Question 5 of 6: Constant Volts-per-Hertz Induction Motor Drive

Question 5: Constant Volts-per-Hertz Induction Motor Drive (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.

Given. A three-phase, four-pole induction motor with a total leakage inductance of 1.5 mH and negligible resistance runs from a constant volts-per-hertz inverter. Two operating points are specified, and the paper supplies the breakdown-torque approximation to be used.

Find. Three undesirable effects of high-frequency PWM drives; the line-to-line supply voltage and line current at point A; and the line-to-line voltage and breakdown torque at point B.

Approach. Invert the supplied torque expression for the line voltage at point A; obtain the current from the fact that at breakdown the referred rotor resistance equals the leakage reactance, so the per-phase impedance magnitude is $\sqrt{2}\,\omega L_T$; then run the same two relations in the opposite order at point B.

Part (a) — Three undesirable effects of high-frequency PWM drives

1. Insulation stress from fast voltage transitions and reflected waves. A modern IGBT inverter switches in tens of nanoseconds, producing $dv/dt$ of several thousand volts per microsecond. On a cable of any appreciable length the motor terminals act as an impedance mismatch, and the incident wave reflects and adds, so the winding can see up to twice the d.c. link voltage on every pulse. The stress falls almost entirely on the first few turns of the first coil, causing partial discharge and premature turn-to-turn failure. Inverter-duty insulation, $dv/dt$ output filters, or terminating networks are the standard countermeasures.

2. Common-mode voltage, shaft voltage and bearing currents. The three inverter pole voltages do not sum to zero instant by instant, so a common-mode voltage appears between the winding neutral and earth. Capacitive coupling across the air gap impresses part of it on the rotor, and when the resulting shaft voltage exceeds the breakdown strength of the bearing lubricant film an electric-discharge-machining current flows through the race. The result is fluting, roughening and early bearing failure. Insulated bearings, shaft grounding rings and common-mode chokes are used to control it.

3. Switching losses and electromagnetic interference. Device losses are proportional to switching frequency, so raising the carrier frequency to reduce audible noise and current ripple directly reduces inverter efficiency and increases the heatsink burden. At the same time the fast edges generate conducted and radiated emissions across a wide spectrum, which interfere with instrumentation, encoder feedback and communications, and drive high-frequency earth-leakage current that can cause nuisance tripping of ground-fault protection.

Parts (b) and (c) — Quantitative solution

1. Part (b) — invert the supplied torque expression for the line voltage. Rearranging $T_{max}=[V_{LL}]^2P/(4\omega_i^2L_T)$ gives $$V_{LL}=\sqrt{\frac{4\,T_{max}\,\omega_i^2\,L_T}{P}} .$$ With $\omega_1 = 2\pi(60) = 376.99$ rad/s, $T_{max}=350$ N$\cdot$m, $L_T = 1.5$ mH and $P = 4$: $$V_{LL,1}=\sqrt{\frac{4(350)(376.99)^2(0.0015)}{4}}=\sqrt{74\,614} =\boxed{273.2\ \text{V (line-to-line)}} .$$
2. Establish the current relation at the breakdown point. Neglecting stator resistance, the per-phase equivalent circuit at breakdown satisfies $R_2'/s=\omega L_T$, so the impedance magnitude is $$|Z|=\sqrt{\left(\frac{R_2'}{s}\right)^2+(\omega L_T)^2}=\sqrt{2}\,\omega L_T .$$ The line current, which equals the phase current in a star-connected machine, is therefore $I=V_{ph}/(\sqrt{2}\,\omega L_T)$. Note that this half of the sub-part needs the phase voltage even though the torque formula is written in line volts.
3. Evaluate the line current at point A. With $V_{ph,1}=273.16/\sqrt{3}=157.71$ V and $\sqrt{2}\,\omega_1 L_T = 0.79977\ \Omega$: $$I_1=\frac{157.71}{0.79977}=\boxed{197.2\ \text{A}} .$$ A useful free check comes from computing the torque a second way, from air-gap power: $T=3I^2(\omega L_T)\big/(2\omega/P)=3(197.2)^2(0.5655)/188.50 = 350.0$ N$\cdot$m, which returns the stated value exactly and so validates both the voltage and the current.
4. Part (c) — run the current relation forwards at the new frequency. At $f_2 = 65$ Hz, $\omega_2 = 408.41$ rad/s and $\sqrt{2}\,\omega_2 L_T = 0.86636\ \Omega$, so a stated line current of 200 A requires $$V_{ph,2}=200\times 0.86636=173.27\ \text{V},\qquad V_{LL,2}=\sqrt{3}\times173.27=\boxed{300.1\ \text{V (line-to-line)}} .$$
5. Evaluate the breakdown torque at point B. Substituting back into the supplied expression, $$T_{max,2}=\frac{(300.12)^2(4)}{4(408.41)^2(0.0015)}=\boxed{360.0\ \text{N}\cdot\text{m}} .$$ Checking it independently from air-gap power, $3(200)^2(408.41\times0.0015)/(2\times408.41/4)=360.0$ N$\cdot$m, in exact agreement.
6. Confirm the volts-per-hertz consistency. Rewriting the supplied formula as $T_{max}=(V_{LL}/f)^2P/(16\pi^2L_T)$ shows that breakdown torque depends only on the volts-per-hertz ratio, not on frequency, so the torque ratio must equal the square of the voltage-per-hertz ratio. Here $V/f$ is 4.5526 V/Hz at point A and 4.6172 V/Hz at point B, a 1.42 per cent boost, and $$\left(\frac{4.6172}{4.5526}\right)^2=1.02857=\frac{360.0}{350.0},$$ which is a complete regression check on both answers at once.

The quoted rotor speed of 1800 rpm is not needed by either sub-part, but it is worth using as a consistency check: for a four-pole machine at 60 Hz the synchronous speed is $120f/P = 1800$ rpm, so the figure quoted in the question is the synchronous speed of point A rather than the loaded speed at breakdown. At 65 Hz the synchronous speed rises to 1950 rpm.