Per unit calculation

1. Department of Electrical and Electronic Engineering
Faculty of Sciences and Engineering
Dhaka, Bangladesh
Full Module Specification
Course Name & Course code:
Course Topic: Per Unit Quantity and Related Math.
Module Number:
Academic Year :
Course Teachers :
Contact Address:
Counseling Hour: By Appointment or, See Class Routine
Module Credit 3 Credit Hours.
Co Requisites : Electrical Machines 1 & 2, Numerical Methods
for Engineers.
Grading : As Outlined in the University policy
Teaching Methodology Class Room Lecture, Multimedia Presentation,
Discussion, Group Study, Assignment, Class
Test etc.
Method of Evaluation Attendance =10
Continuous Assessment =30
Mid-Term =30
Final-Term
______________________________
TOTAL =100
Power Systems Engineering [ EEE-3231]
Fall-2018
=40
Northern University Bangladesh
Md. Mahmudul Hasan
Lecturer, Dept. of EEE, NUB

2. DMAM
Per-Unit Quantity
The voltage, current, power and impedance in a power system are often expressed in per-unit or
percent of a specified base values.
Per-unit quantity is the ratio of actual quantity and base value of quantity. It is represented by pu.
quantityofvalueBase
quantityActual
quantityunitPer =− (3.3.1)
Per-unit quantity is dimensionless quantity.
The angle of the per-unit quantity is the same as the angle of the actual uantity since the base value is
always a real number.
For single-phase systems, or three-phase systems where the term current referes to line current, where
the term voltage referes to voltage to neutral, and where the term kilovoltamperes reerers to
kilovoltampere per phase, the following formulas reltae the various quantities:
φφφφ 1
kVAbase
1base1base1base
=== SQP (3.3.2.1)
LNLN
V kVvoltage,base
base
voltage,Base = (3.3.2.2)
LNLN
V
S
I
kVvoltage,base
1
kVAbase
base
1base
A,
base
current,Base
φφ
== (3.2.3)
base
base
basebasebase
impedance,Base
I
LN
V
XRZ === (3.2.4.1)
( ) ( ) ( )
φφφ 1
MVA,base
2kVvoltage,base
1
kVA,base
10002kVvoltage,base
1base
2
base
base
LNLN
S
LN
V
Z =
×
== (3.2.4.2)
base
1
basebasebase
admitance,Base
Z
BGY === (3.2.4.3)
For Balance Three-Phase Systems
Line voltage are selected as a base voltage;
φφφφ 3
kVAbase
3base3base3base
=== SQP (3.3.2.1)
LLLL
V kVvoltage,base
base
voltage,Base = (3.3.2.2)
φ
φ
φφφ 1
kVAbase
3
3base
1base1base1base
====
S
SQP (3.3.2.1)
LN
LL
V
LN
V kVvoltage,base
3
base
base
== (3.3.2.2)
LLLL
V
S
LN
V
S
I
kVvoltage,base3
3
kVAbase
base
3
3base
base
1base
A,
base
current,Base
×
===
φφφ
(3.2.3)
For Balanced Three Phase Systems
Page-01

3. DMAM
( )
φ1base
2
base
base
base
basebasebase
impedance,Base
S
LN
V
I
LN
V
XRZ ==== (3.2.4.1)
( ) ( )
⎟
⎠
⎞
⎜
⎝
⎛
×
==
3/
3
kVA,base
1000
2
3/kVvoltage,base
3base
2
base
base
φφ
LL
S
LL
V
Z (3.2.4.2)
( ) ( )
φφ 3
MVA,base
2kVvoltage,base
3
kVA,base
10002kVvoltage,base
base
LLLLZ =
×
= (3.2.4.2)
base
1
basebasebase
admitance,Base
Z
BGY === (3.2.4.3)
Single Phase Three Phase
Base VA (Sbase1φ) in single phase
and Base voltage line to neutral
(VbaseLN)
Base VA (Sbase3φ) in three phase
and Base voltage line to line
(VbaseLL)
Base current, Ibase
LN
V
S
base
1base φ
or
LN
kVvoltage,base
1
kVAbase
φ
LL
V
S
base
3
3base φ
or
LL
kVvoltage,base3
3
kVAbase
×
φ
Base impedance, Zbase=
Rbase= Xbase
( )
φ1base
2
base
S
LN
V
or
( )
φ1
kVA,base
10002kVvoltage,base ×
LN
or
( )
φ1
MVA,base
2kVvoltage,base
LN
( )
φ3base
2
base
S
LL
V
or
( )
φ3
kVA,base
10002kVvoltage,base ×
LL
or
( )
φ3
MVA,base
2kVvoltage,base
LL
1000kVvoltage,base
)(
actual
3
1000kVvoltage,base
)(
actual
)(
base
)(
actual
pu ×
×
=
×
==
LL
VV
LN
VV
VV
VV
V
)(
base
)(
actual
pu AI
AI
I =
( ) ( ) ( )2kVvoltage,base
1
MVA,base
actual
10002kVvoltage,base
1
kVA,base
actual
2
base
1baseactual
base
actual
pu
LN
Z
LN
Z
LN
V
SZ
Z
Z
Z
φφφ
×
=
×
×
=
×
==
( ) ( ) ( )2kVvoltage,base
3
MVA,base
actual
10002kVvoltage,base
3
kVA,base
actual
2
base
3baseactual
base
actual
pu
LL
Z
LL
Z
LL
V
SZ
Z
Z
Z
φφφ
×
=
×
×
=
×
==
Page-02

4. DMAM
Example of Generator in Per-Unit System
Example: A generator rated 1000 VA and 200 V has
internal impedance j10 Ω as shown in figure.
Considering the ratings of the generator are base
values, calculate the base current, base impedance and
per-unit of internal impedance of generator.
Solutions: Sbase=1000 VA, Vbase=200 V
Egpu = Eg/Vbase=200/200 = 1 pu
Spu = S/Sbase=1000/1000 = 1 pu
Base current, Ibase= Sbase / Vbase =1000/200=5A
Base impedance, Zbase= ( Vbase)2
/ Sbase
Zbase =2002
/1000= 40 Ω
Per-unit of internal impedance, Zpu= Zactual/ Zbase=j10/40=j0.25 pu
Example: The above generator is short circuited
at its terminals. Find the short circuit curent and
the short circuited power delivered by the
generator in pu and in actual unit.
Solutions: puj
jZ
E
I 0.4
25.0
1
Spu
gpu
SCpu
−===
pujjIES 0.4*)4(1
SCpugpuSCpu
=−×==
AjjIII 0.200.50.4
baseSCpuSC
−=×−==
VAjjSSS 400010000.4
baseSCpuSC
=×==
Example of Transformers and Per Unit System
Example: A transformer is rated 2000 VA, 200V/400V, and has an internal impedance of j4.0 Ω as
seen from the low voltage side.
The internal impedance of the transformer as seen from the high voltage side is
Ω=⎟
⎠
⎞
⎜
⎝
⎛
×=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
== 16
2
200
400
0.4
2
1
2
LV
2
LVHV
jj
V
V
ZKZZ
The internal impedance of the transformer as seen from the low voltage side is
Ω=⎟
⎠
⎞
⎜
⎝
⎛
×=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
== 4
2
400
200
0.16
2
2
1
HV
2/
HVLV
jj
V
V
ZKZZ
(a) Transformer impedance referred to the
low voltage side
(b) Transformer impedance referred to the
high voltage side
Example of Transformer & Per Unit System
Example of Generator in Per Unit System
Example of Transformer in Per Unit System
Page-03

5. DMAM
The rated values for power and voltage are used as the bases for the calculations. It means that the
voltage base is different on each side of the transformer. Comparison of the base and the per unit value
on both sides of the tarnsformer in the following Table.
Low Voltage Side High Voltage Side
Sbase 2000 VA 2000 VA
Vbase 200 V 400 V
Ibase A
V
S
10
200
2000
base
base == A
V
S
5
400
2000
base
base ==
Zbase Ω== 20
2000
2)200(
base
2
base
S
V
Ω== 80
2000
2)400(
base
2
base
S
V
Zpu pu2.0
20
4
base
LV j
j
Z
Z
== pu2.0
80
16
base
HV j
j
Z
Z
==
Notice in above Table that the transformer per
unit impedance is the same, regardless of to which
side of the transformer it is referred. Again, the
conversion complications are absorbed into the
base relationships. In the transformer equivalent
circuit the different voltage levels disappear and
the transformer equivalent circuit is reduced to a
single impedance:
Equivalent Circuit for a Transformer in Per Unit
Analysis
Example 6.4 [1, p.147] A single phase transformer is rated 110/440 V, 2.5 kVA. Leakage reactance
measured from the low-tension side is 0.06 Ω. Determine leakage reactance in per-unit.
Solution:
Base voltage in low-tension side, Vbase(LT) = 0.11 kV
Base kVA, Sbase = 2.5 kVA
Base impedance in low-tension side, Ω=
×
== 84.4
5.2
10002)11.0(
base
2
base(LT)
base(LT) S
V
Z
Leakage reactance in per-unit, 0124.0
84.4
06.0
base(LT)
actual(LT)
pu(LT)
===
Z
X
X pu
If leakage reactance had been measred on the high voltage side, the actual alue would be
Ω=×⎟
⎠
⎞⎜
⎝
⎛== 96.006.0
2
110
440
actual(LT)
2
actual(HT)
XaX
Base impedance in low-tension side, Ω=
×
== 5.77
5.2
10002)44.0(
base
2
base(LT)
base(LT) S
V
Z
0124.0
5.77
96.0
base(HT)
actual(HT)
pu(HT)
===
Z
X
X
Page-04

6. DMAM
Example 6.5 [1, p.147] Three parts of a single-phase electrical system are designated A, B, and C and
are connected to each other through transformer, as shown in Fig. 6.19.
The transformer are rated as follows:
A – B 10,000 kVA, 138/13.8 kV, leakage reactance 10%
B – C 10,000 kVA, 138/69 kV, leakage reactance 8%
If the base in circuit B is chosen as 10,000 kVA, 138 kV, find the per-unit impedance of the 300 Ω
resistive load in the circuit referred to circuits C, B, and A.
Draw the ipedance diagram neglecting magnetizing current, transformer resistances, and line
impedances.
Determine the voltage regulation if the load is 66 kV with the assuption that the voltage input to circuit
A remains costant.
Soltion: Given, the base in circuit B is chosen as 10,000 kVA, 138 kV
Base voltage for circuit A = (1/10)×138=13.8 kV
Base voltage for circuit C = (1/2)×138=69 kV
Per-unit impedance calculation in circuit C:
Base impedance of circuit C, Ω=
×
== 476
10000
10002)69(
base,
2
base,
base,
C
S
C
V
C
Z
Per-unit impedanc of load in circuit C: Ω=== 63.0
476
300
base,
actual,
pu,
C
Z
C
Z
CZ
Per-unit impedance calculation in circuit B:
Base impedance of circuit B, Ω=
×
== 1900
10000
10002)138(
base,
2
base,
base,
B
S
B
V
B
Z
Turns ratio for transformer B – C: abc = 2
Impedance of load referred to B = abc
2
R=22
×300= 1200 Ω
Per-unit impedanc of load referred to circuit B: Ω=== 63.0
1900
1200
base,
actual,
pu,
B
Z
B
Z
BZ
Per-unit impedance calculation in circuit A:
Base impedance of circuit A, Ω=
×
== 19
10000
10002)8.13(
base,
2
base,
base,
A
S
A
V
A
Z
Turns ratio for transformer A – B: aab = 1/10=0.1
Impedance of load referred to A = aab
2
abc
2
R=22
×(0.1)2
×300= 12 Ω
Page-05

7. DMAM
Per-unit impedanc of load referred to circuit A: Ω=== 63.0
19
12
base,
actual,
pu,
A
Z
A
Z
AZ
From above calculation, it is clear that the per-
unit impedance of the load referred to any part of
the system is same since the selection of base in
various parts of the system is determined by the
tns ratio of the transformer.
Fig. 6.20 is the required impedance diagam with
impedances marked in per-unit.
Calculation of voltage regulation:
Voltage at load = 957.0
69
66
= pu
Load current = 52.1
63.0
957.0
= pu
Input voltage = 995.0274.0957.0957.0)08.01.0(52.1 =+=++ jjj pu
Regulation = %97.3100
957.0
957.0995.0
=×
−
Three Phase Systems and Per Unit Calculations
The base values are related through the same relationships as the actual quantities:
base
3
base
base3
3/
base
base
3
base
base
base Y
Z
L
I
LN
V
L
I
LN
V
I
LL
V
Z ===
∆
=
∆
The per unit values of the ∆ connected impedance and the Y connected impedance are
base
pu
∆
∆=
∆ Z
Z
Z
baseY
Y
puY Z
Z
Z =
From that it is easy to show that the per unit value for the ∆ connection is the same as the per unit
value for the Y connection.
puY
baseY
Y
baseY
3
Y
3
base
pu
Z
Z
Z
Z
Z
Z
Z
Z ===
∆
∆=
∆
Let us consider an example of Y-Y transformer composed of three single phase transformers each
rated 25 MVA, 38.1/3.81 kV. The rating as a three-phase transformer is, therefore, 75 MVA, 66/6.6
kV as shwon in Fig. 6.21.
The high voltage side the impedance measured from line to neutral is
2
LT(LL)
HT(LL)
LT
2
LT(LN)
HT(LN)
LTHT ⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
=
V
V
Z
V
V
ZZ
Three Phase Systems and Per Unit Calculations
Page-06

8. DMAM
Ω=⎟
⎠
⎞
⎜
⎝
⎛
=⎟
⎠
⎞
⎜
⎝
⎛
= 60
2
6.6
66
6.0
2
81.3
1.38
6.0
HT
Z
For Fig. 6.22(a) Ω=⎟
⎠
⎞
⎜
⎝
⎛
= 180
2
81.3
66
6.0
HT
Z
For Fig. 6.22(b) Ω=⎟
⎠
⎞
⎜
⎝
⎛
= 180
2
2.2
1.38
6.0
HT
Z
Example 6.6 [1, p.151] The transformers rated 25 MVA, 38.1/3.81 kV are connected Y-∆ as shown in
Fig. 6.22(a) with the balanced load of three 0.6 Ω, Y-connceted resistors. Choose a base of 75 MVA,
66 kV for high tension side of the transformer and specify the base for the low tension side. Determine
the per-unit resistance of the load on the base for the low-tension side. Then determine the load
resistance RL referred to high-tention side and the per-unit value of this resistance on the chosen base.
Solution: therating of the transformer as a three-phase bank is 75 MVA, 66Y/3.81∆ kV. So base for
the low-tension side is 75 MVA, 3.81 kV.
Base impedance on the low-tension side,
( ) Ω==
⎥⎦
⎤
⎢⎣
⎡
= 1935.0
75
281.3
MVAbase
2
)(
KVbase
(LT)base
LTLL
Z
Load resistance in per-unit, 1.3
0.1935
6.0
(LT)base
(LT)actualL,
pu(LT)L,
===
R
R
R pu
Page-07

9. DMAM
Base impedance on the high-tension side,
( ) Ω==
⎥⎦
⎤
⎢⎣
⎡
= 1.58
75
266
MVAbase
2
)(
KVbase
(HT)base
HTLL
Z
Actual load resistance in high-tension-side,
Ω=⎟
⎠
⎞
⎜
⎝
⎛
=
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
= 180
2
81.3
66
6.0
2
LT(LL)
HT(LL)
(LT)actualL,(LT)actualL, V
V
RR
Load resistance in per-unit, 1.3
58.1
180
(HT)base
(HT)actualL,
pu(HT)L,
===
R
R
R pu
Changing the Base of Per-Unit Quantities
When only one component, such as transformer, is considered, the nameplate ratings of that
component are usually selected as base values. When several components (such as generator,
transformer, transmision line, load etc.) are involved in circit, however, the system base values may be
different from the name plate ratings of any particular component. It is then necessary to convert the
per-unit impedance of a component from its nameplate rating to the system base values. To convert a
per-unit impedance from “old or given” to “new” base values, use
newbase,
oldbase,oldpu,
newbase,
actual
newpu, Z
ZZ
Z
Z
Z == (3.3.10)
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
=
oldbase,
newbase,
newbase,
oldbase,
newpu,
2
oldpu,
S
S
V
V
ZZ
If newbase,oldbase, VV = then
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
=
oldbase,
newbase,
newpu, oldpu,
S
S
ZZ
If newbase,oldbase, SS = then
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
=
newbase,
oldbase,
newpu, oldpu,
V
V
ZZ
(3.3.11)
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
old
kVAvoltage,base
new
kVAvoltage,base
new
kVvoltage,base
old
kVvoltage,base
newpu,
2
oldpu,ZZ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
old
MVAvoltage,base
new
MVAvoltage,base
new
kVvoltage,base
old
kVvoltage,base
newpu,
2
oldpu,ZZ
(2.52)
Example 2.5: The reactance of a generator is given as 0.25 pu based on the generator nameplate rating
of 18 kV, 500 MVA. Calculate the new pu of the reactance if the new base are as 20 kV, 100 MVA.
0405.0
500
100
20
18
25.0
newpu,
2
=⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=X
Advantages of per-unit quantity
Changing the Base of Per Unit Quantities
Page-08

10. DMAM
1. Normalized values against a base have more uniform values in the same system. Thus, it is
easier to spot errors. For example, 1000% p.u. will trigger a warning and thus is worth looking
into for potential errors.
2. Power base is the same through the system; voltage bases are changing according to the
transformer turn ratio. Accordingly, the transformer equivalent circuit can be simplified by
using the pu quantity. The ideal transformer winding can be eliminated, such that voltages,
currents, and impedances and admitances expressed in pu do not change when they are refered
from one side to the other side of transformer. the conversion complications are absorbed by
the base relationships.
3. ∆ base and Y base quantities have the same p.u. values in these two different bases; again the
conversion complications are absorbed by the base relationships.
4. Three-phase and single-phase quantities have the same p.u. values in these two different bases;
again the conversion complications are absorbed by the base relationships.
5. Abnormal operating conditions can be easily spotted from the p.u. values.
6. The pu impedances of electrical equipment of similar type usually lie in a narrow neumarical
range when the equipment ratings are used as base values.
7. The pu system allows us to avoid the possibiliy of making serious calculation error.
8. Manufacturers usually specify the impedances of machines and transformer in pu or percent of
nameplate rating.
9. The bases for different sections in the system can be calculated, carefully verified and stored
once for all. When generation/load changes, we only need to change its per unit values and
calculated the per unit values accordingly. The conversion will become more reliable. The
advantages are more pronounced for large system applications and systems that have many
load/generation changes.
Example 1 [1, p.159]
A power system consists of one synchronous generator and one synchronous motor connected by two
transformers and a transmission line as shown in the following figure. Create a per-phase, per-unit
equivalent (simplified impedance) 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, X = 0.05 pu;
T2 ratings: 50 MVA, 120/14.4 kV, R = 0.01 pu, X = 0.05 pu;
M ratings: 50 MVA, 13.8 kV, R = 0.1 pu, Xs = 1.1 pu;
L1 impedance: R = 15 Ω, X = 75 Ω.
Solution:
3Region
2Region
1Region
−−−−−−−−==
−−−−−−−−==
−−−−−−−−=
kV2.13
120
4.14
kV110
8.13
110
kV8.13
2,3,
1,2,
1,
basebase
basebase
base
VV
VV
V
Advantages of Per Unit Quantity
Page-09

11. DMAM
3Region
2Region
1Region
−−−−−−−−Ω===
−−−−−−−−Ω===
−−−−−−−−Ω===
374.1
100
)2.13(
121
100
)110(
904.1
100
)8.13(
2
,3
2
,
3,
2
,3
2
,
2,
2
,3
2
,
1,
MVA
kV
S
V
Z
MVA
kV
S
V
Z
MVA
kV
S
V
Z
base
baseLL
base
base
baseLL
base
base
baseLL
base
φ
φ
φ
Base impedance calculation in region 1 is not required, since the rated value of region 1 is considered
as base value and resistance and reactance values are given in pu.
For G1: R = 0.1 pu, Xs = 0.9 pu
For T1: R = 0.01 pu, Xs = 0.05 pu
For L1: pu124.0
121
15
1 ==LR ; pu62.0
121
75
1 ==LX
Base impedance calculation in region 3 is not required, since the given data in this region is given in
pu. Thus, the values of pu are needed to update.
For T2:
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
=
oldbase,
newbase,
newbase,
oldbase,
newpu, oldpu,
S
S
V
V
ZZ
pu238.0
50
100
2.13
4.14
01.0
newpu,
=⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=R ; pu119.0
50
100
2.13
4.14
05.0
newpu,
=⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=X
For M2: pu219.0
50
100
2.13
8.13
01.0
newpu,
=⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=R ; pu405.2
50
100
2.13
8.13
1.1
newpu,
=⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=X
Page-10

12. DMAM
Example 6.10 [1, p.159] A 300 MVA, 20 kV three-phase generator has a subtransient reactance of
20%. The generator supplies a number of synchronous motors over a 64 km (4- mi) transmission line
having transformers at both ends, as shown on the one-line diagram of Fig. 6.29. The motors, all rated
13.2 kV, are represented by just two equivalent motors. The neutral of one motor M1 is grounded
through reactance. The neural of the second motor M2 is not connected to the ground (an unusual
condition). Rated inputs to the motors are 200 MVA and 100 MVA for M1 and M2, respectively. For
both motors X” =20%. The three-phase transformer T1 is rated 350 MVA, 230/20kV with the leakage
reactance of 10%. Transformer T2 is composed of three single-phase transformer each rated 127/13.2
kV, 100 MVA with leakage reactance of 10%. Series reactance of the transmission line is 0.5 Ω/km.
Draw the reactance diagram with all reactance marked in per-unit. Select the generator rating as base
in the generator circuit.
Solution:
Base MVA, Sbase, 3φ = 300 MVA
Base kV, VbaseLL = 20 kV
Generator:
Given, A 300 MVA, 20 kV three-phase generator has a subtransient reactance of 20%
Base MVA, Sbase, 3φ = 300 MVA
Base kV, VbaseLL = 20 kV
Transformer T1:
Given, transformer T1 is rated 350 MVA, 230/20kV with the leakage reactance of 10%
Base MVA, Sbase, 3φ = 300 MVA
Base kV, VbaseLL = 20 kV
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
old
MVAvoltage,base
new
MVAvoltage,base
2
new
kVvoltage,base
old
kVvoltage,base
oldpu,newpu,
XX
0857.0
350
300
2
230
230
1.0 =⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=X pu
Fig. 6.29 One-line diagram for Example 6.10.
Transmission Line:
Given, series reactance of the transmission line is 0.5 Ω/km.
Base MVA, Sbase, 3φ = 300 MVA
Base kV, VbaseLL = 230 kV (since T1 is rated 230/20 kV)
Reactance of transmission line, Ω×= 640.5X
Page-11

13. DMAM
Base impedance of transmission line, Ω=== 3.176
300
2)230(
2)(
MVAbase
LL
kVbase
base
Z
Per-unit impedance of transmission line, 1815.0
3.176
645.0
=
×
==
base
Z
X
X pu
Transformer T2:
Given, transformer T2 is composed of three single-phase transformer each rated 127/13.2 kV, 100
MVA with leakage reactance of 10%
The three-phase rating is 3×100 = 300 MVA
Line to line voltage ratio is √3× 127/13.2 = 220/13.2 kV
Base kV, VbaseLL = 230×(13.2/220)=13.8 kV
0915.0
300
300
2
13.8
13.2
1.0 =⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=X pu
Motor M1:
Given, the neutral of one motor M1 is grounded through reactance.
Rated inputs to the motors (M1 and M2) are 200 MVA and 100 MVAwith X” =20%.
Base MVA, Sbase, 3φ = 300 MVA
Base kV, VbaseLL = 230×(13.2/220)=13.8 kV
2745.0
200
300
2
13.8
13.2
2.0
newpu,
=⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=X pu
Motor M2:
Given, the neutral of one motor M2 is not grounded.
Rated inputs to the motors (M1 and M2) are 200 MVA and 100 MVA with X” =20%.
Base MVA, Sbase, 3φ = 300 MVA
Base kV, VbaseLL = 230×(13.2/220)=13.8 kV
5490.0
100
300
2
13.8
13.2
2.0
newpu,
=⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=X
Fig. 6.30 Reactance diagram for Example 6.10. Reactances are in per unit n the specified base.
Page-12

14. DMAM
Example 6.11 [1, p.160] If the motors M1 and M2 of Example 6.10 have inputs of 120 and 60 MW
respectively at 13.2 kV, and both opeate at unity power facto, find the voltage at the terminal of the
generator.
Solution:
Together the motors take 180 MW. Or (180/300) = 0.6 pu
Therefore with V and I at the motors in pu |V||I| = 0.6 pu
And since °∠== 09565.0
13.8
13.2
V pu
°∠== 06273.0
9565.0
0.6
I
At the generator
°∠=+=+++= 2.139826.02250.09565.0)0857.01815.00915.0(6273.09565.0 jjjjV pu
The generator terminal voltage is
kV
t
V 65.20199826.0 ×=
Example 3: Prepare a per phase schematic of the system shown in the figure and show all impedances
in per unit on a 100 MVA, 132 kV base in the transmission line circuit
G1 : 100 MVA, 11 kV, X= 0.15 p.u
G2 : 200 MVA, 13.8 kV, X= 0.2 p.u
T1 : 120 MVA, 11/132 kV, X= 0.1 p.u
T2: 250 MVA, 13.8/161 kV, X= 0.1 p.u
Load: 250 MVA, 0.8 Lagging, operating at 132 kV
Determine the per unit impedance of the load for the following cases: (i) load modeled as a series
combination of resistance and reactance, and (ii) load modeled as a parallel combination of resistance
and reactance.
Solution:
Transmission Line: Base kV in the transmission line =132 kV
Base impedance in the transmission line is = (132)2
/ 100 = 174.24 Ω
quantityofvalueBase
quantityActual
quantityunitPer =−
Ztransmission-line = (50+j100) / 174.24 = 0.287 + j1.1478
Ztransmission-line = (25+j100) / 174.24 = 0.1435 + j0.5739
Generator G1: Base kV in generator circuit G1 =132 × 11/132 = 11 kV
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
⎟⎟
⎟
⎠
⎞
⎜⎜
⎜
⎝
⎛
=
oldbase,
newbase,
newbase,
oldbase,
newpu, oldpu,
S
S
V
V
ZZ
X = 0.15 × (11/11)2
× (100/100) = 0.15 p.u
Page-13

15. DMAM
Transformer T1:
X = 0.1 × (11/11)2
× (100/120) = 0.0833 p.u
Transformer T2: X = 0.1 × (13.8/11.31)2
× (100/250) = 0.05955 p.u
Generator 2: Base kV in generator circuit G2 =132 × 13.8/161= 11.31 kV
X = 0.2 × (13.8/11.31)2
× (100/200) = 0.1489 p.u
Load: The base impedance in the load circuit is same as the base impedance in the base impedance in
the transmission line.
Load is specified as: 250 MVA, 0.8 p.f lagging, 132 kV
cosθ = 0.8; θ = cos-1
(0.8) = 36.87o
; sinθ = 0.6
So the load is S = VI( cosθ + jsinθ) = 250 × (0.8 + j0.6) = 200 + j150
Series Connection:
S = VI*
= V(V*
/Z*
) = V2
/ Z*
Z*
=V2
/S
Zload
*
=(132)2
/ (200+j150) = 55.7568 - j41.8176Ω
Zload = 55.7568 + j41.8176 Ω
Zload,pu = (55.7568 + j41.8176 )/ 174.24 =0.32 + j0.24 p.u
Parallel Connection:
As it is parallel, the impact effect on the load is going to be separated
Rload =V2
/P= (132)2
/200 = 87.12 Ω;
Xload =V2
/Q= (132)2
/150 = 116.16 Ω;
Rload,pu = 87.12/ 174.24 = 0.5 pu
Xload,pu = 116.16/ 174.24 = 0.66 pu
Zload,pu = 0.5 + j0.66 pu
Page-14

16. DMAM
Practice Math
Element of Power System Analysis 4th
Edition (Stevenson):
Example: 6.4; 6.5; 6.6; 6.10; 6.11
Exercise: 6.13; 6.15; 6.16
Power System Analysis and Design (Glover and Sharma):
3rd Edition
Example: 3.3; 3.4; 3.7
Exercise: 3.18; 3.19; 3.29; 3.30; 3.33; 3.34
4th Edition
Example: 3.3; 3.4; 3.7
Exercise: 3.23; 3.24; 3.28; 3.41; 3.42; 3.45; 3.46
References
[1] Willaim D. Stevenson, Elements of Power System Analysis, Fouth Edition, McGraw-Hill
International Editions, Civil Engineering Series, McGraw-Hill Inc.
[2] John J. Grainger, William D. Steevnson, Jr., Power System Analysis, McGraw-Hill Series in
Electrical and Conputer Engineering, McGraw-Hill Inc.
[3] J. Duncan Glover, Mulukutla S. Sharma, Thomas J. Overbye, Power System Analysis and Design,
Fouth Edition (India Edition), Course Technology Cengage Learning
[4] V. K. Mehta, Rohit Mehta, Principles of Power System, Multicolor Illustrative Edition, S. Chand
and Company Limited
***************************
Practice Maths for Class Test and Mid term Exams
Page-15