Tutorial Questions and Solutions: The Per-Unit System
Q1. A Y connected load consisting of three equal impedances of 10∠60º Ω is connected to an
11kV busbar at a substation by a 3Φ overhead line of impedance 0.5∠75º Ω per phase.
Calculate the line-line voltage at the terminals of the load in per-unit and in kV if the base
line-line voltage is 11kV and 3-phase “MegaVoltAmpere” is 10MVA. Finally calculate the
current in each phase of the line in per-unit and kA.
S1:
( base kVLL )
Base Impedance =
2
base MVA 3Φ
112
=
= 12.1Ω
10
Load Impedance =
10∠60D
= 0.826∠60D p.u.
12.1
Line Impedance =
0.5∠75D
= 0.0413∠75D p.u.
12.1
Source Voltage = 1.0 p.u.
Terminal Voltage = 1.0 ×
0.826∠60D
0.826∠60D + 0.0413∠75D
= 0.954∠ − 0.7D p.u.
= 0.954∠ − 0.7D ×11kV = 10.49∠ − 0.7D kVLL
Base kA =
Current =
base MVA3Φ
3 × base kVLL
=
10
= 0.525 kA
3 ×11
1.0
= 1.155∠ − 60.7D p.u.
D
0.826∠60 + 0.0413∠75
D
= 1.155∠ − 60.7D × 0.525 = 0.606∠ − 60.7 D kA
Q2. The sub-transient reactance of a generator (X”) is given as 0.2 per-unit based on
the generator’s name-plate rating of 11.8kV, 200MVA. Calculate the per-unit
impedance of the generator referred to the power system base of 11.0kV, 100MVA.
S2:
Znew ( p.u.) = Zold, p.u . × (
base
old base
MVA 3new
kVLL
2
Φ
×
)
(
)
new base
kVLL
MVA3oldΦ base
11.8 2 100
= 0.115 p.u.
X '' (new p.u.) = 0.2 × (
) ×
11.0
200
Q3: A 3Φ two winding transformer is rated at 75MVA, 11.8/145kV, 50Hz and the
equivalent leakage impedance referred to the 11.8kV winding is Zeq = 0.22 Ω. Using
the transformer ratings as base values, determine the per-unit leakage impedances
referred to the 11.8kV winding and the 145kV winding.
S3:
Referred to the 11.8kV winding:
Base Z11.8kV
=
Z11.8kV
p .u .
(base kVLL ) 2 11.82
=
=
= 1.86Ω
base MVA 3Φ
75
0.22
= 0.12 p.u.
1.86
Calculate the per-unit impedance referred to the 145kV winding:
(Theory: the p.u. impedance of an ideal transformer is the same regardless of
whether it is referred to the LV or HV winding)
Proof:
E HV
p .u .
N HV
× E LV (kV)
LV
E HV (kV)
=
= N
base kV HV
base kV HV
where (N HV /N LV ) is the turns ratio of the ideal transformer.
Using the transformer HV and LV voltage ratings as the HV and LV per-unit
voltage base, i.e.:
base kV HV V rating HV N HV
=
=
base kV LV V rating LV N LV
N HV
× E LV (kV)
LV
then: E HV (p.u.) = N HV
× base kV LV
N / N LV
=
E LV (kV)
= E LV ( p.u.)
LV
base kV
Illustration of theory - impedance referred to the 145kV winding:
base Z145kV
(base kVLL ) 2 1452
=
=
= 280Ω
base MVA3Φ
75
The leakage impedance Zeq referred to the 145kV winding:
2
Zeq-145kV
=
Hence: Z145kV
p .u .
⎛ 145 ⎞
=⎜
⎟ × Zeq-11.8kV = 33.22Ω
⎝ 11.8 ⎠
33.22
= 0.12 p.u.
280
Q4: The power network shown below consists of three zones connected to each other
using transformers.
Transformer 1-2 is rated at 5000kVA, 11kV/66kV and its leakage reactance = 12%
Transformer 2-3 is rated at 5000kVA, 66kV/33kV and its leakage reactance = 10%
1
2
2
11kV
3
66kV
33kV
If the zone-2 base is 5000kVA, 66kV find per-unit impedance in 3, 2 & 1 of a 250Ω
resistive load (full-load) connected to 3. Draw the impedance diagram in per-unit
neglecting mag currents, transformer resistances and line impedances. Determine the
voltage regulation at the load if the voltage across the 250Ω load is 32kV and the
voltage at 1 remains constant during full-load and no-load conditions
S4:
Zone 2: Base kV2 = 66kV, Base MVA 2 = 5 MVA,
Base Z 2 = 662 / 5 = 871Ω
Zone 1: Base kV1 = (11/ 66) × 66 = 11kV, Base MVA1 = 5 MVA,
Base Z1 = 112 / 5 = 24.2Ω
Zone 3: Base kV3 = (33 / 66) × 66 = 33kV, Base MVA 2 = 5 MVA,
Base Z3 = 332 / 5 = 218Ω
Per-unit impedance of load in Zone 3: Z3 = 250 / 218 = 1.15 p.u.
Impedance of load referred to Zone 2: Z2 = (66 / 33)2 × 250 = 1000 Ω
Per-unit impedance of load referred to Zone 2: Z2 = 1000 / 871 = 1.15 p.u.
Impedance of load referred to Zone 1: Z1 = (11/ 66) 2 × 1000 = 27.8 Ω
Per-unit impedance of load referred to Zone 1: Z1 = 27.8 / 24.2 = 1.15 p.u.
Note: The voltages, currents and external impedances expressed in per-unit
do not change when they are referred from one side of a transformer to the
other.
Impedance diagram in per-unit:
leakage reactance transformer 1-2 = 12% = 0.12 p.u.
leakage reactance transformer 2-3 = 10% = 0.10 p.u.
Voltage Regulation:
Percent regulation =
VR,NL -VR,FL
VR,FL
×100
where:
VR,NL is the magnitude of the no-load voltage at the load.
VR,FL is the magnitude of the full-load voltage at the load.
Note: full-load = 250Ω
Voltage across full-load = 32kV
Voltage across full-load = 32kV/base kV3 = 32 / 33 = 0.970 p.u.
Load current = (0.970 + j 0) / (1.15 + j 0) = 0.843 + j 0
Voltage input to network = (0.843 + j 0) × ( j 0.12 + j 0.10) + 0.970
= 0.987∠11D p.u.
No-load voltage at the load = voltage input to network.
0.987 − 0.970
Therefore: Regulation =
× 100 = 1.75%
0.970
Q5: For the power network shown below, calculate, using per-unit quantities, the
voltage in kV, the current in kA and the power in kW received by the load.
3
L1
G1
T1
132kV
T2
11kV
G1 = synchronous generator = 75MVA, 11.8kV, sync reactance Xd = j1.83 p.u.
T1 = generator transformer = 75MVA, 11.8kV/145kV X = j0.125 p.u.
L1 = 132kV transmission line: Z = 0.18 + j0.40 Ω/km, length = 20km.
T2 = step-down transformer = 45MVA, 132kV/11kV, X = j0.125 p.u.
ZL = 11kV load = 10MVA, cos φ = 0.8 (lagging).
S5:
The above power system should be divided into three zones interconnected by
transformers T1 and T2.
Zone 1 refers to the 11.8kV generator and 11.8kV winding of T1.
Zone 2 refers to the 145kV T1 winding, L1 & the T2 132kV winding.
Zone 3 refers to the 11.0kV winding of T2 and the load ZL .
Assume a per-unit base of 100MVA and 132kV in zone 2.
Zone 2: base kV2 = 132kV
1322
= 174.2Ω
100
11.8
Zone 1: base kV1 =
×132 = 10.74kV,
145
10.7 2
base Z1 =
= 1.15Ω
100
11
Zone 3: base kV3 =
×132 = 11.0kV,
132
11.02
base Z3 =
= 1.21Ω
100
base Z2 =
2
G1: Xd new
⎛ kV ⎞ MVA 3Φnew
= Xd old × ⎜ old ⎟ ×
⎝ kVnew ⎠ MVA 3Φold
2
⎛ 11.8 ⎞ 100
= j1.83 × ⎜
= j 2.95 p.u.
⎟ ×
⎝ 10.74 ⎠ 75
Note: compare Xd new calculated at the LV side of T1 with Xd new calculated at
the HV side.
2
HV side: Xd new
⎛ 145 ⎞ 100
= j1.83 × ⎜
= j 2.95 p.u.
⎟ ×
⎝ 132 ⎠ 75
2
⎛ 11.8 ⎞ 100
T1: X = 0.125 × ⎜
= j 0.20 p.u.
⎟ ×
⎝ 10.74 ⎠ 75
20 × (0.18 + j 0.40)
L1: Z =
= (0.021 + j 0.046) p.u.
174.2
2
⎛ 11.0 ⎞ 100
T2: X = 0.125 × ⎜
= j 0.278 p.u.
⎟ ×
⎝ 11.0 ⎠ 45
Load: ZL
( kVLL )
=
ZL =
2
MVA 3Φ
=
11.02
= 12.1∠37D
10.0
12.1∠37D
= 10.0∠37D p.u. = (7.99 + j 6.02) p.u
1.21
Hence: Current in the load in p.u.
I Load =
11.8∠0D /10.7
( 0.021 + 7.99 ) + j (2.95 + 0.20 + 0.046 + 0.278 + 6.02)
1.10∠0D
=
= 0.088∠ − 50D p.u.
D
12.43∠50
Current in load in kA (referred to 11kV winding):
base MVA 3Φ
I Load = I Load, p.u . ×
3 × base kV3
= 0.088∠ − 50D ×100 /( 3 ×11.0) = 0.46∠ − 50D kA
Line-line voltage at the load:
VLoad = 3 × 0.46∠ − 50D ×12.1∠37D = 9.7∠ − 13D kV
Power in the load:
PLoad3Φ = 3 × VLL × I L × cos θ = 3 × 9.7 × 0.46 × 0.8 = 6.2 MW
MVA in the load:
MVA Load3Φ = 3 × VLL × I L = 7.7 MVA
MVA Load3Φ
VLL 2 9.7 2
=
=
= 7.7MVA
ZL
12.1
---end of the per unit tutorial questions and solutions---