chapter 2 the per unit system new
TRANSCRIPT
CHAPTER 2
The Per-Unit System
Per Unit System
In power systems there are so many different elements such as Motors, Generators and Transformers with very different sizes and nominal values.
To be able to compare the performances of a big and a small element, per unit system is used.
Per Unit System
The voltage, current and impedance values are divided by base values and expressed in per unit or percentage values.
The percent impedance,
e.g. in a synchronous generator with 13.8 kV as its nominal voltage, instead of saying the voltage is 12.42 kV, we say the voltage is 0.9 p.u.
Value Base
Value ActualValue SystemPer Unit
100% x in
in Z
base
actual%
Z
Z
base
base
base
basebase
base
basebase
basebase
S
V
I
VZ
V
SI
VS
2
,
How Are the Base Values Defined
For an electric element, we have : Power, Voltage, Current and Impedance.
Usually, the nominal apparent power (S) and nominal voltage (V) are taken as the base values for power and voltage.
The base values for the current and impedance can be calculated.
Per Unit values using the base values are:
basepu S
SS
base
pu II
I
basepu V
VV
base
pu ZZ
Z
ZZ
2base
base
basepu V
S
ZZ pu
base
2base
pubase ZS
VZZ Z
Conversion from per unit value to ohm & vise versa
Per Unit System
Transformers Voltage Base
V1/V2
Vb1 Vb2
11
22 bb V
V
VV
Per Unit in 3- Circuits
Simplified: Concerns about using phase or line voltages
are removed in the per-unit system Actual values of R, XC and XL for lines, cables,
and other electrical equipment typically phase values.
It is convenient to work in terms of base VA (base volt-amperes)
Per Unit System (3 Phase)
,3
3
B
BB
BBB
V
SI
IVS
• Usually, the 3-phase SB or MVAB and line-to-line VB or kVB
are selected
• IB and ZB dependent on SB and VB
B
B
B
BB
BBB
S
V
I
VZ
ZIV23/
3
Change of Base
The impedance of individual generators & transformer, are generally in terms of percent/per unit based on their own ratings.
Impedance of transmission line – ohmic value When pieces of equipment with various
different ratings are connected to a system, it is necessary to convert their impedances to a per unit value expressed on the same base.
ZV
S
Z
ZZ
oldB
oldB
oldB
oldpu 2
ZV
S
Z
ZZ
newB
newB
newB
newpu 2
oldB
oldB
V base voltage&
S basepower on the impedanceunit per thebe oldpuZ
newB
newB
V base voltagenew &
S basepower new on the impedanceunit per new thebe newpuZ
1
2
Change of Base
2
newB
oldB
oldB
newBold
punewpu V
V
S
SZZ
oldB
newBold
punewpu S
SZZ
From (1) and (2), the relationship between the old and the new per unit value
If the voltage base are the same,
Change of Base
Example
The one-line diagram of three-phase power system is shown below. Select a common base of 100 MVA and 22 kV on the generator side. Draw an impedance diagram with all impedance including the load impedance marked in per-unit.
Example
G M
1 2 3 4
5 6
T1 T2
T3 T4
Line 1220 kV
Line 2110 kV Load
90 MVA22 kV
X = 18%
50 MVA22/220 kVX = 10%
40 MVA22/110 kVX = 6.4%
40 MVA110/11 kVX = 8.0%
40 MVA220/11 kVX = 6.0%
66.5 MVA10.45 kV
X = 18.5%
57 MVA0.6 pf lag10.45 kV
X = 48.4 Ω
X = 64.43 Ω65.43 Ω
Voltage base for all sections of the network. SB = 100 MVA, VB = 22 kV on Generator side
G M
1 2 3 4
5 6
T1 T2
T3 T4
Line 1220 kV
Line 2110 kV Load
90 MVA22 kV
X = 18%
50 MVA22/220 kVX = 10%
40 MVA22/110 kVX = 6.4%
40 MVA110/11 kVX = 8.0%
40 MVA220/11 kVX = 6.0%
66.5 MVA10.45 kV
X = 18.5%
57 MVA0.6 pf lag10.45 kV
X = 48.4 Ω
X = 64.43 Ω
Example
65.43 Ω
G M
1 2 3 4
5 6
T1 T2
T3 T4
Line 1220 kV
Line 2110 kV Load
90 MVA22 kV
X = 18%
50 MVA22/220 kVX = 10%
40 MVA22/110 kVX = 6.4%
40 MVA110/11 kVX = 8.0%
40 MVA220/11 kVX = 6.0%
66.5 MVA10.45 kV
X = 18.5%
57 MVA0.6 pf lag10.45 kV
X = 48.4 Ω
X = 64.43 Ω
VB1 on the LV of T1 = 22 kV
50 MVA, 22/220 kV, 10%
40 MVA, 22/110 kV, 6.4%
Example
G M
1 2 3 4
5 6
T1 T2
T3 T4
Line 1220 kV
Line 2110 kV Load
90 MVA22 kV
X = 18%
50 MVA22/220 kVX = 10%
40 MVA22/110 kVX = 6.4%
40 MVA110/11 kVX = 8.0%
40 MVA220/11 kVX = 6.0%
66.5 MVA10.45 kV
X = 18.5%
57 MVA0.6 pf lag10.45 kV
X = 48.4 Ω
X = 64.43 Ω40 MVA, 22 kV, 6.4%
50 MVA, 22/220 kV, 10%
VB2 on the HV of T1 = kVkVV
VVV BB 220
22
22022
1
212
VB3 on the HV of T2 = VB2 = 220 kV
Example
G M
1 2 3 4
5 6
T1 T2
T3 T4
Line 1220 kV
Line 2110 kV Load
90 MVA22 kV
X = 18%
50 MVA22/220 kVX = 10%
40 MVA22/110 kVX = 6.4%
40 MVA110/11 kVX = 8.0%
40 MVA220/11 kVX = 6.0%
66.5 MVA10.45 kV
X = 18.5%
57 MVA0.6 pf lag10.45 kV
X = 48.4 Ω
X = 64.43 Ω40 MVA, 22/110 kV, 6.4%
VB5 on the LV of T3 = kVkVV
VVV BB 110
22
11022
1
215
VB6 on the HV of T4 = VB5 = 110 kV
Example
65.43 Ω
G M
1 2 3 4
5 6
T1 T2
T3 T4
Line 1220 kV
Line 2110 kV Load
90 MVA22 kV
X = 18%
50 MVA22/220 kVX = 10%
40 MVA22/110 kVX = 6.4%
40 MVA110/11 kVX = 8.0%
40 MVA220/11 kVX = 6.0%
66.5 MVA10.45 kV
X = 18.5%
57 MVA0.6 pf lag10.45 kV
X = 48.4 Ω
X = 64.43 Ω
VB4 on the LV of T2 =
kVkVV
VVV BB 11
220
11220
1
234
VB4 on the LV of T4 =
40 MVA, 220/11kV, 6.0%
40 MVA, 110/11kV, 8.0%
or
kVkVV
VVV BB 11
110
11110
1
264
Example
Since generator & transformer voltage base are the same as their rated values, their p.u reactance on a 100 MVA
oldB
newBold
punewpu S
SZZ
upX
upX
upX
upX
upX
T
T
T
T
G
.2.040
10008.0
.16.040
100064.0
.15.040
10006.0
.2.050
10010.0
.2.090
10018.0
4
3
2
1
Generator & Transformer
upV
V
S
SXX
newB
oldB
oldB
newBold
punewpu .25.0
11
45.10
5.66
100185.0
22
MVAS
MVAS
X
oldB
newB
oldpu
5.66
100
185.0%5.18
kVV
kVVoldB
newB
45.10
11
Motor
Line 1 & 2
MVAS
kVV
B
B
100
220
MVAS
kVV
B
B
100
110
121100
110
484100
220
22
22
2
1
MVA
kV
S
VX
MVA
kV
S
VX
B
BlB
B
BlB
Line 1 Line 2
upX
XX
upX
XX
lB
actual
lup
lB
actual
lup
.54.0121
43.65
.10.0484
4.48
2
2
1
1
.
.
Base Impedance, XB P.U Impedance, Xpu
Load
laggingfpkVVMVAS 6.0. ,45.10 ,57
upjj
Z
ZZ
MV
kV
S
VZ
jS
VZ
MVAS
Therefore
BaseL
actualLupL
B
BBaseL
oL
LLactualL
oL
o
.2667.195.021.1
53267.11495.1
21.1100
11
53267.11495.113.5357
45.10
13.5357
,
13.536.0cos
)(
)().(
22
)(
2
*3
2
)(
3
1
Per Unit Equivalent Circuit
G M
1 2 3 4
5 6
XG= 0.2p.u
XT1= 0.2p.u XT2= 0.15p.u
XT3= 0.16p.u XT4= 0.20p.u
XL1= 0.10p.u
ZL1= 0.54p.uXM= 0.25p.u
ZLoad= 0.95+j1.2667
Advantages
Give a clear idea of relative magnitudes of various quantities, such as V, I, P & Z.
The per unit values of Z, V & I of transformer are the same whether they are referred to the primary or secondary side.
Ideal for the computerized analysis and simulation of complex power system problems.
The circuit laws are valid in per unit systems, and the power and voltage equation are simplified since the factor √3 and 3 are eliminates in the p.u systems.