Two semi converters for field and armature

Data: $V_s = 220$ V, $V_m = \sqrt2 \times 220 = 311.13$ V, $R_f = 250\ \Omega$, $R_a = 0.2\ \Omega$, $T_L = 150$ N·m at 1500 rpm, $K_v = K_t = 0.8$.

For a single-phase semi-converter (continuous current): $V_o = \dfrac{V_m}{\pi}(1+\cos\alpha)$.

$$\omega = \frac{2\pi \times 1500}{60} = 157.08\ \text{rad/s}$$

i) Field current

With $\alpha_f = 0$:

$$\begin{aligned} V_f &= \frac{V_m}{\pi}(1+1) = \frac{2\times311.13}{\pi} = 198.07\ \text{V} \\ I_f &= \frac{V_f}{R_f} = \frac{198.07}{250} = 0.792\ \text{A} \end{aligned}$$

ii) Firing angle of armature converter

$$\begin{aligned} T &= K_t I_f I_a \Rightarrow I_a = \frac{150}{0.8 \times 0.792} = 236.66\ \text{A} \\ E_b &= K_v I_f \omega = 0.8 \times 0.792 \times 157.08 = 99.56\ \text{V} \\ V_a &= E_b + I_a R_a = 99.56 + 236.66\times0.2 = 146.89\ \text{V} \\ 1 + \cos\alpha_a &= \frac{\pi V_a}{V_m} = \frac{\pi\times146.89}{311.13} = 1.4832 \\ \cos\alpha_a &= 0.4832 \Rightarrow \alpha_a = 61.10^\circ \end{aligned}$$

iii) Input power factor of armature converter

For a semi-converter with ripple-free current $I_a$, the supply current is a quasi-square wave of height $I_a$ flowing for $(\pi - \alpha)$ in each half cycle:

$$I_s = I_a\sqrt{\frac{\pi-\alpha}{\pi}}$$

Neglecting losses, input power = output power $= V_a I_a$:

$$\begin{aligned} PF &= \frac{V_a I_a}{V_s I_s} = \frac{\sqrt2(1+\cos\alpha)}{\sqrt{\pi(\pi-\alpha)}} \\ &= \frac{\sqrt2 \times 1.4832}{\sqrt{\pi \times (\pi - 1.0664)}} = 0.822\ \text{(lagging)} \end{aligned}$$

($\alpha = 61.10^\circ = 1.0664$ rad.)

Answer: $I_f$ = 0.792 A, $\alpha_a$ = 61.1°, PF = 0.82 lagging.