The three basic elements of electrical engineering are resistor, inductor and
capacitor. The resistor consumes ohmic or ... ELECTRICAL POWER SYSTEMS.
1 FUNDAMENTALS OF POWER SYSTEMS
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FUNDAMENTALS OF POWER SYSTEMS
Chapter
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 the plates.
1 2
CV 2 where C is the capacitance and V is the voltage across
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)
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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.
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.
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(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.
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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. In one C R I 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 the Vm sin wt capacitor is fully discharged. Fig. 1.5 Relationship between electric field energy and reactive power.
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). From Fig. 1.5 V
I=
2
R + (1/ωC)
VωC
=
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=
∴ where
I /ωC
sin φ = Now
. 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
(1.8)
Substituting for I and sin φ, we have V ωC
Q=V. ∴
Reactive power =
R2ω 2C 2 + 1
.
1 R2ω 2C 2 + 1
=
V 2 ωC
R2ω 2C 2 + 1
(1.9)
V 2 ωC R2ω 2C 2 + 1
Now this can be related with the electric energy stored by the capacitor. The energy stored by the capacitor. W= Now
v=
∴
W=
1 2
Cv2
(1.10)
1 1 idt = Ú C C 1 C 2
.
Vm wC 2
2
2
R w C +1
Vm2 cos2 ( wt + f) R2 w 2C 2 + 1
=
-
V cos ( wt + f) cos ( wt + f) =- m w R2 w2C 2 + 1
V 2C cos2 ( wt + f) R2 w 2C 2 + 1
(1.11) (1.12)
dW V2 = 2 2 2 . 2 cos (ωt + φ) . sin (ωt + φ) . ωC dt R ω C +1
=
V 2 ωC . sin 2(ωt + φ) R2ω 2C 2 + 1
= Q sin 2(ωt + φ)
(1.13)
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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 ωt Im 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.
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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 unit is 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 (1.19)
= P – jQ
1.4 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
70%
(ii) Heating and lighting equipment
25%
(iii) Electronic devices
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.
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1.3 COMPLEX POWER
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For an impedance load i.e. lumped load P= and
Q=
V2 R 2 + (2πfL) 2 V2 2
R + (2πfL) 2
.R
. (2πfL)
(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|]
(1.21)
Q = Q[f, |V|]
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 ΔP ~ − .| ΔV |+ ∂|V | ∂Q ΔQ ~ − .|ΔV |+ ∂|V |
∂P . Δf ∂f ∂Q . Δf ∂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 under-frequency relays which automatically disconnect blocks of loads or sectionalise the transmission system depending upon the system requirements.
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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. = The base current =
Vactual Vb
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 Actual impedance = Base impedance
= p.u. impedance
=
Z . kVAb × 1000 Vb 2
=
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 .
F GH
Vold kVAnew . kVAold Vnew
I JK
2
(1.23)
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1.5 THE PER UNIT SYSTEM
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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
FV I GH V JK p
2
s
where Vp and Vs are the primary and secondary voltages of the transformer. Now
Zp p.u. =
ZpI p Vp
= Zs .
= Zs
ZpI p Vs
2
FV I GH V JK p
2
.
s
= Zs .
Ip Vp
Vs I s 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.
