2 Three-Phase Circuits
2.5 POWER FACTOR CORRECTION
In general, loads on electric utility systems have two components: active power (measured in kilo-watts) and reactive power (measured in kilovars). Active power has to be generated at power plants, whereas reactive power can be supplied by either power plants or capacitors. If reactive power is supplied only by power plants, each system component, including generators, transformers, and transmission and distribution lines, has to be increased in size accordingly. However, by using capacitors, the reactive power demand as well as line currents are reduced from the capacitor loca-tions all the way back to power plants.* As a result, losses and loadings are reduced in distribution lines, substation transformers, and transmission lines. The power-factor correction generates sav-ings in capital expenditures and fuel expenses through a release of power capacity and a decrease in power losses in all the equipment between the point of installation of the capacitors and the source power plants.
The economic power factor is the power factor at which the economic benefits of using capaci-tors equals the cost of capacicapaci-tors. However, the correction of power factor to unity becomes more expensive with respect to the marginal cost of the capacitors installed. It has been found in practice that the economic power factor is about 0.95. In distribution systems, including industrial applica-tions, shunt capacitors are used and are connected in delta or wye. However, in transmission sys-tems, the capacitors are connected in series with the line involved.
Example 2.8
Assume that a 2.4 kV, single-phase circuit supplies a load of 294 kW at lagging power factor and that the load current is 175 A. To improve the power factor, determine the following:
(a) The uncorrected power factor and reactive load
(b) The new corrected power factor after installing a shunt capacitor unit with a rating of 200 kvar
Solution
(a) Before the power factor correction,
Sold=VI=(2 4 kV)(175 A) 42 kVA. = 0
Therefore, the uncorrected power factor can be found as
PF P
old S
old
= =
= =
cos
. θ
294
420kW 0 7
*For further information, see Chapter 8 of Electric Power Distribution System Engineering of Gönen (2008).
and the reactive load is
(b) After the installation of the 200 kvar capacitors, Qnew=Qold−Qcap
=(300kvar) (2− 00kvar) 1= 00kvar
and therefore the new (or corrected) power factor is
PF P
A three-phase, 400 hp, 60 Hz, 4.16 kV wye-connected induction motor has a full-load efficiency of 86%, and a lagging power factor of 0.8. If it is necessary to correct the power factor of the load to a lagging power factor of 0.95 by connecting three capacitors at the load, find the following:
(a) The rating of such a capacitor bank in kvar
(b) The capacitance of each single-phase unit, if the capacitors are connected in delta in μF (c) The capacitance of each single-phase unit, if the capacitors are connected in wye in μF Solution
(a) The input power of the induction motor is
P=( )( . )=
. .
400hp 07457 kW/hp 346 84 kW 0 86
The reactive power of the motor at the uncorrected power factor is Qold= ×P
The reactive power of the motor at the corrected power factor is Qnew= ×P new
= ×
= ×
−
tan
(346 84 kW) tan(cos 95) (346 84 kW) 3287
Thus, the reactive power provided by the capacitor bank is Qcap Qold Qnew
(26 13 kvar) (114 kvar) 146 13 kvar
Therefore, the rating of the capacitor bank is 146.13 kvar.
(b) If the capacitors are connected in delta, the line current is
IL VL
and thus, the current in each capacitance of the delta-connected capacitor bank is
I I
Therefore, the reactance of each capacitor is
X V
and the capacitance of each unit is
C= X = f X
Note that the aforementioned equation gives the capacitance in μF. This equation is modi-fied from the previous equation, which gives the answer in F, by dividing both sides by 106, as it can be seen easily.
(c) If the capacitors are connected in wye in the capacitor bank, Icap=IL= 2 28A0.
Therefore, the capacitance of each unit is
C= system with line-to-line voltage of 220 V and the impedance of the motor is 6.3 + j3.05 Ω per phase. Determine the following:
(a) The magnitude of the line current (b) The power factor of the motor
(c) The three-phase average power consumed by the motor (d) The current in the neutral wire.
2.2 A balanced three-phase load of 15 MVA with a lagging load factor of 0.85 is supplied by a 115 kV subtransmission line. If the line impedance is 50 + j100 Ω per phase, determine the following:
(a) The line current of the load
(b) The power loss due to the line impedance (c) The power factor angle of the load
(d) The line-to-neutral voltage of the line at the receiving end (e) The voltage drop due to the line impedance
(f) The line-to-line voltage of the line at the sending end
2.3 A balanced, three-phase, delta-connected load is supplied by a balanced, three-phase, wye-connected source over a balanced three-phase line. The source voltages are in abc phase sequence in which Van is 19.94∠0° kV and the line impedance is 10 + j80 Ω per phase. If the balanced load consists of three equal impedances of 60 + j30 Ω, determine the following:
(a) The line currents Ia, Ib, and Ic
(b) The phase voltages Vab, Vbc, and Vca of the load (c) The phase currents Iab, Ibc, and Ica of the load
2.4 Assume that the impedance of a power line connecting buses 1 and 2 is 50∠70° Ω, and that the bus voltages are 7560∠− 10° and 7200∠0° V per phase, respectively. Determine the following:
(a) The real power per phase that is being transmitted from bus 1 to bus 2 (b) The reactive power per phase that is being transmitted from bus 1 to bus 2
(c) The complex power per phase that is being transmitted
2.5 Solve Problem 2.4 assuming that the line impedance is 50∠26° per phase.
2.6 An unbalanced three-phase, three-wire, wye-connected load is connected to a balanced, three-phase, three-wire, wye-connected source, as shown in Figure P2.6. If the line-to-neutral source voltages Va, Vb, and Vc are 220∠30°, 220∠270°, and 220∠150° V, respectively, and the load impedances Za, Zb, and Zc are 4∠0°, 5∠90°, and 8∠30° Ω per phase, respectively.
Determine the following:
(a) The mesh currents I1 and I2 using determinants and Cramer’s rule (b) The line currents Ia, Ib, and Ic.
(c) The potential difference between the source neutral NS and the common node of the load, that is, NL
(d) Whether or not a neutral wire connecting the neutral point NS and NL is required
2.7 An unbalanced, three-phase, three-wire, wye-connected load is connected to a balanced, three-phase, three-wire, wye-connected source, as shown in Figure P2.6. If the line-to-neutral source voltages Va, Vb, and Vc are 220∠30°, 220∠270°, and 220∠150° V, respectively, and the load impedances Za, Zb, and Zc are 4∠0°, 5∠90°, and 8∠30° Ω per phase, respectively, as given in Problem 2.6. Assume that three wattmeters are connected to measure the total power received by the unbalanced three-phase load, as shown in Figure P2.7. Ignore the small impedance of the current coils in the wattmeters and determine the following:
(a) The power recorded on each wattmeter (b) The total power absorbed by the load
2.8 An unbalanced, three-phase, three-wire, wye-connected load is connected to a balanced, three-phase, three-wire, wye-connected source, as shown in Figure P2.6. If the line-to-neutral
C C
FIGURE P2.6 Circuit of Problem 2.6.
a
FIGURE P2.7 Circuit of Problem 2.7.
source voltages Va, Vb, and Vc are 220∠30°, 220∠270°, and 220∠150° V, respectively, and the load impedances Za, Zb, and Zc are 4∠0°, 5∠90°, and 8∠30° Ω per phase, respectively, as given in Problem 2.6. Assume that three wattmeters are connected to measure the total power received by the unbalanced three-phase load, as shown in Figure P2.7. Ignore the small imped-ance of the current coils in the wattmeters. An unbalimped-anced, three-phase, three-wire, wye-connected load is wye-connected to a balanced three-phase, three-wire, wye-wye-connected source, as shown in Figure P2.6. If the line-to-neutral source voltages Va, Vb, and Vc are 220∠30°, 220∠270°, and 220∠150° V, respectively, and the load impedances Za, Zb, and Zc are 4∠0° Ω, 5∠90° Ω, and 8∠30° Ω per phase, respectively, as given in Problem 2.6. Assume that three wattmeters are connected to measure the total power received by the unbalanced three-phase load, as shown in Figure P2.7. Ignore the small impedance of the current coils in the wattme-ters and assume that only two wattmewattme-ters are used and connected, as shown in Figure P2.8.
Determine the following:
(a) The power recorded on each wattmeter (b) The total power absorbed by the load
2.9 If the impedance of a transmission line connecting buses 1 and 2 is 50∠80° Ω, and the bus voltages are 70∠215° and 68∠0° kV per phase, respectively, determine the following:
(a) The complex power per phase that is being transmitted from bus 1 to bus 2 (b) The active power per phase that is being transmitted
(c) The reactive power per phase that is being transmitted
2.10 A three-phase motor is connected to a three-phase line that has an abc phase sequence and is supplied by 15 A current at a 0.85 lagging power factor. If a single-phase motor withdrawing 5 A current at a 0.707 lagging power factor is connected across lines a and b of the three-phase power line, determine the total current in each line.
2.11 Three loads are connected to a 208 V, three-phase power source that has an abc phase sequence. The first load is a wye-connected, three-phase motor withdrawing a line current of 20 A at a 0.8 lagging power factor. The second load is a single-phase load between lines a and b and withdraws a 10 A current at a 0.8 leading power factor. The third load is also a single-phase motor connected between lines b and c and withdraws a 7 A current at a 0.707 lagging power factor. Use the voltage Vab as the reference phasor and determine the following:
(a) All line and phase voltages (b) All line currents
(c) The total input power in watts
2.12 A three-phase, 60 Hz, wye-connected synchronous generator is providing power to two balanced three-phase loads. The first load is delta-connected and made up of three 12∠45°
a
b B
NL A
c C
ZA ZC WA
WC
+ +
+ +
IC IB IA
ZB
FIGURE P2.8 Circuit of Problem 2.8.
impedances, while the second load is wye connected and made up of three 5∠60° Ω imped-ances. Determine the following:
(a) Total (i.e., equivalent) load impedance per phase (i.e., line to neutral).
(b) The line current Ia at the generator terminal. Use Va=
(
208 3/)
∠ ° ≅0 120 0∠ ° V.(c) The total complex power provided by the generator.
2.13 A three-phase, 60 Hz, wye-connected synchronous generator has balanced line-to-line volt-ages of 480 V at its terminals. The generator is supplying power to two balanced and delta-connected, three-phase loads. The first load is made up of three 15∠−30° Ω impedances, while the second load is made up of three 18∠50° Ω impedances. Determine the following:
(a) Total (i.e., equivalent) load impedance per phase (i.e., line to neutral).
(b) The line current Ia at the generator terminal. Use a phase voltage of Va = 277.1281∠0° V (since 48 /0 3V 277 1281 V= . ).
(c) The total complex power provided by the generator.
2.14 If a balanced, three-phase, 15 MW total load is fed by a 138 kV power line at a 0.85 lagging