Flux plot of a magnetic bearing for hollow shaft-when upper coils carry more cur
ID: 3162801 • Letter: F
Question
Flux plot of a magnetic bearing for hollow shaft-when upper coils carry more current than the lower coils. The cross-section of each pole measures 25 mm by 120 mm, and the average values of flux density in the airgaps are as follows: A: 0.90 T; B,C: 0.63 T; D: 0.36 T E,F: 0.27 T. the force exerted by one pole is kBz where k is a constant for this device. The value of k is 1.194 kN/Tz calculate the magnitudes of all the forces. By symmetry, the horizontal components of the forces will cancel out. Calculate the total magnetic force acting on the shaft from the vertical components of forces from individual poles. The angle between adjacent poles is 609. Identify the correct result for the magnitude of the total force in the following list of values: (a) 2.15 kN; (b) 1.48 kN; (c) 1.20 kN; (d) 1.59 kN.Explanation / Answer
cross sectional area A=25*120 m-6 (as 1mm=10-3m)
given f=kB2
value of k=1.194kN/T2
given the angle between adjacent pole=600
The force at A=3000*10-6 * 0.90*0.90= 0.81*3*10-3 =2.43*10-3 kN
the force at D=3000*10-6 *0.36*0.36 =0.3888*10-3 kN
hence the force is force at A-force at D