Page 12 - Solutions Manual to accompany Electric Machinery Fundamentals

SOLUTION This core can be divided up into four regions. Let R 1 be the reluctance of the left-hand
portion of the core, R 2 be the reluctance of the center leg of the core, R 3 be the reluctance of the center
air gap, and R 4 be the reluctance of the right-hand portion of the core. Then the total reluctance of the
core is the reluctance of the left-hand leg plot the parallel combination of the reluctances of the right-hand
and center legs:

R  R  R
 R  R 2 3 4
TOT 1
2  R 3  R R 4
l 1.08 m

R 1  1   95.5 kA t/Wb
 A 1  0 2000 4  10  7   H/m 0.09 m  0.05 m
 r
l 0.34 m

R 2  2   18.0 kA t/Wb
 A 2  0 2000 4  10  7   H/m 0.15 m  0.05 m
 r
l 0.0005 m

R 3  3   51.0 kA t/Wb




 0 A 3  10  7  4   H/m0.15 m0.05 m 1.04 
l 1.08 m

R 4  4   95.5 kA t/Wb


 A 4  0 2000 4  10  7   H/m 0.09 m 0.05 m 
 r
The total reluctance is
  R 18.0 51.0 95.5 R R 

 R  R 2 3 4  95.5   135.5 kA t/Wb
TOT 1  R  R R 18.0 51.0 95.0


2 3 4
The total flux in the core is equal to the flux in the left leg:

 F 100 t 2.0 A 
      0.00148 Wb
left

TOT
R TOT 135.5 kA t/Wb
The fluxes in the center and right legs can be found by the “flux divider rule”, which is analogous to the
current divider rule.
 center  R 4  TOT  95.5  0.00148 Wb  0.00086 Wb


2  R 3  R R 4 18.0 51.0 95.5
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