1
FUNDAMENTALS OF POWER SYSTEMS
1
Fundamentals of Power Systems
INTRODUCTION
The three basic elements of electrical engineering are resistor, inductor and capacitor. The
resistor consumes ohmic or dissipative energy whereas the inductor and capacitor store in the
positive half cycle and give away in the negative half cycle of supply the magnetic field and
electric field energies respectively. The ohmic form of energy is dissipated into heat whenever
a current flows in a resistive medium. If I is the current flowing for a period of t seconds
through a resistance of R ohms, the heat dissipated will be I 2Rt watt sec. In case of an inductor
the energy is stored in the form of magnetic field. For a coil of L henries and a current of I
amperes flowing, the energy stored is given by
1
2
LI 2. The energy is stored between the metallic
plates of the capacitor in the form of electric field and is given by
capacitance and V is the voltage across the plates.
1
2
CV 2 where C is the
We shall start with power transmission using 1-φ circuits and assume in all our analysis
that the source is a perfect sinusoidal with fundamental frequency component only.
1.1
SINGLE-PHASE TRANSMISSION
Let us consider an inductive circuit and let the instantaneous voltage be
v = Vm sin ωt
(1.1)
Then the current will be i = Im sin (ωt – φ) where φ is the angle by which the current lags
the voltage (Fig. 1.1).
The instantaneous power is given by
p = vi = Vm sin ωt . Im sin (ωt – φ)
= VmIm sin ωt sin (ωt – φ)
=
Vm I m
[cos φ – cos (2ωt – φ)]
2
2
(1.2)
3
FUNDAMENTALS OF POWER SYSTEMS
The value of ‘p’ is positive when both v and i are either positive or negative and represents
the rate at which the energy is being consumed by the load. In this case the current flows in
the direction of voltage drop. On the other hand power is negative when the current flows in
the direction of voltage rise which means that the energy is being transferred from the load
into the network to which it is connected. If the circuit is purely reactive the voltage and
current will be 90° out of phase and hence the power will have equal positive and negative half
cycles and the average value will be zero. From equation (1.2) the power pulsates around the
average power at double the supply frequency.
p
v
i
IVIIII cos f
f
Fig. 1.1 Voltage, current and power in single phase circuit.
Equation (1.2) can be rewritten as
p = VI cos φ (1 – cos 2ωt) – VI sin φ sin 2ωt
I
(1.3)
II
We have decomposed the instantaneous power into two components (Fig. 1.2).
p
I
II
p = VI cos f
VI sin f
Fig. 1.2 Active, reactive and total power in a single phase circuit.
(i) The component P marked I pulsates around the same average power VI cos φ but
never goes negative as the factor (1 – cos 2ωt) can at the most become zero but it will never go
negative. We define this average power as the real power P which physically means the useful
power being transmitted.
4
ELECTRICAL POWER SYSTEMS
(ii) The component marked II contains the term sin φ which is negative for capacitive
circuit and is positive for inductive circuit. This component pulsates and has zero as its average value. This component is known as reactive power as it travels back and forth on the line
without doing any useful work.
Equation (1.3) is rewritten as
p = P(1 – cos 2ωt) – Q sin 2ωt
(1.4)
Both P and Q have the same dimensions of watts but to emphasise the fact that Q
represents a nonactive power, it is measured in terms of voltamperes reactive i.e., V Ar.
The term Q requires more attention because of the interesting property of sin φ which is
– ve for capacitive circuits and is +ve for inductive circuits. This means a capacitor is a generator
of positive reactive V Ar, a concept which is usually adopted by power system engineers. So it
is better to consider a capacitor supplying a lagging current rather than taking a leading
current (Fig. 1.3).
+
+
V
V
C
–
C
–
I leads V by 90
I lags V by 90°
Fig. 1.3 V-I relations in a capacitor.
Consider a circuit is which an inductive load is shunted by a capacitor. If Q is the total
reactive power requirement of the load and Q′ is the reactive power that the capacitor can
generate, the net reactive power to be transmitted over the line will be (Q – Q′). This is the
basic concept of synchronous phase modifiers for controlling the voltage of the system. The
phase modifier controls the flow of reactive power by suitable excitation and hence the voltage
is controlled. The phase modifier is basically a synchronous machine working as a capacitor
when overexcited and as an inductor when underexcited.
It is interesting to consider the case when a capacitor and an inductor of the same
reactive power requirement are connected in parallel (Fig. 1.4).
IC
IC
IL
V
V
IL
Fig. 1.4 Power flow in L-C circuit.
5
FUNDAMENTALS OF POWER SYSTEMS
The currents IL and IC are equal in magnitude and, therefore, the power requirement is
same. The line power will , therefore, be zero. Physically this means that the energy travels
back and forth between the capacitor and the inductor.
C
R
I
In one half cycle at a particular moment the capacitor is
fully charged and the coil has no energy stored. Half a
voltage cycle later the coil stores maximum energy and
Vm sin wt
the capacitor is fully discharged.
The following example illustrates the relationship
between the reactive power and the electric field energy
stored by the capacitor. Consider an RC circuit (Fig. 1.5).
Fig. 1.5 Relationship between electric
field energy and reactive power.
From Fig. 1.5
V
I=
2
R + (1/ωC)
Vω C
=
2
(1.5)
2
R ω 2C 2 + 1
and if voltage is taken as reference i.e., v = Vm sin ωt the current
i = Im sin (ωt + φ)
∴
Vm ωC
i=
I /ωC
sin φ =
where
. sin (ωt + φ)
R2ω 2C 2 + 1
2
2
I R + ( I /ωC)
2
=
(1.6)
1
2
(1.7)
2
R ω C2 + 1
reactive power Q = VI sin φ
Now
(1.8)
Substituting for I and sin φ, we have
VωC
Q=V.
∴ Reactive power =
R 2ω 2C 2 + 1
.
1
R2ω 2C 2 + 1
=
V 2 ωC
R2ω 2C 2 + 1
(1.9)
V 2 ωC
R 2ω 2 C 2 + 1
Now this can be related with the electric energy stored by the capacitor. The energy
stored by the capacitor.
W=
1
2
Now
v=
1
C
∴
W=
Cv2
z
idt =
1
C.
2
(1.10)
1
C
VmωC
cos (ωt + φ) Vm cos (ωt + φ)
=
ω
R ω C +1
R2 ω 2 C 2 + 1
2
2
2
.
Vm 2 cos 2 (ωt + φ) V 2 cos 2 (ωt + φ)
=
R2ω 2C 2 + 1
R2 ω 2 C 2 + 1
(1.11)
(1.12)
dW
V2
= 2 2 2
. 2 cos (ωt + φ) . sin (ωt + φ) . ω . C
dt
R ω C +1
=
V 2 ωC
. sin 2(ωt + φ)
R2 ω 2 C 2 + 1
= Q sin 2(ωt + φ)
(1.13)
6
ELECTRICAL POWER SYSTEMS
From this it is clear that the rate of change of electric field energy is a harmonically
varying quantity with a frequency double the supply frequency and has a peak value equal
to Q.
In an R-L circuit the magnetic field energy and reactive power in a coil are similarly
related.
1.2
THE 3-PHASE TRANSMISSION
Assuming that the system is balanced which means that the three-phase voltages and currents
are balanced. These quantities can be expressed mathematically as follows:
Va = Vm sin ωt
Vb = Vm sin (ωt – 120°)
Vc = Vm sin (ωt + 120°)
(1.14)
ia = Im sin (ωt – φ)
ib = Im sin (ωt – φ – 120°)
ic = Im sin (ωt – φ + 120°)
The total power transmitted equals the sum of the individual powers in each phase.
p = Vaia + Vbib + Vcic
= Vm sin ωtIm sin (ωt – φ) + Vm sin (ωt – 120°)Im sin (ωt – 120° – φ)
+ VmIm sin (ωt + 120°) sin (ωt + 120° – φ)
= VI[2 sin ωt sin (ωt – φ) + 2 sin (ωt – 120°) sin (ωt – 120° – φ)
+ 2 sin (ωt + 120°) sin (ωt + 120° – φ)]
= VI[cos φ – cos (2ωt – φ) + cos φ – cos (2ωt – 240° – φ)
+ cos φ – cos (2ωt + 240° – φ)]
= 3VI cos φ
(1.15)
This shows that the total instantaneous 3-phase power is constant and is equal to three
times the real power per phase i.e., p = 3P, where P is the power per phase.
In case of single phase transmission we noted that the instantaneous power expression
contained both the real and reactive power expression but here in case of 3-phase we find that
the instantaneous power is constant. This does not mean that the reactive power is of no
importance in a 3-phase system.
For a 3-phase system the sum of three currents at any instant is zero, this does not
mean that the current in each phase is zero. Similarly, even though the sum of reactive power
instantaneously in 3-phase system is zero but in each phase it does exist and is equal to VI sin
φ and, therefore, for 3-φ the reactive power is equal to Q3φ = 3VI sin φ = 3Q, where Q is the
reactive power in each phase. It is to be noted here that the term Q3φ makes as little physical
sense as would the concept of three phase currents I3φ = 3I. Nevertheless the reactive power in
a 3-phase system is expressed as Q3φ. This is done to maintain symmetry between the active
and reactive powers.
7
FUNDAMENTALS OF POWER SYSTEMS
1.3
COMPLEX POWER
Consider a single phase network and let
V = |V|e jα and I = |I|e jβ
(1.16)
where α and β are the angles that V and I subtend with respect to some reference axis. We
calculate the real and reactive power by finding the product of V with the conjugate of I i.e.
S = VI* = |V|e jα |I|e–jβ = |V| |I|e j(α–β)
= |V| |I| cos (α – β) + j|V| |I| sin (α – β)
(1.17)
Here the angle (α – β) is the phase difference between the phasor V and I and is normally
denoted by φ
∴
S = |V| |I| cos φ + j|V| |I| sin φ
= P + jQ
(1.18)
The quantity S is called the complex power. The magnitude of S =
P 2 + Q 2 is termed
as the apparent power and its units are volt-amperes and the larger units are kVA or MVA.
The practical significance of apparent power is as a rating unit of generators and transformers,
as the apparent power rating is a direct indication of heating of machine which determines the
rating of the machines. It is to be noted that Q is positive when (α – β) is positive i.e. when V
leads I i.e. the load is inductive and Q is –ve when V lags I i.e. the load is capacitive. This
agrees with the normal convention adopted in power system i.e. taking Q due to an inductive
load as +ve and Q due to a capacitive load as negative. Therefore, to obtain proper sign for
reactive power it is necessary to find out VI* rather than V*I which would reverse the sign for
Q as
V*I = |V|e–jα |I|e jβ = |V| |I|e–j(α–β)
= |V| |I| cos (α – β) – j|V| |I| sin (α – β)
= |V| |I| cos φ – j|V| |I| sin φ
= P – jQ
1.4
(1.19)
LOAD CHARACTERISTICS
In an electric power system it is difficult to predict the load variation accurately. The load
devices may vary from a few watt night lamps to multi-megawatt induction motors. The
following category of loads are present in a system:
(i) Motor devices
(ii) Heating and lighting equipment
(iii) Electronic devices
70%
25%
5%
The heating load maintains constant resistance with voltage change and hence the power
varies with (voltage)2 whereas lighting load is independent of frequency and power consumed
varies as V 1.6 rather than V 2.
8
ELECTRICAL POWER SYSTEMS
For an impedance load i.e. lumped load
and
P=
V2
.R
R2 + (2πfL) 2
Q=
V2
. (2πfL)
R 2 + (2πfL) 2
(1.20)
From this it is clear that both P and Q increase as the square of voltage magnitude. Also
with increasing frequency the active power P decreases whereas Q increases.
The above equations are of the form
P = P[f, |V|]
Q = Q[f, |V|]
(1.21)
Composite loads which form a major part of the system load are also function of voltage
and frequency and can in general be written as in equation (1.21). For this type of load, however,
no direct relationship is available as for impedance loads. For a particular composite load an
empirical relation between the load, and voltage and frequency can be obtained. Normally we
are concerned with incremental changes in P and Q as a function of incremental changes in
| V | and f. From equation (1.21)
and
∂P
∂P
∆P ~
.| ∆V |+
. ∆f
−
∂|V |
∂f
∂Q
∂Q
∆Q ~
.|∆V |+
. ∆f
−
∂|V |
∂f
(1.22)
The four partial derivatives can be obtained empirically. However, it is to be remembered
that whereas an impedance load P decreases with increasing frequency, a composite load will
increase. This is because a composite load mostly consists of induction motors which always
will experience increased load, as frequency or speed increases.
The need for ensuring a high degree of service reliability in the operation of modern
electric systems can hardly be over-emphasized. The supply should not only be reliable but
should be of good quality i.e., the voltage and frequency should vary within certain limits,
otherwise operation of the system at subnormal frequency and lower voltage will result in
serious problems especially in case of fractional horse-power motors. In case of refrigerators
reduced frequency results into reduced efficiency and high consumption as the motor draws
larger current at reduced power factor. The system operation at subnormal frequency and
voltage leads to the loss of revenue to the suppliers due to accompanying reduction in load
demand. The most serious effect of subnormal frequency and voltage is on the operation of the
thermal power station auxiliaries. The output of the auxiliaries goes down as a result of which
the generation is also decreased. This may result in complete shut-down of the plant if corrective
measures like load shedding is not resorted to. Load shedding is done with the help of underfrequency relays which automatically disconnect blocks of loads or sectionalise the transmission
system depending upon the system requirements.
9
FUNDAMENTALS OF POWER SYSTEMS
1.5
THE PER UNIT SYSTEM
In a large interconnected power system with various voltage levels and various capacity
equipments it has been found quite convenient to work with per unit (p.u.) system of quantities
for analysis purposes rather than in absolute values of quantities. Sometimes per cent values
are used instead of p.u. but it is always convenient to use p.u. values. The p.u. value of any
quantity is defined as
the actual value of the quantity (in any unit)
the base or reference value in the same unit
In electrical engineering the three basic quantities are voltage, current and impedance.
If we choose any two of them as the base or reference quantity, the third one automatically will
have a base or reference value depending upon the other two e.g., if V and I are the base
voltage and current in a system, the base impedance of the system is fixed and is given by
V
I
The ratings of the equipments in a power system are given in terms of operating voltage
and the capacity in kVA. Therefore, it is found convenient and useful to select voltage and kVA
as the base quantities. Let Vb be the base voltage and kVAb be the base kilovoltamperes, then
Z=
Vp.u. =
Vactual
Vb
The base current
=
kVAb × 1000
Vb
∴ p.u. current
=
Actual current Actual current
=
× Vb
Base current
kVAb × 1000
Base impedance
=
Base voltage
Base current
Vb 2
=
kVAb × 1000
p.u. impedance
=
=
Actual impedance
Base impedance
Z . kVAb × 1000
Vb2
=
Z . MVAb
(kVb ) 2
This means that the p.u. impedance is directly proportional to the base kVA and inversely
proportional to square of base voltage. Normally the p.u. impedance of various equipments
corresponding to its own rating voltage and kVA are given and since we choose one common
base kVA and voltage for the whole system, therefore, it is desired to find out the p.u. impedance
of the various equipments corresponding to the common base voltage and kVA. If the individual
quantities are Zp.u. old, kVAold and Vold and the common base quantities are Zp.u. new, kVAnew and
Vnew, then making use of the relation above,
Zp.u. new = Zp.u. old .
FG
H
kVAnew
Vold
.
kVAold
Vnew
IJ
K
2
(1.23)
10
ELECTRICAL POWER SYSTEMS
This is a very important relation used in power system analysis.
The p.u. impedance of an equipment corresponding to its own rating is given by
Zp.u. =
IZ
V
where Z is the absolute value of the impedance of the equipment. It is seen that the p.u.
representation of the impedance of an equipment is more meaningful than its absolute value
e.g., saying that the impedance of a machine is 10 ohm does not give any idea regarding the
size of the machine. For a large size machine 10 ohms appears to be quite large whereas for
small machines 10 ohms is very small. Whereas for equipments of the same general type the
p.u. volt drops and losses are in the same order regardless of size.
With p.u. system there is less chance of making mistake in phase and line voltages,
single phase or three phase quantities. Also the p.u. impedance of the transformer is same
whether referred on to primary or secondary side of the transformer which is not the case
when considering absolute value of these impedances. This is illustrated below:
Let the impedance of the transformer referred to primary side be Zp and that on the
secondary side be Zs, then
Zp = Zs
FG V IJ
HV K
2
p
s
where Vp and Vs are the primary and secondary voltages of the transformer.
Now
Zp p.u. =
ZpI p
Vp
= Zs .
= Zs
FG V IJ
HV K
p
2
.
s
ZpI p
Vs
2
= Zs .
Ip
Vp
Vs Is
Vs
2
=
Zs I s
Vs
= Zs p.u.
From this it is clear that the p.u. impedance of the transformer referred to primary side
Zp p.u. is equal to the p.u. impedance of the transformer referred to the secondary side Zs p.u..
This is a great advantage of p.u. system of calculation.
The base values in a system are selected in such a way that the p.u. voltages and currents
in system are approximately unity. Sometimes the base kVA is chosen equal to the sum of the
ratings of the various equipments on the system or equal to the capacity of the largest unit.
The different voltage levels in a power system are due to the presence of transformers.
Therefore, the procedure for selecting base voltage is as follows: A voltage corresponding to
any part of the system could be taken as a base and the base voltages in other parts of the
circuit, separated from the original part by transformers is related through the turns ratio of
the transformers. This is very important. Say if the base voltage on primary side is Vpb then on
the secondary side of the transformer the base voltage will be Vsb = Vpb(Ns/Np), where Ns and
Np are the turns of the transformer on secondary and primary side respectively.
11
FUNDAMENTALS OF POWER SYSTEMS
The following example illustrates the procedure for selecting the base quantities in
various parts of the system and their effect on the p.u. values of the impedances of the various
equipments.
Example 1.1: A 100 MVA 33 kV 3-phase generator has a subtransient reactance of
15%. The generator is connected to the motors through a transmission line and transformers
as shown in Fig. E1.1a. The motors have rated inputs of 30 MVA, 20 MVA and 50 MVA at 30
kV with 20% subtransient reactance. The 3-phase transformers are rated at 110 MVA, 32 kV
∆/110 kVY with leakage reactance 8%. The line has a reactance of 50 ohms. Selecting the
generator rating as the base quantities in the generator circuit, determine the base quantities
in other parts of the system and evaluate the corresponding p.u. values.
j 50 W
100 MVA, 33 kV
15%
Fig. E1.1a
Solution: Assuming base values as 100 MVA and 33 kV in the generator circuit, the
p.u. reactance of generator will be 15%. The base value of voltage in the line will be
33 ×
110
= 113.43 kV
32
In the motor circuit,
32
= 33 kV
110
The reactance of the transformer given is 8% corresponding to 110 MVA, 32 kV. Therefore,
corresponding to 100 MVA and 33 kV the p.u. reactance will be (using Eq. 1.23).
113.43 ×
0.08 ×
FG IJ
H K
100
32
×
110
33
The p.u. impedance of line =
2
= 0.06838 p.u.
50 × 100
(113.43) 2
= 0.3886 p.u.
FG IJ
H K
100 F 30 I
motor 2 = 0.2 ×
×G J
20 H 33 K
100 F 30 I
motor 3 = 0.2 ×
×G J
50 H 33 K
The p.u. reactance of motor 1 = 0.2 ×
100
30
×
30
33
2
= 0.5509 p.u.
2
= 0.826
2
= 0.3305 p.u.
12
ELECTRICAL POWER SYSTEMS
The reactance diagram for the system is shown in Fig. E1.1b.
j 0.06838 W
j 0.3888 W
j 0.06838 W
j 0.826 W
j 0.15 W
j 0.5509 W
j 0.3305 W
1
2
3
Fig. E1.1b Reactance diagram for Example 1.1.
PROBLEMS
1.1. Two generators rated at 10 MVA 13.2 kV and 15 MVA 13.2 kV are connected in parallel to a
busbar. They feed supply to two motors of inputs 8 MVA and 12 MVA respectively. The operating
voltage of motors is 12.5 kV. Assuming base quantities as 50 MVA and 13.8 kV draw the reactance diagram. The per cent reactance for generators is 15% and that for motors is 20%.
1.2. Three generators are rated as follows: Generator 1—100 MVA, 33 kV, reactance 10%; Generator
2—150 MVA, 32 kV, reactance 8%; Generator 3—110 MVA, 30 kV, reactance 12%. Determine
the reactance of the generator corresponding to base values of 200 MVA 35 kV.
1.3. A 3-bus system is given in Fig. P1.3. The ratings of the various components are listed below:
Generator 1 = 50 MVA, 13.8 kV, X″ = 0.15 p.u.
Generator 2 = 40 MVA, 13.2 kV, X″ = 0.20
Generator 3 = 30 MVA, 11 kV, X″ = 0.25
Transformer 1 = 45 MVA, 11 kV, ∆/110 kV Y, X = 0.1 p.u.
Transformer 2 = 25 MVA, 12.5 kV ∆/115 kV Y, X = 0.15 p.u.
Transformer 3 = 40 MVA, 12.5 kV ∆/115 kV Y, X = 0.1 p.u.
The line impedances are shown in Fig. P1.3. Determine the reactance diagram based on 50 MVA
and 13.8 kV as base quantities in Generator 1.
j 50 W
G1
G2
j 25 W
j 25 W
G3
Fig. P1.3
FUNDAMENTALS OF POWER SYSTEMS
13
1.4. Explain clearly the concept of reactive power in single phase and three phase circuits.
1.5. Explain clearly how the magnetic field energy and the reactive power in an inductive circuit are
related.
1.6. Explain clearly what you mean by good quality supply and discuss the effect of bad supply on the
performance of the system.
1.7. Explain the p.u. system of analysing power system problems. Discuss the advantages of this
method over the absolute method of analysis.
REFERENCES
1.
Electric Energy System Theory—An Introduction, O. I. Elgord, McGraw-Hill, 1971.
2.
Elements of Power System Analysis, W. D. Stevenson Jr., McGraw-Hill, 1962.
3.
Electric Power Systems, B. M. Weedy, John Wiley & Sons, 1974.