exammen

1. UNIVERSIDAD TÉCNICA DE COTOPAXI
Nombre: FELIX CHADAN
Curso: NOVENO fecha:05/12/2020
EJERCICIO
336,400 30/7 ACSR: GMR = 0.0255 ft
Resistance = 0.306 Ω/mile
3/0 6/1 ACSR: GMR = 0.006 ft
Resistance = 0.723 Ω/mile
𝑑1 = 0 + 𝑗29 𝑑2 = 2.5 + 𝑗29 𝑑3 = 7.0 + 𝑗29 𝑑4 = 4.0 + 𝑗25
𝐷12 = | 𝑑1 − 𝑑2| 𝐷23 = | 𝑑2 − 𝑑3| 𝐷31 = | 𝑑3 − 𝑑1|
𝐷14 = | 𝑑1 − 𝑑4| 𝐷24 = | 𝑑2 − 𝑑4| 𝐷34 = | 𝑑3 − 𝑑4|
𝐷𝑎𝑏 = 2.5 𝑓𝑡 𝐷𝑏𝑐 = 4.5 𝑓𝑡 𝐷𝑐𝑎 = 7.0 𝑓𝑡
𝐷𝑎𝑛 = 5.6569 𝑓𝑡 𝐷𝑏𝑛 = 4.272 𝑓𝑡 𝐷𝑐𝑛 = 5.0 𝑓𝑡
𝐷𝑎𝑎 = 𝐷𝑏𝑏 = 𝐷𝑐𝑐 = 0.0255
𝐷𝑛𝑛 = 0.006

2. 𝑧 𝑎𝑎 = 0.0953 + 0.306 + 𝑗0.12134 ∗ (ln
1
0,00255
+ 7,93402) = 0,4013 + 𝑗1,431 Ω/𝑚𝑖𝑙𝑙𝑎
𝑧 𝑎𝑏 = 0.0953 + 𝑗0.12134 ∗ (ln
1
2,5
+ 7,93402) = 0,0953 + 𝑗0,8515 Ω/𝑚𝑖𝑙𝑙𝑎
𝑧 𝑏𝑐 = 0.0953 + 𝑗0.12134 ∗ (ln
1
4,5
+ 7,93402) = 0,0953 + 𝑗0,7802 Ω/𝑚𝑖𝑙𝑙𝑎
𝑧 𝑐𝑎 = 0.0953 + 𝑗0.12134 ∗ (ln
1
7
+ 7,93402) = 0,0953 + 𝑗0,7266 Ω/𝑚𝑖𝑙𝑙𝑎
𝑧 𝑎𝑛 = 0.0953 + 𝑗0.12134 ∗ (ln
1
5,6569
+ 7,93402) = 0,0953 + 𝑗0,7524 Ω/𝑚𝑖𝑙𝑙𝑎
𝑧 𝑏𝑛 = 0.0953 + 𝑗0.12134 ∗ (ln
1
4,272
+ 7,93402) = 0,0953 + 𝑗0,7865 Ω/𝑚𝑖𝑙𝑙𝑎
𝑧 𝑐𝑛 = 0.0953 + 𝑗0.12134 ∗ (ln
1
5
+ 7,93402) = 0,0953 + 𝑗0,7674 Ω/𝑚𝑖𝑙𝑙𝑎
𝑧 𝑛𝑛 = 0.0953 + 0.723 + 𝑗0.12134 ∗ (ln
1
0,006
+ 7,93402) = 0,8183 + 𝑗1,61 Ω/𝑚𝑖𝑙𝑙𝑎
𝑍 = [
0,4013 + 𝑗1,43 0,0953 + 𝑗0,8515
0,0953 + 𝑗0,8515 0,4013 + 𝑗1,4133
0,0953 + 𝑗0,7266
0,0953 + 𝑗0,7524
0,0953 + 𝑗0,7802
0,0953 + 𝑗0,7865
0,0953 + 𝑗0,7266 0,0953 + 𝑗0,7524
0,0953 + 𝑗0,7802 0,0953 + 𝑗0,7865
0,4013 + 𝑗1,4133
0,0953 + 𝑗0,7674
0,0953 + 𝑗0,7674
0,8183 + 𝑗1,61
]
𝑍𝑖𝑗 = [
0,4013 + 𝑗1,43 0,0953 + 𝑗0,8515 0,0953 + 𝑗0,7266
0,0953 + 𝑗0,8515 0,4013 + 𝑗1,4133 0,0953 + 𝑗0,7865
0,0953 + 𝑗0,7266 0,0953 + 𝑗0,7802 0,4013 + 𝑗1,4133
]
𝑍𝑖𝑛 = [
0,0953 + 𝑗0,7524
0,0953 + 𝑗0,7524
0,0953 + 𝑗0,7674
]
𝑍 𝑛𝑛 = [0,8183 + 𝑗1,61]
𝑍 𝑛𝑗 = [0,0953 + 𝑗0,7524 0,0953 + 𝑗0,7865 0,0953 + 𝑗0,7674]
[𝑧 𝑎𝑏𝑐] = [𝑧𝑖𝑗] − [𝑧𝑖𝑛] ∗ [𝑧 𝑛𝑛]−1
∗ [𝑧 𝑛𝑗]

3. 𝑍 𝑎𝑏𝑐 = [
0,4576 + 𝑗1,0780 0,1560 + 𝑗0,5017 0,1535 + 𝑗0,3849
0,1560 + 𝑗0,5017 0,4666 + 𝑗1,0482 0,1580 + 𝑗0,4236
0,1535 + 𝑗0,3849 0,0953 + 𝑗0,7802 0,4615 + 𝑗1,0651
]
[𝑡 𝑛] = −([𝑧 𝑛𝑛]−1
∗ [𝑧 𝑛𝑗])
𝑡 𝑛 = [−0,4292 − 𝑗0,1291 − 0,4476 − 𝑗0,1373 − 0,4373 − 𝑗0,1327]
[𝑧012] = [𝐴 𝑠]−1
∗ [𝑧 𝑛𝑗] ∗ [𝐴 𝑠]
𝑍012 = [
0,8523 + 𝑗2,04932 0,0143 + 𝑗0,0186 −0,0293 + 0,0159
−0,0361 + 0,002591 0,3061 + 𝑗0,620 −0,0723 − 𝑗0,0060
0,0361 + 𝑗0,002951 0,0723 − 𝑗0,0059 0,3061 + 𝑗0,620
]
𝑍0 = [0,8523 + 𝑗2,04932]
𝒁 𝟎 = [𝟐, 𝟐𝟏 < 𝟔𝟕, 𝟒𝟏] Ω/𝑚𝑖𝑙𝑙𝑎 (secuencia cero)
𝑍1 = [0,3061 + 𝑗0,620] = 𝑍2 = [0,3061 + 𝑗0,620]
𝒁 𝟏 = 𝒁 𝟐 = [𝟎, 𝟔𝟗𝟏 < 𝟔𝟑, 𝟕𝟐] Ω/𝑚𝑖𝑙𝑙𝑎 (secuencia positiva y negativa)
Calcular el RMG de cada uno de los conductores no convencionales sabiendo que el radio de
cada conductor es 2cm
𝑹𝑴𝑮 = √(𝟎, 𝟕𝟕𝟗) 𝟑 ∗ (𝟖 ∗ 𝟎, 𝟎𝟐) 𝟐(𝟖 ∗ 𝟎, 𝟎𝟐) 𝟐(𝟖 ∗ 𝟎, 𝟎𝟐) 𝟐
𝟗
𝑹𝑴𝑮 = 𝟎, 𝟎𝟑𝟗𝟎𝟖

4. Calcular el RMG de cada uno de los conductores no convencionales sabiendo que el radio de
cada conductor es 2cm
𝑹𝑴𝑮 = √𝟎. 𝟕𝟕𝟗 ∗ 𝟎, 𝟎𝟐
𝟔
∗ √(𝟒 ∗ 𝟒 ∗ 𝟐 ∗ 𝟐 ∗ 𝟐√𝟑
𝟐
) 𝟑(𝟐 𝟒 ∗ 𝟐√𝟑
𝟐
) 𝟑(𝟎, 𝟎𝟐) 𝟑𝟎
𝟑𝟔
𝑹𝑴𝑮 = 𝟎, 𝟎𝟐 ∗ 𝟐 ∗ 𝟑(
𝟏
𝟏𝟐)
∗ 𝟎, 𝟕𝟕𝟗(
𝟏
𝟔)
𝑹𝑴𝑮 = 𝟎, 𝟎𝟒𝟐𝟎 //