Question 5 of 5: Synchronous Motor and Synchronous-Condenser Operation

Question 5: Synchronous Motor and Synchronous-Condenser Operation (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 600 V (line), 8-pole, Y-connected synchronous motor with $X_s=9\ \Omega$/phase and $R_a\approx0$. At no load it draws $I_a=8$ A at 0.08 pf leading. Per phase $V_\phi=600/\sqrt3=346.4$ V.

Find. No-load induced emf, power angle and rotational losses; then, as a synchronous condenser, the active and reactive power, armature current and excitation emf needed to bring the whole plant to unity power factor.

Approach. Use the motor phasor relation $E_a=V_\phi-jX_sI_a$ for the no-load condition; the no-load real power (with $R_a=0$ and no shaft output) is exactly the rotational loss. For the condenser duty, resolve every load into real and reactive power, then size the synchronous machine to cancel the net lagging vars (active power = its own losses).

1. Induced voltage, no load (part a). With a leading current $I_a=8\angle 85.4^\circ$ A and the motor convention $E_a=V_\phi-jX_sI_a$,$$E_a=346.4-j9(8\angle 85.4^\circ)=346.4-72\angle175.4^\circ=418.2-j5.75\ \text{V},$$$$\boxed{|E_a|=418.2\ \text{V per phase}}\;(\approx 724\ \text{V line}).$$ Because the current leads strongly, $E_a>V_\phi$: the machine is over-excited even at no load.
2. Power angle (part b). The angle of $E_a$ relative to $V_\phi$ is$$\delta=\angle E_a=\boxed{-0.79^\circ}.$$ It is small and negative — the internal voltage lags the terminal voltage by less than one degree because the machine, as a motor, absorbs only a trickle of real power.
3. Rotational losses (part c). With $R_a=0$ there is no armature copper loss and no shaft output at no load, so the entire electrical input is consumed as rotational (friction, windage and core) loss:$$P_{rot}=\sqrt3\,V_LI_a\cos\theta=\sqrt3(600)(8)(0.08)=\boxed{665\ \text{W}}.$$
4. Resolve the plant loads. The induction motor absorbs $P_{IM}=P_{out}/\eta=15/0.90=16.67$ kW at 0.85 lag, so $Q_{IM}=P_{IM}\tan(\cos^{-1}0.85)=10.33$ kVAR (lagging). The heating/lighting load at the assumed $P_{HL}=10$ kW, 0.95 lag, gives $Q_{HL}=10\tan(\cos^{-1}0.95)=3.29$ kVAR (lagging).
5. Synchronous-motor active and reactive power (part d). Run as a condenser the machine does no shaft work, so its active draw is just its rotational loss, $P_{sm}=0.665$ kW. To bring the plant to unity power factor it must supply (lead) reactive power equal to the total lagging vars of the other loads:$$\boxed{P_{sm}=0.67\ \text{kW},\qquad Q_{sm}=Q_{IM}+Q_{HL}=13.6\ \text{kVAR (leading)}}.$$
6. Armature current (part e). The synchronous machine now carries $S_{sm}=\sqrt{P_{sm}^2+Q_{sm}^2}=13.6$ kVA, hence$$I_a=\frac{S_{sm}}{\sqrt3\,V_L}=\frac{13{,}640}{\sqrt3(600)}=\boxed{13.1\ \text{A}}\ (\text{pf}=0.049\ \text{leading}).$$
7. Excitation emf (part f). With the strongly leading current $I_a=13.1\angle 87.2^\circ$ A,$$E_a=V_\phi-jX_sI_a=346.4-j9(13.1\angle 87.2^\circ)=464\ \text{V per phase},$$$$\boxed{|E_a|=464\ \text{V/phase}\;(\approx 804\ \text{V line})}.$$ The higher excitation (up from 418 V) is what makes the machine supply the extra leading vars.

*Parts (d)–(f) use the assumed heating/lighting power $P_{HL}=10$ kW (see check note); parts (a)–(c) are exact.