To calculate the starting torque of the induction motor, we will follow the key steps once more. Here's a structured solution based on the information provided:

Given Data:

• Rated power $P = 150 \, \text{kW}$
• Stator line voltage $V_s = 3000 \, \text{V}$
• Supply frequency $f = 50 \, \text{Hz}$
• Number of poles $p = 6$
• Stator/rotor transformation ratio $k = 3.6$
• Rotor resistance $R_r = 0.1 \, \Omega$
• Rotor leakage inductance $L_r = 3.61 \, \text{mH} = 3.61 \times 10^{-3} \, \text{H}$
• Stator impedance: Neglected

Step 1: Calculate synchronous speed

The synchronous speed $N_s$ in RPM is given by:

$N_s = \frac{120 \times f}{p}$

For $f = 50 \, \text{Hz}$ and $p = 6$:

$N_s = \frac{120 \times 50}{6} = 1000 \, \text{RPM}$

Now, convert this to angular velocity ($\omega_s$) in radians per second:

$\omega_s = \frac{2 \pi N_s}{60} = \frac{2 \pi \times 1000}{60} = \frac{2000 \pi}{60} \approx 104.72 \, \text{rad/s}$

Step 2: Calculate rotor voltage at standstill

The rotor voltage at standstill $E_r$ is related to the stator voltage by the transformation ratio $k$. The stator line voltage $V_s = 3000 \, \text{V}$, and for a star connection, the phase voltage $V_{s\_ph}$ is:

$V_{s\_ph} = \frac{V_s}{\sqrt{3}} = \frac{3000}{\sqrt{3}} \approx 1732.05 \, \text{V}$

The rotor phase voltage at standstill $E_r$ is:

$E_r = \frac{V_{s\_ph}}{k} = \frac{1732.05}{3.6} \approx 481.13 \, \text{V}$

Step 3: Rotor reactance at standstill

The rotor reactance at standstill $X_r$ is calculated as:

$X_r = 2 \pi f \cdot L_r = 2 \pi \cdot 50 \cdot 3.61 \times 10^{-3} \approx 1.135 \, \Omega$

Step 4: Formula for starting torque

The starting torque $T_{start}$ is given by:

$T_{start} = \frac{3 \cdot E_r^2 \cdot R_r}{\omega_s \cdot (R_r^2 + X_r^2)}$

Substitute the known values:

• $E_r = 481.13 \, \text{V}$
• $R_r = 0.1 \, \Omega$
• $X_r = 1.135 \, \Omega$
• $\omega_s = 104.72 \, \text{rad/s}$

We now calculate each term.

• $E_r^2 = (481.13)^2 \approx 231486.08 \, \text{V}^2$
• $R_r^2 + X_r^2 = (0.1)^2 + (1.135)^2 = 0.01 + 1.288225 = 1.298225 \, \Omega^2$
• Numerator: $3 \cdot E_r^2 \cdot R_r = 3 \cdot 231486.08 \cdot 0.1 = 69445.82$

Now, the denominator:

$\omega_s \cdot (R_r^2 + X_r^2) = 104.72 \cdot 1.298225 \approx 135.94$

Step 5: Final torque calculation

Now, calculate the starting torque:

$T_{start} = \frac{69445.82}{135.94} \approx 510.9 \, \text{N-m}$

Final Answer:

The starting torque is approximately 510.9 N-m.

Let me know if you want further explanation or a breakdown of any specific part of this process!

Here are 5 related questions for further understanding:

1. What would happen to the starting torque if the rotor resistance increased?
2. How does the transformation ratio influence the rotor voltage and torque?
3. Why is the leakage inductance relevant in torque calculations?
4. How would you calculate the rotor current at starting?
5. What happens to torque as slip decreases during motor operation?

Tip: Slip-ring induction motors are advantageous for applications requiring high starting torque because you can externally control the rotor resistance during startup.