You are asked to design (on paper) a 60 Hz, 5 KVA, 480V/120V power transformer. With the following specifications: 1. The transformer core is made of laminated steel sheets (refer to course slides to find the B-H curve of this core material. Choose the flux density B to be equal to 1 Tesla (peak value). 2. No load loss (or core loss) = 35 W 3. Excitation current 96%. This a design problem, which obviously has no unique solution. The problem is to determine an appropriate core size, and number of turns of the copper coils. The weight of the transformer should be taken into account as well. Consider a uniform core that is square in shape with width equal depth. Let the dimentions be defined by x and y as shown below. SI units are used throughout this study. Cross section area: A = x2 Average length of flux path: (= 4(x+y) Core material: Sheet Steel B (max) = 1 Tesla H (max) (obtained from B- H curve) = 180 At/m Preliminary Analysis: 1. Voltage ratio requirement: Vp/Vs=N1/N2 = 480/120 = 4. Let N2 be that variable (i.e., N1 = 4N2), and the total number of turns: Nt =N1+N2 = 5N2 2. Maximum flux density as its relates to frequency and peak voltage: O=B/A=V/(@N1) = N2 * x2 = 0.45 3. Peak value of magnetizing current requirement: I = (H * ( )/N1 96%. This a design problem, which obviously has no unique solution. The problem is to determine an appropriate core size, and number of turns of the copper coils. The weight of the transformer should be taken into account as well. Consider a uniform core that is square in shape with width equal depth. Let the dimentions be defined by x and y as shown below. SI units are used throughout this study. Cross section area: A = x2 Average length of flux path: (= 4(x+y) Core material: Sheet Steel B (max) = 1 Tesla H (max) (obtained from B- H curve) = 180 At/m Preliminary Analysis: 1. Voltage ratio requirement: Vp/Vs=N1/N2 = 480/120 = 4. Let N2 be that variable (i.e., N1 = 4N2), and the total number of turns: Nt =N1+N2 = 5N2 2. Maximum flux density as its relates to frequency and peak voltage: O=B/A=V/(@N1) = N2 * x2 = 0.45 3. Peak value of magnetizing current requirement: I = (H * ( )/N1