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 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.

Part (a) — Lowering the frequency at constant rated voltage

The air-gap flux of an induction machine is set by the ratio of applied voltage to frequency, because the stator winding is essentially a flux-linkage integrator: neglecting the small stator impedance drop, $V\approx4.44\,N\,k_w\,f\,\Phi$, so $\Phi\propto V/f$. Holding $V$ at its rated value while reducing $f$ therefore raises the flux in inverse proportion to the frequency, and every consequence follows from that one fact.

The immediate consequence is magnetic saturation. The iron of a well-designed machine already operates near the knee of its B–H curve at rated flux, so even a 20 per cent reduction in frequency drives the core well into saturation. Once saturated, the magnetising inductance collapses and the magnetising current rises far faster than linearly — it becomes large and strongly peaked rather than sinusoidal, drawing heavy third-harmonic and higher-order components. The stator carries this current on top of the load current, so copper loss rises steeply, the winding overheats, and the machine must be derated or will fail thermally.

Core loss behaves less obviously. Hysteresis loss scales roughly as $f B^{1.6}$ and eddy-current loss as $f^{2}B^{2}$; with $B\propto1/f$ the eddy component is roughly constant while the hysteresis component rises, so total iron loss does not fall in step with the frequency and the loss density in the core increases. Meanwhile the cooling fan, which is usually shaft-mounted, is turning more slowly and moving less air, so the machine's ability to reject the extra loss is reduced exactly when the loss goes up.

Electrically, the power factor deteriorates badly, because the enlarged magnetising current is almost purely reactive; the supply and the inverter must be rated for a current that produces no torque. The synchronous speed falls in proportion to frequency, $n_s=120f/P$, so the operating speed drops as intended, and the breakdown torque nominally rises, since $T_{\max}\propto(V/f)^{2}$ — but saturation prevents that increase from being realised, and the torque actually available is limited by the current the machine and inverter can tolerate. Finally, at a fixed slip frequency the rotor sees the same conditions, so the useful torque-per-ampere does not improve to compensate.

The correct practice, and the reason constant V/Hz control exists, is to reduce the voltage in proportion to the frequency so that the flux stays at its rated value. At low frequencies a small voltage boost is added to offset the stator resistance drop, which is no longer negligible compared with the reduced reactances. Above base speed the voltage cannot be raised further, so the flux is allowed to fall and the machine enters the constant-power field-weakening region.

Parts (b) and (c) — Slip, rotor speed, terminal voltage and power factor

Given. A three-phase, eight-pole, Y-connected induction motor is fed from a current-source inverter that holds the input current constant at 40 A, and at 50 Hz the machine develops 180 N·m.

Find. The slip and the rotor speed, then the terminal voltage per phase and the power factor at that operating point.

[Figure not reproduced: Figure 5.1 — The approximate constant-current equivalent circuit of Fig. (1), redrawn. The inverter current $I_i$ divides between the magnetising reactance and the series branch that carries the rotor current. See the official exam paper.]

Approach. Multiply the torque expression through by $s^{2}$ to turn it into a quadratic in slip, expect two roots, and choose between them on air-gap flux rather than on algebra: the low-slip root diverts nearly all the injected current into the magnetising branch and implies an impossible terminal voltage.

1. Synchronous speed and mechanical synchronous angular velocity. With $P=8$ poles at 50 Hz, $$n_s=\frac{120f}{P}=\frac{120\times50}{8}=750\ \text{rev/min},\qquad \omega_s=\frac{2\pi f}{P/2}=\frac{2\pi(50)}{4}=78.540\ \text{rad/s}.$$
2. Turn the torque equation into a quadratic in slip. Writing $X_T=X_m+X_s+X_r=13.017\ \Omega$ and multiplying the correct torque expression through by $s^{2}$, $$T\,\omega_s\left[(R_s s+R_r)^{2}+X_T^{2}s^{2}\right]=3\left(X_mI_i\right)^{2}R_r\,s ,$$ which with the numbers $T\omega_s=14137.2$ and $(X_mI_i)^{2}=173622$ becomes $$2.3960\times10^{6}\,s^{2}-1.13347\times10^{5}\,s+684.24=0 .$$
3. Both roots satisfy the torque requirement. $$s_1=0.007103\quad\text{and}\quad s_2=0.040203 ,$$ and substituting either into the torque expression returns 180.000 N·m. The quadratic cannot choose; the breakdown slip $s_{\max}=R_r/\sqrt{R_s^{2}+X_T^{2}}=0.016899$ lies between them, so one root is on each side of the torque peak.
4. Select the root on air-gap flux. The current divider gives the rotor current $I_r=I_i\,jX_m/(R_s+R_r/s+jX_T)$ and hence the magnetising current $I_m=I_i-I_r$:

   Root screening

   Root	$|I_r|$	$|I_m|$	$V_{ph}$	Verdict
   $s=0.007103$	12.34 A	37.04 A	385.8 V	rejected
   $s=0.040203$	29.35 A	17.58 A	183.1 V	selected

   At the low-slip root, 37.0 A of the 40 A injected goes into the magnetising branch and the terminal voltage implies roughly twice rated air-gap flux — the machine would be deeply saturated and the linear model would not apply. The high-slip root is the physical operating point.
5. Slip and rotor speed. $$\boxed{s=0.04020},\qquad n=(1-s)\,n_s=(1-0.040203)\times750,\qquad \boxed{n=719.85\ \text{rev/min}} .$$ The slip frequency is $sf=2.01$ Hz.
6. Terminal voltage per phase. The terminal voltage is the injected current times the parallel combination of the magnetising branch and the series branch — not the series branch alone. With $Z_{br}=(R_s+R_r/s)+j(X_s+X_r)=(5.671+j2.600)\ \Omega$, $$Z_{\text{term}}=\frac{jX_m\,Z_{br}}{jX_m+Z_{br}},\qquad V_{ph}=I_i\left|Z_{\text{term}}\right| ,$$ $$\boxed{V_{ph}=183.11\ \text{V}}\quad\text{(line-to-line }317.2\text{ V)} .$$
7. Power factor. The power factor is the cosine of the angle of that same parallel impedance, $$\cos\theta=\cos\left(\angle Z_{\text{term}}\right)=\boxed{0.667\ \text{lagging}} .$$
8. Close the power balance as an independent check. The electrical input must equal the mechanical output plus the stator copper loss: $$3V_{ph}I_i\cos\theta=T\omega_s+3I_r^{2}R_s:\qquad 3(183.11)(40)(0.6669)=14137.2+3(29.35)^{2}(0.2)=14654.0\ \text{W},$$ which closes to five significant figures. This single check validates the chosen root, the current divider and the parallel combination all at once.