Electromechanical Systems 1 Basilio Bona DAUIN Politecnico di - - PowerPoint PPT Presentation

Electromechanical Systems 1

Basilio Bona

DAUIN – Politecnico di Torino

Semester 1, 2016-17

• B. Bona (DAUIN)

Electromechanical Systems 1 Semester 1, 2016-17 1 / 39

Introduction

In a system that includes interacting mechanical and electromagnetic components, or that has an intrinsic electromechanical behaviour, as piezoelectric, electro-rheological, and magnetostrictive materials, magnetic shape memory alloys, and others, it is possible to define an electromechanical Lagrangian function Lem as the sum of the Lagrangian functions related to the different components. The Lagrangian function of the electromagnetic part of the system may be defined using either the charge or the flux approach. We the index ‘e’ to indicate the electromagnetic quantities, and the index ‘m’ to indicate the mechanical quantities.

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Electromechanical Systems 1 Semester 1, 2016-17 2 / 39

Generalized charge coordinates

In this approach we choose the electrical charges qe as the generalized coordinates for electromagnetical subsystems, and the linear/angular displacements qm as the generalized coordinates for mechanical subsystems. Global generalized coordinates, global generalized velocities, and global generalized forces are therefore q =

• qm

qe

• ,

˙ q =

• ˙

qm ˙ qe

• ,

F =

• F m

F e

• The total kinetic co-energy is the sum

K ∗

em(q, ˙

q) = K ∗

m(qm, ˙

qm)+W ∗

i (˙

qe,qm) and the total potential energy is the sum Pem(q) = Pm(qm)+Wc(qe,qm) Observe that the kinetic co-energy and the potential energy of the electrical subsystem are influenced by the mechanical coordinates.

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Electromechanical Systems 1 Semester 1, 2016-17 3 / 39

The total Lagrange function is Lem(q, ˙ q) = K ∗

em(q, ˙

q)−Pem(q) = K ∗

m(qm, ˙

qm)−Pm(qm)+W ∗

i (˙

qe,qm)−Wc(qe,qm) and the Lagrange equations are d dt ∂Lem ∂ ˙ qm

• − ∂Lem

∂qm = F m i = 1,...,nm d dt ∂Lem ∂ ˙ qe

• − ∂Lem

∂qe = F e i = 1,...,ne

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Electromechanical Systems 1 Semester 1, 2016-17 4 / 39

Mechanical equations d dt ∂K ∗

m(qm, ˙

qm) ∂ ˙ qm T − ∂K ∗

m(qm, ˙

qm) ∂qm T − ∂W ∗

i (˙

qe,qm) ∂qm T + ∂Pm(qm) ∂qm T + ∂Wc(qe,qm) ∂qm T = F m Electrical equations d dt ∂W ∗

i (˙

qe,qm) ∂ ˙ qe T + ∂Wc(qe,qm) ∂qe T = F e where ∂W ∗

i (˙

qe,qm) ∂qm and ∂Wc(qe,qm) ∂qm are respectively the change of magnetic/inductive kinetic co-energy due to a change in the mechanical coordinates, and the change of capacitive potential energy due to a change in the mechanical coordinates.

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Electromechanical Systems 1 Semester 1, 2016-17 5 / 39

Generalized flux coordinates

In this approach we choose the magnetic flux linkages λ as the generalized coordinates for electromagnetical subsystems, and the linear/angular displacements qm as the generalized coordinates for mechanical subsystems. Global generalized coordinates, global generalized velocities, and global generalized forces are q =

• qm

λ

• ,

˙ q = ˙ qm ˙ λ

• ,

F =

• F m

F e

• The total kinetic co-energy is the sum

K ∗

em(q, ˙

q) = K ∗

m(qm, ˙

qm)+W ∗

c ( ˙

λ,qm) and the total potential energy is the sum Pem(q) = Pm(qm)+Wi(λ,qm) Observe that the kinetic co-energy and the potential energy of the electrical subsystem are influenced by the mechanical coordinates.

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Electromechanical Systems 1 Semester 1, 2016-17 6 / 39

The total Lagrange function is Lem(q, ˙ q) = K ∗

em(q, ˙

q)−Pem(q) = K ∗

m(qm, ˙

qm)−Pm(qm)+W ∗

c ( ˙

λ,qm)−Wi(λ,qm) and the Lagrange equations are d dt ∂Lem ∂ ˙ qm

• − ∂Lem

∂qm = F m i = 1,...,nm d dt ∂Lem ∂ ˙ λ

• − ∂Lem

∂λ = F e i = 1,...,ne

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Electromechanical Systems 1 Semester 1, 2016-17 7 / 39

Mechanical equations d dt ∂K ∗

m(qm, ˙

qm) ∂ ˙ qm T − ∂K ∗

m(qm, ˙

qm) ∂qm T −

• ∂W ∗

c ( ˙

λ,qm) ∂qm T + ∂Pm(qm) ∂qm T + ∂Wi(λ,qm) ∂qm T = F m Electrical equations d dt

• ∂W ∗

c ( ˙

λ,qm) ∂ ˙ λ T + ∂Wi(λ,qm) ∂λ T = F e where ∂W ∗

c ( ˙

λ,qm) ∂qm and ∂Wi(λ,qm) ∂qm are respectively the change of magnetic/inductive kinetic co-energy due to a change in the mechanical coordinates, and the change of capacitive potential energy due to a change in the mechanical coordinates.

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Electromechanical Systems 1 Semester 1, 2016-17 8 / 39

Electromechanical Two-port Networks

We consider a particular class of electromechanical systems, the two-port networks, whose inner structure may be represented by characteristics that can be either mainly inductive or mainly capacitive. An electrical two-port network is a network that presents two ports; one is the input port the other is the output port. When the ports can be reversed, i.e., the input becomes the output and vice-versa, the network is called a reversible two-port network. If inside the network no electrical power sources are present, the two-port is called passive two-port network, otherwise it is an active two-port network. The physical quantities at the two ports are always a quantity called effort s(t) and a quantity called flux φ(t). The effort-flux couples (s(t),φ(t)) may be electrical or mechanical at both ports, or different at the two ports; we consider this last class of two-port systems.

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The electrical port is characterized by a voltage s(t) = e(t) and a current φ(t) = i(t), while the mechanical port is characterized by a force s(t) = f (t) and a velocity φ(t) = v(t). The port power is the product P(t) = s(t)φ(t). Figure shows a two-port systems, where the input port is located on the left side and the output port on the right side. Pi(t) is the inflowing input power, Po(t) is the

• utflowing output power.
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Electromechanical Systems 1 Semester 1, 2016-17 10 / 39

Inductive two-port networks

In this type of two-port network the power conversion is obtained by an inductive element characterized by a flux linkage λ(t).

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The two-port system is characterized by the following constitutive relations i(t) = i(λ(t),x(t)) f (t) = f (λ(t),x(t)) and e(t) = dλ(t) dt = ˙ λ(t) v(t) = dx(t) dt = ˙ x(t) where x(t) represents a displacement. These relations are generic; they must be specified according to the type

• f electromagnetic interactions of the considered system.
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The power absorbed by the system is given by the difference between the input and the output power Pλ(t) = Pi(t)−Po(t) = e(t)i(t)−f (t)v(t) = ˙ λ(t)i(t)−f (t) ˙ x(t) Considering the energy, we can write Pλdt = dWi(λ,x) = i dλ −f dx and, since dWi(λ,x) = ∂Wi(λ,x) ∂λ dλ + ∂Wi(λ,x) ∂x dx, comparing the above relations one obtains i(λ,x) = ∂Wi(λ,x) ∂λ and f (λ,x) = −∂Wi(λ,x) ∂x

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Considering the co-energy, we can write W ∗

i (i,x) = λi −Wi(λ,x)

and taking the differential work, we have dW ∗

i (i,x) = i dλ +λ di −dWi(λ,x)

Since we can write i dλ −dWi(λ,x) = f dx we obtain dW ∗

i (i,x) = λ di +f dx

Recalling that dW ∗

i (i,x) = ∂W ∗ i (i,x)

∂i di + ∂W ∗

i (i,x)

∂x dx it follows that λ(i,x) = ∂W ∗

i (i,x)

∂i ; f (i,x) = ∂W ∗

i (i,x)

∂x

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We can summarize the above relations as follows dWi(λ,x) = i dλ −f dx dW ∗

i (i,x) = λ di +f dx

i(λ,x) = ∂Wi(λ,x) ∂λ λ(i,x) = ∂W ∗

i (i,x)

∂i f (λ,x) = −∂Wi(λ,x) ∂x f (i,x) = ∂W ∗

i (i,x)

∂x The constitutive relations are now defined as the partial derivatives of the energy; the flux linkage is the partial derivative of the co-energy, and the force f can be expressed as a function of λ(t) or i(t).

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Linear flux

Let us assume that the flux is linear with respect to the current i(t) λ(i,x) = L(x)i(t) where L(x) represents the magnetic circuit inductance, in general function

• f a mechanical displacement x(t). We can write

W ∗

i (i,x) = 1

2L(x)i2(t) and f (λ,x) = 1 2 λ 2 L2(x) d dx L(x)

• r

f (i,x) = 1 2 i2 d dx L(x)

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The voltage e(t) is given by e(t) = d dt [L(x)i ] = L(x) di dt +i ∂L(x) ∂x ˙ x

• r

e = L(x) di dt +e′ where the voltage e′ = i ∂L(x) ∂x ˙ x is due to the variation of the auto-inductance when the circuit is subject to a mechanical deformation.

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Capacitive two-port networks

In this type of two-port network the power conversion is obtained by a capacitive element characterized by an electrical charge q(t).

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The two-port system is characterized by the following constitutive relations e(t) = e(q(t),x(t)) f (t) = f (q(t),x(t)) and i(t) = dq(t) dt = ˙ q(t) v(t) = dx(t) dt = ˙ x(t) where x(t) represents a displacement. These relations are generic; they must be specified according to the type

• f electromagnetic interactions of the considered system.
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Electromechanical Systems 1 Semester 1, 2016-17 19 / 39

The power absorbed by the system is given by the difference between the input and the output power Pq(t) = Pi(t)−Po(t) = e(t)i(t)−f (t)v(t) = e(t) ˙ q(t)−f (t) ˙ x(t) Considering the energy, we can write Pq dt = dWc(q,x) = e dq −f dx and, since dWc(q,x) = ∂Wc(q,x) ∂q dq + ∂Wc(q,x) ∂x dx, comparing the above relations to one obtains e(q,x) = ∂Wc(q,x) ∂q and f (q,x) = −∂Wc(q,x) ∂x

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Electromechanical Systems 1 Semester 1, 2016-17 20 / 39

considering the co-energy, we can write W ∗

c (e,x) = eq −Wc(q,x)

and, considering the virtual work, we have dW ∗

c (e,x) = e dq +q de −dWc(q,x)

Since we can write e dq −dWc(q,x) = f dx we have at the end dW ∗

c (e,x) = q de +f dx

Recalling that, in general dW ∗

c (e,x) = ∂W ∗ c (e,x)

∂e de + ∂W ∗

c (e,x)

∂x dx it follows immediately that q(e,x) = ∂W ∗

c (e,x)

∂e ; f (e,x) = ∂W ∗

c (e,x)

∂x

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We can summarize the above relations as follows dWc(q,x) = edq −f dx dW ∗

c (e,x) = qde +f dx

e(q,x) = ∂Wc(q,x) ∂q q(e,x) = ∂W ∗

c (e,x)

∂e f (q,x) = −∂Wc(q,x) ∂x f (e,x) = ∂W ∗

c (e,x)

∂x The constitutive relations are now defined as the partial derivatives of the energy; the charge is the partial derivative of the co-energy, and the force f can be expressed as a function of q(t) or e(t).

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Electromechanical Systems 1 Semester 1, 2016-17 22 / 39

Linear charge

Let us assume that the charge is linear with respect to the voltage e(t) q(e,x) = C(x)e(t) where C(x) represents the capacity related to the electrostatic phenomenon, in general function of a mechanical displacement x(t). We can write Wc(q,x) = 1 2 q2(t) C(x) and f (q,x) = −q2 2 d dx

• 1

C(x)

• r

f (e,x) = e2 2 d dx C(x)

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The current i(t) is given by i(t) = d dt [C(x)e ] = C(x)de dt +e ∂C(x) ∂x ˙ x

• r

i = C(x)de dt +i′ where the current i′ = e ∂C(x) ∂x ˙ x is due to the capacity variation when the circuit is subject to a mechanical deformation.

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Example – Magnetic suspension

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A magnetic suspension consists of an electromagnet (a cylindric core of ferromagnetic material) with height ℓc and constant area S = πr2, with N winded coils, connected to an electrical voltage source E(t). The electromagnet maintains suspended, against gravity or other forces, a sphere of ferromagnetic material, with radius r and mass m. The current flowing into the coils is i(t); the air gap is x(t). The resulting magnetic circuit presents a magnetomotive force equal to Ni(t) = Rtotφ(t) the magnetic flux φ(t) goes through the magnetic nucleus (the cylinder), the air gap, the sphere and closes its lines in the air going back to the nucleus, so Rtot = Rcyl +Rgap +Rsphere +Rair

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the circuit is made of ferromagnetic material with µ ≫ µ0, the two reluctances Rcyl and Rsphere are negligible with respect to Rgap and Rair If the sphere is sufficiently near to the nucleus, the length of the remaining path through air back to the nucleus is approximately constant with respect to the sphere motion. Since Rgap = x µ0S where µ0 is the air permeability, one can write Rgap +Rair = x +ℓ0 µ0S where ℓ0 = Rairµ0S is the equivalent length of the flux in air.

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It follows that Ni(t) = x +ℓ0 µ0S φ(t) and then λ = Nφ(t) = N2µ0S ℓ0 +x i(t) In conclusion, the circuit inductance is L(x) = N2µ0S ℓ0 +x = kL ℓ0 +x with kL = N2µ0S. The electrical circuit is powered by an ideal voltage generator E(t) in series with a resistance R, and a negligible capacitance; the dynamical equations of the magnetic suspension can be computed assuming as generalized coordinates the electrical charge q(t) and the mechanical displacement x(t) of the sphere.

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Applying the charge approach for the electrical part, where no capacitive elements are present, we have K ∗

m

= 1 2m ˙ x2 K ∗

e

= W ∗

i ( ˙

q,x) = 1 2L(x) ˙ q2 Pm = −mGx Pe = Wc(q) = 1 2 q2 C = 0 Dm = 1 2β ˙ x2 ≈ 0 if β ≈ 0 De = 1 2R ˙ q2 Fm = Fe = E(t) where G is the gravity acceleration value and the viscous friction is negligible.

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Electromechanical Systems 1 Semester 1, 2016-17 29 / 39

Lagrange Equation 1) d dt ∂K ∗

m

∂ ˙ x − ∂(K ∗

e −Pm)

∂x + ∂Dm ∂ ˙ x = 0 whence d dt m ˙ x + k 2(ℓ0 +x)2 ˙ q2 −mG +β ˙ x = 0 and, neglecting the friction force, m¨ x = mG − k 2(ℓ0 +x)2 ˙ q2

• r

m¨ x(t) = mG − k 2(ℓ0 +x(t))2 i2(t)

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Lagrange Equation 2) d dt ∂K ∗

e

∂ ˙ q + ∂De ∂ ˙ q = E(t) whence d dt [L(x) ˙ q ]+R ˙ q = E(t) therefore d dt [L(x)] ˙ q +L(x) d dt [ ˙ q ]+R ˙ q = E(t) and − k (ℓ0 +x)2 ˙ x ˙ q +L(x)¨ q +R ˙ q = E(t) that can also be written as L(x) d dt i(t)+Ri(t) = E(t)+ k ˙ x(t) (ℓ0 +x(t))2 i(t)

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Example – Voice coil

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A voice coil consists of a fixed magnetic circuit, composed by a permanent magnet suitably inserted into a ferromagnetic circuit. A mobile element with several coils (moving coils), interacts with the magnetic circuit and is free to move along one direction. The interaction between the permanent magnet circuit field and the electromagnetic field generated by the current in the coils, causes the motion of the moving coil. During its motion, the number of coils that “cut” the magnetic flux lines varies in function of the coils displacement x. The mechanical part is characterized by the mass m of the moving coil and attachments, by the friction coefficient β that models the dissipative effects due to the air motion, and by the elastic coefficient k that models the elastic effect of the membrane suspensions. The electrical part of the actuator is characterized by a fixed auto-inductance L due to the coils, by a resistance R and by a variable voltage source E(t); the capacitive effects are negligible.

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The permanent magnet generates a magnetic field whose closed flux lines divide into the superior and inferior part of the magnetic circuit; the field density B is constant in the air gap. The magnetic flux Φm that interacts the electrical windings depends on the surface interested by the phenomenon, that is Φm = 2πrx B where we have indicate with x the position of the coil with respect to the external surface of the magnetic circuit. The flux linked to the coils depends on two terms: the first one, Φa, is due to the auto-induced flux produced by the current that transverses the coil, the second one, Φm, is that produced by the permanent magnet and linked with the “active” coils: λ(i,x) = Φa +Φm = Li +Kex (1) where L is the auto-inductance of the coil with N windings and Ke = 2πrN B.

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Applying the charge approach for the electrical part, where no capacitive elements are present, we have K ∗

m

= 1 2m ˙ x2 K ∗

e

= W ∗

i ( ˙

q,x) Pm = 1 2kx2 Pe = Wc(q) = 1 2 q2 C = 0 Dm = 1 2β ˙ x2 De = 1 2R ˙ q2 Fm = Fe = E(t)

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In this case the term K ∗

e = W ∗ i (i,x) is more complex. We recall that

dW ∗

i (i,x) = λ di +f dx

hence W ∗

i (i,x) =

• i,x dW ∗

i (i,x) =

• i,x λ di′ +f dx′

where x′ and i′ are the generic integration variables. If in the plane (i,x) we chose as integration path from (0,0) to (0,x), and then from (0,x) to (i,x), we can express the previous integral as W ∗

i (i,x) =

x

0 f (i = 0,x′)dx′ +

i

0 λ(i′,x)di′

Since it is reasonable to assume that when the flux linkage is zero, the force is also zero, i.e., f (i = 0,x′) = 0 for any value of x, we have W ∗

i (i,x) =

x

0 f (0,x′)dx′

• =0

+

i

0 λ(i′,x)di′ =

i

0 (Li′ +Kex)di′ = 1

2Li2 +Kexi

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Lagrange Equation 1) d dt ∂K ∗

m

∂ ˙ x − ∂(K ∗

e −Pm)

∂x + ∂Dm ∂ ˙ x = 0 gives d dt m ˙ x +kx −Ke ˙ q +β ˙ x = 0 and, considering the friction force,one obtains m¨ x +β ˙ x +kx = Ke ˙ q

• r

m¨ x(t)+β ˙ x(t)+kx(t) = Kei(t)

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Lagrange Equation 2) d dt ∂K ∗

e

∂ ˙ q + ∂De ∂ ˙ q = E(t) gives d dt [L ˙ q +Kex]+R ˙ q = E(t)

• r

L¨ q +Ke ˙ x +R ˙ q = E(t) that we can write as L d dt i(t)+Ri(t) = E(t)−Ke ˙ x(t)

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Notice the term appearing in Equation 1), namely Kei(t) When a two-port network contains an inductive element and the flux linkage is linear in the current, one can write f (i,x) = ∂W ∗

m(i,x)

∂x that in our case becomes f (i,x) = Kei. Similarly, in Equation 2) there is the term Ke ˙ x, that can be associated to the voltage e(t) = ˙ λ(t) = d dt [Li +Kex] = L di dt +e′(t) with e′(t) = Ke ˙ x(t)

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