Question 6 of 6: Three-Phase Bridge-Fed Separately Excited D.C. Drive

Problem 6: Three-Phase Bridge-Fed Separately Excited D.C. 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. Three-phase full-wave (six-pulse) bridge rectifier supplied at $V_{LL} = 230\ \text{V}$ line-to-line, feeding the armature of a separately excited d.c. motor; armature current held at $I_a = 200\ \text{A}$ at all times; operating points (39°, 1720 rpm), (57°, 1000 rpm) and a third at 65°. Field flux constant, so $E = k_e n$.

Find. (a) the classification of d.c. drives by supply and the variables a variable-speed d.c. drive controls; (b) the armature voltage at 39 degrees; (c) the armature circuit resistance, output power and torque at 1000 rpm; (d) the speed at 65 degrees.

Approach. The bridge output is $V_a = (3\sqrt{2}/\pi)V_{LL}\cos\alpha$. Because $I_a$ is the same at every operating point, the $I_aR_a$ drop is identical too, so differencing the two given operating points cancels it and isolates the back-e.m.f. constant; one loop equation then yields $R_a$, and the third firing angle needs no new data.

Part (a) — classification and controlled variables. D.c. drives are classified by the nature of the input supply. A.c.-fed (phase-controlled) drives take power from the a.c. mains through a line-commutated thyristor converter: single-phase half-wave, semi-converter, or full converter for fractional and small ratings, and three-phase semi-converters, full converters or dual converters for larger ratings, the dual converter giving full four-quadrant operation. D.c.-fed (chopper) drives take power from a fixed d.c. source — a battery, a traction third rail or an uncontrolled rectifier — and interpose a chopper of the class A to E families, class E again giving four-quadrant capability. A third family, the Ward–Leonard motor–generator set, is the historical rotating equivalent of the controlled rectifier and is now obsolete except in retrofits.

The variables to be controlled are those that appear in the two governing equations V_a = E + I_aR_a with E = kΦω, and T = kΦI_a. They are: the armature voltage, which sets the speed below base speed at constant flux; the field current (flux), which is weakened to extend the speed above base speed at constant power; and the armature current, which sets the torque and must be held within the commutation and thermal limits. A practical drive therefore uses a nested structure — an inner armature-current (torque) loop inside an outer speed loop, with a separate field-current loop — because current must be limited during every transient regardless of what the speed loop demands.

1. Part (b) — armature voltage at 39 degrees. For a three-phase full-wave bridge in continuous conduction the mean output referred to the line voltage 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 $\alpha = 39^{\circ}$, $\cos 39^{\circ} = 0.77715$, so $$\boxed{V_{a1} = 310.61 \times 0.77715 = 241.39\ \text{V}}$$ This is the terminal voltage applied to the armature while the machine runs at 1720 rpm.
2. Part (c) — second operating point and the back-e.m.f. constant. At $\alpha = 57^{\circ}$, $\cos 57^{\circ} = 0.54464$, so $V_{a2} = 310.61(0.54464) = 169.17\ \text{V}$. Writing the armature loop at each point, $V_a = k_e n + I_a R_a$, and subtracting — legitimate because $I_a$ is the same 200 A at both points, so the $I_aR_a$ terms cancel identically: $$k_e = \frac{V_{a1} - V_{a2}}{n_1 - n_2} = \frac{241.39 - 169.17}{1720 - 1000} = \frac{72.22}{720} = 0.10030\ \text{V/rpm}.$$
3. Part (c) — armature circuit resistance. Back at the first operating point, $E_1 = k_e n_1 = 0.10030(1720) = 172.52\ \text{V}$, so $$\boxed{R_a = \frac{V_{a1} - E_1}{I_a} = \frac{241.39 - 172.52}{200} = 0.344\ \Omega}$$ Checking at the second point, $E_2 = 0.10030(1000) = 100.30\ \text{V}$ and $E_2 + I_aR_a = 100.30 + 68.91 = 169.21\ \text{V}$, which reproduces $V_{a2}$.
4. Part (c) — output power and torque at 1000 rpm. Mechanical power is developed by the back e.m.f., not by the terminal voltage: $$P_{out} = E_2 I_a = 100.30 \times 200 = 20\,060\ \text{W} \;\Rightarrow\; \boxed{P_{out} = 20.06\ \text{kW}}$$ and with $\omega = 2\pi(1000)/60 = 104.72\ \text{rad/s}$, $$\boxed{T = \frac{P_{out}}{\omega} = \frac{20\,060}{104.72} = 191.6\ \text{N}\cdot\text{m}}$$ Equivalently $T = k_t I_a$ with $k_t = k_e(60/2\pi) = 0.9578\ \text{N}\cdot\text{m/A}$, giving the same 191.6 N·m.
5. Part (d) — speed at 65 degrees. The armature current is unchanged, so the resistive drop is still $I_aR_a = 68.87\ \text{V}$. With $\cos 65^{\circ} = 0.42262$, $$V_{a3} = 310.61(0.42262) = 131.27\ \text{V}, \qquad E_3 = 131.27 - 68.87 = 62.40\ \text{V},$$ $$\boxed{n_3 = \frac{E_3}{k_e} = \frac{62.40}{0.10030} = 622\ \text{rpm}}$$ Torque is unchanged at 191.6 N·m because the armature current and flux are unchanged; only the speed, and hence the output power (12.48 kW), have fallen.