Question 3 of 6: Problem 2 — single-phase a.c. voltage controller feeding a motor

Question 3: Problem 2 — single-phase a.c. voltage controller feeding a motor (15 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.

Given.

Find. (b) the delay angle; (c) the rms output voltage of the controller; (d) the mean current carried by each thyristor.

Approach. The motor power factor fixes the load angle; the extinction condition converts the stated conduction angle into the delay angle; the rms of the truncated sine over the real conduction window gives the output voltage; and the motor power balance gives the load rms current, which is converted to a device mean by the shape factor of the normalised conduction pulse.

1. Part (b) — fix the load angle and set up the extinction condition. The motor is an inductive load of angle $$\phi=\arccos(0.85)=31.788^{\circ},\qquad \tan\phi=0.61974$$ In a full-wave a.c. voltage controller each thyristor of the back-to-back pair is fired at $\alpha$ in its own half cycle and conducts until its current reaches zero at $\beta=\alpha+\gamma$. The zero-current condition is the same one as for the half-wave rectifier, written with $\beta$ eliminated: $$\sin(\alpha+\gamma-\phi)=\sin(\alpha-\phi)\,e^{-\gamma/\tan\phi}$$
2. Solve it numerically. With $\gamma=165^{\circ}=2.87979$ rad the exponential factor is $e^{-2.87979/0.61974}=e^{-4.6467}=0.009593$, so the right-hand side is small but not negligible. Solving, $$\boxed{\;\alpha=46.65^{\circ}\;}\qquad \beta=\alpha+\gamma=211.65^{\circ}$$ The closed-form approximation $\alpha\approx 180^{\circ}+\phi-\gamma=46.79^{\circ}$, which is exact in the limit of a vanishing exponential factor, is high by 0.14°. Its error tracks $e^{-\gamma/\tan\phi}$, so at this good power factor and long conduction angle it is a usable estimate — but only as a check, since at a poor power factor the same formula can be wrong by ten degrees.
3. Part (c) — integrate the truncated sine over the real window. Each half cycle contributes conduction from $\alpha$ to $\beta$, so $$\begin{aligned} V_o&=V_s\sqrt{\frac{1}{\pi}\int_{\alpha}^{\beta}2\sin^2\theta\,\mathrm{d}\theta}\ &=V_s\sqrt{\frac{1}{\pi}\Bigl[\gamma-\frac{\sin 2\beta-\sin 2\alpha}{2}\Bigr]} \end{aligned}$$ with the angles in radians. Substituting $\sin 2\beta=\sin 423.30^{\circ}=0.89333$ and $\sin 2\alpha=\sin 93.29^{\circ}=0.99835$ gives a bracket of $2.87979+0.05251=2.93230$ rad, hence $$V_o=230\sqrt{\frac{2.93230}{\pi}}=230\times 0.96611=\boxed{\;222.21\ \text{V}\;}$$ The controller therefore delivers 96.6 per cent of the supply voltage. That is what a 165° conduction angle should give: only 15° of each half cycle is removed, and it is removed near the zero crossing where the sine contributes least to the mean square.
4. Part (d) — get the load rms current from the motor power balance. The motor delivers 100 hp at an efficiency of 0.95, so the electrical input is $$P_{in}=\frac{P_{out}}{\eta}=\frac{100\times 745.7}{0.95}=78\,494.7\ \text{W}$$ and at a power factor of 0.85 the apparent power and the rms current are $$\begin{aligned} S&=\frac{P_{in}}{\cos\phi}=92\,346.8\ \text{VA}\ I_{rms}&=\frac{S}{V_o}=\frac{92\,346.8}{222.21}=415.59\ \text{A} \end{aligned}$$ The load impedance is never given and is never needed: it cancels in the next step.
5. Build the normalised conduction pulse. Over $\alpha\le\omega t\le\beta$ the current has the same shape as before, $$u(\omega t)=\sin(\omega t-\phi)-\sin(\alpha-\phi)\,e^{(\alpha-\omega t)/\tan\phi}$$ which is the true waveform divided by $V_m/Z$. Numerical integration of this pulse gives $$\begin{aligned} \frac{1}{2\pi}\int_{\alpha}^{\beta}u\,\mathrm{d}\theta&=0.287936\ \sqrt{\frac{1}{\pi}\int_{\alpha}^{\beta}u^2\,\mathrm{d}\theta}&=0.664599 \end{aligned}$$ the first averaged over $2\pi$ because each device conducts once per supply cycle, the second over $\pi$ because both devices contribute to the load rms.
6. Convert to the mean device current. The unknown scale factor $V_m/Z$ appears in both integrals and divides out, so $$\begin{aligned} \frac{I_{T,avg}}{I_{rms}}&=\frac{0.287936}{0.664599}=0.433247\ I_{T,avg}&=0.433247\times 415.59=\boxed{\;180.05\ \text{A}\;} \end{aligned}$$
7. Check against the half-sine device shortcut. If the device current were a perfect half sine, the mean would be $\sqrt2\,I_{rms}/\pi=187.08$ A. That is 3.9 per cent above the exact figure, and high in the safe direction, because the real pulse is truncated at both ends by the firing delay and the finite extinction angle. Quoting it alongside the exact answer is worth doing: it is the number a device-selection calculation would use.