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Figure 3-26.  Calculating Effective Diameter

 
  
 
T.O. 33B-1-1 3-40 Example: What  would  the  effective  diameter  be  for  a  cylindrical  part  10  inches  long  that  has  a  2  inch outside diameter with a 0.125 inch wall thickness (see Figure 3-26). Figure 3-26.  Calculating Effective Diameter To find At, calculate the cross-sectional area of the outside diameter of the part as follows: At = pr2 At = p(1)2 At = 3.1416 sq. inches To find Ah, calculate the cross-sectional area for the inside diameter of the part (the part’s hollow portion) as follows: Ah = pr2 Ah = p(0.875)2 Ah = 2.40 sq. inches Insert the results for At and Ah into the formula to find Deff. Deff = - 231416 2 3 1416 1 2 . .40 . Deff = 0.97 inch To calculate the current required to longitudinally magnetize the part in the above example, use the formula from  paragraph  3.3.12.6.3.1  (for  the  part  in  the  bottom  of  a  12  inch  diameter  coil  with  5  turns), except replace D with Deff. (0.97): I KD NL = I = 45000 0 97 5 10 . I = 873 amperes


   


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