1.

INTRODUCTION TO POWER SYSTEMS

The structure of the electric power or energy system is very large & complex.
Nevertheless, it can be divided into five basic stages/components/subsystems.

Energy Energy Transmission Distribution Load

source converter system system (Energy
(Fuel) (Generator) sink)

• Energy source may be - coal, gas or oil (fossil fuel)

- fissionable material (nuclear)
- water in a dam (hydro)
- renewable sources e.g. solar, wind, tidal, biofuels,
geothermal
• Generator that transforms non-electrical energy to electrical energy; usually
rotating-machinery type; power output from few kilowatts to few thousand MW;
voltage levels 440 V to 25 kV.
• Transmission system transports generated energy from generating stations to
major load centres; voltage levels 115 kV to 765 kV (less than 138 kV usually
referred to as sub-transmission system); overhead lines & underground cables.
1
• Transformers used to change voltage levels (to high over transmission system &
low over distribution system, etc)
• Distribution system transports transmitted energy from transmission system to
users; voltage levels typically 1 kV to 33 kV.
• Loads : industrial, commercial, residential, farm, etc.

2
 Schematic of a Coal-fired Generating Station
2. REVIEW OF SINGLE-PHASE & THREE-PHASE CIRCUITS
Before we embark on a detailed study of the generators & lines, it would help to
quickly review the basic concepts of :
• 3-phase systems (as most power systems operate as 3- systems)
• Single-phase equivalent of 3- systems (since for balanced 3- systems, we can
use one phase of the star(Y)-equivalent)
• Per unit (p.u.) systems

3
 Recall that all the above have been covered elsewhere (EE2005/EE3010), but
since we will be using them extensively, it is worth taking a look back in time!
 First, a review of the one-phase system.

I Load represented
V Z by impedance Z

Let V = V0o (reference) & Z = Z, where  is power factor angle.

Z = R + jX (assuming inductive load)
R = Z cos  & X = Z sin 
V V 0
o
V
I=     I 
Z Z  Z
Apparent power S (VA, kVA, MVA and S= VI)
 S = VI* = (V 0o)(I) =VI = S
= VI cos  + jVI sin 
= P + jQ
where P : Real (active) power (W, kW, MW)
Q: Imaginary (reactive) power (var, kvar, Mvar)

4
 For  = 0o (unity power factor load)

I V Z=R S=P
= VI
 For lagging P.F. loads,
jX jQ
V
 Z S

 o
90  o
90
I R P

 For leading P.F. loads,

I R P
 

Z S
 -jX -jQ
V

5
What are Real & Reactive Powers?

By convention,

• Lagging vars  Inductive load  Q (+VE)

• Leading vars  Capacitive load  Q (–VE)

P = Real or active power, consumed/dissipated in resistance R

= VI cos  = (IZ)Icos  = I2Z cos 
= I2R

Q = Reactive power absorbed by inductive reactance X (or generated by

capacitive reactance)
= VI sin  = IZIsin  = I2Z sin 
= I2X

Represents the energy exchange between source & reactor/capacitor i.e. represents
stored energy.

6
Convention for Real & Reactive Powers
1) Generators

I S = VI* = P + jQ

V  If P is positive, real power is delivered/supplied

 If P is negative, real power is absorbed by source
 If Q is positive, reactive power is supplied/delivered
 If Q is negative, reactive power is absorbed by source
2) Loads

I S = VI* = P + jQ

 If P is positive, real power is absorbed by the load

V  If P is negative, real power is supplied by the load
 If Q is positive, reactive power is absorbed by the load (lagging
PF load  inductive load)
 If Q is negative, reactive power is supplied by the load (leading
PF load  capacitive load)

7
 Inductive load absorbs reactive power.
 Capacitive load generates reactive power, or absorbs negative reactive power.
Review Exercises (1 & 2)
It
(1) I1 I2
o
Z2 = 5030 
o
1000 V Z1 =
1010 
o

Find the a) Real power P (1158.014 W)

b) Reactive power Q (273.657 VAr lag)
c) Apparent power S (1189.91 VA)
d) Power factor of the Network (0.973 lag)
(2) It I2
10  I1 20 

2000 V
o
j10  100 F
50 Hz supply

Find the Network a) Power factor (0.7 lag)

b) Real power (1601.16 W)
c) Reactive power (1628.2 VAr)
d) Apparent power (2283.6 VA)
8
Review of Balanced 3-phase Systems
• Mesh or Delta () connection
• Star or Wye (Y) connection
Usually (By default if not specified), 3- systems specified in terms of
 total 3- S, P, Q
 line-to-line voltages
 line currents
Remember : cos  = power factor
= power factor angle, where  is always between phase
voltage & phase current of the same phase.
Delta Connection
Ia a
Zab = Zca = Zbc = Z
Vab Vbc Za b Zc a
Iab = , Ibc = &
Zab Zbc
V Ib b c
Ica = ca Zbc
Zca Ic
`
9
If Vab = Reference = Vab0o V
Then Vbc = Vab-120o V
& Vca = Vab120o V
Line currents Ia = Iab – Ica
Ib = Ibc – Iab & Ic = Ica - Ibc

 Ia = 3Iab--30o ILINE = 3IPHASE

Ib = 3Ibc-120--30o 
Ic = 3Ica120--30o VLINE= VPHASE
Apparent power S = 3 VPHASE I*PHASE
= 3 VPHASEIPHASE
1
= 3 VLINE ILINE
3
= 3 VLINEILINE VA

 S = 3VLINEILINEcos  + j3VLINEILINEsin 
= P + jQ

10
Phasor Diagram : (Vab : Reference)

Review Exercise (3)

Ia a
(3) 220-V
3- Z Z
System
Ib b
c
Z
Ic
Z = 30-15o  11
Find phase & line currents. Sketch phasor diagram. (Ia = 12.7-15o A)
Y-connection
Ia
Van a
Ia = Ian =
Z Z Van
Van
= 
Z n
Z Z = Z
Reference = Van = Van0
Ib b c
Vab = Van + Vnb
Vbc = Vbn + Vnc
Vca = Vcn + Vna Ic
Here ILINE = IPHASE
& VLINE = 3VPHASE
S = 3 VPHASE I*PHASE
= 3 VPHASE 0  I PHASE 
1
= 3 VLINEILINE
3
= 3VLINEILINE
= 3VLINEILINE cos  + j3 VLINEILINE sin 
= P + jQ
12
Review Exercise (4)

VAB = 208 30 V

o
(4)
Load Z = (4 + j3)  in each phase.

 Find the line currents & power dissipated in the loads (total power).
 Also find Q & S.
[Y-connected system, phase seq. A-B-C]

Ia = 24.02 -36.87 A, P = 6923 W, Q = 5192.17 VAr lag, S = 8653.7VA

o

Review of Transformer’s Equivalent CKT.

 Equivalent circuit (per phase)
X '2 R '2 I '
I1 R1 X1 2
Io
'
V1 Xm Rc V2

(All quantities referred to the primary)

 Approximate equivalent circuit (neglecting magnetizing current)

13
 ' '
I1 X1 R1 X2 R 2 I'
2

V1 '
V2

which reduces to
'
I1 X R I2

'
V1 V2

where R = R1 + R '2 = total equivalent resistance & X = X1 + X '2 = total equivalent

reactance.
 Frequently R is neglected, in which case

14
 '
I1 X I2

'
V1 V2

In other words, the transformer can be replaced by an equivalent impedance, or just

an equivalent reactance.
3. PER-UNIT (P.U.) SYSTEM
• Power transmission lines are operated at very high voltage levels (kilovolts). Due
to the large amount of power transmitted, megawatts & megavoltamps are
commonly-used terms!
• It therefore would be more meaningful to scale down all physical values of , A,
kV, MVA, MW using scaling factors called based values. For example, if a base
voltage of 100 kV is selected, then physical system voltages of 80 kV, 110 kV &
100 kV become 80 , 110 & 100 per unit respectively!
100 100 100

15
• The per-unit value of any quantity is defined as the ratio of the actual quantity to
an “arbitrarily” chosen value (base or reference) of the same dimensions.
physical quanity
 Quantity in per unit =
base value
physical quantity
& Quantity in percent =  100
base value
• Percent system should be used with caution (due to mult. factor of 100)
• Per-unit system is preferred in power system calculations as it offers the
following advantages :

Advantages of Per-unit System

• Analysis is greatly simplified, e.g. all impedances of a given equivalent CKT can
be directly added without considering system voltages.
• Use of “3” is eliminated! The base values account for these easily.
• Manufacturers of electrical equipment usually specify the impedance in per unit
or percent of nameplate ratings.

16
• Electrical machines & transformers have widely varying internal impedances with
size & rating. However, it turns out that in the p.u. system, these impedances fall
within a fairly narrow range. Hence, if the actual impedance of a machine is not
known, its per unit value can be easily assigned!
• Circuit analyst is relieved of the worry of referring quantities to one side or other
side of transformer, especially in large networks containing many transformers of
different turns ratios.
Eliminates the possible cause of making serious calculation mistakes!
• P.U. values are more convenient in simulating machine systems on digital
computers.
Significant advantage : Per unit impedance of transformer is the same on both sides
of the transformer!
Per Unit (p.u.) Quantities in 3-phase Power System
 There are 4 base values
1. Power base Sb, usually in MVA.
2. Impedance base Zb, usually in OHMS.
3. Current base Ib, usually in A.
4. Voltage base Vb, usually in kV.
17
 In 3-phase systems, usually
 S, P, Q are three-phase powers in MVA, MW & MVAr respectively
 Voltages considered are line-to-line values
 Currents considered are line values
 Impedances considered are phase values of equivalent star (Y) configuration
(This means that all impedances in  must be converted to equivalent Y
value in , before converting to its p.u. value)
 The 4 base values (Sb, Zb, Ib & Vb) are related as follows :
Vb I b
Sb  3 MVA (1)
1000
where Vb is line-to-line base voltage in kV & Ib is the line current in AMPS.
Sb  1000
 Ib  (2)
3  Vb

(Vb / 3)
Next, Zb   1000  (3)
Ib

18
 Vb
Note : is the phase voltage of the equivalent Y system.
3
Substituting (2) in (3), we get :
(Vb / 3)
Zb   1000
 Sb  1000 
 
 3  V b 

Vb 2
 Zb   (4)
Sb
• From the above, it is clear that we need to select only 2 base values, instead of all 4.
For example, if Sb & Vb are selected, then Ib & Zb can be calculated using equations
(2) & (4) respectively!
• Vb (and consequently Zb & Ib too) changes from transformer primary to secondary
as follows :
Vb (primary) No. of turns (primary)

Vb (sec ondary) No. of turns (secondary)
N1
 (line  to  line turns ratio, a)
N2
19
• Per unit values of Z, R, X & I are the same on either side of the transformer, but the
actual values are not!
 Per unit quantities can now be evaluated as :
Z() I(AMPS)
Zp.u.  ; I p.u.  ;
Z b () I b (AMPS)
V(kV) P(MW)
Vp.u.  ; Pp.u. 
Vb (kV) Sb (MVA)
 If necessary, actual voltages/currents/ohms etc. can also be obtained as :
Actual value = Base value  p.u. value
General Guidelines for Obtaining P.U. Values
Objective : To reduce the number of computations by selecting suitable values for
Vb & Sb.
• Base MVA (Sb) is the same for all parts of the system. Normally Sb in MVA is used.
• Base kV (Vb) is selected in one part of the system; for other parts base kV is
obtained according to the line-to-line voltage ratios of transformers.

20
• Base impedances will be different in different parts of the system.
• In general, the p.u. (or %) impedances of electrical equipment are specified in
terms of their own MVA & kV ratings. These values need to be converted to the
system bases selected in (1) & (2) above. This conversion is done using the
following formula :
Z
Zp.u. NEW  ACTUAL
Zb NEW

ZACTUAL
Zp.u. OLD 
Zb OLD

Manufacturer’s base (Equipment base)
where : Zb NEW = Base impedance selected in that part of the system where this
component is placed
(Vb NEW ) 2
=  these are base values in that section
Sb NEW
(Vb OLD ) 2
& Zb OLD   these are component bases/ratings
Sb OLD

21
 Zp.u. NEW Zb OLD (Vb OLD ) 2 / Sb OLD
  
Zp.u. OLD Zb NEW (Vb NEW ) 2 / Sb NEW
2
V  S 
 Zp.u. NEW  Zp.u. OLD   b OLD    b NEW  (*)
 Vb NEW   Sb OLD 
Example 1 : A component rated for 13.2 kV, 30 MVA & with Z = 0.2 p.u. (on its
own ratings) is placed in a power system portion where Vb = 13.8 kV
& Sb = 50 MVA. What is the new p.u. Z of the component?
Here, Sb NEW = 50 Sb OLD = 30
Vb NEW = 13.8 Vb OLD = 13.2
Zp.u. OLD = 0.2 Zp.u. NEW = ?
2
13.2   50 
 Zp.u. NEW  0.2        0.306 p.u.
13.8   30 
• If a transformer impedance is given in p.u., then this value is based on the MVA
rating of the transformer.
 Use equation in (*) above to convert this given p.u. to the new p.u. on the
system base kVA, Sb.

22
• If the Sb (base MVA) is not specified (and that is usually the case), the system
component that has the largest MVA rating is chosen to give us the base MVA.
Occasionally, a nice round number such as 100 MVA is selected as the base MVA!
Example 2 : For the power system shown below,
(i) Find appropriate voltage bases by selecting Vb = 18 kV at the
generator terminals.
(ii) Find all impedances in p.u. Use Sb = 100 MVA.
TR 2
b
TR 1 400 MVA
250 MVA 345/230 kV
18 kV GEN a
X = 18% TR 3 X = 10%
250 MVA c
18/345 kV 200 MVA
X = 15% 345/138 kV
X = 10%
Solution : Given Vb, GEN = 18 kV (generator circuit)
Base voltage at point a = Vb, a
345
= Vb, GEN   345 kV 23
18
 230
Base voltage at point b = Vb, b  Vb, a  = 230 kV
345
138
Finally, base voltage at point c = Vb, c  Vb, a  = 138 kV
345
Next, assuming a power base of Sb = 100 MVA, let us find all the impedances in p.u.
2
V  S
ZGEN,NEW p.u.  ZGEN,OLD p.u.   b OLD,GEN   b NEW
 Vb,GEN  Sb OLD,GE
2
 18   100 
 0.18        0.072 p.u.
 18   250 
2
V   S 
ZTR1,NEW p.u.  ZTR1,OLD p.u.   b OLD,TR1    b NEW 
 Vb,GEN   Sb OLD,TR1 

Calculated on the primary side
2
18   100 
 0.15     
18   250 
 0.06 p.u.
24
 2
 Vb OLD,TR2   Sb NEW 
ZTR 2,NEW p.u. = ZTR2,OLD p.u.     
 V b,a   Sb OLD,TR2 

Calculated on the primary side

2
 345   100 
 0.10   
 345   400 

 0.025 p.u.

2
V   100 
ZTR3,NEW p.u.  ZTR3,OLD p.u.   b OLD,TR3    
 Vb,a   200 

Calculated on the primary side

2
 345   1 
 0.10     
 345  2
 0.05 p.u.

25
 Vb = 18 kV Vb = 345 kV Vb = 230 kV
b

X = 0.072 p.u. X = 0.025 p.u.

GEN a Vb = 138 kV
X = 0.06 p.u. c

X = 0.05 p.u.
Vb = 345 kV
Example 3 :
Transformer
Load
x y Feeder c 10 MVA
Supply in ZF = (0.726 + j0.847)  0.8 pf lag
per phase 11 kV
60 MVA
33/11 kV
Y/
ZT = (0.2723 + j0.726)  per phase (Referred to H.V. side)
Calculate the input & output line voltages of the 3- transformer i.e. voltages at x and y.
Solution : Let base MVA = Sb = 60 MVA (constant throughout the system)`

26
Transformer H.V. side
Base kV, Vb x = 33 kV (assumed)
Vb x 2 332
Zb primary of trans. = Zb x =   18.15   Zb pri
Sb 60
ZT actual pri 0.2723  j0.726
 ZT p.u.    0.015  j0.04  0.042769.44 p.u.
Zb pri 18.15
Transformer L.V. side
11
Base kV = Vb y = Vb x =11 kV
33
(Vb y ) 2 112
 Zb y    2.017 
Sb 60
Z () 0.726  j0.847
 ZF p.u.  F  = 0.36 + j0.42 = 0.5532 49.4 p.u.
Zb y 2.017
 Total impedance
ZTOT p.u.= ZT p.u. + ZF p.u. = 0.593550.81 p.u.
SLoad = 10 MVA, 0.8 pf lag = 1036.87o MVA
27
 SLoad 1036.87
 SLoad p.u.    0.166736.87 p.u.
Sb 60
VLoad 11
VLoad  11 kV  VLoad p.u.    1.00o p.u.
V b c 11
 Load current
SLoad p.u. 0.166736.87
*
I Load p.u.    0.166736.87 p.u.
VLoad p.u. 10
 I Load p.u. = 0.1667 -36.87o p.u.
 Voltage at the L.V. side of transformer
Vy p.u. = (ILoad p.u.)(ZF p.u.) + VLoad p.u.
= (0.1667-36.87)(0.553249.4) + 10 = 1.091.05 p.u.
Actual voltage Vy = Vy p.u. x Base voltage at y = 1.09 x 11 kV ~ 12 kV
Finally, voltage at H.V. side of transformer
Vx p.u. = (ZT p.u.)(ILoad p.u.) + Vy p.u.
= (0.042769.44)(0.1667-36.87) + 1.091.05 = 1.09631.24 p.u.
Actual voltage, Vx = Vx p.u. x Base voltage at x
= 1.0963 x 33 = 36.18 kV
28
Single-line Diagram (SLD) & Impedance (Reactance) Diagram
• You have already learnt the circuit models for transformers. Very soon, you will
learn circuit models for synchronous machines & transmission lines & loads.

• Our present interest is in “how to portray the assemblage of these components to

model a power system in its entirety”.

• Since a balanced 3- system is always solved as a single-phase equivalent circuit

composed of one of the three lines & the neutral return, it is seldom necessary to
show more than 1-phase & neutral return when drawing a diagram of the system.

• Often the diagram is simplified further by omitting the completed CKT thru the
neutral, and by indicating the components by standard symbols rather than by
equiv. CKTs.

 Such a simplified diagram is called one-line diagram or single-line diagram.

29
Standard Symbols Used in SLD

 Most transformer neutrals in transmission systems are solidly grounded.

 Generator neutrals are usually grounded thru fairly high R or L to limit the
flow of current to ground during a fault (abnormal condition).

30
Example of SLD :

 The amount of information presented on the SLD depends on the purpose

for which the diagram is intended.
 As stated earlier, p.u. system is used to solve for the unknowns.

31
Impedance & Reactance Diagrams
 When the individual components of a SLD are represented by their equivalent
circuits, then the resultant drawing is called the per-phase impedance diagram.
 The impedance diagram of the system shown on the previous page is as follows :

      

 Since the shunt current of a transformer is usually insignificant compared with

the full-load current, the shunt admittance is usually omitted from the equiv.
CKT of transformer.
 Resistance is often omitted since X >> R.
 Transmission-line capacitances may also be omitted.
 Resultant diagram is called reactance diagram.

32
Note : Consistent base kVs must be specified in different sections of the system to
analyze using the p.u. system.
Example 4 : Consider the power system shown :

G1 : 20 MVA, 6.6 kV, X = 0.655 

G2 : 10 MVA, 6.6 kV, X = 1.31 
G3 : 30 MVA, 3.81 kV, X = 0.1452 
T1 : 10 MVA, 6.7/38 kV, X = 14.52  per phase (on 38 kV side)
T2 : 12 MVA, 38/3.8 kV, X = 14.52  per phase (on 38 kV side)
XL = 17.4  per phase
33
Using a 30 MVA base & a 6.6 kV base in G1 circuit, obtain the p.u. impedance diagram.
Solution : Base MVA = 30 for all sections.
Base kV in G1 & G2 CKT = 6.6 kV = Vb X
(Vb X ) 2 6.62
Zb X  ZBase X    1.452 
Sb 30
X G1 () 0.655
 X G1 p.u.    0.451 p.u.
Zb X 1.452
X G 2 () 1.31
X G 2 p.u.    0.9022 p.u.
Zb X 1.452
Base kV in T1 Secondary (38 kV side) + Line
38 38
 Vb Y  Vb X   6.6   37.4328 kV
6.7 6.7
(Vb Y ) 2 (37.4328) 2
 Zb Y    46.707 
Sb 30
X T1 Y () 14.52
 X T1 p.u.    0.3109 p.u.
Zb Y 46.707
34
 X L () 17.4
X L p.u.    0.3725 p.u.
Zb Y 46.707
X T2 Y () 14.52
Now, X T2 p.u.    0.3109 p.u.
Zb Y 46.707
Base kV on Secondary (3.8 kV) side of T2
3.8 3.8
 Vb Z  Vb Y   37.4328   3.7433 kV
38 38
(Vb Z ) 2 3.74332
 Zb Z    0.4671 
Sb 30
X G 3 ()
0.1452
 X G3 p.u.  
 0.3109 p.u.
Zb Z 0.4671
P.U. Impedance (Reactance) Diagram

35
Example 5 : (Last example on p.u./reactance diagram)

T1 T2
A B ZL = (10 + j100) C
G
ZLoad = 300 (resistive)
13.2 kV

Bus 1 Bus 2
T1 : 5 MVA, 13.2 /132 Y kV, X = 10%
T2 : 10 MVA, 138 Y/69  kV, X = 8%
Assume that generator terminal voltage magnitude is 13.2 kV (L-L). (Ignore the
generator reactance).
Find actual values of : Generator current, line current, load current, load voltage &
MVA.
Solution : Assume Sb = Base MVA = 10 MVA & Vb A = 13.2 kV
 Vb OLD T1   Sb NEW 
2 2
 X T1 p.u. NEW  X T1 p.u. OLD   13.2  10 
  S   0.10        0.20 p.u.
 V bA   b T1  13.2   5 

(Calculated on the primary side)
36
 132
Base kV in Section B  Vb B  Vb A   132 kV
13.2
2
132
 Zb B   1742.4 
10
Z () 10  j100
 ZL p.u.  L  = 0.00574 + j0.0574 = 0.057784.29o p.u.
Zb B 1742.4
Now,
 b OLD T2   Sb NEW 
2
V
X T2 p.u. NEW  X T2 p.u. OLD     
 VbB   b T2 
S

(Calculated on primary side)
2
138  10 
 0.08        0.087438 p.u.
132  10 
Next, base kV in section C
69 69
 Vb C  Vb B   132   66 kV
138 138

37
 (Vb C ) 2 662
 Zb C    435.6 
Sb 10

Zload () 300

 ZLoad p.u.   = 0.6887 p.u.
Zb C 435.6
4. IMPEDANCE DIAGRAM (P.U.)

10
I Load p.u. 
j0.2  (0.00574  j0.0574)  j0.0874  0.6887
10 10
   1.2898  26.4 o
p.u.
0.694445  j0.3448 0.775326.404 o

Note : ILoad p.u. represents different actual currents in sections A, B & C.

38
Now,
VLoad p.u.  I Load p.u.  0.68870o  0.88826  26.4 p.u.

SLoad p.u.  (VLoad p.u. )(I Load p.u. )*  1.14570o

 Load power (actual) = SLoad p.u. x Sb = 11.458 MVA 

VLoad Actual = VLoad p.u. x base kV in section C = 0.88826 x 66 = 58.625 kV 

Last step : Find base currents in all sections.
Sb 10  103
Ib A    437.386 Amps
3 Vb A 3  13.2

 IGen = Ib A  ILoad p.u.= 437.386  1.2898 = 564.14 Amps

Sb 10  103
Ib B    43.7387 Amps
3 Vb B 3  132

39
 ILine = Ib B  ILoad p.u.
= 43.7387  1.2898 = 56.414 Amps
Sb 10  103
Ib C    87.4773 Amps
3 Vb C 3  66

 ILoad = Ib C  ILoad p.u.

= 87.4773  1.2898 = 112.828 Amps
Load Representations
 Actual P & Q demands of load depend on system frequency & voltage.
 Residential loads behave differently than (say) commercial or industrial loads.
They are less affected by frequency and are usually resistive – kW demand
sensitive to voltage of supply.
 Industrial loads e.g. induction motor loads draw reactive power which is sensitive
to voltage level (real power demand, kW, does not vary significantly).
 Accurate estimate of load very difficult. Following two representations quite
reasonable/satisfactory :

40
1) Constant Power Model (e.g. Air-conditioning loads)

S S P2  Q2
S  VI*  I*   I    
V0 V V
where  is load power factor angle (assumed lagging)
 Here load power (S) remains constant, although load voltage (V) & current (I)
will change; e.g. air-conditioning loads.
2) Constant-impedance Model (e.g. Water heaters & light bulbs)

V V V2
Zs    
I S  S
   
V  41
 P P
P  I RS  RS  2  2
2
I  P  Q2 
 2 
 V 
P V2
 RS  2 
P Q 2

Q Q Q V2
Q  I XS  XS  2  2
2
 2 
I P Q  P Q
2 2

 2 
 V 
In p.u. system : ZS p.u. = RS p.u. + jXS p.u.
 For a constant-impedance load, the load impedance is kept constant but the
current & power drawn at various voltages will be different.
Note : Constant-impedance load may also be represented by R & X in parallel.
Rp (R p )( jX p )
V  Zp  
R p  jX p
V2
Xp where R p = 
P
V2
& Xp =  42
Q
REVIEW EXERCISES (5)
(5) Draw the impedance diagram for the 3-phase system shown in the following figure indicating the pu values of
impedances.

Tr line : Z = 12.8 + j64 

Tr 1 : 1  3-phase 120 MVA Transformer, /, 34.5 kV/345 kV, Z = (1 + j8)%
Tr 2 : 3  1-phase 30 MVA Transformer, 200 kV/20 kV, Z = (1 + j7)%
Take the base values at bus 2 as 100 MVA and 345 kV.
(Hint : List the base values for all buses and then convert parameters to the common base)

Tr 1 : (0.0083 + j0.0667) pu, Tr. line : (0.0108 + j0.0538) pu, Tr 2 : (0.0112 + j0.0784) pu

(6) The system in problem (5) delivers a load of 60 MW at bus 4 at the rated voltage of 20 kV. Indicate the load terminal
conditions in pu and calculate : (a) the voltage at bus 1, (b) phase angle of voltage at bus 1, for the following cases :
(i) 0.8 (lag), and (ii) 0.8 (lead)
o o o
(i) VL = 1.0040 pu, IL = 0.7469-36.87 pu, 38.51 kV, 5.41
o o o
(ii) VL = 1.0040 pu, IL = 0.746936.87 pu, 32.51 kV, 8.08

43
REVIEW EXERCISES (7 & 8)
(7)(a) Draw the impedance diagram for the system shown in the figure indicating the p.u. values of impedance, for the
following system data.
Tr. Lines : TL1 and TL2 : R = 16 , X = 120 .
Load : (12 + j9) MVA at 13.2 kV.
Tr 1 : 20 MVA, 11/120 kV, Z = (2 + j10)%
Tr 2 : 20 MVA, 120/13.8 kV, Z = (1 + j8)%

Represent the load by a constant series impedance and take the base values for the transmission line as 60 MVA,
and 120 kV.
(b) If the voltage at Bus 1 is maintained at rated value (11 kV), calculate the load terminal voltage.
(a) TL1 & TL2 : 0.067 + j0.50 p.u.; Load impdance : 2.928 + j2.196 p.u.;
Tr 1 : 0.06 + j0.3 p.u.; Tr 2 : 0.03 + j0.24 p.u.
(b) VL = 11.83 kV
Note : Load voltage for (b) will NOT be 13.2 kV as in (a). The load is fixed but the load voltage can vary!
(8) Consider the power system shown below.
(i) Draw the per unit reactance diagram using a 30 MVA, 13.8 kV base in the generator circuit.
(ii) If motor loads M1 and M2 draw 16 MW and 8 MW respectively at unity power factors and 12.5 kV, find the
internally generated generator voltage (i.e., voltage before the generator reactance itself).
(i) XG1 = j0.15; XT1 = j0.078; XLine = j0.166; XT2 = j0.0915; XM1 = j0.2744; XM2 = j0.549
(ii) 1.0391 p.u. (14.339 kV)
44
G1 : 30 MVA, 13.8 kV, 3- generator, XG1 = 15%
M1 : 20 MVA, 12.5 kV, 3- motor load, XM1 = 20% (synchronous motor)
M2 : 10 MVA, 12.5 kV, 3- motor load, XM2 = 20% (synchronous motor)
T1 : 35 MVA, 13.2/115 kV, Transformer, XT1 = 10%
T2 : 30 MVA, 12.5/115 kV, Transformer, XT2 = 10%
L1 : Line XL = 80 

45
Basic Equations for Real/Reactive Powers (Please see Appendix A for details)
 Consider transfer of power from point “A” in a power system to point “B”,
thru an impedance Z = R + jX = Z

 Voltage at “A” = E

 Voltage at “B” = V0
E  V0 E V
I  (  )    
R  jX Z Z
 Complex power delivered at “B”
S = VI* (in per unit)
E  VE
2
V V
 (V0)  (  )     (  )  
Z Z  Z Z

46
 VE V2
 Real power P = cos(  )  cos 
Z Z
VE V2
Reactive power Q = sin(  )  sin 
Z Z
Special case : R ~ 0  Z = 0 + jX = X90o
VE V2
S  (90  )  90
X X
VE VE V2
P sin  & Q = cos  
X X X
 Point “A” is usually
(1) Internally-generated voltage E for a synchronous generator or
(2) Sending-end voltage for a transmission line
 Point “B” is usually
(1) Load voltage V for a generator connected to a load or
(2) Receiving-end voltage for a transmission line
47
 Impedance Z = R + jX (or Z ~ jX) is usually

(1) Synchronous impedance of a synchronous generator or

(2) Series impedance of a trans. line

 Note [for R ~ 0]

 P flows from “A” to “B” for angle  taking positive values; else “B” to “A”;
if  = 0, P = 0 (no real power).
P depends largely on relative angle difference of voltages at “A” & “B”.

 For small  : Q flows from “A” to “B” if E> V; else “B” to “A”; if
E~ V, then Q ~ 0 (no reactive power).
Q depends largely on voltage magnitudes at “A” or “B”.

48