102年高考電機機械第五題 - 高考
By Victoria
at 2013-07-10T22:19
at 2013-07-10T22:19
Table of Contents
※ 引述《Q66 (阿華)》之銘言:
: ※ 引述《letterstar (letterstar)》之銘言:
: : [考題] 國考歷屆考題與考題觀念討論(書裡看到的選這個)請附上想法、出處
: : 題目:設有一50HP, 460V, 60Hz, 865rpm的三相感應
: : 電動機,連至三相, 460v, 60Hz電源,於額定輸出
: : 下運轉,此時輸入功率和輸入電流分別為43.75KW
: : 和61A. 已知換算至定子側每相等效電路R1=0.1
: : 歐姆,請額定轉述時的旋轉損失1050W,試求此感
: : 應電動機滿載時的功率因數及效率, 轉子銅損,
: : 鐵損?
: : SOL:
: : 功因=43.75KW/(3*265.58*61)=0.9
: : 效率=(746*50)/43.75K=0.8526
: : 可是題目沒有極數,就不能求轉速差,
: : 還是有其他方法可以求轉子銅損和鐵損?
: : 請各位高手解惑一下
: : THX
: pag = pi - (i^2R1) -Pcore ---(1)
: pe = po + 外損
: pag = pe + Prcu ---(2)
: 由(1)和(2)將pag聯立 得(3)
: 效率 = po / po + Prcu + Pscu +Pcore + 外損 ---(4)
: (3)和(4)聯立就可以得解。
: 剛想的,不知對不對。
Pinv=(1-s)Pag-------------------(1)
Pinv=P損失+Po(Po=746*50)--------(2)
=>解(1),(2) S可求解,後面用功率潮流可解
Hint:1.P銅=S*Pag
2.定子銅損計得是=3*i^2 r1(重點)
今年題目好活陷阱好多!!
希望我日行一善可以上高考!!YA
--
: ※ 引述《letterstar (letterstar)》之銘言:
: : [考題] 國考歷屆考題與考題觀念討論(書裡看到的選這個)請附上想法、出處
: : 題目:設有一50HP, 460V, 60Hz, 865rpm的三相感應
: : 電動機,連至三相, 460v, 60Hz電源,於額定輸出
: : 下運轉,此時輸入功率和輸入電流分別為43.75KW
: : 和61A. 已知換算至定子側每相等效電路R1=0.1
: : 歐姆,請額定轉述時的旋轉損失1050W,試求此感
: : 應電動機滿載時的功率因數及效率, 轉子銅損,
: : 鐵損?
: : SOL:
: : 功因=43.75KW/(3*265.58*61)=0.9
: : 效率=(746*50)/43.75K=0.8526
: : 可是題目沒有極數,就不能求轉速差,
: : 還是有其他方法可以求轉子銅損和鐵損?
: : 請各位高手解惑一下
: : THX
: pag = pi - (i^2R1) -Pcore ---(1)
: pe = po + 外損
: pag = pe + Prcu ---(2)
: 由(1)和(2)將pag聯立 得(3)
: 效率 = po / po + Prcu + Pscu +Pcore + 外損 ---(4)
: (3)和(4)聯立就可以得解。
: 剛想的,不知對不對。
Pinv=(1-s)Pag-------------------(1)
Pinv=P損失+Po(Po=746*50)--------(2)
=>解(1),(2) S可求解,後面用功率潮流可解
Hint:1.P銅=S*Pag
2.定子銅損計得是=3*i^2 r1(重點)
今年題目好活陷阱好多!!
希望我日行一善可以上高考!!YA
--
Tags:
高考
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