Three Phase

THREE-PHASE CIRCUITS: Review

Power systems are all three-phase systems. Electric power is generated by the generator which is then transmitted to the load over transmission lines. The configuration looks as follows:

Generator

Transmission Lines

Load

Fig. 1 A three-phase power system configuration

Balanced Three Phase Circuits Three-phase circuits are just like three single-phase circuits,

• But with a phase difference between them How much is the phase difference?

• 1200

1200

Fig. 2 Phasor diagram for three-phase voltage with phase sequence abc

For a balanced three-phase system:

•

Va  Vb  Vc Magnitudes are equal.

• Angle between the phasors is 1200 For an unbalanced system

• Any or all of the above two conditions not being satisfied

The Generator Three-phase synchronous generators are normally connected in Y with the neutral grounded. However, they may be connected in Δ also. The connection diagrams look as follows:

Fig. 3 Y-connected generator

Fig. 4 Δ-connected generator

The Load Similar to the generators, the loads also may be Y or Δ connected.

Fig. 5 Y-connected load

Fig. 6 Δ-connected load

For a three phase Y-Y connection, circuit diagram will look like:

Fig. 7 Circuit diagram for generator and load connected in Y

For a balanced three phase system, the neutrals of both generator and load circuit are at zero potential. No current flows between the neutrals or ground (Ig=0). Hence, we can perform the voltage current calculations for only one phase. The quantities for the other phases can be obtained by simply shifting them by 1200. The equivalent circuit is then,

Fig. 8 Equivalent circuit for balanced three phase Y-Y circuit

Y- Δ CIRCUITS If the load is connected in Δ, it can be converted to a Y by the relationship,

ZY 

Z 3

And the circuit can be analyzed as Y-Y.

Voltage-Current-Power of Y-connected System Consider a Y-connected load. We will derive the relationships of voltage, current and power for this connection. IAN

IBN

ICN

θ

IAN

Fig. 9 Three phase Y-connection and phasor diagram

Assume that we are given the phase voltages (sequence ABC): VAN  V 00 ; V is the magnitude of phase voltage VBN  V   1200 VcN  V 1200 We want to find the line voltages VAB, VBC and VCA. Using KVL, VAB = VAN+VNB = VAN - VBN 0 0 =V 0  V 60 0 = 3V 30

This can be seen in the phasor diagram. Similarly, you can find the other line voltages as, VBC  VBN  VCN  V   1200  V 1200  3V   900 VCA  VCN  VAN  V 1200  V 00  3V 1500 See the phasor diagram above. For the Y-connected three phase system, we observe that: 1) Line voltage =√3 Phase Voltage 2) Line current, IL = Phase current, Iφ 3) Line voltage VAB is ahead of phase voltage VAN by 300 4) Total power, PT = 3 Power per phase = 3(Vφ Iφ cosθ) = √3 VL IL cosθ

5) Similarly, total reactive power QT= √3 VL IL sinθ 6) The apparent power (or VA)= S 

PT2  QT2  3VL I L

Note: The power factor angle θ is the angle between phase voltage VAN and phase current IAN. Voltage-Current-Power of Δ -connected System Consider now a Δ-connected load. The circuit connection and phasor diagram showing the voltages and currents for the balanced circuit is shown below.

VAB

θ

Fig. 10 A Δ-connected load

We assume the phase currents, I AB  I00 , I BC  I  1200 , I CA  I1200 Applying KCL, the line currents are found as, I aA  I AB  I CA  3I   300 Similarly, I bB  3I   1500 , I cC  3I 900 For the Δ -connected three phase system, we observe that 1) Line current =√3 Phase current, Iφ

2) Line voltage= Phase voltage 3) Line current IaA is behind phase current IAB by 300 4) Total power, PT = 3 Power per phase = 3(Vφ Iφ cosθ) = √3 VL IL cosθ 5) Similarly, total reactive power QT= √3 VL IL sinθ 6) The apparent power (or VA)= S  PT2  QT2  3VL I L Note again: The power factor angle θ is the angle between phase voltage VAB and phase current IAB.

Summary of Voltage-Current-Power Relationships The total real power, reactive power, complex power and apparent power, respectively for a three phase circuit (Y or Δ connected) are given as: PT=√3 VL IL cos θ QT=√3 VL IL sin θ ST= P+jQ= 3 (Vφ Iφ*) |ST|=√3 VL IL where, VL : line voltage IL: line current θ: Angle between phase voltage and phase current Vφ: phase voltage Iφ: phase current For Y-connected circuits: IL=Iφ ; VL=√3Vφ 300 (Note the exact voltages from the phasor diagram) For Δ-connected circuits VL= Vφ ; IL=√3Iφ   300 (Note the exact currents from the phasor diagram)

Problems 1. Three impedances each having resistance of 20 Ω and inductive reactance of 15 Ω are connected in Y across a 400 V, 3-φ supply. Calculate a. The line current b. The power factor c. The total power, reactive power and apparent power absorbed by the load. 2. A 220-V, 3-φ voltage is applied to a Δ-connected 3-φ load of phase impedance (15+j20) Ω. Find a. Current in each phase b. Current in each line c. The sum of line currents d. The real, reactive and apparent power absorbed 3. A Δ-connected load consists of three identical impedances of 45600  per phase. It is connected to a 3-φ, 208 V source by a three phase feeder with impedance of (1.2+j1.6) Ω per phase. a. Calculate line to line voltage at the load terminals b. A Δ-connected capacitor bank of 60 Ω is connected in parallel with

the load at its terminals. Find the resulting line to line voltage when the capacitor is added. 4. A three-phase motor draws 50KVA at 0.8pf (lagging) from a 220-V source. It is desired to make the power factor 0.9 lagging by adding a capacitor bank across the terminal of the motor. a. Calculate the line current before and after addition of the bank b. Determine the KVA rating of the capacitor and the value of the capacitance per phase. Consider 60 Hz operation.

Power Measurements The wattmeter is a device that measures average (real) power absorbed by a two-terminal load. The device is depicted in the following diagram.

Fig. 11 A watt-meter

The connection diagram is shown in the following. It contains two coils. One coil called the current coil is designed to carry a current proportional to the load current (like ammeter). The second coil, called the potential coil carries a current proportional to the load voltage (like a voltmeter). The reading of the meter is proportional to the product of the effective value of current, the voltage impressed and cosine of the angle between voltage and current.

P=V I cos angle between V and I V and I are the effective values of v and I, respectively.

Fig. 12 Wattmeter connection for single phase circuit

Power Measurements in Three Phase circuits For measuring the power of a balanced three phase circuit, the power for each phase can be measured. The total power will be 3 times the individual readings. The following 2 connection diagrams show the wattmeter connection for Y and Δ-connected loads.

Fig.13 Wattmeter connections for balanced circuits

For measurement of power for unbalanced circuits, three watt-meters could be connected to the three phases. However, a better alternative is to use two watt-meters to measure the total three phase power. This can be used for balanced as well as unbalanced loads. Two Wattmeter Method of Measurement The connection and phasor diagram are shown for an assumed abc phase sequence and lagging power factor.

Fig. 14 Two-wattmeter method- connection diagram and phasor diagram

The watt-meter readings are given by, W1  Vab I a cos (Vab , I a ) =VL I L cos(300   ) W2  Vcb I c cos (Vcb , I c ) =VL I L cos(300   )

(1)

(2)

The sum of the two watt-meter readings gives the total three phase power, PT  W1  W2  VL I L [cos(300   )  cos(300   )] = 3VL I L

(3)

The difference of the two watt-meter readings is W2  W1  VL I L [cos(300   )  cos(300   )] =VL I L sin 

(4)

The total reactive power is, then, QT  3(W2  W1 )

(5)

The power factor angle can also be found from,

  tan 1 (

QT 3(W2  W1 ) )  tan 1[ ] PT W2  W1

(6)

Note: When θ=00

power factor=1

W1=W2 [From eq. (1) and (2)]

θ=600

pf=0.5

W1=0, W2>0~

θ=900

pf=0

W1=-W2

~

For 600    900 , one of the watt-meters will give negative readings. In the laboratory, when you have made the proper watt-meter connections, you will observe that one of the watt-meters is trying to read backwards. After switching the power supply off, reverse the connection of the voltage coil or the current coil (not both). The meter will now read upscale. Assign a negative sign to this reading. Problems 1) Two watt-meters are used to measure power input to a 3-φ load. If the watt-meter readings are 12.5 KW and -4.8 KW, calculate the input power and power factor of the load. What will be the line current if the supply voltage is 440 V? 2) A 220-V, 3-φ, voltage is applied to a Δ-connected load of phase impedance (15+j20) Ω. What will be the 2 watt-meter readings?