22-Elec-A6 Power Systems and Machines · December 2017
Question 4 of 5: Delta-Connected Synchronous Generator
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. Professional Engineers Ontario — 16-Elec-A6 Power Systems and Machines, December 2017. Closed-book, formula sheet supplied; three hours. FIVE questions constitute a complete paper and all are of equal value (20 marks). All AC voltages and currents are rms; three-phase voltages are line-to-line and power is total real power unless stated. All five questions are solved below.
Reference texts. S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill) — DC machines ch. 8–9, transformers ch. 2, synchronous machines ch. 4–5, induction machines ch. 6. P. C. Sen, Principles of Electric Machines and Power Electronics, 3rd ed. (Wiley). T. Wildi, Electrical Machines, Drives, and Power Systems, 6th ed. (Pearson). Formula sheet as printed on pages 5–6 of the exam.
Given. A 480 V, 6-pole, 60 Hz, delta-connected cylindrical-rotor alternator with $X_s=0.95\ \Omega$/phase and $R_a\approx 0$; the field is fixed so the no-load terminal (and hence internal) voltage is 480 V per phase.
Given data — synchronous generator
Quantity
Symbol
Value
Line/phase voltage (Δ)
$V_\phi=V_L$
480 V
Poles / frequency
$p,f$
6, 60 Hz
Synchronous reactance
$X_s$
0.95 Ω/phase
Armature (winding) current
$I_a$
55 A
Internal emf (field fixed)
$E_a$
480 V
Friction+windage / core loss
$P_{fw},P_{core}$
1.5 kW, 1.25 kW
Find. Synchronous speed; terminal voltage at 0.8 leading; voltage regulation at 0.85 lagging; efficiency at 0.8 lagging; and prime-mover input torque at full load.
Check / convention. "Field adjusted so no-load terminal voltage = 480 V" fixes the internal emf magnitude at $E_a=480\ \text{V}$ per phase (at no load $E_a=V_t$). This excitation is held constant for all parts, so the terminal voltage varies with load and power factor. Voltage regulation is then taken in its physical sense, $\mathrm{VR}=(E_a-V_{fl})/V_{fl}$ (no-load minus full-load terminal voltage, referred to the loaded value). For a Δ machine the winding (phase) quantities equal the line voltage, and the stated 55 A is the armature/phase current ($I_L=\sqrt3\,I_a=95.3$ A).
Figure 4.1 — Per-phase phasor diagram at full load, 0.8 pf lagging: $\mathbf{E}_a=\mathbf{V}_t+jX_s\mathbf{I}_a$, with $E_a=480$ V fixed and $V_t=446.8$ V resulting.
Approach. Get synchronous speed from poles and frequency; for each power factor solve $|E_a|=|V_t+jX_sI_a|$ (a quadratic in the unknown $V_t$) with $E_a=480$ fixed; then form the electrical output and add the fixed losses to get input power and torque.
(ii) Terminal voltage, 0.8 pf leading. With $\mathbf V_t$ as reference and current leading, $\mathbf I_a=55\,\angle{+}36.87^\circ$. Then $jX_s\mathbf I_a=52.25\,\angle 126.87^\circ = -31.35+j41.80$, and $\mathbf E_a=(V_t-31.35)+j41.80$ with $|E_a|=480$: $$(V_t-31.35)^2+41.80^2 = 480^2 \;\Rightarrow\; V_t=\boxed{509.5\ \text{V}}.$$ A leading (capacitive) load makes the terminal voltage rise above $E_a$.
(iii) Voltage regulation, 0.85 pf lagging. Now $\mathbf I_a=55\,\angle{-}31.79^\circ$, so $jX_s\mathbf I_a=52.25\,\angle 58.21^\circ = 27.51+j44.42$ and $\mathbf E_a=(V_t+27.51)+j44.42$. Solving $|E_a|=480$ gives $V_{fl}=450.4\ \text{V}$, hence $$\mathrm{VR}=\frac{E_a-V_{fl}}{V_{fl}}=\frac{480-450.4}{450.4}=\boxed{6.57\%}.$$
(iv) Efficiency, 0.8 pf lagging. At 0.8 lagging the same construction gives $V_t=446.8\ \text{V}$. The electrical output is $$P_{out}=3V_tI_a\cos\theta=3(446.8)(55)(0.8)=58.98\ \text{kW}.$$ With $R_a\approx0$ the only losses are friction+windage and core: $P_{in,mech}=58.98+1.5+1.25=61.73\ \text{kW}$, so $$\eta=\frac{58.98}{61.73}=\boxed{95.55\%}.$$
(v) Prime-mover input torque. The mechanical input power equals the electrical output plus all losses, $P_{in,mech}=61.73\ \text{kW}$, delivered at synchronous mechanical speed: $$T_{in}=\frac{P_{in,mech}}{\omega_m}=\frac{61{,}731}{125.66}=\boxed{491.2\ \text{N}\cdot\text{m}}.$$
