CH2

CHAPTER 2

SINGLE-PHASE CIRCUITS

An electric power system is a three-phase system that contains generators, trans-formers, transmission lines, and loads. Transmission lines and cables are linearelements. Generators and transformers are also linear devices if iron saturation isneglected. Loads are generally nonlinear. A typical load in the power system drawsconstant power at a constant power factor. The proper operation of electric powerdevices requires that the load voltage be kept within 5% of the rated voltage.

Most three-phase loads are balanced, which permits the representation of thethree-phase system by a single-phase circuit. This emphasizes the importance ofthe discussion of single-phase circuits in an electric power textbook. Although it isassumed that the reader is familiar with single-phase circuit analysis techniques,in this chapter we review the fundamental concepts and establish the nomen-clature and terminology utilized in this book. After an overview of the basicprinciples, we teach the use of PSpice, Mathcad and MATLAB programs forcircuit analysis. The principal analysis concepts are presented through interactivederivations and numerical solutions using computer tools.

2.1 CIRCUIT ANALYSIS FUNDAMENTALS

2.1.1 Basic Definitions and Nomenclature

The basic electrical quantities include current (I), voltage (V) and power (P).Voltage is an electromotive force or potential. Voltage is the difference in energy

Electrical Energy Conversion and Transport: An Interactive Computer-Based Approach, by GeorgeG. Karady and Keith E. HolbertISBN 0-471-47652-8 Copyright 2005 The Institute of Electrical and Electronics Engineers

49

50 SINGLE-PHASE CIRCUITS

b

Circuit 1 __Element.

Figure 2.1 Passive sign convention, current enters the circuit element at the positivevoltage terminal.

level of a unit charge located at each of two points in a circuit, and therefore,represents the energy required to move the unit charge from one point to the other.Current is time rate of change of electric charge (q) at a given location, that is,i (t) = ~;, and is measured in amperes (A). Normally we talk about the movementof positive charges although we know that, in general, for metallic conductors,current results from electron motion (conventionally positive flow). The sign ofthe current indicates its direction of flow relative to some pre-chosen referencedirection. In selecting the current direction, it is useful to follow the passivesign convention, which designates that current should enter the positive voltageterminal of a circuit element (see Figure 2.1). If the passive sign convention isobeyed, then determining whether an element supplies or absorbs power is madeeasier. In particular, if power (which is the product of the current and voltage)is positive, then the element is absorbing power; conversely, if the power isnegative, the circuit element is supplying power; in summary:

P 0

Element supplies power

Element absorbs power

The conventions prescribed by the Institute for Electrical and Electronics Engi-neers (IEEE) are employed in this book (ANSI/IEEE Std 280-1985). The instan-taneous values of time-varying quantities such as voltage and current are repre-sented by lowercase letters, for example, v(t) and i (t). Uppercase letters charac-terize values associated with time varying quantities such as the root-mean-squarevoltage (Vrms ) and average power (P). Bold uppercase letters denote frequency-dependent variables such as complex power (8) and phasor voltages and currents(e.g., Vrms and I). Figure 2.1 illustrates the equivalency of three designationsfor voltage polarity: (1) the use of explicit +/- labeling, (2) an arrow that pointstoward the positive terminal, and (3) the notation Vab for which the first letterof the subscript indicates the positive terminal and the second letter denotes thenegative terminal.

2.1.2 Voltage and Current PhasorsThe time domain functions are used for transient analysis. Single- and three-phase analyses are more easily performed in the frequency domain using phasorrepresentation. A phasor is a complex representation of a truly real quantity. A

IMPEDANCE 51

general sinusoidal voltage may be transformed from its time domain expression:

v(t) = VM COS(wt + 8) (2.1)

to an equivalent phasor expression by merely considering the magnitude (VM )and phase shift (8) of the original signal, that is, the phasor representation is

(2.2)

The supply voltage is often selected as the reference with a phase angle of 8 = O.Engineering calculations utilize the amplitude (rms value) and phase angle of bothvoltage and current. The amplitude and phase angle can be calculated using apolar (or complex) notation.

The root-mean-square (rms) value is calculated from

Vrms = ! (T v(t)2 dtT 10

For a sinusoidal signal the magnitude and rms values are related by

Hence, the time-domain voltage equation becomes

v(t) = J2vrms cos(wt + 8)

(2.3)

(2.4)

(2.5)

(2.6)

2nwhere co = 2n f = - is the angular frequency in radians per second; f is the

Tcyclic frequency, 60 Hz in the United States, and 50 Hz in Europe and otherparts of the world; and T is the period of the sinusoidal waveform.

Similarly, the current may be represented as

i (t) = 1M cos(wt + ) = J2lrms cos(wt + )or

I = IML = -J2lrms LOften it is necessary to describe the phase shift between current and voltage.

For alternating current (ac) waveforms at the same frequency, v(t) is said to leadi (t) by 8-, or equivalently i (t) lags v(t) by the same amount. If 8 = , thenthe two waveforms are said to be in phase; if 8 i= , then the two waveformsare out ofphase.

2.2 IMPEDANCE

Ohm's Law provides a linear relationship between voltage and current:

V=IZ (2.7)where Z is the impedance, having units of ohms, Q. Unlike voltage and current,impedance is a truly complex quantity that can be expressed in rectangular or


polar notation. Impedance consists of a resistive term, R, and an imaginary term,the reactance, X. The reactance is a function of frequency, which generally makesthe impedance a function of frequency:

Z(w) = R + jX(w) (2.8)

The relation between the magnitude of the impedance, and resistance andreactance can be visualized by the impedance triangle shown in Figure 2.2. Thepolar representation of the impedance is

Z = IZIL8 = IZle j O = IZI[cos(8) + j sin(8)] = R + j Xand

IZI = JR2 + X2 () = arctan (~)(2.9)

Addition and subtraction of complex numbers is easiest using rectangular repre-sentations; whereas phasor (or polar) notation leads to simplified multiplicationand division of complex quantities.

Most calculators and computer software have built-in functions for performingthe complex math associated with phasor and impedances; however, the resultsdepend upon the user's proper utilization of such engineering tools. As a simpleexample, consider the conversion of the impedance Z = - 3 - j 4 Q to a polarrepresentation using MATLAB. The magnitude is easily found using the absolutevalue function (abs)

Z = -3-4j; abs(Z)ans

5

An inexperienced user might simply use the atan function to find the angle

atan(imag(Z)/real(Z *180/pians =

53.1301

x

R

Figure 2.2 Impedance triangle.

IMPEDANCE 53

But the answer above is incorrect, in that the impedance lies in the third quadrant,not the first. MATLAB provides a different function, angle, for properly calcu-lating the angle of a complex variable, but the resultant angle is in radians:

I.

angle(Z) *180/pians =

-126.8699

The reciprocal of the impedance is the admittance, Y = liZ. The impedance andadmittance relations for the resistor, capacitor and inductor are given in Table 2.1.The reactance of an inductor and a capacitor is Xind = col. and X cap = -1/(wC),respectively.

2.2.1 Series ConnectionWhen M impedances are connected in series, the resulting equivalent impedance is

(2.10)

Example: The equivalent impedance of a circuit with a resistor, inductor andcapacitor connected in series is

1Ze = ZR + ZL + Zc = R + jwLind + -.--jWCcap

= IZle j O = IZI[cos(8) + j sin(8)]

where

IZI= R2 + (WLind __1_)2wCcap

e = arctan

1wLind---

wCcapR

The phase angle (e) is positive in an inductive circuit and negative in a capaci-tive circuit.

TABLE 2.1 Passive Circuit Elements

Element

Capacitor (C)Inductor (L)Resistor (R)

Impedance

Zc = l/(jwC) = -1/(wC)L90ZL = jcol. = wLL90ZR = R = RLO

Admittance

Yc = jwCYL = l/(jwL)YR = l/R


Figure 2.3 Parallel impedances with equivalent.

2.2.2 Parallel ConnectionCircuit elements which share the same two end nodes are said to be in parallel.When N impedances are connected in parallel (see Figure 2.3), the resultingequivalent impedance is

1Ze= ---

N 1L Zkk=l

(2.11)

In the case of two impedances connected in parallel the above equation canbe simplified as follows:

1r; = 1 1

-+-z, Z2Alternatively, the admittance of a circuit with N components connected in parallelcan be calculated by adding the admittances of the components together.

(2.12)

ExampleDetermine the impedance of a circuit with a resistor, inductor and capacitorconnected in parallel. The admittance of a circuit with resistance, inductance andcapacitance connected in parallel is

1 1y = - + . + jwCcap

R JwLind

The impedance of this circuit is

1 1Z=y= 1 1

-+--+jwCR jwLind cap

1111-+--+--R j Xind j Xcap

2.2.3 Mathcad ExamplesThe practical aspects of series and parallel circuit impedance calculations willbe presented through numerical examples. The analysis of a simple circuit with

IMPEDANCE 55

complex impedances using a hand calculator is a complicated and lengthy pro-cedure. The circuit analysis process can be simplified by using general-purposecalculation programs like MATLAB, Mathcad or Mathematica. In this book Math-cad and MATLAB are used.

This first example is performed using Mathcad. Particular advantages of thisprogram are that the equations appear as the hand written equations, and theprogram calculates and converts the units automatically. In Appendices A andB, respectively, short tutorials for Mathcad and MATLAB are presented. Thetutorials present the basic knowledge necessary to get started.

Series Connection of ImpedancesUtilities can connect a capacitor in series with a transmission line to compensatethe line impedance. This reduces the voltage drop in cases of heavy load on theline. Figure 2.4 shows the one-line circuit diagram, and Figure 2.5 shows theequivalent circuit.

The system frequency data are f := 60 Hz ro := 2nfThe transmission line resistance and reactance per mile are

nRmi:= 0.32~

rm

nXmi := 0.75~rm

The transmission line length is dline:= 3miThe transmission line resistance and reactance are

R1ine = 0.96 n

Xline = 2.25 n

TransmissionLine

Capacitor

Figure 2.4 One-line diagram of a transmission line compensated by a capacitor con-nected in series.

----1 t--

Figure 2.5 Equivalent circuit of a transmission line compensated with a series capacitor.


Note that Mathcad has two equal signs: the first equal sign (:=) is used when avariable is defined; the second equal sign (=) is used to obtain the value of avariable. The capacitor value is Ccomp := 1572 flF

The reactance of the capacitance is

-1Xcomp := ro,Ccomp

Xcomp = -1.687 11

The line resistance and reactance, and the reactance of the capacitor are con-nected in series. The equivalent impedance of the compensated line is

Zser := j ,Xcomp + Rline+ j eXline Zser = 0.96 + 0.563j 11

Parallel Connection of ImpedancesOperators connect a capacitor in parallel with motors to improve the power factor.Figure 2.6 shows the one-line diagram for a motor and capacitor connected inparallel. The motor is represented by an inductance and resistance connected inparallel, as shown in Figure 2.7. The values are

Xmot := 23 11 Rmot := 20 11

The compensating capacitor is Ccomp_M := 500 JlFThe reactance of the capacitor is

-1Xcomp M := Xcomp_M = -5.30511

- ro-Ccomp_M

Motor

M

1---tE-Capacitor

Figure 2.6 One-line diagram of a motor and capacitor connected in parallel.

Figure 2.7 Equivalent circuit for a motor and capacitor connected in parallel.

IMPEDANCE 57

The three impedances are connected in parallel. The equivalent impedance is

~ar:= 1 1 1-- +--- +-----Rmot j .Xmot j .Xcomp_M

~ar = 2.125 - 6.163j Q

Combined Series and Parallel Connection of ImpedancesWe assume that the compensated transmission line supplies the motor and capac-itor load. Figure 2.8 shows the one-line circuit diagram of the combined system,and Figure 2.9 shows the equivalent circuit, which is a series combination of thecompensated line and motor.

The total impedance of the system can be obtained immediately by adding thetwo equivalent circuit impedances together.

Zequ = 3.085 - 5.6j Q

2.2.4 MATLAB ExamplesThe impedance calculation of the three series and/or parallel networks used abovein the Mathcad examples is repeated here using MATLAB. For all three examples,the system cyclic and angular frequency must first be established as shown inImpedance.m below.

% Impedance.mf = 60;wfreq = 2 * pi * f;

% system frequency (Hz)% angular frequency (rad/sec)

Capacitor

TransmissionLine

Motor

M

Capacitor

Figure 2.8 Compensated transmission line supplies a motor with capacitor load.

C comp X line R1ine Zser

---1Ccomp_M

Figure 2.9 Equivalent circuit of the system in Figure 2.8.

% farads% ohms


Series Connection of ImpedancesThe equivalent impedance of the series circuit of Figure 2.5 is readily seen to bethe summation of the individual component impedances:

1Zser = ZUne + Zcomp = RUne + jXUne + -.---

JWCcomp

The uncompensated line impedance formed by the series resistor and inductoris first computed.

% 1. Series connection of impedances

% establish transmission line parametersRmi = 0.32; % line resistance (ohm/mile)Xmi = 0.75; % line reactance (ohm/mile)length = 3; % line length (miles)Zline = (Rmi + j*Xmi) * length % uncompensated line

% impedance (ohms)

Next, the reactance (impedance) of the compensating capacitor is calculated.

% compute impedance of compensating capacitor for% transmission lineCcomp 1572e-6;Zcomp = 1 / (j*wfreq*Ccomp)

The capacitor impedance is added to the line impedance.

% compute series equivalent impedance of compensated% lineZser = Zline + Zcomp

The results as computed by MATLAB are:

Zline =0.9600 + 2.2500i

Zcomp =o - 1.6874i

Zser =0.9600 + 0.5626i

IMPEDANCE 59

Parallel Connection of ImpedancesThe equivalent impedance of the parallel circuit shown in Figure 2.7 is alsocalculated in a straightforward manner:

1 1 1-- = -- + -.-- + jwCcomp MZpar Rmot ] Xmot -

After setting the motor resistance and reactance, the impedance of the parallelcapacitor is computed.

% 2. Parallel connection of impedances

% establish motor resistance and reactanceRmot 20; % ohmsXmot = 23; % ohms

% compute capacitor reactanceCcompM 500e-6;ZcompM = 1 / (j*wfreq*CcompM)

% farads% ohms

The equivalent parallel impedance of the capacitor, inductor and resistoris calculated

% combine the parallel motor elements and capacitorZpar = 1 / (l/Rmot + l/(j*Xmot) + l/ZcompM) % ohms

The MATLAB computed results are:

ZcompMo - 5.3052i

Zpar =2.1249 - 6.1631i

Combined Series and Parallel Connection of Impedances

The overall equivalent impedance from combining the compensated transmissionline and motor in series (see Figure 2.9) is simply the sum of the their impedances

% 3. Combined series and parallel connection of% impedancesZequ = Zser + Zpar % ohms


The MATLAB result below matches the answer found using Mathcad.

Zequ =3.0849 - 5.6005i

2.2.5 Delta-wye TransformationA particular configuration of impedances that does not lend itself to the usingseries and parallel combination techniques is that of a delta (~) connection. Insuch cases, the delta (~) connection is converted to a wye (Y) configuration, asshown in Figure 2.10.

ZlZ2Za=-----Zl + Z2 + Z3

ZlZ3Zb=-----r, + Z2 + Z3

Z2Z3Zc=-----Zl + Z2 + Z3

(2.13)

The above formula is more easily remembered as the impedance next to a par-ticular Y node is the product of the two delta impedances connected to that nodedivided by the sum of the three delta impedances. The reverse transformation (Yto ~) can also be performed:

ZaZb + ZbZc + ZcZaZl = --------Zc

ZaZb + ZbZc + ZcZaZ2 = --------Zb

ZaZb + ZbZc + ZcZaZ3 = --------Za

(2.14)

For the balanced case in which all the delta (Z~ = Zl = Z2 = Z3) and wye(Zy = Za = Zb = Zc) impedances are equal, the.transformation formula reduces

a b a b

c c

Figure 2.10 Delta-wye impedance transformation.

POWER 61

to simply(2.15)

2.3 POWER

The instantaneous power is the product of the instantaneous values of the voltageand current. The instantaneous power, pet), is

pet) = v(t)i(t) = ,J2vrms cos(wt + 8),J2lrms cos(wt + 4

Using the trigonometric identity

cos(a) cos(fJ) = ~ cos(a - fJ) + ~ cos(a + fJ)the expression for instantaneous power may be rewritten as

pet) = Vrmslrms[cos(8 - 4 + cos(2wt + 8 + 4]

(2.16)

(2.17)

(2.18)

The first term is a constant, and the second term has a cyclic frequency at twicethe applied frequency. These facts make calculation of the average power, P,easier, since we simply integrate a constant and a cosine wave over exactlytwice its period (which is equal to zero):

p = ~ (T p(t) dt = Vrmslrms cos(8 -


The derivation shows that the real part of the complex power is the averagepower. The imaginary part of 8 is called the reactive power, Q. The reactive orquadrature power indicates temporary energy storage rather than any real powerloss in an element. To distinguish Q from P and 8, reactive power is expressedin units of VOlts-amperes reactive (VAR).

The reactive power is positive in an inductive load circuit, where the loadphase angle is positive and the current is lagging the voltage;

The reactive power is negative in a capacitive load circuit, where the loadphase angle is negative and the current is leading the voltage.

The relation between the magnitude of the complex or apparent power, and thereal power and reactive power can be visualized by the power triangle shown inFigure 2.11.

The powerfactor, pI, is defined as the ratio of average power to apparent powerp p

pf= = - = cos(8 -

POWER 63

I rms = Inns A

Z =IZI ~

(2.23)

Figure 2.12 General load.

Because the cosine of an angle between -900 to +900 ranges from zero tounity, it is necessary to supplement the actual pf value with a designation as towhether the current is leading or lagging the voltage as illustrated in Table 2.2.

If Vrms and I rms are the voltage and current at the load impedance Z (seeFigure 2.12), then

Z = Vrms = VrmsL8 = Vrms L8 - 4J = IZILOI rms i.: L /rms

The load phase angle, e, then describes the amount by which the voltage leadsthe current. In this case, the power factor angle is equivalent to the load angle,that is, e = 8 - . Then in a purely resistive load, the power factor is unity(Pf = 1); whereas in a purely reactive (capacitive or inductive) load the powerfactor is zero (Pf = 0).

If the power factor is unity, then the power is simply the product of the rmscurrent and voltage

(2.24)

(2.25)

The above formula demonstrates one reason for utilizing the effective (or rms)values, since they allow us to write an equation for the average power like thoseused in de circuits (i.e., P = /2R); in fact, that relation is ultimately the basisfor defining the rms value. For a resistor, Ohm's Law can be combined with theabove formula to find the power dissipated in a resistor

V2P = /2 R = rmsrms R

Note that since the resistance is always positive, then the resultant power valuein the above expression must also always be positive. Recall that for P > 0that power is absorbed by an element, which means that the resistor alwaysabsorbs power.

For ac steady-state circuits, each of the following quantities is conserved:

instantaneous power L Pk(t)k

real power L r,k


reactive power L a.k

complex power L s.k

where the summations are over all the circuit elements in the network. Apparentpower is not conserved.

2.4 AC CIRCUITS

The typical linear single-phase ac circuit contains voltage sources and impedances(resistance, inductance and capacitance). The components may be connectedin series or parallel, or combinations of the two can occur. As an example,Figure 2.13 shows a simple single-phase circuit. In this circuit, an ac voltagesource generates a sinusoidal ac voltage that drives current through a resistanceand a reactance. An actual generator used in a power system is more complexthan a simple voltage source; however, to a first approximation a generator canbe represented as a simple voltage source.

Figure 2.14 shows the sinusoidal voltage waveform generated by the voltagesource. The voltage direction (or polarity) is indicated on the circuit diagramof Figure 2.13 by an arrow from g to 3. This means during the positive halfcycle the potential of point 3 is larger than g. The generator current flows fromg to a in the positive half cycle; The generator current and voltage are in thesame direction. The load current in the positive half cycle flows from b to g;The load current and voltages are in opposite direction.

The sinusoidal voltage drives a sinusoidal current through the circuit. For avoltage source with a reference phase shift of zero (8 = 0), the current equation is

i(t) = hlrms cos(wt - e) (2.26)where e = 8 - 4J is the phase shift between the voltage and current waveforms.The current can be lagging or leading the voltage. The current is lagging (e > 0)

a i(t)-+

L

R

b

g

Figure 2.13 Example single-phase circuit.

AC CIRCUITS 65

~IT

v.

.:


The phase angle between the resistance and reactance is calculated from

X dtan()) = ---.!.!!.-

R(2.29)

System Efficiency and Voltage RegulationA one-line diagram is often used to provide a concise representation of thenetwork under study. The one-line diagram is a simplified schematic that showsonly the major components and major interconnection(s) between components.As an example, Figure 2.16 shows a three-phase generator supplying a networkthrough a transformer. The shorthand notation of the one-line diagram is thenexpanded into a circuit schematic for analysis. The one-line connection diagramof the system shown in Figure 2.16 can be converted to the three-phase circuitgiven in Figure 2.17, which shows all the components in each phase. Further,in the case of a balanced load, the single-phase equivalent ac circuit diagram,shown on Figure 2.18, could be used for analysis.

The performance of a power system can be quantified in terms of its efficiencyand the voltage drop through the system. The system efficiency is the ratio ofthe output, in most cases the load power, and the input, in most cases the supplypower. For the system of Figure 2.16, the system efficiency can be expressed as

. P net P gen - Plossefficiency = -- = -----

Pgen Pgen(2.30)

Transformer

&-----3 t-------OGenerator ~-Y Network

Figure 2.16 One-line diagram of a simple three-phase system.

Networkr---------I

ru 1111111111... _

TransformerGenerator1----------11 ru1111111 11 11 1L... J

Figure 2.17 Three-phase connection diagram.

AC CIRCUITS 67

Figure 2.18 Single-phase equivalent circuit for balanced load.

Generator

TransmissionLine

Load

Figure 2.19 One-line diagram for load served via transmission or feeder line.

Figure 2.20 Circuit schematic for load served via transmission or feeder line.

where Pgen and Pnet are the supplied generator power and consumed networkpower, respectively, and Ploss is the power loss in the system. An efficiency foran individual component, such as a transformer or a transmission line, can bedefined in a similar manner.

The power engineer is also concerned with the reduction in voltage through thesystem, and ultimately the voltage provided to the customer. Consider the one-line diagram of Figure 2.19 which shows a generator supplying a load througha transmission line. The equivalent single-phase circuit is shown in Figure 2.20.The voltage regulation is defined based upon the input voltage (Vgen in the caseof the network of Figure 2.20) being held constant; the voltage regulation isthe difference between the magnitudes of the voltage at the load terminals forthe no-load (Vno-load) and (full) load conditions divided by the load voltage(Vload) magnitude:

IVno-loadl - IVloadlVoltage Regulation = x 100%IVloadl

(2.31)


The no-load voltage is equal to the generator supply voltage (Vgen) if the loadcurrent is zero. Figure 2.20 shows a single load, if this load is switched off, theno-load current is zero and the load voltage is equal to the supply voltage. Hence,the formula for the voltage regulation can be expressed simply as the differencebetween the magnitudes of the supply (sending end) and load (receiving end)voltages divided by the load voltage magnitude:

. IVgenl - IVloadlVoltage Regulation = x 100%IVloadl

(2.32)

This latter expression for the voltage regulation is frequently applicable in theexamples of this book, and requires a single circuit analysis in contrast toEquation (2.31), which requires analyses at both loaded and unloaded conditions.

MATLAB Power Factor Example

The electrical network of a utility supplies an industrial load 100 kW, 60 Hz,480 V (rms) through a feeder (see Figure 2.19). The feeder impedance is 0.15 +jO.6 ohms. Vary the load pi from 0.6 to 0.9 lagging, and plot the supply realpower versus pi to determine the generation requirements ("cost") for the utilityto serve this industrial customer.

Referring to Figure 2.20, the rms current through the line to the load can becalculated using

PLoad[Line = ---plVLoad

The real power supplied by the utility is that necessary to serve the load aswell as the real power loss in the transmission line:

PSupply = P Load + RLineI'iine

Note that this calculation is independent of the system frequency and whetherthe load power factor is leading or lagging.

Begin the MATLAB script file by establishing the system circuit data:

% PowerFactor.m

% Establish the circuit parametersPload 100e3; % wattsVload 480; % volts-rms

Rline 0.15; % ohms

Next, we vary the power factor as requested using a step size of 0.001, whichwill eventually generate about 300 points to be plotted.

AC CIRCUITS 69

% Vary the power factor from 0 .6 to 0 .9 , with steps% of 0 .001pf = 0 .6 : 0.001 : 0 .9 ;

The line-load current and power supplied by the utility are then computedusing the equations above. Note the use of the MATLAB " .1" and ".~" element-by-element division and power operators, respectively.

% Compute a vector of line-load currents for the pf% valuesIline = Pload . / (pf * Vload);

% Compute the power supplied by utility for each pf% valuePsupply = Pload + Rline * Iline . A2;

MATLAB can then be used to easily plot the results, which are shown inFigure 2.2I.

.) Figure No. I '~1Eile f:dot ~ Insert IOOs I!indow tjelp

Power "Cost" to Utility for Uncompensated Load120,.---,.----.----.----.------,-----r---,--------,

118

~1166."

.!!!]: 114:::J(f)

ia. 112

110

0.6 0.65 0.7 0.75 0.8Power Facto r (PQ

0.85 0.9 0.95

Figure 2.21 MATLAB plot of utility supplied power as a function of power factor.


% Plot the power supplied v. the power factorplot(pf, Psupply/1000);xlabel('Power Factor (pf) ');ylabel('Power Supplied (kW) ');title('Power "Cost" to Utility for Uncompensated Load');

The graph shows that if the load has a power factor of 0.6, then the utilitymust generate about 118 kW to supply the 100 kW load. Whereas if the loadpower factor is raised to 0.9, then the utility need only generate 108 kW to supplythe load.

2.5 BASIC LAWS

2.5.1 Kirchhoff's Current LawKirchhoff's current law (KCL) states that the sum of the currents entering anynode is zero. The other KCL formulation is that the current flowing toward thenode is equal to the current flowing away from the node (see Figure 2.22). ForN currents entering a node, KCL is written as

or (2.33)

A network with three nodes is presented in Figure 2.23. The nodal (KCL)equations for nodes A, Band C are

node A: IA + ICA - lAB = 0node B: IB + lAB - IBC = 0node C: Ic + IBC - ICA = 0

2.5.1.1 Current Division. The combination of KCL and Ohm's Law to circuitelements that are in parallel results in the well-known current divider equation.

Figure 2.22 Illustration of Kirchhoff's current law.

BASIC LAWS 71

---... A

~Ac

..

Figure 2.23 Example network for applying KCL nodal equations.

(2.34)

Figure 2.24 Network for applying current division.

Referring to Figure 2.24, the current (Ik ) through the parallel impedance Zk is

Ik = IT ZparZk

where IT is the total current through all the parallel elements and Zpar is theequivalent impedance for the N parallel elements. For the special case of onlytwo elements in parallel (Z. and Z2) the above formula reduces to

(2.35)

where this expression is normally read as the current through one branch is thetotal current times the impedance in the other branch divided by the sum of thetwo branch impedances.

2.5.1.2 Nodal Analysis. Nodal analysis systematically applies KCL to thenodes of a circuit in order to solve for the voltages at each node. There are threebasic steps followed in nodal analysis:

1. Choose a reference (ground or neutral) node where the voltage is deemedzero, and then assign (label) unique voltages to the other nodes.


2. Apply KCL to each node other than the reference node while expressingthe currents in terms of the node voltages using Ohm's Law (I = t::,.V jZ).

3. Solve the resulting system of KCL equations for the nodal voltages.

A practical example is two electric distribution networks interconnected througha transmission line. Figure 2.25(a) shows the one-line diagram (arrangement) ofthe system. A shunt capacitor is placed in the middle of the line to improve volt-age regulation.

Figure 2.25(b) shows the equivalent circuit diagram. Each of the distributionsystems is represented by a voltage source. The magnitudes of the two voltagesources are equal, but there is a 60 phase difference between them. The voltageof Source 1 is the reference voltage, with a zero degree (0) phase angle. Source2 has a phase shift of 60 with respect to Source 1, that is, V2 = VI L - 60.The transmission line is represented with a resistance and reactance connectedin series. A capacitance is placed at the middle of the line. This requires thedivision of the line into two parts: Line 1 and Line 2.

The example here presents the practical application of the KCL equationsmethod, referred to as nodal analysis. First, a solution is obtained by hand. Thisis followed by a computer solution using Mathead.

Method 1: Hand SolutionThe hand solution follows the nodal analysis procedure above as follows:

Step 1: The nodes have already been labeled in Figure 2.25(b). The nodalvoltages for nodes 1 and 2 are simply the source voltages since each source isdirectly connected to the neutral (ground) node. The remaining node 3 is assignedthe voltage, V3.

Step 2: Since the voltages at nodes 1 and 2 are already known, it is onlynecessary to write a single KCL expression, specifically at node 3:

Network 1

I INetwork 2

Line 1 _Line_2_I IC . Iapacitor I

(a)

tR2 ~ 2

v1t tV2Source 1 n Source 2

(b)

Figure 2.25 One-line diagram and equivalent circuit for the interconnection of two elec-tric distribution networks. (a) One-line diagram; (b) Equivalent circuit.

BASIC LAWS 73

For each current in the KCL expression we now substitute an equivalent Ohm'sLaw expression in terms of the node voltages.

Vt - V3 _~ + V2 - V3 = 0R1+jX1 jXcap R2+jX2

Step 3: Here there is a single equation to be solved. Algebraic manipulationleads to an expression for the only unknown nodal voltage, that of V3:

V t V2R1+jX1 + R2+jX21 1 1

R 'X + -:--X + R 'X1 + ] 1 ] cap 2 + ] 2V3 = -------------

Substituting in the circuit parameter values yields a value for V3. The values ofthe nodal voltages can then be used to find the three currents: 11, 12 and 13

V t - V3It = ----R1+jX1

Method 2: Mathcad SolutionHere we again solve the circuit of Figure 2.25, but using a computer-based nodalanalysis. The reader is encouraged to utilize his/her computer and follow thecalculation using the Mathcad program.

Mathcad Hint: Electrical engineers prefer to use a lowercase j to denote theimaginary number rather than a lowercase i to avoid confusion with currenti (t). However, Mathcad defaults to using the letter i for the imaginary number.This can be changed to j in the Display Options tab of the Result Formatdialog box. In this book, we have chosen to follow the standard conventionand use j for the imaginary unit.

The physical circuit data are first defined. The -600 phase angle for voltagesource V2 can be incorporated using the complex exponential, that is, Lf) = ej e .

VI := 7.2 kVV

2:= 7.2.e- j60 deg kV (0:= 21t60 Hz

Xl := 11 Q

Ccap := 100 W-1

Xcap := - - -(O,Ccap


Immediate results are computed for

V2 =3.6 - 6.235j kV Xcap =-26.526 Q

Mathcad has an equation solver that simplifies the circuit analysis. However,for comparison we will first solve this problem using the equations derived in thehand calculation above. After this, we use the Mathcad equation solver to demon-strate the advantages of the computer use for elementary circuit calculations.

Typing and dimensional errors may prohibit Mathcad from performing thecalculations. To reduce such errors, we display the results of intermediate calcu-lations as we progress through the problem. To do so, we select a guess (seed)value for the variable; this causes Mathcad to compute a numerical result assoon as each new equation is entered. The reasonableness of the numerical val-ues obtained at each step helps to assure us that the equations are solvable andthat we have not erred in typing the formula. In order to verify the equations inthis problem, we select a guess value of V3 := 7 kV.

Approach #1: The nodal current equation (KCL) for node 3 is

The currents can be calculated by dividing the voltage difference by theimpedance, that is, Ohm's Law. The voltage difference between node 1 and3 is VI - V3. The impedance is R1 + jX 1. The current is

11 = 5.839- 16.058j A

Current 12 is calculated similarly. The result is

V2- V312 := - - -R2 + jX2

12 =-599.921+ 90.938j A

The capacitance current 13 is the voltage V3 divided by the capacitive impedance.

13 = 263.894j A

The numerical values help to assure the validity of the equations, but these arenot the actual current values since V3 is a guess value. The substitution of thecurrent values in the nodal equation results in the following equation:

VI - V3 V3 V2 - V3--- + =0

RI + jXI jXcap R2 + jX2

BASIC LAWS 75

where V3 is the unknown in this equation. The rearrangement of the equationresults in

The equation is solved for V3 . The result isVI V2

---+---R1 + jX1 R2 + jX2 V3 = 6.381- 4.54j kV

Approach #2 : Another way is to solve the nodal equation is using the MathcadFind equation solver. The Mathcad equation solvers are more fully explained inAppendix A. The use of the Find equation solver requires a guess value for V3,which was selected before.

GivenVI - V3 V3 V2 - V3------+ =0R1 + jX1 jXcap R2 + jX2

the equation-solver immediately computes the voltage:

Find(V3) = 6.381- 4.54j kV arg(V3) = -35.434 deg

We can see that the two approaches result in the same voltage. The magnitudeoperator (II) and arg function are utilized above to change the complex num-ber expression for V3 into a phasor quantity. The corresponding values of thecurrents are

VI - V311 := - - -R1 + jX111 = 388.46+ 66.775j A

V2 - V312 := - - -R2 + jX212 = -217.3 + 173.771j A

V313 := - -jXcap13 = 171.16+ 240.546j A


2.5.2 Kirchhoff's Voltage LawKirchhoffs voltage law (KVL) states that the sum of the voltages around a loopis zero. The other formulation is that the supply voltage is equal with the loadvoltages or voltage drops in a closed loop. KVL is written as

or (2.36)

It is imperative that the correct voltage polarity be used in the above formulaeas the loop is traversed. Often the engineer prefers to begin at a voltage source,which is taken to be a voltage increase (positive voltage), and then follow theloop using that source as the reference polarity.

Figure 2.26 presents a simple circuit to demonstrate application of KVL loopequations. In this circuit the voltage source, Vs. drives the current I throughthe resistance and reactance. The current produces a voltage drop VR across theresistance and a voltage drop Vx across the inductance. KVL states that the sumof the three voltages is zero. The loop equation is

Vs - VR - Vx = 0or

Vs = IR+ljXwhere VR = I R and Vx = I jX are the voltage drops across the resistor andinductor, respectively.

2.5.2. 1 Voltage Division. The combination of KVL and Ohm's Law to circuitelements that are in series and have the same current flowing through them (i.e.,in a single loop), results in the well-known voltage divider equation. Referringto Figure 2.27, the voltage (Vk) across the series impedance Zk is

(2.37)

I__... a

L

R

b

g

Figure 2.26 Simple circuit for demonstration of loop equation.

BASIC LAWS 77

Figure 2.27 Network for applying voltage division.

where VT is the total voltage across all the series impedance elements and Zseris the equivalent impedance for the N series elements.

2.5.2.2 Loop Analysis. Loop (or mesh) analysis systematically applies KVLto the loops of a circuit in order to solve for the currents around each loop.Similar to nodal analysis, there are three basic steps followed in loop analysis:

1. Identify the loops and assign a current to each mesh or loop.2. Apply KVL around each loop to obtain an equation in terms of the loop

currents. Note that a loop/mesh current is not a branch current, specifically,a branch current may be made up of one or more loop currents.

3. Solve the resulting system of KVL equations for the loop currents.

Example: The examples below repeat the analysis of the circuit of Figure 2.25except that loop analysis is used here. Again, a solution is first obtained by hand,and then Mathcad is used.

Method 1: Hand SolutionThe hand solution follows the loop analysis procedure above as follows:

Step 1 : Figure 2.28 shows that this circuit contains two independent meshes.We assume two loop currents: the current in Loop 1 is ILl and in Loop 2 isI L2 . The directions of the loop currents are marked on Figure 2.28. The chosendirections are advantageous since the loop current directions match the polarityof the voltage sources.

Step 2: KVL is first applied to the left-hand loop (Loop 1):

Figure 2.28 Circuit for using loop equations.


A KVL expression is next written for Loop 2:

Step 3: In this example the application of loop analysis results in two coupledequations, which may be written in a matrix formulation:

[R1 + j(X1 + X cap )

jXcap

This matrix equation is of the form 'l!.. ! = Ywhich has a solution of! = 'l!..-ly.- -

Method 2: Mathcad SolutionHere we solve the circuit of Figure 2.28 using a Mathcad-based loop analysis.Again two approaches are taken using Mathcad: (1) following the hand solutionabove and (2) using the Mathcad Find equation solver.

Using KVL, the loop equations are

The verification of the validity of the equations requires the selection of guessvalues, which are

IL1 := 300 A IL2 := 300 A

Approach #1: First, rearrange the KVL equation for Loop 1 to solve for IL2 .

The numerical result using the guess values is IL2 =-175.593 + 226.195j AThe I L2 is substituted in the second KVL equation:

The further rearrangement of that equation results in

BASIC LAWS 79

Solving for ILl yields

(R2 + jX2 + jXcap)V2 - Vr - - -.- - - -JXcap

I .=-----------------Lt.

X(R1 + jX1 + jXcapHR2 + jX2 + jXcap)

J cap - XJ cap

ILl = 388.46 + 66.775j A

The Loop 2 current is

VI - IL r (R1 + jX1 + jXcap)IL2 := .JXcap

IL2 = -217.3 + 173.771j A IIL21 = 278.237 A arg(IL2) = 141.351 deg

The capacitance (branch) current 13 is the sum of the two loop currents.13 := ILl + IL2

13 = 171.16+ 240.546j A

Approach #2: The Mathcad equation solver can solve the loop equationsdirectly. The method is illustrated below. The guess values are

ILl := 300 A LL2:=250 A

Given the two KVL expressions already developed

VI - ILr (R1 + jX1 + jXcap) - IL2jXcap =0

the Mathcad Find equation solver immediately computes

(388.46+ 66.775j )

Find(1 l' I 2) = AL L -217.3 + 173.771j

The obtained results are the same as those from the first approach.

2.5.3 Thevenln's TheoremThevenin's theorem states that an electric network can be represented by a voltagesource and an impedance connected in series (see Figure 2.29). The voltage of the


Figure 2.29 Thevenin equivalent circuit.

source is the open circuit voltage, Voc, of the network. The Thevenin equivalentimpedance (ZTh) is calculated by removing and replacing the voltage sourcesin the network by a short circuit and calculating the input impedance acrossthe terminals.

In a power network the impedance is calculated from the short circuit current,I short . The short circuit current is computed by the operating utility and is anavailable datum. In a power network the short circuit current is usually induc-tive. Practically, the resistance is often negligible; consequently, the network cangenerally be represented by an inductive reactance. The open circuit voltage ofa power network is normally within 5% of the rated line-to-ground (neutral)voltage, Vln. The Thevenin reactance is

2.5.4 Loads

XNet

= IVlnlIIshortI

(2.38)

In an electric power network the typical load consumes constant power at aconstant power factor if the load voltage is within the 5% limit. The loadcurrent calculation requires the load voltage. The first step is the calculation ofthe phase angle from the power factor value. The phase angle is

eLoad = arccos(p!Load) (2.39)

The phase angle (of the complex power) is positive if the load power factor islagging, and negative if the power factor is leading. The complex power is

S V 1* II I j8LoadLoad = Load Load = Y rms _load rms _loade (2.40)'8 PLoad '8

= ISLoadleJ Load = --eJ LoadP!Load

The real part of the complex power is the real power; the imaginary part isthe reactive power. From Table 2.2, the reactive power is positive if the powerfactor is lagging (inductive load) and negative if the power actor is leading(capacitive load).

APPLICATIONS OF SINGLE-PHASE CIRCUIT ANALYSIS 81

TABLE 2.3 Sign Convention of Exponential

Power Factor

Leading(Capacitive)Lagging(Inductive)

Current, I

Ille+jf)positive exponent

Ille-jf)negative exponent

Complex Power, S

ISle-jf)negative exponent

ISle+jf)positive exponent

Where () == arccos(pj) is the load power factor phase angle.

The load current is the conjugate of the complex power and load voltage ratio:

I _ [SLoad]*Load - VLoad

(2.41)

The three equations above can be combined together to obtain an expression forthe direct calculation of the load current. The result is

I - PLoad =fj arccoscpj; )Load - e LoadVLoad piLoad

(2.42)

In this equation the plus (+) sign is used if the power factor is leading and theminus (-) sign is used if pi is lagging. In other words, the reactive componentof the current is negative if the power factor is lagging (the load is inductive).The reactive component of the current is positive if the power factor is leading(the load is capacitive). The sign of the exponential is summarized in Table 2.3.

Industrial, commercial and domestic loads are typically motor and lightingloads. Motors are a major component of these loads. Consequently, most loadsare inductive and the power factor lagging. However, the use of compensatingcapacitors may produce a leading power factor at night when the loads are light.Another source of leading power factors are the lightly loaded, long transmissionlines, where the line capacitor over compensates for the load inductance.

2.6 APPLICATIONS OF SINGLE-PHASE CIRCUIT ANALYSIS

The practical application of the complex phasor analysis of a single-phase circuitis presented through three simple numerical examples. The reader is encour-aged to use a computer and follow the calculation by entering the equations andevaluating the results. This interactive method of learning will lead to a betterunderstanding of the concepts.

2.6.1 Analysis of Transmission Line OperationIn the first example, the operation of a transmission line is analyzed in threedifferent operating conditions:


1. No-load (open circuit) condition;2. Short-circuit condition; and3. Loaded condition.

The input current, real and reactive power, and the voltage and current at thereceiving end are calculated for each condition.

The transmission line is represented by a n (pi) circuit and supplied by avoltage source (Vs).

The system frequency data are f := 60 Hz co := 2n+' 6The M is introduced into Mathcad to represent the mega prefix: M := 10The supply voltage is: vs := 76 kVThe line impedance is: Z line:= (8 + Ij40) nThe sending end and receiving end capacitances are:

Cs := 0.8 JlF

No-load Condition

Figure 2.30 shows the equivalent circuit for an open circuit (unloaded) condi-tion. Inspection of the equivalent circuit shows that the voltage source suppliestwo parallel current paths. The first (right side) path contains X1ine , Rline and Crconnected in series. The second (left) path contains the line capacitance Cs.

The impedance of the first path is:

1Zcircuit := Zline + ---j mCr

Zcircuit = 8 - 3275.728j n

The voltage source supplies the first path directly. The current through thefirst path is calculated by Ohm's Law.

VsIline_open :=

ZcircuitIline_open = 0.06 + 23.2j A

Supply Transmission Line OpenCircuit

Figure 2.30 Transmission line in an open circuit condition.


The current through the second path is:

ICs:= --- ICs = 22.92j A

The total source current in the open circuit condition is the sum of the twocurrents above.

IS_open = 0.06 + 46.12j A

The generator complex power is the product of the supply voltage and thesource current conjugate, where the complex conjugate operation is denoted byMathcad using the overline mark.

Sopen = 0 - 3.51j MVA

The result shows that the real power is negligible. The negative reactive powerindicates that the voltage source absorbs reactive power. The passive sign con-vention is not obeyed at the supply of Figure 2.30 since the current Is open exits(rather than enters) from the positive voltage terminal; hence, a negative powervalue indicates reactive power is absorbed rather than supplied.

The open circuit voltage on the line end is the line current times the impedanceof the receiving capacitance (Cr) . The open circuit voltage is:

Vr_open = 76.93 - 0.19j kV

arg(Vr_open) = -0.14 deg

The supply voltage, Vs. is 75 kV. The open-circuit voltage is slightly higherthan the supply voltage. In fact, the open-circuit voltage can be dangerouslyhigh in a long line, where the high voltage can endanger the transmission lineinsulation. This is referred to as the Ferranti effect. Power companies load theline with an inductance, when required, to eliminate the overvoltage in an opencircuit condition.

Short-circuit Condition

Figure 2.31 shows the equivalent circuit in the short circuit condition. It isassumed that the short circuit occurs at the receiving end of the line. The equiv-alent circuit shows that the short circuit eliminates the receiving capacitance Cr.


Is short-~

Supply Transmission Line ShortCircuit

Figure 2.31 Transmission line short-circuited.

The line current is the ratio of the supply voltage and line impedance;

VsIshort := --

ZlineIshort = 0.37 - 1.83j kA IIshortl = 1.86 kA

The active (real) part of the short circuit current is small. The large reactivepart is inductive.

Large short circuit currents can produce arcing and may damage line conduc-tors. Utilities protect the transmission lines with circuit breakers that open theline automatically when a short circuit occurs. Typically, the short circuit currentis interrupted within 50-200 msec (3-12 cycles).

Using KCL, the supply current is:

IS short := Ishort + ICs IS_short = 0.37 - 1.8j kA IIS_shortI= 1.84 kACalculate the complex input power:

Ss short := VS.IS short- -

Ss short = 27.77 + 137.1j MVA

Both the real and the reactive powers are significant for the short circuit condition.

Loaded ConditionLoading produces a voltage drop, which varies with the load. To evaluate theload effect, we load the line and plot the supply voltage versus the load power.

The practical question is the calculation of the load, when the supply voltageis an established value and a certain load voltage is to be maintained. Here wecompute the load power when the supply voltage is 80 kV, the load voltage is75 kV, and the pI is 0.75 lagging.

The load data are: Vload := 75 kV pfjoad := 0.75The load is variable, but we test the calculation by using Pload := 50 MW as

a guess value.Figure 2.32 shows the equivalent circuit of the loaded line.


I sup I X . R . I II d----. I line line I 08

vsupl $[ i-~tVIOOdSupply I Transmission Line I Load

Figure 2.32 Equivalent circuit of the loaded line.

Referring to Equation (2.40), the absolute value of the load complex poweras a function of the variable load is:

( )Pload

Sload Pload :=--P~oadUsing the expression from Table 2.3 for a lagging load, the vector value of

the load complex power is:

() () j acos (pfload)Sload v Pload := Sload Pload e

Had the load been leading (capacitive), the sign of the above exponential wouldhave been negative. From Equation (2.41), the load current is:

Iload(Pload) = 666.67 - 587.94j A

Recall that the overline in the above equation is the manner with which Math-cad designates the complex conjugate operation. The capacitive current at thereceiving end is:

Vload1

j mer

Icap_r = 22.62j A

The line current is the sum of the load and receiving-end capacitance currents:

Iline(Pload) = 666.67 - 565.33j A

The required supply voltage is the sum of the load voltage and the voltagedrop on the line impedance:


The capacitive current at the supply side is the supply voltage divided by thesending-end capacitive impedance:

( ) Vsup(Pload)leap_s Pload := 1j -ro-C,

The supply current is the sum of the line current and sending-end capaci-tive current:

The supply complex power is:

Ssup(Pload) := Vsup(Pload)Isup(Pload) Ssup(Pload) = 56.11 + 69.62j MVA

The proper operation of the power system requires that the load voltage bewithin 5% limits. This entails the calculation of the regulation, which is thepercentage value of the voltage drop. The regulation should be less than 5%. Herewe define the voltage regulation as the difference between the absolute valuesof the supply and load voltages divided by the load voltage absolute value. Theregulation versus load is:

Reg(Pload) = 40.4%

Note that the more than 40% regulation is unacceptable.To plot the supply voltage versus load, we vary the load power according to:

Pload:= 0 MW, 1 MW .. 15 MW

Figure 2.33 shows that the required supply voltage varies linearly with theload. For the Mathcad plots presented in this book, the units are given as thedenominator of a fraction having the particular variable as the numerator. Specif-ically in Figure 2.33, the units of P10ad and Vsup are MW and kV, respectively.

A practical question is the determination of the load voltage at a specifiedsupply voltage. This problem calls for the calculation of the load when the supplyvoltage is 80 kV. An approximate value of 10 MW is determined from the plot.The exact value is calculated by solving

IVsup(PZoad) I = 80 kV


15105

75

80

P10ad := 0 MW, 1 MW.. 15 MW85 ...----------------------.

Figure 2.33 Supply voltage versus load power.

for PZoad using the Mathcad root equation solver. A guess value for the equationsolver is Pload := 20 MW and the result is:

root( IVsup(Pload)!- 80 kV,Pload) = 10.01 MW

More information on the use of the Mathcad root equation solver is given inAppendix A.

2.6.2 Generator Supplies a Constant Impedance Network Through aLineFigure 2.34 shows the one-line diagram of a system where a generator suppliesa constant impedance load through a transmission line. The generator can berepresented by a voltage source and impedance. In this example, we neglect thegenerator impedance and represent the generator with a voltage source only. A ncircuit represents the line. The load is represented by a resistance and inductivereactance connected in parallel. The equivalent circuit is shown in Figure 2.35.The circuit data are given below.

The generator voltage rating and frequency are:

Vg := 15 kV tg := 60 Hz

TransmissionLine

Generator Load

Figure 2.34 Generator supplies an impedance load through a line.


Generator Transmission Line Load

Figure 2.35 Equivalent circuit of the system in Figure 2.34.

The transmission line parameters are:

RUne:= 1.1 Q Xline:= 6 Q

Cs := 30~

The load impedance is: XL:= 60Q

Equivalent Impedance CalculationIf the system frequency is 60 Hz, the capacitive impedances at the sending andreceiving ends of the line are:

1Zcs:= ---j o-C, Zcs = -88.419j Q Zcr := Zcs

The load is represented by a resistance and reactance connected in parallel. Theequivalent load impedance is:

ZL = 28.8 + 21.6j Q1

1 1-+--RL j.XL

ZL:=----

The load impedance and the capacitance at the line end are also connected inparallel. The equivalent impedance is:

ZL Cr = 42.529 + 10.252j Q

The transmission line impedance is connected in series with the combinedimpedance of the load and capacitor. The equivalent impedance is:

Zline L c-:- Rline + j .Xline + ZL Cr- - -

Zline L Cr f 43.629 + 16.252j Q

The combination of the impedances results in a simplified circuit, shown inFigure 2.36.


Generator I CombinedI Impedance

Figure 2.36 Simplified equivalent circuit of Figure 2.35.

Current CalculationThe currents in this circuit can be calculated directly using Ohm's Law. Furthercircuit simplification is unnecessary. The sending-end capacitor current is:

VgICs:= -

Zcs

The line current is:

VgIline := ----

Zline L Cr

ICs = 169.646j A

Iline= 301.916-112.466jA

The generator current is the sum of the capacitor and line currents. Theresult is:

Ig := ICs + Iline

IIgl = 307.283 AIg = 301.916 + 57.18j A

g = 10.724 deg

Voltage and Voltage Drop CalculationThe next step is the calculation of the load voltage, which is the product of theline current and the combined impedance of the load and receiving-end capacitor.

IVloadl = 14.095 kV

Vload = 13.993 - 1.688j kV

arg( Vload) = -6.878 deg

An alternative method for calculating the load voltage is subtraction of thevoltage drop on the line impedance from the generator voltage:

Vload alt = 13.993 - 1.688j kV


The load current is the ratio of the load voltage and load impedance.

IL = 282.828 - 270.725j A

arg(IL) = -43.747 deg

An alternate method for load current calculation is the use of the currentdivision equation.

ZcsIL alt:= Iline----

- Zcs + ZLIL_alt = 282.828 - 270.725j A

The proper operation of household and other equipment requires the loadvoltage be within about 5% of the rated voltage. The standards for electricutilities specify that the voltage be within 5% for 95% of the time, and within8% for 99% of the time.

The most frequently used method to evaluate the operating conditions is thecalculation of the percentage voltage drop or voltage regulation. The percentagevoltage drop is defined as:

Voltage_drop= 6.424%

Note that Mathcad automatically converts the fractional voltage drop to a per-centage value; therefore, the usual 100% multiplier is not needed in the aboveformula. The voltage drop in the investigated circuit is slightly more than 5%-itis not acceptable.

2.6.3 Power Factor Improvement

Consider a factory that operates with a poor (low) power factor. The utilitypenalizes the factory operator with a higher electricity rate. The operator caninstall a capacitor connected in parallel with the load to improve the powerfactor and reduce the power cost. Figure 2.37 shows the system one-line dia-gram.

The factory load is:

Pload:= 150 kW P~oad := 0.65 inductive Vload:= 7.2 kV

The desired power factor is: P~up := 0.9 (lagging).

110ad = 20.833 - 24.357j A


E--

Figure 2.37 Power factor improvement by capacitor.

The capacitor does not affect the real" power but reduces the reactive power.Consequently the load current and the desired supply current are calculated usingthe given load value and the appropriate power factor. Also the load and supplyvoltages are equal. Using Equation (2.42) for a lagging power factor, the loadcurrent is:

0_ P10ad - j -acos(pfload)110ad .- eP~oadVload

The desired supply current is:

"_ P10ad - j -acos(pfsup)Isup'- ep~upVload

Isup = 20.833 - 10.09j A

The capacitance current is the difference of the supply and load currents. Thecapacitance current is:

leap = 14.267j AThe capacitor-provided reactive power is:

The impedance of the capacitor is:

Z 0_ Vloadeap .-leap

Qeap = -102.721j kVA

Zeap = -504.667j Q

Ceap = 6.307 JlF

Assuming 50 Hz frequency, the capacitor needed to improve the power factor is:ro := 21t 50 Hz

1Ceap := .Jro.Zeap

This capacitor is connected in parallel with the load.


2.7 SUMMARY

It has been assumed that the reader was already familiar with much of the basiccircuit analysis laws, conventions and techniques presented in this chapter. Thegoal of this chapter has been to review those basics while preparing the readerto apply computer tools, such as Mathcad and MATLAB, to network analysis. Inaddition, several quantities that are possibility new to the reader have been intro-duced; for example, system efficiency and voltage regulation which are definedin Equations (2.30) and (2.31), respectively. Further, we note that the engineershould be prepared to utilize interrelations between real power, reactive power,complex power, voltage, current and power factor to solve for the unknown vari-able(s) based on those quantities that are known. Specifically, the reader will findthe following relations to be particularly useful throughout the remainder of thistextbook. For Vrms and Inns obeying the passive sign convention, as shown inFigure 2.38:

8 V 1* IT j 8 I - j = nns nns = Yrms e rms e

= Vrmslrms cos(8 -l/J) + jVrmslrms sin(8 -l/J)= P+jQ

If that circuit element is an impedance, then

Z = V rms = VrmsLa = Vrms La - = IZIL8Inns Irms Ll/J i.:

8 - V 1* - IT I j8 - 18 I j8 - P j8Z - rms rms - Yrms rms e - Z e - pte

(2.43)

(2.44)

(2.45)

where () = arccos(pj) and the proper sign can be determined by referring toTable 2.4.

CircuitElement

Figure 2.38 General circuit element obeying passive sign convention.

TABLE 2.4 Important Interrelations

Power Factor Reactive ComplexLoad Angle Power Power

Capacitive e < 0 (leading) Q 0 (lagging) Q > 0 ISle+ j O

PROBLEMS 93

2.8 EXERCISES

1. What is the rms value of a sinusoidal voltage or current? Present theequation.

2. Describe the impedance triangle, and present the equations for complexand polar representation of an impedance consisting of a resistance andreactance connected in series.

3. Describe the rules for the calculation of an equivalent impedance whenimpedances are connected in series and in parallel.

4. How do you calculate the instantaneous power?5. What is the complex power? What is the unit of measure for com-

plex power?6. Present the equations for the complex power calculation.7. What is the active power? What is the unit of real power?8. What is the reactive power? What is the unit of reactive power?9. What is the power triangle?

10. What is the power factor? How it is used? Give the basic equations forthe power factor calculation.

11. What is the sign of the reactive power if the current is leading the voltage?Draw a sketch of the current and voltage waveforms.

12. What is the sign of the reactive power if the current is lagging the voltage?Draw a sketch of the current and voltage waveforms.

13. Describe Kirchhoff's current law.14. Describe Kirchhoff's voltage law.15. Discuss the Thevenin equivalent of a circuit. Illustrate the concept with

a diagram.16. Present the equations for the current calculation in the case of loads with

lagging and leading power factors.17. Draw an equivalent circuit for a transmission line. Identify the components.18. What is the definition of voltage regulation? Present the equation and

describe the meaning of the variables.19. Discuss the concept of power factor correction.

2.9 PROBLEMS

Problem 2.1Household lighting voltage is 120 Vrms. Determine the voltage magnitude. Whatis range of the rms voltage if the voltage is within 5% of the nominal value.

Problem 2.2Calculate the impedance of the network shown in Figure 2.39. The 60 Hz networkcircuit element values are


Branch 1:Branch 2:Branch 3:Branch 4:

R 1 = 25 QR2 = 25 QR3 = 48 QC4 = 25 J-LF

L1 =0.1 HL2 = 0.1 HC3 = 0.1 J-LFL 4 = 0.3 H

Problem 2.3A single-phase 3 hp, 240 V pool pump (motor) runs at full load with 92% effi-ciency and a power factor of 0.73 lagging. Calculate the capacitor needed toimprove power factor to 0.98 lagging.

Problem 2.4Two voltage sources are connected together through an impedance of Z = 12 +j21 Q. Determine which source operates as the load and which is the generator.The voltage sources are:

and

Problem 2.5The electrical network shown in Figure 2.40 has a voltage source of 140 V andthe values of the impedances are as follows: Zl = 5 - j8 Q, Z2 = 10 + j5 Q,Z3 = 5 - j5 Q, Z4 = 15 + jl0 Q. Determine: (a) the real power absorbed byeach impedance, and (b) the reactive power of each impedance.

Figure 2.39 Network for Problem 2.2.

Figure 2.40 Electric network for Problem 2.5.

PROBLEMS 95

Problem 2.6

A single-phase voltage source of Vs = 220 VL10 supplies a load impedance ofZ = 100 QL50. Determine: (a) the resistance and reactance of the load, (b) thereal and reactive power absorbed by the load, and (c) the power factor of thecircuit as viewed from the source, and state whether the circuit is lagging orleading.

Problem 2.7

Two ideal voltage sources are connected to each other through a feeder with acombination of impedances, as shown in Figure 2.41. The voltage source valuesare VI = 100 V and V2 = 120 VL - 25, and the impedances are ZI = 10 Q,Z2 = 5j Q, Z3 = -25j Q, and Z4 = 3 Q. Determine: (a) the real power of eachvoltage source, and state whether the source is supplying or absorbing real power;(b) the reactive power of each source, and state whether the source is supplying orabsorbing reactive power; and (c) the real and reactive power of the impedances,and state whether the power is supplied or absorbed. (d) Plot the real powerconsumed by the feeder against the power angle, assuming that the angle variesbetween 0 to 360. Find the maximum power consumed. Explain the trend ofthe power consumed versus the power angle.

Note: The power angle is the phase angle difference between phasors VI andV2 In this part of the problem, the phase angle of VI is zero, and replace the-25 by a variable angle for V2.

Problem 2.8

An industrial plant consists of several single-phase motors. The plant absorbs apower of 600 kW with a power factor of 0.75 lagging from the substation bus.The supply voltage is 12.47 kV. (a) Find the required kVAR rating of a capacitorconnected across the load to raise the power factor to 0.95 lagging. (b) Assuminga synchronous motor rated at 250 hp, with an 80% efficiency is operated from

Figure 2.41 Circuit for Problem 2.7.


the same bus at rated conditions and a power factor of 0.85 .leading, calculatethe resultant supply power factor.

Problem 2.9

A 220 V bus supplies a motor and capacitor. The single-phase motor has thefollowing rating:

v = 220 V S = 65 kVA pI= 0.8 (lagging)

The motor is loaded to its rated capability. The capacitor is connected inparallel with the motor to obtain a total power factor of 0.99 leading. Draw theequivalent circuit. Determine the required kVA rating of the capacitor, and thecapacitance value. Plot the pI as a function of capacitance. Find the capacitanceneeded to obtain a final pI = 1.

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