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The bus voltages are now back in tolerance, and the apparent powers flowing in the transmission lines have decreased The power system can not function properly even with transmission line 3 open circuited 11-7 Assume that the power system is restored to its original configuration A new plant consuming 20 MW at 095 PF lagging is to be added to Bus 4 Will this new load cause any problems for this power system If this new load will cause problems, what solution could you recommend SOLUTION We can solve this problem using program power_flowm The real power added to Bus 4 is 20 MW, and the reactive load added to Bus 4 is
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Q = P tan = P tan cos 1 PF = ( 20 MW ) tan cos 1 ( 095) = 66 MVAR
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The input file prob_11_7_input to calculate the power flows in this power system is given below, with the changes highlighted
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% File describing the base case plus adding loads to Bus 4 % % System data has the form: %SYSTEM name baseMVA Voltage Tolerance SYSTEM Plus_loads 100 005 % % Bus data has the form: %BUS name type volts Pgen Qgen Pload Qload Qcap BUS One SL 100 0 0 0 0 0 BUS Two PQ 100 0 0 60 35 0 BUS Three PQ 100 0 0 70 40 0 BUS Four PQ 100 0 0 100 566 0 BUS Five PV 100 190 0 40 30 0 % % Transmission line data has the form: %LINE from to Rse Xse Gsh Bsh Rating(MVA) LINE One Two 00210 01250 0000 0000 50 LINE One Four 00235 00940 0000 0000 100 LINE Two Three 00250 01500 0000 0000 50 LINE Two Five 00180 00730 0000 0000 100 LINE Three Five 00220 01100 0000 0000 70 LINE Four Five 00190 00800 0000 0000 100
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>> power_flow prob_11_7_input Input summary statistics: 22 lines in system file 1 SYSTEM lines 5 BUS lines 6 LINE lines
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Results for Case Plus_loads |====================Bus Information=======================================|=======Line Information======| Bus Bus Volts / angle |--Generation--|------Load-----|--Cap--| To |----Line Flow---| no Name Type (pu) (deg) (MW) (MVAR) (MW) (MVAR) (MVAR) Bus (MW) (MVAR)| |========================================================================================================| 1 One SL 1000/ 000 8308 4659 000 000 000 Two 3201 1893 Four 5107 2765 2 Two PQ 0970/ -213 000 000 6000 3500 000 One -3172 -1720 Three 1715 995 Five -4530 -2777 3 Three PQ 0951/ -357 000 000 7000 4000 000 Two -1705 -932 Five -5290 -3069 4 Four PQ 0963/ -247 000 000 10000 5660 000 One -5028 -2448 Five -4965 -3213 5 Five PV 1000/ -047 19000 13031 4000 3000 000 Two 4584 2996 Three 5381 3524 Four 5036 3515 |========================================================================================================| Totals 27308 17689 27000 16160 000 |========================================================================================================|
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Line Losses |=====================================================| | Line From To Ploss Qloss | | no Bus Bus (MW) (MVAR) | |=====================================================| 1 One Two 029 173 2 One Four 079 317
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3 Two Three 010 063 4 Two Five 054 219 5 Three Five 091 455 6 Four Five 072 302 |=====================================================| Totals: 335 1528 |=====================================================|
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This new load caused the voltage to decrease at Bus 4 and the load on the transmission lines to increase, but neither limit was exceeded The system could operate indefinitely under these conditions 11-8 Write your own program to solve for the voltages and currents in the power system instead of using program power_flow You may write the program in any programming language with which you are familiar How do your answers compare to the ones produced by power_flow SOLUTION To solve this problem, we must first generate the bus admittance matrix Ybus , and then iterate to a solution using Equation (11-8) for the load busses, and Equations (11-8) and (11-21) for the generator busses N 1 P jQ (11-8) Vi = i * i Yik Vk Yii Vi k =1 ( k i ) N * (11-21) Qi = Im Vi Yik Vk k =1 The admittance equivalent to the series impedance of each transmission line is given below Transmission Line From / To Series impedance Series admittance Number (bus to bus) Z, pu Y, pu 1 1-2 00210 + j01250 13071 j77804 2 1-4 00235 + j00940 25031 j10013 3 2-3 00250 + j01500 10811 j64865 4 2-5 00180 + j00730 31842 j12914 5 3-5 00220 + j01100 17483 j87413 6 4-5 00190 + j00800 28102 j11833 Now the bus admittance matrix Ybus must be built The on-diagonal elements are the sum of the admittances of all lines attached to the particular bus, while the off-diagonal elements are the negative of the admittances between the two busses For example, the on-axis admittance Y11 is
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