Wide Area Protection System

1. Underground cables
EE 449T ELECTRIC POWER IN SHIPS Prof. Hussein Eldisouky
ENG. Amin Hanafy

2. Underground Cables (UGC):
Resistance:
Capacitance:
𝑅 =
𝜌
2𝜋𝑙
∗ ln(
𝑟2
𝑟1
) 𝑅 =
𝜌
2𝜋𝑙
∗ log𝑒(
𝑟2
𝑟1
)
𝐶𝑡 =
2𝜋𝜀0𝜀𝑟
ln(
𝑟2
𝑟1
)
∗ 𝑙 𝐶𝑡 =
𝜀𝑟
41.4 ∗ log10(
𝑟2
𝑟1
)
∗ 𝑙 ∗ 10−9
𝐼𝐶ℎ=
𝑉𝑝ℎ
𝐶
=𝑉𝑝ℎ∗ 𝑌 =𝑉𝑝ℎ∗ (2π ∗𝑓 ∗𝐶𝑡)
𝑄𝐶ℎ= 3∗𝑉𝑝ℎ∗𝐼𝐶ℎ (V
AR)
Stress:
𝑟1
𝑟2
𝑡
𝑟2=𝑟1+𝑡
ε0= 8.85∗10−12
𝐸𝑚𝑎𝑥 =
𝑉𝑝ℎ
𝑟1∗ln(
𝑟2
𝑟1
)
(kv/cm) 𝐸𝑚𝑖𝑛 =
𝑉𝑝ℎ
𝑟2∗ln(
𝑟2
𝑟1
)
(kv/cm) 𝐸𝑚𝑖𝑛
𝐸𝑚𝑎𝑥
=
𝑟2
𝑟1

3. Sheet 5 : Cables
1) A single-core cable has aconductor diameter of 1cm and insulation thickness of 0.4cm. If the specific resistance of insulation
is 5 ×1014-cm, calculate the insulation resistance and cable capacitance for a2km length of the cable.
𝑟1
𝑡
𝑟1 =
1
2
= 0.5𝑐𝑚
𝑡=0.4𝑐𝑚 𝑟2
=𝑟
1
+𝑡=0.5+0.4=0.9𝑐𝑚
𝜌 = 5 ∗ 1014
Ω. 𝑐𝑚 = 5 ∗ 1014
∗ 10−2
Ω. 𝑚 = 5 ∗ 1012
Ω. 𝑚
𝑙 = 2𝑘𝑚 = 2000𝑚
𝑅 =
𝜌
2𝜋𝑙
∗ ln(
𝑟2
𝑟1
) =
5 ∗ 1012
2𝜋 ∗ 2000
∗ ln
0.9
0.5
= 233.87∗106=233.87M
𝑅 =
𝜌
2𝜋𝑙
∗ log𝑒(
𝑟2
𝑟1
) =
5 ∗ 1012
2𝜋 ∗ 2000
∗ log𝑒(
0.9
0.5
) 233.87∗106=233.87M

4. 2) The insulation resistance of asingle-core cable is 495 M/km. If the core diameter is 2.5cm and resistivity of insulation is
4·5 ×1014-cm, find the insulation thickness. (𝑡)
𝑟1
𝑟2
𝑡
𝜌 = 4.5∗1014−cm=4.5∗1014∗10−2−m
𝑅 =495𝑀/km 𝑙 =1𝑘𝑚 =1000𝑚
ln
𝑟2
=0.691
1.25
𝑟2=2.495cm
(.m)
()
2π∗1000
(m)
14 −2
495∗106 =4.5∗10 ∗10
∗ln
𝑟2
1.25
𝑟1=
2
=1.25𝑐𝑚
2.5
𝑅 = ∗ln
2π𝑙 𝑟1
𝜌 𝑟2
𝑡=𝑟2−𝑟1=2.495−1.25
=1.245𝑐𝑚
1.25
=1.996
𝑟2
=𝑒
𝑒 1.25
ln 0.691
𝑟2

5. 3) A 33 kV, 50 Hz, 3-phase underground cable 4 km long has aconductor diameter of 2.5 cm and insulation thickness of 0.2 cm. If
the specific resistance of insulation is 4.5 ×1014Ω.cm and the relative permittivity of insulation is 3; Calculate :
• The insulation resistance. (𝑅)
• Capacitance of the cable/phase.
• Charging current/phase. (𝐼𝑐ℎ)
• Total charging kVAR .(𝑄𝑐ℎ)
• Maximax and minimum stress
(𝐶)
(𝐸𝑚𝑎𝑥&𝐸𝑚𝑖𝑛)
𝑉𝑝ℎ =
33
√3
∗103𝑉 ,𝑓 =50 ,3Φ ,𝑙
=4𝑘𝑚 =4000𝑚
𝑟1
𝑟2
𝑡
1
2
2.5
𝑟 = =1.25𝑐𝑚 ,𝑡=0.2𝑐𝑚 𝑟2=𝑟1+𝑡=1.25+0.2=1.45 𝑐𝑚
𝜌 = 4.5∗1014−cm=4.5∗1014∗10−2−m ,ε𝑟=3
𝑅 =
𝜌
2π𝑙
∗ln
14 −2
𝑟2
=4.5∗10 ∗10
∗ln
𝑟1 2π∗4000
1.45
1.25
=26.57M
𝐶𝑡=
2πε0ε𝑟
𝑟
𝑟1
ln( 2)
∗𝑙
2π∗8.85∗10−12∗3
ln
1.45
1.25
∗4000=4.496 ∗10−6𝐹
=

6. 𝑝ℎ
33
√3
𝑉 = ∗103𝑉
𝐶𝑡 =4.496∗10−6 𝐹
𝑓 =50
𝑟1=1.25𝑐𝑚
𝑟2=1.45𝑐𝑚
𝐼𝐶ℎ 𝑝ℎ
=𝑉 ∗ 2π∗𝑓 ∗𝐶 𝑡 √3
=33
∗103 ∗ 2π∗50∗4.496∗10−6
=26.9𝐴
Q𝐶ℎ 𝑝ℎ 𝐶ℎ 3
=3∗𝑉 ∗𝐼 =3∗33
∗103∗26.9=1538.118∗103VAR=1538.118 𝑘VAR
𝑚 𝑎 𝑥
𝐸 =
𝑉𝑝ℎ
𝑟2
𝑟1∗ln(𝑟1
)
33
1.25∗ln(
1.45
1.25
)
= √3 =102.7(𝑘𝑉/𝑐𝑚)
𝑚 𝑖 𝑛
𝐸 =
𝑉𝑝ℎ
𝑟2
𝑟2∗ln(𝑟1
)
33
1.45∗ln(
1.45
1.25
)
= √3 =88.53 (𝑘𝑉/𝑐𝑚)