Examples Distance Protection
Gustav Steynberg © Siemens AG 2008 Energy Sector
Calculation examples of distance protection
1.
Ph-Ph fault loc location ion
2.
Ph-G fault location ion
3.
Dete Determ rmin ine e Faul Faultt Loop oop direction
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November 09
Energy Automation
© Siemens AG 2008 Energy Sector
Example 1: Calculate fault location for L2-L3 fault (in km) Line length = 50km, 50km, ZL=50 (0.0195 + j0.15) j0.15) Ohm
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November 09
Energy Automation
© Siemens AG 2008 Energy Sector
Example 1: Solution: IL1 ZL IL2
U L2-L3
Z L
=
Z L
=
Z L
=
IL3 IE L1
L2
L3
ZE
Z L
= Z L ( I L2 – I L3)
U L 2 − U L 3 I L 2 − I L 3 (104 + j91.7) − (124 − j14.2) kV (8.59 + j 4.71) − (−8.60 − j 4.68)kA
(−20 + j105.9) kV (17.19 + j 9.39)kA
= 5.5e
j 72.1
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107.8e 19.59e
.
j 28.6°
= 1.69 + j 5.23Ω
Fault _ loc _ km =
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=
5.23Ω 0.15 Ω
= 34.9km
km
© Siemens AG 2008 06.08.97 dtgerdis3 Energy Sector
Example 2: Calculate fault location for L1-G fault (in km) Line length = 50km, ZL=50 (0.0195 + j0.15) Ohm
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November 09
5 ZE = R L 3
+ j
2.07 3
Energy Automation
X L
© Siemens AG 2008 Energy Sector
Example 2: Solution: IL1 ZL
V L1
= I L1 ⋅
( R L + jX L ) − I E ⋅ ( R E +
IL2 IL3 IE ZE U U U V L1
= − 5.12 + j8.98 −
2.07 5 R L + j X L 3 3
V L1
= I L1 ⋅ ( R L + jX L ) − I E
V L1
= I L1 −
5 3
jX E )
5 3
I E ⋅ R L
(5.13 − j9.01) ⋅ R L
+ j I L1 −
2.07 3
+ j − 5.12 + j8.98 −
(
)
I E X L 2.07 3
(5.13 − j 9.01) X L
(
− 76.4 − j14.3 = − 13.67 + j 24.00 ⋅ R L + − 15.20 − j 8.66
) X L
− 76.4 = −13.67 ⋅ R L − 15.20 ⋅ X L − 14.3 =
24.00 ⋅ R L − 8.66 ⋅ X L
Fault _ loc _ km =
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November 09
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4.20Ω 0.15 Ω
=
X L
=
4.20Ω
28.0km
km © Siemens AG 2008 06.08.97 dtgerdis3 Energy Sector
Example 2: Determine direction of all 6 loops (memory and actual voltage) [K0 = 1] Pre-Fault
Fault
Pre-Fault
Fault
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November 09
Energy Automation
© Siemens AG 2008 Energy Sector
Example 2: Solution: Example L1-G Pre-Fault
Dir _ L1G ∝
Angle _ U L1 Angle _( I L1 − K 0 ⋅ IE )
First calculate the loop current angle
Angle _( I L1 − IE ) = Angle _((4.67 + 1.88) + j (−9.86 − 9.91)) = Angle _(6.55 −
j19.77) Fault
= −71.7°
Calculate the direction angle (here actual fault voltage)
Dir _ angle _ L1G = Angle _ U L1 − Angle _( I L1 − IE ) = −7.4 − ( −71.7) =
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64.3° ≡ forward
November 09
Energy Automation
© Siemens AG 2008 Energy Sector
Example 2: Solution Pre-Fault Mem Volt Loop
Mag
Angle
Actual Vo Volt
Loop Current
Mag
Mag
Angle
Angle
L1-G
63,5
0,0 0,
20,8
-7,4
20,8
-71,7
L2-G
63,5 -120,0
63,3
-1 -119,7
12,4
-86,7
L3-G
63,5
120,0
63,3
11 119,7
7,5
-87,9
L1-L2
110
30,0
73,7
45,2
9,5
-51,8
L2-L3
110
-90,0
110
-90,0
4,8
-84,8
L3-L1
110
150,0
77,6
132,0 13
13,8
117,2
Fault
Loop
Mem Dir Angle
Actual Direction Dir Angle
L1-G
71,7
64.3
Forward
L2-G
-33,3
-33.0
Reverse
L3-G
207.9
207.6
Reverse
L1-L2
81.8
97.0
Forward
L2-L3
-5.2
-5.2
Forward
L3-L1
32.8
14.8
Forward
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© Siemens AG 2008 Energy Sector
