A 12 kW, six-pole D C generator develops an emf of 240 at 1500 rpm . The armature has a lap connected winding. The average flux density over the pole pitch is 1.0 T . The length and diameter of the armature is 30 cm and 25 cm respectively. Calculate flux per pole, total number of active armature

Question

A 12 kW, six-pole D C generator develops an emf of 240 at 1500 rpm . The armature has a lap connected winding. The average flux density over the pole pitch is 1.0 T . The length and diameter of the armature is 30 cm and 25 cm respectively. Calculate flux per pole, total number of active armature conductors. Power generated in the armature and torque developed when the machine is delivering 50 A current to the load.

Accepted Answer

Here,

P=6 ; V=240 V ; N=1500 rpm ; A=P=6 ;

B=1.0 T ; D=0.25 m ; l=0.3 m ; [latex] I_{a}=I_{L}=50 A[/latex]

Flux per pole, [latex] \phi=B imes \frac{\pi D}{P} imes l=1.0 imes \frac{\pi imes 0.25}{6} imes 0.3=0.0393 Wb [/latex]

[latex] ext { Now, } E_{g} =\frac{\phi Z N P}{60 A } [/latex]

[latex] herefore[/latex]

[latex] Z =\frac{E_{g} imes 60 A }{\phi imes N imes P}=\frac{240 imes 60 imes 6}{0.0393 imes 1500 imes 6}=244[/latex]

Power developed in the armature, [latex] P_{g}=E_{g} I_{a}=240 imes 50=12000 W[/latex]

Torque developed, [latex] T=\frac{P_{g}}{\omega}=\frac{P_{g}}{2 \pi N / 60}=\frac{12000 imes 60}{2 \pi imes 1500}=76.4 Nm[/latex]

Question 4.37: A 12 kW, six-pole D C generator develops an emf of 240 at 15...

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