# Loadability Limit of a Three-Phase Induction Motor with Quadratic Mechanical Torque This analysis investigates the **loadability limit**—the maximum load torque the motor can sustain—of a three-phase induction motor where the mechanical load torque is a quadratic function of speed. --- ## 1. **Equilibrium Condition** From standard induction motor theory, the **torque equilibrium** is: \[ T_e(s) = T_m(s) \] where: - \( T_e \): Electrical torque (developed by the motor) - \( T_m \): Mechanical load torque - \( s \): Slip (\( s = \frac{\omega_s - \omega_r}{\omega_s} \)) --- ## 2. **Electrical Torque Expression** From eqn (4.19), for a standard induction motor (per-phase equivalent circuit): \[ T_e(s) = \frac{3 V_1^2 R_2 / s}{\omega_s \left[ (R_1 + R_2/s)^2 + (X_1 + X_2)^2 \right]} \] where: - \( V_1 \): Applied stator voltage (phase) - \( R_1 \), \( R_2 \): Stator and rotor resistances - \( X_1 \), \( X_2 \): Stator and rotor leakage reactances - \( \omega_s \): Synchronous speed (rad/s) --- ## 3. **Quadratic Mechanical Torque** From eqn (4.26), assume: \[ T_m(s) = K (1-s)^2 \] where: - \( K \): Proportionality constant, depends on the load. --- ## 4. **Equilibrium: Substitution** Set \( T_e(s) = T_m(s) \): \[ \frac{3 V_1^2 R_2 / s}{\omega_s \left[ (R_1 + R_2/s)^2 + (X_1 + X_2)^2 \right]} = K (1-s)^2 \] --- ## 5. **Independent Loading Parameter \( T_2 \)** Let \( T_2 \) denote the **loading parameter** (the load torque at the shaft; at a given slip, it's proportional to \( K \)). Rearranging for \( T_2 \) (or \( K \)): \[ T_2 = K = \frac{3 V_1^2 R_2}{\omega_s s (1-s)^2 \left[ (R_1 + R_2/s)^2 + (X_1 + X_2)^2 \right]} \] --- ## 6. **Loadability Limit Condition** The **loadability limit** is an extremum of \( T_2(s) \). The necessary condition is: \[ \frac{dT_2}{ds} = \] --- ## 7. **Symbolic Differentiation** Let: - \( A = 3 V_1^2 R_2 / \omega_s \) - \( B = (R_1 + R_2/s)^2 + (X_1 + X_2)^2 \) So, \[ T_2(s) = \frac{A}{s (1-s)^2 B} \] Differentiate with respect to \( s \): \[ \frac{dT_2}{ds} = \] This is a non-trivial expression; a symbolic processor (e.g., SymPy, Mathematica, Maple) would be used to solve: \[ \frac{d}{ds} \left( \frac{A}{s (1-s)^2 B} \right) = \] --- ## 8. **Numerical Example** Take parameters from the first 4 rows of Table 4.3 (assume a typical example): - \( V_1 = 400\,\text{V} \) - \( R_1 = .5\,\Omega \) - \( R_2 = .4\,\Omega \) - \( X_1 = 1.2\,\Omega \) - \( X_2 = 1.5\,\Omega \) - \( f = 50\,\text{Hz} \) - Number of poles \( p = 4 \implies \omega_s = 2\pi f / (p/2) = 2\pi \times 50 / 2 = 157.08\, \text{rad/s} \) Plug these into the formula for \( T_2(s) \): \[ A = \frac{3 \times 400^2 \times .4}{157.08} = \frac{3 \times 160000 \times .4}{157.08} = \frac{192000}{157.08} \approx 1223.13 \] \[ B = (.5 + .4/s)^2 + (1.2 + 1.5)^2 = (.5 + .4/s)^2 + (2.7)^2 \] So, \[ T_2(s) = \frac{1223.13}{s (1-s)^2 \left[ (.5 + .4/s)^2 + 7.29 \right]} \] --- ## 9. **Finding the Loadability Limit** Set \[ \frac{d}{ds} T_2(s) = \] This equation can be solved numerically in Python or Mathematica. The critical slip \( s_c \) that solves this equation gives the **maximum load torque** the motor can sustain for the quadratic torque load. --- ## 10. **Summary Steps for Symbolic/Numeric Evaluation** 1. Plug in parameters into \( T_2(s) \) formula. 2. Use a symbolic processor to compute \( \frac{dT_2}{ds} \). 3. Find value(s) of \( s \) where derivative is zero (\( s_c \)), in physical range \( < s < 1 \). 4. Substitute \( s_c \) back into \( T_2(s) \) for the maximum loadability limit. --- ## 11. **Illustrative Plot** A plot of \( T_2(s) \) vs \( s \) would show a maximum at the loadability limit.  *Alt text: Plot of loading parameter \( T_2 \) versus slip \( s \), showing a peak at the loadability limit.* --- ### **References** - P.C. Krause, "Analysis of Electric Machinery" - Table 4.3 (referenced motor data) --- If you need a worked-out symbolic or numerical solution for a specific motor from Table 4.3, please provide the complete data.