22-Elec-B8 Power Electronics and Drives · December 2013

Question 6 of 6: Converter-Fed Separately Excited DC Drive

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

Question 6: Converter-Fed Separately Excited DC 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.

Part (a) — classification of dc drives and the controlled variables

Classified by input supply, dc drives fall into two families. Ac-fed converter drives take their power from an ac system through a phase-controlled rectifier, and subdivide by the number of supply phases and by the converter topology: single-phase half-wave, semi-converter, full converter and dual converter for fractional and small horsepower machines; three-phase half-wave, semi-converter, full converter and dual converter for larger ratings, where the higher ripple frequency of a six-pulse bridge gives smoother armature current and better form factor. A semi-converter operates in one quadrant, a full converter in two (it can invert, returning energy to the supply while the current direction is fixed), and a dual converter in all four, since the second bridge supplies reverse current. Dc-fed chopper drives take their power from an existing dc source — a battery, a traction third rail, or a fixed rectified dc bus — and control the armature by pulse-width modulation; class A through class E choppers provide one-, two- and four-quadrant operation respectively. Traction and battery-vehicle drives are the natural home of the chopper drive.

The variables to be controlled follow from the machine equations $V_a = E + I_aR_a$, $E = K\phi\,\omega_m$ and $T = K\phi I_a$. Below base speed the field is held at its rated value and the armature voltage is varied; because the flux is constant, this gives a constant-torque region in which speed follows $V_a$ almost linearly. Above base speed the armature voltage has reached its ceiling and the field current is reduced instead — field weakening — giving a constant-power region in which speed rises but available torque falls as $1/\omega$. Underneath both sits the inner control variable, armature current, which is regulated directly because it is proportional to torque and because it must be limited during acceleration and reversal to protect the commutator and the converter. A practical drive therefore closes a fast current loop inside a slower speed loop, and manipulates armature voltage, armature current and field current as its three control handles.

Parts (b), (c) and (d) — the drive operating points

Given.

Find. (b) the armature voltage at 44.5°; (c) the armature-circuit resistance, output power and torque at 995 rpm; and (d) the speed reached at a firing angle of 64.5°.

Approach. Compute the bridge output voltage from the cosine control law, then exploit the fact that a constant armature current makes the $I_aR_a$ drop identical at every operating point — so differencing two points cancels it and isolates the back-emf constant.

1. Part (b) — apply the six-pulse bridge control law. For a three-phase full-wave bridge the mean output voltage in continuous conduction is $$V_a = \frac{3\sqrt{2}}{\pi}V_{LL}\cos\alpha = 1.35047\,V_{LL}\cos\alpha = 310.61\cos\alpha \ \text{V}.$$ At $\alpha_1 = 44.5^\circ$, with $\cos 44.5^\circ = 0.71325$, $$\boxed{\;V_{a1} = 310.61 \times 0.71325 = 221.5 \ \text{V}\;}$$ The constant armature current guarantees continuous conduction, so the law applies without correction.
2. Evaluate the other two operating points now, since they are needed below. With $\cos 55.5^\circ = 0.56641$ and $\cos 64.5^\circ = 0.43051$, $$V_{a2} = 310.61(0.56641) = 175.9 \ \text{V}, \qquad V_{a3} = 310.61(0.43051) = 133.7 \ \text{V}.$$
3. Part (c) — difference the two known operating points. Each point satisfies the armature loop equation $V_a = E + I_aR_a$ with $E = k_e n$, and because $I_a$ is the same 118 A at both, the resistive drop is identical and cancels on subtraction: $$V_{a1} - V_{a2} = k_e(n_1 - n_2) \quad\Rightarrow\quad k_e = \frac{221.54 - 175.93}{1700 - 995} = \frac{45.61}{705} = 0.064697 \ \text{V/rpm},$$ equivalently $0.6178$ V·s/rad. This is the step that makes the problem solvable at all with only two measurements.
4. Recover the armature-circuit resistance. Back-substituting into the first operating point, the back-emf at 1700 rpm is $E_1 = 0.064697 \times 1700 = 109.98$ V, so $$\boxed{\;R_a = \frac{V_{a1} - E_1}{I_a} = \frac{221.54 - 109.98}{118} = 0.945 \ \Omega\;}$$ and the resistive drop is $I_aR_a = 111.6$ V at every operating point. Checking the second point closes the loop: $E_2 = 0.064697 \times 995 = 64.37$ V and $64.37 + 111.56 = 175.93$ V, which is exactly $V_{a2}$.
5. Compute the output power at 995 rpm. Mechanical output is developed by the back-emf, not by the terminal voltage — the difference between them is dissipated in the armature circuit: $$\boxed{\;P_{\text{out}} = E_2I_a = 64.37 \times 118 = 7596 \ \text{W} = 7.60 \ \text{kW}\;}$$
6. Convert power to torque. With $\omega_m = 2\pi(995)/60 = 104.20$ rad/s, $$\boxed{\;T = \frac{P_{\text{out}}}{\omega_m} = \frac{7596}{104.20} = 72.9 \ \text{N}\cdot\text{m}\;}$$ A useful check: since both the flux and the armature current are constant, $T = k\phi I_a$ must be the same at every operating point. Evaluating it at 1700 rpm gives $E_1I_a/\omega_{m1} = 109.98(118)/178.02 = 72.9$ N·m, identical as required.
7. Part (d) — find the speed at the new firing angle. Because $I_a$ is unchanged, the resistive drop is still 111.56 V, so the back-emf at $\alpha_3 = 64.5^\circ$ is $$E_3 = V_{a3} - I_aR_a = 133.72 - 111.56 = 22.16 \ \text{V},$$ $$\boxed{\;n_3 = \frac{E_3}{k_e} = \frac{22.16}{0.064697} = 342.6 \ \text{rpm}\;}$$ No new data are needed — $k_e$ and $R_a$ were both fixed by the first two operating points, which is the whole point of the constant-current condition.
8. Note the loss balance. At the 995 rpm point the converter delivers $V_{a2}I_a = 20.8$ kW while the shaft receives only 7.6 kW, the remaining 13.2 kW being dissipated in the armature circuit — an efficiency of 37 percent. That is the arithmetic consequence of the numbers as printed, and it is flagged below.