# (3 )Three Phase Alternating Voltage and Current

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1 EEE 2015 EECTRCS (3) Monophase 1

2 Three phase Three phase electric power is a common method of alternating current electric power generation, transmission, and distribution. t is a type of polyphase system and is the most common method used by electrical grids worldwide to transfer power. t is also used to power large motors and other heavy loads. Athree phase system is usually more economical than an equivalent single phase or twophase system at the same voltage because it uses less conductor material to transmit electrical power Definition Athree phase (3φ) system is a combination of three single phase systems. n a 3φ balanced system, power comes from a 3φ AC generator that produces three separate and equal voltages, each of which is 120 out of phase with the other voltages Athree phase AC system consists of three phase generators, transmission lines, and loads. Three hase Generation 2

3 Three hase Waveform V f 50 [Hz ] V V V rms A A A 220 [V ] (t) sin( 2 50 t) (t) sin( 2 50 t 120 ) (t) sin( 2 50 t 240 ) OneCriticalAdvatageof Three hase Waveform ower delivered to a three phase load is constant at all time, instead of pulsing as it does in a single phase system. Concerning that the power is proportional to square of the voltage signal Monophase ower: Three hase ower: ower transfer into a linear balanced load is constant, which helps to reduce generator and motor vibrations. 3

4 Three hase Distribution ines We can connect the negative (ground) ends of the three single phase generators and loads together, so they share the common return line (neutral). Then the three phase system reduces to necessary number of cable to transfer power from 6 to 4. Three hase Distribution ines Three phase systems may have a neutral wire. A neutral wire allows the three phase system to use a higher voltage while still supporting lower voltage single phase loads. n high voltage distribution situations, it is common not to have a neutral wire as the loads can simply be connected between phases (phase phase connection). 4

5 Three hase Currents f each of the phases transfers power to loads that have same impedance, such threephase power systems (equal magnitude, phase differences of 120 0, identical loads) are called balanced. The phase currents tend to cancel out one another, summing to zero in the case of a linear balanced load. This makes it possible to reduce the size of the neutral conductor because it carries little to no current; all the phase conductors carry the same current and so can be the same size, for a balanced load. Star (Wye or Y) Connected Three hase Systems t is not necessary to have six wires from the three phase windings to the three loads. Each winding will have a start (S) and a fi nish (F) end. The star or wye (Y) connection mentioned is achieved by connecting the corresponding ends of the three phases together. 5

6 Star (Wye or Y) Connected Three hase Systems V ph = hase Voltage, V = ine Voltage, h = hase Current, = ine Current Delta (Mesh or ) Connected Three hase Systems t is not necessary to have six wires from the three phase windings to the three loads. Each winding will have a start (S) and a fi nish (F) end. The star or wye (Y) connection mentioned is achieved by connecting the corresponding ends of the three phases together. 6

7 Delta (Mesh or ) Connected Three hase Systems V ph = hase Voltage, V = ine Voltage, h = hase Current, = ine Current Three hase ower Total power ( T ) is equal to three times the single phase power. Then the mathematical representation for total power in a balanced delta or wye load is 3 3V 3 For Star or Y connected systems V 3V, 3 3 3V 3V V 3 3 For Delta or connected systems V V, 3 V = hase Voltage, V = ine Voltage, = hase Current, = ine Current 3 3 3V 3V 3V 3 3 3V 7

8 Three hase ower Triangle f the 3 system is connected to a balanced load with impedance: Z Z Then the power triangle and the power values of the 3 system is similar to monophase systems: S Q V 3 V sin 3V cos [VA] [VAR] [W] S Q V 3 V 3V sin cos [VA] [VAR] [W] V = hase Voltage, V = ine Voltage, = hase Current, = ine Current roblem: An asynchronous electric that runs using three phase grid has a power factor of 0.85 and measured that it gets 7 [A] from grid. What are the powers it gets from R S T phases if the phases balanced? Solution: : Real ower[watt] ; Q : Apparent ower [VAR]; S : Reactiveower [VA] U : 380 [V] The voltage between the phases is 380 [V] : t is the current taken from one phase [A] = 3 x U x x cos = 3 x 380 x 7 x 0,85 085= 3916 [W] Q = 3 x U x x sin = 3 x 380 x 7 x 0,5268 = 2427 [VAR] S = 3 x U x = 3 x 380 x 7 = 4607 [VA] 8

9 To do per phase analysis 1. Convert all load/sources to equivalent Y s 2. Solve phase a independent of the other phases 3. Total system power S = 3 V a * a 4. f desired, phase b and c values can be determined by inspection (i.e., ±120 degree phase shifts) 5. f necessary, go back to original circuit to determine line line values or internal values Delta to wye conversion The Y Δ transform is known by a variety of other names, mostly based upon the two shapes involved, listed in either order. The Y, spelled out as wye, can also be called T or star; the Δ, spelled out as delta, can also be called triangle, Π (spelled out as pi), or mesh. Thus, common names for the transformation include wye delta or delta wye, star delta, starmesh, or T Π. 9

10 roblem: An asynchronous electric that runs using three phase grid has a power factor of 0.85 and measured that it gets 7 [A] from grid. What are the powers it gets from R S T phases if the phases balanced? Solution: : Real ower[watt] ; Q : Apparent ower [VAR]; S : Reactiveower [VA] U : 380 [V] The voltage between the phases is 380 [V] : t is the current taken from one phase [A] = 3 x U x x cos = 3 x 380 x 7 x 0,85 085= 3916 [W] Q = 3 x U x x sin = 3 x 380 x 7 x 0,5268 = 2427 [VAR] S = 3 x U x = 3 x 380 x 7 = 4607 [VA] For a 208 V three phase ideally balanced system, find: a) the magnitude of the line current ; b) The magnitude of the load s line and phase voltages V and V ; c) The real, reactive, and the apparent powers consumed by the load; d) The power factor of the load. 10

11 er phase circuit : a) The line current: s the load inductive or capacitive?? b) The phase voltage on the load: The magnitude of the line voltage on the load: er phase circuit : c) The real power consumed by the load: The reactive power consumed by the load: The apparent power consumed by the load: d) The loadpower factor: 11

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Survey of Harmonics Measurements in Electrical Distribution Systems Leon M. Tolbert, Member, EEE Oak Ridge ational Laboratory* P.O. Box 8, Bldg Oak Ridge, T 3783-6334 Alexandria, VA 35-386 Phone: (43)