Lista de Hidráulica - A3 - Respostas - Monalisa

Ana Carolina Ribeiro Tinto RA: 20635170\nAna Claudia Julio Padoani RA: 20620669\nDiogo de Oliveira Saga RA: 21470172\nJoão Lucas de Carvalho Conia RA: 20873019\nThiago Silva Do Nascimento RA: 20338328\nEx1.\n\nTraço 3\n\nJ: 10,65, Q = 10,65. 0,050 = 0,210 m/m.\nC = 4,85 D = 0,20\n\nΔH = J. L3 = 0,210. 300 = 6,3 m.\n\nPonto B\n\nHc = Hc + ΔH6 = J. 3 + 6,3 = 10,9,3 m.\n\nTraço 1\n\nJ: ΔH = 117 = 0,1463 m/mm.\n\nLT = 80\nQ = 0,279 . 20, 0,10 = 0,1463 = 0,208 m3/s.\n\nTraço 2.\n\nQ3.Q1 = 0,050 - 0,208 = 0,022 m3/s.\n\nLT = 160 + 40 = 200 m. Jz = ΔH = 117 = 0,05385 m/mm.\n\nLT 200 = 0,05385m/mm.\n\nQ = 0,279 . z,63 . 0,54\n\nD = (0,279 . C . D )/(0,279 . 120 . b . 0,0585) = 0,125 m.\n\nD= 0,125 m.\n\nEx2\n\ny1 - y2 = Hf + ΔH = 50-45 = 5 m.\n\nk = 12,3 m\nD = 50 mm = 0,05 m\nL real = 45 m\nLT = 57,9 m.\n\nΔH = 10,65 . Q. L => 5 = 10,65 . Q, 2,35 5,79\n\nQ = 5 . 125 . 0,05 / 10,65 . 5,79 = 0,00348 m3/s\nou 3,48 L/s. Ex3\n\nQ = A.V Q75 = Q150\nV150 = 3 m/seg.\n\n75 mm\nA = π/4. D² = π/4. (0,75)² = 0,49\n\n150 mm\nA = π/4. (1,5)² = 1,76\n\n0,49.V = 1,76.3\nV75 = 12 m/Δ\n\nEnergia A = 41,12 + (1,193 m - 1,12 m) / 1,5 1,8 = 0,60 m\nEnergia B = 6,71 + (6,9 - 6,71) / 3 . 6 = 7,03 m\n\nyA + PA + v²A /2g = yB + PB + v²B /2g + ΔH\n\n0,60 + 1,12/ 2 = 7,03 + 3² / 2 + ΔH\n\n7,94 = 7,51 + ΔH\nΔH = 0,331 m.\n\nΔH = k. v² / 2g => 0,331 = k. 1,2/ z.981. Ex 4. \\u0394h = 15 m\nL1 = 1000 m\nD1 = 400 mm = 0,4m\nL2 = 800 mm\nD2 = 300 mm = 0,3m\nf = 0,0 20.\n\n\\u0394h = 8.f = 8.0,020 = 0,00165.\n\\u03b3 = 9,81\n\nQ1 = Q2 = Q.\n15 = 0,0065\\u03bc2/(1000 + 800)/(0,40 + 0,30) = 0,146 m3/s.\n\n15 = 0,0065.\\u0394\\u20322 \n(B.L) = 0,40. = 0,30.0,25 =\n1/2 \n(B.L)\n15 = 0.0065.\\u0394\\u20322 (600 +132)\nQ = 0,200 m3/s/\n\nAna Carolina Ribeiro Timbo RA:20635170\nAna Claudia Julio Padaromi RA:20626069\nDiego da Oliveira Sampaio RA:21470127\nJo\\u00e3o Lucas do Carvallo Conia RA:20873019\nThiago Silas Do Nascimento RA:20383283 Ex 5.\n\\u0394h = 6 mm\nL1 = 1200 mm\nD1 = 0,254m\nC1 = 140.\nL2 = 800 mm\nD2 = 0,2032\nC2 = 120.\nC3 = 130\nL3 = 232m\nQ = 206 l/s = 0,102 m3/s\nQ = 0,2785.140.0,025. J = 0,902 m2/s\nJ = 0,00 326 mm/m.\n\\u0394h =\\u0394h1 + \\u0394h2 + \\u0394h3 = 6 m. \\u0394h3 = 2,256 m.\n\\u03b3 = 9,81.0,2 = 2,56. \n232\n0,2785.130. d.\n\\u03c6 = 0,146 mm\nD = 150 mm = 6\"\n\nEx 6.\nTrucho 1\nL = 300 mm\nD = 200 mm = 0,2 m\n\nTrucho 2\nL = 370 mm\nD = 300 mm = 0,3 mm\n\nTrucho 3\nL = 1220\nD = 450 mm = 0,45 m\n\\u0394h = 11,619+1,387+0,213 = 14,325m\n\nAna Carolina Ribeiro Timbo\nRA: 20635170.\nAna Claudia Julio Padaromi\nRA: 20626069\nDiego da Oliveira Sampaio\nRA: 21470127\nJo\\u00e3o Lucas do Carvallo Conia\nRA: 20873019\nThiago Silas Do Nascimento\nRA: 20383283 Ex 7\nR1 = 613mm\nR2 = 648 mm\n\\u0394h = 45 m\nC = 320\nTrucho 1\nL1 = 1000 m\nD1 = 0,25 m\nTrucho 2\nL2 = 500 mm\nD2 = 0,15m\nTrucho 3\nL3 = 360\nD3 = 0,1m\nTrucho 4\nL4 = 140 mm\nD4 = 0,20 m\n\n\\u0394h = 10,65. L\n\\u2206L8 = 0,04 1\n\\u0394h = E1 + E2 + E3\nQ4 = Q1 = Q2 + Q3\n\\u0394h = \\u0394h3.\n\nQ4 = Q1 + Q2 + Q3\n0,041 = 1516 + Q3\nQ3 = 0,063 m3/s\n\\u03c3 = 16,31 l/s.\n\nAna Carolina Ribeiro Timbo\nRA: 20635170\nAna Claudia Julio Padaromi\nRA: 20626069\nDiego da Oliveira Sampaio\nRA: 21470127\nJo\\u00e3o Lucas do Carvallo Conia\nRA: 20873019\nThiago Silas Do Nascimento\nRA: 20383283 C) ΔH = 10,65. Q.L = 10,65.0,041.1000 = 5,93m\nΔH = 5,93m = 6gm.\n\nAna Carolina Ribeiro Timbo RA: 20695170\nAna Claudia Julio Padorni RA: 20626069\nDiego d’Oliveira Sousa RA: 21470172\nJosé Lucas de Carvalho Coria RA: 20873019\nThiago Silva Do Nascimento RA: 20383328 Ex8.\nTraço AD\nl = 450mm\nQ = 260l/Δ\nL = 600mm\nC = 130.\n\nΔHAD = JAD = 10,65. Q 1,85 = 10,65. 0,276 = 0,0052\nC 1,85.D1,37. \n\nTraço DB\nl = 300mm\nQ = ?\nL = 450m\nC = 130.\n\nD = 400mm = √6m.\n\nAna Carolina Ribeiro Timbo RA: 20695170\nAna Claudia Julio Padorni RA: 20626069\nDiego d’Oliveira Sousa RA: 21470172\nJosé Lucas de Carvalho Coria RA: 20873019\nThiago Silva Do Nascimento RA: 20383328 Ex3\n549+1=550m.\nQmín = Q1+Q2.\nQmín = Q1.\nTraço A\nC = 1.10\nL = 850m\nD = 12 = 0,3m\n\nTraço B\nC = 1.00\nL = 450m\nD = 8 = 0,20m.\nVazão mínima.\nQ = Q1\nΔH = 554-552 = 2m\nQ = 2.110.0,3 = 0,049 m³/s.\n\nAna Carolina Ribeiro Timbo RA: 20695170\nAna Claudia Julio Padorni RA: 20626069\nDiego d’Oliveira Sousa RA: 21470172\nJosé Lucas de Carvalho Coria RA: 20873019\nThiago Silva Do Nascimento RA: 20383328 Ex 10.\n\nTraco 1\n\\u0394H = 7m\n\nL = 1400mm\nC = 105\nD = 400mm = 0,4m\n\nTraco 2\nL = 2000mm \\u0394H = 8m\nj = 0,0269\nD = 300mm = 0,3m\n\nTraco 3\nL = 1.000m\nC = 30\nD ?\n\\u0394H = 13mm.\n\n\\u0394H = 10,65. Q^{1.85} L\n\nQ^{1.85} = \\frac{1.85}{1.87}\n\nQ_1 = 7 . 105 = 0,15 m\\u00b3/b.\n\n10,65 . 1400\n\nV\\u00e1g\\u00e3o Traco 1 = 0,15 m3/b.\n\nV\\u00e1g\\u00e3o Traco 2\n\nHA : PA + ZA = 505 + 10 = 515 m.\nj\n\\u0394H = 523 - 515 = 8m.\n\nQ = \\frac{(7.5^{0.85} \\u0394H)^{\\frac{1}{2}}}{8 . j . L^{2}} = \\frac{\\pi . 0.0269 . 3.18}{8.0.0263.2000} = 0,066 m3/b.\n\nV\\u00e1g\\u00e3o Traco 2 = 0,66 m3/b.\n\nV\\u00e1g\\u00e3o Traco 3.\n\n\\u0394H = 13mm.\n\nQ_1 = Q_1 + Q_3 = 0,15 + 0,066 + Q_3 = 0,089 m3/\\u0394.\n\n\\u0394H = 10,65 . Q^{1.85} \n\n\\u0394H = 10,65 . 0,084\n\n\\frac{10,65}{0,07}\n\nD = 80mm\n\nou 0,28m.\n\nAna Carolina Ribeiro Sim\\u00f5s\nRA: 20635170.\nAna Claudia Julio Padorni\nRA: 20626069.\nDiego de Oliveira Souza\nRA: 21470172.\nJo\\u00e3o Lucas do Carvalho Cont\\u00edn\nRA: 20873019.\nThiago Silva Do Nascimento\nRA: 20933228.