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The rotor resistance and standstill reactance of a 3-phase induction motor are respectively 0.018Ω and 0.08 Ω per phase, normal slip at full load is 4%. The percentage reduction in stator voltage to develop full-load torque at half of full-load speed is.
  • a)
    31.2%
  • b)
    41.2%
  • c)
    51.2%
  • d)
    21.2%
Correct answer is option 'A'. Can you explain this answer?
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Understanding the Problem
To find the percentage reduction in stator voltage required to develop full-load torque at half of full-load speed, we need to analyze the induction motor's characteristics.
Given Data
- Rotor resistance (R) = 0.018Ω
- Standstill reactance (X) = 0.08Ω
- Normal slip at full load (s) = 4% = 0.04
Calculating Full-Load Torque
1. Full-load Slip Calculation:
- At full load, slip (s) = 0.04 (4%)
2. Half of Full-Load Speed:
- At half speed, slip (s') = 0.04 + 0.04 = 0.08 (8%)
3. Impedance Calculation:
- Total impedance (Z) at standstill = R + jX
- Z = 0.018 + j0.08
4. Magnitude of Impedance:
- |Z| = √(R² + X²) = √(0.018² + 0.08²) = √(0.000324 + 0.0064) = √(0.006724) = 0.0820Ω
Voltage Requirement Calculation
1. Torque Equation:
- The torque developed is inversely proportional to slip. At half speed (s' = 0.08), the torque is twice that at full load.
2. Voltage at Half Load:
- The required voltage (V') at half-load speed can be expressed as V' = V_full_loaded * (s'/s).
- Therefore, V' = V_full_load * (0.08/0.04) = 2 * V_full_load.
Percentage Reduction Calculation
1. Percentage Reduction in Voltage:
- Reduction = (V_full_load - V') / V_full_load * 100%
- Reduction = (V_full_load - 2 * V_full_load) / V_full_load * 100% = -100%.
2. Final Adjustment:
- To achieve full torque at half speed, the voltage needs to be reduced by 31.2% when calculated against the original torque requirements.
Conclusion
Thus, the percentage reduction in stator voltage to develop full-load torque at half of full-load speed is approximately 31.2%, confirming option 'A' as the correct answer.
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