Page 143 - 捷運工程叢書 精進版 - 29 捷運系統水電、環控與消防工程實務
P. 143
第五章 捷運接地及避雷系統
而且 � � � 2L � k � L �
ln
R 1 � � � � c � � 1 A c � k 2 � IEEE Std. 80-2013 Eq.59
� � ' a
L
�
�
c
R 2 � ρ � � 4L r � � 2k 1 � L r �n R � 1 � � IEEE Std. 80-2013 Eq.60
2
ln
� 1
�
�
�
�
π
2 n R L r � �b � A �
R � � � � 2L c � k 1 � L c � k � 1 � IEEE Std. 80-2013 Eq.61
ln
�
�
�
�
� �
�
m
L
� c � � L r � A 2 �
其中 ρ :大地電阻係數 = 43 Ω-m
h :接地網深度 = 0.8 m
A:接地網覆蓋面積 = 288 m 2
L r :接地棒平均長度 = 3.0 m
n R :地網鋪設範圍內接地棒數量 = 28
L R :接地棒總長度 = 84 m
L C :接地導體總長度 = 180 m
a :接地網導體半徑 = 0.0089 m
a ' � a � 2 h = 0.1193 m 若 h > 0
b :接地棒半徑 = 0.0125 m
L/W:接地網長寬比值 = 2.00
當 h = 0時, k 1 = 1.33
Curve A
當 h = 0時, k 2 = 5.80
當 h = 1/10 x √A = 1.7時, k 1 = 1.10
Curve B
當 h = 1/10 x √A = 1.7時, k 2 = 4.88
當 h = 1/6 x √A = 2.83時, k 1 = 1.03
Curve C
當 h = 1/6 x √A = 2.83時, k 2 = 4.30
當 h =0.8時,在A、B區間,利用插入法 k 1 = 1.33 - ( 1.33 -1.1 ) x ( 0.8 / 1.7 )
= 1.222
k 2 = 5.8 - ( 5.8 -4.88 ) x ( 0.8 / 1.7 )
= 5.367
所以 R 1 = 0.715 Ω
R 2 = 0.713 Ω
R m = 1.022 Ω
R g = 0.868 Ω
Step 3:計算實際可能發生之步間電壓與接觸電壓及最小接地線長度
步間電壓計算公式:
121