Mention four methods of reducing leakage fluxes in a transformer.

Here are the solutions to the questions: QUESTION 1: TRANSFORMERS 1.1: Four methods of reducing leakage fluxes in a transformer are: Interleaving of primary and secondary windings. Using concentric windings. Sectionalizing the windings. Using shell-type core construction. 1.2: Given: Primary voltage V_1 = 230 V Load resistance R_L = 200 Load reactance X_L = 100 Turns ratio N_2/N_1 = 4 Equivalent primary resistance R_eq1 = 0.1 Equivalent primary leakage reactance X_eq1 = 0.5 Step 1: Calculate the load impedance referred to the primary side. The turns ratio a = N_1/N_2 = 1/4. The load impedance is Z_L = R_L + jX_L = 200 + j100 . The load impedance referred to the primary is Z_L1 = a^2 Z_L. Z_L1 = ((1)/(4))^2 (200 + j100) = (1)/(16) (200 + j100) = 12.5 + j6.25 The total equivalent impedance referred to the primary is Z_total,1 = R_eq1 + jX_eq1 + Z_L1. Z_total,1 = (0.1 + j0.5) + (12.5 + j6.25) = (0.1 + 12.5) + j(0.5 + 6.25) = 12.6 + j6.75 The magnitude of the total equivalent impedance is: |Z_total,1| = sqrt(12.6^2 + 6.75^2) = sqrt(158.76 + 45.5625) = sqrt(204.3225) ≈ 14.294 1.2.1: The primary circulating current (I_1). The primary current is I_1 = V_1 / Z_total,1. I_1 = 230 V12.6 + j6.75 I_1 = (230)/(14.294 (6.75)12.6) = (230)/(14.294 28.18^) ≈ 16.09 -28.18^ A The magnitude of the primary circulating current is: 16.09 A 1.2.2: The per unit regulation of the transformer. The power factor angle is the angle of Z_total,1, so = 28.18^. = (28.18^) ≈ 0.8815 = (28.18^) ≈ 0.4722 The voltage regulation (VR) is given by: VR = I_1 (R_eq1 + X_eq1 )V_1 VR = (16.09 (0.1 × 0.8815 + 0.5 × 0.4722))/(230) VR = (16.09 (0.08815 + 0.2361))/(230) = (16.09 × 0.32425)/(230) = (5.218)/(230) ≈ 0.02268 The per unit regulation is: 0.02268 pu 1.2.3: The secondary terminal voltage (V_2). The voltage across the load referred to the primary is V_L1 = I_1 Z_L1. Z_L1 = 12.5 + j6.25 = sqrt(12.5^2 + 6.25^2) ((6.25)/(12.5)) = 13.975 26.565^ V_L1 = (16.09 -28.18^) × (13.975 26.565^) V_L1 = (16.09 × 13.975) (-28.18^ + 26.565^) = 224.86 -1.615^ V The secondary terminal voltage V_2 is related to V_L1 by the turns ratio N_2/N_1 = 4. V_2 = V_L1 × (N_2)/(N_1) = 224.86 × 4 -1.615^ = 899.44 -1.615^ V The magnitude of the secondary terminal voltage is: 899.44 V QUESTION 2: MEASUREMENT OF POWER IN THREE PHASE SYSTEMS 2.1: Given: Line voltage V_L = 440 V Frequency f = 50 Hz Star connected load Impedance per phase z = 5.2 + j6.7 / phase Step 1: Calculate phase voltage, phase current, and power factor. For a star-connected load, the phase voltage V_p = V_L / sqrt(3). V_p = (440)/(sqrt(3)) ≈ 254.03 V The magnitude of the impedance per phase is |z| = sqrt(5.2^2 + 6.7^2). |z| = sqrt(27.04 + 44.89) = sqrt(71.93) ≈ 8.481 The phase current I_p = V_p / |z|. I_p = (254.03)/(8.481) ≈ 29.95 A For a star connection, the line current I_L = I_p ≈ 29.95 A. The power factor angle = (X/R) = (6.7/5.2) ≈ 52.17^. The power factor = (52.17^) ≈ 0.613. 2.1.1: The total power of the load (P_T). The total power for a three-phase system is P_T = sqrt(3) V_L I_L . P_T = sqrt(3) × 440 V × 29.95 A × 0.613 P_T = 13979.8 W ≈ 13.98 kW The total power of the load is: 13.98 kW 2.1.2: The readings on watt-meters, if two-wattmeter method is used to measure the power taken by the load. The two wattmeter readings are W_1 = V_L I_L (30^ - ) and W_2 = V_L I_L (30^ + ). Here = 52.17^. 30^ - = 30^ - 52.17^ = -22.17^ 30^ + = 30^ + 52.17^ = 82.17^ W_1 = 440 × 29.95 × (-22.17^) = 440 × 29.95 × 0.926 ≈ 12200 W W_2 = 440 × 29.95 × (82.17^) = 440 × 29.95 × 0.136 ≈ 1792 W The readings on the wattmeters are: W_1 = 12200 W, W_2 = 1792 W 2.2: Calculate the resistance per phase of a three phase l = 28 km, overhead line with solid conductors of D = 1.8 cm diameter, with a symmetrical spacing of d = 0.8 m between centres. Take resistivity of the conductor material as = 1.7 cm. Step 1: Convert units and calculate the cross-sectional area. Length l = 28 km = 28 × 10^5 cm. Diameter D = 1.8 cm, so radius r = D/2 = 0.9 cm. Resistivity = 1.7 cm = 1.7 × 10^-6 cm. The cross-sectional area A = r^2. A = (0.9 cm)^2 = 0.81 cm^2 Step 2: Calculate the resistance per phase. The resistance R_phase is given by the formula R = (l)/(A). R_phase = (1.7 × 10^-6 cm) × 28 × 10^5 cm0.81 cm^2 R_phase = (1.7 × 28)/(0.81 ) = (47.6)/(2.54469) ≈ 18.705 The resistance per phase is: 18.71 QUESTION 3: AC MACHINES 3.1: Given: Synchronous impedance per phase Z_s = 1 + j6.5 / phase. Open circuit line voltage of one alternator E_1,line = 2.2 kV = 2200 V. Circulating current I_C = 80 A. Machines are star-connected and in exact phase opposition. Step 1: Calculate phase values and magnitude of synchronous impedance. The phase voltage of the first alternator is E_1,phase = E_1,line / sqrt(3). E_1,phase = (2200)/(sqrt(3)) ≈ 1270.17 V The magnitude of the synchronous impedance is |Z_s| = sqrt(1^2 + 6.5^2). |Z_s| = sqrt(1 + 42.25) = sqrt(43.25) ≈ 6.576 3.1.1: The open circuit voltage of the other machine (E_2) assuming it is bigger than that of the first machine. When two alternators are in parallel and in exact phase opposition, the circulating current I_C is given by: I_C = E_2,phase - E_1,phase2Z_s This formula assumes E_1 and E_2 are in phase, and E_2 > E_1. However, "exact phase opposition" means E_1 and E_2 are 180^ out of phase. Let E_1 = E_1,phase 0^ and E_2 = E_2,phase 180^. The voltage difference driving the circulating current is E_1 - E_2 = E_1,phase - (-E_2,phase) = E_1,phase + E_2,phase. The magnitude of the circulating current is |I_C| = E_1,phase + E_2,phase2|Z_s|. 80 A = 1270.17 V + E_2,phase2 × 6.576 80 × (2 × 6.576) = 1270.17 + E_2,phase 1052.16 = 1270.17 + E_2,phase E_2,phase = 1052.16 - 1270.17 = -218.01 V This result is negative, which contradicts the assumption that E_2 is bigger than E_1. This indicates that the problem statement "exact phase opposition" might be interpreted differently in the context of the given circulating current value, or there's an inconsistency in the problem data. A common scenario for circulating current in parallel alternators is due to a phase difference between EMFs, or a difference in magnitudes when EMFs are in phase. If they are in "exact phase opposition", it usually means they are 180^ out of phase. If the circulating current is I_C = (E_1 - E_2)/(2Z_s) where E_1 and E_2 are phasors, and E_1 = E_1,phase 0^ and E_2 = E_2,phase 180^, then E_1 - E_2 = E_1,phase - (-E_2,phase) = E_1,phase + E_2,phase. The magnitude of this voltage difference is E_1,phase + E_2,phase. The magnitude of I_C is 80 A. |I_C| = E_1,phase + E_2,phase2|Z_s|. This leads to E_2,phase = -218.01 V, which is not physically possible for a magnitude. Let's assume "exact phase opposition" means that the resultant voltage causing the circulating current is E_2 - E_1 (phasor difference), and the problem implies that E_1 and E_2 are in phase but with different magnitudes, and the "phase opposition" refers to the current flow. This is a common source of confusion. If E_1 and E_2 are in phase, and E_2 > E_1, then I_C = E_2,phase - E_1,phase2Z_s. |I_C| = E_2,phase - E_1,phase2|Z_s|. 80 = E_2,phase - 1270.172 × 6.576 1052.16 = E_2,phase - 1270.17 E_2,phase = 1052.16 + 1270.17 = 2322.33 V This value is positive and greater than E_1,phase, consistent with the problem statement. The open circuit line voltage of the other machine is E_2,line = E_2,phase × sqrt(3). E_2,line = 2322.33 × sqrt(3) ≈ 4021.9 V The open circuit voltage of the other machine is: 4.022 kV 3.1.2: The terminal voltage (line value). The terminal voltage V_T (phase value) is the voltage at the busbar. It can be calculated using either alternator's EMF and circulating current. Let' ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._