Chapter 8

Application of Capacitors to Distribution Systems

Who neglects learning in his youth, loses the past and is dead for the future.

Euripides, 438 BC

Where is there dignity unless there is honesty?

Cicero

8.1 Basic Definitions

Capacitor element: an indivisible part of a capacitor consisting of electrodes separated by a dielectric material

Capacitor unit: an assembly of one or more capacitor elements in a single container with terminals brought out

Capacitor segment: a single-phase group of capacitor units with protection and control system Capacitor module: a three-phase group of capacitor segments

Capacitor bank: a total assembly of capacitor modules electrically connected to each other

8.2 Power Capacitors

At a casual look, a capacitor seems to be a very simple and unsophisticated apparatus, that is, two metal plates separated by a dielectric insulating material. It has no moving parts but instead functions by being acted upon by electric stress.

In reality, however, a power capacitor is a highly technical and complex device in that very thin dielectric materials and high electric stresses are involved, coupled with highly sophisticated processing techniques. Figure 8.1 shows a cutaway view of a power factor correction capacitor. Figure 8.2 shows a typical capacitor utilization in a switched pole-top rack.

In the past, most power capacitors were constructed with two sheets of pure aluminum foil separated by three or more layers of chemically impregnated kraft paper. Power capacitors have been improved tremendously over the last 30 years or so, partly due to improvements in the dielectric materials and their more efficient utilization and partly due to improvements in the processing techniques involved. Capacitor sizes have increased from the 15–25 kvar range to the 200–300 kvar range (capacitor banks are usually supplied in sizes ranging from 300 to 1800 kvar).

Nowadays, power capacitors are much more efficient than those of 30 years ago and are available to the electric utilities at a much lower cost per kilovar. In general, capacitors are getting more attention today than ever before, partly due to a new dimension added in the analysis: changeout economics. Under certain circumstances, even replacement of older capacitors can be justified on the basis of lower-loss evaluations of the modern capacitor design.

Capacitor technology has evolved to extremely low-loss designs employing the all-film concept; as a result, the utilities can make economic loss evaluations in choosing between the presently existing capacitor technologies.

8.3 Effects of Series and Shunt Capacitors

As mentioned earlier, the fundamental function of capacitors, whether they are series or shunt, installed as a single unit or as a bank, is to regulate the voltage and reactive power flows at the point where they are installed. The shunt capacitor does it by changing the power factor of the load, whereas the series capacitor does it by directly offsetting the inductive reactance of the circuit to which it is applied.

8.3.1 Series Capacitors

Series capacitors, that is, capacitors connected in series with lines, have been used to a very limited extent on distribution circuits due to being a more specialized type of apparatus with a limited range of application. Also, because of the special problems associated with each application, there is a requirement for a large amount of complex engineering investigation. Therefore, in general, utilities are reluctant to install series capacitors, especially of small sizes.

As shown in Figure 8.3, a series capacitor compensates for inductive reactance. In other words, a series capacitor is a negative (capacitive) reactance in series with the circuit’s positive (inductive) reactance with the effect of compensating for part or all of it. Therefore, the primary effect of the series capacitor is to minimize, or even suppress, the voltage drop caused by the inductive reactance in the circuit.

At times, a series capacitor can even be considered as a voltage regulator that provides for a voltage boost that is proportional to the magnitude and power factor of the through current. Therefore, a series capacitor provides for a voltage rise that increases automatically and instantaneously as the load grows.

Also, a series capacitor produces more net voltage rise than a shunt capacitor at lower power factors, which creates more voltage drop. However, a series capacitor betters the system power factor much less than a shunt capacitor and has little effect on the source current.

Consider the feeder circuit and its voltage phasor diagram as shown in Figure 8.3a and c. The voltage drop through the feeder can be expressed approximately as

$\begin{array}{cc}\text{VD}=IR\cos \theta +I{X}_L\sin \theta \hfill & (8.1)\hfill \end{array}$

where

R is the resistance of the feeder circuit

XL is the inductive reactance of the feeder circuit

cos θ is the receiving-end power factor

sin θ is the sine of the receiving-end power factor angle

As can be observed from the phasor diagram, the magnitude of the second term in Equation 8.1 is much larger than the first. The difference gets to be much larger when the power factor is smaller and the ratio of R/XL is small.

However, when a series capacitor is applied, as shown in Figure 8.3b and d, the resultant lower voltage drop can be calculated as

$\begin{array}{cc}\text{VD}=IR\cos \theta +I(X_L-X_C)\sin \theta \hfill & (8.2)\hfill \end{array}$

where Xc is the capacitive reactance of the series capacitor.

8.3.1.1 Overcompensation

Usually, the series-capacitor size is selected for a distribution feeder application in such a way that the resultant capacitive reactance is smaller than the inductive reactance of the feeder circuit. However, in certain applications (where the resistance of the feeder circuit is larger than its inductive reactance), the reverse might be preferred so that the resultant voltage drop is

$\begin{array}{cc}\text{VD}=IR\cos \theta -I(X_C-X_L)\sin \theta \hfill & (8.3)\hfill \end{array}$

The resultant condition is known as overcompensation. Figure 8.4a shows a voltage phasor diagram for overcompensation at normal load. At times, when the selected level of overcompensation is strictly based on normal load, the resultant overcompensation of the receiving-end voltage may not be pleasing at all because the lagging current of a large motor at start can produce an extraordinarily large voltage rise, as shown in Figure 8.4b, which is especially harmful to lights (shortening their lives) and causes light flicker, resulting in consumers’ complaints.

8.3.1.2 Leading Power Factor

To decrease the voltage drop considerably between the sending and receiving ends by the application of a series capacitor, the load current must have a lagging power factor. As an example, Figure 8.5a shows a voltage phasor diagram with a leading-load power factor without having series capacitors in the line. Figure 8.5b shows the resultant voltage phasor diagram with the same leading-load power factor but this time with series capacitors in the line. As can be seen from the figure, the receivingend voltage is reduced as a result of having series capacitors.

When cos θ = 1.0, sin θ ≅ 0, and therefore,

$I(X_L-X_C)\sin \theta \cong 0$

hence, Equation 8.2 becomes

$\begin{array}{cc}VD\cong IR\hfill & (8.4)\hfill \end{array}$

Thus, in such applications, series capacitors practically have no value.

Because of the aforementioned reasons and others (e.g., ferroresonance in transformers, subsynchronous resonance during motor starting, shunting of motors during normal operation, and difficulty in protection of capacitors from system fault current), series capacitors do not have large applications in distribution systems.

However, they are employed in subtransmission systems to modify the load division between parallel lines. For example, often a new subtransmission line with larger thermal capability is parallel with an already existing line. It may be very difficult, if not impossible, to load the subtransmission line without overloading the old line. Here, series capacitors can be employed to offset some of the line reactance with greater thermal capability. They are also employed in subtransmission systems to decrease the voltage regulation.

8.3.2 Shunt Capacitors

Shunt capacitors, that is, capacitors connected in parallel with lines, are used extensively in distribution systems. Shunt capacitors supply the type of reactive power or current to counteract the outof-phase component of current required by an inductive load. In a sense, shunt capacitors modify the characteristic of an inductive load by drawing a leading current that counteracts some or all of the lagging component of the inductive load current at the point of installation. Therefore, a shunt capacitor has the same effect as an overexcited synchronous condenser, generator, or motor.

As shown in Figure 8.6, by the application of shunt capacitor to a feeder, the magnitude of the source current can be reduced, the power factor can be improved, and consequently the voltage drop between the sending end and the load is also reduced. However, shunt capacitors do not affect current or power factor beyond their point of application. Figure 8.6a and c shows the single-line diagram of a line and its voltage phasor diagram before the addition of the shunt capacitor, and Figure 8.6b and d shows them after the addition.

Voltage drop in feeders, or in short transmission lines, with lagging power factor can be approximated as

$\begin{array}{cc}\text{VD}=I_{R}R+I_X{X}_L\hfill & (8.5)\hfill \end{array}$

where

R is the total resistance of the feeder circuit, Ω

XL is the total inductive reactance of the feeder circuit, Ω

IR is the real power (or in-phase) component of the current, A

IX is the reactive (or out-of-phase) component of the current lagging the voltage by 90°, A

Example 8.1

Consider the right-angle triangle shown in Figure 8.7b. Determine the power factor of the load on a 460 V three-phase system, if the ammeter reads 100 A and the wattmeter reads 70 kW.

Solution

$\begin{array}{lll}S\hfill & =\hfill & \frac{\sqrt{3}(V)(I)}{1000}\hfill \\ \hfill & =\hfill & \frac{\sqrt{3}\text{(460}\text{V)(100}\text{A)}}{1000}\hfill \\ \hfill & \cong \hfill & 79.67\text{kVA}\hfill \end{array}$

Thus,

$\begin{array}{lll}\text{PF}\hfill & =\hfill & \cos \theta =\frac{P}{S}\hfill \\ \hfill & =\hfill & \frac{70\text{kW}}{79.67\text{kVA}}\hfill \\ \hfill & \cong \hfill & 0.88\text{or}\text{88\%}\hfill \end{array}$

When a capacitor is installed at the receiving end of the line, as shown in Figure 8.6b, the resultant voltage drop can be calculated approximately as

$\begin{array}{r}\hfill \\ \begin{array}{cc}\text{VD}=I_R{R}_R+I_X{X}_L-I_C{X}_L\hfill & (8.6)\hfill \end{array}\hfill \end{array}$

where Ic is the reactive (or out-of-phase) component of current leading the voltage by 90°, A.

The difference between the voltage drops calculated by using Equations 8.5 and 8.6 is the voltage rise due to the installation of the capacitor and can be expressed

$\begin{array}{cc}\text{VR}=I_C{X}_L\hfill & (8.7)\hfill \end{array}$

8.4 Power Factor Correction

8.4.1 General

A typical utility system would have a reactive load at 80% power factor during the summer months. Therefore, in typical distribution loads, the current lags the voltage, as shown in Figure 8.7a. The cosine of the angle between current and sending voltage is known as the power factor of the circuit. If the in-phase and out-of-phase components of the current I are multiplied by the receiving-end voltage VR, the resultant relationship can be shown on a triangle known as the power triangle, as shown in Figure 8.7b. Figure 8.7b shows the triangular relationship that exists between kilowatts, kilovoltamperes, and kilovars.

Note that, by adding the capacitors, the reactive power component Q of the apparent power S of the load can be reduced or totally suppressed. Figures 8.8a and 8.9 illustrate how the reactive power component Q increases with each 10% change of power factor. Figure 8.8a also illustrates how a portion of lagging reactive power Qold is cancelled by the leading reactive power of capacitor Qc.

Note that, as illustrated in Figure 8.8, even an 80% power factor of the reactive power (kilovar) size is quite large, causing a 25% increase in the total apparent power (kilovoltamperes) of the line. At this power factor, 75 kvar of capacitors is needed to cancel out the 75 kvar of the lagging component.

As previously mentioned, the generation of reactive power at a power plant and its supply to a load located at a far distance is not economically feasible, but it can easily be provided by capacitors (or overexcited synchronous motors) located at the load centers. Figure 8.10 illustrates the power factor correction for a given system. As illustrated in the figure, capacitors draw leading reactive power from the source; that is, they supply lagging reactive power to the load. Assume that a load is supplied with a real power P, lagging reactive power Q1, and apparent power S1 at a lagging power factor of

$\cos {\theta }_1=\frac{P}{S_1}$

or

$\begin{array}{cc}\cos {\theta }_1=\frac{P}{{(P^2+Q_1^2)}^{1/2}}\hfill & (8.8)\hfill \end{array}$

When a shunt capacitor of Qc kVA is installed at the load, the power factor can be improved from cos θ1 to cos θ2, where

$\begin{array}{lll}\cos {\theta }_2\hfill & =\hfill & \frac{P}{S_2}\hfill \\ \hfill & =\hfill & \frac{P}{{(P^2+Q_2^2)}^{1/2}}\hfill \end{array}$

or

$\begin{array}{cc}\cos {\theta }_2=\frac{P}{{[P^2+{(Q_1-Q_C)}^2]}^{1/2}}\hfill & (8.9)\hfill \end{array}$

8.4.2 Concept of Leading and Lagging Power Factors

Many consider that the terms “lagging” and “leading” power factor are somewhat confusing, and they are meaningless, if the directions of the flows of real and reactive powers are not known. In general, for a given load, the power factor is lagging if the load withdraws reactive power; on the other hand, it is leading if the load supplies reactive power.

Hence, an induction motor has a lagging power factor since it withdraws reactive power from the source to meet its magnetizing requirements. But a capacitor (or an overexcited synchronous motor) supplies reactive power and thus has a leading power factor, as shown in Figure 8.11 and indicated in Table 8.1.

On the other hand, an underexcited synchronous motor withdraws both the real and reactive power from the source, as indicated. The use of varmeters instead of power factor meters avoids the confusion about the terms “lagging” and “leading.” Such a varmeter has a zero center point with scales on either side, one of them labeled “in” and the other one “out.”

8.4.3 Economic Power Factor

As can be observed from Figure 8.10b, the apparent power and the reactive power are decreased from S1 to S2 kVA and from Q1 to Q2 kvar (by providing a reactive power of Q), respectively. The reduction of reactive current results in a reduced total current, which in turn causes less power losses.

Thus, the power factor correction produces economic savings in capital expenditures and fuel expenses through a release of kilovoltamperage capacity and reduction of power losses in all the apparatus between the point of installation of the capacitors and the power plant source, including distribution lines, substation transformers, and transmission lines.

The economic power factor is the point at which the economic benefits of adding shunt capacitors just equal the cost of the capacitors. In the past, this economic power factor was around 95%. Today’s high plant and fuel costs have pushed the economic power factor toward unity.

However, as the corrected power factor moves nearer to unity, the effectiveness of capacitors in improving the power factor, decreasing the line kilovoltamperes transmitted, increasing the load capacity, or reducing line copper losses by decreasing the line current sharply decreases. Therefore, the correction of power factor to unity becomes more expensive with regard to the marginal cost of capacitors installed.

8.4.4 Use of a Power Factor Correction Table

Table 8.2 is a power factor correction table to simplify the calculations involved in determining the capacitor size necessary to improve the power factor of a given load from original to desired value. It gives a multiplier to determine the kvar requirement. It is based on the following formula:

$\begin{array}{cc}Q=P(\tan {\theta }_{\text{orig}}-\tan {\theta }_{\text{new}})\hfill & (8.10)\hfill \end{array}$

or

$\begin{array}{cc}Q=P\left(\sqrt{\frac{1}{{\text{PF}}_{\text{orig}}^2}-1}-\sqrt{\frac{1}{{\text{PF}}_{\text{new}}^2}-1}\right)\hfill & (8.11)\hfill \end{array}$

where

Q is the required compensation in kvar

P is the real power kW

PForig is the original power factor

PFnew is the desired power factor

8.4.5 Alternating Cycles of a Magnetic Field

Furthermore, in order to understand how the power factor of a device can be improved, one has to understand what is taking place electrically. Consider an induction motor that is being supplied by the real power P and the reactive power Q. The real power P is lost, whereas the reactive power Q is not lost. But instead it is used to store energy in the magnetic field of the motor.

Since the current is alternating, the magnetic field undergoes cycles of building up and breaking down. As the field is building up, the reactive current flows from the supply or source to the motor. As the field is breaking down, the reactive current flows out of the motor back to the supply or source. In such application, what is needed is some type of device that can be used as a temporary storage area for the reactive power when the magnetic field of the motor breaks down.

The ideal device for this is a capacitor that also stores energy. However, this energy is stored in an electric field. By connecting a capacitor in parallel with the supply line of the load, the cyclic flow of reactive power takes place between the motor and the capacitor. Here, the supply lines carry only the current supplying real power to the motor. This is only applicable for a unity power factor condition. For other power factors, the supply lines would carry some reactive power.

8.4.6 Power Factor of a Group of Loads

In general, the power factor of a single load is known. However, it is often that the power factor of a group of various loads needs to be determined. This is accomplished based on the known power relationship.

Example 8.2

Assume that a substation supplies three different kinds of loads, mainly, incandescent lights, induction motors, and synchronous motors, as shown in Figure 8.12. The substation power factor is found from the total reactive and real powers of the various loads that are connected. Based on the given data in Figure 8.12, determine the following:

1. The apparent, real, and kvars of each load
2. The total apparent, real, and reactive powers of the power that should be supplied by the substation
3. The total power factor of the substation

Solution

1. 1. For a 100 kVA lighting load

      Since incandescent lights are basically a unity power factor load, it is assumed that all the current is kilowatt current. Hence,

      $\begin{array}{ccc}\hfill S_1& =\hfill & P_1\hfill \\ 100\text{kVA}\hfill & \cong \hfill & 100\text{kW}\hfill \end{array}$

   2. For 400 hp of connected induction motor loads

      Assume that for the motor loads,

      $\text{kVA}\text{load}=0\text{.75×(Connected}\text{motor}\text{horse}\text{power)}$

      with an opening power factor of 85% lagging:

      $S_2=\left(0.75\frac{\text{kW}}{\text{hp}}\right)(400\text{hp})=300\text{kVA}$

      $P_3=(0.75\times 400)\times 0.85=255\text{kW}$

      $\begin{array}{lll}{Q}_2\hfill & =\hfill & \sqrt{{(300)}^2-{(255)}^2}\hfill \\ \hfill & =\hfill & \sqrt{90,000-65,025}\hfill \\ \hfill & =\hfill & 24,975\hfill \\ \hfill & \cong \hfill & 158\text{k}\mathrm{var}\hfill \end{array}$

   3. 200 hp motor with a 0.8 leading power factor

      At full load, assume kVA = motor-hp rating = 200 kVA:

      $\begin{array}{lll}{P}_3\hfill & =\hfill & (200\text{kVA})\cos \theta \hfill \\ \hfill & =\hfill & 200\times 0.8\hfill \\ \hfill & =\hfill & 160\text{kW}\hfill \end{array}$

      $\begin{array}{lll}{Q}_2\hfill & =\hfill & \sqrt{{(200\text{kVA})}^2-{(160\text{kW})}^2}\hfill \\ \hfill & =\hfill & \sqrt{40,000-25,600}\hfill \\ \hfill & =\hfill & \sqrt{14,400}\hfill \\ \hfill & =\hfill & 120\text{kavr}\hfill \end{array}$

2. At the substation, the total real power is

   $\begin{array}{lll}{P}_{\text{total}}\hfill & =\hfill & P_{\text{lights}}+P_{\text{ind}\text{.mot}\text{.}}+P_{\text{sync}\text{.mot}\text{.}}\hfill \\ \hfill & =\hfill & 100+255+160\hfill \\ \hfill & =\hfill & 515\text{kW}\hfill \end{array}$

   The total reactive power is

   $\begin{array}{lll}{Q}_{\text{total}}\hfill & =\hfill & Q_{\text{lights}}+Q_{\text{ind}\text{.mot}\text{.}}\hfill \\ \hfill & =\hfill & 0+158\hfill \\ \hfill & =\hfill & 158\text{kvar}\hfill \end{array}$

   Thus, an overexcited synchronous motor operating without the mechanical load connected to its shaft can supply the leading reactive power. Hence, the net lagging reactive power that must be supplied by the substation is the difference between the reactive power supplied by the synchronous motor and the reactive power required by the induction motor loads:

   Induction motor load required = 158

   kvar Synchronous motor supplied = 120

   kvar Substation must supply = 38 kvar

3. The kVA of the substation is

   $\begin{array}{cc}{S}_{\text{total}}=\sqrt{P_{\text{tot}}^2+Q_{\text{tot}}^2}\hfill & (8.12)\hfill \end{array}$

   or

   $\begin{array}{lll}{S}_{\text{total}}\hfill & =\hfill & \sqrt{{515}^2+{38}^2}\hfill \\ \hfill & =\hfill & \sqrt{266,669}\hfill \\ \hfill & =\hfill & 516.4\text{kVA}\hfill \end{array}$

   The power factor of the substation is

   $\begin{array}{lll}\text{PF = power}\text{factor}\hfill & =\hfill & \frac{P}{S}\hfill \\ \hfill & =\hfill & \frac{515\text{kW}}{516.4\text{kVA}}\hfill \\ \hfill & =\hfill & 0.997\text{lagging}\hfill \end{array}$

8.4.7 Practical Methods Used by the Power Industry for Power Factor Improvement Calculations

It is often that the formulas that are used by the power industry contains kW, kVA, or kvar instead of the symbols of P, S, Q, which are the correct form and used in the academia. However, there are certain advantages of using them since one does not have to think which one is P, S, or Q.

From the right-triangle relationship, several simple and useful mathematical expressions may be written as

$\begin{array}{cc}\text{PF}=\cos \theta =\frac{\text{kW}}{\text{kVA}}\hfill & (8.13)\hfill \end{array}$

$\begin{array}{cc}\tan \theta =\frac{\text{kvar}}{\text{kW}}\hfill & (8.14)\hfill \end{array}$

$\begin{array}{cc}\sin \theta =\frac{\text{kvar}}{\text{kVA}}\hfill & (8.15)\hfill \end{array}$

Since the kW component normally stays the same (the kVA and kvar components change with power factor), it is convenient to use Equation 8.11 involving the kW component. The relationship can be reexpressed as

$\begin{array}{cc}\text{kvar = kW}\times \tan \theta \hfill & (8.16)\hfill \end{array}$

For instance, if it is necessary to determine the capacitor rating to improve the load’s power factor, one would use the following relationships:

$\begin{array}{cc}\text{kvar at}\text{original}\text{PF}=\text{kW}\text{×}\text{tan}{\theta }_1\hfill & (8.17)\hfill \end{array}$

$\begin{array}{cc}\text{kvar at}\text{improved}\text{PF}=\text{kW}\text{×}\text{tan}{\theta }_2\hfill & (8.18)\hfill \end{array}$

Thus, the capacitor rating required to improve the power factor can be expressed as

$\begin{array}{cc}\text{ckvar*}=\text{kW}\text{×}(\text{tan}{\theta }_1-\tan {\theta }_2)\hfill & (8.19)\hfill \end{array}$

or

$\Delta \begin{array}{cc}\tan \theta =\tan {\theta }_1-\tan {\theta }_2\hfill & (8.20)\hfill \end{array}$

then

$\begin{array}{cc}\text{ckvar*}=\text{kW}\times \Delta \text{tan}\theta \hfill & (8.21)\hfill \end{array}$

Table 8.2 has a “kW multiplier” for determining the capacitor based on the previously mentioned expression. Also, note that the prefix “c” in ckvar is employed to designate the capacitor kvar in order to differentiate it from load kvar.

To find irrespective currents of kVA, kW, and kvar, use the following relationships:

$\begin{array}{cc}\text{kVA}=\sqrt{{(\text{kW})}^2+{(\text{kvar})}^2}\hfill & (8.22)\hfill \end{array}$

$\begin{array}{cc}\text{kW}=\sqrt{{(\text{kVA})}^2+{(\text{kvar})}^2}\hfill & (8.23)\hfill \end{array}$

$\begin{array}{cc}\text{kvar}=\sqrt{{(\text{kVA})}^2+{(\text{kW})}^2}\hfill & (8.24)\hfill \end{array}$

Example 8.3

Assume that a load withdraws 80 kW and 60 kvar at a 0.8 power factor. It is required that its power factor is to be improved from 80% to 90% by using capacitors. Determine the amount of the reactive power to be provided by using capacitors.

Solution

Without capacitors at PF = 0.8

$\begin{array}{lll}\text{kW}\hfill & =\hfill & 80\hfill \\ \text{kvar}\hfill & =\hfill & 60\hfill \end{array}$

Thus, the kVA requirement of the load is

${\text{kVA = (80}}^2{\text{ + 60}}^2{\text{)}}^{\text{1/2}}\text{ = 100 kVA}$

With capacitors at PF = 0.9

$\begin{array}{lll}\text{kW}\hfill & =\hfill & 80\hfill \\ \text{kVA}\hfill & \cong \hfill & 88.9\hfill \end{array}$

$\text{Line kvar = (88}{.9}^2{\text{ – 80}}^2{\text{)}}^{\text{1/2}}{\text{ (7903–s6400)}}^{\text{1/2}}\text{ = 38}.7$

Hence, the line supplies 38.7 kvar and the load needs 60 kvar, and the capacitor supplies the difference, or

$\text{ckvar}=60-38.7=21.3\text{kvar}$

as it is illustrated in Figure 8.13.

Example 8.4

Determine the capacitor rating in Example 8.3 by using Table 8.2.

Solution

From Table 8.2, the "kW multiplier" or Δ tan θ is read as 0.266. Therefore,

$\begin{array}{lll}\text{ckvar}\hfill & =\hfill & kW\times \Delta \tan \theta \hfill \\ \hfill & =\hfill & (80\text{kW})\times (0.266)\hfill \\ \hfill & =\hfill & 21.3\text{kvar}\hfill \end{array}$

which is the same value determined by the calculation in Example 8.3.

Example 8.5

Assume that a certain load withdraws a kilowatt current of 2 A and kilovar current of 2 A. Determine the amount of total current that it withdraws.

Solution

The answer is not the following!

$\text{(2}\text{A)}+\text{(2}\text{A)}=4\text{A}$

The correct answer can be found from the following right-triangle relationships:

${\text{(kvar current)}}^2{\text{ + (kW current)}}^2{\text{ = (total current)}}^2$

${\text{(2 A)}}^2{\text{ + (2 A)}}^2{\text{ = (total current)}}^2$

or

${\text{4 + 4 = (total current)}}^2$

Hence,

${\text{Total current = 8}}^{\text{1/2}}\text{ = 2}\text{.83 A}$

Thus,

$\text{2 + 2 }\ne 4!$

Figure 8.14 shows the component current diagram.

Example 8.6

Assume that a 460 V cable circuit is rated at 240 A but is carrying a load of 320 A at 0.65 power factor. Determine the kvar of capacitor that is needed to reduce the current to 240 A.

Solution

$\begin{array}{lll}\text{kVA}\hfill & =\hfill & \sqrt{3}\text{×}\text{(460}\text{V)}\text{×}\text{(320}\text{A)}\hfill \\ \hfill & =\hfill & 254.96\text{kVA}\hfill \\ \text{kW}\hfill & =\hfill & \text{(254}.96\text{kVA)0}.65\hfill \\ \hfill & =\hfill & 165.72\text{kW}\hfill \end{array}$

The kVA corresponding to 240 A is

$\begin{array}{lll}\text{kVA}\hfill & =\hfill & \frac{\sqrt{3}\times (460\text{V})(240\text{A})}{1000}\hfill \\ \hfill & =\hfill & 191.2\text{kVA}\hfill \end{array}$

Thus, the operating power factor corresponding to the new load is

$\begin{array}{lll}{\text{PF}}_2=\cos {\theta }_2\hfill & =\hfill & \frac{P}{S_2}\hfill \\ \hfill & =\hfill & \frac{165.72\text{kW}}{191.2\text{kVA}}\hfill \\ \hfill & =\hfill & 0.8667\hfill \end{array}$

The capacitor kvar required is

$\begin{array}{lll}\text{ckvar}\hfill & =\hfill & (165.72\text{kW})\tan ({\cos }^{-1}0.8667)\hfill \\ \hfill & =\hfill & (165.72\text{kW})\tan (29.92°)\hfill \\ \hfill & \cong \hfill & (165.72)(0.5755)\hfill \\ \hfill & \cong \hfill & 95.38\hfill \end{array}$

8.4.8 Real Power-Limited Equipment

Certain equipments such as turbogenerator (i.e., turbine generators) and engine generator sets have a real power (P) limit of the prime mover as well as a kVA limit of the generator. Usually the real power limit corresponds to the generator S rating, and the set is rated at that P value at unity power factor operation.

Other real power (P) values that correspond to the lesser power factor operations are determined by the power factor and real power (S) rating at the generator in order that the P and S ratings of the load do not exceed the S rating of the generator. Any improvement of the power factor can release both P and S capacities.

Example 8.7

Assume that a 1000 kW turbine unit (turbogenerator set) has a turbine capability of 1250 kW. It is operating at a rated load of 1250 kVA at 0.85 power factor. An additional load of 150 kW at 0.85 power factor is to be added. Determine the value of capacitors needed in order not to overload the turbine nor the generator.

Solution

Original load

$\begin{array}{lll}P\hfill & =\hfill & 1000\text{kW}\hfill \\ Q\hfill & =\hfill & \sqrt{{(\text{kVA})}^2-{(\text{kW})}^2}\hfill \\ \hfill & =\hfill & \sqrt{{(1250)}^2-{(1000)}^2}\hfill \\ \hfill & =\hfill & 750\text{kvar}\hfill \end{array}$

Additional load

$P=\text{kW}=150\text{kW}$

$\begin{array}{lll}S\hfill & =\hfill & \text{kVA}\hfill \\ \hfill & =\hfill & \frac{150\text{kW}}{0.85}\hfill \\ \hfill & =\hfill & 200\text{kVA}\hfill \end{array}$

$\begin{array}{lll}Q\hfill & =\hfill & \sqrt{{(200)}^2-{(1000)}^2}\hfill \\ \hfill & =\hfill & 132.29\text{kvar}\hfill \end{array}$

Total load

$\begin{array}{lll}{P}_{\text{tot}}=\text{kW}\hfill & =\hfill & 1000+150\hfill \\ \hfill & =\hfill & 1150\text{kW}\hfill \end{array}$

$\begin{array}{lll}{Q}_{\text{tot}}\hfill & =\hfill & 750+132.29\hfill \\ \hfill & =\hfill & 882.29\text{kvar}\hfill \end{array}$

The minimum operating power factor for a load of 1150 kW and not exceeding the kVA rating of the generator is

$\begin{array}{lll}\text{PF}=\cos \theta \hfill & =\hfill & \frac{1150\text{kW}}{1250\text{kVA}}\hfill \\ \hfill & =\hfill & 0.92\hfill \end{array}$

The maximum load kvar for this situation is

$\begin{array}{lll}Q\hfill & =\hfill & (1150\text{kW}){\tan }^{-1}\theta \hfill \\ \hfill & =\hfill & 1150\times {\tan }^{-1}(23.073918°)\hfill \\ \hfill & \cong \hfill & 489.9\text{kvar}\hfill \end{array}$

where 0.426 is the tangent corresponding to the maximum power factor of 0.935.

Thus, the capacitors must provide the difference between the total load kvar and the permissible generator kvar, or

$\begin{array}{lll}\text{ckvar}\hfill & =\hfill & 882.29-489.9\hfill \\ \hfill & =\hfill & 392.39\text{kvar}\hfill \end{array}$

Example 8.8

Assume that a 700 k VA load has a 65% power factor. It is desired to improve the power factor to 92%. Using Table 8.2, determine the following:

1. The correction factor required.
2. The capacitor size required.
3. What would be the resulting power factor if the next higher standard capacitor size is used? Solution

Solution

1. From Table 8.2, the correction factor required can be found as 0.74.
2. The 700 kVA load at 65% power factor is

   $\begin{array}{llll}{P}_L\hfill & =\hfill & S_L\times \cos \theta \hfill & \hfill \\ \hfill & =\hfill & 700\times 0.65\hfill & \hfill \\ \hfill & =\hfill & 455\text{kW}\hfill & (8.25)\hfill \end{array}$

   The capacitor size necessary to improve the power factor from 65% to 92% can be found as

   $\begin{array}{llll}\text{Capacitor}\text{size}\hfill & =\hfill & P_L(\text{correlation}\text{factor})\hfill & \hfill \\ \hfill & =\hfill & 455(0.74)\hfill & \hfill \\ \hfill & =\hfill & 336.7\text{kvar}\hfill & (8.26)\hfill \end{array}$

3. Assume that the next higher standard capacitor size (or rating) is selected to be 360 kvar. Therefore, the resulting new correction factor can be found from

   $\begin{array}{llll}\text{New correction factor}\hfill & =\hfill & \frac{\text{Standard capacitor rating}}{P_L}\hfill & \hfill \\ \hfill & =\hfill & \frac{\text{360 kvar}}{\text{455 kW}}\hfill & \hfill \\ \hfill & =\hfill & 0.7912\hfill & (8.27)\hfill \end{array}$

   From the table by linear interpolation, the resulting corrected percent power factor, with an original power factor of 65% and a correction factor of 0.7912, can be found as

   $\begin{array}{lll}\text{New corrected\%power factor}\hfill & =\hfill & 93+\frac{172}{320}\hfill \\ \hfill & \cong \hfill & 93.5\hfill \end{array}$

8.4.9 Computerized Method to Determine the Economic Power Factor

As suggested by Hopkinson [1], a load flow digital computer program can be employed to determine the kilovoltamperes, kilovolts, and kilovars at annual peak level for the whole system (from generation through the distribution substation buses) as the power factor is varied.

As a start, shunt capacitors are applied to each substation bus for correcting to an initial power factor, for example, 90%. Then, a load flow run is performed to determine the total system kilovoltamperes, and kilowatt losses (from generator to load) at this level and capacitor kilovars are noted. Later, additional capacitors are applied to each substation bus to increase the power factor by 1%, and another load flow run is made. This process of iteration is repeated until the power factor becomes unity.

As a final step, the benefits and costs are calculated at each power factor. The economic power factor is determined as the value at which benefits and costs are equal. After determining the economic power factor, the additional capacitor size required can be calculated as

$\begin{array}{cc}\Delta Q_c=P_{PK}(\tan \varphi -\tan \theta )\hfill & (8.28)\hfill \end{array}$

where

ΔQc is the required capacitor size, kvar

PPK is the system demand at annual peak, kW

tan ɸ is the tangent of original power factor angle

tan θ is the tangent of economic power factor angle

An illustration of this method is given in Example 8.12.

8.5 Application of Capacitors

In general, capacitors can be applied at almost any voltage level. As illustrated in Figure 8.15, individual capacitor units can be added in parallel to achieve the desired kilovar capacity and can be added in series to achieve the required kilovolt voltage. They are employed at or near rated voltage for economic reasons.

The cumulative data gathered for the whole utility industry indicate that approximately 60% of the capacitors is applied to the feeders, 30% to the substation buses, and the remaining 10% to the transmission system [1].

The application of capacitors to the secondary systems is very rare due to small economic advantages. Zimmerman [3] has developed a nomograph, shown in Figure 8.16, to determine the economic justification, if any, of the secondary capacitors considering only the savings in distribution transformer cost.

Example 8.9

Assume that a three-phase 500 hp 60 Hz 4160 V wye-connected induction motor has a fullload efficiency (η) of 88% and a lagging power factor of 0.75 and is connected to a feeder. If it is desired to correct the power factor of the load to a lagging power factor of 0.9 by connecting three capacitors at the load, determine the following:

1. The rating of the capacitor bank, in kilovars
2. The capacitance of each unit if the capacitors are connected in delta, in microfarads
3. The capacitance of each unit if the capacitors are connected in wye, in microfarads

Solution

1. The input power of the induction motor can be found as

   $\begin{array}{lll}P\hfill & =\hfill & \frac{(\text{HP})(0.7457\text{kW/hp})}{\eta }\hfill \\ \hfill & =\hfill & \frac{(500\text{hp})(0.7457\text{kW/hp})}{0.88}\hfill \\ \hfill & =\hfill & 423.69\text{kW}\hfill \end{array}$

   The reactive power of the motor at the uncorrected power factor is

   $\begin{array}{lll}{Q}_1\hfill & =\hfill & P\tan {\theta }_1\hfill \\ \hfill & =\hfill & 423.69\tan ({\cos }^{-1}0.75)\hfill \\ \hfill & =\hfill & 423.69\times 0.8819\hfill \\ \hfill & =\hfill & 373.7\text{kvar}\hfill \end{array}$

   The reactive power of the motor at the corrected power factor is

   $\begin{array}{lll}{Q}_2\hfill & =\hfill & P\tan {\theta }_2\hfill \\ \hfill & =\hfill & 423.69\tan ({\cos }^{-1}0.90)\hfill \\ \hfill & =\hfill & 423.69\times 0.4843\hfill \\ \hfill & =\hfill & 205.2\text{kvar}\hfill \end{array}$

   Therefore, the reactive power provided by the capacitor bank is

   $\begin{array}{lll}{Q}_C\hfill & =\hfill & Q_1-Q_2\hfill \\ \hfill & =\hfill & 373.7-205.2\hfill \\ \hfill & =\hfill & 168.5\text{kvar}\hfill \end{array}$

   Hence, assuming the losses in the capacitors are negligible, the rating of the capacitor bank is 168.5 kvar.

2. If the capacitors are connected in delta as shown in Figure 8.17a, the line current is

   $\begin{array}{lll}{I}_L\hfill & =\hfill & \frac{Q_C}{\sqrt{3}\times V_L{{}_{-}}_L}\hfill \\ \hfill & =\hfill & \frac{168.5}{\sqrt{3}\times 4.16}\hfill \\ \hfill & =\hfill & 23.39\text{A}\hfill \end{array}$

   and therefore,

   $\begin{array}{lll}{I}_C\hfill & =\hfill & \frac{I_L}{\sqrt{3}}\hfill \\ \hfill & =\hfill & \frac{23.39}{\sqrt{3}}\hfill \\ \hfill & =\hfill & 13.5\text{A}\hfill \end{array}$

   Thus, the reactance of each capacitor is

   $\begin{array}{lll}{X}_C\hfill & =\hfill & \frac{V_{L-L}}{I_C}\hfill \\ \hfill & =\hfill & \frac{4160}{13.5}\hfill \\ \hfill & =\hfill & 308.11\Omega \hfill \end{array}$

   and hence, the capacitance of each unit,* if the capacitors are connected in delta, is

   $C=\frac{{10}^6}{\omega X_C}\text{μF}$

   or

   $\begin{array}{lll}C\hfill & =\hfill & \frac{{10}^6}{\omega X_C}\hfill \\ \hfill & =\hfill & \frac{{10}^6}{2\pi \times 60\times 308.11}\hfill \\ \hfill & =\hfill & 8.61\text{μF}\hfill \end{array}$

3. If the capacitors are connected in wye as shown in Figure 8.17b,

   $I_C=I_L=23.39\text{A}$

   and therefore,

   $\begin{array}{lll}{X}_C\hfill & =\hfill & \frac{V_{L-N}}{I_C}\hfill \\ \hfill & =\hfill & \frac{4160}{\sqrt{3\times 23.39}}\hfill \\ \hfill & =\hfill & 102.70\Omega \hfill \end{array}$

   Thus, the capacitance of each unit, if the capacitors are connected in wye, is

   $\begin{array}{lll}C\hfill & =\hfill & \frac{{10}^6}{\omega X_C}\hfill \\ \hfill & =\hfill & \frac{{10}^6}{2\pi \times 60\times 102.70}\hfill \\ \hfill & =\hfill & 25.82\text{μF}\hfill \end{array}$

   $C=1/(\omega X_C)F.$

   $C={10}^6/(\omega X_C)({10}^{-6}\text{F})$

   $C={10}^6/(\omega X_C)\text{μF:}{\text{C/10}}^6=(1/\omega X_C)/{10}^6\text{F}$

Example 8.10

Assume that a 2.4 kV single-phase circuit feeds a load of 360 kW (measured by a wattmeter) at a lagging load factor and the load current is 200 A. If it is desired to improve the power factor, determine the following:

1. The uncorrected power factor and reactive load.
2. The new corrected power factor after installing a shunt capacitor unit with a rating of 300 kvar.
3. Also write the necessary codes to solve the problem in MATLAB®.

Solution

1. Before the power factor correction,

   $\begin{array}{lll}{S}_1\hfill & =\hfill & V\times I\hfill \\ \hfill & =\hfill & 2.4\times 200\hfill \\ \hfill & =\hfill & 480\text{kVA}\hfill \end{array}$

   therefore, the uncorrected power factor can be found as

   $\begin{array}{lll}\cos {\theta }_1\hfill & =\hfill & \frac{P}{S_1}\hfill \\ \hfill & =\hfill & \frac{360\text{kW}}{480\text{kVA}}\hfill \\ \hfill & =\hfill & 0.75\hfill \end{array}$

   and the reactive load is

   $\begin{array}{lll}{Q}_1\hfill & =\hfill & S_1\times \sin ({\cos }^{-1}{\theta }_1)\hfill \\ \hfill & =\hfill & 480\times 0.661\hfill \\ \hfill & =\hfill & 317.5\text{kvar}\hfill \end{array}$

2. After the installation of the 300 kvar capacitors,

   $\begin{array}{lll}{Q}_2\hfill & =\hfill & Q_1-Q_C\hfill \\ \hfill & =\hfill & 317.5-300\hfill \\ \hfill & =\hfill & 17.5\text{kvar}\hfill \end{array}$

   and therefore, the new power factor can be found from Equation 8.9 as

   $\begin{array}{lll}\cos {\theta }_2\hfill & =\hfill & \frac{P}{{[P^2+{(Q_1-Q_C)}^2]}^{1/2}}\hfill \\ \hfill & =\hfill & \frac{360}{{({360}^2+{17.5}^2)}^{1/2}}\hfill \\ \hfill & =\hfill & 0.9989\text{or}99\text{.89\%}\hfill \end{array}$

3. Here is the MATLAB script:

   clc

   clear

   % System parameters

   HP = 500;

   PFold = 0.75;

   PFnew = 0.90;

   effi = 0.88;

   kVL = 4.16;

   VL = 4160;

   % Solution for part a

   % Solve for input power of motor in kW

   P = (HP*0.7457)/effi

   % Solve for reactive power of motor in kvar

   Q1 = P*tan(acos(PFold))

   % Reactive power of motor at corrected power factor in kvar

   Q2 = P*tan(acos(PFnew))

   % Reactive power provided by capacitor bank in kvar

   Qc = Q1 - Q2

   % Solution for part b when capacitors are in delta

   % Line current when capacitors are in delta, in Amps

   IL = Qc/(sqrt(3)*kVL)

   % Capacitor current in Amps

   Ic = IL/sqrt(3)

   % Reactance of each capacitor in ohms

   Xc = VL/Ic

   % Capacitance of each unit in micro farads

   C = (10^6)/(120*pi*Xc)

   % Solution for part c when capacitors are in wye

   % Capacitance current is equal to line current when capacitors are in wye

   % Reactance of each capacitor in ohms

   Xcwye = VL/(sqrt(3)*IL)

   % Capacitance of each unit in micro farads

   Cwye = (10^6)/(120*pi*Xcwye)

Example 8.11

Assume that the Riverside Substation of the NL&NP Company has a bank of three 2000 kVA transformers that supplies a peak load of 7800 kVA at a lagging power factor of 0.89. All three transformers have a thermal capability of 120% of the nameplate rating. It has already been planned to install 1000 kvar of shunt capacitors on the feeder to improve the voltage regulation.

Determine the following:

1. Whether or not to install additional capacitors on the feeder to decrease the load to the thermal capability of the transformer
2. The rating of the additional capacitors

Solution

1. Before the installation of the 1000 kvar capacitors,

   $\begin{array}{lll}P\hfill & =\hfill & S_1\times \cos \theta \hfill \\ \hfill & =\hfill & 7800\times 0.89\hfill \\ \hfill & =\hfill & 6942\text{kW}\hfill \end{array}$

   and

   $\begin{array}{lll}{Q}_1\hfill & =\hfill & S_1\times \sin \theta \hfill \\ \hfill & =\hfill & 7800\times 0.456\hfill \\ \hfill & =\hfill & 3556.8\text{kvar}\hfill \end{array}$

   Therefore, after the installation of the 1000 kvar capacitors,

   $\begin{array}{lll}{Q}_2\hfill & =\hfill & Q_1-Q_C\hfill \\ \hfill & =\hfill & 3556.8-1000\hfill \\ \hfill & =\hfill & 2556.8\text{kvar}\hfill \end{array}$

   and using Equation 8.9,

   $\begin{array}{lll}\cos {\theta }_2\hfill & =\hfill & \frac{P}{{[P^2+{(Q_1-Q_C)}^2]}^{1/2}}\hfill \\ \hfill & =\hfill & \frac{6942}{{({6942}^2+{2556.8}^2)}^{1/2}}\hfill \\ \hfill & =\hfill & 0.938\text{or}93\text{.8\%}\hfill \end{array}$

   and the corrected apparent power is

   $\begin{array}{lll}{S}_2\hfill & =\hfill & \frac{P}{\cos {\theta }_2}\hfill \\ \hfill & =\hfill & \frac{6942}{0.938}\hfill \\ \hfill & =\hfill & 7397.9\text{kVA}\hfill \end{array}$

   On the other hand, the transformer capability is

   $\begin{array}{lll}{S}_T\hfill & =\hfill & 6000\times 1.20\hfill \\ \hfill & =\hfill & 72000\text{kVA}\hfill \end{array}$

   Therefore, the capacitors installed to improve the voltage regulation are not adequate; additional capacitor installation is required.

2. The new or corrected power factor required can be found as

   $\begin{array}{lll}{\text{PF}}_{2,\text{new}}=\cos {\theta }_{2,\text{new}}\hfill & =\hfill & \frac{P}{S_T}\hfill \\ \hfill & =\hfill & \frac{6942}{7200}\hfill \\ \hfill & =\hfill & 0.9642\text{or}96.42\%\hfill \end{array}$

   and thus, the new required reactive power can be found as

   $\begin{array}{lll}{Q}_{\text{2,new}}\hfill & =\hfill & P\times \tan {\theta }_{\text{2,new}}\hfill \\ \hfill & =\hfill & P\times \tan ({\cos }^{-1}{\text{PF}}_{\text{2,new}})\hfill \\ \hfill & =\hfill & 6942\times 0.2752\hfill \\ \hfill & =\hfill & 1910\text{kvar}\hfill \end{array}$

   Therefore, the rating of the additional capacitors required is

   $\begin{array}{lll}{Q}_{C,\text{add}}\hfill & =\hfill & Q_2-Q_{\text{2,new}}\hfill \\ \hfill & =\hfill & 2556.8-1910\hfill \\ \hfill & =\hfill & 646.7\text{kvar}\hfill \end{array}$

Example 8.12

If a power system has 10,000 kVA capacity and is operating at a power factor of 0.7 and the cost of a synchronous capacitor (i.e., synchronous condenser) to correct the power factor is $10 per kVA, find the investment required to correct the power factor to

1. 0.85 lagging power factor
2. Unity power factor

Solution

At original cost

${\theta }_{\text{old}}={\cos }^{-1}\text{PF}={\text{cos}}^{-1}0.7=45.57°$

$P_{\text{old}}=S\cos {\theta }_{\text{old}}=(10,000\text{kVA})0.7=7,000\text{kW}$

${\text{Q}}_{\text{old}}=S\sin {\theta }_{\text{old}}=(10,000\text{kVA})\sin 45.57°=7,141.43\text{kvar}$

1. For PF = 0.85 lagging

   $\begin{array}{lll}{P}_{\text{new}}=P_{\text{old}}\hfill & =\hfill & 7000\text{kW}\text{(as before)}\hfill \\ S_{\text{new}}=\frac{P_{\text{new}}}{\cos {\theta }_{\text{new}}}\hfill & =\hfill & \frac{7000\text{kW}}{0.85}=8235.29\text{kVA}\hfill \end{array}$

   $Q_{\text{new}}=S_{\text{new}}\sin ({\cos }^{-1}\text{PF})=(8235.29\text{kVA})\sin ({\cos }^{-1}0.85)=4338.21\text{kvar}$

   $Q_C\text{ = }{Q}_{\text{required}}\text{ = }{Q}_{\text{old}}-Q_{\text{new}}\text{ = 7141}\text{.43 }-\text{ 4338}\text{.21= 2803}\text{.22 kvar correction needed}$

   Hence, the theoretical cost of the synchronous capacitor is

   ${\text{Cost}}_{\text{capacitor}}=(2,803.22\text{kVA})\left(\frac{\$10}{\text{kVA}}\right)=\$28,032.20$

   Note that it is customary to give the cost of capacitors in dollars per kVA rather than in dollars per kvar.

2. For PF = 1.0

   $Q_C=Q_{\text{required}}=Q_{\text{old}}-Q_{\text{new}}=7141.43-0.0=7141.43\text{kvar}$

   Thus, the theoretical cost of the synchronous capacitor is

   ${\text{Cost}}_{\text{capacitor}}=(7,141.43\text{kVA})\left(\frac{\$10}{\text{kVA}}\right)=\$71,414.30$

   Note that Pnew = 7000 kW is the same as before.

Example 8.13

If a power system has 15,000 kVA capacity, operating at a 0.65 lagging power factor, and the cost of synchronous capacitors to correct the power factor is $12.5/kVA, determine the costs involved and also develop a table showing the required (leading) reactive power to increase the power factor to

1. 0.85 lagging power factor
2. 0.95 lagging power factor
3. Unity power factor

Solution

At original power factor or 0.65

$P =S\text{ cos}\theta \text{= (15,000 kVA)0}\text{.65 = 9750 kW at a power factor angle of 49}.46°$

$Q=S\text{ sin}\theta {\text{ = (15,000 kVA) sin(cos}}^{-1}0\text{.65) = 11,399 kvar}$

The following table shows the amount of reactive power that is required to improve the power factor from one level to the next at 0.05 increments.

1. For PF = 0.85 lagging

   P = S cos θ = (15,000 kVA) × 0.65 = 9,750 kW. It will be the same at a power factor of 0.85.

   $S=\frac{P}{\cos \theta }=\frac{9,750\text{kW}}{0.85}=11,470\text{kVA}$

   and

   $Q =S\text{sin}\theta \text{ = (11,470 kVA) sin(}{\cos }^{-1}0\text{.85) = 6,042 kvar}$

   amount of additional reactive power correction required is

   $\text{Additional var correction = 11,399 }-\text{ 6,042 = 5,357}\text{kvar}$

   The cost of this correction is

   $\text{Cost of correction =}\text{(5,357 kVA) }\left(\frac{\$12.5}{\text{kVA}}\right)\text{= \$66,962}.50$

2. For PF = 0.95 lagging:

   $S\text{ = }\frac{9,750\text{kW}}{0.95}\text{=10,263 kVA}$

   and

   $Q{\text{ = (10,263 kVA)sin(cos}}^{-1}0\text{.95)}\text{= 3,204 kvar}$

   The amount of additional reactive power correction required is

   $\text{Additional var correction 11,399 }-\text{ 3,204 = 8,195}\text{kvar}$

   The cost of this correction is

   $\text{Cost of correction =}\text{(8,195 kVA) }\left(\frac{\$12.5}{\text{kVA}}\right)\text{= \$102,438}$

3. For unity PF

   The amount of additional reactive power correction required is

   $\text{Additional var correction =}\text{11,399 kvar}$

   The cost of this correction is

   $\text{Cost of correction =}\text{(11,399 kVA) }\left(\frac{\$12.5}{\text{kVA}}\right)\text{= \$142,487}.50$

8.5.1 Capacitor Installation Types

In general, capacitors installed on feeders are pole-top banks with necessary group fusing. The fusing applications restrict the size of the bank that can be used. Therefore, the maximum sizes used are about 1800 kvar at 15 kV and 3600 kvar at higher voltage levels. Usually, utilities do not install more than four capacitor banks (of equal sizes) on each feeder.

Figure 8.18 illustrates the effects of a fixed capacitor on the voltage profiles of a feeder with uniformly distributed load at heavy load and light load. If only fixed-type capacitors are installed, as can be observed in Figure 8.18c, the utility will experience an excessive leading power factor and voltage rise at that feeder. Therefore, as shown in Figure 8.19, some of the capacitors are installed as switched capacitor banks so they can be switched off during light-load conditions.

Thus, the fixed capacitors are sized for light load and connected permanently. As shown in the figure, the switched capacitors can be switched as a block or in several consecutive steps as the reactive load becomes greater from light-load level to peak load and sized accordingly.

However, in practice, the number of steps or blocks is selected to be much less than the ones shown in the figure due to the additional expenses involved in the installation of the required switchgear and control equipment.

A system survey is required in choosing the type of capacitor installation. As a result of load flow program runs or manual load studies on feeders or distribution substations, the system’s lagging reactive loads (i.e., power demands) can be determined and the results can be plotted on a curve as shown in Figure 8.19. This curve is called the reactive load–duration curve and is the cumulative sum of the reactive loads (e.g., fluorescent lights, household appliances, and motors) of consumers and the reactive power requirements of the system (e.g., transformers and regulators). Once the daily reactive load–duration curve is obtained, then by visual inspection of the curve, the size of the fixed capacitors can be determined to meet the minimum reactive load. For example, from Figure 8.19 one can determine that the size of the fixed capacitors required is 600 kvar.

The remaining kilovar demands of the loads are met by the generator or preferably by the switched capacitors. However, since meeting the kilovar demands of the system from the generator is too expensive and may create problems in the system stability, capacitors are used. Capacitor sizes are selected to match the remaining load characteristics from hour to hour.

Many utilities apply the following rule of thumb to determine the size of the switched capacitors: Add switched capacitors until

$\begin{array}{cc}\frac{\text{kvar from switched}\text{+ fixed capacitors}}{\text{kvar of peak reactive feeder load}}\ge 0.70\hfill & (8.29)\hfill \end{array}$

From the voltage regulation point of view, the kilovars needed to raise the voltage at the end of the feeder to the maximum allowable voltage level at minimum load (25% of peak load) are the size of the fixed capacitors that should be used. On the other hand, if more than one capacitor bank is installed, the size of each capacitor bank at each location should have the same proportion, that is,

$\begin{array}{cc}\frac{\text{kvar of load center}}{\text{kvar of total feeder}}=\frac{\text{kVA of load center}}{\text{kVA of total feeder}}\hfill & (8.30)\hfill \end{array}$

However, the resultant voltage rise must not exceed the light-load voltage drop. The approximate value of the percent voltage rise can be calculated from

$\begin{array}{cc}\%\text{VR}=\frac{Q_{c,3}{}_{\varphi }\times x\times l}{10\times V_{L-L}^2}\hfill & (8.31)\hfill \end{array}$

where

%VR is the percent voltage rise

Qc,3ϕ is the three-phase reactive power due to fixed capacitors applied, kvar x is the line reactance, O/min

l is the length of feeder from sending end of feeder to fixed capacitor location, min

VL–L is the line-to-line voltage, kV

The percent voltage rise can also be found from

$\begin{array}{cc}\%\text{VR}=\frac{I_c\times x\times l}{10\times V_{L-L}}\hfill & (8.32)\hfill \end{array}$

where

$\begin{array}{llll}{I}_C\hfill & =\hfill & \frac{Q_{c,3\varphi }}{\sqrt{3}\times V_{L-L}}\hfill & \hfill \\ \hfill & =\hfill & \text{current drawn by fixed-capacitor bank}\hfill & (8.33)\hfill \end{array}$

If the fixed capacitors are applied to the end of the feeder and if the percent voltage rise is already determined, the maximum value of the fixed capacitors can be determined from

$\begin{array}{cc}\text{Max}{Q}_{c,3\varphi }=\frac{10(\%\text{VR})V_{L-L}^2}{x\times l}\text{kvar}\hfill & (8.34)\hfill \end{array}$

Equations 8.31 and 8.32 can also be used to calculate the percent voltage rise due to the switched capacitors. Therefore, once the percent voltage rises due to both fixed and switched capacitors, the total percent voltage rise can be calculated as

$\begin{array}{cc}{\displaystyle \sum \%\text{VR}={\text{\%VR}}_{\text{NSW}}+\%{\text{VR}}_{\text{SW}}}\hfill & (8.35)\hfill \end{array}$

where

${\displaystyle \sum \%\text{VR}}$ is the total percent voltage rise

%VRNSW is the percent voltage rise due to fixed (or nonswitched) capacitors

%VRSW is the percent voltage rise due to switched capacitors

Some utilities use the following rule of thumb: The total amount of fixed and switched capacitors for a feeder is the amount necessary to raise the receiving-end feeder voltage to maximum at 50% of the peak feeder load.

Once the kilovars of capacitors necessary for the system are determined, there remains only the question of proper location. The rule of thumb for locating the fixed capacitors on feeders with uniformly distributed loads is to locate them approximately at two-thirds of the distance from the substation to the end of the feeder.

For the uniformly decreasing loads, fixed capacitors are located approximately halfway out on the feeder. On the other hand, the location of the switched capacitors is basically determined by the voltage regulation requirements, and it usually turns out to be the last one-third of the feeder away from the source.

8.5.2 Types of Controls for Switched Shunt Capacitors

The switching process of capacitors can be done by manual control or by automatic control using some type of control intelligence. Manual control (at the location or as remote control) can be employed at distribution substations. The intelligence types that can be used in automatic control include time–switch, voltage, current, voltage–time, voltage–current, and temperature.

The most popular types are the time-switch control, voltage control, and voltage––current control. The time–switch control is the least–expensive one. Some combinations of these controls are also used to follow the reactive load–duration curve more closely, as illustrated in Figure 8.20.

8.5.3 Types of Three-Phase Capacitor-Bank Connections

A three-phase capacitor bank on a distribution feeder can be connected in (1) delta, (2) grounded wye, or (3) ungrounded wye. The type of connection used depends upon the following:

1. System type, that is, whether it is a grounded or an ungrounded system
2. Fusing requirements
3. Capacitor-bank location
4. Telephone interference considerations

A resonance condition may occur in delta and ungrounded-wye (floating neutral) banks when there is a one- or two-line open-type fault that occurs on the source side of the capacitor bank due to the maintained voltage on the open phase that backfeeds any transformers located on the load side of the open conductor through the series capacitor. As a result of this condition, the singlephase distribution transformers on four-wire systems may be damaged. Therefore, ungrounded-wye capacitor banks are not recommended under the following conditions:

1. On feeders with light load where the minimum load per phase beyond the capacitor bank does not exceed 150% of the per phase rating of the capacitor bank
2. On feeders with single-phase breaker operation at the sending end
3. On fixed capacitor banks
4. On feeder sections beyond a sectionalizing-fuse or single-phase recloser
5. On feeders with emergency load transfers

However, the ungrounded-wye capacitor banks are recommended if one or more of the following conditions exist:

1. Excessive harmonic currents in the substation neutral can be precluded.
2. Telephone interferences can be minimized.
3. Capacitor-bank installation can be made with two single-phase switches rather than with three single-pole switches.

Usually, grounded-wye capacitor banks are used only on four-wire three-phase primary systems. Otherwise, if a grounded-wye capacitor bank is used on a three-phase three-wire ungrounded-wye or delta system, it furnishes a ground current source that may disturb sensitive ground relays.

8.6 Economic Justification for Capacitors

Loads on electric utility systems include two components: active power (measured in kilowatts) and reactive power (measured in kilovars). Active power has to be generated at power plants, whereas reactive power can be provided by either power plants or capacitors. It is a well-known fact that shunt power capacitors are the most economical source to meet the reactive power requirements of inductive loads and transmission lines operating at a lagging power factor.

When reactive power is provided only by power plants, each system component (i.e., generators, transformers, transmission and distribution lines, switchgear, and protective equipment) has to be increased in size accordingly. Capacitors can mitigate these conditions by decreasing the reactive power demand all the way back to the generators. Line currents are reduced from capacitor locations all the way back to generation equipment. As a result, losses and loadings are reduced in distribution lines, substation transformers, and transmission lines.

Depending upon the uncorrected power factor of the system, the installation of capacitors can increase generator and substation capability for additional load at least 30% and can increase individual circuit capability, from the voltage regulation point of view, approximately 30%–100%.

Furthermore, the current reduction in transformer and distribution equipment and lines reduces the load on these kilovoltampere-limited apparatus and consequently delays the new facility installations. In general, the economic benefits force capacitor banks to be installed on the primary distribution system rather than on the secondary.

It is a well-known rule of thumb that the optimum amount of capacitor kilovars to employ is always the amount at which the economic benefits obtained from the addition of the last kilovar exactly equal the installed cost of the kilovars of capacitors.

The methods used by the utilities to determine the economic benefits derived from the installation of capacitors vary from company to company, but the determination of the total installed cost of a kilovar of capacitors is easy and straightforward.

In general, the economic benefits that can be derived from capacitor installation can be summarized as follows:

1. Released generation capacity
2. Released transmission capacity
3. Released distribution substation capacity
4. Additional advantages in distribution system
   1. Reduced energy (copper) losses
   2. Reduced voltage drop and consequently improved voltage regulation
   3. Released capacity of feeder and associated apparatus
   4. Postponement or elimination of capital expenditure due to system improvements and/ or expansions
   5. Revenue increase due to voltage improvements

8.6.1 Benefits Due to Released Generation Capacity

The released generation capacity due to the installation of capacitors can be calculated approximately from

$\begin{array}{cc}\Delta S_G=\{\underset{Q_c\times \sin \theta \text{when}{Q}_c\le 0.10S_G}{\overset{\left[{\left(1-\frac{Q_c^2\times {\text{cos}}^2\theta }{S_G^2}\right)}^{1/2}+\frac{Q_c\times \sin \theta }{S_G}-1\right]S_G\text{when}{Q}_c>0.10S_G}{}}\hfill & (8.36)\hfill \end{array}$

where

ΔSG is the released generation capacity beyond maximum generation capacity at original power factor, kVA

SG is the generation capacity, kVA

Qc is the reactive power due to corrective capacitors applied, kvar

cos θ is the original (or uncorrected or old) power factor before application of capacitors

Therefore, the annual benefits due to the released generation capacity can be expressed as

$\begin{array}{cc}\Delta {\$}_G=\Delta S_G\times C_G\times i_G\hfill & (8.37)\hfill \end{array}$

where

Δ$G is the annual benefits due to released generation capacity, $/year

ΔSG is the released generation capacity beyond maximum generation capacity at original power factor, kVA

CG is the cost of (peaking) generation, $/kW

iG is the annual fixed charge rate* applicable to generation

8.6.2 Benefits Due to Released Transmission Capacity

The released transmission capacity due to the installation of capacitors can be calculated approximately as

$\begin{array}{cc}\Delta S_T=\{\underset{Q_c\times \sin \theta \text{when}{Q}_c\le 0.10S_T}{\overset{\left[{\left(1-\frac{Q_c^2\times {\text{cos}}^2\theta }{S_T^2}\right)}^{1/2}+\frac{Q_c\times \sin \theta }{S_T}-1\right]S_T\text{when}{Q}_c>0.10S_T}{}}\hfill & (8.38)\hfill \end{array}$

where

ΔST is the released transmission capacity† beyond maximum transmission capacity at original power factor, kVA

ST is the transmission capacity, kVA

Thus, the annual benefits due to the released transmission capacity can be found as

$\begin{array}{cc}\Delta {\$}_T=\Delta S_T\times C_T\times i_T\hfill & (8.39)\hfill \end{array}$

where

Δ$T is the annual benefits due to released transmission capacity, $/year

ΔST is the released transmission capacity beyond maximum transmission capacity at original power factor, kVA

CT is the cost of transmission line and associated apparatus, $/kVA

iT is the annual fixed charge rate applicable to transmission

8.6.3 Benefits Due to Released Distribution Substation Capacity

The released distribution substation capacity due to the installation of capacitors can be found approximately from

$\begin{array}{cc}\Delta S_S=\{\underset{Q_c\times \sin \theta \text{when}{Q}_c\le 0.10S_S}{\overset{\left[{\left(1-\frac{Q_c^2\times {\text{cos}}^2\theta }{S_S^2}\right)}^{1/2}+\frac{Q_c\times \sin \theta }{S_S}-1\right]S_S\text{when}{Q}_c>0.10S_S}{}}\hfill & (8.40)\hfill \end{array}$

where

ΔSS is the released distribution substation capacity beyond maximum substation capacity at original power factor, kVA

SS is the distribution substation capacity, kVA

Hence, the annual benefits due to the released substation capacity can be calculated as

$\begin{array}{cc}\Delta {\$}_S=\Delta S_S\times C_S\times i_S\hfill & (8.41)\hfill \end{array}$

where

Δ$S is the annual benefits due to the released substation capacity, $/year

ΔSS is the released substation capacity, kVA

CS is the cost of substation and associated apparatus, $/kVA

iS is the annual fixed charge rate applicable to substation

8.6.4 Benefits Due to Reduced Energy Losses

The annual energy losses are reduced as a result of decreasing copper losses due to the installation of capacitors. The conserved energy can be expressed as

$\begin{array}{cc}\Delta \text{ACE}=\frac{Q_{c,3\varphi }R(2S_{L,3\varphi }\sin \theta -Q_{c,3\varphi })8760}{1000\times V_{L-L}^2}\hfill & (8.42)\hfill \end{array}$

where

ΔACE is the annual conserved energy, kWh/year

Qc,3ϕ is the three-phase reactive power due to corrective capacitors applied, kvar

R is the total line resistance to load center, O

QL,3ϕ is the original, that is, uncorrected, three-phase load, kVA

sin θ is the sine of original (uncorrected) power factor angle

VL–L is the line-to-line voltage, kV

Therefore, the annual benefits due to the conserved energy can be calculated as

$\begin{array}{cc}\Delta {\$}_{\text{ACE}}=\Delta \text{ACE}\times \text{EC}\hfill & (8.43)\hfill \end{array}$

where

ΔACE is the annual benefits due to conserved energy, $/year

EC is the cost of energy, $/kWh

8.6.5 Benefits Due to Reduced Voltage Drops

The following advantages can be obtained by the installation of capacitors into a circuit:

1. The effective line current is reduced, and consequently, both IR and IXL voltage drops are decreased, which results in improved voltage regulation.
2. The power factor improvement further decreases the effect of reactive line voltage drop.

The percent voltage drop that occurs in a given circuit can be expressed as

$\begin{array}{cc}\%\text{VD}=\frac{S_{L,3\varphi }(r\cos \theta +x\sin \theta )l}{10\times V_{L-L}^2}\hfill & (8.44)\hfill \end{array}$

where

%VD is the percent voltage drop

SL,3ϕ is the three-phase load, kVA

r is the line resistance, 0/min

x is the line reactance, 0/min

l is the length of conductors, min

VL-L is the line-to-line voltage, kV

The voltage drop that can be calculated from Equation 8.44 is the basis for the application of the capacitors. After the application of the capacitors, the system yields a voltage rise due to the improved power factor and the reduced effective line current. Therefore, the voltage drops due to IR and IXL are minimized. The approximate value of the percent voltage rise along the line can be calculated as

$\begin{array}{cc}\%\text{VR}=\frac{Q_{c,3\varphi }\times x\times l}{10\times V_{L-L}^2}\hfill & (8.45)\hfill \end{array}$

Furthermore, an additional voltage-rise phenomenon through every transformer from the generating source to the capacitors occurs due to the application of capacitors. It is independent of load and power factor of the line and can be expressed as

$\begin{array}{cc}\%{\text{VR}}_T=\left(\frac{Q_{c,3\varphi }}{S_{T,3\varphi }}\right)x_T\hfill & (8.46)\hfill \end{array}$

where

%VRT is the percent voltage rise through the transformer

ST,3ϕ is the total three-phase transformer rating, kVA

xT is the percent transformer reactance (approximately equal to the transformer’s nameplate impedance).

8.6.6 Benefits Due to Released Feeder Capacity

In general, feeder capacity is restricted by allowable voltage drop rather than by thermal limitations (as seen in Chapter 4). Therefore, the installation of capacitors decreases the voltage drop and consequently increases the feeder capacity.

Without including the released regulator or substation capacity, this additional feeder capacity can be calculated as

$\begin{array}{cc}\Delta S_F=\frac{(Q_{c,3\varphi })x}{x\sin \theta +r\cos \theta }\text{kVA}\hfill & (8.47)\hfill \end{array}$

Therefore, the annual benefits due to the released feeder capacity can be calculated as

$\begin{array}{cc}\Delta {\$}_F=\Delta S_F\times C_F\times i_F\hfill & (8.48)\hfill \end{array}$

where

Δ$F is the annual benefits due to released feeder capacity, $/year

ΔSF is the released feeder capacity, kVA

CF is the cost of installed feeder, $/kVA

iF is the annual fixed charge rate applicable to the feeder

8.6.7 Financial Benefits Due to Voltage Improvement

The revenues to the utility are increased as a result of increased kilowatthour energy consumption due to the voltage rise produced on a system by the addition of the corrective capacitor banks. This is especially true for residential feeders.

The increased energy consumption depends on the nature of the apparatus used. For example, energy consumption for lighting increases as the square of the voltage used. As an example, Table 8.3 gives the additional kilowatthour energy increase (in percent) as a function of the ratio of the average voltage after the addition of capacitors to the average voltage before the addition of capacitors (based on a typical load diversity).

Thus, the increase in revenues due to the increased kilowatthour energy consumption can be calculated as

$\begin{array}{cc}\Delta {\$}_{\text{BEC}}=\Delta \text{BEC}\times \text{BEC}\times \text{EC}\hfill & (8.49)\hfill \end{array}$

where

Δ$BEC is the additional annual revenue due to increased kWh energy consumption, $/year

ΔBEC is the additional kWh energy consumption increase

BEC is the original (or base) annual kWh energy consumption, kWh/year

8.6.8 Total Financial Benefits Due to Capacitor Installations

Therefore, the total benefits due to the installation of capacitor banks can be summarized as

$\begin{array}{llll}{\displaystyle \sum \Delta \$}\hfill & =\hfill & \text{(Demand reduction) +}\text{(Energy reduction) +}\text{(Revenue increase)}\hfill & \hfill \\ \hfill & =\hfill & (\Delta {\$}_G+\Delta {\$}_T+\Delta {\$}_S+\Delta {\$}_F)+\Delta {\$}_{\text{ACE}}+\Delta {\$}_{\text{BEC}}\hfill & (8.50)\hfill \end{array}$

The total benefits obtained from Equation 8.50 should be compared against the annual equivalent of the total cost of the installed capacitor banks. The total cost of the installed capacitor banks can be found from

$\begin{array}{cc}\Delta {\text{EIC}}_c=\Delta Q_c\times {\text{IC}}_c\times i_c\hfill & (8.51)\hfill \end{array}$

where

ΔEICc is the annual equivalent of the total cost of installed capacitor banks, $/year

ΔQc is the required amount of capacitor-bank additions, kvar

ICc is the cost of installed capacitor banks, $/kvar

ic is the annual fixed charge rate applicable to capacitors

In summary, capacitors can provide the utility industry with a very effective cost reduction instrument. With plant costs and fuel costs continually increasing, electric utilities benefit whenever new plant investment can be deferred or eliminated and energy requirements reduced.

Thus, capacitors aid in minimizing operating expenses and allow the utilities to serve new loads and customers with a minimum system investment. Today, utilities in the United States have approximately 1 kvar of power capacitors installed for every 2 kW of installed generation capacity in order to take advantage of the economic benefits involved [4].

Example 8.19*

Assume that a large power pool is presently operating at 90% power factor. It is desired to improve the power factor to 98%. To improve the power factor to 98%, a number of load flow runs are made, and the results are summarized in Table 8.4.

Assume that the average fixed charge rate is 0.20, average demand cost is $250/kW, energy cost is $0.045/kWh, the system loss factor is 0.17, and an average capacitor cost is $4.75/kvar. Use responsibility factors of 1.0 and 0.9 for capacitors installed on the substation buses and on feeders, respectively. Determine the following:

1. The resulting additional savings in kilowatt losses at the 98% power factor when all capacitors are applied to substation buses.
2. The resulting additional savings in kilowatt losses at the 98% power factor when some capacitors are applied to feeders.
3. The total additional savings in kilowatt losses.
4. The additional savings in the system kilovoltampere capacity.
5. The additional capacitors required, kvars.
6. The total annual savings in demand reduction due to additional capacitors applied to substation buses and feeders, $/year.
7. The annual savings due to the additional released transmission capacity, $/year.
8. The total annual savings due to the energy loss reduction, $/year.
9. The total annual cost of the additional capacitors, $/year.
10. The total net annual savings, $/year.
11. Is the 98% power factor the economic power factor?

Solution

1. From Table 8.4, the resulting additional savings in kilowatt losses due to the power factor improvement at the substation buses is

   $\Delta P_{\text{LS}}=495,165-491,738=3427\text{kW}$

2. From Table 8.4 for feeders,

   $\begin{array}{lll}\Delta P_{\text{LS}}\hfill & =\hfill & 85,771-75,342\hfill \\ \hfill & =\hfill & 10,429\text{kW}\hfill \end{array}$

3. Therefore, the total additional kilowatt savings is

   $\begin{array}{lll}\Delta P_{\text{LS}}\hfill & =\hfill & 3,427-10,429\hfill \\ \hfill & =\hfill & 13,856\text{kW}\hfill \end{array}$

   As can be observed, the additional kilowatt savings due to capacitors applied to the feeders is more than three times that of capacitors applied to the substation buses. This is due to the fact that power losses are larger at the lower voltages.

4. From Table 8.4, the additional savings in the system kilovoltampere capacity is

   $\begin{array}{lll}\Delta S_{\text{sys}}\hfill & =\hfill & 22,506,007-21,172,616\hfill \\ \hfill & =\hfill & 1,333,391\text{kVA}\hfill \end{array}$

5. From Table 8.4, the additional capacitors required are

   $\begin{array}{lll}\Delta Q_c\hfill & =\hfill & 9,810,141-4,213,297\hfill \\ \hfill & =\hfill & 5,596,844\text{kvar}\hfill \end{array}$

6. The annual savings in demand reduction due to capacitors applied to distribution substation buses is approximately

   $\text{(3427 kW)(1}\text{.0)(\$250/kW)(0}\text{.20/year) = \$171,350/year}$

   and due to capacitors applied to feeders is

   $\text{(10,429 kW)(0}\text{.9)(\$250/kW)(0}\text{.20/year) = \$469,305/year}$

   Therefore, the total annual savings in demand reduction is

   $\text{\$171,350 + \$469,305 = \$640,655/year}$

7. The annual savings due to the additional released transmission capacity is

   $\text{(1,333,391 kVA)(\$27/kVA)(0}\text{.20/year) = \$7,200,311/year}$

8. The total annual savings due to the energy loss reduction is

   $\text{(\$13,856 kW)(8760 h/year)(0}\text{.17)(\$0}\text{.045/kWh) = \$928,546/year}$

9. The total annual cost of the additional capacitors is

   $\text{(5,596,844 kvar)(\$4}\text{.75/kvar)(0}\text{.20/year) = \$5,317,002/year}$

10. The total annual savings is summation of the savings in demand, capacity, and energy:

    $\text{\$640,655 + \$7,200,311 + \$928,5466 = \$8,769,512/year}$

    Therefore, the total net annual savings is

    $\text{\$8,769,512 }-\text{ \$5,317,002 = \$3,452,510/year}$

11. No, since the total net annual savings is not zero.

8.7 Practical Procedure to Determine the Best Capacitor Location

In general, the best location for capacitors can be found by optimizing power loss and voltage regulation. A feeder voltage profile study is performed to warrant the most effective location for capacitors and the determination of a voltage that is within recommended limits.

Usually, a 2 V rise on circuits used in urban areas and a 3 V rise on circuits used in rural areas are approximately the maximum voltage changes that are allowed when a switched capacitor bank is placed into operation. The general iteration process involved is summarized in the following steps:

1. Collect the following circuit and load information:
   1. Any two of the following for each load: kilovoltamperes, kilovars, kilowatts, and load power factor
   2. Desired corrected power of circuit
   3. Feeder-circuit voltage
   4. A feeder-circuit map that shows locations of loads and presently existing capacitor banks
2. Determine the kilowatt load of the feeder and the power factor.
3. From Table 8.2, determine the kilovars per kilowatts of load (i.e., the correction factor) necessary to correct the feeder-circuit power factor from the original to the desired power factor. To determine the kilovars of capacitors required, multiply this correction factor by the total kilowatts of the feeder circuit.
4. Determine the individual kilovoltamperes and power factor for each load or group of loads.
5. To determine the kilovars on the line, multiply individual load or groups of loads by their respective reactive factors that can be found from Table 8.2.
6. Develop a nomograph to determine the line loss in W/1000 ft due to the inductive loads tabulated in steps 4 and 5. Multiply these line losses by their respective line lengths in thousands of feet. Repeat this process for all loads and line sections and add them to find the total inductive line loss.
7. In the case of having presently existing capacitors on the feeder, perform the same calculations as in step 6, but this time subtract the capacitive line loss from the total inductive line loss. Use the capacitor kilovars determined in step 3 and the nomograph developed for step 6 and find the line loss in each line section due to capacitors.
8. To find the distance to capacitor location, divide the total inductive line loss by capacitive line loss per thousand feet. If this quotient is greater than the line section length
   1. Divide the remaining inductive line loss by the capacitive line loss in the next line section to find the location
   2. If this quotient is still greater than the line section length, repeat step 8a
9. Prepare a voltage profile by hand calculations or by using a computer program for voltage profile and load analysis to determine the circuit voltages. If the profile shows that the voltages are inside the recommended limits, then the capacitors are installed at the location of minimum loss. If not, then use engineering judgment to locate them for the most effective voltage control application.