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This last statement is easily veri ed by sketching the phasor diagram The sequence of phasor voltages shown in Figure 747 is usually referred to as the positive (or abc) sequence Consider now the lines connecting each source to the load and observe that it is possible to also de ne line voltages (also called line-to-line voltages) by considering the voltages between the lines aa and bb , aa and cc , and bb and cc Since the line voltage, say, between aa and bb is given by Vab = Van + Vnb = Van Vbn (753)
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Figure 747 Positive, or abc, sequence for balanced three-phase voltages
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the line voltages may be computed relative to the phase voltages as follows: Vab = V 0 V 120 = 3V 30 (754) Vbc = V 120 V 120 = 3V 90 Vca = V 120 V 0 = 3V 150 It can be seen, then, that the magnitude of the line voltages is equal to 3 times the magnitude of the phase voltages It is instructive, at least once, to point out that the circuit of Figure 746 can be redrawn to have the appearance of the circuit of Figure 748 One of the important features of a balanced three-phase system is that it does not require a fourth wire (the neutral connection), since the current In is identically zero (for balanced load Za = Zb = Zc = Z) This can be shown by applying KCL at the neutral node n: In = (Ia + Ib + Ic ) = 1 (Van + Vbn + Vcn ) Z (755)
Figure 748 Balanced three-phase AC circuit (redrawn)
Part I
Circuits
Another, more important characteristic of a balanced three-phase power system may be illustrated by simplifying the circuits of Figures 746 and 748 by replacing the balanced load impedances with three equal resistances, R With this simpli ed con guration, one can show that the total power delivered to the balanced load by the three-phase generator is constant This is an extremely important result, for a very practical reason: delivering power in a smooth fashion (as opposed to the pulsating nature of single-phase power) reduces the wear and stress on the generating equipment Although we have not yet discussed the nature of the machines used to generate power, a useful analogy here is that of a single-cylinder engine versus a perfectly balanced V-8 engine To show that the total power delivered by the three sources to a balanced resistive load is constant, consider the instantaneous power delivered by each source: pa (t) = pb (t) = pc (t) = V2 (1 + cos 2 t) R V2 [1 + cos(2 t 120 )] R (756)
V2 [1 + cos(2 t + 120 )] R The total instantaneous load power is then given by the sum of the three contributions: p(t) = pa (t) + pb (t) + pc (t) = 3V V + [cos 2 t + cos(2 t 120 ) R R + cos(2 t + 120 )] 2 2
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a + _ ~ Vab _ ~ Vca b + ~ Vbc _ + c
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3V 2 = a constant! R You may wish to verify that the sum of the trigonometric terms inside the brackets is identically zero It is also possible to connect the three AC sources in a three-phase system in a so-called delta (or ) connection, although in practice this con guration is rarely used Figure 749 depicts a set of three delta-connected generators =
A delta-connected three-phase generator with line voltages Vab, Vbc, Vca
Figure 749 Delta-connected generators
EXAMPLE 715 Per-Phase Solution of Balanced Wye-Wye Circuit
Problem
~ Van _~+ ~ Vbn _~+ ~ Vcb _~+ a R line a b R line b c R line c Rneutral
Compute the power delivered to the load by the three-phase generator in the circuit shown in Figure 750
Zy Zy Zy n
Solution
Known Quantities: Source voltage, line resistance, load impedance Find: Power delivered to the load, PL
7
AC Power
Schematics, Diagrams, Circuits, and Given Data: Van = 480 (0) V; Vbn = 480 ( 2 /3) V; Vcn = 480 (2 /3) V; Zy = 2 + j 4 = 447 (1107) Rline = 2 ; Rneutral = 10
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