25-Nav-B6 Ocean Engineering and Offshore Structures · December 2017
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 National Examinations, December 2017 —
98-Mar-B6, printed for Electrical & Electronics Engineering
candidates. Three hours, closed book, one approved Casio or
Sharp calculator. Eight questions of equal value; any five constitute a
complete paper, and only the first five appearing in the answer book are
marked. Constants supplied on the front page: $\pi = 3.14159$,
$1\ \text{hp} = 746\ \text{W}$, $\mu_0 = 4\pi\times10^{-7}\ \text{H}\,\text{m}^{-1}$.
All eight questions are solved here so the solutions cover whichever five a candidate chooses.
Subject note
This paper is listed under 25-Nav-B6 “Ocean Engineering and Offshore Structures”, but the printed paper is headed 98-Mar-B6 and every question is Electrical & Electronics Engineering content (BJT amplifier bias/load lines, XOR/NOR gated-logic synthesis, homopolar disc dc machine, transformer magnetic-circuit equivalent, op-amp Bode/frequency response, induction-motor slip and motor-load operating point, RC transient, parallel R-L-C phasor network) — zero naval-architecture or ocean-engineering content. Solved as the exam actually printed.
Reference texts
A. S. Sedra and K. C. Smith, Microelectronic Circuits, 8th ed.
— BJT biasing and load lines (Ch. 5), op-amp frequency response (Ch. 2).
M. M. Mano and M. D. Ciletti, Digital Design, 6th ed. —
Boolean algebra and DeMorgan’s theorems (Ch. 2), NOR/NAND universal-gate
synthesis (Ch. 3).
S. J. Chapman, Electric Machinery Fundamentals, 5th ed. —
dc machine emf and torque (Ch. 7–8), induction motors and slip (Ch. 6),
transformers (Ch. 2).
C. K. Alexander and M. N. O. Sadiku, Fundamentals of Electric
Circuits, 7th ed. — phasor analysis, ac power and power-factor
correction (Ch. 9–11), first-order transients (Ch. 7).
W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.
— magnetic circuits and induced emf.
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.
