Three Phase System Outline

third Phase System Out foldSingle strain dusts ar define by having an AC source with only wizard emf waveform. Figure 1 is a simple AC circuit. Single- arrange spring distribution is widely used especially in rural argonas, because the cost of a single- descriptor distribution nedeucerk is low.Figure 1- Single contour system schematic diagramToday well-nigh of the electrical world power generated in the world is trey- bod. Three- cast 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 beginning has three separate but identical windings that atomic number 18 1200 electrical apart from one another.5.2 Three Phase CircuitThree-phase potential 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 radians) alon g the time axis. The plotted borders show the variation of insburningtaneous voltage (or ongoing) over time. This power wave cycle leave alone repeat usually50 (50Hz), 60 (60Hz), or 400 (400Hz)times per second, depending on the power systemfrequency (Hz).The colors of the gillyflowers are in theAmerican Color Code for 3-phase wiring. It is black=VL1red=VL2blue=VL3.Figure 2- Three phase waveformsThree phase systems may or may not throw away a neutral wire. The neutral wire allows 3 phase systems to use a higher voltage while still supporting lower voltage 1 phase appliances. Inhigh voltage 3 phase distributionsituations it is common not to have a neutral wire as the thins can just be affiliated betwixt phases (phase-phase nexus).5.2.1 Advantage over Single Phase systemThree 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 gets and the KVA (kilo-volt-amp) rating of three -phase transformers is ab appear 150% great than for single-phase motors or transformers with a convertible frame size.Figure 3- Single-phase power falls to naught three times for for all(prenominal) one one 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 severally 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 cut is the identical at any instant. This produces superior operating characteristics for three-phase motors.In a fit three-phase system, the conductors need be only about 75% the size of conductors for a single-phase both-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 nonmoving 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 ClassificationThree-phase supply voltages and incubus systems have two basic configurationsa). wye or star connection andb). delta connection.5.3 hotshot and Delta connectionThe Wye is a 4-wire system. Wye configurations typically include a neutra l plication (N) affiliated 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 create 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 un commonwealthed 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 signal 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 connection5.4 Phasor diagrams5.4.1 Star connectionThe voltage measured crosswise a single winding or phase is known as the phase voltage, as shown in Figure 9. The voltage measured between the cables lengths is known as the dema rcation line-to-line voltage or simply as the line voltage. The actuals 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 rule 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 emfHowever, 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 define 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 reve rse 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 VPhaseThe two phasors VB and VC are 600 apart.V2-3 = VLine = VB-VC= 3 VPhaseThe two phasors VC and VA are 600 apart.V3-1 = VLine = VC-VA= 3 VPhase V1-2 = V2-3 = V3-1 = line voltage = VLine =3 VPhase watercourseOn 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 = IPhasePowerTotal power P = 3 Power in each phase= 3 VPhase IPhase cos= 3 (VLine/3) ILine cos for Wye connection= 3 VLine ILine cosWhere VLine and ILine are the line voltage and the line current of a star connected supply. The term cos is called power part of the circuit and its apprise is given bycos = R/ZWhere R and Z are the ohmic resistance and impedance of a circuit.5.4.2 DELTA CONNECTIONSIn Fig ure 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 connectionThe delta connection is convertible 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 connectionVoltageIn 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 = VphaseCurrentIn 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 differe nce of IB and IA. The line current I3 is equal to the phasor difference of IC and IB.The line currents are delimitate 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 IPhaseThe two phasors IB and IA are 600 apart.I2 = ILine = IB IA= 3 IPhaseThe two phasors IC and IB are 600 apart.I3 = ILine = IC IB= 3 IPhase I1 = I2 = I3 = ILine = line current = 3 IPhaseHowever, the line current of a delta connection is higher than the phase current by a factor of the square fore of 3 (1.732).PowerTotal power P = 3 Power in each phase= 3 VPhase IPhase cos= 3 VLine- (ILine/3) cos for delta connection= 3 VLine ILine cosWhere VLine, ILine and cos are the line voltage, the line current and power factor of a delta connected supply. 5.5 Relationship between line and phase quantities5.5.1 Star connectionOn a Wye system, the line current is equal to the phase current i.e.ILine = IPhaseWhere 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 guess the voltage in a wye connected system areVLine = 3 VPhase= 1.732 VPhase VPhase = VLine / 1.732Where VLine and VPhase are the line voltage and phase voltage of Wye connection.5.5.2 Delta connectionIn the delta connection, line voltage and phase voltage are the same.VLine = VphaseWhere VLine and VPhase are the line voltage and phase voltage of delta connection.Formulas for determining the current in a delta connection areWhere ILine and IPhase are the line current and phase current of delta connection.5.6 Power measurement by two watt meters methodIn 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 + W2Figure 13- Wye connected loadFigure 14- Phasor DiagramThe power factor angle of load impedance being lag. The currents will lag behind their respective phase voltages by 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 + ). W1 = V1-2 IA cos(300 + )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 ). W2 = V2-3 IB cos(300 )Here load is balanced, V1-2 = V2-3 = VLine = line voltage and IA = IB = ILine = line current. W1 = VLine ILine cos(300 + ) W2 = VLine ILine cos(300 ) W1 + W2 = VLine ILine cos(300 + ) + cos(300 )= VLine ILine(2cos300cos)= 3VLine ILine cos W2 W1 = V Line ILine cos(300 ) cos(300 + )= VLine ILine(2sin300sin)= VLine ILine sintan = 3 (W2 W1) / (W1 + W2) therefrom from the two wattmeter method, we can find .PROBLEM1. 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. encipher (i) the line current (ii) power factor and (iii) power supplied. origin-VPhase = VLine / 1.732 = 400/1.732 = 231VZPhase = (202 + 152) = 25-(i) IPhase = VPhase/ ZPhase = 231/25 = 9.24A = ILine(ii) Power factor = cos = RPhase/ ZPhase = 20/25 = 0.8 lag(iii) P = 3VLine ILine cos = 3 400 9.24 0.8 = 5121W2. A balanced star-connected load of impedance (6 + j8)- per phase is connected to a 3-phase, 230V, 50Hz supply. picture the line current and power absent-minded by each phase.Solution-ZPhase = (62 + 82) = 10-VPhase = VLine / 1.732 = 230/1.732 = 133VPower factor = cos = RPhase/ ZPhase = 6/10 = 0.6 lagIPhase = VPhase/ ZPhase = 133/10 = 13.3A = ILineP = 3VLine ILine cos = 3 23 0 13.3 0.6 = 1061W3. 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 = 231VP = 3VLine ILine cos ILine = P / (3VLine cos) = 1500 / (1.732 400 0.2) = 10.83A = IPhaseZPhase= VPhase/ IPhase = 231 / 10.83 = 21.33-RPhase = ZPhase cos = 21.33 0.2 = 4.27-XPhase = (21.332 4.272) = 20.9-LPhase = XPhase/ 2f =20.9 / (2 50) = 0.0665H4. 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) = 267VVLine= 3 VPhase=1.732-267 = 462VInput kVAR = (kVA2 kW2) = (202 112) = 16.7kVARPower factor = cos = kW/kVA = 11/20ZPhase= VPhase/ IPhase = 267 / 25 = 10.68-RPhase = ZPhase cos = 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 = 14.7APower factor = cos = RPhase/ ZPhase = 25/47.17 = 0.53 lagP = 3VLine ILine cos = 3 400 14.7- 0.53 = 5397.76W6. 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-cos = RPhase/ ZPhase = 6/10 = 0.6 lagVPhase = VLine = 400VIPhase= VPhase/ ZPhase = 400 / 10 = 40AILine= 3 IPhase=1.732-40 = 69.28AP = 3VLine ILine cos = 3 400 69.28 0.6 = 28799W7. Two-wattmeter method is used to measure the power abs orbed by a 3-phase induction motor. The wattmeter take awayings 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 = -4.8kWPower absorbed = W2 + W1 = 12.5 + (-4.8) = 7.7kWtan = 3 (W2 W1) / (W1 + W2) = (12.5+4.8) / 7.7 = 3.89 = tan-13.89 = 75.60Power factor = cos = cos75.60 = 0.2487lagReactive power = 3 (W2-W1) = 3 (12.5 + 4.8) = 29.96kVARP O I N T S TO call up1. 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 FORMULAE1. On a wye system, the relation between line and phase current isILine = IPhase2. On a wye system, the line voltages are defined asV1-2 = VA VB, V2-3 = VB-VC, and V3-1 = VC-VA.3. In the delta connection, the relation between line and phase voltage isVLine = Vphase4. In the delta connection, the line currents are defined asI1 = IA IC, I2 = IB IA and I3 = IC IB5. On a wye system, the relation between line and phase voltage isVPhase = VLine / 1.7326. In the delta connection, the relation between line and phase current isOBJECTIVE QUESTIONS1. In a two phase generator, the electrical displacement between the two phases or winding is(a) 1200 (b) 900 (c) 1800 (d) none of these2. 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 arranged3. In a balanced star-connected system, line voltage are ahead of their respective phase voltages.( a) 300 (b) 600 (c) 1200 (d) none of these4. 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 these5. 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 these6. 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 these7. 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 these8. 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 theseREVIEW QUESTIONS1. How many degrees out of phase with each other are the voltages of a three-phase system?2. What are the two main(prenominal) 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 practically 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 PROBLEMS1. 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 20F 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 winning 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 lo ad. 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.