Firing angles for 200 V, 875 rpm DC motor

Data: $V = 200$ V, $N = 875$ rpm, $I_a = 150$ A, $R_a = 0.06\ \Omega$; supply 220 V, 50 Hz, $V_m = 311.13$ V.

For a single-phase full converter (continuous conduction):

$$V_a = \frac{2V_m}{\pi}\cos\alpha = 198.07\cos\alpha$$

Motor constant

At rated condition:

$$\begin{aligned} E &= V - I_a R_a = 200 - 150\times0.06 = 191\ \text{V} \\ K &= \frac{E}{N} = \frac{191}{875} = 0.21829\ \text{V/rpm} \end{aligned}$$

Rated torque means rated current, so $I_a R_a = 9$ V in every part.

i) Rated torque, 750 rpm

$$\begin{aligned} E &= 0.21829 \times 750 = 163.71\ \text{V} \\ V_a &= 163.71 + 9 = 172.71\ \text{V} \\ \cos\alpha &= \frac{172.71}{198.07} = 0.8720 \Rightarrow \alpha = 29.31^\circ \end{aligned}$$

ii) Rated torque, −500 rpm

Speed reversed, torque positive: regenerative braking in quadrant IV (e.g. an overhauling load).

$$\begin{aligned} E &= 0.21829 \times (-500) = -109.14\ \text{V} \\ V_a &= -109.14 + 9 = -100.14\ \text{V} \\ \cos\alpha &= \frac{-100.14}{198.07} = -0.5056 \Rightarrow \alpha = 120.37^\circ \end{aligned}$$

iii) Speed at $\alpha = 160^\circ$ and rated torque

$$\begin{aligned} V_a &= 198.07\cos160^\circ = -186.12\ \text{V} \\ E &= V_a - I_a R_a = -186.12 - 9 = -195.12\ \text{V} \\ N &= \frac{-195.12}{0.21829} = -893.9\ \text{rpm} \end{aligned}$$

Answer: (i) 29.3°, (ii) 120.4°, (iii) −894 rpm (reverse direction, regenerative braking).