Atividade Hidráulica

1) L=1100m\nD=0,10m\nf=0,000000101m2/s\nJ=AN=620-600=0,018mm\n\nJ=8f*Q2/(T2*g*D5) \n\nμ2=J*T*g*D5/ (8*f*D2)\n\nμ=√(2*J*g*D/8)\n\n/ por cota nsok - white\n1/sqrt(8) = -2 * log( (0,06*10^-3) + 2,51/\n3,7,01 \nRe\n\n1/sqrt(8) = -2 * log( (0,06*10^-3) + 2,51 / 1,0*10^-6 )\n0,37 μ.D\n\n1/sqrt(8) = -2 * log( 0,06*10^-3 + 2,51.1,0*10^-6)\n0,37√(2.0,018.9,81.0,1)\n\n1/sqrt(8) = -2 ( -3,53) => 1/f = 7,06^2 => f = 1/49,84 => f=0,02006\n\n0,018=8.0,02006*a2 => Q=0,010102 m3/s\n\n(π2.9,81.0,15) Verificação de tipo de jorna dos condutores.\n\nδ = 32,5.D\nRe = μ.D/v \nQ = A.μ\n0,010102 = π.0,01^2.μ/4\n\nδ = 32,5.0,1\n131356.√0,02006\n\nRe = 1,3367.0,1 \n1,04.10^-6\n\nδ = 0,000175 m α 0,175mm\nRe=133356\n\ne=0,06mm δ=0,058mm\n3.8=14,9mm\n\nδ<3 e δ< 85 = conduta de transição. OK\n\nConfirmado Q = 0,0042 m3/s ou 10,42 L/s