Chapter 3

FUNDAMENTAL THEORY OF LOADCOMPENSATION(Lectures 19-28)

3.1 Introduction

In general, the loads which have poor power factor, unbalance, harmonics, and dc componentsrequire compensation. These loads are arc and induction furnaces, sugar plants, steel rolling mills(adjustable speed drives), power electronics based loads, large motors with frequent start and stopetc. All these loads can be classified into three basic categories.

1. Unbalanced ac load2. Unbalanced ac + non linear load3. Unbalanced ac + nonlinear ac + dc component of load.

The dc component is generally caused by the usage of half-wave rectifiers. These loads, par-ticularly nonlinear loads generate harmonics as well as fundamental frequency voltage variations.For example arc furnaces generate significant amount of harmonics at the load bus.Other serious loads which degrade power quality are adjustable speed drives which include powerelectronic circuitry, all power electronics based converters such as thyristor controlled drives, rec-tifiers, cyclo converters etc.. In general, following aspects are important, while we do provide theload compensation in order to improve the power quality [1].

1. Types of load (unbalance , harmonics and dc component)2. Real and Reactive power requirements3. Rate of change of real and reactive power etc.

In this unit, we however, discuss fundamental load compensation techniques for unbalanced linearloads such as combination of resistance, inductance and capacitance and their combinations. Theobjective here will be to maintain currents balanced and unity factor with their voltages.

69

3.2 Fundamental Theory of Load Compensation

We shall find some fundamental relation ship between supply system the load and the compensator.We shall start with the principle of power factor correction, which in its simplest form, and can bestudied without reference to supply system [2]–[5].

The supply system, the load and the compensator can be modeled in various ways. The supplysystem can be modeled as a Thevenin’s equivalent circuit with an open circuit voltage and a se-ries impedance, (its current or power and reactive power) requirements. The compensator can bemodeled as variable impedance or as a variable source (or sink) of reactive current. The choice ofmodel varied according to the requirements. The modeling and analysis done here is on the basisof steady state and phasor quantities are used to note the various parameters in system.

3.2.1 Power Factor and its Correction

Consider a single phase system shown in 3.1(a) shown below. The load admittance is represented

VsI

lI

VRI

lIXI

l

Fig. 3.1 (a) Single line diagram of electrical system (b) Phasor diagram

by Yl = Gl + jBl supplied from a load bus at voltage V = V ∠0. The load current is I l is given as,

I l = V (Gl + jBl) = V Gl + jV Bl (3.1)= IR + jIX

According to the above equation, the load current has a two components, i.e. the resistive or inphase component and reactive component or phase quadrature component and are represented byIR and IX respectively. The current, IX will lag 90o for inductive load and it will lead 90o forcapacitive load with respect to the reference voltage phasor. This is shown in 3.1(b). The loadapparent power can be expressed in terms of bus voltage V and load current Il as given below.

Sl = V (Il)∗

= V (IR + j IX)∗

= V (IR − j IX)

= V (Il cosφl − j Il sinφl)= V Il cosφl − j V Il sinφl= Sl cosφl − j Sl sinφl (3.2)

70

From (3.1), I l = V (Gl + jBl) = V Gl + jV Bl, equation (3.2) can also be written as following.

Sl = V (Il)∗

= V (V Gl + jV Bl)∗

= V (V Gl − jV Bl)

= V 2Gl − j V 2Bl

= Pl + jQl (3.3)

From equation (3.3), load active (Pl) and reactive power (Ql) are given as,

Pl = V 2Gl

Ql = −V 2Bl (3.4)

Now suppose a compensator is connected across the load such that the compensator current, Iγ isequal to −IX , thus,

Iγ = V Yγ = V (Gγ + jBγ) = −IX= −j V Bl (3.5)

The above condition implies that Gγ = 0 and Bγ = −Bl. The source current Is, can thereforegiven by,

Is = I l + Iγ = IR (3.6)

Therefore due to compensator action, the source supplies only in phase component of the loadcurrent. The source power factor is unity. This reduces the rating of the power conductor andlosses due to the feeder impedance. The rating of the compensator is given by the followingexpression.

Sγ = Pγ + j Qγ = V (Iγ)∗

= V (−j V Bl)∗

= jV 2Bl (3.7)

Using (3.4), the above equation indicates the Pγ = 0 and Qγ = −Ql. This is an interestinginference that the compensator generates the reactive power which is equal and opposite to theload reactive and it has no effect on active power of the load. This is shown in Fig. 3.2. Using(3.2) and (3.7), the compensator rating can further be expressed as,

Qγ = −Ql = −Sl sinφl = −Sl√

1− cos2 φl VArs (3.8)

From (3.8),

|Qγ| = Sl√

1− cos2 φl (3.9)

If |Qγ| < |Ql| or |Bγ| < |Bl|, then load is partially compensated. The compensator of fixedadmittance is incapable of following variations in the reactive power requirement of the load. In

71

XI

I

s RI IVsI

lIIlI

l

XI

Fig. 3.2 (a) Single line diagram of compensated system (b) Phasor diagram

practical however a compensator such as a bank of capacitors can be divided into parallel sections,each of switched separately, so that discrete changes in the reactive power compensation can bemade according to the load. Some sophisticated compensators can be used to provide smooth anddynamic control of reactive power.

Here voltage of supply is being assumed to be constant. In general if supply voltage varies, theQγ will not vary separately with the load and compensator error will be there. In the followingdiscussion, voltage variations are examined and some additional features of the ideal compensatorwill be studied.

3.2.2 Voltage Regulation

Voltage regulation can be defined as the proportional change in voltage magnitude at the load busdue to change in load current (say from no load to full load). The voltage drop is caused due tofeeder impedance carrying the load current as illustrated in Fig. 3.3(a). If the supply voltage isrepresented by Thevenin’s equivalent, then the voltage regulation (VR) is given by,

V R =

∣∣E∣∣− ∣∣V ∣∣∣∣V ∣∣ =

∣∣E∣∣− |V ||V |

(3.10)

for V being a reference phasor.In absence of compensator, the source and load currents are same and the voltage drop due to thefeeder is given by,

∆V = E − V = ZsIl (3.11)

The feeder impedance, Zs = Rs + jXs. The relationship between the load powers and its voltageand current is expressed below.

Sl = V (I l)∗ = Pl + jQl (3.12)

Since V = V , the load current is expressed as following.

I l =Pl − jQl

V(3.13)

72

Substituting, Il from above equation into (3.11), we get

∆V = E − V = (Rs + jXs)

(Pl − jQl

V

)=

RsPl +XsQl

V+ j

XsPl −RsQl

V= ∆VR + j∆VX (3.14)

Thus, the voltage drop across the feeder has two components, one in phase (∆VR) and another isin phase quadrature (∆VX). This is illustrated in Fig. 3.3(b).

E

V RV

Xj V

lIlR I l ljX I

l

lI l l lS P jQ

l l lY G jB

Load

sI

V

E

Fee

der

s s sZ R jX

Fig. 3.3 (a) Single phase system with feeder impedance (b) Phasor diagram

From the above it is evident that load bus voltage (V ) is dependent on the value of the feederimpedance, magnitude and phase angle of the load current. In other words, voltage change (∆V )depends upon the real and reactive power flow of the load and the value of the feeder impedance.

Now let us add compensator in parallel with the load as shown in Fig. 3.4(a). The question is:whether it is possible to make

∣∣E∣∣ =∣∣V ∣∣, in order to achieve zero voltage regulation irrespective of

change in the load. The answer is yes, if the compensator consisting of purely reactive components,has enough capacity to supply to required amount of the reactive power. This situation is shownusing phasor diagram in Fig. 3.4(b).The net reactive at the load bus is now Qs = Qγ + Ql. The compensator reactive power (Qγ) hasto be adjusted in such a way as to rotate the phasor ∆V until

∣∣E∣∣ =∣∣V ∣∣.

From (3.14) and Fig. 3.3(b),

E∠δ =

(V +

Rs Pl +XsQs

V

)+ j

(Xs Pl −RsQs

V

)(3.15)

The above equation implies that,

E2 =

(V +

Rs Pl +XsQs

V

)2

+

(Xs Pl −RsQs

V

)2

(3.16)

73

E

V

lI

l

lI l l lS P jQ

l l lY G jB

Load

sI V

E

Fee

der

s s sZ R jX

Co

mp.

I

Q

VI sI

s sR I

s sjX I

Fig. 3.4 (a) Voltage with compensator (b) Phasor diagram

The above equation can be simplified to,

E2V 2 = (V 2 +Rs Pl)2 +X2

s Q2s + 2(V 2 +Rs Pl)XsQs

+X2s P

2l +R2

s Q2s −(((((

((2Xs PlRsQs (3.17)

Above equation, rearranged in the powers of Qs, is written as following.

(R2s +X2

s )Q2s + 2V 2XsQs + (V 2 +Rs Pl)

2 + (Xs Pl)2 − E2 V 2 = 0 (3.18)

Thus the above equation is quadratic inQs and can be represented using coefficients ofQs as givenbelow.

aQ2s + bQs + c = 0 (3.19)

Where a = R2s +X2

s , b = 2V 2Xs and c = (V 2 +Rs Pl)2 +X2

s P2l − E2 V 2.

Thus the solution of above equation is as following.

Qs =−b±

√(b2 − 4ac)

2a(3.20)

In the actual compensator, this value would be determined automatically by control loop. Theequation also indicates that, we can find the value of Qs by subjecting a condition such as E = Virrespective of the requirement of the load powers (Pl, Ql). This leads to the following conclusionthat a purely reactive compensator can eliminate supply voltage variation caused by changes inboth the real and reactive power of the load, provided that there is sufficient range and rate of Qs

both in lagging and leading pf. This compensator therefore acts as an ideal voltage regulator. Itis mentioned here that we are regulating magnitude of voltage and not its phase angle. In fact itsphase angle is continuously varying depending upon the load current.

It is instructive to consider this principle from different point of view. We have seen that com-pensator can be made to supply all load reactive power and it acts as power factor correction device.If the compensator is designed to compensate power factor, then Qs = Ql + Qγ = 0. This im-plies that Qγ = −Ql. Substituting Qs = 0 for Ql in (3.14) to achieve this condition, we get the

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following.

∆V =(Rs + jXs)

VPl (3.21)

From above equation, it is observed that ∆V is independent of Ql. Thus we conclude that a purelyreactive compensator cannot maintain both constant voltage and unity power factor simultaneously.Of course the exception to this rule is a trivial case when Pl = 0.

3.2.3 An Approximation Expression for the Voltage Regulation

Consider a supply system with short circuit capacity (Ssc) at the load bus. This short circuit capac-ity can be expressed in terms of short circuit active and reactive powers as given below.

Ssc = Psc + jQsc = E I∗sc = E

(E

Zsc

)∗=E2

Z∗sc(3.22)

Where Zsc = Rs + jXs and Isc is the short circuit current. From the above equation

|Zsc| =E2

Ssc

Therefore, Rs =E2

Ssccosφsc

Xs =E2

Sscsinφsc

tanφsc =Xs

Rs

(3.23)

Substituting above values of Rs and Xs, (3.14) can be written in the following form.

∆V

V=

(Pl cosφsc +Ql sinφsc

V 2+ j

Pl sinφsc −Ql cosφscV 2

)E2

Ssc

∆V

V=

∆VRV

+ j∆VXV

(3.24)

Using an approximation that E ≈ V , the above equation reduces to the following.

∆V

V=

(Pl cosφsc +Ql sinφsc

Ssc+ j

Pl sinφsc −Ql cosφscSsc

)(3.25)

The above implies that,

∆VRV≈ Pl cosφsc +Ql sinφsc

Ssc∆VXV≈ Pl sinφsc −Ql cosφsc

Ssc

75

Often (∆VX/V ) is ignored on the ground that the phase quadrature component contributes negli-gible to the magnitude of overall phasor. It mainly contributes to the phase angle. Therefore theequation (3.25) is simplified to the following.

∆V

V=

∆VRV

=Pl cosφsc +Ql sinφsc

Ssc(3.26)

Implying that the major change in voltage regulation occurs due to in phase component, ∆VR.Although approximate, the above expression is quite useful in terms of short circuit level (Ssc),(Xs/Rs, active and reactive power of the load.

On the basis of incremental changes in active and reactive powers of the load, i.e., ∆Pl and ∆Ql

respectively, the above equation can further be written as,

∆V

V=

∆VRV

=∆Pl cosφsc + ∆Ql sinφsc

Ssc. (3.27)

Further, feeder reactance (Xs) is far greater than feeder resistance (Rs), i.e., Xs >> Rs. Thisimplies that, φsc → 90o, sinφsc → 1 and cosφsc → 0. Using this approximation the voltageregulation is given as following.

∆V

V≈ ∆VR

V≈ ∆Ql

Sscsinφsc ≈

∆Ql

Ssc. (3.28)

That is, per unit voltage change is equal to the ratio of the reactive power swing to the short circuitlevel of the supply system. Representing ∆V approximately by E−V and assuming linear changein reactive power with the voltage, the equation (3.28) can be writtwn as,

E − VV

≈ Ql

Ssc. (3.29)

The above leads to the following expression,

V ' E

(1 + QlSsc

)' E(1− Ql

Ssc) (3.30)

with the assumption that, Ql/Ssc << 1. Although above relationship is obtained with approxi-mations, however it is very useful in visualizing the action of compensator on the voltage. Theabove equation is graphically represented as Fig. 3.5. The nature of voltage variation is droopingwith increase in inductive reactive power of the load. This is shown by negative slope −E/Ssc asindicated in the figure.

The above characteristics also explain that when load is capacitive, Ql is negative. This makesV > E. This is similar to Ferranti effect due to lightly loaded electric lines.Example 3.1 Consider a supply at 10 kV line to neutral voltage with short circuit level of 250MVA and Xs/Rs ratio of 5, supplying a star connected load inductive load whose mean power is25 MW and whose reactive power varies from 0 to 50 MVAr, all quantities per phase.

76

0lQ

V

sc

E

S

E

Fig. 3.5 Voltage variation with reactive power of the load

(a) Find the load bus voltage (V ) and the voltage drop (∆V ) in the supply feeder. Thus determineload current (I l), power factor and system voltage (E ).

(b) It is required to maintain the load bus voltage to be same as supply bus voltage i.e. V =10 kV.Calculate reactive power supplies by the compensator.

(c) What should be the load bus voltage and compensator current if it is required to maintain theunity power factor at the supply?

Solution: The feeder resistance and reactance are computed as following.Zs = E2

s/Ssc = (10 kV)2/250 = 0.4 Ω/phaseIt is given that, Xs/Rs = tanφsc = 5, therefore φsc = tan−1 5 = 78.69o. From this,

Rs = Zs cosφsc = 0.4 cos(78.69o) = 0.0784 Ω

Xs = Zs sinφsc = 0.4 sin(78.69o) = 0.3922 Ω

(a) Without compensation Qs = Ql, Qγ = 0To know ∆V , first the voltage at the load bus has to be computed. This is done by rearranging(3.18) in powers of voltage V . This is given below.

(R2s +X2

s )Q2l + 2V 2XsQl + (V 2 +Rs Pl)

2 +X2s P

2l − E2 V 2 = 0

(R2s +X2

s )Q2l︸ ︷︷ ︸

III

+ 2V 2XsQl︸ ︷︷ ︸II

+ V 4︸︷︷︸I

+R2s P

2l︸ ︷︷ ︸

III

+ 2V 2Rs Pl︸ ︷︷ ︸II

+X2s P

2l︸ ︷︷ ︸

III

−E2 V 2︸ ︷︷ ︸II

= 0

Combining the I, II and III terms in the above equation, we get the following.

V 4 +

2(RsPl +XsQl)− E2V 2 + (R2

s +X2s )(Q2

l + P 2l ) = 0 (3.31)

Now substituting values of Rs, Xs, Pl, Ql and E in above equation, we get,

V 4 +

2 [0.0784× 25 + 0.3922× 50]− 102V 2 + (0.07842 + 0.39222)(252 + 502) = 0

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After simplifying the above, we have the following equation.

V 4 − 56.86V 2 + 500 = 0

Therefore

V 2 =56.86±

√56.86− 4× 500

2= 45.985, 10.875

and V = ±6.78 kV, ±3.297 kV

Since rms value cannot be negative and maximum rms value must be a feasible solution, thereforeV = 6.78 kV.Now we can compute ∆V using (3.14), as it is given below.

∆V =Rs Pl +XsQl

V+ j

Xs Pl −RsQl

V

=0.0784 25 + 0.392 50

6.78+ j

0.3922 25− 0.0784 50

6.78= 3.1814 + j0.8677 kV = 3.2976∠15.25o kV

Now the line current can be found out as following.

I l =Pl −Ql

V=

25− j50

6.782= 3.86− j7.3746 kA= 8.242∠− 63.44o kA

The power factor of load is cos (tan−1(Ql/Pl)) = 0.4472 lagging. The phasor diagram for thiscase is similar to what is shown in Fig. 3.3(b).

(b) Compensator as a voltage regulator

Now it is required to maintain V = E = 10.0 kV at the load bus. For this let their be reac-tive power Qγ supplied by the compensator at the load bus. Therefore the net reactive power at theload bus is equal to Qs, which is given below.

Qs = Ql +Qγ

Thus from (3.18), we get,

(R2s +X2

s )Q2s + 2V 2XsQs + (V 2 +RsPl)

2 +X2sP

2l − E2V 2 = 0

(0.7842 + 0.39222)2Q2s + 2× 102 × 0.3922×Qs +

(102 + 0.784× 25)2 + 0.39222 × 252 − 104

= 0

From the above we have,

0.16Q2s + 78.44Qs + 491.98 = 0.

78

Solving the above equation we get,

Qs =−78.44±

√78.442 − 4× 0.16× 491.98

2× 0.16= −6.35 or− 484 MVAr.

The feasible solution is Qs = −6.35 MVAr because it requires less rating of the compensator.Therefore the reactive power of the compensator (Qγ) is,

Qγ = Qs −Ql = −6.35− 50 = −56.35 MVAr.

With Qs = −6.35 MVAr, the ∆V is computed by replacing Qs for Ql in (3.14) as given below.

∆V =RsPl +XsQs

V+ j

XsPl −RsQs

V

=0.0784× 25 + 0.39225×−6.35

10+ j

0.39225× 25− 0.0784× (−6.35)

10

=1.96− 2.4

10+ j

9.805 + 0.4978

10= −0.0532 + j1.030 kV = 1.03137∠92.95o kV

Now, we can find supply voltage E as given below.

E = ∆V + V

= 10− 0.0532 + j1.030

= 9.9468 + j1.030 = 10∠5.91o kV

The supply current is,

Is =Pl − jQs

V=

25− j(−6.35)

10= 2.5 + j0.635 kA = 2.579∠14.25o kA.

This indicates that power factor is not unity for perfect voltage regulation i.e., E = V . For thiscase the compensator current is given below.

Iγ =−jQγ

V=−j(−56.35)

10Iγ = j5.635 kA

The load current is computed as below.

I l =Pl − jQl

V=

25− j50

10= 2.5− j5.0 = IlR + jIlX = 5.59∠63.44o kA

The phasor diagram is similar to the one shown in Fig. 3.4(b). The phasor diagram shown hasinteresting features. The voltage at the load bus is maintained to 1.0 pu. It is observed that thereactive power of the compensator Qγ is not equal to load reactive power (Ql). It exceeds by 6.35

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MVAr. As a result of this compensation, the voltage regulation is perfect, however power factor isnot unity. The phase angle between V and Is is cos−1 0.969 = 14.25o as computed above. There-fore the angle between E and Is is (14.25o − 5.91o = 8.34o). Thus, source power factor (φs) iscos(8.340) = 0.9956 leading.

(c) Compensation for unity power factor

To achieve unity power factor at the load bus, the condition Qγ = −Ql must be satisfied, whichfurther implies that the net reactive power at the load bus is zero. Therefore substituting Ql = 0 in(3.31), we get the following.

V 4 +

2(RsPl − E2)V 2 + (R2

s +X2s )(P 2

l +Q2l ) = 0

V 4 + (2× 0.0784× 25− 102)V 2 + (0.07842 + 0.39222) 252 = 0

From the above,

V 4 + 96.08V 2 + 99.79 = 0

The solution of the above equation is,

V 2 =96.08± 93.97

2= 95.02, 1.052

V = ±9.747 kV,±1.0256 kV.

Since rms value cannot be negative and maximum rms value must be a feasible solution, thereforeV = 9.747 kV. Thus it is seen that for obtaining unity power factor at the load bus does not ensuredesired voltage regulation. Now the other quantities are computed as given below.

I l =Pl − jQ

V=

25− j50

9.747= 2.5648− j5.129 = 5.7345∠−63.43o kA

SinceQγ = −Ql, this implies that Iγ = −jQγ/V = jQl/V = j5.129 kA. The voltage drop acrossthe feeder is given as following.

∆V =Rs Pl +XsQl

V+ j

Xs Pl −RsQl

V

=(0.784× 25 + j0.3922× 25)

9.747= 0.201 + j1.005 = 1.0249∠5.01o kV

The phasor diagram for the above case is shown in Fig. 3.6.The percentage voltage change = (10 − 9.748)/10 × 100 = 2.5. Thus we see that power factorimproves voltage regulation enormously compared with uncompensated case. In many cases, de-gree of improvement is adequate and the compensator can be designed to provide reactive powerrequirement of load rather than as a ideal voltage regulator.

80

5.1

3kA

Ij

10 kVE

V =9.75 kV

5.1

3kA

lXIj

= 2.56 kAlRI RV

XV

5.73 63.43 kAlI

5.77

sI

Fig. 3.6 Phasor diagram for system with compensator in voltage regulation mode

3.3 Some Practical Aspects of Compensator used as Voltage Regulator

In this section, some practical aspects of the compensator in voltage regulation mode will be dis-cussed. The important parameters of the compensator which play significant role in obtainingdesired voltage regulation are: Knee point (V k), maximum or rated reactive power Qγmax and thecompensator gain Kγ .

The compensator gain Kγ is defined as the rate of change of compensator reactive power Qγ

with change in the voltage (V ), as given below.

Kγ =dQr

dV(3.32)

For linear relationship between Qγ and V with incremental change, the above equation be writtenas the following.

∆Qγ = ∆V Kγ (3.33)

Assuming compensator characteristics to be linear with Qγ ≤ Qγmax limit, the voltage can berepresented as,

V = Vk +Qγ

Kγ

(3.34)

This is re-written as,

Qγ = Kγ(V − Vk) (3.35)

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Flat V-Q characteristics imply that Kγ → ∞. That means the compensator which can absorb orgenerate exactly right amount of reactive power to maintain supply voltage constant as the loadvaries without any constraint. We shall now see the regulating properties of the compensator,when compensator has finite gain Kr operating on supply system with a finite short circuit level,Ssc. The further which are made in the following study are: high Xs/Rs ratio and negligible loadpower fluctuations. The net reactive power at the load bus is sum of the load and the compensatorreactive power as given below.

Ql +Qγ = Qs (3.36)

Using earlier voltage and reactive power relationship from equation (3.30), it can be written as thefollowing.

V ' E(1− Qs

Ssc) (3.37)

The compensator voltage represented by (3.34) and system voltage represented by (3.37) are shownin Fig. 3.7(a) and (b) respectively.

kV

Q

k

QV V

K

V

E

sQ

1- s

SC

QV E

S

(a) (b)

Fig. 3.7 (a) Voltage characterstics of compensator (b) System voltage characteristics

Differentiating V with respect to Qs, we get, intrinsic sensitivity of the supply voltage withvariation in Qs as given below.

dV

dQs

= − E

Ssc(3.38)

It is seen from the above equation that high value of short circuit level Ssc reduces the voltagesensitivity, making voltage variation flat irrespective of Ql. With compensator replacing Qs =Qγ +Ql in (3.37), we have the following.

V ' E

(1− Ql +Qγ

Ssc

)(3.39)

82

Substituting Qγ from (3.35), we get the following equation.

V ' E

[1 +KγVk/Ssc1 + EKγ/Ssc

− Ql/Ssc1 + EKγ/Ssc

](3.40)

Although approximate, above equation gives the effects of all the major parameters such as loadreactive power, the compensator characteristics Vγ and Kγ and the system characteristics E andSsc. As we discussed, V-Q characteristics is flat for high or infinite value Kγ . However the highervalue of the gain Kγ means large rating and quick rate of change of the reactive power with varia-tion in the system voltage. This makes cost of the compensator high.

The compensator has two effects as seen from (3.40), i.e., it alters the no load supply voltage(E) and it modifies the sensitivity of supply point voltage to the variation in the load reactive power.Differentiating (3.40) with respect to Ql, we get,

dV

dQl

= − E/Ssc1 +KrE/Ssc

(3.41)

which is voltage sensitivity of supply point voltage to the load reactive power. It can be seen thatthe voltage sensitivity is reduced as compared to the voltage sensitivity without compensator asindicated in (3.38).

It is useful to express the slope (−E/Ssc) by a term in a form similar to Kγ = dQγ/dV , as givenbelow.

Ks = −SscE

Thus,1

Ks

= − E

Ssc(3.42)

Substituting V from (3.39) into (3.35), the following is obtained.

Qγ = Kγ

[E

(1− Ql +Qγ

Ssc

)− Vk

](3.43)

Collecting the coefficients of Qγ from both sides of the above equation, we get

Qγ =Kγ

1 +Kγ (E/Ssc)

[E

(1− Ql

Ssc

)− Vk

](3.44)

Setting knee voltage Vk of the compensator equal to system voltage E i.e., Vk = E, the aboveequation is simplified to,

Qγ = − Kγ (E/Ssc)

1 +Kγ (E/Ssc)Ql

= −(

Kγ/Ks

1 +Kγ/Ks

)Ql. (3.45)

83

From the above equation it is observed that, when compensator gain Kγ → ∞, Qγ → −Ql. Thisindicates perfect compensation of the load reactive power in order to regulate the load bus voltage.

Example 3.2 Consider a three-phase system with line-line voltage 11 kV and short circuit capacityof 480 MVA. With compensator gain of 100 pu determine voltage sensitivity with and withoutcompensator. For each case, if a load reactive power changes by 10 MVArs, find out the change inload bus voltage assuming linear relationship between V-Q characteristics. Also find relationshipbetween compensator and load reactive powers.

Solution: The voltage sensitivity can be computed using the following equation.

dV

dQl

= − E/Ssc1 +KγE/Ssc

Without compensator Kγ = 0, E = (11/√

3) = 6.35 kV and Ssc = 480/3 = 160 MVA.Substituting these values in the above equation, the voltage sensitivity is given below.

dV

dQl

= − 6.35/160

1 + 0× 6.35/160= −0.039

The change in voltage due to variation of reactive power by 10 MVArs, ∆V = −0.039 × 10 =−0.39 kV.With compensator, Kγ = 100

dV

dQl

= − 6.35/160

1 + 100× 6.35/160= −0.0078

The change in voltage due to variation of reactive power by 10 MVArs, ∆V = −0.0078 × 10 =−0.078 kV.Thus it is seen that, with finite compensator gain their is quite reduction in the voltage sensitivity,which means that the load bus is fairly constant for considerable change in the load reactive power.The compensator reactive power Qγ and load reactive power Ql are related by equation (3.45) andis given below.

Qγ = − Kγ (E/Ssc)

1 +Kγ (E/Ssc)Ql = − 100× (6.35/160)

1 + 100× (6.35/160)Ql

= −0.79Ql

It can be observed that when compensator gain, (Qγ) is quite large, then compensator reactivepowerQγ is equal and opposite to that of load reactive power i.e.,Qγ = −Ql. It is further observedthat due to finite compensator gain i.e., Kγ = 100, reactive power is partially compensated Thecompensator reactive power varies from 0 to 7.9 MVAr for 0 to 10 MVAr change in the loadreactive power.

3.4 Phase Balancing and Power Factor Correction of Unbalanced Loads

So far we have discussed voltage regulation and power factor correction for single phase systems.In this section we will focus on balancing of three-phase unbalanced loads. In considering unbal-anced loads, both load and compensator are modeled in terms of their admittances and impedances.

84

3.4.1 Three-phase Unbalanced Loads

Consider a three-phase three-wire system suppling unbalanced load as shown in Fig. 3.8.

anV

aZ

bZ

cZ

bnV

cnV

N n

aI

1I

2I

bI

cI

Fig. 3.8 Three-phase unbalanced load

Applying Kirchoff’s voltage law for the two loops shown in the figure, we have the followingequations.

−V an + Za I1 + Zb (I1 − I2) + V bn = 0

−V bn + Zb I2 + Zb (I2 − I1) + V cn = 0 (3.46)

Rearranging above, we get the following.

V an − V bn = (Za + Zb) I1 − Zb I2

V bn − V cn = (Zb + Zc) I2 − Zb I1 (3.47)

The above can be represented in matrix form as given below.[V an − V bn

V bn − V cn

]=

[(Za + Zb) −Zb−Zb (Zb + Zc)

] [I1

I2

](3.48)

Therefore the currents are given as below.[I1

I2

]=

[(Za + Zb) −Zb−Zb (Zb + Zc)

]−1 [V an − V bn

V bn − V cn

]=

1

∆Z

[(Zb + Zc) Zb

Zb (Za + Zb)

] [V an − V bn

V bn − V cn

]Therefore,

[I1

I2

]=

1

∆Z

[(Zb + Zc) Zb

Zb (Za + Zb)

] [V an − V bn

V bn − V cn

]. (3.49)

Where, ∆Z = (Zb + Zc)(Za + Zb)− Z2b = Za Zb + Zb Zc + Zc Za. The current I1 is given below.

I1 =1

∆Z

[(Zb + Zc)(V an − V bn) + Zb(V bn − V cn)

]=

1

∆Z

[(Zb + Zc)V an − ZcV bn − ZbV cn

](3.50)

85

Similarly,

I2 =1

∆Z

[Zb(V an − V bn) + (Za + Zb)(V bn − V cn)

]=

1

∆Z

[ZbV an + ZaV bn − (Za + Zb)V cn

](3.51)

Now,

Ia = I1 =1

∆Z

[(Zb + Zc)V an − ZcV bn − ZbV cn

]Ib = I2 − I1

=1

∆Z

[

ZbV an + ZaV bn − (Za + Zb)V cn − (Zb + Zc)V an + ZcV bn + ZbV cn

]=

[(Zc + Za)V bn − ZaV cn − ZcVan

]∆Z

=(Zc + Za)V bn − ZcV an − ZaV cn

∆Z

(3.52)

and

Ic = −I2 = −Ib − Ia =(Za + Zb)V cn − ZbV an − ZaV bn

∆Z

(3.53)

Alternatively phase currents can be expressed as following.

Ia =V an − V Nn

Za

Ib =V bn − V Nn

Zb(3.54)

Ic =V cn − V Nn

Zc

Applying Kirchoff’s current law at node N , we get Ia + Ib + Ic = 0. Therefore from the aboveequation,

V an − V Nn

Za+V bn − V Nn

Zb+V cn − V Nn

Zc= 0. (3.55)

Which implies that,

V an

Za+V bn

Zb+V cn

Zc=

[1

Za+

1

Zb+

1

Zc

]V Nn =

ZaZb + ZbZc + ZcZaZaZbZc

V Nn (3.56)

From the above equation the voltage between the load and system neutral can be found. It is givenbelow.

V Nn =ZaZbZc

∆Z

[V an

Za+V bn

Zb+V cn

Zc

]=

11Za

+ 1Zb

+ 1Zc

[V an

Za+V bn

Zb+V cn

Zc

](3.57)

86

Some interesting points are observed from the above formulation.

1. If both source voltage and load impedances are balanced i.e., Za = Zb = Zc = Z, thenV Nn = 1

3

(V an + V bn + V cn

)= 0. Thus their will not be any voltage between two neutrals.

2. If supply voltage are balanced and load impedances are unbalanced, then V Nn 6=0 and isgiven by the above equation.

3. If supply voltages are not balanced but load impedances are identical, then V Nn = 13

(V an + V bn + V cn

).

This equivalent to zero sequence voltage V 0.

It is interesting to note that if the two neutrals are connected together i.e., V Nn = 0, then eachphase become independent through neutral. Such configuration is called three-phase four-wiresystem. In general, three-phase four-wire system has following properties.

V Nn = 0

Ia + Ib + Ic = INn 6= 0

(3.58)

The current INn is equivalent to zero sequence current (I0) and it will flow in the neutral wire.For three-phase three-wire system, the zero sequence current is always zero and therefore followingproperties are satisfied.

V Nn 6= 0

Ia + Ib + Ic = 0

(3.59)

Thus, it is interesting to observe that three-phase three-wire and three-phase four-wire system havedual properties in regard to neutral voltage and current.

3.4.2 Representation of Three-phase Delta Connected Unbalanced Load

A three-phase delta connected unbalanced and its equivalent star connected load are shown in Fig.3.9(a) and (b) respectively. The three-phase load is represented by line-line admittances as givenbelow.

cI

bI

aIab

lY

bclY

ca

lY

c

ab

cI

bI

aI

clZ

c

ab

alZ

blZ n

(a) (b)

Fig. 3.9 (a) An unbalanced delta connected load (b) Its equivalent star connected load

87

Y abl = Gab

l + jBabl

Y bcl = Gbc

l + jBbcl

Y cal = Gca

l + jBcal

(3.60)

The delta connected load can be equivalently converted to star connected load using followingexpressions.

Zal =

Zabl Zca

l

Zabl + Zbc

l + Zcal

Zbl =

Zbcl Z

abl

Zabl + Zbc

l + Zcal

Zcl =

Zcal Zbc

l

Zabl + Zbc

l + Zcal

(3.61)

Where Zabl = 1/Y ab

l , Zbcl = 1/Y bc

l and Zcal = 1/Y ca

l . The above equation can also be written inadmittance form

Y al =

Y abl Y bc

l + Y bcl Y ca

l + Y cal Y ab

l

Y bcl

Y bl =

Y abl Y bc

l + Y bcl Y ca

l + Y cal Y ab

l

Y cal

Y cl =

Y abl Y bc

l + Y bcl Y ca

l + Y cal Y ab

l

Y abl

(3.62)

Example 3.3 Consider three-phase system supply a delta connected unbalanced load with Z la =

Ra = 10 Ω, Z lb = Rb = 15 Ω and Z l

c = Rc = 30 Ω as shown in Fig. 3.8. Determine the voltagebetween neutrals and find the phase currents. Assume a balance supply voltage with rms value of230 V. Find out the vector and arithmetic power factor. Comment upon the results.

88

Solution: The voltage between neutrals V Nn is given as following.

VNn =RaRbRc

RaRb +RbRc +RcRa

[V an

Ra

+V bn

Rb

+V cn

Rc

]=

10× 15× 30

10× 15 + 15× 30 + 30× 10

[V ∠0o

10+V ∠−120o

15+V ∠120o

30

]=

4500

900

[3V ∠0o + 2V ∠−120o + V ∠120o

30

]=

4500

900

1

30V

[3 + 2

(−1

2− j√

3

2

)+

(−1

2+ j

√3

2

)]

=V

6

[3− 1− 1

2− j2×

√3

2+ j

√3

2

]

=V

6

[3

2− j√

3

2

]= V

[1

4− j 1

4√

3

]= V [0.25− j0.1443]

VNn =V

2√

3∠−30o = 66.39∠−30o Volts

Knowing this voltage, we can find phase currents as following.

Ia =V an − V Nn

Ra

=V ∠0o − V/(2

√3)∠−30o

10

=V [1− 0.25 + j0.1443]

10= 230× [0.075 + j0.01443]

= 17.56∠10.89o Amps

Similarly,

Ib =V bn − V Nn

Rb

=V ∠− 120o − V/(2

√3)∠−30o

15= 230× [−0.05− j0.04811]

= 15.94∠−136.1o Amps

and

Ic =V cn − V Nn

Zc=V ∠120o − V/(2

√3)∠−30o

30= 230× [−0.025 + j0.03367]

= 9.64∠126.58o Amps

89

It can been seen that Ia + Ib + Ic = 0. The phase powers are computed as below.

Sa = V a (Ia)∗ = Pa + jQa = 230× 17.56∠− 10.81o = 3976.12− j757.48 VA

Sb = V b (Ib)∗ = Pb + jQb = 230× 15.94∠(−120o + 136.1o = 3522.4 + j1016.69 VA

Sc = V c (Ic)∗ = Pc + jQc = 230× 9.64∠(120o − 126.58o) = 2202.59− j254.06 VA

From the above the total apparent power SV = Sa + Sb + Sc = 9692.11 + j0 VA. Therefore,SV =

∣∣Sa + Sb + Sc∣∣ = 9692.11 VA.

The total arithmetic apparent power SA =∣∣Sa∣∣ +

∣∣Sb∣∣ +∣∣Sc∣∣ == 9922.2 VA. Therefore, the

arithmetic and vector apparent power factors are given by,

pfA =P

SA=

9692.11

9922.2= 0.9768

pfV =P

SV=

9622.11

9622.11= 1.00.

It is interesting to note that although the load in each phase is resistive but each phase has somereactive power. This is due to unbalance of the load currents. This apparently increases the ratingof power conductors for given amount of power transfer. It is also to be noted that the net reactivepower Q = Qa + Qb + Qc = 0 leading to the unity vector apparent power factor . However thearithmetic apparent power factor is less than unity showing the effect of the unbalance loads on thepower factor.

3.4.3 An Alternate Approach to Determine Phase Currents and Powers

In this section, an alternate approach will be discussed to solve phase currents and powers directlywithout computing the neutral voltage for the system shown in Fig.3.9(a). First we express three-phase voltage in the following form.

V a = V ∠0o

V b = V ∠− 120o = α2V (3.63)V c = V ∠1200 = αV

Where, in above equation, α is known as complex operator and value of α and α2 are given below.

α = ej2π/3 = 1∠120o = −1/2 + j√

3/2

α2 = ej4π/3 = 1∠240 = 1∠− 120 = −1/2− j√

3/2 (3.64)

Also note the following property,

1 + α + α2 = 0. (3.65)

Using the above, the line to line voltages can be expressed as following.

90

V ab = V a − V b = (1− α2)V

V bc = V b − V c = (α2 − α)V (3.66)V ca = V c − V a = (α− 1)V

Therefore, currents in line ab, bc and ca are given as,

Iabl = Y abl V ab = Y ab

l (1− α2)V

Ibcl = Y bcl V bc = Y bc

l (α2 − α)V (3.67)Ical = Y ca

l V ca = Y cal (α− 1)V

Hence line currents are given as,

Ial = Iabl − Ical = [Y abl (1− α2)− Y ca

l (α− 1)]V

Ibl = Ibcl − Iabl = [Y bcl (α2 − α)− Y ab

l (1− α2)]V (3.68)Icl = Ical − Ibcl = [Y ca

l (α− 1)− Y bcl (α2 − α)]V

Example 3.4 Compute line currents by using above expressions directly for the problem in Exam-ple 3.3.Solution: To compute line currents directly from the above expressions, we need to compute Y ab

l .These are given below

Y abl =

1

Zabl

=Zcl

Zal Z

bl + Zb

l Zcl + Zc

l Zal

Y bcl =

1

Zbcl

=Zal

Zal Z

bl + Zb

l Zcl + Zc

l Zal

(3.69)

Y cal =

1

Zcal

=Zbl

Zal Z

bl + Zb

l Zcl + Zc

l Zal

Substituting, Z la = Ra = 10 Ω, Z l

b = Rb = 15 Ω and Z lc = Rc = 30 Ω into above equation, we get

the following.

Y abl = Gab =

1

30Ω

Y bcl = Gbc =

1

90Ω

Y cal = Gca =

1

60Ω

Substituting above values of the admittances in (3.68) , line currents are computed as below.

91

Ia =

[1

30

1− (−1

2− j√

3

2)

− 1

60

(−1

2+ j

√3

2)− 1

]V

= V (0.075 + j0.0144)

= 0.07637V ∠10.89o

= 17.56∠10.89o Amps, for V=230 V

Similarly for Phase-b current,

Ib =

[1

90

(−1

2− j√

3

2)− (−1

2+ j

√3

2)

− 1

30

1− (−1

2− j√

3

2)

]V

= V (−0.05− j0.0481)

= 0.06933V ∠− 136.1o

= 15.94∠− 136.91o Amps, for V=230 V

Similarly for Phase-c current,

Ic =

[1

60

(−1

2+ j

√3

2)− 1)

− 1

90

(−1

2− j√

3

2)− (−1

2+ j

√3

2)

]V

= V (−0.025 + j0.0336)

= 0.04194V ∠126.58o

= 9.64∠126.58o Amps, for V=230 V

Thus it is found that the above values are similar to what have been found in previous Example3.3. The other quantities such as powers and power factors are same.

3.4.4 An Example of Balancing an Unbalanced Delta Connected Load

An unbalanced delta connected load is shown in Fig. 3.10(a). As can be seen from the figure thatbetween phase-a and b there is admittance Y ab

l = Gabl and other two branches are open. This is

an example of extreme unbalanced load. Obviously for this load, line currents will be extremelyunbalanced. Now we aim to make these line currents to be balanced and in phase with their phasevoltages. So, let us assume that we add admittances Y ab

γ , Y bcγ and Y ca

γ between phases ab, bc andca respectively as shown in Fig. 3.10(b) and (c). Let values of compensator susceptances are givenby,

Y abγ = 0

Y bcγ = jGab

l /√

3

Y caγ = −jGab

l /√

3

92

cI

bI

aIab ab

l lY G

c

b

0cal

Y

0bc

lY

a

cI

bI

aIab ab

l lY G

c

ab

/3

ca

abl

Y

jG

/

3

bc

abl

YjG

0abY

0calY0

bclY

cI

bI

aI

c

ab

/3

ca

abl

Y

jG/

3

bc

abl

YjG

ab ablY G

(a) (b) (c)

Fig. 3.10 (a) An unbalanced three-phase load (b) With compensator (c) Compensated system

Thus total admittances between lines are given by,

Y ab = Y abl + Y ab

γ = Gabl + 0 = Gab

l

Y bc = Y bcl + Y bc

γ = 0 + jGabl /√

3 = jGabl /√

3

Y ca = Y cal + Y ca

γ = 0− jGabl /√

3 = −jGabl /√

3.

Therefore the impedances between load lines are given by,

Zab =1

Y ab=

1

Gabl

Zbc =1

Y bc=−j√

3

Gabl

Zca =1

Y ca=j√

3

Gabl

Note that Zab + Zbc + Zca = 1/Gabl − j

√3/Gab

l + j√

3/Gabl = 1/Gab

l .The impedances, Za, Zb and Zc of equivalent star connected load are given as follows.

Za =Zab × Zca

Zab + Zbc + Zca

= (1

Gabl

× j√

3

Gabl

)/(1

Gabl

)

=j√

3

Gabl

93

Zb =Zbc × Zab

Zab + Zbc + Zca

= (1

Gabl

× −j√

3

Gabl

)/(1

Gabl

)

=−j√

3

Gabl

Zc =Zca × Zbc

Zab + Zbc + Zca

= (−j√

3

Gabl

× j√

3

Gabl

)/(1

Gabl

)

=3

Gabl

The above impedances seen from the load side are shown in Fig. 3.11(a) below. Using (3.57),

cI

bI

aI

3cl ab

l

ZG

c

ab

3al ab

l

Z jG

n

3bl ab

l

Z jG

cI

bI

aI

c

ab

N1ablG

1ablG

1ablG

Source

Load

Fig. 3.11 Compensated system (a) Load side (b) Source side

the voltage between load and system neutral of delta equivalent star load as shown in Fig. 3.11, iscomputed as below.

VNn =1

1Za

+ 1Zb

+ 1Zc

[V an

Za+V bn

Zb+V cn

Zc

]=

11

(j√

3/Gabl )+ 1

(−j√

3/Gabl )+ 3

Gabl

[V ∠0o

j√

3/Gabl

+V ∠− 120o

−j√

3/Gabl

+V ∠120o

3/Gabl

]=

3V

Gabl

(Gabl

3− jG

abl√3

)= 2V

(1

2− j√

3

2

)= 2V ∠− 60o

94

Using above value of neutral voltage the line currents are computed as following.

Ia =V an − V Nn

Za

=[V ∠0o − 2V ∠− 60o]

j√

3Gabl

= Gabl V = Gab

l V a

Ib =V bn − V Nn

Zb

=[V ∠− 120o − 2V ∠− 60o]

−j√

3Gabl

= Gabl V ∠240o

= Gabl V b

Ic =V cn − V Nn

Zc

=[V ∠120o − 2V ∠− 60o]

3Gabl

= Gabl V ∠120o

= Gabl V c

From the above example, it is seen that the currents in each phase are balanced and in phase withtheir respective voltages. This is equivalently shown in Fig. 3.11(b). It is to be mentioned here thatthe two neutrals in Fig. 3.11 are not same. In Fig. 3.11(b), the neutral N is same as the systemneutral as shown in Fig. 3.8, whereas in Fig. 3.11(a), V Nn = 2V ∠−60o. However the readermay be curious to know why Y ab

γ = 0, Y bcγ = jGab

l /√

3 and γca = −jGabl /√

3 have been chosenas compensator admittance values. The answer of the question can be found by going followingsections.

3.5 A Generalized Approach for Load Compensation using SymmetricalComponents

In the previous section, we have expressed line currents Ia, Ib and Ic, in terms load admittancesand the voltage V for a delta connected unbalanced load as shown in Fig 3.12(a). For the sake ofcompleteness, theses are reproduced below.

95

Ial = Iabl − Ical = [Y abl (1− α2)− Y ca

l (α− 1)]V

Ibl = Ibcl − Iabl = [Y bcl (α2 − α)− Y ab

l (1− α2)]V (3.70)Icl = Ical − Ibcl = [Y ca

l (α− 1)− Y bcl (α2 − α)]V

cI

bI

aI

ca

Y

ablYabY

bcY bclY

ca

lY

c

ab

clI

blI

alI

ablY

bclY

ca

lY

c

ab

(a) (b)

Fig. 3.12 (a) An unbalanced delta connested load (b) Compensated system

Since loads currents are unbalanced, these will have positive and negative currents. The zerosequence current will be zero as it is three-phase and three-wire system. These symmetrical com-ponents of the load currents are expressed as following.

I0l

I1l

I2l

=1√3

1 1 11 a a2

1 a2 a

IalIblIcl

(3.71)

In equation (3.71), a factor of 1/√

3 is considered to have unitary symmetrical transformation.From the above equation, zero sequence current is given below.

I0l =(Ial + Ial + Ial

)/√

3

The positive sequence current is as follows.

I1l =1√3

[Ial + αIbl + α2Icl

]=

1√3

[Y abl (1− α2)− Y ca

l (α− 1) + αY bcl (α2 − α)− Y ab

l (1− α2)

+α2Y cal (α− 1)− Y bc

l (α− α2)

]V

=1√3

[Y abl − α2Y ab

l + Y cal − αY ca

l + α3Y bcl − α2Y bc

l − αY abl − α3Y ab

l + α3Y cal − α2Y ca

l

−α4Y bcl + α3Y bc

l ]V

=(Y abl + Y bc

l + Y cal

)V√

3

96

Similarly negative sequence component of the current is,

I2l =1√3

[Ial + α2Ibl + αIcl

]=

1√3

[Y abl (1− α2)− Y ca

l (α− 1) + α2Y bcl (α2 − α)− Y ab

l (1− α2)

+αY cal (α− 1)− Y bc

l (α2 − α)

]V

=1√3

[Y abl − α2Y ab

l − αY cal + Y ca

l + α4Y bcl − α3Y bc

l − α2Y abl

+α4Y abl + α2Y ca

l − αY cal − α3Y bc

l + α2Y bcl ]V

=1√3

[−3α2Y abl − 3Y bc

l − 3αY cal ]V

= −[α2Y abl + Y bc

l + αY cal ]√

3V

From the above, it can be written that,

I0l = 0

I1l =(Y abl + Y bc

l + Y cal

)√3V (3.72)

I2l = −(α2Y ab

l + Y bcl + αY ca

l

)√3V

When compensator is used, three delta branches Y abγ , Y bc

γ and Y caγ are added as shown in Fig.

3.12(b). Using above analysis, the sequence components of the compensator currents can be givenas below.

I0γ = 0

I1γ =(Y abγ + Y bc

γ + Y caγ

) √3V (3.73)

I2γ = −(α2Y ab

γ + Y bcγ + αY ca

γ

) √3V

Since, compensator currents are purely reactive, i.e., Gabγ = Gbc

γ = Gcaγ = 0,

Y abγ = Gab

γ + jBabγ = jBab

γ

Y bcγ = Gbc

γ + jBbcγ = jBbc

γ (3.74)Y caγ = Gca

γ + jBcaγ = jBca

γ .

Using above, the compensated sequence currents can be written as,

I0γ = 0

I1γ = j(Babγ +Bbc

γ +Bcaγ

) √3V (3.75)

I2γ = −j(α2Babγ +Bbc

γ + αBcaγ )√

3V

Knowing nature of compensator and load currents, we can set compensation objectives as fol-lowing.

1. All negative sequence component of the load current must be supplied from the compensatornegative current, i.e.,

I2l = −I2γ (3.76)

97

The above further implies that,

Re(I2l

)+ j Im

(I2l

)= −Re

(I2γ

)− j Im

(I2γ

)(3.77)

2. The total positive sequence current, which is source current should have desired power factorfrom the source, i.e.,

Im (I1l + I1γ)

Re (I1l + I1γ)= tanφ = β (3.78)

Where, φ is the desired phase angle between the line currents and the supply voltages. The aboveequation thus implies that,

Im (I1l + I1γ) = β Re (I1l + I1γ) (3.79)

Since Re (I1γ) = 0, the above equation is rewritten as following.

Im (I1l)− β Re (I1l) = −Im (I1γ) (3.80)

The equation (3.77) gives two conditions and equation (3.79) gives one condition. There arethree unknown variables, i.e., Bab

γ , Bbcγ and Bca

γ and three conditions. Therefore the unknownvariables can be solved. This is described in the following section. Using (3.75), the current I2γ isexpressed as following.

I2γ = −j[α2Babγ +Bbc

γ + αBcaγ ]√

3V

= −j

[(−1

2− j√

3

2

)Babγ +Bbc

γ +

(−1

2+ j

√3

2

)Bcaγ

]√

3V

=

[(−√

3

2Babγ +

√3

2Bcaγ

)− j

(−1

2Babγ +Bbc

γ −1

2Bcaγ

)] √3V (3.81)

= Re(I2γ

)− j Im

(I2γ

)Thus the above equation implies that(√

3

2Babγ −

√3

2Bcaγ

)= − 1√

3VRe(I2γ

)=

1√3V

Re(I2l

)(3.82)

and, (−1

2Babγ +Bbc

γ −1

2Bcaγ

)= − 1√

3VIm(I2γ

)=

1√3V

Im(I2l

)(3.83)

Or (−Bab

γ + 2Bbcγ −Bca

γ

)=

1√3V

2 Im(I2l

)(3.84)

98

From (3.75), Im (I1γ) can be written as,

Im (I1γ) =(Babγ +Bbc

γ +Bcaγ

) √3V. (3.85)

Substituting Im (I1γ) from above equation into (3.85), we get the following.

−(Babγ +Bbc

γ +Bcaγ ) = − 1√

3VIm (I1γ) =

1√3V

Im (I1l)− β Re (I1l)

(3.86)

Subtracting (3.86) from (??), the following is obtained,

Bbcγ =

−1

3√

3V[Im (I1l)− 2 Im (I2l)− β Re (I1l)]. (3.87)

Now, from (3.82) we have

−1

2Babγ −

1

2Bcaγ =

1√3V

Im (I2l)−Bbcγ

=1√3V

Im (I2l)−[−1

3√

3V

Im (I1l)− β Re (I1l)− 2 Im (I2l)

]=

1

3√

3V

Im (I2l) + Im (I1l)− β Re (I1l)

(3.88)

Reconsidering (3.88) and (3.83), we have

−1

2Babγ −

1

2Bcaγ =

1

3√

3V

Im (I1l) + Im (I2l)− β Re (I1l)

1

2Babγ −

1

2Bcaγ =

1

3√

3V[√

3Re (I2l)]

Adding above equations, we get

Bcaγ =

−1

3√

3V[Im (I1l) + Im (I2l) +

√3 Re (I2l)− β Re (I1l)]. (3.89)

Therefore,

Babγ = Bca

γ +2

3√

3V[√

3 Re (I2l)]

=−1

3√

3V[Im (I1l) + Im (I2l) +

√3 Re (I2l)− β Re (I1l)− 2

√3 Re (I2l)]

=−1

3√

3V[Im (I1l) + Im (I2l)−

√3 Re (I2l)− β Re (I1l)] (3.90)

Similarly, Bbcγ can be written as in the following.

Bbcγ = − 1

3√

3V

[Im(I1l)− 2Im(I2l)− β Re(I1l)

](3.91)

(3.92)

99

From the above, the compensator susceptances in terms of real and imaginary parts of the loadcurrent can be written as following.

Babγ =

−1

3√

3V[Im (I1l) + Im (I2l)−

√3 Re (I2l)− β Re (I1l)]

Bbcγ = − 1

3√

3V[Im(I1l)− 2Im(I2l)− β Re(I1l)] (3.93)

Bcaγ =

−1

3√

3V[Im (I1l) + Im (I2l) +

√3 Re (I2l)− β Re (I1l)]

In the above equation, the susceptances of the compensator are expressed in terms of real andimaginary parts of symmetrical components of load currents. It is however advantageous to expressthese susceptances in terms of instantaneous values of voltages and currents from implementationpoint of view. The first step to achieve this is to express these susceptances in terms of loadcurrents, i.e., Ial, Ibl and Icl, which is described below. Using equation (3.71), the sequencecomponents of the load currents are expressed as,

I0l =1√3

[Ial + Ibl + Icl

]I1l =

1√3

[Ial + αIbl + α2Icl

](3.94)

I2l =1√3

[Ial + α2I1l + αIcl

].

Substituting these values of sequence components of load currents, in (3.93), we can obtain com-pensator susceptances in terms of real and inaginary components of the load currents. Let us startfrom the Bbc

γ , as obtained following.

Bbcγ = − 1

3√

3V

[Im(I1l)− 2Im(I2l)− β Re(I1l)

]= − 1

3√

3V

[Im(Ial + αIbl + α2Icl√

3

)− 2 Im

(Ial + α2Ibl + αIcl√

3

)− β Re

(Ial + αIbl + α2Icl√

3

)]= − 1

9V

[Im

(−Ial + (2 + 3α)Ibl + (2 + 3α2)Icl− β Re (Ial + αIbl + α2Icl)

]= − 1

9V

[Im−Ial + 2Ibl + 2Icl + 3αIbl + 3α2Icl

− β Re (Ial + αIbl + α2Icl)

]By adding and subracting Ibl and Icl in the above equation we get,

Bbcγ = − 1

9V

[Im

(−Ial − Ibl − Icl) + 3Ibl + 3Icl + 3α Ibl + α2 Icl− β Re(Ial + αIbl + α2Icl)

]We know that Ial+Ibl+Icl=0, therefore Ial + Ibl = −Icl.

Bbcγ = − 1

3V

[−Im

(Ial)

+ Im(αIbl

)+ Im

(α2Icl

)− β

3Re(Ial + αIbl + α2Icl

])(3.95)

100

Similarly, it can be proved that,

Bcaγ = − 1

3V

[Im(Ial)− Im

(αIbl

)+ Im

(α2Icl

)− β

3Re(Ial + αIbl + α2Icl

])(3.96)

Babγ = − 1

3V

[Im(Ial)

+ Im(αIbl

)− Im

(α2Icl

)− β

3Re(Ial + αIbl + α2Icl

])(3.97)

The above expressions forBcaγ andBab

γ are proved below. For convenience, the last term associatedwith β is not considered. For the sake simplicity in equations (3.96) and (3.97) are proved to thosegiven in equations (3.93).

Bcaγ =

−1

3V

[Im(Ial)− Im

(αIbl

)+ Im

(α2Icl

)]= − 1

3V

[Im

(I0l + I1l + I2l

)√

3− α

(I0l + α2I1l + αI2l

)√

3+ α2

(I0l + αI1l + α2I2l

)√

3

]

Since I0l=0

Bcaγ = − 1

3√

3VIm[I1l − 2α2I2l

]= − 1

3√

3VIm

[I1l − 2

(−1

2− j√

3

2

)I2l

]= − 1

3√

3VIm[I1l + I2l + j

√3I2l

]= − 1

3√

3V

[Im(I1l

)+ Im

(I2l

)+√

3Re(I2l

)]Note that Im(jI2l) = Re(I2l). Adding β term, we get the following.

Bcaγ = − 1

3√

3V

[Im(I1l) + Im(I2l) +

√3Re

(I2l

)− βRe(I1l)

]Similarly,

Babγ = − 1

3V

[Im(Ial)

+ Im(αIbl

)− Im

(α2Icl

)]= − 1

3V

[Im

(I0l + I1l + I2l

)√

3+ α

(I0l + α2I1l + αI2l

)√

3− α2

(I0l + αI1l + α2I2l

)√

3

]101

Babγ = − 1

3√

3VIm[I1l − 2αI2l

]= − 1

3√

3VIm

[I1l − 2

(−1

2+ j

√3

2

)I2l

]= − 1

3√

3VIm[I1l + I2l − j

√3I2l

]= − 1

3√

3V

[Im(I1l

)+ Im

(I2l

)−√

3Re(I2l

)]Thus, Compensator susceptances are expressed as following.

Babγ = − 1

3V

[Im(Ial) + Im(αIbl)− Im(α2Icl)−

β

3Re(Ial + αIbl + α2Icl)

]Bbcγ = − 1

3V

[−Im(Ial) + Im(αIbl) + Im(α2Icl)−

β

3Re(Ial + αIbl + α2Icl)

](3.98)

Bcaγ = − 1

3V

[Im(Ial)− Im(αIbl) + Im(α2Icl)−

β

3Re(Ial + αIbl + α2Icl)

]An unity power factor is desired from the source. For this cosφl = 1, implying tanφl = 0 henceβ = 0. Thus we have,

Babγ = − 1

3V

[Im(Ial)

+ Im(αIbl

)− Im

(α2Icl

)]Bbcγ = − 1

3V

[−Im

(Ial)

+ Im(αIbl

)+ Im

(α2Icl

)](3.99)

Bcaγ = − 1

3V

[Im(Ial)− Im

(αIbl

)+ Im

(α2Icl

)]The above equations are easy to realize in order to find compensator susceptances. As mentionedabove, sampling and averaging techniques will be used to convert above equation into their timeequivalents. These are described below.

3.5.1 Sampling Method

Each current phasor in above equation can be expressed as,

Ial = Re(Ial) + jIm(Ial)

= Ial,R + jIal,X (3.100)

An instantaneous phase current is written as follows.

ial(t) =√

2 Im (Ial ejωt)

=√

2 Im[(Ial,R + jIal,X) ejωt

]=√

2 Im [(Ial,R + jIal,X)(cosωt+ j sinωt)]

=√

2 Im [(Ial,R cosωt− Ial,X sinωt) + j(Ial,R sinωt+ Ial,X cosωt)]

=√

2 [(Ial,R sinωt+ Ial,X cosωt)] (3.101)

102

Im (Ial) = Ial,X =ial(t)√

2at sinωt = 0, cosωt = 1 (3.102)

From equation (3.63), the phase voltages can be expressed as below.

va(t) =√

2V sinωt

vb(t) =√

2V sin(ωt− 120o) (3.103)vc(t) =

√2V sin(ωt+ 120o)

From above voltage expressions, it is to be noted that, sinωt = 0, cosωt = 1 implies that thephase-a voltage, va(t) is going through a positive zero crossing, hence, va(t) = 0 and d

dtva(t) = 0.

Therefore, equation (3.102), can be expressed as following.

Ia,l =ial(t)√

2when, va(t) = 0, dva/dt > 0 (3.104)

Similarly,

Ib,l = Ibl,R + jIbl,X (3.105)

Therefore,

α (Ib,l) = α(Ibl,R + jIbl,X)

=

(−1

2+ j

√3

2

)(Ibl,R + jIbl,X)

=

(−1

2Ibl,R −

√3

2Ibl,X

)+ j

(√3

2Ibl,R −

1

2Ibl,X

)(3.106)

From the above,

Im α (Ibl) =

√3

2Ibl,R −

1

2Ibl,X (3.107)

103

Similar to equation, (3.101), we can express phase-b current in terms Im (α Ibl), as given below.

ibl(t) =√

2 Im (Ibl ejωt)

=√

2 Im (α Ibl ejwtα−1)

=√

2 Im (αIbl ej(wt−1200))

=√

2 Im

[(−1

2+ j

√3

2

)(Ibl,R + jIbl,X) ej(wt−1200)

]

=√

2 Im [

(−1

2Ibl,R −

√3

2Ibl,X

)+ j

(√3

2Ibl,R −

1

2Ibl,X

)

(cos(wt− 1200) + j sin(wt− 1200))

] (3.108)

=√

2

[(−1

2Ibl,R −

√3

2Ibl,X) sin(wt− 1200) + (

√3

2Ibl,R −

1

2Ibl,X) cos(wt− 1200)

]From the above equation, we get the following.

Im (αIbl) =ibl(t)√

2when, vb(t) = 0, dvb/dt > 0 (3.109)

Similarly for phase-c, it can proved that,

Im (α2Icl) =icl(t)√

2when, vc(t) = 0, dvc/dt > 0 (3.110)

Substituting Im (Ial) , Im (αIbl) and Im (α2Icl) from (3.104), (3.109) and (3.110) respectively, in(3.99), we get the following.

Babr = − 1

3√

2V

[ia|(va=0, dva

dt>0) + ib|(vb=0,

dvbdt>0)− ic|(vc=0, dvc

dt>0)

]Bbcr = − 1

3√

2V

[−ia|(va=0, dva

dt>0) + ib|(vb=0,

dvbdt>0)

+ ic|(vc=0, dvcdt>0)

](3.111)

Bcar = − 1

3√

2V

[ia|(va=0, dva

dt>0) − ib|(vb=0,

dvbdt>0)

+ ic|(vc=0, dvcdt>0)

]Thus the desired compensating susceptances are expressed in terms of the three line currents sam-pled at instants defined by positive-going zero crossings of the line-neutral voltages va, vb, vc. Anartificial neutral at ground potential may be created measuring voltages va, vb and vc to implementabove algorithm.

3.5.2 Averaging Method

In this method, we express the compensator susceptances in terms of real and reactive power termsand finally expressed them in time domain through averaging process. The method is described

104

below.From equation (3.99), susceptance, Bab

γ , can be re-written as following.

Babγ = − 1

3V

[Im(Ial)

+ Im(αIbl

)− Im

(α2Icl

)]= − 1

3V 2

[V Im

(Ial)

+ V Im(αIbl

)− V Im

(α2Icl

)]= − 1

3V 2

[Im(V Ial

)+ Im

(V αIbl

)− Im

(V α2Icl

)](3.112)

Note the following property of phasors and applying it for the simplification of the above expres-sion.

Im(V I)

= −Im(V∗I∗)

(3.113)

Using above equation (3.112) can be written as,

Babγ =

1

3V 2

[Im (V Ial)

∗ + Im (αV a Ibl)∗ − Im (α2V Icl)

∗]=

1

3V 2

[Im (V

∗a I∗al) + Im (α∗V

∗a I∗bl)− Im ((α2)∗V

∗a I∗cl)]

=1

3V 2

[Im (V

∗a I∗al) + Im (α2V

∗a I∗bl)− Im (αV

∗a I∗cl)]

Since V a = V ∠0o is a reference phasor, therefore V a = V∗a = V , α2 V a = V

∗b and αV a = V

∗c .

Using this, the above equation can be written as following.

Babγ =

1

3V 2

[Im (V a I

∗al) + Im (V b I

∗bl)− Im (V c I

∗cl)]

Similarly, Bbcγ =

1

3V 2

[−Im (V a I

∗al) + Im (V b I

∗bl) + Im (V c I

∗cl)]

(3.114)

Bcaγ =

1

3V 2

[Im (V a I

∗al)− Im (V b I

∗bl) + Im (V c I

∗cl)]

It can be further proved that,

Im (V a I∗a) =

1

T

∫ T

0

va(t)∠(−π/2) ial(t) dt

Im (V b I∗b) =

1

T

∫ T

0

vb(t)∠(−π/2) ibl(t) dt (3.115)

Im (V c I∗c) =

1

T

∫ T

0

vc(t)∠(−π/2) icl(t) dt

105

In (3.115), the term va(t)∠(−π/2) denotes the voltage va(t) shifted by −π/2 radian in time do-main. For balanced voltages, the following relationship between phase and line voltages are true.

va(t)∠(−π/2) = vbc(t)/√

3

vb(t)∠(−π/2) = vca(t)/√

3 (3.116)

vc(t)∠(−π/2) = vca(t)/√

3

From (3.115) and (3.116), the following can be written.

Im (V a I∗a) =

1√3T

∫ T

0

vbc(t) ial(t) dt

Im (V b I∗b) =

1√3T

∫ T

0

vca(t) ibl(t) dt (3.117)

Im (V c I∗c) =

1√3T

∫ T

0

vab(t) icl(t) dt

Substituting above values of Im (V a I∗a), Im (V b I

∗b) and Im (V c I

∗c) into (3.114), we get the fol-

lowing.

Babγ =

1

(3√

3V 2)

1

T

∫ T

0

(vbc ial + vca ibl − vab icl) dt

Bbcγ =

1

(3√

3V 2)

1

T

∫ T

0

(−vbc ial + vca ibl + vab icl) dt (3.118)

Bcaγ =

1

(3√

3V 2)

1

T

∫ T

0

(vbc ial − vca ibl + vab icl) dt

The above equations can directly be used to know the the compensator susceptances by performingthe averaging on line the product of the line to line voltages and phase load currents. The term∫ T

0=∫ t1+T

t1can be implemented using moving average of one cycle. This improves transient

response by computing average value at each instant. But in this case the controller responsewhich changes the susceptance value, should match to that of the above computing algorithm.

3.6 Compensator Admittance Represented as Positive and Negative SequenceAdmittance Network

Recalling the following relations from equation (3.93) for unity power factor operation i.e. β = 0,we get the following.

Babγ =

−1

3√

3V[Im (I1l) + Im (I2l)−

√3 Re (I2l)]

Bbcγ =

−1

3√

3V[Im (I1l)− β Re (I1l)− 2Im (I2l)] (3.119)

Bcaγ =

−1

3√

3V[Im (I1l) + Im (I2l) +

√3 Re (I2l)]

106

From these equations, it is evident the the first terms form the positive sequence suceptance as theyinvolve I1l terms. Similarly, the second and third terms in above equation form negative sequencesusceptance of the compensator, as these involve I2l terms. Thus, we can write,

Babγ = Bab

γ1 +Babγ2

Babγ = Bab

γ1 +Babγ2 (3.120)

Babγ = Bab

γ1 +Babγ2

Therefore,

Babγ1 = Bbc

γ1 = Bcaγ1 = − 1

3√

3V

[Im(I1l)

](3.121)

And,

Babγ2 = − 1

3√

3V

(Im(I2l)−

√3Re(I2l)

)Bbcγ2 = − 1

3√

3V

(−2 Im(I2l)

)(3.122)

Bcaγ2 = − 1

3√

3V

(Im(I2l) +

√3Re(I2l)

)Earlier in equation, (3.123), it was established that,

I0l = 0

I1l =(Y abl + Y bc

l + Y cal

)√3V

I2l = −(α2Y ab

l + Y bcl + αY ca

l

)√3V

Noting that,

Y abl = Gab

l + jBabl

Y bcl = Gbc

l + jBbcl

Y cal = Gca

l + jBcal

Therefore,

Im (I1l) = Im ((Y abl + Y bc

l + Y cal )√

3V )

= (Babl +Bbc

l +Bcal )√

3V (3.123)

Thus equation (3.121) is re-written as following.

Babγ1 = Bbc

γ1 = Bcaγ1 = −1

3

(Babl +Bab

l +Bcal

)(3.124)

107

Now we shall compute Babγ2, Bbc

γ2 and Bcaγ2 using equations (3.122) as following. Knowing that,

I2l = −(α2Y ab

l + Y bcl + αY ca

l

)√3V

= −

[(−1

2− j√

3

2

)(Gabl + jBab

l

)+(Gbcl + jBbc

l

)+

(−1

2+j√

3

2

)(Gca

l + jBcal )

]√

3V

= −

[−G

abl

2+

√3

2Babl +Gbc

l −Gcal

2−√

3

2Bcal − j

(√3

2Gabl +

Babl

2−Bbc

l −√

3

2Gcal +

Bcal

2

)]√

3V

=

[Gabl

2−√

3

2Babl −Gbc

l +Gcal

2+

√3

2Bcal + j

(√3

2Gabl +

Babl

2−Bbc

l −√

3

2Gcal +

Bcal

2

)]√

3V

(3.125)

The above implies that,

Im (I2l) =

(√3

2Gabl +

Babl

2−Bbc

l −√

3

2Gcal +

Bcal

2

)√

3V

−√

3 Re(I2l) =

(−√

3

2Gabl +

3

2Babl +√

3Gbcl −√

3

2Gcal −

3

2Bcal

)√

3V

Thus, Babγ2 can be given as,

Babγ2 = − 1

3√

3V

[2Bab

l −Bbcl −Bca

l +√

3Gbcl −√

3Gcal

]√3V

= −1

3

[2Bab

l −Bbcl −Bca

l +√

3(Gbcl −Gca

l

)]=

1√3

(Gcal −Gbc

l

)+

1

3

(Bbcl +Bca

l − 2Babl

)(3.126)

Similarly,

Bcaγ2 = − 1

3√

3V

[Im(I2l) +

√3 Re(I2l)

]= − 1

3√

3V

[√3

2Gabl +

Babl

2−Bbc

l −√

3

2Gcal +

Bcal

2+

√3

2Gabl −

3

2Babl −√

3Gbcl +

√3

2Gcal +

3

2Bcal

]√

3V

= −1

3

[√3Gab

l −√

3Gbcl −Bab

l −Bbcl + 2Bca

l

]=

1√3

(Gbcl −Gab

l

)+

1

3

(Babl +Bbc

l − 2Bcal

)(3.127)

And, Bbcγ2 is computed as below.

108

Bbcγ2 = − 1

3√

3V

[−2Im(I2l)

]=

2

3√

3V

(√3

2Gabl +

Babl

2−Bbc

l −√

3

2Gcal +

Bcal

2

)V√

3

=1√3

(Gabl −Gca

l

)+

1

3

(Babl +Bca

l − 2Bbcl

)(3.128)

Using (3.121), (3.126)-(3.128), We therefore can find the overall compensator susceptances asfollowing.

Babγ = Bab

γ1 +Babγ2

= −1

3

(Babγ +Bbc

γ +Bcaγ

)+

1√3

(Gcal −Gbc

l

)+

1

3

(Bcal +Bbc

l − 2Babl

)= −Bab

l +1√3

(Gcal −Gbc

l

)Similarly,

Bbcγ = Bbc

γ1 +Bbcγ2

= −1

3

(Babγ +Bbc

γ +Bcaγ

)+

1√3

(Gabl −Gca

l

)+

1

3

(Babl +Bca

l − 2Bbcl

)= −Bbc

l +1√3

(Gabl −Gca

l

)And,

Bcaγ = Bca

γ1 +Bcaγ2

= −1

3

(Babγ +Bbc

γ +Bcaγ

)+

1√3

(Gbcl −Gab

l

)+

1

3

(Bbcl +Bab

l − 2Bcal

)= −Bca

l +1√3

(Gbcl −Gab

l

)Thus, the compensator susceptances in terms of load parameters are given as in the following.

Babγ = −Bab

l +1√3

(Gcal −Gbc

l

)Bbcγ = −Bbc

l +1√3

(Gabl −Gca

l

)(3.129)

Bcaγ = −Bca

l +1√3

(Gbcl −Gab

l

)109

It is interesting to observe above equations. The first parts of the equation nullifies the effect of theload susceptances and the second parts of the equations correspond to the unbalance in resistiveload. The two terms together make source current balanced and in phase with the supply voltages.The compensator’s positive and negative sequence networks are shown in Fig. 3.13 .What happens if we just use the following values of the compensator susceptances as given below?

Babγ = −Bab

l

Bbcγ = −Bbc

l (3.130)Bcaγ = −Bca

l

In above case, load suceptance parts of the admittance are fully compensated. However the sourcecurrents after compensation remain unbalanced due to unbalance conductance parts of the load.

Load

1,abcI

sai

sci

sbi

lai

lci

lbi

2,abcI

1abB

1caB

1bcB

2abB

2caB

2bcB

Sou

rce

, ,ab bc caB B B

Fig. 3.13 Sequence networks of the compensator

Example 3.5 For a delta connected load shown in 3.14 , the load admittances are given as follow-ing,

Y abl = Gab

l + jBabl

Y bcl = Gbc

l + jBbcl

Y cal = Gca

l + jBcal

Given the load parameters:

Zabl = 1/Y ab

l = 5 + j12 Ω

Zbcl = 1/Y bc

l = 3 + j4 Ω

Zcal = 1/Y ca

l = 9− j12 Ω

Determine compensator susceptances (Babγ , B

bcγ , B

caγ ) so that the supply sees the load as balanced

and unity power factor. Also find the line currents and source active and reactive powers beforeand after compensation.

110

caY

abY

a

bc

bcY

abB

bcB

caB

N

av

cv

bv

ci

ai

bi

Fig. 3.14 A delta connected load

Solution:

Zabl = 5 + j12 Ω ⇒ Y ab

l = 0.03− j0.0710 fZbcl = 3 + j4 Ω ⇒ Y bc

l = 0.12− j0.16 fZcal = 9− j13 Ω ⇒ Y ca

l = 0.04 + j0.0533 f

Once we know the admittances we know,

Gabl = 0.03, Bab

l = −0.0710

Gbcl = 0.12, Bbc

l = −0.16

Gcal = 0.04, Bca

l = 0.0533

Babγ = −Bab

l +1√3

(Gcal −Gbc

l ) = 0.0248 f

Bbcγ = −Bbc

l +1√3

(Gabl −Gca

l ) = 0.1540 f

Bcaγ = −Bca

l +1√3

(Gbcl −Gab

l ) = −0.0011 f

Total admittances are:

Y ab = Y abl + Y ab

γ = 0.03− j0.0462 fY bc = Y bc

l + Y bcγ = 0.12− j0.006 f

Y ca = Y cal + Y ca

γ = 0.04 + j0.0522 f

Knowing these total admittances, we can find line currents using following expressions.Current Before Compensation

111

Ia = Iabl − Ical =[(1− α2)Y ab

l − (α− 1)Y cal

]V = 0.2150V ∠−9.51oA

Ib = Ibcl − Iabl =[(α2 − α)Y bc

l − (1− α2)Y abl

]V = 0.4035V ∠−161.66oA

Ic = Ical − Ibcl =[(α− 1)Y ca

l − (α2 − α)Y bcl

]V = 0.2358V ∠43.54oA

Powers Before Compensation

Sa = V a (Ial)∗ = Pa + jQa = V (0.2121 + j0.0355)

Sb = V b (Ibl)∗ = Pb + jQb = V (0.3014 + j0.2682)

Sc = V c (Icl)∗ = Pc + jQc = V (0.0552 + j0.2293)

Total real power, P = Pa + Pb + Pc = V 0.5688W

Total reactive power, Q = Qa +Qb +Qc = V 0.5330 VArPower factor in phase-a, pfa = cosφa = cos(9.51o) = 0.9863 lagPower factor in phase-b, pfb = cosφb = cos(41.63o) = 0.7471 lagPower factor in phase-c, pfc = cosφc = cos(76.45o) = 0.2334 lag

Thus we observe that the phases draw reactive power from the lines and currents are unbalancedin magnitude and phase angles.

After Compensation

Ia = Iab − Ica =[(1− α2)Y ab − (α− 1)Y ca

]V = 0.1896V ∠0oA

Ib = Ibc − Iab =[(α2 − α)Y bc − (1− α2)Y ab

]V = 0.1896V ∠−120oA

Ia = Ica − Ibc =[(α− 1)Y ca

l − (α2 − α)Y bc]V = 0.1896V ∠120oA

Powers After Compensation

Sa = V a (Ial)∗ = Pa + jQa = V (0.1986 + j0.0)

Sb = V b (Ibl)∗ = Pb + jQb = V (0.1986 + j0.0)

Sc = V c (Icl)∗ = Pc + jQc = V (0.1986 + j0.0)

Total real power, P = Pa + Pb + Pc = (V × 0.5688 )W

Total reactive power, Q = Qa +Qb +Qc = 0 VArPower factor in phase-a, pfa = cosφa = cos(0o) = 1.0

Power factor in phase-b, pfb = cosφb = cos(0o) = 1.0

Power factor in phase-c, pfc = cosφc = cos(0o) = 1.0

From above results we observe that after placing compensator of suitable values as calculatedabove, the line currents become balanced and have unity power factor relationship with their volt-ages.

112

Example 3.6 Consider the following 3-phase, 3-wire system. The 3-phase voltages are balancedsinusoids with RMS value of 230 V at 50 Hz. The load impedances are Za = 3 + j 4 Ω, Zb =5 + j 12 Ω, Zc = 12− j 5 Ω. Compute the following.

1. The line currents I la, I lb, I lc.

2. The active (P ) and reactive (Q) powers of each phase.

3. The compensator susceptance ( Babγ , B

bcγ , B

caγ ), so that the supply sees the load balanced and

unity power factor.

4. For case (3), compute the source, load, compensator active and reactive powers (after com-pensation).

Na

c bZ

laI

lbI

lcI

saV

scV

sbV

Fig. 3.15 An unbalanced three-phase three-wire star connected load

Solution:

Given that Za = 3 + j 4 Ω, Zb = 5 + j 12 Ω, Zc = 12− j 5 Ω.

1. Line currents I la, I lb, andI lc are found by first computing neutral voltage as given below.

V nN =1

( 1Za

+ 1Zb

+ 1Zc

)(V a

Za+V b

Zb+V c

Zc)

= (ZaZbZc

ZaZb + ZbZc + ZcZa)(V a

Za+V b

Zb+V c

Zc)

=Zabc∆Z

(V a

Za+V b

Zb+V c

Zc)

=50∠97.82o

252.41∠55.5o(24.12∠−99.55o)

= 43.79− j 67.04V

= 80.75∠−57.15o V

Now the line currents are computed as below.

113

I la =V a − V nN

Za

=230∠0o − 80.75∠−57.15o

3 + j 4= 33.2− j 21.65

= 39.63∠−33.11oA

I lb =V b − V nN

Zb

=230∠−120o − 80.75∠−57.15o

5 + j 12= 15.85∠152.21oA

I lc =V c − V nN

Zc

=230∠120o − 80.75∠−57.15o

12− j 5= 23.89∠143.35oA

2. Active and Reactive Powers

For phase a

Pa = VaIa cosφa = 230× 39.63× cos(33.11o) = 7635.9W

Qa = VaIa sinφa = 230× 39.63× sin(33.11o) = 4980 VAr

For phase b

Pb = VbIb cosφb = 230× 15.85× cos(−152.21o − 120o) = 140.92W

Qb = VbIb sinφb = 230× 15.85× sin(−152.21o − 120o) = 3643.2 VAr

For phase c

Pc = VcIc cosφc = 230× 23.89× cos(−143.35o + 120o) = 5046.7W

114

Qc = VcIc sinφc = 230× 23.89× sin(−143.35o + 120o) = −2179.3 VAr

Total three phase powers

P = Pa + Pb + Pc = 12823W

Q = Qa +Qb +Qc = 6443.8 VAr

3. Compensator Susceptance

First we convert star connected load to a delta load as given below.

Zabl = ZaZb+ZbZc+ZcZa

Zc= 4 + j 19 = 19.42∠77.11o Ω

Zbcl = ∆Z

Za= 50.44 + j 2.08 = 50.42∠2.36o Ω

Zcal = ∆Z

Zb= 19.0− j 4.0 = 19.42∠− 11.89o Ω

The above implies that,

Y abl = 1/Zab

l = Gabl + j Bab

l = 0.0106− j0.050f

Y bcl = 1/Zbc

l = Gbcl + j Bbc

l = 0.0198− j0.0008f

Y cal = 1/Zca

l = Gcal + j Bca

l = 0.0504 + j0.0106f

From the above, the compensator susceptances are computed as following.

Babγ = −Bab

l +(Gcal −G

bcl )√

3= 0.0681f

Bbcγ = −Bbc

l +(Gabl −G

cal )√

3= −0.0222f

Bcaγ = −Bca

l +(Gbcl −G

abl )√

3= −0.0053f

Zabγ = −j 14.69 Ω (capacitance)

Zbcγ = j 45.13 Ω (inductance)

Zcaγ = j 188.36 Ω (inductance)

115

4. After Compensation

Zab′

l = Zabl ||Zab

γ = 24.97− j 41.59 = 48.52∠− 59.01o Ω

Zbc′

l = Zbcl ||Zbc

γ = 21.52− j 24.98 = 32.97∠49.25o Ω

Zca′

l = Zcal ||Zca

γ = 24.97− j 41.59 = 19.7332∠− 6.0o Ω

Let us convert delta connected impedances to star connected.

Za =Zab′ × Zca′

Zab′ + Zbc′ + Zca′= 9.0947− j 10.55

= 13.93∠− 49.25o Ω

Zb =Zbc′ × Zab′

Zab′ + Zbc′ + Zca′= 23.15 + j 2.43

= 23.28∠6.06o Ω

Zc =Zca′ × Zbc′

Zab′ + Zbc′ + Zca′= 4.8755 + j 8.12

= 9.47∠59.01o Ω

The new voltage between the load and system neutral after compensation is given by,

V′

nN =1

( 1Za

+ 1Zb

+ 1Zc

)(V a

Za+V b

Zb+V c

Zc)

= 205.42∠−57.15o V

Based on the above, the line currents are computed as following.

Ia =V a − V

′

nN

Za= 18.58∠0o A

Ib =V b − V

′

nN

Zb= 18.58∠− 120o A

Ic =V c − V

′

nN

Zc= 18.58∠120o A

116

Thus, it is seen that after compensation, the source currents are balanced and have unity powerfactor with respective supply voltages.

Source powers after compensation

Pa = Pb = Pc = 230× 18.58 = 4272.14 W

P = 3Pa = 12820.2 W

Qa = Qb = Qc = 0

Q = 0 VAr

Compensator powers

Sab

γ = V ab(Iab

γ

∗)

= V ab(V ab∗.Y abγ

∗)

= V2

abYabγ

∗

= (230×√

3)2 × (−j 0.0068)

= −j 10802 VA

Sbc

γ = V 2bcY

bcγ

∗

= (230×√

3)2 × (j 0.0222)

= j 3516 VASca

γ = V 2caY

caγ∗

= (230×√

3)2 × (j 0.0053)

= j 842 VA

References

[1] A. Ghosh and G. Ledwich, Power quality enhancement using custom power devices. KluwerAcademic Pub, 2002.

[2] T. J. E. Miller, Reactive power control in electric systems. Wiley, 1982.

[3] L. Gyugyi, “Reactive power generation and control by thyristor circuits,” IEEE Transactionson Industry Applications, no. 5, pp. 521–532, 1979.

117

[4] R. Otto, T. Putman, and L. Gyugyi, “Principles and applications of static, thyristor-controlledshunt compensators,” IEEE Transactions on Power Apparatus and Systems, no. 5, pp. 1935–1945, 1978.

[5] N. Hingorani and L. Gyugyi, Understanding FACTS: Concepts and Technology of Flexible ACTransmission Systems, 2000. Wiley-IEEE Press, 1999.

118