4. The magnetic core in Fig. P4.1 is made from laminations of M-5 grain-oriented
ID: 2248553 • Letter: 4
Question
4. The magnetic core in Fig. P4.1 is made from laminations of M-5 grain-oriented electrical steel. The winding is excited with a 60-Hz voltage to produce a flux density in the steel of B(1)-1.5 sin ar T, 2n(60) 377 rad/sec. The steel occupies 0.94 of the core cross-sectional area. The mass-density of the steel is 7.65 g/cm. The magnetizing curve for the M-5 grain-oriented electrical stell 0.012 in thick is shown in Fig. P4.2. The curve for the exciting rms volt-amperes per kilogram at 60 Hz for M-5 grain-oriented electrical steel 0.012 in thick is shown in Fig. P4.3. The core-loss density at 60 Hz in watts per kilogram of M-5 grain-oriented electrical steel 0.012 in thick is shown in Fig. P4.4. [47x 10 H/ml (a) Determine cross-sectional area of core, Ac (b) Determine the applied voltage (c) (i) Find the maximum magnetic intensity H 20 cm (ii) Based on H obtain from 3(b)), determine the relative permeability of the steel. (iii) Determine the mean length l of flux path. (iv) Determine the ms exciting current 5 cm cm N- 200 (c) (i) Find the rms VA/kg. (ii) Determine the volume of core (ii) Determine the weight of the magnetic core (iv) Determine the total volt-amperes (v) Determine the rms exciting current. cm 5cm (d)() Find the core-loss W/kg Fig P4.1 (ii) Determine total core losses
Explanation / Answer
As seen from the diagram,
(a) Cross sectional area of the core = 5 cm * 5 cm = 25 cm2.
(b) Let us first calculate Ampere turns per metre from the first curve. Corresponding to Bmax = 1.5 Wb/m2 , H = 55 A turns per metre. N = 200 and mean path flux length = 2 * 10 cm + 2*15 cm + 5 cm + 5 cm = 60 cm. Thus, current I = H*le/N = 0.165 A. From the second curve we can get VA/kg for Bmax. VA/kg is 2. To find the mass, we need the volume of the core which is Area times le . Since, steel occupies 0.94 times area of core. The volume is 0.94 times 25 cm2 times 60 cm or 1410 cm3. The density of steel is 7.65 g/cm3. The mass of stell is 7.65 * 1410 g or 10786.5 g or 10.787 kg. Now to find voltage, I = 0.165 A , mass is 10.786 kg and VA/kg is 2. Therefore Vmax = 2 * 10.786/0.165 = 131 V.
(c) (i) Maximum Magnetising field intensity H = 55 A turns/m.
(ii) B =uourH . Therefore ur= B/(uo * H) = 1.5/(4*10-7*55) = 21702.
(iii) Mean flux path length = 60 cm
(iv) RMS exciting current = 0.165 A
(d) (i) RMS VA/kg is 2
(ii) Volume of the core is 1410 cm3.
(iii) Weight of the core = 10.787 kg.
(iv) Total Volt Amperes = 2 VA/kg * 10.787 kg = 21.57 VA
(v) RMS exciting current = 131 A.
(e) (i) As seen from the third graph, core loss in W/kg is 2 for B = 1.5 Wb/m2.
(ii) Total core loss = 2 W/kg * 10.787 kg = 21.57 W
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