Three-phase full converter at 50% of maximum output

A three-phase full converter (six-thyristor bridge) gives $V_{dc} = \frac{3\sqrt3V_m}{\pi}\cos\alpha$, where $V_m$ is the peak phase voltage. With a resistive load this holds for $\alpha \le 60^\circ$ (continuous conduction).

        P (+) o-----+------+------+-----+
                    |      |      |     |
                   T1     T3     T5     |
                    |      |      |     R
 a o----------------+      |      |     |
 b o-----------------------+      |     |
 c o------------------------------+     |
                    |      |      |     |
                   T4     T6     T2     |
                    |      |      |     |
        N (-) o-----+------+------+-----+
 (a joins T1-T4, b joins T3-T6, c joins T5-T2)

Assumption: 230 V is the line voltage of the star-connected supply (as in the standard textbook problem). Then

$$\begin{aligned} V_{ph} &= \frac{230}{\sqrt3} = 132.79\ \text{V}, \quad V_m = \sqrt2 \times 132.79 = 187.79\ \text{V} \\ V_{dm} &= \frac{3\sqrt3V_m}{\pi} = \frac{3\sqrt3 \times 187.79}{\pi} = 310.61\ \text{V}\ (= 1.35 \times 230) \end{aligned}$$

(i) Firing angle

$$\begin{aligned} V_{dc} &= 0.5V_{dm} = 155.30\ \text{V} \\ \cos\alpha &= \frac{V_{dc}}{V_{dm}} = 0.5 \quad\Rightarrow\quad \alpha = 60^\circ \end{aligned}$$

At $\alpha = 60^\circ$ the R-load current is just continuous, so the continuous-conduction formulas are valid.

(ii) Average and RMS load current

$$I_{dc} = \frac{V_{dc}}{R} = \frac{155.30}{10} = 15.53\ \text{A}$$ $$\begin{aligned} V_{rms} &= \sqrt3V_m\left[\frac{1}{2} + \frac{3\sqrt3}{4\pi}\cos2\alpha\right]^{1/2} = 325.27\left[0.5 + 0.4135 \times (-0.5)\right]^{1/2} \\ &= 325.27 \times 0.5415 = 176.14\ \text{V} \\ I_{rms} &= \frac{176.14}{10} = 17.61\ \text{A} \end{aligned}$$

(iii) Rectification efficiency

$$\begin{aligned} P_{dc} &= V_{dc}I_{dc} = 155.30 \times 15.53 = 2412.0\ \text{W} \\ P_{ac} &= V_{rms}I_{rms} = 176.14 \times 17.61 = 3102.6\ \text{W} \\ \eta &= \frac{P_{dc}}{P_{ac}} = \frac{2412.0}{3102.6} = 77.74\ \% \end{aligned}$$

Answer: $\alpha = 60^\circ$; $I_{dc} = 15.53$ A, $I_{rms} = 17.61$ A; $\eta = 77.74$ %.

(If 230 V were taken as the phase voltage, $\alpha$ is still $60^\circ$ and $\eta$ is still 77.74%, but $I_{dc} = 26.90$ A and $I_{rms} = 30.51$ A.)