Synchronous Machine: 22-Elec-A6 Power Systems and Machines, December 2014, Q5

Question 5 — Synchronous Machine

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 round-rotor synchronous generator rated 10 kV, 50 MVA, on an infinite bus at 10 kV, delivering 2000 A at 0.9 pf leading, with $R_a=0.1$ pu and $X_s=1.65$ pu on its own rating.

Given data
Rating	10 kV, 50 MVA	Base current $I_{base}$	$50\ \text{MVA}/(\sqrt3\cdot10\ \text{kV})=2887\ \text{A}$
$R_a$	0.1 pu	$X_s$	1.65 pu
Terminal voltage $V_t$	10 kV = 1.0 pu	Armature current	2000 A, 0.9 pf leading

Find. The internal (excitation) voltage $E_a$ and power angle $\delta$; the phasor diagram; the open-circuit voltage; and the steady-state short-circuit current, all at fixed excitation.

Approach. Work in per-unit. Express the current in pu, take $V_t=1.0\angle0^\circ$ as reference with the current leading, and apply the generator relation $E_a=V_t+I(R_a+jX_s)$. The magnitude of $E_a$ fixes both the open-circuit voltage and, divided by $|Z_s|$, the sustained short-circuit current.

Phasor diagram (generator convention, 0.9 pf leading). The current $I$ leads $V_t$; the drop $jX_sI$ (plus a small $R_aI$) adds to $V_t$ to give the internal voltage $E_a=1.20$ pu at power angle $\delta=62^\circ$.

Per-unit current. $I_{base}=\dfrac{50\times10^6}{\sqrt3\,(10\times10^3)}=2887\ \text{A}$, so $I=\dfrac{2000}{2887}=0.693\ \text{pu}$. Leading pf $\Rightarrow$ $I=0.693\angle{+25.84^\circ}$ pu (current leads $V_t$).
Synchronous impedance drop. $Z_s=R_a+jX_s=0.1+j1.65=1.653\angle86.5^\circ$ pu, so $IZ_s=0.693\angle25.84^\circ\times1.653\angle86.5^\circ=1.145\angle112.4^\circ=-0.436+j1.059$ pu.
Internal voltage and power angle. $$E_a=V_t+IZ_s=1.0+(-0.436+j1.059)=0.564+j1.059 =\boxed{1.20\ \text{pu}\ \angle62.0^\circ}$$ i.e. $E_a=12.0\ \text{kV}$ (line-to-line) and power angle $\delta=62.0^\circ$. (Check: $P=\tfrac{V_tE_a}{X_s}\sin\delta=0.642$ pu $\approx$ the delivered $0.624$ pu, the difference being the $R_a$ term.)
Open-circuit voltage. Removing the load at fixed excitation leaves no armature drop, so the terminal voltage rises to the internal voltage: $V_{OC}=E_a=\boxed{12.0\ \text{kV}}$ (1.20 pu).
Steady-state short-circuit current. Shorting the terminals at the same $E_a$ drives $$I_{SC}=\frac{E_a}{|Z_s|}=\frac{1.20}{1.653}=0.726\ \text{pu} =0.726\times2887=\boxed{2096\ \text{A}}$$ (about $2.10$ kA, i.e. $1.05\times$ rated).

Question 5 — results (fixed excitation)
Quantity	Value
(a) Internal voltage $E_a$	1.20 pu = 12.0 kV (L–L)
Power angle $\delta$	$62.0^\circ$
(c) Open-circuit voltage	12.0 kV (1.20 pu)
(d) Steady-state short-circuit current	0.726 pu = 2096 A ($\approx$ 2.10 kA)

Topic. Round-rotor synchronous generator on an infinite bus; per-unit phasor analysis.

Key relations. $E_a=V_t+I(R_a+jX_s)$; $P=\tfrac{V_tE_a}{X_s}\sin\delta$; $V_{OC}=E_a$; $I_{SC}=E_a/|Z_s|$.

Why this works. The internal voltage is set by field excitation; once the operating point gives $E_a$, holding excitation constant means $E_a$ is fixed regardless of load. Open circuit removes the armature drop (terminal = $E_a$), while a bolted short leaves only $Z_s$ to limit the current — both read off the same $E_a$.

Common pitfall. Using the wrong sign for a leading power factor (the current leads, so $+\theta$), and taking $V_t$ off the 10 kV rating incorrectly — since the bus equals the rating, $V_t=1.0$ pu exactly. Neglecting $R_a$ changes $\delta$ by a couple of degrees.

Source. Chapman, Electric Machinery Fundamentals, 5th ed., §4.6–4.9 (synchronous generator, phasor diagrams, per-unit); Glover, §6.

Question 5 of 6: Synchronous Machine

Question 5 — Synchronous Machine