vb.net barcode reader code Assumptions: Use rms values for all phasor quantities in the problem in Software

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Assumptions: Use rms values for all phasor quantities in the problem
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through the neutral line is zero, ie, Vn n = 0 The resulting per-phase circuit is shown in Figure 751 Using phase a for the calculations, we look for the quantity Pa = |I|2 RL
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Analysis: Since the circuit is balanced, we can use per-phase analysis, and the current
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where |I| = 480 0 480 0 Va = = Zy + Rline 2 + j4 + 2 566 4 = 8485 A
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Figure 751 One phase of the three-phase circuit
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and Pa = (8485)2 2 = 144 kW Since the circuit is balanced, the results for phases b and c are identical, and we have: PL = 3Pa = 432 kW
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Comments: Note that, since the circuit is balanced, there is zero voltage across neutrals
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This fact is shown explicitly in Figure 751, where n and n are connected to each other directly Per-phase analysis for balanced circuits turn three-phase power calculations into a very simple exercise
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Balanced Wye Loads In the previous section we performed some power computations for a purely resistive balanced wye load We shall now generalize those results for an arbitrary balanced complex load Consider again the circuit of Figure 746, where now the balanced load consists of the three complex impedances Za = Zb = Zc = Zy = |Zy | (758)
From the diagram of Figure 746, it can be veri ed that each impedance sees the corresponding phase voltage across itself; thus, since the currents Ia , Ib , and Ic have the same rms value, I , the phase angles of the currents will differ by 120 It is therefore possible to compute the power for each phase by considering the phase voltage (equal to the load voltage) for each impedance, and the associated line current Let us denote the complex power for each phase by S: S = V I so that S = P + jQ = V I cos + j V I sin (760) (759)
where V and I denote, once again, the rms values of each phase voltage and line current Consequently, the total real power delivered to the balanced wye load is 3P , and the total reactive power is 3Q Thus, the total complex power, ST , is given by ST = PT + j QT = 3P + j 3Q = (3P )2 + (3Q)2 (761)
Part I
Circuits
and the apparent power is |ST | = 3 (V I )2 cos2 + (V I )2 sin2 = 3V I and the total real and reactive power may be expressed in terms of the apparent power: PT = |ST | cos QT = |ST | sin Balanced Delta Loads In addition to a wye connection, it is also possible to connect a balanced load in the delta con guration A wye-connected generator and a delta-connected load are shown in Figure 752 (762)
Ia a
+ Van ~ _
n Ic
+ _ c Z Vab
_ +~
_ ~ Vcn +
Iab Ibc
+ b
Figure 752 Balanced wye generators with balanced delta load
It should be noted immediately that now the corresponding line voltage (not phase voltage) appears across each impedance For example, the voltage across Zc a is Vca Thus, the three load currents are given by the following expressions: 3V ( /6) Vab = Iab = Z |Z | Vbc 3V ( /4) (763) Ibc = = Z |Z | 3V (5 /6) Vca Ica = = Z |Z | To understand the relationship between delta-connected and wye-connected loads, it is reasonable to ask the question, For what value of Z would a deltaconnected load draw the same amount of current as a wye-connected load with
7
AC Power
impedance Zy for a given source voltage This is equivalent to asking what value of Z would make the line currents the same in both circuits (compare Figure 748 with Figure 752) The line current drawn, say, in phase a by a wye-connected load is (Ian )y = V Van = Z |Zy | (764)
while that drawn by the delta-connected load is (Ia ) = Iab Ica = = = = Vab Vca Z Z 1 (Van Vbn Vcn + Van ) Z 1 (2Van Vbn Vcn ) Z 3Van 3V = Z |Z | (765)
One can readily verify that the two currents (Ia ) and (Ia )y will be equal if the magnitude of the delta-connected impedance is 3 times larger than Zy : Z = 3Zy (766)
This result also implies that a delta load will necessarily draw 3 times as much current (and therefore absorb 3 times as much power) as a wye load with the same branch impedance
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