Question

can i get the right solution please

Question 1: (25 marks) A ring of steel has an external diameter of 24cm and a square cross section of 3cm by 3cm, as she m, as shown in Figure 1. A gap of 3mm exists on the ring. A copper bar of 18cm x 3cm x 3cm is fitted inside Titted inside and across the ring with negligible gap. Account for fringing by using the effective area of the air approximated by: A = (a + Leap (b + Igap), where a and b are the cross-sectional dimensions of the core and low is the length of the air gap. The B-H magnetization characteristic for the cast stoel is given in Table 1. B, Tesla 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 7A/m 125 295 500 770 1140 1670 2500 4000 Table 1: Magnetization curve of cast steel for Problem 1 • Draw the equivalent magnetic circuit next to Figure 1. Calculate the number of turns required to produce a flux density of I Tesla in the air gap if the current in the coil is /= 1.5 A. Fill your answers in the fields of Table 2. The necessary calculations should fit on the next (blank) page. 0-1 Rsteel Figure 1: Magnetic circuit for problem 1 Bsteel- Asteel 20-06X0-045 Value 0.045m2 ango.o meone 24x10-2m DIXELS6Ł 210.30 395 2367.314 2832.314 1821 Table 2: Solution for Problem 1

Answer 1

18+ 24 HN S 3 nm. 24cm A core = 343 = 9 cm² A air gap = (3 +0.3)(3+0.3). = 10.89 cm? A copper bar = 3x3 = 9 cm? Equivalent ùruut. Betee! 19_ w - NI IR Given Bg = 1T I flux dersity in air gap). Bg = By Aailer 1 x 10.89 x10-4. = 1.089 mW flux dersity in steel cove Beted = 0g/ Acore 21.21 TL > H correspording to Bstrel :- 1670. Am >>> Usled = 1:21 -7.24 x 10-4 Muspel 1670 ith CamScan
]
HI Hair gap 4TX 10 - 7 = 7.95 X 104/1824 y 10 Risteel - l = 271 4 )? Uit 7.24710-4 y 9X10-4 . = 1.02 X106 air gap = 0.3 x10 . 47x107x10.89x1044 = 2.192 x 106 NI = % ( Risteet + Bairgap) = 1.089x10-3 ( 3.212 x106) = 3.497 x 10 I-15 A >> N = 3.499 x 10 = 2:331 x 10 N = 2331, Fairgap = 1.089 x10-x 2.192x106 • 21387 x10° Esteel = 1.089 x 1.02 x 103 ACS Scanned with = 1:1107 x 10 CamScanner