Question 4 of 6: Three-Phase Synchronous Generator

Question 4: Three-Phase Synchronous Generator (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.

Given. A wye-connected round-rotor synchronous generator with negligible armature resistance; the field is fixed so that the no-load terminal voltage is 480 V line-to-line.

Find. Speed; terminal voltage and phasor diagram for 0.8 lag and 0.8 lead; percent regulation for each; efficiency for the lagging case.

Approach. No load with $R_A\approx0$ means $E_A = V_{\phi,nl}$; the field is then held constant, so $|E_A|$ is fixed for both loads. For each power factor, solve $E_A = V_\phi + jX_s I_A$ for the unknown terminal voltage.

1. Speed (a). $n_s = \dfrac{120 f}{P} = \dfrac{120(60)}{6}$. $$\boxed{n_s = 1200\ \text{rpm}}$$
2. Internal voltage from no-load. At no load $I_A = 0$, so with $R_A\approx0$ the internal generated voltage equals the phase voltage: $$E_A = V_{\phi,nl} = \frac{480}{\sqrt{3}} = 277.1\ \text{V (held constant, field fixed)}.$$
3. Terminal voltage, 0.8 pf lagging (b-i). Take $V_\phi$ as reference; $I_A = 60\,\angle{-36.87^\circ}$. Then $jX_sI_A = 60\,\angle 53.13^\circ = 36 + j48$, and $|E_A|^2 = (V_\phi + 36)^2 + 48^2$. Solving with $|E_A| = 277.1$, $$V_\phi = \sqrt{277.1^2 - 48^2} - 36 = 236.9\ \text{V},\qquad V_{L} = \sqrt{3}\,V_\phi.$$ $$\boxed{V_{L} = 410.4\ \text{V (0.8 lagging)}}$$
   Figure 4.1 — phasor diagram, 0.8 pf lagging: $E_A$ leads $V_\phi$, and $V_\phi \lt E_A$ (voltage drops under lagging load).
4. Terminal voltage, 0.8 pf leading (b-ii). Now $I_A = 60\,\angle{+36.87^\circ}$, so $jX_sI_A = -36 + j48$ and $|E_A|^2 = (V_\phi - 36)^2 + 48^2$: $$V_\phi = \sqrt{277.1^2 - 48^2} + 36 = 308.9\ \text{V},\qquad V_{L} = \sqrt{3}\,V_\phi.$$ $$\boxed{V_{L} = 535.1\ \text{V (0.8 leading)}}$$
   Figure 4.2 — phasor diagram, 0.8 pf leading: $V_\phi \gt E_A$ (leading load raises the terminal voltage).
5. Voltage regulation (c). The no-load terminal voltage is 480 V for both, so $$\mathrm{VR}_{lag} = \frac{480 - 410.4}{410.4}\times 100 = 17.0\%,\qquad \mathrm{VR}_{lead} = \frac{480 - 535.1}{535.1}\times 100 = -10.3\%.$$ $$\boxed{\mathrm{VR}_{lag} = +17.0\%,\quad \mathrm{VR}_{lead} = -10.3\%}$$
6. Efficiency, lagging load (d). With $R_A\approx0$ the copper loss is negligible, so the only losses are the 2.5 kW. Output power is $$P_{out} = 3\,V_\phi I_A\cos\theta = 3(236.9)(60)(0.8) = 34.1\ \text{kW},$$ $$\eta = \frac{P_{out}}{P_{out}+P_{loss}} = \frac{34\,119}{34\,119 + 2500}.$$ $$\boxed{\eta = 93.2\%}$$

The sign of the regulation tells the whole story: a lagging (inductive) load pulls the terminal voltage down 17 % below its no-load value, while a leading (capacitive) load pushes it up — a negative regulation. This is the armature-reaction behaviour every synchronous machine exhibits, and it is why excitation must be raised to hold voltage under lagging load.