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Example 1.1

Example 1.1.

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Example 1.1

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  1. Example 1.1 a). Determine the voltage that must be applied to the magnetizing coil in order to produce a flux density of 0.2T in the air gap. Flux fringing will be assumed negligible. Assume that the magnetization curve for the core material (which is homogeneous) is that given in Fig 1.5. The coil has 80 turns and a resistance of 0.05 Ω. The cross-sectional area of the core material is 0.0400 m2. ECE 441

  2. Flux “Fringing” • All lines of flux must • leave and arrive ECE 441

  3. Flux Distribution ECE 441

  4. Procedure • Determine Φgap and Fgap • Determine Hbcde, Bbcde, and Φbcde • Determine Φefab, Befab, Hefab, and Fefab • Determine FT and the required current • Using Ohm’s Law, determine the required voltage ECE 441

  5. For the Gap • Φgap = BgapAgap = (0.2)(0.04) = 0.008Wb • Flux Density to establish 0.2 T in the center leg is determined from the magnetization curve in Fig. 1.5. (Next slide) • From the curve, H0.30=H0.60=0.47 oersteds • Multiply by 79.577 for A-t/m • H30 = H60 = 37.4 A-t/m ECE 441

  6. 0.2 0.47 ECE 441

  7. Magnetic Potential Differences • F0.30 = H·l = (37.4)(0.30) = 11.22 A-t • F0.69 = H·l = (37.4)(0.69) = 25.81 A-t • Fgap = Hgap·lgap • To get Hgap, • μgap = Bgap/Hgap 4πx10-7 = 0.2/Hgap • Hgap = 159,155 A-t/m • Fgap = (159,155)(0.005) = 795.77 A-t • Fbghe = 11.22 + 25.81 + 795.77 = 833 A-t ECE 441

  8. For the bcde leg • Fbghe = Fbcde • Hbcde=Fbcde/lbcde=833/(1+1+1)=277.67 A-t/m • In oersteds • Hbcde = 277.67/79.577 = 3.49 oersteds • Determine Bbcde from the magnetization curve in Fig. 1.5. (Next slide) • Bbcde = 1.45 T • Φbcde = BA = (1.45)(0.04) = 0.058 Wb ECE 441

  9. 1.45 3.49 ECE 441

  10. For the efab leg • Φefab=Φgap + Φbcde=0.08+0.058=0.066 Wb • Befab = Φ/A = 0.066/0.04 = 1.65 T • Determine H to establish this field intensity from the magnetization curve in Fig. 1.5. (Next slide) • Hefab = 37 oersteds • Hefab = (37)(79.577) = 2944.35 A-t/m • Fefab= H·l = (2944.35)(1+0.8+0.8) • Fefab =7655.31A-t ECE 441

  11. 1.65 37 ECE 441

  12. Total mmf to be supplied by the coil • FT = Fbghe + Fefab = 7655.31 + 833 • FT = 8488.31 A-t • FT = N·I = 8488.31 • I = 106.1 A • V = I·R =(106.1)(0.05) • V = 5.30 V ECE 441

  13. Example 1.1 Part b • Using equations 1-5 and 1-7, determine the relative permeability of each of the legs of the core, and compare the calculated values with corresponding values obtained from the permeability curve in Fig. 1.5. ECE 441

  14. Combining Eq (1-5) and (1-7) • μ = B/H • μr = μ/μ0 • μr = (B/H)/4π·10-7 = B/(4π·10-7·H) • μrleft = 1.65/(4π·10-7·2944) = 446 • μrcenter = 0.20/(4π·10-7·37.4) = 4256 • μrright = 1.45/(4π·10-7·277.67) = 4156.1 ECE 441

  15. 1.65 450 37 ECE 441

  16. 4000 0.2 0.47 ECE 441

  17. 1.45 4100 3.49 ECE 441

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