22-Mec-A5 Electrical and Electronics Engineering · December 2019
Question 6 of 8: Induction Machines — Slip Relations and the Motor–Load Operating Point
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. PEO / Engineers Canada National Examinations, December 2019 — 16-Mec-A5, Electrical & Electronics Engineering (Mechanical Engineering discipline). Three hours, closed book, one approved Casio or Sharp calculator. Eight questions; any five constitute a complete paper and each question is of equal value, so each counted question carries 20 marks. All eight questions are solved below, because this set is a study resource rather than an exam script. The front page prints the constants $\pi = 3.14159$, $1\text{ hp} = 746\text{ W}$ and $\mu_0 = 4\pi\times10^{-7}\text{ H m}^{-1}$; all three are printed clearly on the front page and are used exactly as given.
Reference texts. A. S. Sedra and K. C. Smith, Microelectronic Circuits, 8th ed. (BJT biasing and load lines, op-amp frequency response); M. M. Mano and M. D. Ciletti, Digital Design, 6th ed. (universal-gate synthesis, DeMorgan reduction); S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (dc machines, transformer magnetic circuits, induction-motor testing and slip); C. K. Alexander and M. N. O. Sadiku, Fundamentals of Electric Circuits, 7th ed. (phasor analysis, power factor, first-order transients); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (magnetic circuits and induced emf).
the gated-logic array, the transformer magnetic circuit, the op-amp band-pass Bode question, the RC transient and the parallel ac network as printed in December 2014; and the induction-motor slip / speed-torque question as printed in December 2016. The full working is transcribed in place below rather than cross-referenced, so this paper stands on its own as a study document.
Question 6: Induction Machines — Slip Relations and the Motor–Load Operating Point (20 marks)
Given. Two independent induction machines: a 12-pole wound-rotor motor whose full-load speed is measured, and an 8-pole cage motor characterised by one point on its linear torque–slip region and driven against a load characterised by one point on its linear torque–speed line.
Given data
Part
Quantity
Symbol
Value
I
Supply frequency
$f$
60 Hz
I
Number of poles
$P$
12
I
Rated output
—
300 hp
I
Per-phase rotor resistance
$r_2$
0.04 $\Omega$
I
Full-load rotor speed
$n_m$
582 rev/min
II
Supply frequency and poles
$f$, $P$
60 Hz, 8
II
Motor test point
$T$, $n$
3 N·m at 810 rev/min
II
Load test point
$T$, $n$
0.5 N·m at 435 rev/min
Find. Part I: the seven quantities [a]–[g], the last four expressed as angular velocities. Part II: the sketched characteristics and the speed and torque at which the motor and load lines intersect.
Approach. Part I rests on one idea — the rotor field and the stator field are locked together in space at synchronous speed, no matter how fast the rotor itself turns — so every one of [d] to [g] follows from $n_s$ and the slip speed $s\,n_s$. Part II turns the two single test points into two straight lines and equates them.
Part I — wound-rotor slip relations
Synchronous speed from poles and frequency. This is the speed of the rotating stator field, which is what [a] asks for.
$$n_s = \frac{120 f}{P} = \frac{120(60)}{12} = \boxed{600\ \text{rev/min}}$$
Slip from the measured full-load speed. Slip is the fractional shortfall of the rotor behind the field.
$$s = \frac{n_s - n_m}{n_s} = \frac{600 - 582}{600} = \boxed{0.030 \;(3.0\%)}$$
Rotor-current frequency. The rotor conductors are cut by the field at the slip speed, so they are excited at the slip frequency.
$$f_r = s\,f = (0.030)(60) = \boxed{1.8\ \text{Hz}}$$
[d] Stator field with respect to the stator. The field sweeps past the stationary stator at synchronous speed.
$$\omega_s = \frac{2\pi n_s}{60} = \frac{2\pi(600)}{60} = \boxed{62.83\ \text{rad/s}}$$
[e] Stator field with respect to the rotor. Subtract the rotor's own rotation; what remains is the slip speed, $n_s - n_m = 18$ rev/min.
$$\omega_{s/r} = \frac{2\pi (n_s - n_m)}{60} = \frac{2\pi (18)}{60} = \boxed{1.885\ \text{rad/s}}$$
Parts [f] and [g] are where candidates most often go wrong, and the reasoning is worth stating carefully. The rotor currents are three-phase currents of frequency $f_r = s f$ flowing in a winding with the same 12 poles, so relative to the rotor structure they set up a field rotating at $120 f_r / P = s\,n_s$ — exactly the slip speed. But the rotor structure is itself turning at $n_m$, so an observer on the stator sees that field travelling at $n_m + s\,n_s = n_s$. The rotor field and the stator field therefore rotate in perfect step, which is the necessary condition for a steady, non-pulsating torque.
[f] Rotor field with respect to the rotor.
$$\omega_{r/r} = \frac{2\pi\,(s\,n_s)}{60} = \frac{2\pi\,(0.030 \times 600)}{60} = \frac{2\pi(18)}{60} = \boxed{1.885\ \text{rad/s}}$$
identical to [e], as it must be.
[g] Rotor field with respect to the stator. Add the rotor's own motion back on.
$$\omega_{r/s} = \frac{2\pi\,(n_m + s\,n_s)}{60} = \frac{2\pi\,(582 + 18)}{60} = \boxed{62.83\ \text{rad/s}}$$
which equals [d]: the two fields are stationary with respect to one another.
Check — the unused datum. The per-phase rotor resistance $r_2 = 0.04\ \Omega$ and the 300 hp rating are not required by any of parts [a] to [g]; they are context (and would be needed only if the question went on to ask for rotor copper loss or developed torque). Reporting them as unused is the correct engineering response — it is not a sign that something has been missed.
Part II — graphical and algebraic operating point
Synchronous speed of the 8-pole machine.
$$n_s = \frac{120(60)}{8} = 900\ \text{rev/min}$$
Turn the motor test point into a torque–slip line. At 810 rev/min the slip is $s = (900-810)/900 = 0.10$, and torque is stated to be proportional to slip in the normal operating region.
$$k_m = \frac{T}{s} = \frac{3}{0.10} = 30\ \text{N}\cdot\text{m per unit slip} \;\Rightarrow\; T_m(n) = 30\,\frac{900 - n}{900}$$
Turn the load test point into a torque–speed line. A load torque that is a linear function of speed and vanishes at standstill passes through the origin.
$$T_L(n) = \frac{0.5}{435}\,n = \frac{n}{870}$$
Equate the two and solve for the operating speed. The system settles where the torque the motor develops equals the torque the load demands.
$$30\,\frac{900 - n}{900} = \frac{n}{870} \;\Longrightarrow\; \frac{900 - n}{30} = \frac{n}{870} \;\Longrightarrow\; 29(900 - n) = n$$
$$26100 = 30n \;\Rightarrow\; \boxed{n = 870\ \text{rev/min}}$$
Evaluate the common torque and the operating slip.
$$T = \frac{870}{870} = \boxed{1.00\ \text{N}\cdot\text{m}}, \qquad s = \frac{900 - 870}{900} = 3.33\%$$
$$P_{\text{shaft}} = T\,\omega = (1.00)\frac{2\pi(870)}{60} = 91.1\ \text{W}$$
Figure 6 — Part II [a]: the motor characteristic (falling with speed, reaching zero torque at the synchronous speed of 900 rev/min) and the load characteristic (rising through the origin). The intersection is the operating point.
The sketch also answers a question the examiner does not ask explicitly but always rewards: the intersection is stable. To the left of it the motor develops more torque than the load absorbs, so the system accelerates; to the right the load demands more than the motor develops, so it decelerates. Any disturbance therefore drives the speed back towards 870 rev/min, which is the general stability criterion $\mathrm{d}T_m/\mathrm{d}n < \mathrm{d}T_L/\mathrm{d}n$ at the crossing.
