(Q-3] The cross section of the magnetic system of a two pole rotating machine is
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(Q-3] The cross section of the magnetic system of a two pole rotating machine is shown in Figure attached to this assignment. Rotor diameter = 4 cm. Each pole covers an arc of 160°. Axial length of rotor is 2.5 cm. Air gaps are 0.5 mm each. Yoke has a outer diameter of 3.5 cm (assume circular yoke) and a uniform thickness of 0.25 cm. Each pole has a coil of 360 turns and the coils are connected in series. Assume that the material of the core is M36 0.356 mm laminates. Using Slemon's curve for the material and assuming we aim for a flux density of 0.8 T under the poles in the air gaps, determine a) flux density in the yokeExplanation / Answer
Total no.of turns=360.We know Z represents no.of conductors which is twice the no.of turns.
Hence Z=2T=2 X 360=720.
No.of Poles=2, Flux density(B)=0.8T.
Flux density of yoke=1/2(Flux density under the pole tips)=1/2 X 0,8=0.4T because flux becomes half in the yoke as shown in the figure.This gives our answer to part a.
From above we have denoted the no.of turns as 'T' or we can also write it as 'N'.
Now,from the expression of flux density we can use this as; B=(MU)NI/L where MU=(4pi)x 10-7=Permeability of Free Space,N=no.of turns,I= current in the coil,L=length of coil in metre.
So, B={(4pi)x 10-7x360xI}/L and L=pi x d=0.1099 m as the outer diameter of yoke is 3.5cm
By substituting the values of B and L we will get the answer to option b that is the current required to set up the flux=I=194.34 A is the required answer.
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