22-Elec-B8 Power Electronics and Drives · December 2014

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

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2014 — 07-Elec-B8 Power Electronics and Drives. Open Book, 3 hours. Six problems, all of equal value (20 marks each); the rubric states that any five constitute a complete paper and only the first five presented are marked. All six are solved below, in full and including every sub-part, because the set is a study resource rather than an examination script.

Reference texts. M. H. Rashid, Power Electronics: Circuits, Devices and Applications, 4th ed. (primary EGBC reference for this code); N. Mohan, T. M. Undeland and W. P. Robbins, Power Electronics: Converters, Applications and Design, 3rd ed.; C. W. Lander, Power Electronics, 3rd ed.; B. K. Bose, Modern Power Electronics and AC Drives; R. Krishnan, Electric Motor Drives: Modeling, Analysis and Control. Angles are in degrees unless a formula requires radians, and every conduction integral below is written over the real conduction window rather than over an assumed half cycle.

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.

Preamble — Undesirable effects of high-frequency PWM drives

Raising the carrier frequency of a PWM drive buys a cleaner current waveform and quieter operation, but it does so by increasing both the number of switching events per second and the steepness of every voltage edge, and each of those brings its own penalty.

The first and most damaging is stress on the motor winding insulation. A modern device switches several hundred volts in a few tens of nanoseconds, and a cable more than a few metres long behaves as a transmission line: the fast edge reflects at the motor's high surge impedance and the terminal voltage can reach twice the dc link voltage. Worse, a steep edge does not distribute itself evenly through the winding — most of it appears across the first few turns of the first coil. The result is partial discharge in the turn-to-turn insulation and premature winding failure, which is why inverter-duty motors specify reinforced magnet wire and why $dv/dt$ filters or output reactors are fitted on long cable runs.

The second is bearing damage from shaft and common-mode currents. The three inverter output voltages do not sum to zero instant by instant, so a common-mode voltage appears between the winding neutral and earth at every switching edge. Stray capacitance from stator to rotor couples this onto the shaft; when the resulting shaft voltage exceeds the breakdown strength of the bearing lubricant film it discharges through the rolling elements, pitting the races and producing the characteristic fluting pattern. Insulated bearings, shaft grounding rings and common-mode chokes are the usual countermeasures.

The third is electromagnetic interference. Fast edges produce broadband conducted and radiated emission; the same capacitive paths that carry current into the shaft also return common-mode current through the earth conductor, which can trip residual-current devices, corrupt encoder and 4–20 mA instrumentation signals, and require screened cable, EMC filters and disciplined single-point grounding to meet the relevant CSA and IEC emission limits.

Two further effects deserve mention. Switching loss in the semiconductors is directly proportional to carrier frequency, so a higher carrier forces either derating of the inverter or a larger heat sink, reducing overall drive efficiency at exactly the point where efficiency is being sold. And although a carrier above about 15 kHz moves the magnetic noise out of the audible band, the harmonic currents that remain still produce additional stator and rotor heating, so a machine fed from an inverter must generally be derated relative to its sinusoidal rating.

Parts (a) and (b) — Voltage, current and breakdown torque

Given. A three-phase, four-pole induction motor with 1.4 mH of total leakage inductance and negligible winding resistance is fed from a constant volts-per-hertz drive, and two operating points are specified: 245 N·m of breakdown torque at 60 Hz, and a line current of 180 A at 63 Hz.

Given data
Quantity	Symbol	Value
Number of poles	$P$	4
Total leakage inductance	$L_T$	1.4 mH
Stator/rotor resistance	$R$	negligible
Point (a): frequency	$f_1$	60 Hz
Point (a): maximum torque	$T_{max,1}$	245 N·m
Point (a): rotor speed at $T_{max}$	$n$	1500 rpm
Point (b): frequency	$f_2$	63 Hz
Point (b): line current	$I_2$	180 A

Find. The line-to-line supply voltage and the line current at the first operating point, and the line-to-line supply voltage and breakdown torque at the second.

Figure 5.1 — The two operating points against the constant volts-per-hertz line drawn through point (a). Point (b) sits above the line: the 180 A stated in the question implies a voltage boost of about 5.4 per cent, and it is that boost which raises the breakdown torque from 245 to 272 N·m.

Approach. Rearrange the supplied breakdown-torque expression for the voltage, then obtain the current from the fact that at breakdown, with resistance neglected, the reflected rotor resistance divided by slip equals the leakage reactance — so the per-phase impedance magnitude is $\sqrt{2}\,\omega L_T$. Reverse the two steps for part (b), and check both answers independently through the air-gap power.

Part (a) — invert the torque expression for the voltage. With $\omega_1 = 2\pi(60) = 376.99$ rad/s, $$V_{LL}^2 = \frac{4\,\omega_1^2 L_T\,T_{max}}{P} = \frac{4(376.99)^2(1.4\times10^{-3})(245)}{4}$$ $$V_{LL} = \sqrt{48\,748} = \boxed{220.79\ \text{V}}$$ corresponding to $V_{ph} = 220.79/\sqrt{3} = 127.47$ V and a drive setting of $220.79/60 = 3.680$ V/Hz.
Part (a) — find the impedance at the breakdown point. Neglecting resistance, the standard result for maximum torque is that the reflected rotor resistance divided by slip equals the total leakage reactance, $R_2'/s = \omega L_T$. The per-phase circuit at that point therefore has equal resistive and reactive parts: $$|Z| = \sqrt{(\omega L_T)^2 + (\omega L_T)^2} = \sqrt{2}\,\omega L_T$$ With $\omega_1 L_T = 376.99 \times 1.4\times10^{-3} = 0.5278\ \Omega$, $|Z| = 0.7464\ \Omega$.
Part (a) — the line current follows. The machine is star-connected for this calculation, so $$I_{line} = \frac{V_{ph}}{\sqrt{2}\,\omega_1 L_T} = \frac{127.47}{0.7464} = \boxed{170.78\ \text{A}}$$ At breakdown the power factor is exactly $1/\sqrt{2} = 0.707$ lagging, which is a useful signature of this operating point.
Part (a) — cross-check through the air-gap power. Torque is air-gap power divided by synchronous mechanical speed, $$T = \frac{3 I^2 (R_2'/s)}{\omega_{s,mech}} = \frac{3(170.78)^2(0.5278)}{376.99/2} = 245.0\ \text{N}\cdot\text{m}$$ which reproduces the given torque to four figures by a route that does not use the supplied formula at all. As a further consistency note, the stated 1500 rpm against a synchronous speed of $120(60)/4 = 1800$ rpm implies a breakdown slip of 0.1667 and hence $R_2' = s\,\omega_1 L_T = 0.088\ \Omega$ — a plausible rotor resistance, so the data are mutually consistent even though neither answer needs the speed.
Part (b) — run the current relation backwards. At 63 Hz, $\omega_2 = 2\pi(63) = 395.84$ rad/s and $\omega_2 L_T = 0.5542\ \Omega$. Using the same breakdown impedance, $$V_{ph} = I_{line}\sqrt{2}\,\omega_2 L_T = 180 \times \sqrt{2} \times 0.5542 = 141.07\ \text{V}$$ so $$V_{LL} = \sqrt{3}\,(141.07) = \boxed{244.34\ \text{V}}$$
Part (b) — evaluate the breakdown torque at the new point. Substituting back into the supplied expression, $$T_{max,2} = \frac{(244.34)^2(4)}{4(395.84)^2(1.4\times10^{-3})} = \boxed{272.16\ \text{N}\cdot\text{m}}$$ and the air-gap route agrees exactly: $3(180)^2(0.5542)/(395.84/2) = 272.16\ \text{N}\cdot\text{m}$.
Interpret the two points together. Written in terms of the drive setting, the supplied formula becomes $$T_{max} = \frac{P}{16\pi^2 L_T}\left(\frac{V_{LL}}{f}\right)^2$$ so under a strictly constant volts-per-hertz law the breakdown torque is independent of frequency. Point (a) runs at 3.680 V/Hz and point (b) at $244.34/63 = 3.878$ V/Hz; the torque ratio $272.16/245 = 1.1109$ is exactly $(3.878/3.680)^2 = 1.1109$, confirming both answers against each other.

Check: two points in the question statement need comment. First, the preamble calls $L_T$ a total leakage inductance of 1.4 mH while the legend under the formula calls it a total leakage reactance; only the inductance reading is dimensionally consistent, because $V^2/(\omega^2 L)$ has units of newton-metres whereas $V^2/(\omega^2 X)$ does not. The solution therefore takes $L_T = 1.4$ mH throughout. Second, a drive holding a strictly constant volts-per-hertz ratio would deliver 220.79(63/60) = 231.83 V at 63 Hz and draw the same 170.78 A, not the 180 A stated in part (b). The question's own data therefore describe a point with about 5.4 per cent of voltage boost applied — entirely normal in a real drive, which boosts the volts-per-hertz ratio to compensate stator resistance. Part (b) has been answered from the data as given, and the discrepancy is reported rather than smoothed over.

Final results
Quantity	Symbol	Result
Point (a): line-to-line voltage	$V_{LL,1}$	220.79 V
Point (a): phase voltage	$V_{ph,1}$	127.47 V
Point (a): leakage reactance	$\omega_1 L_T$	0.5278 $\Omega$
Point (a): line current	$I_1$	170.78 A
Point (b): line-to-line voltage	$V_{LL,2}$	244.34 V
Point (b): leakage reactance	$\omega_2 L_T$	0.5542 $\Omega$
Point (b): maximum torque	$T_{max,2}$	272.16 N·m
Implied drive settings	$V_{LL}/f$	3.680 and 3.878 V/Hz

Concept & Theory

Topic. Breakdown (pull-out) torque of an induction machine under variable frequency, and the constant volts-per-hertz control law.

Key relations. $T_{max} = V_{LL}^2 P/(4\omega_i^2 L_T)$, equivalently $T_{max} = P(V_{LL}/f)^2/(16\pi^2 L_T)$; at breakdown with negligible resistance $R_2'/s = \omega L_T$, so $|Z| = \sqrt{2}\,\omega L_T$ and $I = V_{ph}/(\sqrt{2}\,\omega L_T)$; $T = 3I^2(R_2'/s)/\omega_{s,mech}$ with $\omega_{s,mech} = \omega_i/(P/2)$.

Why this works. Air-gap flux in an induction machine is set by the ratio of applied voltage to frequency, because the stator EMF is $4.44 N\Phi f$. Holding $V/f$ constant therefore holds the flux constant, and with constant flux the torque-slip curve simply slides along the speed axis as frequency changes, keeping the same peak. That is the whole point of the control law: full torque capability is available at any speed below base speed. The breakdown condition itself comes from maximising $3V^2(R_2'/s)/[\omega_{s,mech}((R_2'/s)^2 + X^2)]$ with respect to $R_2'/s$, which occurs when $R_2'/s = X$ — giving the equal real and imaginary impedance parts used above.

Common pitfall. Using the rotor speed rather than the synchronous speed when converting air-gap power to torque; the air-gap power always divides by $\omega_{s,mech}$, and dividing by the mechanical speed of 1500 rpm here would inflate the torque by 20 per cent. A second pitfall is dividing $\omega_i$ by the pole count instead of by pole pairs when forming $\omega_{s,mech}$.

Source. Bose, Modern Power Electronics and AC Drives, Ch. 2 and 8 (induction machine torque-slip relations, volts-per-hertz control); Krishnan, Electric Motor Drives, Ch. 5.

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