Question 6 of 6: Problem 5 — bridge-fed separately excited d.c. motor

Question 6: Problem 5 — bridge-fed separately excited d.c. 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. (a) the firing angle at 1750 rpm; (b) the firing angle at 95 per cent of that speed; (c) the speed at a firing angle of 55°.

Approach. The bridge fixes the armature voltage as a cosine of the firing angle; the machine equation, with the armature drop given as a constant, makes the back e.m.f. and hence the speed a linear function of that armature voltage. One operating point calibrates the machine constant and the remaining two parts follow.

1. Part (a) — write the converter and machine equations. For a three-phase fully controlled bridge the mean output voltage is $$V_a=\frac{3\sqrt2}{\pi}V_{LL}\cos\alpha = 1.35047\times 230\times\cos\alpha =310.61\cos\alpha$$ and the armature loop, with the drop given as a constant 20 V, is $$V_a=E+I_aR_a=E+20,\qquad E=k_e\,n$$ Note the coefficient: $3\sqrt2/\pi=1.35047$ goes with the line-to-line voltage. The alternative $3\sqrt3/\pi$ belongs with the peak phase voltage and would give an answer high by a factor of $\sqrt3$.
2. Solve part (a) for the firing angle. With $V_{a1}=220$ V, $$\cos\alpha_1=\frac{220}{310.61}=0.70829 \quad\Longrightarrow\quad \boxed{\;\alpha_1=44.90^{\circ}\;}$$ The angle sits comfortably below 90°, so the converter is rectifying and delivering power to the machine, as a motoring problem requires.
3. Calibrate the machine constant. At 1750 rpm the back e.m.f. is $$E_1=V_{a1}-20=200\ \text{V} \quad\Longrightarrow\quad k_e=\frac{200}{1750}=0.114286\ \text{V/rpm}$$ Because the field is separately excited and held constant, and because the armature drop is given as a constant, this one constant serves all three parts.
4. Part (b) — drop the speed to 95 per cent. The new speed and back e.m.f. are $$n_2=0.95\times 1750=1662.5\ \text{rpm},\qquad E_2=0.114286\times 1662.5=190.0\ \text{V}$$ so the armature must be held at $V_{a2}=E_2+20=210.0$ V and $$\cos\alpha_2=\frac{210.0}{310.61}=0.67609 \quad\Longrightarrow\quad \boxed{\;\alpha_2=47.46^{\circ}\;}$$ A five per cent speed reduction therefore needs the firing angle retarded by only 2.56°. That non-linearity — the same speed change costs more angle near $\alpha=0$ and less near $\alpha=90^{\circ}$ — is the reason a practical drive closes a speed loop around the converter rather than scheduling firing angles open loop.
5. Part (c) — work the same chain backwards. At $\alpha_3=55^{\circ}$, $$V_{a3}=310.61\cos 55^{\circ}=310.61\times 0.573576=178.16\ \text{V}$$ $$E_3=V_{a3}-20=158.16\ \text{V}$$ $$n_3=\frac{E_3}{k_e}=\frac{158.16}{0.114286}=\boxed{\;1383.9\ \text{rpm}\;}$$
6. Check the three points for consistency. All three lie on one straight line in the $V_a$–n plane, since $n=(V_a-20)/k_e$: the pairs (220 V, 1750 rpm), (210 V, 1662.5 rpm) and (178.16 V, 1383.9 rpm) give slopes of exactly $1/k_e=8.75$ rpm per volt between any two of them. Speed falls monotonically as the firing angle is retarded, as it must, and at 55° the machine has been slowed to 79.1 per cent of its part (a) speed.
7. Note the assumption that the constant drop conceals. Stating $I_aR_a=20$ V as a constant is equivalent to assuming constant armature current, and therefore constant torque, at all three operating points: torque is $T=k_e(60/2\pi)I_a$, which depends on current alone. The problem is consistent with that reading, and no separate value of $R_a$ or $I_a$ is needed or deducible from the data given.