Capítulo 11 - Rodrigo Porto - Hidráulica 2 - Exercícios 1 à 18

∆y = 0.50 m\n∆H = 0.05 m\nq = 7\n y1 = ?\ny2 = ?\ny2 = 0.5 + y1\n y2 = 0.5 + 0.579\n y2 = 1.08 m\nFr^2 = \\frac{3.25^2}{(2 + 1 - 0.95)} = 0 Fr = 0.75\n9.81 (1.0 - 0.95 - 2) forco especifico: \\frac{q^2}{ga} + y^2 . a.\nconus retangulares:\ny^2 = \\frac{q^2}{gy y^2} + \\frac{1}{2y^2}s.y^2 .\nFr^{2/3} = \\frac{q}{g y^3}\nFr^{1/3} = \\frac{q^{2/3}}{g^{2/3}y^{2/3}}\nF_{n}^{1/3} = (2 F_{r2}^{2} + 1) . q^{1/3}\n\\frac{2F_{r}^{2}+1}{Fr^{1/3}} = \\frac{2F_{r2}^{2} + 1}{Fr^{1/3}} Fc = yC . Ec\nFc = \\frac{q^2}{g yC^2}\nFc = \\frac{yC^3}{yC}\nFc = \\frac{3}{2} yC\nFc = yC . c . q . d .\nFr = 5.0\n y2 - y1 = 1.20\ny2 = 1.27 + y1\ny2 = 1.12 + 0.215\ny2 = 1.145 m\n∆E = (y2 - y1)^3 = \\frac{(1.415 - 0.215)^3}{4 y2 - y0}\nFr^2 = \\frac{q^2}{g y1^3} => 25 \\cdot \\frac{q^2}{g y1 . 0.215^3}\nq = 1.56 m^3 / (s . m) 1) b=2.5 m n=0.06 Io: 0.010 m/m Q: 10.0 m3/s \n h1: y2? h2: ?\n q1 = q2 \n q1 = 0.010 m3/s \n 0.06, 10 = 25.0 (25.90) \\frac{3}{2} y0: y1=0.96m\n y2^2-y1^2 = \\frac{1}{g} \n y2 = \\sqrt{y1^2 + 4y2.996} = 24.2 \\frac{2}{981} \n E: 0.96 + \\frac{z^2}{2.981.0.96} = 1.845 m \n Ec=2 y2C = 3.2 1.297 = 1.765\n ∆z = 0.08 m F=F_{r2}= \\frac{q_2}{y_2^2} =\\frac{2}{(1+8F_{r1}-1)} \n F_{r2} = 8 F_{r2}\n (1+8F_{r2}-1)3 =\n \\\\\n F_{r2} = \\frac{(1 + \\sqrt{(1 + 8F_{r2}-1)} - 1)}{2}\\cdots\n (1+8F_{r2}) = F+1 Q=41.8 m3/s y1=0.80 y2 = 15 z=15 \n v1 = \\frac{Q}{A} =? V1 = 5 m/s \n Hm1 = 0.96 = 0.4 m = \\frac{1}{2.94} F_1 = 5 \\sqrt{(9.81)} \n D1 F1: 2.52 e M = ∞ \n y2=2 y1 0.18 \n y2 = \\sqrt{16} Y? DZ:? YZ:?\n* Y=YC (en un vaciamiento crítico)\nJ5: \nΔZ = Y + DZ\n2.0g. 1,5²\n2.0g. Y²\nJ5: YC = YC + DZ\n2. \n0.15²\nE0: E2 + DH\nJ5: q² = Y + q² + (Yz. 0.3)³\n2.0g. 1.5²\n2.0g. 4.0.3. YZ\nY2 = 0.8 m\nE0: E1\nJ5: q² = 0.3 + q²\nq² = 1.485 m³/(s.m)\nY: 0.6 m\nDZ: 0.638 m J1.12\nq = 7 m³/(s.m) ΔZ:? \nE0: 650 - 641.702 = 58.7 m\n5.8: Y = YZ = 0.7 m\n2.981.Y²\nFr = 7/v921.0.3³ = 3.82\nYZ = 1/2 [√(J + 8.F₁.2)]\nY2 = 0.7 = 2 [√(J + 8.(3.2)²)] YZ = 3.45 m\nE0 = 3.121 m = 2.56 m\nDZ: 0.10 m\nb = 120\nq = 24\nY1 = 0.48 m YZ = 1.44 m + bY²/2 + bY²³/3 = q²\nq = (bY + ZY)² /2\nq = (bY + ZY)²/(5.6782)\nq = 3.23 m·s J1.16\nq = 7.0 m³/(s.m) I0 = 0.000 m³/ml m n: 0.016\n0.0167 = Y0 = 4.26 m = YZ\nFrz = 7/√(9.81.4.26) = 0.254\nYl = 1/4(16) ½ [√(J + 8.0.254²) - J]\nY1 = 0.49 m\nq = 0.5 m³/(s.m) ΔZ = 0.25 m\nYC = (0.5/9.81)¹/³ = 0.294 m\nEC = g/2 = 0.294 = 0.44 m\nE2 = YZ + 0.5Z\n2.5.81.Y²\nFrz = 0.5/√(9.81.0.663) = 0.1297\nY1 = 0.66 = 1/2 [√(J + 8.0.2997²) - J] Y1 = 0.50 m h1 = \\frac{y_2 - y_1}{E_1} = \\frac{y_2 - y_4}{E_1} = \\frac{y_2 + q_2}{-2y_2^2} = \\frac{y_2 - y_1}{y_1 + Fr_1^2} - 2 y_c = \\left( \\frac{q_2}{g} \\right)^{\\frac{1}{3}} \\left( \\frac{2 z_2}{2 g} \\right)^{\\frac{1}{3}} = \\left[ \\frac{1}{2}(y_2 y_2 + y_2 y_4) \\right]^{\\frac{1}{3}} // c.q.d