22-Elec-A6 Power Systems and Machines · December 2018
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format: Closed-book, 3 hours. Five questions constitute a complete paper and all are of equal value (20 marks each); answer all five. All AC voltages and currents are rms, three-phase voltages are line-to-line, and power is total real power unless noted otherwise.
Reference texts (22-Elec-A6 Power Systems and Machines): S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill); B. S. Guru & H. R. Hiziroglu, Electric Machinery and Transformers, 3rd ed. (Oxford); P. C. Sen, Principles of Electric Machines and Power Electronics, 2nd ed. (Wiley).
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) Loss of one supply leg on a delta-fed motor. The motor will keep running, but only marginally and with a serious de-rating; it will not restart once stopped. Opening one of the three supply lines turns the balanced three-phase supply into a single-phase supply across the remaining two lines (the machine is now "single-phasing"). A running motor continues to rotate because its own rotation still produces a rotating-field component, but the current in the two energized lines rises sharply (roughly by $\sqrt{3}$ for the same load), torque pulsates at twice line frequency, and the windings overheat — thermal protection should trip it. Critically, a single-phase supply produces only a pulsating (not rotating) field at standstill, so the motor develops zero average starting torque and cannot start from rest.
If "one of the power supply legs" is instead read as one winding of the delta-connected source (for example, one transformer of a delta bank failing open) while all three lines stay connected, the answer is yes: the two remaining windings form an open-delta (V–V) connection, which still delivers a balanced set of three line-to-line voltages, so the motor starts and runs normally. The catch is capacity: an open-delta bank can carry only $1/\sqrt{3}\approx57.7\%$ of the original three-winding kVA, so the motor load must be reduced accordingly to avoid overloading the two remaining windings.
(b) Generator droop. Droop is the deliberate, near-linear decline of a generator's speed (frequency) as its real-power output increases, set by the prime-mover governor. It is expressed as a percentage, $\text{droop}\% = \dfrac{f_{nl}-f_{fl}}{f_{fl}}\times100\%$, where $f_{nl}$ is the no-load frequency and $f_{fl}$ the full-load frequency. Equivalently the operating characteristic is a straight line $f = f_{nl} - m\,P$ with slope $m = \dfrac{\text{droop}\%\times f_{nom}}{P_{rated}}$ (Hz/kW). Droop lets two or more machines share load stably in parallel: the common bus frequency sets each unit's output on its own droop line, and load changes divide between machines in inverse proportion to their slopes — the "stiffer" (low-droop) machine picks up the larger share.
(c) Why a synchronous motor is not self-starting. A synchronous machine develops steady torque only when its rotor turns in step with the stator's rotating magnetic field. At standstill the stator field sweeps past the stationary DC-excited rotor poles at synchronous speed, so the torque on the rotor reverses every half-cycle and its average value is zero — the rotor, having inertia, cannot follow the field and simply vibrates. It must be brought up to near synchronous speed by another means (most commonly a squirrel-cage "amortisseur" winding that starts it as an induction motor) before the DC field is applied to pull it into synchronism.
(d) Five specifications for selecting an induction motor. (1) Rated power (kW or hp) and duty cycle; (2) rated voltage, number of phases and frequency; (3) synchronous/full-load speed, i.e. the number of poles and rated slip; (4) NEMA/IEC torque–speed design class (starting torque, breakdown torque and starting current, e.g. NEMA A/B/C/D); and (5) enclosure and insulation/thermal class for the environment (TEFC, IP rating, service factor, ambient and altitude). Efficiency class (e.g. IE3/IE4) and frame size are further common selection criteria.
(e) Why low power factor is penalized. A low power factor means the customer draws a large reactive component, so the apparent power (kVA) — and hence the current — is much larger than the billed real power (kW) would suggest. That extra current loads the utility's generators, transformers and lines, increasing $I^2R$ losses and voltage drop and consuming capacity that cannot be sold as energy. The utility must size its equipment for the kVA, not the kW, so it recovers this cost through a power-factor penalty (or a kVA/kVA-demand charge), which also incentivizes the customer to install correction capacitors.
(f) 18 kVA, 20 kV/480 V, 60 Hz transformer supplying 15 kVA to a 415 V load at 50 Hz. The kVA is not the problem — 15 kVA is below the 18 kVA rating. The problem is core flux. Peak flux density is proportional to $V/f$ ($B_{max}\propto V/(f N A)$), so operating at a lower frequency raises the flux unless the voltage is lowered in the same proportion. The safe secondary voltage at 50 Hz is $V \le 480\times\frac{50}{60} = 400\ \text{V}$. The proposed 415 V exceeds this by $\frac{415-400}{400}=3.75\%$, so the core would be driven $\approx3.75\%$ into over-flux — increased magnetizing current, saturation, extra core heating and audible noise. It is not safe. It would be acceptable only if the 50 Hz voltage were held at or below 400 V.
(g) Poor efficiency at high slip. In an induction motor the rotor copper loss equals the slip times the air-gap power, $P_{rcl} = s\,P_{ag}$, while the mechanical power developed is $P_{conv} = (1-s)P_{ag}$. The rotor-circuit efficiency is therefore at most $(1-s)$. At high slip a large fraction of the air-gap power is dissipated as $I^2R$ heat in the rotor instead of being converted to shaft power — at 50% slip half the air-gap power is lost in the rotor. High slip also means large rotor and stator currents and correspondingly large stator copper and stray losses, so overall efficiency collapses.
(h) Three causes of harmonics and their effects. Causes: (1) power-electronic converters — rectifiers, variable-frequency drives, switch-mode supplies — that draw current in non-sinusoidal pulses; (2) saturated magnetic devices such as transformers and reactors operating on the knee of their B–H curve; and (3) other non-linear loads such as arc furnaces, welders and gas-discharge/fluorescent lighting. Effects: additional $I^2R$ and eddy/hysteresis heating of transformers, motors and cables (leading to de-rating), overloaded neutral conductors from triplen (3rd-harmonic) currents, nuisance breaker/fuse operation, resonance with power-factor capacitors, torque pulsation in motors, and interference with metering, protection and communication circuits.
(i) Why the transformer core is laminated, and three causes of core damage. The core is built from thin, individually insulated silicon-steel laminations to break up the paths available to induced eddy currents. Eddy-current loss rises with the square of the lamination thickness, so thin sheets (typically ~0.3 mm) drastically reduce that loss and the associated heating; silicon content also raises resistivity and lowers hysteresis loss. Three causes of core damage: (1) over-fluxing/over-excitation (too high $V/f$) driving the core into saturation and overheating; (2) shorted or degraded interlaminar insulation causing localized eddy-current "hot spots" and burning; and (3) mechanical damage or looseness — clamping failure, vibration and the resulting insulation abrasion, or moisture/contamination corroding the laminations.
