Two semi converters (50 N-m at 1000 rpm)

Data: $V_s = 220$ V, $V_m = \sqrt2\times220 = 311.13$ V; $R_f = 250\ \Omega$, $R_a = 0.2\ \Omega$; $T = 50$ N-m at $N = 1000$ rpm; $K_v = 0.8$ V/A-rad/s, $K_t = 0.8$ N-m/A². For a single-phase semiconverter with continuous current:

$$V_{dc} = \frac{V_m}{\pi}(1 + \cos\alpha)$$

i) Field current

Field firing angle $\alpha_f = 0$:

$$\begin{aligned} V_f &= \frac{311.13}{\pi}(1 + 1) = 198.07\ \text{V} \\ I_f &= \frac{V_f}{R_f} = \frac{198.07}{250} = 0.792\ \text{A} \end{aligned}$$

ii) Armature firing angle

Armature current from $T = K_tI_fI_a$:

$$I_a = \frac{50}{0.8\times0.7923} = 78.89\ \text{A}$$

Speed $\omega = \dfrac{2\pi\times1000}{60} = 104.72$ rad/s. Back emf:

$$E_g = K_v\,\omega\,I_f = 0.8\times104.72\times0.7923 = 66.37\ \text{V}$$

Armature voltage:

$$V_a = E_g + I_aR_a = 66.37 + 78.89\times0.2 = 82.15\ \text{V}$$

Firing angle:

$$\begin{aligned} 82.15 &= \frac{311.13}{\pi}(1 + \cos\alpha_a) \\ \cos\alpha_a &= \frac{82.15\pi}{311.13} - 1 = -0.1705 \\ \alpha_a &= 99.82^\circ \end{aligned}$$

iii) Power factor of armature converter

With ripple-free armature current $I_a$, the supply current is a quasi-square wave of height $I_a$ conducting for $(\pi - \alpha)$ in each half cycle.

$$\begin{aligned} I_s &= I_a\sqrt{\frac{\pi-\alpha}{\pi}} = 78.89\sqrt{\frac{180-99.82}{180}} = 52.65\ \text{A} \\ I_{s1} &= \frac{2\sqrt2}{\pi}I_a\cos\frac{\alpha}{2} = 0.9003\times78.89\times0.6440 = 45.74\ \text{A} \end{aligned}$$

Displacement factor $= \cos(\alpha/2) = \cos 49.91^\circ = 0.644$.

$$PF = \frac{I_{s1}}{I_s}\cos\frac{\alpha}{2} = \frac{45.74}{52.65}\times0.644 = 0.559\ \text{(lagging)}$$

Check by power balance (losses neglected): $PF = \dfrac{V_aI_a}{V_sI_s} = \dfrac{82.15\times78.89}{220\times52.65} = 0.559$.

Answer: (i) $I_f = 0.792$ A; (ii) $\alpha_a = 99.8^\circ$; (iii) PF $= 0.559$ lagging.