Estimate the number of turns needed on each commutating pole of six-pole generator delivering 200 kW at 200V, given that the number of armature conductors is 540 and the winding is lap connected interpole air-gap is 1.0 cm and the flux density in the interpole air-gap is 0.3 Wb/m^2. Neglect the

Question

Estimate the number of turns needed on each commutating pole of six-pole generator delivering [latex]200 kW[/latex] at [latex]200 V[/latex], given that the number of armature conductors is [latex]540[/latex] and the winding is lap connected interpole air-gap is [latex]1.0 cm[/latex] and the flux density in the interpole air-gap is [latex]0.3 Wb/m^{2}[/latex]. Neglect the effect of iron parts of the circuit and of leakage.

Accepted Answer

Here, [latex]Load = 200 kW; V = 200 V; Z = 540; A = P = 6 [/latex]

[latex]l_{ig} = 1.0 cm = 0.01 m; B_{ig} = 0.3 Wb/m^{2}[/latex]

Load current, [latex]I_{L} = \frac{kW imes 1000}{V} = \frac{200 imes 1000}{200} = 1000 A [/latex]

Armature current, [latex]I_{a} = I_{L} = 1000 A [/latex]

Current per conductor, [latex]I_{c} = \frac{I_{a}}{A} = \frac{1000}{6} = 166.7 A [/latex]

[latex]Armature reaction ampere turns per pole = \frac{I_{c} Z}{2P} = \frac{166.7 imes 540}{2 imes 6} = 7500 [/latex]

Ampere-turns for air gap flux per pole, [latex]A T_{ig} = \frac{l_{ig} B_{ig}}{\mu _{0}} = \frac{0.01 imes 0.3}{4\pi imes 10^{-7}} = 2386 [/latex]

Total ampere-turns required for each interpole, [latex]A T_{ip} = 7500 + 2386 = 9886[/latex]

[latex]No. of turns for each interpole = \frac{A T_{ip}}{I_{a}} = \frac{9886}{1000} = \mathbf{9.886} [/latex]

Question 4.33: Estimate the number of turns needed on each commutating pole...

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