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Transcript of Basic Concepts-lecture Note
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Module 2
BASIC CONCEPTS
FOR ELECTRIC POWER SYSTEM ANALYSIS
The intention of this module is to lay the groundwork for the study of electric
power systems. This is done by developing some basic tools involving concepts,
definitions, and some procedures fundamental to electric power system. This module
can be considered as simply as review of topics that will be utilized throughout this
teaching module. We start by introducing the principal electrical quantities that we
will deal with.
2.1. POWER IN SINGLE PHASE AC CIRCUITS
The electric power systems specialist is almost always more concerned with
describing the rate of change of energy with respect to time (which is the definition
of power) rather than voltage and current. As the power into an element is basically
the product of voltage across and current through it, it seems reasonable to swap the
current for power without losing any information in describing the phenomenon. In
treating sinusoidal steady-state behavior of circuits, some further definitions are
necessary. To illustrate the concepts, we will use a cosine representation of the
waveform.
If the terminals of the load are designated a and n, and if the voltage and
current are expressed by
van= Vmaxcos t and ian= Imaxcos (t - )
the instantaneous power isp = van .ian= VmaxImaxcos t cos (t -) (2.1)
The angle in these equations is positive for current lagging the voltage and negative
for leading current. A positive value ofpexpressed the rate at which energy is being
absorbed by the part of the system between the points a and n. The instantaneous
power is obviously positive when both van and ian are positive but will become
negative when van and ian are opposite in sign. Figure 2.1 illustrates this point.
Positive power calculated as vanianresults when current is flowing in the direction of
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a voltage rise and mean energy is being transferred to the load. Conversely, negative
power calculated as van ian results when current is flowing in the direction of a
voltage rise and means energy is being transferred from the load into the system to
which the load is connected. If van and ian are in phase, as they are in a purely
resistive load, the instantaneous power will never becomes negative. If the current
and voltage are out of phase by 900, as in purely inductive or purely capacitive ideal
circuit element, the instantaneous power will have equal positive and negative half
cycles and its average value will be zero.
Figure 2.1 Current, voltage, and power plotted versus time.
By using trigonometric identities the expression of Eq.(2.1) is reduced to
tIV
tIV
p 2sinsin2
2cos1cos2
maxmaxmaxmax (2.2)
where VmaxImax/2may be replaced by the product of the rms voltage and current |Van|
. | Ian|or |V| . | I|.
For tVvan cosmax ,
tIi
Ri
R coscos
max
max
Figure 2.2 shows Raniv plotted versus t.
Similarly,
tIV
iv
ttIViv
xan
xan
2sinsin2
cossinsin
maxmax
maxmax
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Figure 2.2 Voltage, current in phase with the voltage, and the resulting power
plotted versus time.
which the instantaneous power in the inductance and is the second term in Eq.(2.2).
Figure 2.3 shows ,, xan iv and their product plotted versus t.
Figure 2.3 Voltage, current lagging the voltage by 900, and the resulting power
plotted versus time
Examination of Eq.(2.2) shows that the first term, the term which contains
cos , is always positive and has an average value of
cos2
maxmaxIVp
or, when rms value of voltage and current are substituted,
cosIVp (2.3)
P is the quantity to which the word power refers. P, the average power, is also called
the real power. The fundamental unit for both instantaneous and average power is the
watt, but a watt is such a small unit in relation to power quantities that P is usually
measured in kilowatts or megawatts.
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The cosine of the phase angle between the voltage and the current is called
the power factor. An inductive circuit is said to have a lagging power factor, and a
capacitive circuit is said to have a leading power factor. In other words, the term
lagging power factor and leading power factor indicate, respectively, whether the
current is lagging or leading the applied voltage.
The second term of Eq.(2.2), the term containing sin , is alternately positive
and negative and has an average value of zero. This component of the instantaneous
powerpis called the instantaneous reactive powerand expresses the flow of energy
alternately toward the load and away from the load. The maximum value of this
pulsating power, designated Q, is called reactive power or reactive volt-amperes and
is very useful in describing the operation of a power system, as will become
increasingly evident in further discussion. The reactive power is
sin2
maxmaxIVQ
or (2.4)
sinIVQ
The square root of the sum of the squares of Pand Qis equal to the product
of |V| and | I |, for
IVIVIVQP 222 sincos
Of course P and Q have the same dimensional units, but it is usual to designate the
unit for Q as vars (for volt-amperes reactive). The more practical unit for Q are
kilovars or megavars.
We define a quantity called the complex or apparent power, designated S,
which P and Q are components. By definition,
sincos
sincos
jVIS
jVIVIS
jQPS
(2.5)
Using Eulers identity, we thus have
jeVIS
or
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S = VI /
If we introduce the conjugate current defined by the asterisk(*)
I*= | I | /
It becomes immediately obvious that an equivalent definition of complex or apparent
power is
S = VI* (2.6)
The use of concepts of complex power may prove advantageous in solving
problems of power system analysis.
Consider the situation with an inductive circuit in which the current lags the
voltage by the angle . The conjugate of the current will be in the first quadrant in
the complex plane as shown in figure 2.4(a). Multiplying the phasors by V, we obtain
the complex power diagram shown in figure 2.4(b). Inspection of the diagram as well
as the previous development leads to a relation for the power factor of the circuit:
S
Pcos (2.7)
Example 2.1Consider the circuit composed of a series R-L branch in parallel with
capacitance with the following parameters:
R = 0.5 ohms
XL= 0.8 ohms
Bc= 0.6 siemens
Figure 2-4. Complex power diagrams.
-
I*
V
I
S = V I*
Q = VI sin
P = VI cos
(a) (b)
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Assume
V = 100 /0 V
Calculate the input current and the active, reactive, and apparent power into
the circuit.
Solution
The current into the R-L branch is given by
8.05.0
100
jIZ
= 106.00 /-57.99
0A
The power factor (PF) of the R-L branch is
53.0
99.57coscos 0
Z
ZZ
PF
PF
which is a laggingpower factor.
The current into the capacitance is
AjIc090/601006.0
The input current Itis
0
00
01.28/64.63
90/6099.57/00.106
t
t
Zct
I
I
III
The power factor (PF) of the overall circuit is
88.001.28coscos 0
ttPF lagging
Note that the magnitude of It is less than that of IZ, and that cos is higher
than cos z. This is the effect of the capacitor, and its action is called power
factor correction in power system terminology.
The apparent power into the circuit is
VAS
S
VIS
t
t
tt
0
0
*
01.28/00.6364
01.28/64.630/100
In rectangular coordinates we get
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76.298898.5617 jSt
Thus the active and reactive powers are:
var76.2988
98.5617
t
t
Q
WP
2.2 DIRECTION OF POWER FLOW
For an ac system figure 2.5 shown an ideal voltage source (constant
magnitude, constant frequency, zero impedance) with polarity marks which, as
usually, indicate the terminal which is positive during half cycle is positive
instantaneous voltage. Of course, the positively marked terminal is actually the
negative terminal during the negative half cycle of the instantaneous voltage.
Similarly the arrow indicates the direction of current during the positive half cycle of
current.
Figure 2.5 An ac cicuit representation of an emf and current to illustrate
polarity marks.
In figure 2.5a a generator is expected since the current is positive when
flowing away from the positively marked terminal. However, the positively marked
terminal may be negative when the current is flowing away from it. The approach to
understanding the problem is to resolve phasor Iinto a component along the axis of
the phasor Eand a component 900out of phase with E. The product of |E| and the
magnitude of the component of I which is 900 out of phase with E is Q. If the
component of I along the axis of E is in phase with E, the power is generatedpower
which is being delivered to the system, for this component of current is always
flowing away from the positively marked terminal when that terminal is actually
positive (and toward that terminal when the terminal is negative).P, the real part of
EI*, is positive.
+ -I
E
(a)
+ -I
E
(b)
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If the component of current along the axis of E is negative (180 0out of phase
with E), power is being absorbed and the situation is that of a motor.P, the real part
ofEI*, would be negative.
The voltage and current relationship might be as shown in figure 2.5b, and a
motor would be expected. However, an average power absorbedwould occur only if
the component of the phasor I along the axis of the phasor E was found to be in
phase rather than 1800out of phase with E, so that this component of current would
be always in the direction of the drop in potential. In this caseP, the real part ofEI*
would be positive. Negative P here would indicated generated power.
To consider the sign of Q, figure 2.6 is helpful. In figure 2.6a positivereactive power equal to |I|2X is supplied to the inductance since inductance draws
Figure 2.6 Alternating emf applied (a) to apurely inductive element
and (b) to a purely capasitive element.
positive Q. Than I lags E by 900, and Q, the imaginary part of EI*, is positive. In
figure 2.6b negative Q must be supplied to the capacitance of the circuit, or the
source with the emf E is receiving positive Q from the capacitor, I leads E by 900.
2.3. BALANCED THREE-PHASE SYSTEM
A balanced three-phase voltage system is composed of three single phase
voltage having the same magnitude and frequency but time displaced from one
another by 1200. Figure 2.7(a) shows a schematic representation where the three
single phase voltage source appear in a Y connection; a configuration is also
possible. A phasor diagram showing each of the phase voltage is also given in figure
2.7(b). As the phasors revolve at the angular frequency with respect to the
reference line in the counterclockwise (positive) direction, the positive maximum
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value first occurs for phase a and then in succession for phases band c. Stated in a
different way, to an observer in the phasor space, the voltage of phase aarrives first
followed by that of band then of c. For this reason the three phase voltage of figure
2.7 is said to have the phase sequence abc. This is important for certain applications.
For example, in three phase induction motors, the phase sequence determines
whether the motor turn clockwise or counterclockwise.
(a) (b)
Figure 2-7.A Y-connected three-phase system and the corresponding phasor diagram.
Current and Voltage Relation
Balanced three phase system can be studied using technique developed for
single phase circuits. The arrangement of the three single phase voltage into a Y or
configuration requires some modification in dealing with the overall system.
Y Connection
With reference to figure 2.7 the common terminal n is called the neutralor
star(Y)point. The voltage appearing between any two of the line terminals a, b, and
c have different relationships in magnitude and phase to the voltages appearing
between any one line terminal and the neutral point n. The set of voltages Vab, Vbc,
and Vca are called the line voltages, and the set of voltages Van, Vbn, and Vcn are
referred to as thephase voltages. Analysis of phasor diagrams provides the required
relationships.
~
~
~ a
b
120o
120o
120o
cVcn
Vbn
Van
Reference Line
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Figure 2-8. Illustrating relation between phase and line voltages in a Y connection.
The effective values of the phase voltages are shown in figure 2.8 as V an, Vbn,
and Vcn. Each has the same magnitude, and each is displaced 1200from the other two
phasors. To obtain the magnitude and phase angle of the line voltage from a to b (i.e.,
Vab), we apply Kirchhoffs law:
nbanab VVV (2.8)
This equation states that the voltage existing from a to b is equal to the voltage from
a to n (i.e., Van) plus the voltage from n to b. Thus equation 2.8 can be rewritten as
bnanab VVV (2.8 a)
Since for a balanced system, each phase voltage has the same magnitude, let us set
pcnbnan VVVV (2.9)
where Vpdenotes the effective magnitude of the phase voltage. Accordingly we may
write0
pan 0/VV (2.10)
0pbn 120/VV (2.11)
0p
0pcn
0p
0pcn 120/V240/VV120/V240/VV (2.12)
Substituting equations 2.10 and 2.11 in equation 2.8a yield
012011 /VV pab
0303 /VV pab (2.13)
opcn 120VV
Vab
30o
opan 120VV
opan 0VV
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Similarly we obtain
0903 /VV pbc (2.14)
01503 /VV pca (2.15)
The expressions obtained above for the line voltages show that they constitute
a balanced three phase voltage system whose magnitude are 3 times the phase
voltages. Thus we write
pL VV 3 (2.16)
A current flowing of out line terminal a (or b or c) is the same as that flowing
through the phase source voltage appearing between terminal n and a (or n and b, or
n and c). We can thus conclude that for a Y-connection three phase source, the line
current equals the phase current. Thus
pL II (2.17)
In the above equation, ILdenotes the effective value of the line current and Ipdenotes
the effective value for the phase current.
Connection
We consider now the case when the three single phase sources are rearranged
to form a three phase connection as shown in figure 2.9. It is clear from inspection
of the circuit shown that the line and phase voltage have the same magnitude:
pL VV (2.18)
The phase and line currents, however, are not identical, and the relationship between
them can be obtained using Kirchhoffs current law at one of the line terminals.
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Figure 2-9. -connected three-phase source.
In a manner similar to that adopted for the Y connection source, let us
consider the phasor diagram shown in figure 2.10. Assume the phase currents to be
0
0
120/
120/
0/
pca
pbc
pab
II
II
II
Gambar 2-10. Illustrating relation between phase and line currents in a connection.
The current that follows in the line joining a to a is denoted Iaaand is given
by
abcaaa III
~
~
~
a
bc
Iaa
Ibb
Icc
a
b
c
Ica
Iab
Iaa
Iba
Ica
Vca
Vab
Ibb
IccVbc
30o
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As a result, we have
0/1120/1 0 paa II
which simplifies to
0150/3 paa II
similarly,
0
0
90/3
30/3
pcc
pbb
II
II
Note that a set of balanced three phase currents yields a corresponding set of
balanced line currents that are 3 times the phase values:
pL II 3 (2.19)
where ILdenotes the effective value of any of the three line currents.
Power Relationships
Assume that the three phase generator is supplying a balanced load with the
three sinusoidal phase voltages:
0
0
120sin2
120sin2
sin2
tVtv
tVtv
tVtv
pc
pb
pa
With the currents given by
0
0
120sin2
120sin2
sin2
tIti
tIti
tIti
pc
pb
pa
where is the phase angle between the current and voltage in each phase. The total
power in the load is
titvtitvtitvtp ccbbaa 3
This turns out to be
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120sinsin
120sin120sin
sinsin23
tt
tt
ttIVtp pp
Using a trigonometric identity, we get
2402cos2402cos2coscos33 tttIVtp pp
Note that the last three terms in the above equation are reactive power terms and they
add up to zero. Thus we obtained
cos33 ppIVtp (2.20)
When referring to the voltage level of a three phase system, one invariablyunderstands the line voltages. From the above discussion the relationship between
the line and phase voltages in a Y connection system is
pL VV 3
The power equation thus reads in term of line quantities:
cos33 LL IVp (2.21)
We note the total instantaneous power is constant, having a magnitude of
three times the real power per phase. We may be tempted to assume that the reactive
power is of no importance in a three-phase system since the Q cancel out. However,
the situation is analogous to the summation of balanced three-phase currents and
voltages that also cancel out. Although the sum cancels out this quantities are still
very much in evidence in each phase. We thus extend the concept of complex or
apparent power (S) to three phase system by defining
*
3 3 ppIVS (2.22)
as
cosIVp pp33 (2.23)
sinIVQ pp33 (2.24)
In terms of line values, we can assert that
*
3 3 LLIVS (2.25)
and
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cosIVp LL33 (2.26)
sinIVQ LL33 (2.27)
Example 2.2
A Y connected, balanced three phase load consisting of three
impedances of 10 /300 ohms each shown in figure 2.11 is supplied with
balanced line to neutral voltages:
VV
VV
VV
cn
bn
an
0
0
120/220
240/220
0/220
a. Calculate the phasor currents in each line
b. Calculate the line to line phasor voltages
c. Calculate the total active and reactive power supplied to the load.
Figure 2-11. Load connection for example 2.2.
Solution
a. The phase currents are obtained as
2200
a
b
c
Z = 1030oohms
n
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AI
AI
AI
cn
bn
an
0
0
0
90/2230/10
120/220
210/2230/10
240/220
30/2230/10
220
b. The line voltages are obtained as
0
0
0
0
210/3220
90/322012030/3220
30/3220
240/2200/220
ca
bc
ab
ab
bnanab
V
V
V
V
VVV
c. The apparent power into phase ais given by
VAS
S
IVS
a
a
anana
0
0
*
30/4840
30/22220
The total apparent power is three times the phase value:
00.726069.12574
30/00.1452030/34840 00
jS
VAS
t
t
Thus
var00.7260
69.12574
t
t
Q
WP
Example 2.3
Repeat example 2.2 as if the same three impedance were connected in
a connection.
Solution
From example 2.2 we have
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0
0
0
210/3220
90/3220
30/3220
ca
bc
ab
V
V
V
The currents in each of the impedances are
0
0
00
120/322
120/322
0/32230/10
30/3220
ca
bc
ab
I
I
I
The line currents are obtained with reference to figure 2.12 as
0
0
0
0
210/66
90/66
30/66
120/3220/322
c
bccac
b
abbcb
a
a
caaba
I
III
I
III
I
I
III
The apparent power in the impedance between a and b is
0
0
*
30/14520
0/32230/3220
ab
ab
ababab
S
S
IVS
The total three phase power is then
00.2178004.37724
30/43560 0
jS
S
t
t
As a result,
var00.21780
04.37724
t
t
Q
WP
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Figure 2-12. Load connection for example 2.3.
2.4PER PHASE ANALYSIS
The stage has now been set to introduce the powerful method of per phase
analysis. The justification for the method follows directly from the following
theorem.
Balanced Three-Phase Theorem. Assume that we are given a
1. Balanced three-phase (connected) system with
2. All loads and sources wye connected, and
3. In the circuit model there is no mutual inductance between phases.
Then
(a)
All the neutrals are the same potential,
(b)The phases are completely decoupled, and
(c)All corresponding network variables occur in balanced sets of the same sequence
as the sources.
Outline of Proof. The fact that the neutrals are at the same potential and the
network phases are decoupled follows that; using superposition, the number of
sources can be arbitrary. Without mutual inductance between the phases they are
decoupled and with balanced source it is clear that the responses in phases b and c
lag the corresponding responses occur in balanced sets.
a
c
Z
Ic
Ib
b
Z
Z
Ica
Ibc
Iab
Ia
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Method : per Phase analysis. We next turn to the method of per phase
analysis. Given a balanced three-phase network with no mutual inductance between
phases :
1. Convert all delta-connected sources and loads into equivalent wye connections.
2.
Solve for the desired phase a variables using the phase a circuit with all
neutrals connected.
3. The phase b and phase c variables can then be determined by inspection ;
subtract 1200and 2400, respectively, from the phase angles found in step 2, in the
usual case of abc-sequence. Add 1200 and 2400 in the case of acb-sequence
sources.4.
If necessary, go back to the original circuit to find line-line variables or variables
internal to delta connection.
Note: In using per phase analysis we always pick the neutral as datum. In this
case it is simpler to use a single-subscript notation for phase voltage. We will use Va
rather than Van, etc.
Examples 2.4
Given the balanced three-phase system shown in fig. 2.13 (a), find v1(t) and i2(t)
Figure 2. 13 (a)
SolutionReplace delta by equivalent Wye, Za = -j2/3. Using per phase analysis, we
consider the per phase (phase a) circuit shown in fig. 2.13 (b). Note that while v1
j0.1
j1.0
n
+
-
+
-v1
b
1 + j0.01
cbc c b
a
-j2n
a
oa 45
2
350E
a
i2
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appears in the phase circuit, i2 has been suppressed in the process of the -Y
conversion. Carrying out the calculation to find v1, we note that the equivalent
parallel impedance from a to n isj2. Using the voltage-divider law.
Figure 2. 13 (b)
aa EEjj
jv 05,1.
1.02
21
= 045.
2
368
The phasor v1 represents the sinusoid v1(t) = 368 cos ( 45t0). From the
original circuit we see that to find i2(t), we need to find Vab. Thus,
Vab= Va,n-Vbn=''6/3 nVaje
Then using the impedance of the capacitor, we find that
0
'' 165.2
319baI
and it follows that the corresponding sinusoidal is
02 165cos319 tti
Example 2.5
In the circuit shown in figure 2.14, the source phasor voltage is V = 30 /15 0.
Determine the phasor currents I2 and I3 and the impedance Z2. Assume that I1 is
equal to 5 A.
Solution
The voltage V2is given by
+
-
oa 45
2
350E
j0.1
+
-v1 j1.0
32j
a
nn
a
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0
0
2
12
94.17/20.25
0/515/30
1
V
IVV
Figure 2-14. Circuit for Problem 2-1.
The current I3is thus
023 94.17/52.2
10
VI
The current I2is obtained using KCL. Thus
0
312
61.16/72.2
III
Finally we have
ohmsj
I
VZ
26.564.7
56.34/28.9 0
2
2
2
Example 2.6
For the circuit of problem 2.5, calculate the apparent power produced by the
source and individual apparent power consumed by the 1-ohm resistor, the
impedance Z2, and the resistance R3. Show that conservation of power hold true.
Solution
The apparent power produced by the source is
~
b
n
-
+Z2I2V I3 R3= 10
I1
a
V2-
+
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VAj
VA
VISs
82.3889.144
15/150
0/515/30
0
0
*
1
The apparent power taken by the 1-ohm resistor is
WI
II
IVS abt
25
1
2
1
*
11
*
1
The apparent power taken by the impedance Z2is
VAj
IVS
82.3837.56
56.34/44.68
61.16/72.294.17/20.25
0
0
*
222
The apparent power taken by the resistor R3is
W
IVS
52.63
94.17/52.294.17/20.25
*
323
The total power consumed is
VAj
SSSSt
83.3889.144
321
This is equal to the source apparent power, which proves the principle ofconservation of power.
Example 2.7
A three phase transmission link is rated 100 kVA at 2300 V. when operating
at rated load, the total resistive and reactive voltage drop in the link are, respectively,
2.4 and 3.6 percent of the rated voltage. Determine the rated power and power factor
when the link delivers 60 kW at 0.8 PF lagging at 2300 V.
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Solution :
The active voltage drop per phase is
3
2300024.0 IRVr
The reactive drop is
3
2300036.0 IXVx
But the rated current is
AI 1.2532300
10100
3
As a result,
ohmsX
ohmsR
90.1
27.1
For a load of 60 kW at 0.8 PF lagging, the phase current is
AIl 83.18
8.032300
1060 3
The active and reactive power consumed by the link are thus
var32.20203
43.13503
2
2
XIQ
WRIP
llk
llk
As a result, the apparent power consumed by the link is
32.202043.1350 jSlk
The apparent load power is
VAj
St
000,45000,60
8.0cos/8.0
1060 13
Thus the total apparent power is now obtained by
047.37/74.77296
32.4702043.61350
jSt
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24
As a result
kWP
lagging
t
t
35043.61
79.0
47.37coscos 0