SME
Power Systems · Problem type
Single line-to-ground fault
Use this for “X₁ = X₂ = 0.2 pu and X₀ = 0.1 pu. Find the line-to-ground fault current on a 100 MVA, 13.8 kV base.”
Diagram
Iₐ
=
3
×
Vf
X₁ + X₂ + X₀
=
3
×
1
0.2 + 0.2 + 0.1
=
6
pu:
the
three
sequence
networks
carry
one
current
in
series.
Ibase
=
Sbase
√
3
Vbase
=
100
√
3
× 13.8
=
4,180
A,
so
If
=
6
×
4,180
A
=
25.1
kA.
This is an example. Type your own values to start a new problem.
Iₐ (pu)
=
3
×
Vf (pu)
÷
(
X₁ (pu)
+
X₂ (pu)
+
X₀ (pu)
)
Sbase
Base power
example · MVA
Vbase
Base voltage
example · kV
Ibase
Base current
calculated · A
IkA
Fault current in kA
calculated · kA
Formulas
Iₐ
=
3
×
Vf
X₁
+
X₂
+
X₀
6
=
3
×
1
0.2 + 0.2 + 0.1
Ibase
=
Sbase
√
3
× Vbase
4.1837
=
100
√
3
× 13.8
IkA = Iₐ × Ibase
25.1022 = 6 × 4.1837
Formulas are worked in kA. The step-by-step shows the conversions.
Assumptions
•
A bolted fault from phase a to ground; loads are ignored, so no current flows before it.
•
The positive-, negative- and zero-sequence networks connect in series at the fault.
•
One sequence current Iₐ₁ = Vf ÷ (X₁ + X₂ + X₀) flows through all three, and Iₐ = 3Iₐ₁.
•
Reactances only, every one in pu on the common base.
Step-by-step
WE KNOW
Vf = 1 pu, X₁ = 0.2 pu, X₂ = 0.2 pu, X₀ = 0.1 pu, Sbase = 100 MVA, Vbase = 13.8 kV
FIND
fault current (Iₐ), base current (Ibase), fault current in kA (IkA)
Step 1 · Find fault current (Iₐ)
Use Iₐ = 3 × Vf ÷ (X₁ + X₂ + X₀)
The
three
sequence
networks
in
series
carry
Iₐ₁
=
Vf
X₁
+
X₂
+
X₀
,
and
the
fault
current
is
three
times
it.
Iₐ
=
3
×
1
0.2 + 0.2 + 0.1
Iₐ
=
3
×
1
0.5
Iₐ
=
3
0.5
Iₐ = 6 pu
Step 2 · Find base current (Ibase)
Use Ibase = Sbase ÷ (√3 × Vbase)
Three-phase
power
is
√
3
×
line
voltage
×
current,
and
MVA
kV
gives
kA.
Ibase
=
100
√
3
× 13.8
Ibase
=
100
1.732 × 13.8
Ibase
=
100
23.9016
Ibase = 4.1837 kA
Step 3 · Find fault current in kA (IkA)
Use IkA = Iₐ × Ibase
A per-unit current times the base current is the current itself.
IkA = 6 × 4.1837
IkA = 25.1022 kA
Step 4 · Convert the answers
Change each answer back to the unit you chose.
Ibase = 4.1837 kA = 4,183.698 A (1 kA = 1000 A)
Check
Put the numbers back in. Both sides should match.
6
=
3
×
1
0.2 + 0.2 + 0.1
✓
4.1837
=
100
√
3
× 13.8
✓
25.1022 = 6 × 4.1837 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Fault analysis
Main lesson
Thévenin reactance from the short-circuit MVA
Use this for “A bus has 500 MVA of short-circuit power available. Find Xₜₕ on a 100 MVA base.”
Symmetrical components
Use this for “Iₐ = 10∠10° A, Ib = 8∠−100° A and Ic = 6∠140° A. Find the zero-, positive- and negative-sequence currents.”