lecture8

# lecture8 - EE 740 Professor Ali Keyhani Lecture#8 Homework...

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EE 740 Professor Ali Keyhani Lecture #8 Homework Solutions

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Example 5.5: Consider the 3 phase system shown. At bus 2, S 3 φ base = 100 MVA and V = 345 kV. a) Determine all bases at all buses. b) Properly convert all data into per unit, and place results on a per-phase diagram. 1-3 φ 3-1 φ Solution: Sb = 100/3 = 33.3 MVA Vb 2 = 345/ 3 = 199 kV Vb 1 = 199(35/350) = 19.9 kV Vb 4 = 199((20/ 3)/200) = 11.5 kV Zb 1 = (Vb 1 ) 2 /Sb = (19.9) 2 /33.3 = 11.9 Zb 2 = Zb 3 = (199) 2 /33.3 = 1190 Zb 4 = (11.5) 2 /33.3 = 3.97 Zpu = 12.8 + j64/1190 = 0.0108 + j0.0538 pu Y/2 pu = j.280 x 10 -3 /(1/1190) = j.333 pu Bank T1 Bank T2 120 MVA 35kV/350kV Z=(1+j8%) on ratings 30MVA/each 200kV/20kV Z=(1+j7%) on ratings Z=12.8+j64 Y/2=j.280 mS Vb 1 =19.9kV Vb 2 =199kV Vb 4 =11.5kV 1 2 3 4 Y Y
ZpuT 1 = Zold (Vold/Vnew) 2 (Snew/Sold) = (.01+j.08)(350/ 3) 2 (33.3) 199 (120/3) = (.01+j.08)(350 2 /199)(33.3/120) = (.01+j.08)(350 2 /120)(1190) = .0086 + j.0686 pu ZpuT 2 = (.01+j.07)(200/199) 2 (33.3/30) = .0112 + j.0784 pu Sb = 100/3 Vb 1 Vb 2 Vb 3 Vb 4 1 2 3 4

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Example 5.11: Consider the 3 phase system shown. a) Tabulate all four base values(S,V,I,Z) at all four buses (16 total values) for S 3 φ base = 100MVA and V Lbase = 34.5 kV at bus 2. b) Draw the positive equivalent circuit for the given system. Locate all four buses. c) Convert all impedance data provided into per unit, enter values into the circuit in (b). 3-1 φ 1-3 φ Solution: 3 φ Sb = 100 MVA Vb 2 = 34.5 (L-L) = 199 kV Vb

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