CHAPTER 1: PRESENTATION OF POWER SYSTEM COMPOMENTS
Unit - 1_________________________________________________________________
■ Introduction to single phase representation of a balanced three phase power system
network or component
■ Single line diagram (SLD) of power system network
■ Representation or model of
(i)
synchronous machine as a constant voltage source in series with its
impedance or reactance if resistance is neglected
(ii)
three phase two-winding transformer as an impedance
(iii)
transmission line as -model
■ Representation of power system loads: P-V and Q-V characteristics of loads are
drawn and then it is suitable modeled generally in following three ways
(i) constant power load model : basically this model is used for power system
load flow study where both active and reactive powers specified and constant
(ii) constant current load model: here load is represented as constant
P  jQ
I 
   
V*
(iii)
constant impedance load model: this is quite used in stability study as the load
specified in MW and MVAR at nominal voltage is used to compute the load
impedance
V
VV *
 V 
Z  

I
P  jQ P  jQ
■ Graphical representation of above three load in P-V curve
Constant impedance
Constant current
P
Constant power
V
Unit - 2________________________________________________________________
■ Introduction to PER UNIT (pu) system:
In power system analysis instead of using actual values of quantities, it is usual to
express them as fractions of base values, such as rated or full load values. These
fractions are called per unit values and the basic per unit scaling equation may be
written as:
per unit value 
actual value
,
base value
where the base value have the same unit as the actual value, forcing the per unit value
to be dimensionless. Note that the base value is always a real number where as the
actual value may be complex.
■ Normally in power system, volt-ampere and voltage are chosen as base variable and
current and impedance are derived from these for entire system. i.e. VAb or MVAb
and Vb or kVb
■ Per unit system applied to 1- circuits:
MVAb  kVb  kI b
 Zb 
kVb
kVb 2

kI b MVAb
Z actual
MVAb
 Z actual 
Zb
kVb 2
■ Per unit system applied to 3- circuits:
MVAb  3 kVb  kI b
 Z pu 
 Zb 
kVb 2
ohm per phase 
MVAb
3 kI b
kVb
 Z pu 
Z actual
MVAb
 Z actual 
Zb
kVb 2
NOTE: kVb is line to line voltage and MVAb is total three phase complex power.
■ Criterion for selection of base values:
■ Change in base value to determine new pu value:
Z pu (new )  Z pu (old )
MVAb (new )
MVAb (old )

kVb (old )2
kVb (new )2
Unit - 3________________________________________________________________
■ Per unit impedance of a two winding transformer:
To prove that per unit impedance of a two winding transformer referred to primary is
equal to that of secondary. i.e.
Z pu1  Z pu1
■ Rules for use of pu systems:
1. selection of proper base MVA and base kV at starting point
2. Keep base MVA throughout constant and find out per unit values of all the
impedances
3. Note when transformer is encountered that base kV is different at its HV side
and LV side however the same pu impedance is on both side. So when
crossing the transformer, find out a new base kV at the opposite side
(neglecting loading effect of transformer) i.e.
N
kVb ( HV / LV )  2  kVb ( LV / HV )
N1
■ When load is given in form of MW and MVAR at specified kV, how to express such
a load as per unit impedance is as follows:
V
VV *
 V 
Z actual  

I
P  jQ P  jQ
MVAb
 Z pu  Z actual 
kVb 2
■ Problem on PU impedance or reactance diagram:
Convert SLD shown in Fig.1 into per unit reactance diagram.
T
T
M
G
M
Fig.1- single line diagram of power system
Ratings given:
Generator
: 10 kV, 100 MVA, X d = 15 %
Transformers
: 11/110 kV, 150 MVA, X=10%
Transmission line: 10 km length, impedance is (4 + j 15)  per km
Motors
: 10 kV, 100 MVA, X d = 15 %
■ Advantages and disadvantages of per unit system