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Balanced poly phase circuits Two and four phase systems A two phase system is an electrical system in which the voltages of the phases are 90 degree out of time phase. The emf’s Eac and Ebd are 90 degree out of time phase. A two phase system is the equivalent of two separate single-phase systems that are separated 90 degree in time phase. If the connection is made between the two windings at n and n`, the system would be called a four-phase system. The voltages Eda, Eab, Ebc and Ecd are called the line voltages, while voltages E0a, E0b, E0c and E0d are called the phase voltages or voltages to neutral. Eda=Ed0+E0a. Thus, the line voltages= 2 times the phase voltages in the four phase star. Two and four phase systems The line voltages are 45 degree or 135 degree out of time phase from the phase voltages. For four-phase mesh, the aa` line current is Iaa`=Ida+Iba. Thus, the line currents= 2 times the phase currents and 45 degree or 135 degree degree out of time phase in the four phase mesh Three phase four wire system Vcn - Vbn Vab 30 o Van Vbn Vab = Van - Vbn Vab = 3 Van 30o N is the neutral wire. Lighting loads are placed from line to neutral. Motors and other three phase power loads are connected between the three lines. Three phase three wire system Three phase four wire system Vab Van Vbn | V p | 0 | V p | 120 | V p | 1 (cos 120 j sin 120) 1 3 | V | p 1 j 2 2 3 3 | V | p j 2 2 3 | V p | 30 VL 3 | V p | Line Voltage Three phase four wire system The current in any line is the same as the current in the corresponding phase. I L I Current in the neutral wire is obtained through the application of KCL. I nn I na I nb I nc These currents are equal in magnitude and displaced from one another in time phase by 120 degree. Thus the current in the neutral wire is 0 since I nn I na I nb I nc 0 Three phase three wire system VL V Three phase three wire system Balanced Wye loads Given the line voltages as 220 volts balanced three phase and R and X of each phase 6 ohms resistance and 8 ohms inductive reactance. Find the line current, power per phase and total power. Draw the vector diagram VL 127volts 3 127 Ip 12.7 amperes 2 2 6 8 VP IL power per phase I P RP 968watts 2 total power 3 968 2904 watts Vna 127 j 0volts Vnb 127 120 127 cos 120 j sin 120 63.5 j110volts Vnc 127 120 63.5 j110volts Vba Vbn V na 190.5 j110volts I na Vna 12.7 53.13 Z na I nb 12.7 173.13 I nc 12.766.87 Pna vi vi Balanced Delta loads Given the line voltages as 220 volts balanced three phase and R and X of each phase 6 ohms resistance and 8 ohms inductive reactance. Find the line current, power per phase and total power. Draw the vector diagram self Power calculation in balanced systems Pp V p I P cos P Pt nPP nV p I P cos P VL Pt 3V p I P cos P 3 I L cos P 3VL I L cos P 3 IL Pt 3V p I P cos P 3VL cos P 3VL I L cos P 3 Wye connection Dealta connection Volt-Amperes vat 3vap 3V p I P The sine of the angle between phase voltage and phase Wye connection current is called the reactive factor of a balanced system. VL 3 I L 3VL I L 3 IL 3VL 3VL I L Dealta connection 3 Reactive Volt-Amperes P 3V I sin X p P P Wye connection Dealta connection VL PX 3 I L sin P 3VL I L sin P 3 IL PX 3VL sin P 3VL I L sin P 3 Single phase and balanced three phase power Pa Vm I m sin t sin t sin t 240 sin t 240 Pb Vm I m sin t 120 sin t 120 Pa Vm I m P3 Pa Pb Pa Vm I m sin t sin t Vm I m sin t 120 sin t 120 Vm I m sin t 240 sin t 240 1.5Vm I m cos For any phase a P1 Vm I m sin t sin t Vm I m Vm I m cos cos2t 2 2 Balanced three phase power under steady state conditions is constant from instant to instant. But it is a double frequency variation with respect to time for single phase power.. Three wattmeter method It is not feasible to break into the phases of a delta connected load. For the Y load it is necessary to connect to the neutral point. This point is not always accessible. Two wattmeter method Two wattmeter method Watt ratio curve For each value of ϴ, there is a definite ratio of Wa/Wb. If the ratio of the smaller to the larger reading is always taken and plotted against the corresponding cosϴ, a curve called the watt-ratio curve results. At 0.5 power factor, one wattmeter reads 0. At 0, each wattmeter has the same deflection but the readings are of opposite signs. Fig:46. watt-ratio curve for two wattmeter method of measuring power Sign of wattmeter Method 1 Open line a, all power must be transferred to the load over lines b and c. If wattmeter Wb reads upscale, the power is going to the load. Reconnect line a and open line b. then connect Wa so that it reads upscale. Reconnect line b. If at any time after this wither wattmeter needle goes backward against down-scale, power through this wattmeter channel is being transferred to the generator and this power must be of opposite sign to that registered by the other. Either the potential or current coil will have to be reversed to secure an up-scale reading. Method 2 disconnect the potential coil of Wb from line c and connect to line a. It the needle goes against the down-scale stop, the wattmeter reading was negative. Copper required to transmit power under fixed conditions P1 VI1 cos P3 3VI 3 cos P1 P3 VI1 cos 3VI 3 cos I1 3 I 3 I1 R1 2 I 3 R3 3 2 2 2 R1 3I 3 1 2 R3 2 I1 2 copper 3 number of copper1 number of wires 3 wires 1 R1 3 R3 4 Copper required to transmit power under fixed conditions P1 VI1 cos P3 3VI 3 cos P1 P3 VI1 cos 3VI 3 cos I1 3 I 3 I1 R1 2 I 3 R3 3 2 2 2 R1 3I 3 1 2 R3 2 I1 2 copper 3 number of copper1 number of wires 3 wires 1 R1 3 R3 4 Unbalanced system An unbalanced system is due to unbalanced voltage sources or unbalanced load. In a unbalanced system the neutral current is NOT zero. Line currents DO NOT add up to zero. In= -(Ia+ Ib+ Ic) ≠ 0