22-Elec-A6 Power Systems and Machines · May 2017
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (transformers ch. 2, DC machines ch. 8, synchronous machines ch. 4–5, induction machines ch. 6); J. D. Glover et al., Power System Analysis and Design, 6th ed. (per-phase and power-factor analysis, ch. 2).
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.
(a) Power factor correction. Power-factor correction is the deliberate addition of a leading (capacitive) reactive source — shunt capacitor banks or an over-excited synchronous condenser — in parallel with predominantly inductive load so that the reactive power is supplied locally rather than drawn from the source. Because real power is unchanged while the reactive component of current is cancelled, the line current falls for the same delivered power. This lowers $I^2R$ feeder and transformer losses, releases thermal capacity in cables and transformers, improves voltage regulation, and avoids the low-power-factor penalty most utilities levy.
(b) Increasing generator excitation on an infinite bus. The infinite bus fixes terminal voltage and frequency, and the prime-mover throttle fixes the real power. Raising the DC field current increases the internal EMF magnitude $E_f$ but cannot change $P=3V_tE_f\sin\delta/X_s$ (the machine simply reduces $\delta$ so the product $E_f\sin\delta$ stays constant). What changes is reactive power: the machine becomes over-excited and delivers lagging vars to the bus. In short, field current controls reactive power (Q), the prime mover controls real power (P).
(c) Reversing a three-phase induction motor. Interchange any two of the three stator supply leads. This reverses the phase sequence applied to the windings, which reverses the direction of the rotating air-gap flux and therefore the direction of rotation.
(d) Why the rotor cannot reach synchronous speed. Torque in an induction machine relies on relative motion (slip) between the rotating flux and the rotor conductors. If the rotor turned at synchronous speed there would be zero relative velocity, hence no rate of change of flux linkage, no induced rotor EMF, no rotor current, and no torque. Some slip must always exist to induce the current that produces torque — which is why the machine is called asynchronous.
(e) Power-angle relation (derivation). Take the per-phase equivalent with armature resistance neglected, terminal voltage as reference $V_t\angle 0^\circ$ and internal EMF $E_f\angle\delta$. The phasor diagram below places $jX_sI_a$ between $V_t$ and $E_f$.
From $E_f = V_t + jX_sI_a$ the armature current is $I_a=(E_f-V_t)/(jX_s)$. The per-phase real power delivered at the terminals is
$$P_{1\phi}=\operatorname{Re}\!\left[V_t\,I_a^{*}\right]=\operatorname{Re}\!\left[V_t\,\frac{E_f\angle(-\delta)-V_t}{-jX_s}\right].$$
Writing $\tfrac{1}{-jX_s}=\tfrac{j}{X_s}$ and keeping only the real part (the $\cos\delta$ and $V_t^2$ terms are imaginary) leaves $P_{1\phi}=V_tE_f\sin\delta/X_s$. For the three phases,
$$\boxed{\,P=\dfrac{3\,V_t\,E_f}{X_s}\sin\delta\,}$$
The output is maximum at $\delta=90^\circ$ (the steady-state stability limit); beyond that the machine loses synchronism.
(f) Low power factor at light load. The magnetizing current that establishes the air-gap flux is essentially constant and almost purely reactive, set by the applied voltage rather than by load. At light load the in-phase (torque-producing) component of stator current is small, so the total current is dominated by this fixed reactive magnetizing component and the power factor is low. As mechanical load increases the in-phase component grows and the power factor rises.
(g) Why a synchronous motor is not self-starting. At standstill the stator produces a field rotating at synchronous speed while the DC-excited rotor, held back by inertia, is stationary. The field sweeps past the rotor poles so quickly that the torque alternates in direction and averages to zero over each cycle; the heavy rotor cannot accelerate to lock in. Starting therefore requires an auxiliary means — a damper (amortisseur) winding that lets the machine run up as an induction motor before the field is applied, a pony motor, or a variable-frequency drive.
(h) Negative wattmeter reading. In the two-wattmeter method the meters read $W_1=V_LI_L\cos(30^\circ+\theta)$ and $W_2=V_LI_L\cos(30^\circ-\theta)$. When the load power-factor angle $\theta$ exceeds $60^\circ$ (i.e. power factor below $0.5$), the argument $30^\circ+\theta$ passes $90^\circ$ and $\cos(30^\circ+\theta)$ turns negative, so one meter reads down-scale. The total power is still $W_1+W_2$.
(i) Turbo-alternators vs hydro-alternators. Synchronous speed is $n_s=120f/p$. Steam turbines run efficiently only at very high speed (3600 rpm), so at 60 Hz just two poles ($p=2$) give the required frequency — few poles suffice. To survive the centrifugal stress at that speed the rotor is a small-diameter, cylindrical (round, non-salient) forging. Water turbines run slowly, so many poles are needed; their alternators use large-diameter salient-pole rotors turning at low speed.
(j) Poor efficiency at high slip. Rotor copper loss equals $s\,P_{ag}$ while developed mechanical power equals $(1-s)P_{ag}$. At high slip a large fraction of the air-gap power is therefore burned as heat in the rotor rather than converted to shaft work, so efficiency collapses — the reason normal running slip is kept to a few percent.
