SME
Power Systems · Problem type
Per-unit impedance
Use this for “On a 138 kV, 100 MVA base, what is a 19.04 Ω line impedance in per unit?”
Diagram
Zₚᵤ = 0.09998 pu on the top line sits over 19.04 Ω on the bottom line: each 1 above is 190.44 below.
This is an example. Type your own values to start a new problem.
Zbase (Ω)
=
(
Vbase (kV)
)
2
÷
Sbase (MVA)
Zₚᵤ (pu)
=
Z (Ω)
÷
Zbase (Ω)
Formulas
Zbase
=
Vbase
2
Sbase
190.44
=
138
2
100
Zₚᵤ
=
Z
Zbase
0.09998
=
19.04
190.44
Assumptions
•
Vbase is the line-to-line voltage and Sbase the three-phase power; Zbase is then the impedance of one phase.
•
A per-unit value is the value over its base, so Zbase itself is 1 pu.
•
With Vbase in kV and Sbase in MVA, Vbase² ÷ Sbase comes out in ohms: the 10⁶ factors cancel.
Step-by-step
WE KNOW
Vbase = 138 kV, Sbase = 100 MVA, Z = 19.04 Ω
FIND
base impedance (Zbase), per-unit impedance (Zₚᵤ)
Step 1 · Find base impedance (Zbase)
Use Zbase = Vbase² ÷ Sbase
The
impedance
that
draws
the
base
power
at
the
base
voltage:
V
2
S
.
Zbase
=
138
2
100
Zbase
=
19,044
100
Zbase = 190.44 Ω
Step 2 · Find per-unit impedance (Zₚᵤ)
Use Zₚᵤ = Z ÷ Zbase
Per unit means a fraction of the base: divide the ohms by Zbase.
Zₚᵤ
=
19.04
190.44
Zₚᵤ = 0.09998 pu
Check
Put the numbers back in. Both sides should match.
190.44
=
138
2
100
✓
0.09998
=
19.04
190.44
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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