Understanding Coupling Fields in Electromechanical Systems

1. EE 5592016Iowa State University WE generators – Coupling Fields James D. McCalley Harpole Professor of Electrical & Computer Engineering

2. Let’s devise a way so that a similar rotating magnetic field is produced by the windings on the rotor, such that it rotates at the same speed as that from the stator. Thus, we now have two rotating magnetic fields. The two rotating magnetic fields, that from the rotor and the composite field from the armature, are “locked in,” and as long as they rotate in synchronism, a torque (Torque=P/ωm=Force×radius, where Force is tangential to the rotor surface), is developed. This torque is identical to that which would be developed if two magnetic bars were fixed on the same pivot as shown on the next slide. Torque production

3. In the case of generator operation, we can think of bar A (the rotor field) as pushing bar B (the armature field), as in Fig. a. In the case of motor operation, we can think of bar B (the armature field) as pulling bar A (the rotor field), as in Fig. 11b. Torque production To compute the torque, we must make use of coupling fields and co-energy, according to:

4. One can make conceptual use of a version of the Lorentz Force Law, below, in computing torque exerted on a current -carrying conductor in the presence of a magnetic field. Torque production However, most electromechanical energy-conversion devices contain magnetic material; in these systems, forces of the B-field act magnetically on the magnetic material, and also on any current-carrying conductors. So the above relation is not enough for us to compute force. Instead, we turn to an energy method, which makes use of The above two relations will allow us to express torque via energy relations.

5. In electromechanical systems, we have an electrical system, a mechanical system, and a coupling field. We assume in our work that the coupling field is magnetic. Coupling fields Coupling field Most of the coupling field energy is stored in airgap because it has very large “resistance” (reluctance) to the flux. Electrical system Mechanical system Example of an elementary translational-motion electromechanical system with one electrical input, one mechanical input, and one coupling field. P. Krause, O. Wasynczuk, & S. Sudhoff, “Analysis of Electric Machinery,” IEEE Press, 1995.

6. Mechanical system Coupling field Coupling fields Coupling field Electrical system Example of an elementary rotational-motion electromechanical system with one electrical input, one mechanical input, and one coupling field. A. Fitzgerald, C. Kingsley, and A. Kusko, “Electric Machinery, 3rd edition, 1971

7. Mechanical system is the rotational motion of the rotor. Coupling fields Coupling field is in the flux path of the iron path/air gap Electrical systems are the two coils. Example of an elementary rotational-motion electromechanical system with two electrical inputs, one mechanical input, and one coupling field. P. Krause, O. Wasynczuk, & S. Sudhoff, “Analysis of Electric Machinery,” IEEE Press, 1995.

8. The importance of the coupling field is that this is where the energy conversion takes place. If losses in the coupling field (eddy current, hysteresis) are neglected or assumed to be modeled external to the coupling field, then Coupling fields: torque Energy transferred to the coupling field by the electrical system Energy transferred to the coupling field by the mechanical system Energy stored in the coupling field + = OR In differential form, The last equation enables computation of torque. We investigate dWm and dWe next. We consider a rotational system with just 1 electrical input and 1 mechanical input.

9. Generalization of systems with 1 electrical input, 1 mechanical input, and 1 coupling field. Coupling fields: torque This is neglected, or it is represented within the model of the electrical or the mechanical system. A. Fitzgerald, C. Kingsley, and A. Kusko, “Electric Machinery, 3rd edition, 1971

10. From previous slide, we observe the electrical terminals at the coupling field receive current i with voltage e. Thus power is ei and the total energy supplied by the electrical source is: Coupling fields: torque In electromagnetic systems, the coupling element is a machine winding, and e will be an induced voltage according to Faraday’s Law: Therefore: For the following development, see (a) P. Krause, O. Wasynczuk, & S. Sudhoff, “Analysis of Electric Machinery,” IEEE Press, 1995; (b) A. Fitzgerald, C. Kingsley, and A. Kusko, “Electric Machinery, 3rd edition, 1971; (c) J. Meisel, “Principles of electromechanical-energy conversion,” McGraw-Hill, 1966.

11. The mechanical torque Tm and the electromagnetic torque Te are assumed positive in the same rotational direction. We neglect torque due to shaft twist. Coupling fields: torque J: moment of inertia D: damping Then the total energy supplied by the mechanical source is This equation is the rotational analogue to the one for linear motion, which is Term 1 on RHS is stored inertial energy; Term 2 is losses. Only term 3 is energy transferred to coupling field. So…

12. Substituting expressions (b) and (c) into (a) results in Coupling fields: torque Solving for the torque term results in: (*) Assume independent variables of our system are i, θ. Then: Taking the total differential Substitute these last equations into (*)…

13. Coupling fields: torque Gather like terms in di and dθ: Equate coefficients: The first expression provides a way to compute torque: (T1)

14. Define co-energy: Recalling that independent variables of our system are i, θ: Coupling fields: torque Solve for the last term on the right: Substitute into (T1): (T1) 0 (T2)

15. What is co-energy? An expression for which torque computation is convenient (T2) Coupling fields: torque To better understand co-energy, consider the relation for the coupling field energy: Set the system so that initially, Wf=0 (no stored energy), and fix θ so that no energy can be added to the coupling field via mechanical means, i.e., Wm=0. Then increase the current to a value ia, establishing a corresponding flux linkage of λa. There is then energy in the coupling field, but only via electrical means, i.e.

16. But recall that: Coupling fields: torque And since Wm=0, this is also the energy of the coupling field: We observe the corresponding area in the λ-i curve in the figure. Recall the definition of co-energy: Since iλ is the area of the (shaded) box, then Wc must be the area below the curve. Wc is therefore given by

17. If the medium is magnetically linear (no saturation), then the λ-i curve is just a diagonal through the iλ shaded box, as shown. In this case, the area above the λ-i is the same as the area below it, and we have: Coupling fields: torque Recall (T2), which is Therefore, under condition of magnetic linearity, we have: (T3)

18. Recall from basic physics that the energy stored in a winding of self-inductance Lpp carrying current i is given by: Coupling fields: energy with Lpp in henries and defined by Lpq=λp/Iq=NpNq/Rpq for linear medium; Rpqis path reluctance (like resistance). Generalization: For a linear electromagnetic system with J electrical inputs (windings), the total field energy is given by: where Lpq is the winding’s self inductance when p=q and when p≠q, it is the mutual inductance between the two windings. Derivation: Given pp 22-24 of Krause, Wasynchzuk, and Sudhoff, “Analysis of electric machinery,” 1995.

19. This device consists of two conductors, #1 is on the stator; #2 is on the rotor. The magnetic system is assumed linear. Example So Wf is given by (with J=2): The self-inductances, given by Lpp=λp/ip, are constant, independent of θ, because the reluctance of the path seen by the winding does not change as the rotor turns. But the mutual inductances are not constant.

20. The mutual inductances, given by Lpq=λp/iq, i.e., the amount of flux seen by winding p due to a current in winding q, are not constant. The extreme Example conditions of maximum & minimum linkages are given below. θ=0° condition (maximum negative linkage) θ=0° condition (maximum positive linkage) θ=90° condition (zero linkages) The mutual inductance goes from max positive to 0 to max negative to 0 and back to max positive. Thus, L12, L21 are:

21. θ=0° condition (maximum negative linkage) θ=0° condition (maximum positive linkage) θ=90° condition (zero linkages) Example The mutual inductance goes from max positive to 0 to max negative to 0 and back to max positive. Thus, L12, L21 are:

22. Summarizing, we have. L11, L22 are constant. Example Substituting, we obtain Recalling , we obtain

23. The previous procedure can be applied to a three-phase induction machine to obtain an expression for its torque. The effort requires a coordinate transformation which is involved, and so we will not do it. The resulting relation for torque (see Meisel, pp. 321-322, and chapter 12) may be used to derive the torque expression for steady-state conditions. We will use this relation in the next set of material. Summary