A 2 MVA, 13.2 kV/440 V, 60 Hz, 3-phase transformer has a per unit impedance of 0.01 + j0.08. Calculate the percent efficiency at full-load and 0.8 p.f. lagging if core loss is 8 kW. A. 97.25 B. 96.92 C. 98.28 D. 97.72
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The formula for full-load current is: \[ I_{FL} = \frac{S}{\sqrt{3} \times V_{secondary}} \] Given that \( S = 2 \) MVA and \( V_{secondary} = 440 \) V, we can substitute these values into the formula: \[ I_{FL} = \frac{2 \times 10^6}{\sqrt{3} \times 440} \] \[ Show more…
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A three-phase transformer rated 5 MVA, 115/13.2 kV has a per-phase series impedance of (0.007+j0.075) per unit. The transformer is connected to a short distribution line which can be represented by a series impedance per phase of (0.02+j0.10) per unit on a base of 10 MVA, 13.2 kV. The line supplies a balanced three-phase load rated 4 MVA, 13.2 kV, with a lagging power factor 0.85. a. Draw an equivalent circuit of the system indicating all impedances in per unit. Choose 10 MVA, 13.2 kV as the base at the load. b. With the voltage at the primary side of the transformer held constant a 115 kV, the load at the receiving end of the line is disconnected. Find the voltage regulation at the load.
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