26 – j52 26 + j52 = 178.282 + j356.564 – j1571.88 – j23143.8 676 + j1352 – j1352 – j22704 = 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
