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Cursos Gerais ·

Termodinâmica 1

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Lista I\n\nη = Wcido = qenta - qresi , onde qrai = μ1 - μs\ngentra \ngentra \ngentra = 1300 KJ/Kg\n\nPropriedades do ar como gás ideal em Tf 300K:\nμs = 21407 KJ/Kg & Vhs = 62,3 m³\n\nProcesso 1-2:\n1Vr1 = Vh1 = 62,3m³ / 10 = 6,23 m³, logo T2 = 730K\nl μ2 = 536,07\n\nProcesso 2-3:\n\ngentra = μ3 - μ2\n1300 = μ3 - 536,07 → Δμ3 = 1836,07, logo l μ3 = l75\nT3 = 2450K\n\nProcesso 2-4:\nVh4 = M . 85Vr3\n1Vr4 = 10 - 2,175 = 25,75, logo T4 = 1040K & μ4 = 793,36 \ngentra = μ4 - μ3 = 793,36 - 314,07\n\ngentra = 579,09\n\nWcido = qenta - qrai = 1300 - 579,09 = 720,91\n\nη = Wcido / qentra = 720,91 / 1300\nm = 0,554 ≈ 0,6\n\n2) PME = Pot * 1300 , onde Pot em KWL e PME em bar, então:\nDVD: R\nPME = 600 KPa = 6 bar, logo:\n\nPot = PME . DVD:R\n= 2,4 litros e R: = 1800rpm\n\nPot = 6 . 24 . 1800 / 1200 = 31,6 KW\n\n3) como ar a 280K temos:\nT1 = 280K, Vr1 = 738, μ1 = 549,75\nComo ar a 2000K, temos:\nT3 = 2000K & Vr3 = 2,776\nl μ3 = 367,87 Processo 2-3:\n\np3Vr3 / T3 = p2Vr2 / T2 ⟹ p3 = p2 . T3 . Vr2 / T3\nΔp2 = p3 / T2\n\np2 = 2500\n\np1 = p2 . Vr2 / T2 ⟹ p1 = p2 . 1 / T1:\n85.000 / 280\nM = 1500\n303,57\nM = 8,235\n\n4) Já temos μ1, agora precisamos de μf:\nProcesso 3-4:\nVh4 = M . Vr3\n v\\u1d4e4 = 8,235 2,776\n|v\\u1d4e4 = 22,86, logo, T\\u1d4e0 = 1040 K, \\u03bc\\u1d4e4 = 793,36.\nProceso 1-2:\nv\\u1d4e2 = 1/v\\u1d4e3 -> v\\u1d4e2 = v\\u1d4e1 = 738/8,235\n1v\\u1d4e2 = 89,61, logo T\\u1d4e2 = 640K\n\\u03bc\\u1d4e3 = 465,5\n|g\\u1d4e8 = 1223,2\ng\\u1d4e2 = \\u03bc\\u1d4e4 - \\u03bc\\u1d4e3 = 793,36 - 199,75\n|g\\u1d4e2 = 593,61\nv\\u1d4e5 = Weido = g\\u1d4e2 - g\\u1d4e2 \\u2219 g\\u1d4e2\n\\u03bc\\u1d4e5 = 222372 - 593,61\n12,373,2 = 699,59\n\\u1d4e5 = 0,612 = 0,57 T\\u1d4e2 = T\\u1d4e4 - T\\u1d4e1\nT\\u1d4e2 = 1040 - 280\nT\\u1d4e2 = 750K \\u2248 745,3K P\\u03c3v2 = P3 T2\nT3\n\nP2 = P3\nT2\nT3\n\n1746,02 = P3\n683\n2503\n\nP3 = 2,53:2503\n\nP3 = 6,34 Mpa