# Electric Machinery and Power System Fundamentals - Solutions by Stephen Chapman

By Stephen Chapman

* chosen info in a concise, all-in-one structure makes electrical equipment and tool platforms basic* Stresses utilizing beneficial calculations to resolve difficulties instead of spending prolonged time explaining easy methods to truly practice the calculations* References for additional learn and calculations

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Additional info for Electric Machinery and Power System Fundamentals - Solutions Manual

Example text

7; % Equivalent X (ohms) % Calculate the current values for the three % power factors. The first row of I contains % the lagging currents, the second row contains % the unity currents, and the third row contains % the leading currents. * ( re + j*im); % Leading % Calculate secondary phase voltage referred % to the primary side for each current and % power factor. *I); % Refer the secondary phase voltages back to % the secondary side using the turns ratio. % Because this is a delta-connected secondary, % this is also the line voltage.

C) Find the maximum primary and secondary currents under these conditions. SOLUTION (a) For this configuration, the common winding must be the smaller of the two windings, and N SE = 4 N C . 4 A VP 600 V and the maximum secondary current is IS = 3-15. 1 A 120 V VS A 5000-VA 480/120-V conventional transformer is to be used to supply power from a 600-V source to a 480-V load. Consider the transformer to be ideal, and assume that all insulation can handle 600 V. Answer the questions of Problem 3-14 for this transformer.

94 kΩ 48 3-12. 8-kV distribution line. It must be capable of handling 2000 kVA. There are three phases, connected Y-Y with their neutrals solidly grounded. (a) What must the N C / N SE turns ratio be to accomplish this connection? (b) How much apparent power must the windings of each autotransformer handle? (c) If one of the autotransformers were reconnected as an ordinary transformer, what would its ratings be? SOLUTION (a) The transformer is connected Y-Y, so the primary and secondary phase voltages are the line voltages divided by 3 .