Question 5 of 6: Synchronous Generator on an Infinite Bus

Question 5: Synchronous Generator on an Infinite Bus (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. $p=4$, $f=60$ Hz, bus $V_L=600$ V, $T=2650$ N·m, $\delta=34^\circ$, $X_S=1.06\ \Omega$/phase, losses negligible.

Find. The qualitative field-current effect (a); the power-angle derivation (b); and (c) mechanical input, excitation voltage, armature current, complex power and generator pf.

(a) Effect of increased field current. With the machine tied to a stiff grid, the real power is fixed by the prime-mover (turbine) input, so increasing the field current leaves $P$ essentially unchanged. A larger field raises the internal emf $E_f$, which forces the machine to export more reactive power: it moves toward over-excitation, supplying lagging VARs to the grid (the armature current increases and swings to a more lagging angle). Reducing the field would under-excite the machine so it absorbs VARs.

(b) Power-angle derivation. Neglecting armature resistance, per phase $E_f=V_t+jX_SI_a$. The complex power delivered per phase is $S=V_tI_a^*$. Writing $I_a=(E_f\angle\delta - V_t)/(jX_S)$ and taking the real part, $$P_{ph}=\text{Re}\{V_tI_a^*\}=\frac{V_tE_f}{X_S}\sin\delta \ \Rightarrow\ \boxed{P=\frac{3V_tE_f}{X_S}\sin\delta}$$ for the three-phase total. The phasor diagram below shows $V_t$, $E_f$ advanced by $\delta$, and the drop $jX_SI_a$ closing the triangle; the vertical projection $E_f\sin\delta$ is proportional to $P$.

Approach (c). Get the mechanical power from torque and synchronous speed, use the power-angle equation for $E_f$, then solve the phasor equation for $I_a$ and form $S=3V_tI_a^*$.

1. (i) Mechanical power input. Synchronous speed $n_s=120f/p=1800$ rpm, $\omega_m=188.5$ rad/s: $$P_{mech}=T\omega_m=2650(188.5)=\boxed{499.5\ \text{kW}.}$$ With losses neglected this equals the electrical power delivered.
2. (ii) Excitation voltage. Phase voltage $V_t=600/\sqrt3=346.4$ V. From $P=\dfrac{3V_tE_f}{X_S}\sin\delta$, $$E_f=\frac{P\,X_S}{3V_t\sin\delta}=\frac{499\,500(1.06)}{3(346.4)\sin34^\circ}=\boxed{911\ \text{V/phase}.}$$
3. (iii) Armature current. $$I_a=\frac{E_f\angle\delta-V_t}{jX_S}=\frac{911\angle34^\circ-346.4}{j1.06}=616\angle{-38.8^\circ}\ \text{A}.$$ $$\boxed{I_a=616\ \text{A}.}$$
4. (iv) Complex power to the bus. $$S=3V_tI_a^*=3(346.4)(616\angle 38.8^\circ)=\boxed{640.5\ \text{kVA}}$$ i.e. $P=499.5$ kW and $Q=+401$ kVAR (lagging), consistent with the mechanical input.
5. (v) Generator power factor. The armature current lags $V_t$ by $38.8^\circ$, so $$\text{pf}=\cos 38.8^\circ=\boxed{0.780\ \text{lagging}.}$$