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EXAMPLE 716 Parallel Wye-Delta Load Circuit
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Compute the power delivered to the wye-delta load by the three-phase generator in the circuit shown in Figure 753
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Figure 753 AC circuit with delta and wye loads
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Part I
Circuits
Solution
Known Quantities: Source voltage, line resistance, load impedance Find: Power delivered to the load, PL
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) Z = 5 j 2 = 54 ( 0381) ; Rline = 2 ; Rneutral = 10
Assumptions: Use rms values for all phasor quantities in the problem
Analysis: We rst convert the balanced delta load to an equivalent wye load, according to equation 766 Figure 754 illustrates the effect of this conversion
Z = 1667 j 0667 = 18 ( 0381) 3 Since the circuit is balanced, we can use per-phase analysis, and the current through the neutral line is zero, ie, Vn n = 0 The resulting per-phase circuit is shown in Figure 755 Using phase a for the calculations, we look for the quantity Z
Pa = |I|2 RL where ZL = Zy Z
Zy Z Zy + Z
y y
= 162 j 0018 = 162 ( 0011)
and the load current is given by: |I| = Va 480 0 = 1326 A = ZL + Rline 162 + j 0018 + 2
and Pa = (1326)2 Re(ZL ) = 285 kW Since the circuit is balanced, the results for phase b and c are identical, and we have: PL = 3Pa = 855 kW
Zy Zy a Z /3 Z Z Zy b Z /3 Z Zy ~ Va + ~ _ Zy Z /3 a Rline a
c Zy Z /3 n n
Figure 754 Conversion of delta load to equivalent wye load
Figure 755 Per-phase circuit
Comments: Note that per-phase analysis for balanced circuits turns three-phase power calculations into a very simple exercise
7
AC Power
Focus on Computer-Aided Tools: A computer-generated solution of this example may
be found in the accompanying CD-ROM
Check Your Understanding
714 Find the power lost in the lines in the circuit of Example 715 715 Draw the phasor diagram and power triangle for a single phase and compute the
power delivered to the balanced load of Example 715 if the lines have zero resistance and ZL = 1 + j 3
716 Show that the voltage across each branch of the balanced wye load in Exercise 715 is equal to the corresponding phase voltage (eg, the voltage across Za is Va ) 717 Prove that the sum of the instantaneous powers absorbed by the three branches in a balanced wye-connected load is constant and equal to 3V I cos 718 Derive an expression for the rms line current of a delta load in terms of the rms line current for a wye load with the same branch impedances (ie, Zy = Z ) and same source voltage Assume ZS = 0 719 The equivalent wye load of Example 716 is connected in a delta con guration Compute the line currents
RESIDENTIAL WIRING; GROUNDING AND SAFETY
+ ~ VB _ _ ~ VR +
Hot ~ Neutral ~ Hot ~ (Neutral) VW = 0 0 ~ VB = 120 0 (Hot) ~ VR = 120 180 (Hot) ~ ~ or VR = VB
Common residential electric power service consists of a three-wire AC system supplied by the local power company The three wires originate from a utility pole and consist of a neutral wire, which is connected to earth ground, and two hot wires Each of the hot lines supplies 120 V rms to the residential circuits; the two lines are 180 out of phase, for reasons that will become apparent during the course of this discussion The phasor line voltages, shown in Figure 756, are usually referred to by means of a subscript convention derived from the color of the insulation on the different wires: W for white (neutral), B for black (hot), and R for red (hot) This convention is adhered to uniformly The voltages across the hot lines are given by: VB VR = VBR = VB ( VB ) = 2VB = 240 0 (767)
Figure 756 Line voltage convention for residential circuits
Thus, the voltage between the hot wires is actually 240 V rms Appliances such as electric stoves, air conditioners, and heaters are powered by the 240-V rms arrangement On the other hand, lighting and all of the electric outlets in the house used for small appliances are powered by a single 120-V rms line The use of 240-V rms service for appliances that require a substantial amount of power to operate is dictated by power transfer considerations Consider the two circuits shown in Figure 757 In delivering the necessary power to a load, a lower line loss will be incurred with the 240-V rms wiring, since the power loss in the lines (the I 2 R loss, as it is commonly referred to) is directly related to the current required by the load In an effort to minimize line losses, the size of the wires is increased for the lower-voltage case This typically reduces the wire resistance by a factor of 2 In the top circuit, assuming RS /2 = 001 , the current required
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