electrical installation 一問?

你會問呢個問題證明你已經好高級佐，至少應該係 B 牌以上。
380 V 3φ 50 Hz
100 KW
0.7 o Cosφ = 1.02 o tanφ
90% eff
I = P / ( V x 1.732 x Pf ) = 100000 / 461 = 217A
KVA = V x 1.732 x I = 142 KVA
KVAR = KW x o tanφ = 100 x 1.02 = 102KVAR
Ic = KVAR / ( V x 1.732 ) = 102000 / 658 = 155A
Xc = ( V x 1.732 ) / Ic, [ V x 1.732 = 380 x 1.732 = 658 ]
so, Xc = 658 / 155 = 4.245Ω
C = 106 / 2π x F x Xc
= 1000000 / (6.28 x 50 x 4.245) = 750μF
for delta connection (Δ) is 3 x 250μF
for 90% eff is no effected the power factor
the above is improve the overall power factor to 1.00
至於你要將功率因數改爲0.95 o Cosφ，你只需重復照此計算法計多一次 0.95 o Cosφ，然後將 0.7 o Cosφ 之 KVAR 減 0.95 o Cosφ 之 KVAR 你就可以知道所需之電容量啦。
1.732 = √3
以前讀書有學過，已經廿年冇計此數學，有時都會漏佐一個步驟冇計到既。
如有錯漏，祈請指教