高 等 電 機 機 械 報 告 作業討論 第四章 指導教授 謝銘原 M9920105 蘇俊銘 2010 11/09.

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高 等 電 機 機 械 報 告 作業討論 第四章 指導教授 謝銘原 M 蘇俊銘 /09

4.12 The three-phase synchronous machine of Problem 4.9 is to be moved to an application which requires that its operating frequency be reduced from 60 to 50Hz. This application requires that, for the operating condition considered in Problem 4.8, the Vrms generated voltage equal 13.0 kV line-to-line. As a result, the machine armature must be rewound with a different number of turns. Assuming a winding factor of Kw=0.928, calculate the required number of series turns per phase.

Solution 4.12

4.18 A two-pole, 60 Hz, three-phase, laboratory-size synchronous generator has a rotor radius of 5.71 cm, a rotor length of 18.0 cm, and an air-gap length of 0.25 mm. The rotor filed winding consists of 264 turns with a winding factor of kr = The Y-connected armature winding consists of the 45 turns per phase with a winding actor kw = a. Calculate the flux per pole and peak fundamental air-gap flux density which will result in an open-circuit, 60-Hz armature voltage of 120 V rms / phase (line-to-neutral). b. Calculate the dc field current required to achieve the operating condition of part (a). c. Calculate the peak values of the field-winding to armature- phase winding mutual inductance.

Solution 4.18

4.20 A four-pole, 60Hz synchronous generators has a rotor length of 5.2 m, diameter of 1.24 m, and air- gap length of 5.9 cm. The rotor winding consists of a series connection of 63 turns per pole with a winding factor of Kr = The peak values of the fundamental air-gap flux density is limited to 1.1 T and the rotor winding current to 2700 A. Calculate the maximum torque (N.m) and power output (MW) which can be supply by this machine.

Solution 4.20

Thank you Very much!!