Lab 6 Single line to ground Fault The one line diagram of a simple power system is shown in Figure below. The actual of

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Lab 6 Single line to ground Fault The one line diagram of a simple power system is shown in Figure below. The actual of

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Lab 6 Single Line To Ground Fault The One Line Diagram Of A Simple Power System Is Shown In Figure Below The Actual Of 1
Lab 6 Single Line To Ground Fault The One Line Diagram Of A Simple Power System Is Shown In Figure Below The Actual Of 1 (130.48 KiB) Viewed 31 times
plz make report for lab 6 by computer using power system simulator for drawing
Lab 6 Single line to ground Fault The one line diagram of a simple power system is shown in Figure below. The actual of each generator is grounded through a current limiting reactor of 0.2573 pu on a 100 MVA base. The system data expressed in pu on a common 100 MVA base is tabulated below. The generators are running on no load at their rated voltage and rated frequency with their EMF's in phase. a) Determine the fault current and rating of circuit breaker if a L-G fault at bus 3 through a fault impedance Zf=j0.1 pu b) b. Now, simulate the above circuit on powerworld and apply L-G fault on bus 3. Item X1, pu 0.15 X2, pu 0.15 GI G2 T T2 L12 L13 L23 MVA Base 100 100 100 100 100 100 100 Voltage rating, KV 20 20 20/220 20/220 220 220 220 0.15 0.1 0.1 0.125 0.15 0.25 0.15 0.1 0.1 0.125 0.15 0.25 Xo, pu 0.05 0.05 0.1 0.1 0.3 0.35 0.7125 KO G1 G1 T1 T2 3 الصفحة 14 PS1 lab by: Elemary, Yousef, Alhabri % Lab 6 % Unsymmetrical Short Circuit (L-G) z133 = j*0.22; 2033 =j*0.35; zf=j*0.1; disp('(a) Balanced three-phase fault at bus 3") ia3F = 1.0/(z133+zf) disp('(b) Single line-to-ground fault at bus 3') 103 = 1.0/(2033 +3*zf+z133 + zl33); 1012-[i03; 103; 103] % transformation matrix from 012 to actual abc a =cos(2*pi/3)+j+sin(2*pi/3); tt = [1 1 1: 1 a 2 a; 1 a a 2]; iabc3 = tt*i012 disp('(c) Line-to-line fault at bus 3') il3 = 1.0/z133 +z133 + zl); i012 = [0; il 3; -113] iabc3 = tt*i012 disp('(b) Double line-to-ground fault at bus 3') il3 = 1/(2133 +z133*(z033+3*zf)/(z133+z033+3*zf)); i23 = -(1.0 - z133*i13)/z133; i03 = -(1.0 - zl33*i13) (2033+3*zf); 1012 = [103; 13; 123] iabc3 = tti012: %Fpol = abs(Frec) angle(Frec)*180/pi); %labc3p = rec2pol(Labc3) iabc3p= [abs(iabc3) angle(iabc3)*180/pi] Repeat the above analysis if fault occurs at another buses and 2 d) Design a table containing the bus voltage, current flow in each lines, and current flow from each generators. e) From table given from part d, design a suitable circuit breaker f) Comment the results obtained between calculation M-file program and power world simulation
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