- 2073 Shrawan · 12 marks
Air gap and rotor bars, 10 kW motor
Short answer
No-load flux 4.70 mWb, air gap 0.433 mm, stator turns 402. Rotor bar current 744 A (area 148.8 mm²); end-ring current 1776 A (area 355.2 mm²). Bar loss 334.8 W, end-ring loss 133.9 W.
The following design data are provided for an induction motor: Diameter of stator bore (D) = 16 cm, Length of stator core (L) = 8.5 cm, Average flux density (Bav) = 0.44 Wb/m², Power factor = 0.85, Efficiency = 86%, Frequency = 50 Hz, Current density = 5 A/mm², Stator slots = 36, Rotor slots = 30, Length of rotor bar = 15 cm, Mean diameter of end ring = 12 cm, Resistivity of bar conductor = 0.020 Ohm-metre, Power output of 3-phase, 4-pole, 400 V, delta connected = 10 kW. Calculate no-load maximum flux, length of air gap, no. of turns per phase, rotor bar current and area, end ring current and area, losses in bars and end rings.
Data: 10 kW, 400 V, 3-phase, 4 poles, delta, 50 Hz; m, m, Wb/m², , pf , A/mm², , , bar length 0.15 m, end-ring mean diameter 0.12 m. Assumptions: ; resistivity taken as Ω per m per mm² (i.e. Ω·m; "0.020 Ω-m" is a misprint); same current density 5 A/mm² in bars and rings.
Flux per pole
Length of air gap
Turns per phase
Conductors per slot , so and .
Stator phase current
Rotor bar current and area
Rotor mmf of stator mmf:
End-ring current and area
Losses in bars
Losses in end rings
Total rotor copper loss W.
| Quantity | Value |
|---|---|
| Flux per pole | 4.70 mWb |
| Air gap | 0.433 mm |
| Turns per phase | 402 |
| Bar current, area | 744 A, 148.8 mm² |
| End-ring current, area | 1776 A, 355.2 mm² |
| Bar loss | 334.8 W |
| End-ring loss | 133.9 W |
✓ Answer checked by Claude Sonnet 5.5
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