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.
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## This note was uploaded on 07/17/2008 for the course ECE 740 taught by Professor Keyhani during the Fall '07 term at Ohio State.

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

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