Question 5 of 6: Under-frequency operation, and a current-source-inverter-fed induction motor

Question 5: Under-frequency operation, and a current-source-inverter-fed induction motor (20 points)

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.

Part (a) — Reducing frequency below rated at constant rated voltage

The air-gap flux of an induction machine is fixed by the volts-per-hertz ratio, because the stator back-emf is $E \approx 4.44\,f\,N\,k_w\,\Phi$. Holding $V$ at its rated value while lowering $f$ therefore raises $\Phi$ in inverse proportion to frequency, and every consequence below follows from that one fact.

Magnetic saturation. A machine is designed with its rated flux already near the knee of the iron's B-H curve, so even a modest increase drives the core deep into saturation. Beyond the knee the magnetising inductance $L_m$ collapses, and the magnetising current needed to sustain the flux rises far faster than linearly — a 20% frequency reduction can multiply the no-load current several times over. The magnetising current also becomes strongly non-sinusoidal, rich in third and fifth harmonics, which distorts the supply current and produces additional losses in the machine and the supply.

Excessive current and heating. The large magnetising current flows through the stator winding whether or not the shaft is loaded, so stator copper loss rises sharply at no load and the useful load current that the machine can carry within its thermal rating shrinks correspondingly. At the same time the shaft-mounted cooling fan is turning more slowly, because synchronous speed $n_s = 120f/P$ has fallen with frequency, so the machine is producing more loss and rejecting it less effectively. Overheating and insulation ageing are the practical outcome; a totally-enclosed fan-cooled machine is particularly vulnerable.

Core losses and pull-out torque. Hysteresis loss per cycle rises with peak flux and eddy-current loss rises with the square of the flux, though both are partially offset by the lower frequency, so the net core loss change depends on the machine; what is certain is that the loss density in the teeth rises where the saturation is worst. Breakdown (pull-out) torque nominally varies as $(V/f)^2$ and so appears to increase, but in practice saturation of the leakage paths and the collapse of $L_m$ prevent the machine from realising that gain, and the increase is much smaller than the simple formula predicts.

Power factor and supply distortion. The extra magnetising current is almost purely reactive, so the operating power factor falls markedly, increasing the current rating required of the supply, the cabling and the converter for the same shaft output.

Operating speed and torque capability. Synchronous speed falls in proportion to frequency, so the motor runs slower — which is usually the intent — but constant-torque operation is not achieved by frequency reduction alone. The correct practice is constant volts-per-hertz control, in which $V$ is lowered with $f$ so that flux, and therefore torque per ampere, remains at its design value; at very low frequency a small voltage boost is added to offset the stator resistance drop, which is the opposite correction to the problem described here.

In summary: reducing frequency at constant voltage over-fluxes the machine. It saturates, draws a large distorted magnetising current at poor power factor, overheats while cooling less effectively, and does not deliver the torque improvement the idealised formulae suggest. It is a fault condition, not an operating strategy.

Part (b) — Completing the constant-current table

Given.

Find. The three missing slips, the three terminal voltages and the one missing developed torque that complete the table.

[Figure not reproduced: Figure 5.1 — Fig. (1) of the exam paper redrawn: the inverter injects a constant $I_i$, which divides between the magnetising branch $jX_m$ and the series rotor branch $R_s + jX_s + jX_r + R_r/s$. See the official exam paper.]

Approach. Derive the constant-current torque from the current divider in Fig. (1), invert it into a quadratic in $R_r/s$ for the rows where torque is given, select between the two roots on magnetising current, and obtain each terminal voltage as the injected current times the parallel impedance seen by the source.

1. Part (b) — Synchronous quantities and the total reactance. With eight poles at 50 Hz, $$n_s = \frac{120f}{P} = 750\ \text{rpm},\qquad \omega_s = \frac{2\pi f}{P/2} = 78.540\ \text{rad/s},\qquad X_m + X_s + X_r = 12.8\ \Omega.$$ Here $\omega_s$ is the synchronous mechanical speed, which is the one that divides air-gap power to give torque.
2. Derive the torque expression from Fig. (1). The injected current divides between the two parallel branches, so the rotor-branch current is $$I_r = I_i\,\frac{jX_m}{R_s + R_r/s + j(X_m + X_s + X_r)} \;\Longrightarrow\; |I_r|^{2} = \frac{(X_m I_i)^{2}}{(R_s + R_r/s)^{2} + (X_m+X_s+X_r)^{2}}.$$ Air-gap power is $3|I_r|^{2}(R_r/s)$ and torque is air-gap power divided by $\omega_s$, giving $$T = \frac{3(X_m I_i)^{2}(R_r/s)}{\omega_s\left[(R_s + R_r/s)^{2} + (X_m+X_s+X_r)^{2}\right]}.$$
3. Reconcile this with the printed formula. The expression printed on the paper carries an extra $s$ in its denominator ($s\,\omega_s[\cdots]$) on top of the $R_r/s$ already in its numerator, so it cannot be the torque of the circuit in Fig. (1). Taken literally it returns $s = 0.2355$ for the 40 A / 180 N.m row, at which slip the printed circuit develops only 42.4 N.m — less than a quarter of the stated torque. The printed expression is therefore irreconcilable with its own figure, and all the work below uses the circuit-derived form. See the callout following the table.
4. Row 1 — solve for the slip at $I_i = 40$ A, $T = 180$ N·m. Putting $x = R_r/s$ and rearranging the torque equation gives a quadratic in $x$, $$x^{2} + \left(2R_s - \frac{3(X_m I_i)^{2}}{T\omega_s}\right)x + \left(R_s^{2} + (X_m+X_s+X_r)^{2}\right) = 0 \;\Longrightarrow\; x^{2} - 36.2233x + 163.9025 = 0,$$ whose roots $x = 30.923$ and $x = 5.3004$ correspond to $s = 0.008085$ and $s = 0.047167$. The breakdown slip is $s_{max} = R_r/\sqrt{R_s^{2} + (X_m+X_s+X_r)^{2}} = 0.019528$, so the two roots straddle it, as they must.
5. Choose the root on flux, not on stability. Evaluate the magnetising current $I_m = |I_i - I_r|$ at each root. The low-slip root sends only 12.34 A into the rotor branch and forces 37.11 A of the 40 A injected through $X_m$, implying a terminal voltage of 386.0 V per phase — more than twice the flux of the other solution and far beyond saturation. The high-slip root leaves $I_m = 17.34$ A and a sensible 180.3 V per phase. Hence $$\boxed{s = 0.04717 \ \ (\text{rejecting } s = 0.00808)}.$$ The paper corroborates this independently: it hands over $s = 0.04$ for the 45 A row, which is likewise beyond the breakdown slip.
6. Row 1 terminal voltage. The inverter injects current into the parallel combination, so $$V_s = I_i\left|\,jX_m \parallel \left(R_s + \frac{R_r}{s} + j(X_s+X_r)\right)\right| = 40 \times 4.5076 = \boxed{180.30\ \text{V/phase}} = 312.3\ \text{V line}.$$ It is the parallel combination, not the rotor branch alone, that the current is driven into.
7. Row 2 — torque and voltage at $I_i = 45$ A, $s = 0.04$. Here the slip is handed over, so the torque follows by direct substitution with $x = R_r/s = 6.25\ \Omega$: $$T = \frac{3(10.4\times45)^{2}(6.25)}{78.540\left[(6.5)^{2} + (12.8)^{2}\right]} = \frac{4.1067\times10^{6}}{1.61863\times10^{4}} = \boxed{253.7\ \text{N}\cdot\text{m}},$$ and the same parallel-impedance calculation gives $$V_s = 45 \times 5.0196 = \boxed{225.88\ \text{V/phase}} = 391.2\ \text{V line}.$$
8. Row 3 — slip and voltage at $I_i = 50$ A, $T = 125$ N·m. The same quadratic with the new current and torque becomes $x^{2} - 82.1270x + 163.9025 = 0$, giving $x = 80.080$ and $x = 2.04675$, i.e. $s = 0.003122$ and $s = 0.122148$. Applying the same flux test, the low-slip root puts 49.40 A of the 50 A injected through $X_m$ and demands 513.8 V per phase, which is impossible; the high-slip root leaves $I_m = 12.77$ A at 132.8 V per phase. Hence $$\boxed{s = 0.12215},\qquad V_s = 50 \times 2.6566 = \boxed{132.83\ \text{V/phase}} = 230.1\ \text{V line}.$$
9. Check every completed row on power balance. The input apparent power resolved along the terminal power factor must equal the mechanical power plus the stator copper loss: $$3V_sI_i\cos\theta = T\omega_s + 3I_r^{2}R_s.$$ Row 1: $3(180.30)(40)(0.6842) = 14\,804$ W against $180(78.540) + 3(29.817)^{2}(0.25) = 14\,804$ W. Row 2 and row 3 close to the same precision. This single check catches a wrong root, a wrong current divider or a wrong parallel combination in one line.

Final results — completed table.