Let's break down the problem step by step: ### **Given Data** - **Rated voltage** = 460 V (Line-to-Line), Y-connected - **Frequency** = 40 Hz (for this scenario) - **Number of Poles (P)** = 6 - **Synchronous speed at 40 Hz**: \( N_s = \frac{120f}{P} = \frac{120 \times 40}{6} = 800 \) rpm - **Motor parameters (referred to stator):** - \( R_s = 0.19\, \Omega \) - \( R_r = 0.07\, \Omega \) - \( X_s = 0.75\, \Omega \) - \( X_r = 0.67\, \Omega \) - \( X_m = \infty \) (i.e., neglect magnetizing current; simplifies calculations!) - **Braking torque** \( T = 350\, \text{Nm} \) - **Supplied at rated voltage** --- ### **Step 1: Per-phase Voltage** For a Y-connection: - Phase voltage, \( V_{ph} = \frac{V_{LL}}{\sqrt{3}} = \frac{460}{\sqrt{3}} = 265.6\ \text{V} \) *Assume the voltage is not derated for lower frequency (usually, V/Hz is kept constant, but the problem says "rated voltage").* --- ### **Step 2: Synchronous Speed at 40 Hz** \[ N_s = \frac{120f}{P} = \frac{120 \times 40}{6} = 800\ \text{rpm} \] --- ### **Step 3: Equivalent Circuit** With \( X_m = \infty \), the equivalent circuit is: **Stator:** \( R_s + jX_s \) **Rotor (referred):** \( \frac{R_r}{s} + jX_r \) --- ### **Step 4: Output Power and Torque Equation** **Air-gap power per phase:** \[ P_{g,ph} = I_2^2 \cdot \frac{R_r}{s} \] **Total air-gap power:** \[ P_g = 3 I_2^2 \cdot \frac{R_r}{s} \] Where \( I_2 \) is the current through the rotor branch. **Mechanical Power Output:** \[ P_{mech} = P_g (1-s) = 3 I_2^2 \cdot \frac{R_r}{s} (1-s) \] **Electromagnetic Torque:** \[ T_e = \frac{P_g}{\omega_s} = \frac{3 I_2^2 \cdot \frac{R_r}{s}}{\omega_s} \] Where \( \omega_s = 2\pi N_s/60 \) (in rad/s) --- ### **Step 5: Impedance Calculations** The **total impedance per phase** (since \( X_m = \infty \)): \[ Z_{total} = R_s + jX_s + \left( \frac{R_r}{s} + jX_r \right) = (R_s + \frac{R_r}{s}) + j(X_s + X_r) \] \[ Z_{total} = (0.19 + \frac{0.07}{s}) + j(0.75 + 0.67) = (0.19 + \frac{0.07}{s}) + j(1.42) \] --- ### **Step 6: Torque Equation (Per Phase)** \[ T_e = \frac{3}{\omega_s} \cdot \frac{V_{ph}^2 \cdot \frac{R_r}{s}}{ \left[ (R_s + \frac{R_r}{s})^2 + (X_s + X_r)^2 \right] } \] Plug in values: - \( V_{ph} = 265.6\ \text{V} \) - \( R_s = 0.19 \) - \( R_r = 0.07 \) - \( X_s + X_r = 1.42 \) - \( \omega_s = 2\pi N_s / 60 = 2\pi \times 800 / 60 = 83.78\ \text{rad/s} \) So, \[ T_e = \frac{3}{83.78} \cdot \frac{(265.6)^2 \cdot \frac{0.07}{s}}{ \left[ (0.19 + \frac{0.07}{s})^2 + (1.42)^2 \right] } \] Set \(T_e = 350\) Nm, solve for \(s\): --- ### **Step 7: Set up the Equation** Let’s denote \( K = \frac{3}{83.78} (265.6)^2 \approx 3 \times 70,562.56 / 83.78 = 2,532.8 \) So, \[ 350 = \frac{2,532.8 \cdot \frac{0.07}{s} }{ (0.19 + \frac{0.07}{s})^2 + 2.0164 } \] where \( 1.42^2 = 2.0164 \). Let \( x = \frac{0.07}{s} \): \[ 350 = \frac{2,532.8 \cdot x}{ (0.19 + x)^2 + 2.0164 } \] \[ 350 \left[ (0.19 + x)^2 + 2.0164 \right] = 2,532.8 x \] \[ 350 (0.19 + x)^2 + 350 \cdot 2.0164 = 2,532.8 x \] \[ 350 (0.19 + x)^2 + 705.74 = 2,532.8 x \] \[ 350 (0.0361 + 0.38x + x^2 ) + 705.74 = 2,532.8 x \] \[ 12.635 + 133x + 350x^2 + 705.74 = 2,532.8 x \] \[ 718.375 + 133x + 350x^2 = 2,532.8 x \] \[ 718.375 + 350x^2 = 2,532.8 x - 133x = 2,399.8 x \] \[ 350x^2 - 2,399.8x + 718.375 = 0 \] Divide both sides by 350: \[ x^2 - 6.8577x + 2.0525 = 0 \] --- ### **Step 8: Solve the Quadratic Equation** \[ x^2 - 6.8577x + 2.0525 = 0 \] The quadratic formula: \[ x = \frac{6.8577 \pm \sqrt{6.8577^2 - 4 \cdot 1 \cdot 2.0525}}{2} \] \[ = \frac{6.8577 \pm \sqrt{47.0356 - 8.21}}{2} = \frac{6.8577 \pm \sqrt{38.8256}}{2} = \frac{6.8577 \pm 6.2318}{2} \] So, - \( x_1 = \frac{6.8577 + 6.2318}{2} = \frac{13.0895}{2} = 6.5448 \) - \( x_2 = \frac{6.8577 - 6.2318}{2} = \frac{0.6259}{2} = 0.3130 \) Recall \( x = \frac{0.07}{s} \), so \[ s = \frac{0.07}{x} \] So, - For \( x_1 = 6.5448 \): \( s_1 = \frac{0.07}{6.5448} = 0.0107 \) - For \( x_2 = 0.3130 \): \( s_2 = \frac{0.07}{0.3130} = 0.2238 \) --- ### **Step 9: Braking vs Motoring** For **braking torque**, the slip is **negative** (i.e., the rotor is rotating faster than the synchronous speed, so s < 0). But both solutions above are positive. This means, with the parameters and equations, the numbers are for motoring operation. For **braking**, the torque-slip curve is symmetric about s = 0, so \( T(-s) = -T(s) \). If you use negative slip: \[ s = -0.0107 \quad \text{or} \quad s = -0.2238 \] But, physically, only one solution will be near synchronous speed. --- ### **Step 10: Find the Speed** \[ N = N_s (1-s) \] Take the solution close to synchronous speed (since induction braking is typically done just above synchronous speed): \[ s = -0.0107 \] \[ N = 800 \times (1 - (-0.0107)) = 800 \times 1.0107 = 808.6\ \text{rpm} \] Alternatively, for larger slip: \[ s = -0.2238 \] \[ N = 800 \times (1 - (-0.2238)) = 800 \times 1.2238 = 979.0\ \text{rpm} \] But at such a high negative slip, the current would be very high—typically, braking is done close to synchronous speed. **So the answer is:** --- ## **Final Answer** ### **The speed of the motor for a braking torque of 350 Nm at 40 Hz and rated voltage is:** \[ \boxed{808.6\ \text{rpm}} \] **(rounded to one decimal place)** --- **Summary of Steps:** 1. Found synchronous speed at 40 Hz: 800 rpm 2. Set up torque equation for braking (negative slip) 3. Solved quadratic for slip 4. Used the slip near synchronous speed: \( s = -0.0107 \) 5. Calculated actual speed: \( N = 808.6\ \text{rpm} \) --- ### **If you need the other root for a higher slip value, you can use \( N = 979.0\ \text{rpm} \), but typically, the lower slip is correct for normal braking torque levels.** If you need the **slip** or **process explained** in more depth, let me know!