22-Elec-A6 Power Systems and Machines · May 2015
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill) — Ch. 2 transformers, Ch. 6–7 induction machines, Ch. 5 synchronous machines, Ch. 8–9 DC machines; A. E. Fitzgerald, C. Kingsley & S. Umans, Electric Machinery, 6th ed.; P. C. Sen, Principles of Electric Machines and Power Electronics, 3rd ed.
Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.
This is a short-answer conceptual question; each part is answered in turn, with the governing physics stated explicitly.
(a) Why the core is laminated. The alternating flux in the core induces circulating eddy currents in the iron, which is itself a conductor. If the core were solid, these currents would flow in large loops and dissipate significant $I^2R$ heat. Building the core from thin sheets (typically 0.3–0.5 mm) that are insulated from one another with an oxide or varnish film breaks those loops into many small, high-resistance paths. Eddy-current loss scales roughly as $P_e \propto t^2 B_m^2 f^2$, where $t$ is the lamination thickness, so thin laminations sharply reduce the loss while leaving the useful magnetic path essentially unchanged.
(b) Torque–speed curve. The characteristic rises from a finite starting torque at standstill ($n=0$, slip $s=1$), climbs to a breakdown (pull-out) torque near 80 % of synchronous speed, then falls steeply through a nearly linear stable region to zero torque at synchronous speed $n_s$. The machine operates on that steep stable segment just below $n_s$.
(c) Reversing a three-phase induction motor. Interchange any two of the three stator supply leads. Swapping two phases reverses the phase sequence, which reverses the direction of the rotating stator field and hence the direction of rotation.
(d) Two speed-control methods for three-phase induction motors. (1) Variable-frequency control — supply the stator from an inverter that varies the frequency $f$ while holding $V/f$ constant, which shifts the synchronous speed $n_s = 120f/P$ and slides the whole curve along the speed axis. (2) Pole changing — reconfigure the stator winding (consequent-pole or two separate windings) to change the pole number $P$, giving discrete synchronous speeds. (For wound-rotor machines, adding external rotor resistance is a third method.)
(e) Why a DC shunt motor needs a starter. The armature current is $I_a = (V - E_a)/R_a$, where the back-emf $E_a = K\phi\,\omega$. At the instant of starting the rotor is stationary, so $E_a = 0$ and the current is limited only by the small armature resistance: $I_a = V/R_a$, which for this machine would be $250/0.4 = 625$ A — many times rated. A starter inserts external resistance in series with the armature to hold the starting current to a safe value, and the resistance is cut out in steps as the motor accelerates and $E_a$ builds up.
(f) Low power factor at light load. An induction motor always draws a substantial, roughly constant magnetizing current (largely reactive) to establish the air-gap flux across the air gap. The in-phase, torque-producing component of current is proportional to the shaft load. At light load that active component is small while the fixed magnetizing reactive component dominates, so $\text{pf} = P/S = \cos\theta$ is low; as the load increases the active current grows and the power factor improves.
(g) Two ways to reverse a DC shunt motor. The direction depends on the relative sense of armature current and field flux, so reverse one of them (reversing both leaves the direction unchanged): (1) interchange the two armature terminal connections, or (2) interchange the two field winding connections.
(h) Two ways to vary the speed of a DC shunt motor. From $\omega = (V - I_aR_a)/(K\phi)$: (1) Field control — a rheostat in the shunt-field circuit changes $\phi$; weakening the field raises the speed above base speed (used in part e–h below). (2) Armature-voltage control — vary the voltage applied to the armature to obtain speeds below base speed at constant flux. (Armature-circuit series resistance is a third, lossy option.)
(i) Two reasons the synchronous motor is industrially useful. (1) It runs at exactly constant (synchronous) speed $n_s = 120f/P$, independent of load up to pull-out, which is valuable for constant-speed drives. (2) Its power factor is adjustable through field excitation: run over-excited, it draws leading current and can supply reactive power to correct the plant power factor — the synchronous-condenser action exploited in Question 5. Large machines are also highly efficient.
