Question

Q.22 Describe how a variable-reluctance motor could be made to rotate at 10 r/min. A hysteresis loop is plotted against a horizontal axis which scales 1000 A/m 1 cm and a vertical axis which scales 1 T 5 cm. If the area of the loop is 9 cm2 and the overall height is 14 cm, calculate: (a) the hysteresis loss in joules per cubic metre per cycle (b) the maximum flux density (c) the hysteresis loss in watts per kilogram, assuming the density of the material to be 7800 kg/m

Answer 1

Magnetic flux have a tendency to flow through lowest reluctance path therefore rotor always tends to align along minimum reluctance path . This is principle of working of variable reluctance motor. Different 3 phases are excited in sequence and according to excitation rotor will align in axis offering least reluctance path.

Rotor pole can be obtained as = N_S $\pm$ N_S/ q where q is number of phases so if N_S=6 and for 3 phase rotor poles obtained will be 4 and 8 . In case of 4 rotor poles it takes 12 steps to complete one rotation as on one step 30 degree is moved by rotor . Now if we require 10 revolution per minute. 1 revolution need to be completed in 6 second that means 12 steps are desired in 6 second so each step of excitation should be done in 0.5 second i e adjusted to pulse for excitation.

a) Assuming rectangular loop width of loop = 9/14 = 0.6428 cm so magnetic flux density (max) = 14/(5x2) = 1.4 T and corecivity = 1000 x 0.6428/2 = 321.4 A/m so Hysteresis loss = 2.8 x 642.8 = 1800 J / cubic meter in one cycle

b) maximum flux density = 1.4 T

c) Hysteresis loss = 1800/ 7800 = 0.23 J/Kg and if it takes 6 second to complete one cycle then loss = 0.038 W/kg