modeling and analysis of dynamic systems · 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, Switzerland

1 / 21

Outline

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

2 Lecture 4: Hydraulic SystemsWater DuctCompressible Duct Element

2 / 21

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

Lagrange Method with Kinematic ConstraintsBall on Wheel

Outline

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

2 Lecture 4: Hydraulic SystemsWater DuctCompressible Duct Element

3 / 21

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

Lagrange Method with Kinematic ConstraintsBall on Wheel

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).

4 / 21

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

Lagrange Method with Kinematic ConstraintsBall on Wheel

Outline

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

2 Lecture 4: Hydraulic SystemsWater DuctCompressible Duct Element

5 / 21

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

Lagrange Method with Kinematic ConstraintsBall on Wheel

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

6 / 21

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

Lagrange Method with Kinematic ConstraintsBall on Wheel

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

7 / 21

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

Lagrange Method with Kinematic ConstraintsBall on Wheel

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

8 / 21

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

Lagrange Method with Kinematic ConstraintsBall on Wheel

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

9 / 21

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

Lagrange Method with Kinematic ConstraintsBall on Wheel

Modeling the ball on the wheel

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.

10 / 21

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

Lagrange Method with Kinematic ConstraintsBall on Wheel

Modeling the ball on the wheel

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)

11 / 21

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

Water DuctCompressible Duct Element

Outline

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

2 Lecture 4: Hydraulic SystemsWater DuctCompressible Duct Element

12 / 21

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

Water DuctCompressible Duct Element

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.

13 / 21

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

Water DuctCompressible Duct Element

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)

14 / 21

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

Water DuctCompressible Duct Element

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

15 / 21

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

Water DuctCompressible Duct Element

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

16 / 21

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

Water DuctCompressible Duct Element

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

17 / 21

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

Water DuctCompressible Duct Element

Outline

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

2 Lecture 4: Hydraulic SystemsWater DuctCompressible Duct Element

18 / 21

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

Water DuctCompressible Duct Element

∗

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]

.19 / 21

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

Water DuctCompressible Duct Element

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

20 / 21

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

Water DuctCompressible Duct Element

Next lecture + Upcoming Exercise

Next lecture

Pelton Turbine

Electromagnetic systems

Next exercise: Online next Friday

Hydro-electric Power plant, Part I

21 / 21