University:
Massachusetts Institute of Technology
Course:
2.141 | Modeling and Simulation of Dynamic Systems
Academic year:

2019
Views:

246

Pages:

18
Author:

prediguemzhv

CO-ENERGY (AGAIN) In the linear case, energy and co-energy are numerically equal. —the value of distinguishing between them may not be obvious. Why bother with co-energy at all? EXAMPLE: SOLENOID WITH MAGNETIC SATURATION. Previous solenoid constitutive equations assumed electromagnetic linearity. —arbitrarily large magnetic fluxes could be generated. In reality flux cannot exceed saturation flux. For sufficiently high currents behavior is strongly nonlinear. MODEL THAT PHENOMENON. Assume an electrical constitutive equation as follows: λ(i, x) = L(x) i λs L(x)2i2 + λs2 where L(x) is position-dependent inductance as before. For sufficiently small currents the relation is approximately linear. λs i << L(x) λ ≈ L(x) i For sufficiently large currents the flux linkage reaches a limiting value, λs. λs i >> L(x) λ ≈ λs MECHANICAL CONSTITUTIVE EQUATION may be found using the stored energy. —Find the stored energy at a fixed displacement. —Find force as the gradient with respect to displacement. That yields the relation between force and flux linkage. F = F(λ, x) But flux cannot be specified arbitrarily. Realistic boundary conditions require current input. To find the relation between force and current, substitute. λ = λ(i, x) STORED ELECTRICAL ENERGY (at a fixed displacement) E = ⌠⌡i dλ Need to invert the relation between flux linkage and current. —In general, anything but straightforward. In this case, a little algebra yields the following. λs i = L(x) λ λs2 – λ2 |λ| ≤ |λs| USE WITH CAUTION! If λ > λs this expression yields an imaginary number for the current. A LITTLE CALCULUS (and some more algebra) yields an expression for energy. λs2 ⎛ E(λ, x) = L(x) ⎜1 – ⎝ ⎛ λ ⎞2⎞ 1–⎜ ⎟⎟ ⎝λs⎠ ⎠ see Note 1 attached Partial derivative with respect to displacement: 2 ⎛ ⎛ λ ⎞ ⎞⎟ ∂L(x ) λ s2 ⎜ F(λ, x ) = − 1 − 1 − ⎜⎜ ⎟⎟ ⎟ 2 ⎜ ∂x L(x ) ⎜ ⎝ λ s ⎠ ⎟⎠ ⎝ Substitute for flux linkage: ⎛ ⎛ L(x )i ∂L(x ) λ ⎜ ⎜ F(i, x ) = − 1 1 − − ⎜ ⎜ ( )2 2 ∂x L(x )2 ⎜ L x i + λ 2S ⎝ ⎝ 2 s ⎞ ⎟ ⎟ ⎠ 2 Still more algebra simplifies this expression: λS ∂L(x ) λ s2 ⎛⎜ F(i, x ) = − 1 − 2 2 2 ∂x L(x )2 ⎜ ( ) L x i λ + S ⎝ ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎟ ⎠ COMMENT: The assumed electrical constitutive equation λ(i, x ) = L(x )iλ s L(x ) i 2 + λ s2 2 was chosen primarily for pedagogic simplicity —a more realistic relation may be (far) less tractable —usually no simple algebraic form exists Consequently a procedure requiring (a) inversion (b) integration with respect to flux linkage and (c) differentiation with respect to displacement may be impractical. AN ALTERNATIVE APPROACH: use co-energy instead of energy. Total stored energy: E = ⌠⌡ i dλ + ⌠⌡ F dx Electrical co-energy: E* = iλ – E = iλ – ⌠⌡ i dλ – ⌠⌡ F dx (a Legendre transformation with respect to current) Force is the negative gradient of this co-energy. ∂E * ∂E =− = −F ∂x ∂x Electrical co-energy at a fixed displacement: E* = ⌠⌡ λ di λs ⎛ E*(i, x) = L(x) ⎝ L(x)2i2 + λs2 – λs⎞⎠ see Note 2 attached KEY POINT: Co-energy may be found without inverting the relation between flux linkage and current. Partial differentiation with respect to displacement: ∂L(x ) λ s2 ⎛⎜ λs F(i, x ) = − 1 − 2 2 2 ∂x L(x )2 ⎜ ( ) L x i λ + s ⎝ see Note 3 attached No further algebra required. ⎞ ⎟ ⎟ ⎠ REMARKS In general, co-energy functions are useful for multiport and/or nonlinear energy storage elements. Inverting constitutive equations may be avoided. In the linear case, energy and co-energy are numerically equal. the value of distinguishing between them may not be obvious. In the nonlinear case, the two are not equal. Distinguishing between them is important. In this example, energy is upper-bounded co-energy is not. In general, energy is conserved co-energy need not be. NOTE 1: *************************************** Integration to find energy: λ ⌠ λs E(λ, x) = ⎮L(x) ⌡ 0 y dy |λ| ≤ |λs| 2 2 λs – y where y is a “dummy variable”. rearrange: λ 1 E(λ, x) = ⌠L(x) ⎮ ⌡ 0 substitution: define y dy ⎛y⎞ 1 – ⎜ ⎟2 ⎝λs⎠ sin u = y λs ⎛ y ⎞2 1 – ⎜ ⎟ = 1 – sin2 u = cos2 u = cos u ⎝λs⎠ y = λs sin u ⎛y⎞ u = sin-1⎜ ⎟ –π/2 ≤ u ≤ π/2 ⎝λs⎠ dy = λs cos u du (assuming dx = 0) sin-1(λ/λs) ⌠ 1 λs sin u ⎮L(x) cos u λs cos u du E(λ, x) = ⌡ 0 sin-1(λ/λs) ⌠ λs2 ⎮ L(x) sin u du E(λ, x) = ⌡ 0 sin-1(λ/λs) ⌠ λs2 ⎮ L(x) d(–cos u) E(λ, x) = ⌡ 0 λs2 ⎛ ⎛ ⎞ ⎛ λ ⎞⎞ E(λ, x) = –L(x) ⎜cos⎜sin-1⎜ ⎟⎟ – cos (0)⎟ ⎝ ⎝ ⎝λs⎠⎠ ⎠ ⎛ ⎛ λ ⎞⎞ cos⎜sin-1⎜ ⎟⎟ = ⎝ ⎝λs⎠⎠ λs2 ⎛ E(λ, x) = L(x) ⎜1 – ⎝ ⎛λ⎞ 1 – ⎜ ⎟2 ⎝λs⎠ ⎛ λ ⎞2 ⎞ 1–⎜ ⎟⎟ ⎝λs⎠ ⎠ Does that make sense? In the limit as λ approaches λs λs2 E(λ, x) ≈ L(x) Plausible; implies that only a finite amount of energy may be stored. In the limit as λ approaches 0, binomial series expansion of the root yields 1 ⎛ λ ⎞2 ⎛ λ ⎞2 1 – ⎜ ⎟ = 1 – 2 ⎜ ⎟ + higher order terms ⎝λs⎠ ⎝λs⎠ λs2 ⎛ 1 ⎛ λ ⎞2 ⎞ E(λ, x) ≈ L(x) ⎜1 – 1 + 2 ⎜ ⎟ ⎟ ⎝λs⎠ ⎠ ⎝ λ2 E(λ, x) ≈ 2 L(x) as expected for a linear inductor. NOTE 2: *************************************** Integration to find co-energy: i ⌠ L(x) i λs E*(i, x) = ⎮ di 2 2 2 ⌡ L(x) i + λs 0 ⎛ d⎜⎝ L(x)2i2 + ⎞ λs2⎟⎠ = L(x)2 i di L(x)2i2 + λs2 i λs ⌠ ⎛ ⎞ E*(i, x) = L(x) ⎮⌡d⎜⎝ L(x)2i2 + λs2⎟⎠ 0 λs ⎛ E*(i, x) = L(x) ⎜⎝ L(x)2i2 + λs2 – Does this make sense? ⎞ λs2⎟⎠ In the limit as L(x) i >> λs E* ≈ λs i This is the area of a rectangle of sides i and λs. Note that co-energy may increase without bound, whereas energy may not. In the limit as L(x) i << λs series expansion of the square root 1 L(x)2i2 + λs2 = (λs2)1/2 + 2 (λs2) -1/2 L(x)2i2 + higher order terms ⎞ λs ⎛⎜ 1 2 1/2 2 -1/2 2 2 2 1/2 E*(i, x) ≈ L(x) ⎜(λs ) + 2 (λs ) L(x) i – (λs ) ⎟⎟ ⎝ ⎠ 1 E*(i, x) ≈ 2 L(x) i2 as expected for a linear inductor. NOTE 3: *************************************** Partial differentiation with respect to displacement: ∂E * F(i, x ) = − ∂x 2 ⎛ ( ) ∂E * ∂L(x ) λ s ⎛ λ L x i 2 2 2 ⎜ s L(x ) i + λ s − λ s ⎞⎟ + =− 2 ⎜ ⎠ L(x ) ⎜ L(x )2 i 2 + λ 2 ∂x ∂x L(x ) ⎝ s ⎝ 2 2 2 2 2 2 2 2 ∂E * ∂L(x ) λ s ⎛⎜ L(x ) i + λ s − λ s L(x ) i + λ s - L(x ) i ⎞⎟ =− 2 2 ⎟ 2 ∂x ∂x L(x )2 ⎜ ( ) L x i λ + s ⎝ ⎠ ∂L(x ) λ s2 ⎛⎜ λs F(i, x ) = − 1 − 2 2 2 ∂x L(x )2 ⎜ ( ) L x i λ + s ⎝ ⎞ ⎟ ⎟ ⎠ ⎞ ∂L(x ) ⎟ ⎟ ∂x ⎠ An expanded form may be easier to understand physically: 2 2 2 2 2 2 ∂L(x ) λ s2 ⎛⎜ L(x ) i + λ s − λ s L(x ) i + λ s F(i, x ) = − 2 ∂x L(x )2 ⎜ L(x ) i 2 + λ s2 ⎝ ⎞ ⎟ ⎟ ⎠

Solenoid and Co-Energy

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