A \( 210 \mathrm{~V}, 1200 \mathrm{RPM}, 10 \) A separately excited DC motor is controlled by a \( 1- \) phase fully controlled converter with an AC source voltage of \( 230 \mathrm{~V}, 50 \mathrm{~Hz} \). Assume that sufficient inductance is present in the armature circuit to make the motor current continuous and ripple free for any torque greater than \( 25 \% \) of rated torque. Ra \( =1.5 \Omega \) (a) What should be the value of the firing angle to get the rated torque at 800 rpm?

Given data: - DC motor rated voltage: \( V_{\text{rated}} = 210\text{ V} \) - DC motor rated speed: \( N_{\text{rated}} = 1200\text{ rpm} \) - DC motor rated current: \( I_{\text{rated}} = 10\text{ A} \) - Armature resistance: \( R_a = 1.5\,\Omega \) - AC supply: \( V_{\text{AC,rms}} = 230\text{ V},\ f = 50\text{ Hz} \) - Fully controlled converter (single-phase) is used. - We assume enough armature inductance so that the current is continuous. - Desired operating speed: \( N = 800\text{ rpm} \) at rated torque (which implies \( I_a = I_{\text{rated}} = 10\text{ A} \)). The procedure is as follows. --- **Step 1. Find the Back EMF Constant \(K\phi\)** For a separately excited DC motor, at rated conditions the armature circuit equation is \[ V_a = E_b + I_a R_a \] At rated condition: \[ E_{b,\text{rated}} = V_{\text{rated}} - I_{\text{rated}} R_a = 210\text{ V} - 10\text{ A} \times 1.5\,\Omega = 210 - 15 = 195\text{ V}. \] The motor speed in rad/s is \[ \omega_{\text{rated}} = 1200 \times \frac{2\pi}{60} = 40\pi \, \text{rad/s} \approx 125.66\, \text{rad/s}. \] Since \[ E_b = K\phi\,\omega, \] the back EMF constant is \[ K\phi = \frac{E_{b,\text{rated}}}{\omega_{\text{rated}}} = \frac{195}{40\pi} \approx \frac{195}{125.66} \approx 1.55\, \text{V-s}. \] --- **Step 2. Determine the Back EMF at 800 rpm** First, convert \(800\text{ rpm}\) to rad/s: \[ \omega = 800 \times \frac{2\pi}{60} = \frac{1600\pi}{60} = \frac{80\pi}{3}\, \text{rad/s} \approx 83.78\, \text{rad/s}. \] Then the back EMF at \(800\text{ rpm}\) is \[ E_b = K\phi \, \omega = 1.55 \times 83.78 \approx 130\text{ V}. \] --- **Step 3. Calculate the Required Converter Output Voltage** To develop the rated torque (i.e. \(10\text{ A}\) armature current), the converter output \(V_{\text{dc}}\) must overcome both the back EMF and the voltage drop across \(R_a\). Thus, \[ V_{\text{dc}} = E_b + I_a R_a = 130\text{ V} + 10\text{ A} \times 1.5\,\Omega = 130 + 15 = 145\text{ V}. \] --- **Step 4. Relate \(V_{\text{dc}}\) to the Converter Firing Angle \( \alpha \)** For a single-phase fully controlled converter (assuming continuous conduction without commutation overlap), the average output voltage is given by \[ V_{\text{dc}} = \frac{2V_m}{\pi} \cos \alpha, \] where \(V_m\) is the peak value of the AC supply voltage. The AC supply has an RMS value of \(230\text{ V}\), so \[ V_m = \sqrt{2} \times 230 \approx 325.27\text{ V}. \] Solving for \(\cos \alpha\): \[ \cos \alpha = \frac{\pi V_{\text{dc}}}{2V_m} = \frac{\pi \times 145}{2\times 325.27}. \] Compute the numerator: \[ \pi \times 145 \approx 455.53. \] Then, \[ \cos \alpha \approx \frac{455.53}{650.54} \approx 0.700. \] Finally, determining the firing angle: \[ \alpha = \arccos (0.700) \approx 45.6^\circ. \] --- **Answer:** The firing angle required to obtain rated torque at 800 rpm is approximately \(45.6^\circ\).