Trabalho Hidráulica

Dados:\nP\\u{b4} = 0,872 \\cdot 10^{3} Pa\nD_{1} = 400 \\cdot 10^{-3} m\nD_{2} = 600 \\cdot 10^{-3} m\nL_{1} = 8 m\nL_{2} = 225 m\nv_{2} = 1.51 \\cdot 10^{-6} m^{3}/s\n\\mu = 3.51 \\cdot 10^{-3} N. s/m^{2}\nREENTRANTE \\u{2260} K = 0,78\nV\\u{225}LVA DE RETEN\\u{c7}\\u{227}O \\u{2260} 600\\nV\\u{225}LVA ANGULAR \\u{2260} \\frac{K_{e}}{D} = 150\nSUC\\u{c7}\\u{227}O\nL_{S} = 8 m\nK = 0,48\n\\sum \\frac{L_{e}}{D} = 370 + \\frac{L}{400\\cdot 10^{-3}} = 20\nA_{c} = \\frac{\\pi \\cdot D^{2}}{4} = \\frac{\\pi \\cdot (400 \\cdot 10^{-3})^{2}}{4} = 0,125 m^{2}\nV_{j} = \\frac{Q}{A} = \\frac{6 \\cdot 10^{-3} m^{3}/s}{0,125 m^{2}} = 0,048 m/s\nRe = \\frac{P\\u{b4} \\cdot V \\cdot D}{\\mu} = \\frac{1000 \\cdot 0,048 \\cdot 400 \\cdot 10^{-3}}{1,51 \\cdot 10^{-3}} = 5,27 \\cdot 10^{4}\n\\frac{f}{D} = \\frac{0,045 \\cdot 10^{-3}}{400 \\cdot 10^{-3}} = 3,125 \\cdot 10^{-4}\nf = 0,029 PERDA DE CARGA NA SUC\\u{c7}\\u{227}O\nH_{L\\u{223}} = \\left(\\sum K + \\sum f \\cdot \\frac{L_{e}}{D} + f \\cdot \\frac{L}{D}\\right) \\cdot \\frac{V^{2}}{2g}\nH_{L\\u{223}} = \\left[0,78 + (0,029 \\cdot 370) + (0,029 \\cdot 20)\\right] \\cdot (0,048)^{2} \\cdot \\frac{1}{2(9,81)}\nH_{L\\u{223}} = 1,41 \\cdot 10^{-3}\nRECALQUE\n\\frac{L}{D} = \\frac{225}{600 \\cdot 10^{-3}} = 375\n\\sum \\frac{L_{e}}{D} = \\frac{600 + 150}{D} = 750\nA_{2} = \\frac{\\pi \\cdot (600 \\cdot 10^{-3})^{2}}{4} = 0,28 m^{2}\nV_{2} = \\frac{6 \\cdot 10^{-3} m^{3}/s}{0,28 m^{2}} = 0,023 m/s\nRe = \\frac{P\\u{b4} \\cdot V \\cdot D}{\\mu} = \\frac{5000 \\cdot 0,023 \\cdot 600 \\cdot 10^{-3}}{1,51 \\cdot 10^{-3}} = 8,34 \\cdot 10^{3}\n\\frac{f}{D} = \\frac{0,12 \\cdot 10^{-3}}{600 \\cdot 10^{-3}} = 2 \\cdot 10^{-4}\nf = 0,022 PERDA DE CARGA REALQUI\nH_{LR} = \\left(\\sum K + \\sum f \\cdot \\frac{L_{e}}{D} + f \\cdot \\frac{L}{D}\\right) \\cdot \\frac{V^{2}}{2g}\nH_{LR} = (0,032 \\cdot 750) + (0,032 \\cdot 375) \\cdot (0,023)^{2} \\cdot \\frac{1}{2(9,81)}\nH_{LR} = 8,09 \\cdot 10^{-4}\nALTURA MANOM\\u{e9}TRICA SUC\\u{c7}\\u{227}O\n* H_{S}\nP_{S} = \\frac{P_{0} + V_{1}^{2}}{\\rho g} + z_{1} = \\frac{P_{2} + V_{2}^{2}}{\\rho g} + z_{2} + H_{Ls}\nH_{s} = \\frac{10,1325 \: Pa - 3,5 - (0,022)^{2}}{1000 \cdot 9,81} - H_{Ls}\nH_{S} = 6,82 m - ALTURA DA PRESSÃO DE VAPOR\nH\u1EAF V = \u03B1 V\np.g\n= 0,872 . 10 \u00B3\n= 8,8 . 10 \u00B2\nm\n1000 . 9,81\n\n- ALTURA MANOMÉTRICA DA BOMBA\n* HB\nPj + V² + z1 + HB = P2 + V² + z2 + HLT\np.g p.g \n2.g\nHLT = (5,41 . 10 \u00B3) + (8,09 . 10 \u207B \u00B2) = 2,21 . 10 \u207B\n\nHB = 3,5 m\n\n- CÁLCULO DO NPSH\nNPSH d = Hs + V² - H\u1EAF V\n2.g\nNPSH d = 6,82 + (0,048)² - 8,8.10 \u207B \u00B2\n2.(9,81)\nNPSH d = 6,73 m.c.a - SELEÇÃO DA BOMBA\n\nQ = 6 l/s = 23,6 m³/h\nALTURA = 3,5 m\nJ750 npm\n\nFABRICANTE KJB\nKJB ETA: BOMBA CENTRÍFUGA PARA 250 GERAL\n50 - 52 \u03B2 105\n\nNPSH n = 9,81 - Hs + V²\n2.g\nNPSH n = 9,81 - (6,82) + (0,048)² + 0,5\n2.(9,81)\nNPSH n = 3,49 m.c.a\n\nNPSH d > NPSH n\n\"PORTANTO NÃO HAVERÁ CAVITAÇÃO\"