高 等 電 機 機 械 報 告 作業討論 第四章 指導教授

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高 等 電 機 機 械 報 告
作業討論 第四章
指導教授 秦純 教授
M9920102 蔡育銘
2010 11/09
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 = 0.95. The Y-connected
armature winding consists of the 45 turns per phase with a
winding actor kw = 0.93.
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 armaturephase 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 airgap length of 5.9 cm. The rotor winding consists of
a series connection of 63 turns per pole with a
winding factor of Kr = 0.91. 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!!
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