symmetrical fault calculations
TRANSCRIPT
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Symmetrical Fault CalculationsSymmetrical Fault Calculations
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Example 1Example 1
34.5kV
SCC = 1200 MVA
10/12.5 MVA
6%
13.8 kV
F
Calculate the fault current at F
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Example 1Example 1
10/12.5 MVA
6%
F
13.8 kV
Calculate the fault current at F
A55.2898
kV5.34MVA100
kVMVAI
ohms9.11100
5.34MVA
kVZ
5.34kV100MVA
base
basebase
2
base
2base
base
base
base
=
==
==
=
=
=
Step 1: Calculate Base Values34.5kV
SCC = 1200 MVA
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Example 1Example 1
10/12.5 MVA
6%
F
13.8 kV
Calculate the fault current at F
48.05.12
100x06.0
MVA
MVAxZZ
oldbase
basetTpu
==
=
Step 2: Calculate p.u impedances34.5kV
SCC = 1200 MVA
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Example 1Example 1
10/12.5 MVA
6%
F
13.8 kV
Calculate the fault current at F
48.05.12
100x06.0
MVA
MVAxZZ
oldbase
basetTpu
==
=
Step 2: Calculate p.u impedances
0833.0
5.34
5.34x
1200
100kV
kVx
MVA
MVAZ
base
oldbase
oldbase
baseutility
=
=
=
34.5kV
SCC = 1200 MVA
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Example 1Example 1
10/12.5 MVA
6%
F
13.8 kV
Calculate the fault current at F
pu775.10833.048.0
0.1
Z
V
Itotal
puFpu
=
+
=
=
Step 3: Calculate fault current34.5kV
SCC = 1200 MVA
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Example 1Example 1
10/12.5 MVA
6%
F
13.8 kV
Calculate the fault current at F
pu775.10833.048.0
0.1
Z
V
Itotal
puFpu
=
+
=
=
Step 3: Calculate fault current
A7426
5.34x
8.13
II
A29703
55.2898x775.1
3
IxII
FHVFLV
baseFpuFHV
=
=
==
=
34.5kV
SCC = 1200 MVA
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Example 2Example 2 –– Addition of a GeneratorAddition of a Generator
34.5kV
SCC = 1200 MVA
10/12.5 MVA
6%
5MVAXd
”=0.12
13.8 kV
F
Calculate the fault current at F
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Example 2Example 2 –– Addition of a GeneratorAddition of a Generator
0833.0
5.345.34x
1200100
kV
kVx
MVA
MVAZ
base
oldbase
oldbase
baseutility
=
=
=
Step 1: Calculate impedances
34.5kV
SCC = 1200 MVA
10/12.5 MVA
6%
5MVAXd
”=0.12
13.8 kV
F
Calculate the fault current at F
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Example 2Example 2 –– Addition of a GeneratorAddition of a Generator
48.05.12
100x06.0
MVA
MVAxZZ
oldbase
basetTpu
==
=
0833.0
5.345.34x
1200100
kV
kVx
MVA
MVAZ
base
oldbase
oldbase
baseutility
=
=
=
Step 1: Calculate impedances
34.5kV
SCC = 1200 MVA
10/12.5 MVA
6%
5MVAXd
”=0.12
13.8 kV
F
Calculate the fault current at F
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Example 2Example 2 –– Addition of a GeneratorAddition of a Generator
48.05.12
100x06.0
MVA
MVAxZZ
oldbase
basetTpu
==
=
0833.0
5.34
5.34x
1200
100
kV
kVx
MVA
MVAZ
base
oldbase
oldbase
baseutility
=
=
=
Step 1: Calculate impedances
34.5kV
SCC = 1200 MVA
10/12.5 MVA
6%
5MVAXd
”=0.12
13.8 kV
4.25
100x12.0
MVA
MVAxZZ
oldbase
baseGGpu
==
=
F
Calculate the fault current at F
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Example 2Example 2 –– Addition of a GeneratorAddition of a Generator
Step 2: Calculate equivalent impedances
34.5kV
SCC = 1200 MVA 1.0V
10/12.5 MVA
6%
F
13.8 kV
Calculate the fault current at F
5MVAXd
”=0.12
0.0833
Zutility ZGpu2.4
ZTpu 0.48
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Example 2Example 2 –– Addition of a GeneratorAddition of a Generator
Step 2: Calculate equivalent impedances
456.04.25633.0
4.2*5633.0Z
Eq =
+
=
1.0V
0.456
0.0833
Zutility2.4
ZTpu 0.48
ZGpu
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Example 2Example 2 –– Addition of a GeneratorAddition of a Generator
Step 3: Calculate fault current
456.04.25633.0
4.2*5633.0Z
Eq =
+
=
1.0V
0.456
pu192.2
456.0
0.1
Z
VI
Eq
puFpu ===
0.0833
Zutility
2.4
ZTpu 0.48
ZGpu
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Example 2Example 2 –– Addition of a GeneratorAddition of a Generator
Step 3: Calculate fault current
456.04.25633.0
4.2*5633.0Z
Eq =
+
=
1.0V
0.456
pu192.2
456.0
0.1
Z
VI
Eq
puFpu ===
0.0833
Zutility
2.4
ZTpu 0.48
ZGpu
A91705.34x8.13
I
I
A36683
55.2898x192.2
3
IxII
FHV
FLV
baseFpuFHV
=
=
==
=
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Example 1 & 2 ComparisonExample 1 & 2 Comparison
34.5kV
SCC = 1200 MVA
34.5kV
10/12.5 MVA
6%
F
13.8 kV
5MVA
Xd”=0.12
IF
=9170 Amps
SCC = 1200 MVA
10/12.5 MVA
6%
13.8 kV
F
IF =7426 Amps
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Example 3Example 3 ––Presence of transformer & motorPresence of transformer & motor
34.5kV
SCC = 1200 MVA
10/12.5 MVA
6%5MVA
Xd”=0.12
13.8 kV
F 3 MVA
5%
2.4 kV
M 2500 HPXd
”=0.17
Calculate the fault current at F
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Example 3Example 3 ––Presence of transformer & motorPresence of transformer & motor
0833.0
5.34
5.34x1200
100
kV
kVx
MVA
MVAZ
base
oldbase
oldbase
baseutility
=
=
=
Step 1: Calculate impedances
34.5kV
SCC = 1200 MVA
10/12.5 MVA
6%5MVA
Xd”=0.12
13.8 kV
F 3 MVA
5%
2.4 kV
M 2500 HPXd
”=0.17
Calculate the fault current at F
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Example 3Example 3 ––Presence of transformer & motorPresence of transformer & motor
0833.0
5.34
5.34x1200
100
kV
kVx
MVA
MVAZ
base
oldbase
oldbase
baseutility
=
=
=
48.05.12
100x06.0
MVA
MVAxZZ
oldbasebaset1Tpu
==
=
Step 1: Calculate impedances
34.5kV
SCC = 1200 MVA
10/12.5 MVA
6%5MVA
Xd”=0.12
13.8 kV
F 3 MVA
5%
2.4 kV
M 2500 HPXd
”=0.17
Calculate the fault current at F
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Example 3Example 3 ––Presence of transformer & motorPresence of transformer & motor
0833.0
5.34
5.34x1200
100
kV
kVx
MVA
MVAZ
base
oldbase
oldbase
baseutility
=
=
=
48.05.12
100x06.0
MVA
MVAxZZ
oldbasebaset1Tpu
==
=
Step 1: Calculate impedances
34.5kV
SCC = 1200 MVA
10/12.5 MVA
6%5MVA
Xd”=0.12
13.8 kV
F 3 MVA
5%
2.4 kV
4.25
100x12.0
MVA
MVAxZZ
oldbase
baseGGpu
==
=
M 2500 HPXd
”=0.17
Calculate the fault current at F
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Example 3Example 3 ––Presence of transformer & motorPresence of transformer & motor
pu4.2Z
pu48.0Z
pu0833.0Z
Gpu
1Tpu
utility
=
=
=Step 1: Calculate impedances
34.5kV
SCC = 1200 MVA
10/12.5 MVA
6%5MVA
Xd”=0.12
F
13.8 kV
Calculate the fault current at F
667.13
100x05.0
MVA
MVAxZZ
oldbase
baset2Tpu
==
=
3 MVA
5%
2.4 kV
M 2500 HPXd
”=0.17
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Example 3Example 3 ––Presence of transformer & motorPresence of transformer & motor
pu667.1Z
pu4.2Z
pu48.0Z
pu0833.0Z
2Tpu
Gpu
1Tpu
utility
=
=
=
=Step 1: Calculate impedances
34.5kV
SCC = 1200 MVA
10/12.5 MVA
6%5MVA
Xd”=0.12
89.7154.2
100x17.0
MVAMVAxZZ
2154866.0
1865
pf
746.0*2500ratingMotor
oldbase
basemMpu
==
=
==
=13.8 kV
F 3 MVA
5%
2.4 kV
M 2500 HPXd
”=0.17
Calculate the fault current at F
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Example 3Example 3 ––Presence of transformer & motorPresence of transformer & motor
Step 1: Calculate impedances
34.5kV
SCC = 1200 MVA
pu89.7Z
pu667.1Z
pu4.2Z
pu48.0Z
pu0833.0Z
Mpu
2Tpu
Gpu
1Tpu
utility
=
=
=
=
=10/12.5 MVA
6%5MVA
Xd”=0.12
13.8 kV
F 3 MVA
5%
2.4 kV
M 2500 HPXd
”=0.17
Calculate the fault current at F
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Example 3Example 3 ––Presence of transformer & motorPresence of transformer & motor
Step 2: Calculate Equivalent impedances
34.5kV
SCC = 1200 MVA
1.0V
10/12.5 MVA
6%5MVA
Xd”=0.12
F
13.8 kV
Calculate the fault current at F
3 MVA
5%
2.4 kV
M 2500 HPXd
”=0.17
ZTpu1
0.0833
Zutility
0.48
ZGpu
2.4
7.89
1.667
ZMpu
ZTpu2
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Example 3Example 3 ––Presence of transformer & motorPresence of transformer & motor
Step 2: Calculate Equivalent impedances
1.0V 1.0V
ZTpu1
0.0833
Zutility
0.48
ZGpu
2.4
7.89
1.667
ZMpu 0.4354
ZTpu2
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Example 3Example 3 ––Presence of transformer & motorPresence of transformer & motor
Step 3: Calculate fault current
pu297.24354.0
0.1
Z
V
IEq
pu
Fpu ===1.0V
A9609
5.34x8.13
II
A38443
55.2898x297.2
3
IxII
FHVFLV
base
FpuFHV
=
=
==
=
0.4354
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Symmetrical FaultSymmetrical Fault
34.5kV34.5kV34.5kV
SCC = 1200 MVASCC = 1200 MVASCC = 1200 MVA
10/12.5 MVA
6%
10/12.5 MVA
6%
10/12.5 MVA
6%
5MVA
Xd”
=0.12
13.8 kV5MVA
Xd”=0.12
F
IF =7426 Amps
F
13.8 kV
F
13.8 kV
3 MVA
5%
2.4 kV
M 2500 HPXd
”=0.17
IF =9170 Amps IF =9609 Amps