Question 5 of 5: Short-Answer Concepts (a)–(j)

Question 5: Short-Answer Concepts (a)–(j) (20 marks)

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) Three advantages of three-phase over single-phase distribution. First, the total instantaneous power of a balanced three-phase system is constant (the three pulsating phase powers sum to a steady value), so generators and motors deliver smooth, ripple-free torque with less vibration. Second, three-phase transmits more power for a given amount of conductor material — the copper (or aluminium) is used more effectively, giving lower transmission losses and cost per kW delivered. Third, three-phase currents inherently set up a rotating magnetic field, which makes robust, self-starting three-phase induction motors possible and lets machines of a given rating be smaller and lighter. (The exam's trailing sentence about steam-turbine alternators needing "only few poles" is unrelated to this list — it is correct in itself, since 3600-rpm turbo-alternators are 2-pole machines, but it is not one of the three advantages.)

(b) Increasing excitation on an infinite bus. The bus fixes both frequency and terminal voltage, and with the prime-mover throttle unchanged the real power output stays essentially constant. Increasing the field current raises the internal emf $E_a$. Since $P=3V_\phi E_a\sin\delta/X_s$ is fixed, a larger $E_a$ forces the power angle $\delta$ to decrease. The machine becomes over-excited and begins to deliver lagging reactive power to the bus — the armature current grows and shifts to a more lagging angle. In short, over-excitation makes the generator supply vars (act capacitively toward the system) without changing its real-power output; under-excitation would make it absorb vars.

(c) Why the synchronous motor is not self-starting. At standstill the stator produces a magnetic field that rotates at synchronous speed, while the dc-excited rotor field is stationary. The torque between them reverses direction every half-cycle of the rotating field; because the rotor has appreciable inertia it cannot follow these rapid reversals, so the average starting torque over a cycle is zero. The rotor never "catches" the field from rest. Synchronous motors are therefore started by other means — amortisseur (damper) windings that let the machine start as an induction motor and then pull into step, or a variable-frequency drive that ramps the stator frequency up from near zero.

(d) Five specifications for selecting an induction motor. (1) Rated output power (hp or kW) for the driven load. (2) Rated voltage, number of phases and frequency, to match the supply. (3) Rated speed / number of poles (and whether variable speed is needed), since $n_s=120f/p$. (4) Starting torque and starting (locked-rotor) current — the NEMA design class (A, B, C, D) — chosen for the load's inertia and breakaway torque. (5) Full-load efficiency and power factor, which govern operating cost and system loading. Other important items include the enclosure type and insulation/temperature class (TEFC, ODP; Class B/F), service factor, duty cycle, and frame size/mounting.

(e) Why low-power-factor customers pay a penalty. For a given real power $P$, the line current is $I=P/(\sqrt3\,V_L\cdot\text{pf})$, so a low power factor means a proportionally larger current and apparent power (kVA). That extra current increases $I^2R$ losses throughout the utility's generators, lines and transformers and forces the utility to install larger-rated equipment and reactive support — all without delivering any additional billable real energy (kWh). To recover the cost of that idle capacity and to encourage customers to correct their power factor, utilities bill large users on kVA demand or apply a power-factor penalty/surcharge.

(f) 18 kVA, 20 kV/480 V, 60 Hz transformer supplying 15 kVA to a 415 V, 50 Hz load.

Given. Rated 480 V at 60 Hz; proposed operation 415 V at 50 Hz, 15 kVA (< 18 kVA rating). Find. Whether the core flux stays within its design value. The core flux is governed by $V=4.44\,fN\Phi_{max}$, i.e. $\Phi_{max}\propto V/f$. The design ratio is $480/60 = 8.0$ V/Hz; the proposed ratio is $415/50 = 8.3$ V/Hz. Because $8.3 \gt 8.0$, the flux would be driven to $8.3/8.0 = 1.0375$ times rated — about $\boxed{3.75\%}$ over-flux — pushing the core toward saturation, with a sharp rise in magnetizing current and core loss (overheating). The maximum voltage that keeps the flux at its design value at 50 Hz is $V_{max}=480\times(50/60)=\boxed{400\ \text{V}}$. Since 415 V exceeds 400 V, the transformer should not be used to supply the 415 V, 50 Hz load. (The 15 kVA loading is acceptable on a current basis; the problem is purely the volts-per-hertz over-excitation. Had the load voltage been 400 V or less, operation at 50 Hz would be safe.)

(g) Poor induction-motor efficiency at high slip. The rotor copper loss is exactly $P_{rcl}=sP_{ag}$ while the developed mechanical power is $P_{dev}=(1-s)P_{ag}$. The best the rotor conversion can do is therefore a fraction $(1-s)$ of the air-gap power — the fraction $s$ is unavoidably dissipated as rotor $I^2R$ heat. At high slip $s$ is large (e.g. at $s=0.5$ at least half the air-gap power is lost in the rotor), so efficiency is inherently low. Efficient running demands a small slip, which is why cage motors operate at only a few percent slip near full load.

(h) Three means of controlling induction-motor speed. (1) Supply-frequency (V/f) control. A variable-frequency drive changes $f$, hence the synchronous speed $n_s=120f/p$, keeping $V/f$ roughly constant to hold flux; this gives smooth, wide-range speed control and is by far the most common method today. (2) Pole changing. Switching the stator winding connection (consequent-pole) or providing two separate windings changes the pole number $p$ and therefore $n_s$ in discrete steps — simple but only a few fixed speeds. (3) Rotor-resistance control. In a wound-rotor machine, adding external resistance in the rotor circuit increases the slip at which a given torque is produced, lowering the speed; it is simple and gives high starting torque but wastes energy as heat in the added resistance. (Other methods include stator-voltage control and slip-energy recovery.)

(i) Why the transformer core is laminated. The alternating flux induces circulating eddy currents in the electrically conducting iron; these currents dissipate power as $I^2R$ heat and rise with the square of both frequency and flux density (and with the square of the core thickness). Building the core from thin sheets, each coated with an insulating oxide/varnish and stacked in the plane of the flux, breaks the large eddy-current loops into many small, high-resistance ones. This slashes the eddy-current loss (which falls roughly as the lamination thickness squared), reducing core heating and raising efficiency, while leaving the useful magnetic path essentially unchanged.

(j) Two reasons the synchronous motor is industrially useful. First, power-factor control: an over-excited synchronous motor draws leading current, supplying reactive power to the plant. It can drive a mechanical load and correct the facility's power factor at the same time, or run unloaded purely as a synchronous condenser. Second, precise constant speed: it runs at exactly synchronous speed, independent of load up to pull-out, which suits constant-speed drives such as large compressors, pumps and mills; large synchronous motors are also highly efficient and their power factor is adjustable through the field.