the Lagrange Method
Simplified Model of a Gas TurbineLagrange Formalism
Simplified model of a gas turbine
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)
9 / 17
Lecture 3: Modeling Tools for Mechanical SystemsLecture 3: Examples with the Lagrange Method
Simplified Model of a Gas TurbineLagrange Formalism
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
10 / 17
Lecture 3: Modeling Tools for Mechanical SystemsLecture 3: Examples with the Lagrange Method
Simplified Model of a Gas TurbineLagrange Formalism
Lagrange: 1736 -1813
11 / 17
Lecture 3: Modeling Tools for Mechanical SystemsLecture 3: Examples with the Lagrange Method
Simplified Model of a Gas TurbineLagrange Formalism
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
Lecture 3: Modeling Tools for Mechanical SystemsLecture 3: Examples with the Lagrange Method
Nonlinear Pendulum on a Cart
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
13 / 17
Lecture 3: Modeling Tools for Mechanical SystemsLecture 3: Examples with the Lagrange Method
Nonlinear Pendulum on a Cart
Nonlinear Pendulum on a Cart
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.
14 / 17
Lecture 3: Modeling Tools for Mechanical SystemsLecture 3: Examples with the Lagrange Method
Nonlinear Pendulum on a Cart
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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Lecture 3: Modeling Tools for Mechanical SystemsLecture 3: Examples with the Lagrange Method
Nonlinear Pendulum on a Cart
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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Lecture 3: Modeling Tools for Mechanical SystemsLecture 3: Examples with the Lagrange Method
Nonlinear Pendulum on a Cart
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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