ssrs barcodelib . Obtain the power triangles for the branches and combine them into the total power triangle. in Software

Generation QR Code ISO/IEC18004 in Software . Obtain the power triangles for the branches and combine them into the total power triangle.

. Obtain the power triangles for the branches and combine them into the total power triangle.
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V V 7:07 308 A I2 5:66 08 A Z1 Z2    28:28 7:07 p 308 100 308 86:6 j50:0 S1 p 608 2 2    28:28 5:66 p 08 80:0 608 40:0 j69:3 S2 p 608 2 2 ST S1 S2 126:6 j119:3 174:0 43:38 VA I1 The power triangles and their summation are shown in Fig. 10-18.
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Fig. 10-18
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AC POWER
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[CHAP. 10
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10.14 Determine the total power information for three parallel-connected loads: load #1, 250 VA, pf 0:50 lagging; load #2, 180 W, pf 0:80 leading; load #3, 300 VA, 100 var (inductive).
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Calculate the average power P and the reactive power Q for each load. Load #1 Given s 250 VA, cos  0:50 lagging. Then, q P 250 0:50 125 W Q 250 2 125 2 216:5 var (inductive) Load #2 Given P 180 W, cos  0:80 leading. Then,  cos 1 0:80 36:878 and Q 180 tan 36:878 135 var (capacitive) Load #3 Given S 300 VA, Q 100 var (inductive). Then, q P 300 2 100 2 282:8 W
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Combining componentwise: PT 125 180 282:8 587:8 W QT 216:5 135 100 181:5 var (inductive) ST 587:8 j181:5 615:2 17:168 Therefore, ST 615:2 VA and pf cos 17:168 0:955 lagging.
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10.15 Obtain the complete power triangle and the total current for the parallel circuit shown in Fig. 1019, if for branch 2, S2 1490 VA.
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2 From S2 I2;eff Z2 ,
1490 2 I2;eff p 222 A2 32 62 and, by current division, I1 3 j6 I2 2 j3 Then, 32 62 2 45 222 768 A2 I2;eff 13 22 32
whence
2 I1;eff
2 S1 I1;eff Z1 768 2 j3 1536 j2304 2 S2 I2;eff Z2 222 3 j6 666 j1332 ST S1 S2 2202 j3636
that is, PT 2202 W, QT 3636 var (inductive), ST q 2202 2 3636 2 4251 VA and pf 2202 0:518 lagging 4251
CHAP. 10]
AC POWER
Since the phase angle of the voltage is unknown, only the magnitude of IT can be given. By current division, I2 and so
2 IT;eff
2 j3 I 5 j9 T
2 I2;eff
22 32 2 13 2 IT;eff I 106 T;eff 52 92
106 222 1811 A2 13
IT;eff 42:6 A
10.16 Obtain the complete power triangle for the circuit shown in Fig. 10-20, if the total reactive power is 2500 var (inductive). Find the branch powers P1 and P2 .
Fig. 10-20 The equivalent admittance allows the calculation of the total power triangle. Yeq Y1 Y2 0:2488 39:578 S Then, PT 2500 cot 39:578 3025 W ST 3025 j2500 3924 39:578 VA and pf PT =ST 0:771 lagging. The current ratio is I1 =I2 Y1 =Y2 0:177=0:0745. P1 I 2 4 21 1:88 P2 I2 12 from which P1 1975 W and P2 1050 W. and P1 P2 3025 W
10.17 A load of 300 kW, with power factor 0.65 lagging, has the power factor improved to 0.90 lagging by parallel capacitors. How many kvar must these capacitors furnish, and what is the resulting percent reduction in apparent power
The angles corresponding to the power factors are rst obtained: cos 1 0:65 49:468 Then (see Fig. 10-21), Q 300 tan 49:468 350:7 kvar (inductive) Q Qc 300 tan 25:848 145:3 kvar (inductive) whence Qc 205:4 kvar (capacitive). Since S 300 461:5 kVA 0:65 S0 300 333:3 kVA 0:90 cos 1 0:90 25:848
AC POWER
[CHAP. 10
Fig. 10-21
the reduction is 461:5 333:3 100% 27:8% 461:5
10.18 Find the capacitance C necessary to improve the power factor to 0.95 lagging in the circuit shown in Fig. 10-22, if the e ective voltage of 120 V has a frequency of 60 Hz.
Fig. 10-22 Admittance provides a good approach. 1 0:0433 j 0:0250 !C 20 308
Yeq j!C
The admittance diagram, Fig. 10-23, illustrates the next step.  cos 1 0:95 18:198 0:0250 !C 0:0433 tan 18:198 !C 0:0108 C 28:6 mF
Fig. 10-23
CHAP. 10]
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