22-Elec-A6 Power Systems and Machines · December 2014
Question 2 of 6: Induction Motor
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Exam format. Professional Engineers Ontario / Engineers Canada national examination
07-Elec-A6 Power Systems and Machines, Fall (December) 2014. Closed book; approved Casio/Sharp calculator; formula sheet supplied.
Part A is compulsory (20 multiple-choice questions, 1 mark each, −½ per wrong answer).
Part B asks the candidate to solve any 4 of 5 problems; a complete paper is FIVE questions.
All AC quantities are RMS; three-phase voltages are line-to-line and powers are totals unless noted.
Every Part B problem is worked below as a study resource.
Reference texts.
S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill) — transformers, induction motors, synchronous machines, DC machines, magnetic circuits.
J. D. Glover, M. S. Sarma & T. J. Overbye, Power System Analysis and Design, 6th ed. (Cengage) — balanced three-phase circuits, power-factor correction.
C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits, 7th ed. (McGraw-Hill) — AC power, complex phasor methods.
Given. A 460 V (line-to-line), 60 Hz, four-pole, Y-connected induction motor running at
slip $s=0.03$, with the per-phase equivalent-circuit parameters below and 455 W of core loss.
Given data (per phase, referred to the stator)
Line voltage $V_{LL}$
460 V
Stator resistance $R_1$
$0.641\ \Omega$
Rotor resistance $R_2$
$0.300\ \Omega$
Stator reactance $X_1$
$0.750\ \Omega$
Rotor reactance $X_2$
$0.500\ \Omega$
Magnetizing reactance $X_m$
$26.3\ \Omega$
Core loss $P_{core}$
455 W
Slip $s$ / poles $p$
0.03 / 4
Find. Stator current, shaft (mechanical) speed, shaft output power, output torque, and
efficiency at 3% slip.
Approach. Reduce the per-phase circuit to an input impedance (rotor branch $R_2/s+jX_2$
in parallel with $jX_m$, in series with $R_1+jX_1$), find $I_1$, then follow the Chapman power flow
$P_{in}\to P_{SCL}\to P_{AG}\to P_{conv}\to P_{out}$, subtracting the core loss at the shaft.
Per-phase equivalent circuit (referred to the stator). The
rotor branch resistance is $R_2/s=0.300/0.03=10.0\ \Omega$; the lossless $jX_m$ carries the magnetizing
current, and the 455 W core loss is accounted as a rotational loss at the shaft.
Rotor branch and its parallel with the magnetizing reactance.
$Z_2=\dfrac{R_2}{s}+jX_2=10.0+j0.5\ \Omega$, and
$Z_f=\dfrac{jX_m\,Z_2}{jX_m+Z_2}=9.21\angle23.3^\circ=8.45+j3.65\ \Omega$.
Input impedance and stator current.
$Z_{in}=R_1+jX_1+Z_f=9.09+j4.40=10.10\angle25.8^\circ\ \Omega$, so
$$I_1=\frac{V_1}{Z_{in}}=\frac{265.6\angle0^\circ}{10.10\angle25.8^\circ}=\boxed{26.3\ \text{A}\ \angle{-25.8^\circ}}$$
a lagging power factor of $\cos 25.8^\circ=0.900$.
Synchronous and shaft speed. $n_s=\dfrac{120f}{p}=\dfrac{120\cdot60}{4}=1800\ \text{rpm}$;
the shaft runs at $n_m=(1-s)n_s=0.97\times1800=\boxed{1746\ \text{rpm}}$
($\omega_m=182.8\ \text{rad/s}$).
Input power and stator copper loss.
$P_{in}=\sqrt3\,V_{LL}I_L\cos\theta=\sqrt3(460)(26.3)(0.900)=18.86\ \text{kW}$;
$P_{SCL}=3I_1^2R_1=3(26.3)^2(0.641)=1.33\ \text{kW}$.
Air-gap and converted power. Because $jX_m$ absorbs no real power,
$P_{AG}=P_{in}-P_{SCL}=17.53\ \text{kW}$, the rotor copper loss is $P_{RCL}=sP_{AG}=0.526\ \text{kW}$, and the
developed (converted) power is $P_{conv}=(1-s)P_{AG}=17.01\ \text{kW}$.
Shaft output, torque and efficiency. Subtract the 455 W core (rotational) loss:
$$P_{out}=P_{conv}-P_{core}=17.01-0.455=\boxed{16.55\ \text{kW}}$$
$T_{out}=P_{out}/\omega_m=16550/182.8=\boxed{90.5\ \text{N}\cdot\text{m}}$, and
$\eta=P_{out}/P_{in}=16.55/18.86=\boxed{87.7\%}$.
