請求 電機機械高手幫解 - 國營工作討論

By Odelette
at 2014-08-24T13:54
at 2014-08-24T13:54
Table of Contents
※ 引述《bug09182001 (小傲雪)》之銘言:
: 題目出自:97年地方三等特考
: 題目內容:有三個單相50kVA,2300/230V,60HZ變壓器
: 連接成一組三相4000/230V變壓器.已知每一相變壓器換算至低壓側之阻抗為0.012+j0.016
: 而其鐵損及磁化電感不計,假設此三相變壓器供給120KVA,230V,0.85滯後功因之三相負載,
: 試求其高壓側線電流與線電壓.
: 求詳解...有想一起討論電機甲的可加我LINE bug09182001 謝謝
Sbase=120 kVA
Vbase=230 V
Zbase=230^2/120k =0.4408
Zline=0.012+j0.016= 0.02 (53.13 degree) = 0.045 p.u (53.13 degree)
I = 1 p.u. (-31.79 degree)
V = 1 + 0.045 (53.13) * 1 (-31.79)
=1+0.045 (21.34)
=1.042+j0.016
=1.042 (0.88 degree)
I(高壓側線電流)=120k/sqrt(3)/4000=17.32 A
V(高壓側線電壓)=1.042*4000=4168 V
敬請指正
--
: 題目出自:97年地方三等特考
: 題目內容:有三個單相50kVA,2300/230V,60HZ變壓器
: 連接成一組三相4000/230V變壓器.已知每一相變壓器換算至低壓側之阻抗為0.012+j0.016
: 而其鐵損及磁化電感不計,假設此三相變壓器供給120KVA,230V,0.85滯後功因之三相負載,
: 試求其高壓側線電流與線電壓.
: 求詳解...有想一起討論電機甲的可加我LINE bug09182001 謝謝
Sbase=120 kVA
Vbase=230 V
Zbase=230^2/120k =0.4408
Zline=0.012+j0.016= 0.02 (53.13 degree) = 0.045 p.u (53.13 degree)
I = 1 p.u. (-31.79 degree)
V = 1 + 0.045 (53.13) * 1 (-31.79)
=1+0.045 (21.34)
=1.042+j0.016
=1.042 (0.88 degree)
I(高壓側線電流)=120k/sqrt(3)/4000=17.32 A
V(高壓側線電壓)=1.042*4000=4168 V
敬請指正
--
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