P1 2007 - Poli-usp - Engenharia Elétrica

1a Prova de Sistemas de Pot\u00eancia I - PEA 2410 - 2007\n\nnome:\n\nNa solu\u00e7\u00e3o do sistema abaixo, adote Sbase = 4MVA:\n\n1 2\n T1 \n\nlinha Z = 10\u00b0 29\u00b0(Q) \n \n 3 4\n T2\n\nxk = 20%\nSk = 4MVA\n 20/10 kV\n xk = 10%\nVann = 5,77 kV\n\n1a Quest\u00e3o: Considerando apenas o gerador e o transformador T1 (linha desconectada), \u00e9 ligada uma carga monof\u00e1sica indutiva na barra 2 com 40\u03a9. \n\na) Obter as correntes (kA) e tens\u00f5es (kV) na barra 2 (secund\u00e1rio do transformador).\n (Valor 3,0)\nb) Obter as correntes de linha (kA) e tens\u00f5es (kV) na barra 1 (prim\u00e1rio do transformador)\n (Valor 2,0)\n\n2a Quest\u00e3o: Considerando o sistema completo, com a carga em opera\u00e7\u00e3o, aplica-se um curto circuito fase-terra rigido na barra 1. Obter a corrente de curto circuito fase-terra, em kA,\n\n(valor 3,0)\n\n3a Quest\u00e3o: Cada fase de uma linha de transmiss\u00e3o tr\u00edfica \u00e9 constitu\u00eddas por tr\u00eas cabos Hawk espa\u00e7ados em 40 cm. As fases est\u00e3o espa\u00e7adas de 10 m e est\u00e3o a uma mesma altura em rela\u00e7\u00e3o ao solo (disposi\u00e7\u00e3o horizontal). Adotando a resist\u00eancia de 50\u00b0 C, calcule a imped\u00e2ncia s\u00e9rie e a capacit\u00e2ncia da linha de transmiss\u00e3o por quil\u00f4metro, admitindo linha transposta.\n\n(valor 2,0)\n\nTABELA DE CABOS\n\nNome Area Al. N.Fios Ext. Resist\u00eancia\n cmil Al/Apol. pol. DC 20\u00b0C AC 60Hz AC 60Hz GMR\n 1.6cm X_m\nHawk 477.000 267 0.838 0.0357 0.1931 0.2120 0.0289 0.430 0.0998 1a Quest\u00e3o: \n\n \n \n \n \n \n \n \n \n a) No secun dario\n \n \n\n \n \n \n \n \n \n \n j1 = 0,895 kA.\n A \n \n \n \nj0.1\n \n \n \n kuy1 j0.3 = j0.894/30\u00b0 \n \n\n V12 = 0,563/60\u00b0 p.u.\n\n\n j2 = ob. pode ser obtida com 120\u00b0\ne rolar mais no kiter\n\n I2 = 3/20 0,1155 kA.\n \n \n \n \n \n \n \n \n \n I1 2 7,21/0\u00b0 \n \n \n j0.4 j2.2 = j0.0 \n \n \n \n \n I2 = (\n IABC ){ \n \n { \n { \n \n { Vabc = \n \n = [I \n { \n { j1.1\n b) No p\u00faincia -\n \n Escolha dois respectivos planos \n \n\n\n i13 = I1/2/30\u00b0\n \n { I2 = I2/10\u00b0 \n \n \n I2 = I2/30\u00b0\n \n \n Tentos:\n\nOb.\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009\u2009 2ª Questão\n\n16 j0 \n\nLine 2,10. \n\nLine 3, \nTensão de Thevenin no ponto 3\n\nNo ponto 3, resistência:\n\n\n\n\n\n\n\n\n1 = 12ca\n81 j1\n2,2\n(j04) = j0,33G\n\nDiagrama de eqüação zero\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nj0,33\n\nCircuito equivalente no ponto 3\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nI1oId = 0.285 \n\n\nj0.833G\n\n 3ª Questão\n\n\n\nDmg = 3 V/k0\n\n12,6 m.\n\n\n\ncom os valores tab. fatores, \no termois:\n\nrmG = 0,0088687 \n\nfect = 0,0106897 \n\n\nreg = 3/\nsmg² = 0,1104 m.\n\n(11 cm)\n\nrjc = 0,1204 m.\n\nImpedância em série.\n\nZ = 0,044 \n+ 2wo,0\n2ln Dmg = 0,044 + j0.356\n\n\n\n\nCapacitância.\n\n1/C = 17,98 x ln Dmg\n\nregc = 83,62 km\n\nC = 11,96 uF/km\n