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22-Elec-B8 Power Electronics and Drives · May 2018

Question 6 of 6: Problem 5 — three-phase bridge feeding a separately excited dc motor

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

Notes on this paper

Paper format. National Exams, May 2018 — 16-Elec-B8, Power Electronics and Drives. Open book, three hours, any non-communicating calculator permitted. The paper is in two parts: Part 1 is twenty short-answer items (a) to (t) worth 2.5 points each, and Part 2 is five calculation problems worth 15 points each. The rubric says “attempt all parts” and that the maximum total score of 125 points includes a bonus of 25 points, so a candidate scoring 100 of the 125 available marks has a full paper. Every item and every sub-part is worked below.

Reference texts.

Question 6: Problem 5 — three-phase bridge feeding a separately excited dc 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. A six-pulse fully controlled bridge on a 220 V line-to-line supply feeds the armature of a separately excited dc machine, and the armature current is held at 155 A at every operating point.

Given data
QuantitySymbolValue
Supply (line-to-line, rms)$V_{LL}$220 V
Armature current (all cases)$I_{a}$155 A
Case (a)$\alpha$ / $n$45° / 1750 rpm
Case (b)$\alpha$ / $n$55° / 1200 rpm
Case (c)$\alpha$65°

Find. (a) the armature voltage at 45°; (b) the armature-circuit resistance, the output power and the torque at 55° and 1200 rpm; (c) the speed reached when the firing angle is retarded to 65°.

Problem 5: bridge output voltage against firing anglebridge output voltage Va = 1.3505 VLL cos α45° → 1750 rpm55° → 1200 rpm65° → 578 rpm0°15°30°45°60°75°90°firing angle α
The bridge output follows a cosine in the firing angle. Because the armature current is held constant, the IR drop is the same at every point on this curve, so the spacing of the operating points measures the back-EMF constant directly.

Approach. With $I_{a}$ held constant the armature drop $I_{a}R_{a}$ is identical at every operating point, so differencing two armature-voltage/speed pairs cancels it and isolates the back-EMF constant; a single loop equation then yields $R_{a}$, and the third firing angle needs no new data at all.

  1. Write the bridge output voltage. For a six-pulse fully controlled bridge on a line-to-line supply, $$V_{a}=\frac{3\sqrt{2}}{\pi}V_{LL}\cos\alpha=1.3505\times220\cos\alpha=297.10\cos\alpha$$
  2. Evaluate at 45°. $$V_{a1}=297.10\times\cos 45^\circ=\boxed{210.1\ \text{V}}$$
  3. Evaluate at 55°. $$V_{a2}=297.10\times\cos 55^\circ=170.4\ \text{V}$$

Both operating points obey $V_{a}=E+I_{a}R_{a}$ with the same current, hence with the same IR drop. Subtracting them therefore removes $R_{a}$ entirely and leaves only the machine constant, which is the key that unlocks parts (b) and (c).

  1. Difference the two cases to get the back-EMF constant. $$k_{e}=\frac{V_{a1}-V_{a2}}{n_{1}-n_{2}}=\frac{210.08-170.41}{1750-1200}=0.072132\ \text{V/rpm}$$
  2. Substitute back into either loop equation for the resistance. With $E_{1}=k_{e}n_{1}=126.23$ V, $$R_{a}=\frac{V_{a1}-E_{1}}{I_{a}}=\frac{210.08-126.23}{155}=\boxed{0.541\ \Omega}$$ The other case gives the same value: $E_{2}=k_{e}n_{2}=86.56$ V and $(170.41-86.56)/155=0.541\ \Omega$.
  3. Take the output power at 1200 rpm. The mechanical power is the back EMF times the armature current, never the terminal voltage times the current: $$P_{out}=E_{2}I_{a}=86.558\times155=\boxed{13.42\ \text{kW}}$$
  4. And the torque. With $\omega_{2}=2\pi\times1200/60=125.66$ rad/s, $$T=\frac{P_{out}}{\omega_{2}}=\frac{13416.5}{125.664}=\boxed{106.8\ \text{N}\cdot\text{m}}$$

That torque is worth a second look, because it is also $T=k_{e}(60/2\pi)I_{a}=0.072132\times9.5493\times155=106.8$ N·m — a function of armature current alone. Since the current is held at 155 A throughout, every operating point in this problem develops the same torque, and a solution that produced different torques at different firing angles would contain an error.

  1. Retard the firing to 65°. $$V_{a3}=297.10\times\cos 65^\circ=125.56\ \text{V}$$
  2. Subtract the unchanged armature drop. $$E_{3}=V_{a3}-I_{a}R_{a}=125.56-155\times0.540994=41.71\ \text{V}$$
  3. Convert to speed. $$n_{3}=\frac{E_{3}}{k_{e}}=\frac{41.708}{0.072132}=\boxed{578\ \text{rpm}}$$
Final results
QuantitySymbolResult
Armature voltage at 45°$V_{a1}$210.1 V
Armature voltage at 55°$V_{a2}$170.4 V
Armature voltage at 65°$V_{a3}$125.6 V
Back-EMF constant$k_{e}$0.07213 V/rpm
Armature-circuit resistance$R_{a}$0.541 Ω
Back EMF at 1200 rpm$E_{2}$86.56 V
Output power at 1200 rpm$P_{out}$13.42 kW
Developed torque (all cases)$T$106.8 N·m
Speed at 65°$n_{3}$578 rpm
Check: read “resistance of the armature circuit” literally. The deduced 0.541 Ω dissipates $I_{a}^{2}R_{a}=155^{2}\times0.541=13.0$ kW against a mechanical output of 13.4 kW, which no real armature winding could tolerate. The data are internally consistent — both loop equations return the same value to six figures — so the figure is not an arithmetic slip. It is best read as the total circuit resistance seen by the converter: armature winding plus brush drop plus the equivalent commutation resistance of the bridge, and quite possibly a series starting resistor still in circuit. The answers are reported as the data give them, with that reading stated as an assumption.
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