zxing barcode reader java download 0 V = 24 30 A 5 30 A 120 0 V = 24 45 A 5 45 A in Software

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120 0 V = 24 30 A 5 30 A 120 0 V = 24 45 A 5 45 A
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The current I 2 in Load 2 is
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I2 =
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Therefore the total current from the source is
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I = I1 + I 2 = 24 30 A + 24 45 A = 4759 375 A
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The power factor supplied by the source is
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PF = cos = cos ( 375 ) = 0793 lagging
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P = VI cos = (120 V )( 4759 A ) cos ( 375 ) = 4531 W
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The real, reactive, and apparent power supplied by the source are
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Q = VI cos = (120 V )( 4759 A ) sin ( 375 ) = 3477 var S = VI = (120 V )( 4759 A ) = 5711 VA
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(b) With the switch open, all three loads are connected to the source The current in Loads 1 and 2 is the same as before The current I 3 in Load 3 is
I3 =
120 0 V = 24 90 A 5 90 A
Therefore the total current from the source is
I = I1 + I 2 + I3 = 24 30 A + 24 45 A + 24 90 A = 3808 75 A
The power factor supplied by the source is
PF = cos = cos ( 75 ) = 0991 lagging
P = VI cos = (120 V )( 3808 A ) cos ( 75 ) = 4531 W
The real, reactive, and apparent power supplied by the source are
Q = VI cos = (120 V )( 3808 A ) sin ( 75 ) = 596 var
S = VI = (120 V )( 3808 A ) = 4570 VA
(c) The current flowing decreased when the switch closed, because most of the reactive power being consumed by Loads 1 and 2 is being supplied by Load 3 Since less reactive power has to be supplied by the source, the total current flow decireases 1-20 Demonstrate that Equation (1-50) can be derived from Equation (1-49) using simple trigonometric identities:
p (t ) = v (t ) i (t ) = 2VI cos t cos ( t ) p (t ) = VI cos (1 + cos 2 t ) + VI sin sin 2 t
SOLUTION The first step is to apply the following identity from Appendix D:
(1-49) (1-50)
cos cos =
The result is
1 cos ( ) + cos ( + ) 2
(1-49)
p (t ) = v (t ) i (t ) = 2VI cos t cos ( t )
1 p(t ) = 2VI cos ( t t + ) + cos ( t + t ) 2
p(t ) = VI cos + cos ( 2 t )
Now we must apply the angle addition identity from Appendix D to the second term:
cos ( ) = cos cos + sin sin
The result is
p(t ) = VI [ cos + cos 2 t cos + sin 2 t sin ]
Collecting terms yields the final result:
p(t ) = VI cos (1 + cos 2 t ) + VI sin sin 2 t
2: Three-Phase Circuits
2-1 Three impedances of 4 + j3 are -connected and tied to a three-phase 208-V power line Find I , I L , P, Q, S, and the power factor of this load SOLUTION
IL + I
Z = 3 + j4
240 V
Here, VL = V = 208 V , and Z = 4 + j 3 = 5 3687 , so
I =
V Z
208 V = 416 A 5
I L = 3I = 3 ( 416 A ) = 7205 A
5 ( 208 V ) 2 sin 3687 = 1558 kvar Q=3 sin = 3 Z 5 Z 2 V
( 208 V ) 2 cos 3687 = 2077 kW cos = 3
S = P 2 + Q 2 = 2596 kVA PF = cos = 08 lagging
2-2 Figure P2-1 shows a three-phase power system with two loads The -connected generator is producing a line voltage of 480 V, and the line impedance is 009 + j016 Load 1 is Y-connected, with a phase impedance of 25 3687 and load 2 is -connected, with a phase impedance of 5 -20
(a) What is the line voltage of the two loads (b) What is the voltage drop on the transmission lines (c) Find the real and reactive powers supplied to each load (d) Find the real and reactive power losses in the transmission line (e) Find the real power, reactive power, and power factor supplied by the generator SOLUTION To solve this problem, first convert the two deltas to equivalent wyes, and get the per-phase equivalent circuit
0090 j016
Line
277 0 V
Z 1
Z 2
V ,load
Z 1 = 25 3687
Z 2 = 167 20
The phase voltage of the equivalent Y-loads can be found by nodal analysis
=0 009 + j 016 25 3687 167 20 (5443 606 ) (V ,load 277 0 V ) + (04 3687 )V ,load + (06 20 )V ,load = 0
V ,load 277 0 V
V ,load
V ,load
(5955 5334 ) V ,load = 1508 606
V ,load = 2532 73 V
Therefore, the line voltage at the loads is VL 3 V = 439 V (b) The voltage drop in the transmission lines is
Vline = V ,gen V ,load = 277 0 V - 2532 - 73 = 413 52 V
(c) The real and reactive power of each load is
25 ( 2532 V ) 2 sin 3687 = 462 kvar Q1 = 3 sin = 3 Z 25 2 V (2532 V )2 cos (- 20 ) = 1084 kW P2 = 3 cos = 3 Z 167 2 V ( 2532 V ) 2 sin -20 = 395 kvar Q2 = 3 sin = 3 ( ) Z 167 Z 2 V
(d) The line current is
P1 = 3
cos = 3
(2532 V )2 cos 3687 = 616 kW
I line =
Vline 413 52 V = = 225 86 A Z line 009 + j 016
Therefore, the loses in the transmission line are
Pline = 3I line Rline = 3 (225 A ) (009 ) = 137 kW
Qline = 3I line X line = 3 ( 225 A ) ( 016 ) = 243 kvar
The real and reactive power supplied by the generator is
Pgen = Pline + P1 + P2 = 137 kW + 616 kW + 1084 kW = 1837 kW Qgen = Qline + Q1 + Q2 = 243 kvar + 462 kvar 395 kvar = 31 kvar
The power factor of the generator is
Q 31 kvar = 0986 lagging PF = cos tan -1 gen = cos tan 1 Pgen 1837 kW
2-3 The figure shown below shows a one-line diagram of a simple power system containing a single 480 V generator and three loads Assume that the transmission lines in this power system are lossless, and answer the following questions (a) Assume that Load 1 is Y-connected What are the phase voltage and currents in that load (b) Assume that Load 2 is -connected What are the phase voltage and currents in that load (c) What real, reactive, and apparent power does the generator supply when the switch is open (d) What is the total line current I L when the switch is open (e) What real, reactive, and apparent power does the generator supply when the switch is closed (f) What is the total line current I L when the switch is closed (g) How does the total line current I L compare to the sum of the three individual currents I1 + I 2 + I 3 If they are not equal, why not 21
SOLUTION Since the transmission lines are lossless in this power system, the full voltage generated by G1 will be present at each of the loads (a) Since this load is Y-connected, the phase voltage is
V 1 =
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