Question 5 of 5: Synchronous Machines

Question 5: Synchronous Machines (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.

Part (a) — Effect of increasing excitation on an infinite bus

With the machine tied to an infinite bus, both terminal voltage \(V_\phi\) and frequency are fixed, and the real power is set by the prime mover, not by the field. The real power is \(P=\dfrac{3V_\phi E_a}{X_s}\sin\delta\); since \(P\) and \(V_\phi\) are constant, raising the excitation increases \(|E_a|\), so \(\sin\delta\) (and hence the power angle \(\delta\)) must decrease to keep \(P\) fixed. The extra excitation therefore does not change the real power — it changes the reactive power: the machine becomes over-excited and supplies lagging reactive power to the grid (it looks capacitive to the system), and the armature current increases and becomes more lagging. This is the well-known synchronous "V-curve" behaviour, and it is the basis of using an over-excited synchronous machine as a source of vars (a synchronous condenser).

Part (b) — Generator on an infinite bus

Given. Six-pole, Y-connected synchronous machine, 208 V (L-L), 5500 VA, \(X_s=8\ \Omega\)/phase, \(R_a\approx0\); generating onto an infinite bus.

Find. (i) excitation voltage \(E_a\) and power angle \(\delta\) at rated kVA, 0.8 lag, and the phasor diagram; (ii) after a 20% rise in field at constant \(P\): new \(I_a\), power factor and reactive power; (iii) the steady-state stability limit (max power) at the original excitation, with the corresponding \(I_a\), power factor and reactive power.

Approach. Use the round-rotor phasor equation \(E_a=V_\phi+jX_sI_a\) for part (i). For part (ii), hold \(P\) fixed while \(|E_a|\) rises 20%: get the new \(\delta\) from \(P=\dfrac{3V_\phi E_a}{X_s}\sin\delta\), then recompute \(I_a=(E_a-V_\phi)/jX_s\). For part (iii), the steady-state stability limit is \(\delta=90^\circ\) at the original \(|E_a|\).

1. Part (i): excitation voltage and power angle. The armature current at rated kVA and 0.8 lag is \(I_a=15.27\angle{-36.87^\circ}\text{ A}\). Then $$E_a=V_\phi+jX_sI_a=120.1+j8(15.27\angle{-36.87^\circ}) =193.4+j97.7=216.7\angle26.8^\circ\text{ V}.$$ So the excitation voltage is \(E_a=216.7\text{ V/phase}\) and the power angle is \(\delta=26.8^\circ\).
2. Part (ii): new operating point after a 20% field increase. The per-phase converted power is fixed at \(P_\phi=V_\phi I_a\cos\theta=120.1(15.27)(0.8)=1467\text{ W}\). The new excitation is \(E_a'=1.20(216.7)=260.0\text{ V}\), so $$\sin\delta'=\frac{P_\phi X_s}{V_\phi E_a'}=\frac{1467(8)}{120.1(260.0)}=0.376 \;\Rightarrow\;\delta'=22.1^\circ.$$ The new armature current is $$I_a'=\frac{E_a'-V_\phi}{jX_s}=\frac{260.0\angle22.1^\circ-120.1}{j8} =19.4\angle{-51.1^\circ}\text{ A},$$ i.e. \(I_a'=19.4\text{ A}\) at power factor \(\cos51.1^\circ=0.629\) lagging. The reactive power supplied is $$Q=3V_\phi I_a'\sin\theta'=3(120.1)(19.4)(0.778)=5.44\text{ kvar (supplied).}$$ (The real power checks: \(3V_\phi I_a'\cos\theta'=4.40\text{ kW}\), unchanged.)
3. Part (iii): steady-state stability limit at the original field. Restoring \(E_a=216.7\text{ V}\) and increasing the prime-mover power drives \(\delta\) toward \(90^\circ\), where the power-angle characteristic peaks: $$P_{max}=\frac{3V_\phi E_a}{X_s}=\frac{3(120.1)(216.7)}{8}=9.76\text{ kW}.$$ At \(\delta=90^\circ\), \(E_a=216.7\angle90^\circ\), so $$I_a=\frac{E_a-V_\phi}{jX_s}=\frac{216.7\angle90^\circ-120.1}{j8} =30.96\angle{+29.0^\circ}\text{ A}.$$ The current now leads the voltage: power factor \(\cos29.0^\circ=0.874\) leading, and the machine absorbs reactive power $$Q=3V_\phi I_a\sin\theta=3(120.1)(30.96)(-0.485)=-5.41\text{ kvar}$$ (i.e. it draws 5.41 kvar from the grid). Note the armature current 30.96 A is about twice rated — the stability-limit condition is a ceiling, not a continuous rating.