Hidraulica Aplicada

Julia Cristina Ramos da Silva 211 56860\nK=2000.25 m\nL=1870 m\ng=9.81 m/s²\nv=10^-6 m²/s\nΔH=45 cm\n\nFluxo gama p/Q\n1) R_f = (D / K) [ 28 D_h / L ]^0.5 - R_f = 0.3 / 10\nR_f = 376.68137.35\nR_f = 94.004\n\n2) R_f > 800 regime turbulento\n\n3) 1 / (R_f) regime turbulento\nf = [ 2.10^(-k) + 2.51 ]^(-2) D_f [ -7.100.025 + 2.51 ]\n(3.71.03 1129.06)\n\nf = [ -2.10^(-0.00025 / 1684903) ]^(-2)\n\nQ = (π² D^5 g ΔH) / (8 f L)\nQ = (π² 0.3^5 9.81 15) / (8 . 9.0121 . 1870)\nQ = 0.036 m³/s → Q = 0.192 m³/s