Three Phase System Outline

Single phase systems are defined by having an AC source with only one voltage waveform. Figure 1 is a simple AC circuit. Single-phase power distribution is widely used especially in rural areas, because the cost of a single-phase distribution network is low.

Figure 1:- Single phase system schematic diagram

Today most of the electrical power generated in the world is three-phase. Three-phase power was first conceived by Nikola Tesla. Three-phase power was the most efficient way that electricity could be produced, transmitted, and consumed. A three-phase generator has three separate but identical windings that are 1200 electrical apart from one another.

5.2 Three Phase Circuit

Three-phase voltage systems are composed of three sinusoidal voltages of equal magnitude, equal frequency and separated by 120 degrees, as shown in Figure 2.

It is one voltage cycle of a 3 phase system. It is labeled 0 to 360° (2 π radians) along the time axis. The plotted lines show the variation of instantaneous voltage (or current) over time. This power wave cycle will repeat usually 50 (50Hz), 60 (60Hz), or 400 (400Hz) times per second, depending on the power system frequency (Hz).

The colors of the lines are in the American Color Code for 3-phase wiring. It is black=VL1 red=VL2blue=VL3.

Figure 2:- Three phase waveforms

Three phase systems may or may not have a neutral wire. The neutral wire allows 3 phase systems to use a higher voltage while still supporting lower voltage 1 phase appliances. In high voltage 3 phase distribution situations it is common not to have a neutral wire as the loads can simply be connected between phases (phase-phase connection).

5.2.1 Advantage over Single Phase system

Three phase system is better to single phase system. The reason for the advantage over single phase system is given below.

The horsepower rating of three-phase motors and the KVA (kilo-volt-amp) rating of three-phase transformers is about 150% greater than for single-phase motors or transformers with a similar frame size.

Figure 3:- Single-phase power falls to zero three times each cycle.

Figure 4:- Three-phase power never falls to zero.

The power delivered by a single-phase system pulsates, as shown in Figure 3. The power falls to zero three times during each cycle. The power delivered by a three-phase circuit pulsates also, but it never falls to zero, as shown in Figure 4. In a three-phase system, the power delivered to the load is the same at any instant. This produces superior operating characteristics for three-phase motors.

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In a balanced three-phase system, the conductors need be only about 75% the size of conductors for a single-phase two-wire system of the same KVA rating. This helps offset the cost of supplying the third conductor required by three-phase systems.

If a magnetic field is rotate through the conductors of a stationary coil then a single phase alternating voltage can be produced. This explanation is shown in Figure 5.

Figure 5:- A single-phase voltage.

Since alternate polarities of the magnetic field cut through the conductors of the stationary coil, the induced voltage will change polarity at the same speed as the rotation of the magnetic field. The alternator shown in Figure 5 is single phase because it produces only one AC voltage.

Figure 6:- The voltages of a three-phase system are 120° out of phase with each other.

If three separate coils are spaced 120° apart, as shown in Figure 6, three voltages 120° out of phase with each other will be produced when the magnetic field cuts through the coils. This is the manner in which a three-phase voltage is produced.

5.2.2 Classification

Three-phase supply voltages and load systems have two basic configurations:

a). wye or star connection and

b). delta connection.

5.3 Star and Delta connection

The Wye is a 4-wire system. Wye configurations typically include a neutral line (N) connected to the common point (3 phase plus neutral for a total of four wires), as shown in Figure 7.

Figure 7:- A wye connections is formed by joining one end of each of the windings together.

The Delta, as shown in Figure 8, is a 3-wire system which is primarily used to provide power for three-phase motor loads. The system is normally ungrounded and has only one three-phase voltage available. The lack of a system ground makes it difficult to protect for ground faults. Often, a ground detection scheme, employing ground lamps, is used to provide an indication or alarm in the event of a system ground. The Delta System is sometimes corner grounded to protect for ground faults on the other two phases.

Figure 8:- Three-phase delta connection

5.4 Phasor diagrams

5.4.1 Star connection

The voltage measured across a single winding or phase is known as the phase voltage, as shown in Figure 9. The voltage measured between the lines is known as the line-to-line voltage or simply as the line voltage. The currents flowing in the phases are called phase currents and currents flowing in the lines are called line currents.

Figure 9:- Line and phase voltages are different in a wye connection.

The parallelogram method of vector addition for the voltages in a wye-connected three-phase system is shown in Figure 10. Figure 10 shows how the line voltage may be obtained using the normal parallelogram addition.

Figure 10:- Phasor diagram of Star connection

Voltage

However, the line voltage is not equal to the phase voltage.

The line voltage V1-2 is equal to the phasor difference of VA and VB’. The line voltage V2-3 is equal to the phasor difference of VB and VC’. The line voltage V3-1 is equal to the phasor difference of VC and VA’.

The line voltages are defined as: V1-2 = VA – VB’, V2-3 = VB-VC’, and V3-1 = VC-VA’.

Here V1-2, V2-3, V3-1 are the line voltage (VLine) and VA, VB, VC are the phase voltage (VPhase) of Wye connection. VA’, VB’, VC’ are the reverse phase voltage of VA,VB, VC.

The two phasors VA and VB’ are 600 apart.

V1-2 = VLine = VA – VB’

= [VPhase – (-VPhase)] cos(600/2)

= 2 VPhase cos300

= √3 VPhase

The two phasors VB and VC’ are 600 apart.

V2-3 = VLine = VB-VC’

= √3 VPhase

The two phasors VC and VA’ are 600 apart.

V3-1 = VLine = VC-VA’

= √3 VPhase

œ V1-2 = V2-3 = V3-1 = line voltage = VLine =√3 VPhase

Current

On a Wye system or star connected supply, the phase unbalance current is carried by the neutral. On a Wye system, the line current (current in the line) (ILine) is equal to the phase current (current in a phase) (IPhase) i.e.

ILine = IPhase

Power

Total power P = 3 – Power in each phase

= 3 – VPhase IPhase cosΦ

= 3 – (VLine/√3) – ILine – cosΦ [for Wye connection]

= √3 VLine ILine cosΦ

Where VLine and ILine are the line voltage and the line current of a star connected supply. The term cosΦ is called power factor of the circuit and its value is given by;

cosΦ = R/Z

Where R and Z are the resistance and impedance of a circuit.

5.4.2 DELTA CONNECTIONS

In Figure 11, voltmeters have been connected across the lines and across the phase. Ammeters have been connected in the line and in the phase.

Figure 11:- Voltage and current relationships in a delta connection

The delta connection is similar to a parallel connection because there is always more than one path for current flow. Since these currents are 120° out of phase with each other, vector addition must be used when finding the sum of the currents, as shown in Figure 12.

Figure 12:- Phasor Diagram of Delta connection

Voltage

In the delta connection, the three voltages are equal in magnitude but displaced 1200 from one another.

In the delta connection, line voltage (VLine) and phase voltage (Vphase) are the same.

VLine = Vphase

Current

In the delta connection, the line current and phase current are different. The line current is the vector sum of two individual phase currents. The line current I1 is equal to the phasor difference of IA and IC’. The line current I2 is equal to the phasor difference of IB and IA’. The line current I3 is equal to the phasor difference of IC and IB’.

The line currents are defined as: I1 = IA – IC’, I2 = IB – IA’ and I3 = IC – IB’.

Here I1, I2, I3 are the line current (ILine) and IA, IB, IC are the phase current (IPhase) of Wye connection. IA’, IB’, IC’ are the reverse phase current of IA, IB, IC.

The two phasors IA and IC’ are 600 apart.

I1 = ILine = IA – IC’

= [IPhase – (-IPhase)] cos(600/2)

= 2 IPhase cos300

= √3 IPhase

The two phasors IB and IA’ are 600 apart.

I2 = ILine = IB – IA’

= √3 IPhase

The two phasors IC and IB’ are 600 apart.

I3 = ILine = IC – IB’

= √3 IPhase

œ I1 = I2 = I3 = ILine = line current = √3 IPhase

However, the line current of a delta connection is higher than the phase current by a factor of the square root of 3 (1.732).

Power

Total power P = 3 – Power in each phase

= 3 – VPhase IPhase cosΦ

= 3 – VLine- (ILine/√3) – cosΦ [for delta connection]

= √3 VLine ILine cosΦ

Where VLine, ILine and cosΦ are the line voltage, the line current and power factor of a delta connected supply.

5.5 Relationship between line and phase quantities

5.5.1 Star connection

On a Wye system, the line current is equal to the phase current i.e.

ILine = IPhase

Where ILine and IPhase are the line current and phase current of Wye connection.

In a wye connected system, the line voltage is higher than the phase voltage by a factor of the square root of 3 (1.732). Two formulas used to compute the voltage in a wye connected system are:

VLine = √3 VPhase

= 1.732 – VPhase

œ VPhase = VLine / 1.732

Where VLine and VPhase are the line voltage and phase voltage of Wye connection.

5.5.2 Delta connection

In the delta connection, line voltage and phase voltage are the same.

VLine = Vphase

Where VLine and VPhase are the line voltage and phase voltage of delta connection.

Formulas for determining the current in a delta connection are:

Where ILine and IPhase are the line current and phase current of delta connection.

5.6 Power measurement by two watt meters method

In two wattmeters method, current coils of the two wattmeters are connected in any two terminals of Wye system, as shown in Figure 13. The algebraic sum of two wattmeters gives the total power consumed whether the load is balanced or not i.e.

Total power = W1 + W2

Figure 13:- Wye connected load

Figure 14:- Phasor Diagram

The power factor angle of load impedance being Φ lag. The currents will lag behind their respective phase voltages by Φ as shown in Fig. 14.

Current through current coil of W1 = IA.

Potential difference across potential coil of W1, V1-2 = VA – VB’.

The phase angle between V1-2 and IA is (300 + Φ).

œ W1 = V1-2 IA cos(300 + Φ)

Current through current coil of W2 = IB.

Potential difference across potential coil of W2, V2-3 = VB-VC’.

The phase angle between V2-3 and IB is (300 – Φ).

œ W2 = V2-3 IB cos(300 – Φ)

Here load is balanced, V1-2 = V2-3 = VLine = line voltage and IA = IB = ILine = line current.

œ W1 = VLine ILine cos(300 + Φ)

œ W2 = VLine ILine cos(300 – Φ)

œ W1 + W2 = VLine ILine [cos(300 + Φ) + cos(300 – Φ)]

= VLine ILine(2cos300cosΦ)

= √3VLine ILine cosΦ

œ W2 – W1 = VLine ILine [cos(300 – Φ) – cos(300 + Φ)]

= VLine ILine(2sin300sinΦ)

= VLine ILine sinΦ

tanΦ = [√3 (W2 – W1)] / (W1 + W2)

Thus from the two wattmeter method, we can find Φ.

PROBLEM

1. Three coils, each having a resistance of 20- and an inductive reactance of 15-, are connected in star to a 400V, 3-phase, 50Hz supply. Calculate (i) the line current (ii) power factor and (iii) power supplied.

Solution:-

VPhase = VLine / 1.732 = 400/1.732 = 231V

ZPhase = √(202 + 152) = 25-

(i) IPhase = VPhase/ ZPhase = 231/25 = 9.24A = ILine

(ii) Power factor = cosΦ = RPhase/ ZPhase = 20/25 = 0.8 lag

(iii) P = √3VLine ILine cosΦ = √3 – 400 – 9.24 – 0.8 = 5121W

2. A balanced star-connected load of impedance (6 + j8)- per phase is connected to a 3-phase, 230V, 50Hz supply. Find the line current and power absorbed by each phase.

Solution:-

ZPhase = √(62 + 82) = 10-

VPhase = VLine / 1.732 = 230/1.732 = 133V

Power factor = cosΦ = RPhase/ ZPhase = 6/10 = 0.6 lag

IPhase = VPhase/ ZPhase = 133/10 = 13.3A = ILine

P = √3VLine ILine cosΦ = √3 – 230 – 13.3 – 0.6 = 1061W

3. Three similar coils, connected in star, take a total power of 1.5kW at a power factor of 0.2 lagging from 3-phase, 400V, 50Hz supply. Calculate the resistance and inductance of each coil.

Solution:-

VPhase = VLine / 1.732 = 400/1.732 = 231V

P = √3VLine ILine cosΦ

œ ILine = P / (√3VLine cosΦ) = 1500 / (1.732 – 400 – 0.2) = 10.83A = IPhase

ZPhase= VPhase/ IPhase = 231 / 10.83 = 21.33-

RPhase = ZPhase cosΦ = 21.33 – 0.2 = 4.27-

XPhase = √(21.332 – 4.272) = 20.9-

LPhase = XPhase/ 2πf =20.9 / (2π – 50) = 0.0665H

4. The load to a 3-phase supply comprises three similar coils connected in star. The line currents are 25A and kVA and kW inputs are 20 and 11 respectively. Find (i) the phase and line voltages (ii) the kVAR input and (iii) resistance and reactance of each coil.

Solution:-

VPhase = Apparent power / (3 – IPh) = (20-103) / (3 – 25) = 267V

VLine= √3 VPhase=1.732-267 = 462V

Input kVAR = √ (kVA2 – kW2) = √ (202 – 112) = 16.7kVAR

Power factor = cosΦ = kW/kVA = 11/20

ZPhase= VPhase/ IPhase = 267 / 25 = 10.68-

RPhase = ZPhase cosΦ = 10.68 – 11/20 = 5.87-

XPhase = √(10.682 – 5.872) = 8.92-

5. A balanced 3-phase, delta-connected load has per phase impedance of (25+j40)-. If 400V, 3-phase supply is connected to this load, find (i) phase current (ii) line current (iii) power supplied to the load.

Solution:-

ZPhase = √(252 + 402) = 47.17-

IPhase= VPhase/ ZPhase = 400 / 47.17 = 8.48-

ILine= √3 IPhase=1.732-8.48 = 14.7A

Power factor = cosΦ = RPhase/ ZPhase = 25/47.17 = 0.53 lag

P = √3VLine ILine cosΦ = √3 – 400 – 14.7- 0.53 = 5397.76W

6. A balanced 3-phase load consists of three coils, each of resistance 6-, and inductive reactance of 8-. Determine the line current and power absorbed when the coils are delta-connected across 400V, 3-phase supply.

Solution:-

ZPhase = √(62 + 82) = 10-

cosΦ = RPhase/ ZPhase = 6/10 = 0.6 lag

VPhase = VLine = 400V

IPhase= VPhase/ ZPhase = 400 / 10 = 40A

ILine= √3 IPhase=1.732-40 = 69.28A

P = √3VLine ILine cosΦ = √3 – 400 – 69.28 – 0.6 = 28799W

7. Two-wattmeter method is used to measure the power absorbed by a 3-phase induction motor. The wattmeter readings are 12.5kW and -4.8kW. Find (i) the power absorbed by the machine (ii) load power factor (iii) reactive power taken by the load.

Solution:-

W2 = 12.5kW ; W1 = -4.8kW

Power absorbed = W2 + W1 = 12.5 + (-4.8) = 7.7kW

tanΦ = [√3 (W2 – W1)] / (W1 + W2) = (12.5+4.8) / 7.7 = 3.89

Φ = tan-13.89 = 75.60

Power factor = cosΦ = cos75.60 = 0.2487lag

Reactive power = √3 (W2-W1) = √3 (12.5 + 4.8) = 29.96kVAR

P O I N T S TO REMEMBER

1. The voltages of a three-phase system are 120° out of phase with each other.

2. The two types of three-phase connections are wye and delta.

3. Wye connections are characterized by the fact that one terminal of each device is connected together.

4. In a wye connection, the phase voltage is less than the line voltage by a factor of 1.732. The phase current and line current are the same.

5. In a delta connection, the phase voltage is the same as the line voltage. The phase current is less than the line current by a factor of 1.732.

IMPORTANT FORMULAE

1. On a wye system, the relation between line and phase current is:

ILine = IPhase

2. On a wye system, the line voltages are defined as:

V1-2 = VA – VB’, V2-3 = VB-VC’, and V3-1 = VC-VA’.

3. In the delta connection, the relation between line and phase voltage is:

VLine = Vphase

4. In the delta connection, the line currents are defined as:

I1 = IA – IC’, I2 = IB – IA’ and I3 = IC – IB’

5. On a wye system, the relation between line and phase voltage is:

VPhase = VLine / 1.732

6. In the delta connection, the relation between line and phase current is:

OBJECTIVE QUESTIONS

1. In a two phase generator, the electrical displacement between the two phases or winding is:

(a) 1200 (b) 900 (c) 1800 (d) none of these

2. The advantage of star-connected supply system is that:

(a) line current is equal to phase current (b) two voltages can be used

(c) phase sequence can be easily changed (d) it is a simple arranged

3. In a balanced star-connected system, line voltage are ………… ahead of their respective phase voltages.

(a) 300 (b) 600 (c) 1200 (d) none of these

4. In a star connected system, the relationship between the line voltage VL and phase voltage VPh is:

(a) VL = VPh (b) VL = VPh / √3 (c) VL = √3VPh (d) none of these

5. The algebraic sum of instantaneous phase voltages in a three-phase circuit is equal to:

(a) zero (b) line voltage (c) phase voltage (d) none of these

6. If one line conductor of a 3-phase line is cut, the load is then supplied by:

(a) single phase voltage (b) two phase voltage

(c) three phase voltage (d) none of these

7. The resistance between any two terminals of a balanced star-connected load is 12-. The resistance of each phase is:

(a) 12- (b) 24- (c) 6- (d) none of these

8. A 3-phase load is balanced if all the three phases have the same

(a) impedance (b) power factor

(c) impedance and power factor (d) none of these

REVIEW QUESTIONS

1. How many degrees out of phase with each other are the voltages of a three-phase system?

2. What are the two main types of three-phase connections?

3. A wye-connected load has a voltage of 480 V applied to it. What is the voltage drop across each phase?

4. A wye-connected load has a phase current of 25 A. How much current is flowing through the lines supplying the load?

5. A delta connection has a voltage of 560 V connected to it. How much voltage is dropped across each phase?

6. A delta connection has 30 A of current flowing through each phase winding. How much current is flowing through each of the lines supplying power to the load?

7. A three-phase resistive load has a phase voltage of 240 V and a phase current of 18 A. What is the power of this load?

8. If the load in question 7 is connected in a wye, what would be the line voltage and line current supplying the load?

9. An alternator with a line voltage of 2400 V supplies a delta-connected load. The line current supplied to the load is 40 A. Assume the load is a balanced three-phase load, what is the impedance of each phase?

10. If the load is pure resistive, what is the power of the circuit in question 9?

PRACTICE PROBLEMS

1. Three similar coils are star connected to a 3-phase, 400V, and 50Hz supply. If the inductance and resistance of each coil are 38.2mH and 16- respectively, determine (i) line current (ii) power factor (iii) power consumed.

2. Three 50- resistors are connected in star across 400V, 3-phase supply. (i) Find phase current, line current and power taken from the main. (ii) What would be the above value if one of the resistors were disconnected?

3. Calculate the active and reactive components of current in each phase of a star-connected 10,000 volts, 3-phase generator supplying 5,000kW at a lagging power factor 0.8. Find the new output if the current is maintained at the same value but the power factor is raised to 0.9 lagging.

4. Three 20µF capacitors are star-connected across 420V, 50Hz, 3-phase, three wire supplies. (i) Calculate the current in each line.

(ii) If one of the capacitors is short-circuited, calculate the line currents.

(iii) If one of the capacitors is open-circuited, calculate the line currents and potential difference across each of the other two capacitors.

5. If the phase voltage of a 3-phase star connected alternator be 231V, what will be the line voltages (i) when the phases are correctly connected (ii) when the connections of one of the phases are reversed?

6. Calculate the phase and line currents in a balanced delta connected load taking 75kw at a power factor 0.8 from a 3-phase 440V supply.

7. Three identical resistances, each of 18-, are connected in delta across 400V, 3-phase supply. What value of resistance in each leg of balanced star connected load would take the same line current?

8. Three similar resistors are connected in star across a 415V, 3-phase supply. The line current is 10A. Calculate (i) the value of each resistance (ii) the line voltage required to give the same line current if the resistors were delta-connected.

9. Two wattmeters are used to measure power in a 3-phase balanced load. The wattmeter readings are 8.2kW and 7.2kW. Calculate (i) total power (ii) power factor and (iii) total reactive power.

10. A balanced 3-phase load takes 10kW at a power factor of 0.9 lagging. Calculate the readings on each of the two wattmeters connected to read the input power.

11. Three identical coils, each having a resistance of 20- and a reactance of 20- are connected in (i) star (ii) delta across 440v, 3-phase lines. Calculate for each method of connection the line current and readings on each of the two wattmeters connected to measure the power.