Question

In an electric machine F₂ (rotor mmf) = 920 AT. F₁ (stator emf) = 400 AT, alpha = 125° and the resultant air gap flux per pole is 0.107 Wb. Find out permeance per pole.

Note: AT (Ampere Turn) is a unit.
1. 1.408 × 10^-4 Wb/AT
2. 8.408 × 10^-4 Wb/AT
3. 2.480 × 10^-4 Wb/AT
4. 2.408 × 10^-4 Wb/AT

Answer 1

Correct Answer - Option 1 : 1.408 × 10^-4 Wb/AT

Given that, stator emf (F₁) = 400 AT

Rotor emf (F₂) = 920 AT

Included angle (α) = 125°

θ = 180 - α = 180 – 125° = 55°

Resultant emf, \({F_r} = \sqrt {F_1^2 + F_2^2 - 2{F_1}{F_2}\cos \theta }\)

\(= \sqrt {{{400}^2} + {{920}^2} - 2 \times 400 \times 920 \times \cos 55^\circ } = 764.36\;AT\)

The resultant air-gap flux per pole (ϕ) = 0.107 Wb

Permeance, \(P = \frac{\phi }{{{F_r}}} = \frac{{0.107}}{{764.36}} = 1.4 \times {10^{ - 4}}\;Wb/AT\)