By John Hindmarsh
Contemporary years have introduced big advancements in electric force expertise, with the looks of hugely rated, very-high-speed power-electronic switches, mixed with microcomputer keep watch over systems.
This renowned textbook has been completely revised and up to date within the gentle of those adjustments. It keeps its winning formulation of training via labored examples, that are installed context with concise motives of concept, revision of equations and dialogue of the engineering implications. quite a few difficulties also are supplied, with solutions supplied.
The 3rd version comprises superior assurance of power-electronic structures and new fabric on closed-loop keep watch over, as well as thorough therapy of electric machines.
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Extra resources for Electrical Machines and Drives, Third Edition
Sample text
F. lagging. The magnetising current is 4% of rated current; the iron loss being I kW total. 1] value of the primary current and power factor and input kVA when the other two windings are operating on the above loads. Again the leakage impedances will be neglected. See Reference 2 for an exact equivalent circuit for the 3-winding transformer. '. 909 33 x~ - 5 o / 5 33 ( 0 . 9 - j 0 . 4 3 6 ) ffi 0 . 4 5 5 - j 0 . 2 ! 7 T h e following are the light-load test readings on a 3-phase, 100-kVA, 400/6600-V, star/delta transformer: O p e n circuit; supply to low-voltage side 400 V, 1250 W Short circuit; supply to high-voltage side 314 V, 1600 W, full-load current.
A) 3 0 0 r e v / m i n while operating at rated electromagnetic torque; (b) 6 0 0 r e v / m i n at the same torque; (c) 8 0 0 r e v / m i n while operating at the same gross power (Cam " Te) as in condition (b). For each condition, find the appropriate value of the resistor. f. 36 rad/s. /,~ . 16 5 o o - o . 6 x 42 . . . 07 X 42 = 381Nm. 8)" Cam . T~ . e. by inserting extra resistance in the field circuit. C. 12. Hence, for (a), rearranging the equation: R = l_(V-( T ~ e j 9krr .... 10) will have to be used.
4A x 2500 = 1000 At/pole shunt excitation. On load, the total excitation is therefore 1000 + 1375 = 2375 At/pole. 65 Nm/A. E 210 Hence, C0m on load will be . . . 1%. 1000 Note the great increase from the 500 rev/min condition because of the weak shunt field. (d) T~ = /~, I,. 65 x 40 = 146 Nm So although the speed has fallen considerably, due to the series winding, the electromagnetic torque has increased by 66/80 = 82%. for the same reason and the air-gap power is the same. 4 In the shunt motor of the last question, the no-load armature current was neglected.