T.O. 33B-1-13-40Example:What would the effective diameter be for a cylindrical part 10 inches long that has a 2 inchoutside diameter with a 0.125 inch wall thickness (see Figure 3-26).Figure 3-26. Calculating Effective DiameterTo find At, calculate the cross-sectional area of the outside diameter of the part as follows:At = pr2At = p(1)2At = 3.1416 sq. inchesTo find Ah, calculate the cross-sectional area for the inside diameter of the part (the part’s hollow portion)as follows:Ah = pr2Ah = p(0.875)2Ah = 2.40 sq. inchesInsert the results for At and Ah into the formula to find Deff.Deff=-23141623141612. .40.Deff = 0.97 inchTo calculate the current required to longitudinally magnetize the part in the above example, use the formulafrom 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):IKDNL=I=45000097510.I = 873 amperes
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