Atividade Estruturada Hidraulica

Estácio\nENSINO SUPERIOR\n\nENGENHARIA CIVIL\nATIVIDADE ESTRUTURADA\n\nDISCIPLINA: HIDRÁULICA\nPROFESSOR: LEANDRO BERTACO LUCIO\nNOME: RAPHAEL CARLOS VALLE\nMATRÍCULA: 201802033351 Lista de exercícios 7 - Equações de Bernoulli\n\nex 1 - D\nVn = 3,6 m/s\n\nP1 = 302 kN/m²\n\nD1 = 150 mm\nD1 = 0,02571 = 0,004 m²\n\nQe = Qs\n\nQ1 = ?\nV2 = ?\n\nP2 = ?\nQe = Qs\n\nv² = v1 + P1 + h1 = v2² + P2 + h2\n\n2g 2g\n\n3,6² + 302 * 16,2 + P2 - 3\n2,98 10 2,98 10\n\nJ6,96 = 36.376 + P2\n10 10\nP2 = 0,585 kN\n\nP0 = 36.961 - 36.376\nP2 = 5,85 kN ex 2 - D\nA1 = 100 cm² = 0,01 m²\n\nQe = Qs\nA2 = 50 cm² = 0,005 m²\n\nP1 = 0,5 kgf/cm² = 5,03\n\nP2 = 3,38 kgf/cm² = 33,8 x 10³\nh1 = 100 m\nh2 = 70 m\n\nv² + P1 + h1 = v² + P2 + h2\n\n2g 2g\n\nv1² + 5,3² + b0 = 2v1³ + 33,8 x 10³ + 70\n2,98 1000 29,81 1000 Q = V.A\nQ = 1.05 l/s\nD1 = 250 mm = 0.25 m\nQ = 0.105 m3/s\nD2 = 125 mm = 0.125 m\nV1 = 2.139 m/s\nV2 = 0.556 m/s\nV1^2 P1 + h1 = V2^2 P2 + h2\n0.223 + P1 = 3.731\nP1 = 3.731 - 0.223\nh = y - P1 = h = 0.233 + 3.498\nh = 3.731 m\nh = 3.731 m Q = V.A\nQ = 1.05 l/s\nQ = 0.105 m3/s\nD1 = 250 mm = 0.25 m\nD2 = 125 mm = 0.125 m\nV1 = 2.139 m/s\nV2 = 0.556 m/s\nV2^2 P1 + h1 = V2^2 + P2 + h2\n2.139^2 P1 + 0 = 2.556^2 + 0.2 = 2.39 D1= 450 mm = 0.55 m\nD2= 75 mm = 0.075 m\nQe = 0.05\nA1V1 = A2V2\nP1 = 344.8 kN/m2\nP1 = b1 m = 101.39 kN/m2\nV1 = \nQ = \nV2 = 0.0177 V1 = 0.00444 V2\nV2 = 4.02 V1\nQ = A1V1\nQ = 0.0177, 3.08 \nQ = 54.52 l/s\nQ = 0.545436 Ex 8 ->\nD: 300 mm = 0.30 m\nQ: 3600 L/min = 60 L/s = 0.06 m³/s\nρ: 9 kgf/cm² = 20 m = 20 kN/m²\nH: 10 m\n\nQ1 = Q2\n√(𝑦𝑏 + 𝑙1 + 𝑏2 + 𝑕𝑏2) = √𝑗𝑏 + 𝑗𝑏 + 𝑕2\nh1 = h0\nA1 = A2\n\nh2 = 20 + 10\nh2 = 30 m\nH2 = 30 m\n\nEx 9 ->\nA1 = 0.25 m\nA1 = π.(0.25)²\nA1 = 0.049 m²\nA2 = 0.20 m\n\nρ1 = 1 kgf/cm² = 1000 kg/m³\nA1 = 0.049 m²\np0: 1\n\nh1 = 10 m\nh2 = 0\n\n5.20 + p0 = p2 + 0\n\n2.5 = p0 + 0\np0 = 2.5 = 0.5\n\np2 = 2.5 kgf/cm² Ex 10 ->\nA1: 300 mm = 0.30 m = 0.073 m², A1 = π.(0.30)²\n\nh1 = 90 m\np1 = 275 kN/m²\nh2 = 75 m\np2 = 343 kN/m²\n\nΔH: 7.96 m\n\n75 - 90 = 3415 + 75 + ΔH\n\n130.03 - 110.17 = ΔH\nΔH: 7.96 m\n\nEx 11 -> Identico ao Exercicio H\nEx 18 ->\nD: 150 mm = 0.15 m\np1 = 1 atm = 10.3\nΔh = 0.75 m\nΔhBC = 1.25 m\nhA = 4.5 m\n\nQ = J.A\nQ = π.(0.15)².7 = 0.0.1237 m³/s\nQ = 0.0.237 m³/s\n\n-2.5 = p0 + 2.55\np0 = 2.5 = 2.55\np0 = 5.05 m.c.a