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Engenharia Mecânica ·

Termodinâmica 1

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Exercício\n\nη_b = 83%\nη_r = 85%\nT_in = 30°C\nP1 = 5 MPa\nT1 = 40°C\nP2 = 4 MPa\nT2 = 400°C\nP3 = 3 MPa\nT3 = 500°C\nP4 = 1 MPa\n\nVALOR DÁDIVA a 380°C e 3 MPa = 3.185,20 kJ/kg \nS4 = 6,9472 kJ/kg\nS1 = 0,6945\nS2 = 7.500\n\n\n\nx2 = 6.9472 - 6.6955\n7.500 \n\nx2 = 0,8125\n\nh2 = h1 + x * (h4 - h3)\nh2 = 191,87 + 0,8125 * 2372,8 = 2155,19 kJ/kg\n\nw = η_1 (h2 - h3)\nη_1 = (h2 - h3)/h4\nW = η_1 * (h3 - h2)\nw_1 = (h2 - h3)/ (p1 - p2)\n\nw_1 = \u221A((p2- p1)\n(5000-10))\n0,001009\nη2 = 0,80 h4 = h4 - h3 = 32136 - 171.8 = 3041.8\n\nh4 = 3041,8 kJ/kg\n\nη = w_liq = 842,61 - 63 / h4\nη = 842,61 - 63 = 0,13749\n\nη = 27,49%