26 – j52 26 + j52 = 178

26 – j52 26 + j52

= 178.282 + j356.564 – j1571.88 – j23143.8
676 + j1352 – j1352 – j22704

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= 3317.082 – j1215.316
3380

= 0.9814 – j0.3596
To find
= + Y13V1+ Y23V2(2)

Y13 + Y23

=

(10 – j30) + (16 – j32)

= -1.386 + j0.452 x 1.00048 + j0.5363
1.00048 –j0.5363 1.00048 + j0.5363

=
0.965301 – j0.030458 + j0.030458 – j20.2876

= -1.6291 – j0.2911= -1.264 – j0.2259
1.2886

= -1.264–j0.2259+10.5–J31.5 + 15.7024 – j5.7536 – j31.4048 + j211.5072
26 – j62

=-13.4312 – j68 – 8843 x 26 + 62
26 – j62 26 + j62

= 349.2112 + j832.7344 – j1790.9918 – j24270.8266
676 + j1612 – j1612 – j23844
= 4620.0378 – j958.2579
4520

= 1.0221 – j0.2120

To find
= + Y12V1+ Y23V3(2)

Y12 + Y23
=

10 – j20 + 16 – j32

Solving for the numerator conjugate
-2.566 + j1.102 x 0.9814 + j0.3596
0.9814 – j0.3596 0.9814 + j0.3596

0.9631 + j0.3529 – j0.3529 – j20.1293

-2.9146 + j0.1588
1.0924

-2.6681 + j0.1454

= -2.6681 + j0.1454 + 10.5 – j21 +16.3536 – j3.392 – j32.7072 + j26.784
26 – j52

-17.4015 – j56.9538 x 26 + j52
26 – j52 26 + j52

452.439 + j904.878 – j1480.7988 – j22961.5976
676 + j1352 – j22704

= 3414.0366 – j575.9208
3383

= 1.0092 – j0.1702

To find
= + Y13V1+ Y23V3(3)

Y13 + Y23

=

(10 – j30) + 16 – j32

Solving for the numerator conjugate
-1.386 + j0.452 x 1.0221 + j0.212
1.0221 – j0.212 1.0221 + j0.212

1.0447 – j0.2167 + j0.2167 – j20.04494

-1.5124 + j0.1682
1.0896

-1.3880 + j0.1544

= – 1.3880+ j0.1544 + 10.5 – j31.5 + 16.1472 – j2.7232–32.2944+j25.4464
26 – j62

-19.8128 – j66.2688 x 26 + j62
26 – j62 26 + j62

515.1328 + j1228.3936 – j1722.9888 – j24108.6656
676 + j1612 – j1612 – j23844

= 4623.7984 – j494.5952
4520

= 1.02296 – j0.1094

To find in iteration 4
= + Y12V1+ Y23V3(3)

Y12 + Y23

=

(10 – j20) + 16 – j32

Solving for the numerator conjugate
-2.566 + j1.102 x 1.0092 + j0.1702
1.0092 – j0.1702 1.0092 + j0.1702

1.0185 + j0.1718 – j0.1718 – J20.0290

-2.7772 + j1.5488
1.0475

-2.6513 + j1.4786

= – 2.6513+j1.4786+ 0.5 – j21+16.3664 –j1.7504 –j32.734+j21.7504 –j32.734+j23.5008
26 – j52

= 20.7143 – j54.0065 x 26 + j5
26 – j52 26 + j52

= 538.5718 + j1077.1436 – j1404.169 – j22808.338
676 + j1352 – j1352 – j22704

= 3346.9098 – j327.0254
3380

= 0.99021 – j0.0968

To calculate in iteration 4
Then applying formula
= + Y13V1+ Y23V2(4)

Y13 + Y23

=

(10 – j30) + 16 – j32
So, solving for the numerator conjugate
-1.386 + j0.452 x 1.02296 + j0.1094
1.02296 – j0.1094 1.02296 + j0.1094

1.0464 + j1.1191 – j1.1191 – j20.0120

-1.4672 + j0.3108
1.0594

-1.3849 + j0.2934

= – 1.385+j0.2934+10.5-j31.5+15.8434 – j1.549 – j31.687 + j23.0976
26 – j62

= 21.8609 – j64.4421 x 26 + j62
26 – j62 26 + j62

= 568.3834 + j1355.3758 – j1675.4946 – j23995.4102
676 + j1612 – j1612 – j23844

= 4563.7936 – j320.1188
4520

= 1.0097 – j0.0708
The process is continued until the solution is converged as represented from the voltage iterations
= 0.9825 – j0.0310 = 1.00048 – j0.5363
= 0.9814 – j0.3596 = 1.0221 – j0.2120
= 1.0092 – j0.3596 = 1.02296 – j0.1094
= 0.99021 – j0.0968 = 1.0097 – j0.0708
To find bus 2 voltage
V2 = 0.999021 – j0.0968
Then, converting rectangular to polar
tan-1
tan-1
tan-1 0.0978
tan-1 -0.0978
0.9950 L-5.5858o pu
To find bus 3 voltage is 1.0097 – j0.0708 and the polar conversion is
tan-1
tan-1
tan-1 0.0701
tan-1 -0.0701
1.0122 L-4.0099o pu
To find the slack bus power
Applying formula for slack bus power
P1 – jQ1 = V1V1(Y12 + Y13) – (Y12V2 + Y13 V3
P1 – jQ1 = 1.051.05(20 – j50) – (10 – j20)(0.99021-j0.0968) – (10 – j30)(1.0097 – j0.0708)
P1 – jQ1 = 1.0521-j52.5 – (9.9021-j0.968 – j19.8042+J21.936) – (10.097 – j0,708 – j30.291 + J22.124)
P1 – jQ1 = 1.0521-j52.5 – (7.9661 – j20.7722) – (7.973 – j30.999)
P1 – jQ1 = 1.05 21 – j52.5 – 7.9661 + j20.7722 – 7.973 + j30.999
P1 – jQ1= 1.05 5.0609 – j0.7288
P1 – jQ1 = 5.3139 – j0.7652

3.5 To determine the slack bus real and reactive powers
The slack real and reactive power powers are
P1 = 5.3139pu = 5.3139 x 100
P1 ToTAL = 531.39mw
Q1 = 0.7652pu = 0.7652 x 100
Q1 = 76.52Mvar
To calculate the line current when the charging capacitor is neglected
To calculate I12
Applying formula
I12 = Y12 (V1 – V2)
I12 = (10 –j20) (1.05 + j0) – (0.99021 – j0.0968)
I12 = 10 – j20 1.05 + j0 – 0.99021 + j0.0968
I12 = 10 – j20 0.0598 + j0 .0968
I12 = 0.598 + j0.968 – j1.196 – j21.936
I12 = 2.534 – j0.228
To find I21
I21 = -1 x (2.534 – j0.228) = -2.534 + j0.228
To calculate I13
I13 = Y13 (V1 – V3)
I13 = (10 – j30) (1.05 + j0) – (1.0097 – j0.0708)
I13 = (10 – j30) 0.0403 + j0.0708
I13 = 0.403 + j0.708 – j1.209 –J22.124
I13 = 2.527 – j0.501

To find I31
I31= -1 x (2.527 – j0.501) = -2.527 + j0.501
To calculate I23
I23 =Y23 (V2 – V3)
I23 = (16 – j32) 0.99021 – j0.0968) – (1.0097 – j0.6708
I23 = (16 – j32) -0.0195 – j0.026
I23 = -1.144 + j0.312
To find I32
I32 =-1 x I32
I32 = -1 x (-1.144 + j0.312)
I32 = 1.144 – j0.312

To calculate the line flows
To find S12 line flow
S12 = V1I12
S12 = (1.05 + j0) (2.534 + j0.228)
S12 = 2.6607 + j0.2394pu
Change per unit to mega watts
S12 = 100(2.6607 + j0.2394)
S12 = 266.07mw + j23.94Mvar

To find S21 line flow
S21 = V1I21
S21 = (0.99021 – j0.0968) (-2.534 – j0.228)
S21 = -2.5092 – j0.2258 + j0.2453 + J20.02207
S21 = -2.5313 + j0.195pu
Changing per unit to megawatts
S21 = 100 (-2.5313 + j0.0195)
S21 = -253.13mw + j1.95Mvar

To find S13 line flow
S13= V1I13
S13= (1.05 + j0.0)(2.527 + j0.501)
S13= 2.6534 + j0.5261 pu

Changing per unit to megawatts.
S13= 100(2.6534 + j0.5261)
S13= 265.34mw + 52.61Mvar

To find S31 line flow
S31 = V3I31
S31 = (1.0097 – j0.0708) (-2.527 – j0.501)
S31 = -2.5515 – j0.5059 + j0.1789 + j20.03547
S31 = -2.5870 – j0.327 pu
Changing per unit to megawatts
S31 = (100(-2.5870 – j0.327)
S31 = -258.7Mw – j32.7Mvar
To find S23 = line flow
S23= V2I23
S23 = (0.99021 – j0.09869) (-1.144 + j0.312)
S23 = -1.1328 + j0.3089 + j0.1107 – J20.0302
S23 = -1.1026 + j0.4196pu
Changing pu to Mw
S23 =100(-1.1026 = j0.4196)
S23 = -110.26Mw + j41.96Mvar

To find S32line flow
S32 = (1.0097 – j0.0708) (1.144 + j0.312)
S32 = 1.1551 + j0.3150 – j0.0810 – J20.0219
S32 = 1.177 + j0.234pu
Changing per unit to megawatts
S32 = 100(1.177 + j0.234)
S32 = 117.2mw + 123.4Mvar.
P2 total = 117mw
The result gotten are P1 total = 531.31KW and P2 total = 117KW
To optimize a power loss reduction in a transmission network using optimization.
An EEDC company produces two types of power supply in a transmission network A and B that require power P1 and P2. Each unit of type A require 1KW of P1 and 2KW of P2. Type B requires 2KW of P1 and 1KW of P2 (Each unit). The company has only 531.31KW of P1 and 117KW of P2. Each unit of type A brings a profit of #500 Million and each unit of type B brings a profit of #400 Million for 330 KVA. Formulate the optimization problem to maximize profit that will optimize power loss reduction in a transmission network. Shown in table 3.1 of Appendix 2.
To put the data in a tabular form

Table 3.1: Power supply with profit maximization.
Power P1 (KW) P2 (KW) Profit (#)
A 1 2 500
B 2 1 400
531.31 117

The optimization equation becomes
maximize z = 500x + 400y 3.1
Subject to
x + 2y

x

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