電機機械 同步機 - 考試
By Ingrid
at 2017-10-14T20:32
at 2017-10-14T20:32
Table of Contents
1.
題目及解法:http://i.imgur.com/OOsjAaA.jpg
原文書解法:http://i.imgur.com/sSM0CF5.jpg
題目及a小題解答如上,想請問a小題的電動
機電流,(2題只差100kva 跟80kw,但100kva,
0.8超前跟80kw,0.8超前表示是一樣的)
想請問原文書用80kw算Iam是否錯了,根據
P=S*cos(功因角),功因角可變所以"輸出實功
為可變",但是S是額定(如機上銘牌)不變才
對,我的想法是S=P/pf=100kVA,現在功因調為
1,所以Iam=S/(480*根號3)=120.3A,如第一個
解法,但是因為原文書解答很少出錯所以上
來求證一下。
2.
題目:http://i.imgur.com/OEBCJ7A.jpg
解法:http://i.imgur.com/H6qpSbd.jpg
我想請問最後一行是不是要改成減號?因為同
步發電機,功因超前,感應電勢<端電壓才對
(參考:http://i.imgur.com/rYXb1sk.jpg )
煩請各位賜教,謝謝
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