22-Elec-A6 Power Systems and Machines · December 2016

Question 5 of 6: Synchronous Motor on an Infinite Bus

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: Professional Engineers Ontario — 07-Elec-A6 Power Systems and Machines, December 2016. Closed-book; formula sheet supplied. Six questions of equal value; candidates answer any five. All six are solved here as a study resource. All ac voltages and currents are rms; three-phase voltages are line-to-line unless noted.

Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (transformers, induction/synchronous machines, dc motors); J. D. Glover et al., Power System Analysis and Design, 6th ed. (three-phase power, two-wattmeter method, power-factor correction).

Question 5: Synchronous Motor 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. A 10 MVA, 6600 V, 60 Hz, 4-pole, Y-connected synchronous motor, $X_s=10\ \Omega$/phase, $R_a\approx0$.

Given data
V_LL	X_s	No-load current	No-load pf
6600 V	10 Ω/ph	10 A	0.05 leading

Find. Excitation voltage, power angle and real input power at no-load; then—with excitation fixed and the load raised to unity pf—the stator current, power angle, real input power and efficiency.

Phasor diagram (motor convention $E_a=V-jX_sI_a$): over-excited at no-load ($E_a\gt V$, $I_a$ leads); at unity pf $I_a$ aligns with $V$ and $E_a$ swings to a larger power angle.

Approach. Use the motor phasor equation $E_a=V-jX_sI_a$ per phase. At no-load, compute $E_a$, $\delta$ and $P=3V I_a\cos\theta$. With $|E_a|$ then held fixed, the unity-pf current follows from $|E_a|^2=V^2+(X_sI_a)^2$, and efficiency uses the no-load input as the rotational-loss estimate ($R_a\approx0$).

Per-phase voltage. $V=6600/\sqrt3=3810.5$ V.
Excitation voltage at no-load (part b). With $I_a=10\angle{+87.1^\circ}$ A (0.05 leading), $$E_a=V-jX_sI_a=3810.5-j10(10\angle87.1^\circ)=3910.4-j5.0=3910\angle{-0.07^\circ}\ \text{V/ph}.$$ $$\boxed{|E_a|=3910\ \text{V/phase}\;(=6773\ \text{V line-to-line}).}$$
Power angle and input power (parts c, d). $\boxed{\delta=-0.07^\circ\approx0}$ (the machine develops essentially no torque at no-load), and $$\boxed{P_{in,\text{nl}}=3V I_a\cos\theta=3(3810.5)(10)(0.05)=5.72\ \text{kW}.}$$ This small input supplies the rotational (friction, windage, core) losses.
Unity-pf stator current (part e). The field is unchanged, so $|E_a|=3910$ V is fixed. At unity pf $I_a$ is in phase with $V$, so $E_a=V-jX_sI_a$ gives $|E_a|^2=V^2+(X_sI_a)^2$: $$\boxed{I_a'=\frac{\sqrt{|E_a|^2-V^2}}{X_s}=\frac{\sqrt{3910.4^2-3810.5^2}}{10}=87.8\ \text{A}.}$$
New power angle (part f). $E_a=3810.5-j10(87.8)=3810.5-j878.2$, so $$\boxed{\delta'=\tan^{-1}\!\frac{-878.2}{3810.5}=-12.98^\circ.}$$
New real input power (part g). At unity pf, $$\boxed{P_{in}'=3V I_a'=3(3810.5)(87.8)=1.004\ \text{MW},}$$ which the power-angle form confirms: $P=3VE_a\sin|\delta'|/X_s=1.004$ MW.
Efficiency (part h). With $R_a\approx0$ there is no armature copper loss, so the only losses are the rotational losses found at no-load, $P_{rot}=5.72$ kW. Then $P_{out}=P_{in}'-P_{rot}=998.1$ kW and $$\boxed{\eta=\frac{P_{out}}{P_{in}'}=\frac{998.1}{1003.9}=99.4\%.}$$

Check: Efficiency here counts only the rotational losses inferred from the no-load input (armature resistance is given as negligible and field-circuit loss is not specified). A real machine's efficiency would be lower once field-winding and stray losses are included.

Final results — Question 5
Quantity	Value
(b) Excitation voltage	3910 V/phase (6773 V line)
(c) Power angle (no-load)	≈ 0° (−0.07°)
(d) Real input power (no-load)	5.72 kW
(e) Stator current (unity pf)	87.8 A
(f) Power angle (unity pf)	−12.98°
(g) Real input power (unity pf)	1.004 MW
(h) Efficiency	99.4 %

Topic: Cylindrical-rotor synchronous motor on an infinite bus; excitation, power angle and the V-curve idea.

Key relations: $E_a=V-jX_sI_a$ (motor); $P=3VE_a\sin\delta/X_s=3VI_a\cos\theta$; with $|E_a|$ fixed and unity pf, $|E_a|^2=V^2+(X_sI_a)^2$.

Why this works: The excitation (field current) sets $|E_a|$. At no-load an over-excited machine draws a small leading current—it behaves like a capacitor (a synchronous condenser)—and the power angle is essentially zero because it develops no shaft power. Loading the shaft while holding excitation constant forces the rotor to fall back by an angle $\delta$; the real power rises as $\sin\delta$. Because $R_a\approx0$, the armature dissipates nothing and the machine is highly efficient; the only loss is the mechanical/core rotational loss revealed by the no-load input.

Common pitfall: Using the generator sign convention ($E_a=V+jX_sI_a$), which flips the power angle; and forgetting that "excitation unchanged" pins $|E_a|$, not $I_a$—the current is what changes when the load is applied.

Source: Chapman, Electric Machinery Fundamentals, 5th ed., §5–6 (synchronous machines, power angle, V-curves).

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