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

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 4: Modeling Tools for Mechanical SystemsLagrange Method with Kinematic ConstraintsBall on Wheel

2 Lecture 4: Hydraulic SystemsWater DuctCompressible Duct Element

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Lecture 4: Modeling Tools for Mechanical SystemsLecture 4: Hydraulic Systems

Lagrange Method with Kinematic ConstraintsBall on Wheel

Outline



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Lagrange Equations for Constrained Systems

d

dt

{

∂L

∂q̇k

}

−∂L

∂qk−

ν∑

j=1

µjαj,k = Qk, k = 1, . . . , n, (1)

Remarks:

the constraints are included using “Lagrange multipliers”: µjj = 1 . . . ν

Number of constraints: ν with (ν < n)

n may be seen as the number of DOF

In the end, we obtain: n+ ν coupled equations to be solvedfor q̈k and µj (usually requires computing the time derivativeof the constraints, i.e., µ̇).

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Outline



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y(t)

u(t)

r,m, ϑχ(t)

R,Θ

ϕ(t)

ψ(t)

mg

wheel: mass moment of inertia around c.o.g. is Θ, radius R,

ball: mass m, mass moment of inertia around c.o.g. ϑ, radius r

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Ball on Wheel

Control objective

The ball must be kept on top of the wheel.

Model objectives

Build a model for:

system analysis (stability, observability, controllability)

control design

Assumptions

No-slip

no-slip equation

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Modeling the ball on the wheel

Step 1: Inputs and outputs

Input: u(t) torque to the wheel

Output: y(t) = (R+ r)sinχ horizontal distance of the centerof the ball w.r.t. the vertical axis of the wheel

( )u t y( )tModel of theball on the wheel

Torque tothe wheel

Horizontal distanceto the center of the ball

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Step 2: n generalized coordinates (n DOF) and energies

The system has 3 DOF: rotation of

1 the wheel around its center: angle ψ(t)

2 the ball around the center of the wheel: angle χ(t)

3 the ball around its own center: angle ϕ(t)

y(t)

u(t)

r,m, ϑχ(t)

R,Θ

ϕ(t)

ψ(t)mg

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Step 3: Lagrange function

L(t) = T (t)− U(t)

Step 4: Differential equations including constraints

Lagrange equations nonholonomic case (n = 3, ν = 1, q1 = ψ,q2 = χ, q3 = ϕ)

d

dt

(

∂L

∂q̇k

)

−∂L

∂qk− µαk = Qk

with Q1 = u(t), Q2 = Q3 = 0 and α1q̇1 + α2q̇2 + α3q̇3 = 0 withα1 = R, α2 = −(R+ r) and α3 = r.

Finally, we get a set of four equations define the four unknownvariables {ψ̈, χ̈, ϕ̈, µ}, where {ϕ̈, µ} are easy to eliminate.

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Final results

Θ+ ϑR2

r2−ϑR(R+r)

r2

−ϑR(R+r)r2

m(R+ r)2 + ϑ (R+r)2

r2

[

ψ̈

χ̈

]

=

[

u

mg(R+ r) sin(χ)

]

Mass matrix M is positive definite. Therefore

ψ̈(t) =[

(mr2 + ϑ)u(t) +mgRϑ sin(χ(t))]

/Γ

χ̈(t) =[

ϑRu(t) + (Θr2 + ϑR2)mg sin(χ(t))]

/[Γ(r +R)]

where the scalar Γ is given by

Γ = Θϑ +m(ϑR2 +Θr2)

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Water DuctCompressible Duct Element

Outline



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Introduction

In general

they are described by Navier-Stokes equations.

For control purposes

simpler formulations are necessary, to build networks with buildingblocks:

ducts,

compressible nodes,

valves, etc.

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Water duct in gravitational field

l

A

v(t)

v(t)

p1

p2

h

Objective

d

dtv(t) = f (p1(t), p2(t), v(t), h, ρ,A, l)

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Change in momentum: Newton’s law

d~p

dt= m

d~v

dt= ~Fpressure + ~Fgravity + ~Ffriction

= [P1A− P2A] ~x+

∫

tube

~g · dm+ ~Ffriction

The mass m of the fluid in the element of tube of length l is givenby

m = ρ ·A · l → dm = ρ ·A · dl

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Angle of the duct

sinα =dh

dl

Gravity force

∫

tube~g dm = g

∫

tube(− cosα ~y + sinα ~x) ρ · A · dl

= ρ · g ·A[

∫ h

0− cosα

sinαdh ~y +

∫ h

0

sinαsinα

dh ~x]

= −ρ · g · A (tanα)−1 h ~y + ρ g Ah ~x

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Water duct in gravitational field

Dynamics along the ~x axis (because ~v = v~x)

ρ A ldv(t)

dt= A (P1 − P2) + ρ g A h− Ffriction,x(t)

with

Ffriction,x(t) =1

2ρ v2(t) sign [v(t)] · λ (v(t)) ·

A l

d

Remark: shape factor: ld

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Outline



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∗

V in (t)∗

V out (t)p(t)

∆V = 0

∆V (t)

k = 1/(σ0V0)

Definitions of compressability

Property of a body (solid, liquid, gas, etc.) to deform (to changeits volume) under the effect of applied pressure.Defined as:

σ0 =1

V0

dV

dP

V0: nominal volume [m3], P : pressure [Pa]σ0: compressibility [Pa−1]

k0 =1

σ0is called elasticity constant

[

Pa ·m−3]

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Compressibility effects

∗

V in (t)∗

V out (t)p(t)

∆V = 0

∆V (t)

k = 1/(σ0V0)

Modeling

ddtV (t) =

∗

V in(t)−∗

V out(t) = Ainvin(t)−Aoutvout(t)

∆P (t) = k∆V (t) = 1

σ0V0∆V (t)

∆V (t) = V (t)− V0

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Next lecture + Upcoming Exercise

Next lecture

Pelton Turbine

Electromagnetic systems

Next exercise: Online next Friday

Hydro-electric Power plant, Part I

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Documents

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