Modeling of electromechanical systems

Transcription

1 Modeling of electromechanical systems Carles Batlle II EURON/GEOPLEX Summer School on Modeling and Control of Complex Dynamical Systems Bertinoro, Italy, July

2 Contents of this lecture Basic ideas of electromechanical energy conversion from a network modeling point of view Port Hamiltonian formulation of electromechanical systems Very simple examples: lumped parameter only, no complex induction machines, no piezoelectric devices,......but it can be done Variable structure systems: power converters A more complex example

3 Network description of systems Energy-storing elements Impose constitutive relations between efforts and flows e 1 f 1 e 2 f 2 e n f n [e i ][f i ] = power Network Is power continuous nx e i f i = i=1 n+m X i=n+1 e i f i e n+1 f n+1 e n+m f n+m Open ports Can be connected to other systems or to dissipative elements

4 Example: Tellegen s theorem Circuit with b branches and n nodes i 1 u 1 i 2 u 2 u 3 i 3 To each node we assign a voltage u j, j =1,...,n To each branch we assign acurrenti α, α =1,...,b, and this gives an orientation to the branch i b u n i α u l Foreachbranchwedefine the voltage drop v α, α =1,...,b: This is KVL! u j v α = u j u l

5 Mathematically, the circuit, with the orientation induced by the currents, is a digraph (directed graph) We can define its n b adjacency matrix A by 1 if branch α is incident on node i A i α = +1 if branch α is anti-incident on node i 0 otherwise Then, KCL states that bx A i αi α =0, α=1 i =1,...,n Infact,KVLcanalsobestatedintermsofA: v α = nx A i αu i i=1 The sum contains only two terms, because each branch connects only two nodes

6 Tellegen s theorem. Let {v (1)α (t 1 )} α=1,...,b be a set of branch voltages satisfying KVL at time t 1,andlet{i α (2) (t 2)} α=1,...,b be a set of currents satisfying KCL at time t 2.Then Proof: bx v (1)α (t 1 )i α (2) (t 2) hv (1) (t 1 ),i (2) (t 2 )i =0 α=1 P b α=1 v (1)α(t 1 )i α (2) (t 2) = = KVL Ã bx X n! A i αu (1)i (t 1 ) i α (2) (t 2) α=1 KCL i=1 Ã nx bx! A i αi α (2) (t 2) u (1)i (t 1 ) i=1 α=1 = nx 0 u (1)i (t 1 )=0 i=1

7 Notice that {v (1)α (t 1 )} and {i α (2) (t 2)} may correspond to different times and they may even correspond to different elements for the branches of the circuit. The only invariant element is the topology of the circuit i.e. the adjacency matrix. Corollary. Under the same conditions as for Tellegen s theorem, d r À dt r v (1) (t 1 ), ds 1 dt s i (2) (t 2 ) =0 2 for any r, s N. In fact, even duality products between voltages and currents in different domains (time or frequency) can be taken and the result is still zero.

8 In terms of abstract network theory, a circuit can be represented as follows Element in branch 2 v 2 i 2 Element in branch 1 v 1 i 1 Network: KVL+KCL i b v b Element in branch b The kth branch element imposes a constitutive relation between v k and i k. Maybelinearornonlinear, algebraic or differential,...

9 Energy storing elements Consider a system with m power ports e 1 f 1 e 2 f 2 W f (t) W f (t 0 )= Z t t 0 mx e i (τ)f i (τ)dτ i=1 f m e m Assume the state of the system is given by x R n,andthat Internal dynamics ẋ = G(x)+g(x)e Dynamics due to the inputs Assume also that outputs can be computed from the state f = g T (x)φ(x)

10 W f (t) W f (t 0 )= Z t t 0 he(τ),f(τ)idτ = Z t t 0 he(τ),g T (x(τ))φ(x(τ))idτ = Z t t 0 hg(x(τ))e(τ), φ(x(τ))idτ = Z t t 0 hẋ(τ) G(x(τ)), φ(x(τ))idτ Assume the internal dynamics is G(x) =J(x)φ(x) with J T = J W f (t) W f (t 0 )= Z t t 0 hẋ ( τ), φ(x(τ))i dτ = Z γ(x 0,x) φ(z) dz x 0 γ x To avoid violating energy conservation (First Principle of Thermodynamics) we cannot let this depend on the particular γ connecting x 0 and x

11 γ 2 x 0 γ 1 x Suppose W (γ 2 ) > W (γ 1 ) γ 1 γ 2 x 0 x x 0 W (γ 1 ) W (γ 2 ) We input W (γ 1 )andthenobtain W (γ 2 ) while the system returns to the same state free generator of energy! To avoid this φ(x)dx must be an exact form φ i (x) = Φ(x) x i, i =1,...,n where Φ(x) isanstatespacefunction

12 W f (x) W f (x 0 )=Φ(x) W f (x) = Z or, taking appropriate references, γ(x) φ(z)dz φ i (x) = Φ(x) x i, i =1,...,n φ i x j (x) = φ j x i (x), i,j =1,...,n Maxwell s reciprocity relations φ i (x) = W f(x) x i we are using an all-input power convention a sign is used in EE literature to compute the mech force done by the system

13 Electric capacitor n = m =2,g = I 2, G =0 zero E constant E zero E e 1 = i (current), e 2 = V (velocity) q +q varying x F (x, q), φ 1 = F (x, q), φ 2 = v(x, q) v(x, q) = q C(x, q) nonlinear dielectric F q = v x = q x µ 1 C(x, q) q = C 2 (x, q) xc(x, q)

14 F (x, q) = = q2 2 Z q 0 x µ 1 x C(x, ξ) µ 1 C(x, q) ξ dξ Z q in the linear case, C(x, q) = C(x) 0 ξ ξ x µ 1 dξ C(x, ξ) F (x, q) = q2 2 = q x µ 1 C(x) E(x, q) 2 = q 2 x µ q C(x) = q 2 v(x, q) x electric field seen by the + plate Furthermore, going from (0, 0) to (x, q) first with q =0andthentoq with the final x, one gets the energy W f (x, q) = Z q 0 ξ C(x) dξ = 1 2 q 2 C(x)

15 Magnetic stationary system v 1 + i 1 Φ l1 i 2 + v 2 Φ 1 = Φ l1 + Φ m1 + Φ m2 Φ 2 = Φ l2 + Φ m2 + Φ m1 N 1 N 2 Φ l2 λ 1 = N 1 Φ 1, λ 2 = N 2 Φ 2. Φ m1 Φ m2 Forlinearmagneticsystems Φ l1 = N 1i 1 R l1 L l1 i 1, Φ m1 = N 1i 1 R m L m1 i 1, Φ l2 = N 2i 2 R l2 L l2 i 2, Φ m2 = N 2i 2 R m L m2 i 2,

16 λ 1 = L 11 i 1 + L 12 i 2 λ 2 = L 21 i 1 + L 22 i 2 L 11 = N 1 (L l1 + L m1 ), L 22 = N 2 (L l2 + L m2 ) L 12 = N 1 L m2 = N 1N 2 R m = N 2 L m1 = L 21 n = m =2,g = I 2, G =0,x =(λ 1, λ 2 ), e =(v 1,v 2 ), i 1 = φ 1 (λ 1, λ 2 ), i 2 = φ 2 (λ 1, λ 2 ) L 12 = L 21 implies that i = L 1 λ with L 1 symmetric Maxwell s reciprocity relations adding parasitic resistances means that G is no longer zero v 1 = r 1 i 1 + dλ 1 dt, v 2 = r 2 i 2 + dλ 2 dt

17 Assume now that L l1 = L l2 =0 λ 1 = λ 2 = N 1 N 2 R m N 1 N 2 R m µ N1 i 1 + i 2 N 2 µ i 1 + N 2 i 2 N 1 λ 1 N 2 N 1 = λ 2 Furthermore, if r 1 = r 2 = 0, it follows from the dynamics, that v 1 N 2 N 1 = v 2 This is half the ideal transformer relationships! The final assumption is this Let N 1 N 2 /R m become very large while N 1 /N 2 remains finite The only way for λ to remain finite is i 2 = N 1 N 2 i 1 The minus sign is due to the all-input power convention

18 Elementary electromagnet v + r i N Φ l Φ m F v = ri + dλ dt λ = NΦ Φ = Φ l + Φ m Φ l = Ni R l, Φ m = Ni R m but now R m depends on the air gap x: x R m (x) = 1 µ 0 A µ li +2x µ ri

19 This again a 2-port system with state variables λ, x, inputs v (voltage) and V (velocity), and outputs i = φ 1 (λ,x), F = φ 2 (λ,x). λ = µ N 2 + N 2 i =(L l + L m (x))i R l R m (x) L m (x) = N 2 = N 2 µ 0 A R l i m µ ri +2x b c + x From this i = φ 1 (λ,x) can be computed. Exercise: Compute the force F (λ,x) and the electromechanical energy W f (λ,x).

20 DC motor i a current into the page N θ S i a current out of the page a special mechanism with slip rings and brushes is necessary i f i f The main effectisdirectconversionofenergy without storing it in coordinate dependent elements τ = l 1 l 2 Bi a v emf = l 1 l 2 B θ = l 1 l 2 Bω r

21 Assuming linear magnetic materials B = L Af l 1 l 2 i f τ = L Af i f i a v emf = L Af i f ω r λ f = L f i f (the field flux) λ a = L A i a (the armature flux) p m = J m ω r (the mech angular momentum) λ f = r f i f + V f λ a = L Af i f ω r r a i a + V a ṗ m = L Af i f i a B r ω r T L T L is the external mech torque

22 Exercise: Write the bond graph corresponding to the dc motor presented Exercise: Show that, neglecting the field port, in the limit when L A and J m go to zero (define this formally!), and dissipations disapear, the DC motor becomes a pure gyrator.

23 Co-energy for electromech systems Suppose an electromechanical system with electrical state variables x e R n and geometrical state variables θ R m We have W f = W f (x e, θ) associated flows f i (x e, θ) = W f x ei (x e, θ), i =1,...,n Co-energy is defined as W c (f,x e, θ) = nx f i x ei W f (x e, θ) i=1... but W c x ei = f i W f x ei (x e, θ) =f i f i =0, i =1,...,n Hence, W c can be written as a function of only f (and θ)

24 Legendre transformation Widely used in analytical mechanics and thermodynamics While variables x are known as energy or Hamiltonian variables, transformed variables f are the co-energy or Lagrangian variables. Mechanical forces can be computed also from the co-energy W c (f,θ) =f x e (f,θ) W f (x e (f,θ), θ) θ W c (f,θ) = f θ x e xe W f θ x e θ W f = θ W f = F So mechanical forces can be computed as f F = θ W c notice again a reversed sign with respect to standard EE literature

25 For linear electromagnetic systems, q = C(x)v, λ = L(x)i, W c (v, i, x) =W f (q,λ,x) q=q(v,x),λ=λ(i,x) In particular, for linear magnetic systems, W f (λ,x)= 1 2 λt L 1 (x)λ, W c (i, x) = 1 2 it L(x)i Exercise: Prove the above equation. Exercise: Consider an electromechanical system with a nonlinear magnetic material such that λ =(a + bx 2 )i 2, where a and b are constants and x is a variable geometric parameter. Compute W f, W c and f, and check all the relevant relations.

26 Port Hamiltonian modeling of electromechanical systems Cast the previous systems as (explicit) port Hamiltonian models ẋ =(J (x) R(x))( H(x)) T + g(x)u x R n, J is antisymmetric, R is symmetric and positive semi-definite and u R m is the control H(x) is the energy (= W f (x)) y = g T (x)( H(x)) T

27 General electromechanical model H(λ,p,θ) = 1 2 λt L 1 (θ)λ pt J 1 m p λ are the generalized electrical energy variables (they may be charges or magnetic fluxes) p are the generalized mechanical momenta (linear or angular, or associated to any other generalized coordinate) θ are the generalized geometric coordinates λ + R e i = Bv ṗ = R m ω T e (λ, θ)+t m θ = Jm 1 p B indicates how the input voltages are connected to the electrical devices T e is the electrical torque T m is the external applied mechanical torque

28 i = L 1 (θ)λ = λ H, ω = J 1 m p = p H Notice that the port yielding T e is internal (connected to the mechanical inertia) T e = H θ = 1 2 λt θ L 1 (θ)λ = 1 2 λt L 1 θ LL 1 λ λ + R e i = Bv θ L 1 = L 1 θ LL 1 ṗ = R m ω T e (λ, θ)+t m θ = Jm 1 ẋ = R e R m x H + B µ v T m x =(λ p θ) T

29 This general model includes many of the classical electrical machines, as well as linear motors and levitating systems. Exercise: Write the general port Hamiltonian model for the electromagnet and for the variable geometry capacitor The dc motor needs a special formulation x T = λ f λ a p = J m ω r H(x) = 1 2 λl 1 λ T + 1 µ p 2 Lf 0, L = 2J m 0 L A J = L Af i f, R = r f r a 0 0 L Af i f B r g = I 3 u T = V f V a T L

30 The input voltages V f and V a can be obtained from the same source in several ways Shunt connection V a = V f and i T = i a + i f Series connection i a = i f and V T = V a + V f Exercise: Obtain the port Hamiltonian models of the shunt-connected and of the series-connected dc machines. The later is in fact an implicit port Hamiltonian model. Check this by writing the corresponding bond graph and identifying the differential causality assignment.

31 Power converters Control of electromechanical systems Portable battery-powered equipment UPS Power conversion systems (e.g. conditioning from fuel-cell, wind-based,...) Power converters dc-to-dc: to elevate or reduce dc voltages in a load-independent way ac-to-dc: rectifiers dc-to-ac: inverters dc-to-dc store the energy in intermediate ac elements an deliver it as needed trick: use of switches, operated in a periodic manner system alternates between several dynamics

32 Variable Structure Systems (VSS) VSS are non-smooth dynamical systems difficulties in simulation and control design averaging Assume the VSS cycles over several dynamics in a periodic way Compute the average of relevant variables over a cycle State Space Averaging (SSA) hxi(t) = 1 T Z t t T x(τ) dτ

33 VSS in PHDS form ẋ =[J (S, x) R(S, x)] ( H(x)) T + g(s, x)u S is a (multi)-index, with values on a finite, discrete set, enumerating the different structure topologies For simplicity, S {0, 1} S =0 ẋ =(J 0 (x) R 0 (x))( H(x)) T + g 0 (x)u S =1 ẋ =(J 1 (x) R 1 (x))( H(x)) T + g 1 (x)u if x does not vary too much during a given topology S =0 ẋ (J 0 (hxi) R 0 (hxi))( H(hxi)) T + g 0 (hxi)u S =1 ẋ (J 1 (hxi) R 1 (hxi))( H(hxi)) T + g 1 (hxi)u

34 the length of time in a given cycle with the system in a given topology is determined by a function of the state variables in our approximation, a function of the averages, t 0 (hxi), t 1 (hxi), with t 0 + t 1 = T x(t) =x(t T ) + t 0 (hxi) (J 0 (hxi) R 0 (hxi))( H(hxi)) T + g 0 (hxi)u + t 1 (hxi) (J 1 (hxi) R 1 (hxi))( H(hxi)) T + g 1 (hxi)u d dt hxi = d 0(hxi) (J 0 (hxi) R 0 (hxi))( H(hxi)) T + g 0 (hxi)u + d 1 (hxi) (J 1 (hxi) R 1 (hxi))( H(hxi)) T + g 1 (hxi)u d 0,1 (hxi) = t 0,1(hxi) T d 0 + d 1 =1 duty cycle

35 Boost (step-up) dc-to-dc converter i L s 1 v 1 s 1 closed (s 1 =1) i E + L i R s v L R 2 open (s 2 =0) i s 2 i C 1 and viceversa E v 2 v C v i R 2 only a single C boolean variable S = s 2 Two energy-storing elements: H = H C + H L Connected by Kirchoffs laws Two 1-d PHDS i L = i 1 + i 2 dq C dt dφ L dt = i C, v C = H q C = v L, i L = H φ L i 1 = i C + i R v 2 + v L = E v C + v 1 = v 2 v C = v R i E + i L = 0

36 Using the PHD equations, the first four KL are written as H φ L = i 1 + i 2 i 1 = dq C dt + i R v 2 + dφ L dt = E H + v 1 q C = v 2 Internal ports The second and third yield a 2-d PCH d dt µ qc φ L =0( H) T + µ i 1 v 2 i R E True ports

37 Theswitchesimposethefollowingconditions S =0 s 1 =1,s 2 =0 v 1 =0,i 2 =0 S =1 s 1 =0,s 2 =1 i 1 =0,v 2 =0 For S =1wehavethevaluesofv 2 and i 1 For S =0 H φ L = i 1 + i 2 i 1 = dq C dt v 2 + dφ L dt = E H + v 1 q C = v 2 + i R i 1 = H φ L, v 2 = H q C In compact form i 1 = (1 S) H φ L v 2 = (1 S) H q C

38 Substituting i 1 and v 2 back in the 2-d PHS d dt µ qc φ L = µ = µ µ 0 1 S H/ qc (1 S) 0 H/ φ L + (1 S) H/ φ L (1 S) H/ q C i R E µ µ ir E J g Skew symmetric y = µ Not a coincidence! T µ H/ qc H/ φ L = Is a consequence of Kirchoff s laws being an instantiation of Dirac structures and FE T + EF T =0 µ µ vc vr = i L i E last two KL

39 Now we can terminate the resistive port and introduce some dissipation i R = v R R = v C R = 1 R H q C d dt µ qc φ L = µ 0 1 S (1 S) 0 µ 1/R µ H/ qc H/ φ L µ E symmetric, positive semidefinite matrix Only a single natural output y = µ 0 1 T µ H/ qc H/ φ L = i L = i E

40 So the boost converter, a VSS, has a PHDS representation. The three basic second order dc-to-dc converters have this representation ẋ = µ 1/R α βs (α βs) 0 ( H(x)) T + µ 0 1 γs E Converter α β γ buck boost buck-boost x =(q C φ L ) T Assuming linear elements, H(q c, φ L )= 1 2C q2 C + 1 2L φ2 L

41 µ ẋ = 1/R α βs (α βs) 0 ( H(x)) T + µ 0 1 γs E Let us compute the averaged equation µ µ d 1/R α 0 dt hxi = d 0(hxi) ( H(hxi)) T + α 0 1 µ 1/R α β + d 1 (hxi) ( H(hxi)) T + (α β) 0 d 0 + d 1 =1 µ µ 1/R α βd = 1 (hxi) ( H(hxi)) T + (α βd 1 (hxi)) 0 E µ 0 1 γ 0 1 γd 1 (hxi) E E hsi =0 d 0 +1 d 1 d dt hxi = µ 1/R α βhsi (α βhsi) 0 VSSaveragedPHDSmodel µ ( H(hxi)) T γhsi E

42 Better averaged descriptions of VSS can be obtained by considering higher order Fourier-like coefficients, instead of just hxi. k-phasors hxi k (t) = 1 T Z t t T x(τ)e jkωτ dτ The phasor averages of VSS PHDS are also PHDS See printed notes for details

43 A storing and conditioning energy system Many metropolitan electrical-based vehicles use dc-motors which draw their power from segmented dc power lines Vehicles are able to brake in a regenerative way, returning power to the line Power can be reused only if another vehicle is accelerating in the same segment of the power line if not, this power is dissipated in special resistors and lost Problem: install in each power segment a device to store the excess energyandreturnitwhenneededattherequiredvoltage

44 Due to several reasons (technical and economical) the storage elements must be either supercapacitors or mechanical flywheels A power conversion system must be put in the middle to deliver the energy in the required form r 1 L 1 diode r 2 L 2 R IGBT r N L N i l r C S i T i C sc

45 The causal bond graph of the system is (power conventions are not displayed) S f : i l R: R C: C 1 R: r 1 0 S S i S N T 1 T i T N R: r 1 R: r i R: r N I: L 1 1 I: L i 1 I: L N 1 0 C: C sc internal causality of each pair of switches is not fixed by external causality

46 Two combinations of internal causality are possible e a i a e a i a S i 1 S i 1 T i 0 T i 0 closed ideal switch: zero voltage source e b i b e b i b open ideal switch: zero current source If S i =1whenS i is closed and S i =0whenS i is open, then e b = S i e a i a = S i i b

47 Each pair of switches can be modeled as a modulated transformer e a i a S i MTF i e b = S i e a e b i b and then i a = S i i b i = N S i R: r i i = N S f : i l 1 0 MTF i 1 i = 1 R: R 1 R: r I: L i 0 i = 1 C: C sc C: C

48 Exercise: Writethestateequationsfortheabovebondgraph. If you solve the above exercise, you will see that the resulting equations can be written in PHDS form with state variables q (charge of C), Q (charge of C sc )andλ i (flux of L i, i =1,...,N) ẋ =(J R) x H + gi l J = S 1 S N 1 S S N 0 0 H(Q, q, λ i )= 1 Q C sc 2C q2 + This can be obtained also from the interconnection of the individual PHD subsystems, as we did for the boost. R = NX i=1 1 2L i λ 2 i µ N 0 N 2 δ ij r j + rs i S j g = 0 1 rs 1 rs 2. rs N

49 The model can be expanded to include the power line + the electrical vehicles, or the supercapacitor can be substituted by a normal one attached to a dc-motor and a flywheel Important modeling remarks The transformer model obtained is dependent on the external causality imposed upon each pair of switches. It has to be recomputed if something different is connected that changes that. The model has produced four-quadrant switches: there is no constraint on the signs of currents and voltages through them. In practice, the type of switch displayed is two-quadrant: it can sustain any current but only non-zero voltages in one way.