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Per-Unit System
Power System Analysis
Per-Unit System
In the per-unit system, the voltages, currents, powers,
impedances, and other electrical quantities are
expressed on a per-unit basis by the equation:
It is customary to select two base quantities to define
a given per-unit system. The ones usually selected
are voltage and power.
Quantity per unit =Actual value
Base value of quantity
2
Per-Unit System
Assume:
Then compute base values for currents and
impedances:
ratedb VV
ratedb SS
b
bb
V
SI
b
b
b
bb
S
V
I
VZ
2
3
Example 1
An electrical lamp is rated 120 volts, 500 watts. Compute
the per-unit and percent impedance of the lamp. Give the
p.u. equivalent circuit.
Solution:
(1) Compute lamp resistance
if power factor = 1.0 then :
8.28500
)120( 222
P
VR
R
VP
08.28Z
4
Example 1
(2) Select base quantities
(3) Compute base impedance
(4) The per-unit impedance is:
VASb 500
VVb 120
8.28500
)120( 22
b
bb
S
VZ
..018.28
08.28.. up
Z
ZZ
b
up
5
Example 1
(5) Percent impedance:
(6) Per-unit equivalent circuit:
%100% Z
..01 upZ ..01 upVS
6
Example 2
An electrical lamp is rated 120 volts, 500 watts. If the
voltage applied across the lamp is twice the rated value,
compute the current that flows through the lamp. Use the
per-unit method.
Solution:
VVb 120
..02120
240.. up
V
VV
b
up
..01.. upZ up
7
Example 2
The per-unit equivalent circuit is as follows:
..01 upZ ..02 upVS
..0201
02
..
..
.. upZ
VI
up
up
up
AV
SI
b
bb 167.4
120
500
AIII bupactual 0334.8167.402..
8
Three-Phase Systems
For a given single-line (one-line) diagram of a power
network, all component parameters are expressed in 3-
quantity whether it is the rating (capacity) expressed as
MVA or voltage as kV.
Let begin with 3- base quantity of
basebasebase IVS 3
where Vbase = line voltage, Ibase= line or phase current
(i)
9
Three-Phase Systems
Per phase base impedance,
base
base
baseI
V
Z3
(ii) This is line-to-neutral impedance
base
base
base
base
V
S
V
Z
3
3
base
basebase
MVA
kVZ
2
where kVbase and MVAbase are 3- quantities.
Combining (i) and (ii) yields,
10
Changing base impedance (Znew
)
Sometimes the parameters for two elements in the same
circuit (network) are quoted in per-unit on a different base.
The changing base impedance is given as,
2
2
base OLD base NEW
NEW OLD
base OLDbase NEW
kV MVAZ pu Z
MVAkV
11
Example 3
Determine the per-unit values of the following single-line
diagram and draw the impedance diagram.
XT1 = 0.1 p.u
50 MVA
Xg = 16%
100 MVA
275 kV/132 kV50 MVA
132 kV/66 kV
Transmission line
j 3.4
XT2 = 0.04 p.uLoad
40 MVA,
0.8 p.f. lagging
12
XT1 = 0.1 p.u
50 MVA
Xg = 16%
100 MVA
275 kV/132 kV50 MVA
132 kV/66 kV
Transmission line
j 3.4
XT2 = 0.04 p.u
Load
40 MVA,
0.8 p.f. lagging
Solution:
For S , always choose the largest rating, therefore Sbase = 100 MVA
anywhere in the power system.. For V, we have 2 transformers and
three regions with different voltage levels. So choose :
Vbase(Reg1)= 275 kV, Vbase(Reg2)=132 kV and Vbase(Reg3)=66 kV.
Region 1
Vbase=275 kV
Region 2
Vbase=132 kV
Region 3
Vbase=66 kV
13
275132
12 basebase VV132
6623 basebase VVkVVbase 2751
Per-unit calculations:
Generator G1:(Region 1) Transformer T1:
1.0)(1 puXT
2
2
base OLD base NEW
NEW OLD
base OLDbase NEW
kV MVAZ pu Z
MVAkV
32.050
100
275
27516.0)(
MVA
MVA
kV
kV
g puX p.u.
p.u.
MVA
baseT SS 1001
14
)(
)(
2
)1(Re
2
)1(Re)()(
OLD
NEW
base
base
gbase
gbase
OLDgNEWgS
S
V
VpuXpuX
Transmission line TL:
24.174
4.3)(
)(
base
TLactual
TLZ
XpuX
Transformer T2:
MVA
MVA
T puX50
100104.0)(2
p.u.
base
basebase
MVA
kVZ
2
actual
pu
base
ZZ
Z
p.u.
24.17410100
)10132(6
23
baseZ
base
MVA
T SS 502
2
2
baseOLD base NEW
NEW OLD
baseOLDbase NEW
kV MVAZ pu Z
MVAkV
08.0)(2 puXT
0195.0)( puXTL
(Region 2)
15
)(
)(
)3(Re
)3(Re
2
2
22 )()(
OLD
NEW
g
g
OLDNEW
base
base
base
base
TTS
S
V
VpuXpuX
Inductive load:
o
actual LoadZ 87.3612.87
8.0106631040
31066
)(
3
6
3
)2.16.1(87.36256.43
87.3612.87)( jorpuZ o
o
L
p.u.
PFV
S
CosV
SI
.3.3
I
V
Z3
In 3-phase systems :
and
56.4310100
)1066(6
23
baseZ
(Region 3)
16
17
Now, we have all the
impedance values in per-
unit with a common base
and we can now combine
all the impedances and
determine the overall
impedance.
17
Load
G
j 0.32 p.u.
j 0.1 p.u. j 0.0195 p.u.
Transformer
T1
Transformer
T2
Transmission Line
TL
j 0.08 p.u.
1.6 p.u..
j 1.2 p.u.
Generator
17
Example 4
Eg=275 kV T1 T2Transmission line
Load
18
In the previous example , assume that the generator
voltage is Eg=275 kV. Find the values of Ig , Iline , Iload ,Vload
and Pload.
Ig ILineILoad
19
From the previous example we know per-phase, per-unit
equivalent circuit of this power system:
Load
j 0.32 p.u.
j 0.1 p.u. j 0.0195 p.u.
Transformer
T1
Transformer
T2
Transmission Line
TL
j 0.08 p.u.
1.6 p.u..
j 1.2 p.u.
Generator
pupuEg 01275
275)(
puV
EpuE
gbase
g
g 01275
275)(
)1(Re
puj
jpuZeq
4735.27195.16.1
)2.108.00195.01.032.0(6.1)(
pupuZ
puEpuI
eq
g474256.0
4735.2
01
)(
)()(
I(pu)
˚
˚
20
)(94.209102753
10100
.3 3
6
)1(Re
)1(Re AmpV
SI
gbase
basegbase
)(473.8994.209)474256.0()( )1(Re AmpIpuII gbaseg
)(39.437101323
10100
.3 3
6
)2(Re
)2(Re AmpV
SI
gbase
basegbase
)(471.18639.437)474256.0()( )2(Re AmpIpuII gbaseLine
)(77.87410663
10100
.3 3
6
)3(Re
)3(Re AmpV
SI
gbase
basegbase
)(472.37277.874)474256.0()( )3(Re AmpIpuII gbaseLoad
˚
˚
˚
21
1.1043.32
)87.3612.87()472.372(
kV
LoadLoadLoad ZIV
o
actual LoadZ 87.3612.87
8.0106631040
31066
)(
3
6
3
From the previous example, we know :
Moreover:
)(65.98.0)1043.32(2.372 3 MWCosVIP LoadLoadLoad
Power Factor
˚
˚ ˚
Example 5
A power system consists of one synchronous generator and
one synchronous motor connected by two transformers and
a transmission line. Create a per-phase, per-unit equivalent
circuit of this power system using a base apparent power of
100 MVA and a base line voltage of the generator G1 of 13.8
kV. Given that:
G1 ratings: 100 MVA, 13.8 kV, R = 0.1 pu, Xs = 0.9 pu;
T1 ratings: 100 MVA, 13.8/110 kV, R = 0.01 pu, Xs = 0.05 pu;
T2 ratings: 50 MVA, 120/14.4 kV, R = 0.01 pu, Xs = 0.05 pu;
M ratings: 50 MVA, 13.8 kV, R = 0.1 pu, Xs = 1.1 pu;
L1 impedance: R = 15 , X = 75 .
22
The system base apparent power is Sbase = 100 MVA everywhere
in the power system. The base voltage in the three regions will
vary as the voltage ratios of the transformers that delineate the
regions. These base voltages are:
23
The corresponding base impedances in each region are:
24
The impedances of G1 and T1 are specified in per-unit on a base
of 13.8 kV and 100 MVA, which is the same as the system base
in Region 1. Therefore, the per-unit resistances and reactances of
these components on the system base are unchanged:
There is a transmission line in Region 2 of the power system. The
impedance of the line is specified in ohms, and the base impedance
in that region is 121 . Therefore, the per-unit resistance and
reactance of the transmission line are:
25
The impedance of T2 is specified in per-unit on a base of 14.4 kV
and 50 MVA in Region 3. Therefore, the per-unit resistances and
reactances of this component on the system base are:
The impedance of M2 is specified in per-unit on a base of 13.8 kV
and 50 MVA in Region 3. Therefore, the per-unit resistances and
reactances of this component on the system base are:
26
Therefore, the per-phase, per-unit equivalent circuit of this
power system is shown:
27