學術討論(吹水)- Electrical Power Systems之1
話說去年趁sem break時報讀了一個（39小時）的短期課程 “Electrical Power Systems＂，如果你曾經修讀過這個課程，你應該會知道我講緊乜；如果未讀過而又有興趣想知讀乜既話，不妨睇下我既分享!
簡單來說，這個課程函蓋以下內容︰
Chapter 1 – Introduction to Electrical Power Systems
Chapter 2 – Two-port Network
Chapter 3 – Power Quality of T&D Systems and Compensation
Chapter 4 – Tariffs and Power Factor ImprovementChapter 5 – Power System Fault and Protection
課程第一堂主要是介紹一下產電、輸電及配電，包括一些 components例如transmission line、火牛、circuit breaker之類。
第二堂講Two-port Network，包括︰短/中/長 transmission line既定義，有些什麼parameters包括電阻值(resistance, inductance, capacitance)，跟住開始計數….
比一個例子大家參考︰
「A 3-phase, 50Hz, short transmission line, having resistance 0.6Ω/phase and inductance 100 mH/phase, supplies a wye-connected 80 MW load of 0.9 lagging power factor at 215 kV. Determine:
(a)The sending-end current
(b)The sending-end line voltage
」
Solution
P = √3 x V x I x cos Ø
80 x 10E6 = √3 x (215 x 10E3 x I x 0.9)
I = 238.7A
cos Ø = 0.9 Ø = 25.80
I = 238.7∠-25.80A
Z = R + XL = 0.6 + j(2π(50) (100 x 10-3)
VS = Vr + IZ
= (215 x 10E3 / √3 ) + (238.7∠-25.80)【0.6 + j(2π(50) (100 x 10-3)】
= 127,700∠30V or (127,500 + j6,689)V
Vs(line) = 127,700 x (√3) = 221.2 kV
睇完仲有興趣的話，請留意下一篇"Electrical Power Systems之2"
簡單來說，這個課程函蓋以下內容︰
Chapter 1 – Introduction to Electrical Power Systems
Chapter 2 – Two-port Network
Chapter 3 – Power Quality of T&D Systems and Compensation
Chapter 4 – Tariffs and Power Factor ImprovementChapter 5 – Power System Fault and Protection
課程第一堂主要是介紹一下產電、輸電及配電，包括一些 components例如transmission line、火牛、circuit breaker之類。
第二堂講Two-port Network，包括︰短/中/長 transmission line既定義，有些什麼parameters包括電阻值(resistance, inductance, capacitance)，跟住開始計數….
比一個例子大家參考︰
「A 3-phase, 50Hz, short transmission line, having resistance 0.6Ω/phase and inductance 100 mH/phase, supplies a wye-connected 80 MW load of 0.9 lagging power factor at 215 kV. Determine:
(a)The sending-end current
(b)The sending-end line voltage
」
Solution
P = √3 x V x I x cos Ø
80 x 10E6 = √3 x (215 x 10E3 x I x 0.9)
I = 238.7A
cos Ø = 0.9 Ø = 25.80
I = 238.7∠-25.80A
Z = R + XL = 0.6 + j(2π(50) (100 x 10-3)
VS = Vr + IZ
= (215 x 10E3 / √3 ) + (238.7∠-25.80)【0.6 + j(2π(50) (100 x 10-3)】
= 127,700∠30V or (127,500 + j6,689)V
Vs(line) = 127,700 x (√3) = 221.2 kV
睇完仲有興趣的話，請留意下一篇"Electrical Power Systems之2"