Resolução do Exercícios 1 1 1 2 1 3 e 1 4 Cap 01 - Hidráulica Básica Rodrigo de Melo Porto 4ed

Equilibre de Boussinesq - jet de cerceau de quantité de mouvement\n\nβ = ∫ R ρ v² . dA = ∫ R v max (1 - (n/R)²) . 2πdn\n\nj ∫ v² . dA\n\n8/R² = ∫ R [1 - (n/R)²]. n dn\n\n∫ R [ - 1 -n²/R² + n/R¹ ] . n dn = ∫ R [ - 2n³/R² + n 5/R¹ ] dn\n\n(2n² - 2n⁴/2 + 6) | R = (R²/2R² - R²/6) = (R³/6)\n\nβ = 8/R² = 4/3\n\n\nβ = 9/3\n\nb) V(n) = Vmax (1 - n/R)^{7/4} Vmax ( (y/R)^{1/4})\nv̄ = ∫ V(n) . dA = Vmax ∫ (1 - (n/R)^{1/4} 2πr ln dn\n\n= ∫ R²/R . n²/R²\n\n= (y/R) → (y = R - n)\ndn = -dy\n\n∫ ( (R - R + y) / R) (R - y) . dy = ∫ ( (y/R)^{1/4} (y - R) . dy\n\n\n\nA\n\n\n\n∫ ( (y/R)^{1/4} (R - y) . dy Cap 01 - Radially Full\n\n1.6 Problems\n\n1) Consider α Boussinesq B\n\nv̄ = ∫ v max (1 - (n/R)²) . dA\n\n= ∫ R²R² dA (n)\nv(n) . dA = dA = πR²\n\nv̄ = ∫ (1 - (n/R)²) dA = 2/ R² ∫ v · n dn\n\n= (π R²)\n\n= (2/R²)v max ∫ (1 - (n/R)²) . dn\n\n= (2v max/R²) [Integral Value is 2.V(max)]\n\n= ∫ (n - R³) dn = ∫ [2.R2 . dA(n)(y) . R² = (3)(R ́)²\n\n(i) => λ = 2 . V max R² (R²) v max\n\n\n\nα = ∫ A v³ dA = 2ρ ( (v max³) / (R²)) A\n\n1/2 ∫ ∇³ . A\n\n= v max³/(8R)²\n\n= 1/(8) α = 16 ∫ R [ (1 - (x)²)² ( (1 - x)) . (n) )] dn = 16 ( R)⁃ 3R ( R)\n\n(1 - x)³\n\n0 ⇒ 1 - 3 n² + 3 n' n (R²) - n' • 1 ( (R)\n\n∫ ( 1 - 3 n² + 3 n' n R ) [2.n ²/r] dn\n\n∫ (n (0 - 1) (R/R)) dn = ∫ (5/8)\n\n= ∫ (6R² - R²)\n\n/((r = R^{5}) d)\n\n= R - n (3n) - (3n') - (n - R²)\n\n=\n\n(1 - 3) [R² 8R/ R²]\n\n= \n\n1.5 (n)\n\n∫ (z/(4y) + n/R) ± dy = 1/R² n\n\n∫(3 - 1)1/8 = 0∫x dy 2/R² = (R(R²)/8) = ∫ ( 6R²/ 88R²)\n\n= (0 - 1) = 16 ∫ (5=5)\n\nR = 8/R²\n\n= 1/R² √λ = 1/R\n\nR - = ∫ [ R ] [ 1/2 7/8 ] dy\n= ∫ [ 0 ] [ y ( R - y ) R^2 ] dy\n= 0\n7/8 7/8 7/8 = [ -7/8 7/8 ] - 0 = [ R ] [ R ]\n= 7/8 15 R^7 ( -7 )\n= 49 V m^2\nR^1 1/30\n\nV = 2 V m^3.\n( 49 R^2 / 30 )\n= V = 49 V \n\n\nC = \n= ∫ 0 ^ [ y ] y R^3 dy\n= \n(49 / 60) V man^3 ∫( y - R ) dy\n α = ∫ [ 0 ] R^3 R^3 R^3 dy\n0 0 1/2 = \n( 60 ) 0\n7 7/\nint = ( R )\n= R^3 R^3 R^3/( R^3 ) + ( R^3 )\n= ( 49 / 60 ) × ( 1/2 )\n= [ 49 R^2 ]\n= 3. 49 / 17\n= 49 R^2\n= \nB = 60 R^2\n= α = 49\n\nα = \n= 2 ∫ R R R^3 R^2 dy\n= V^3/\n0 1 R^2 ((y - R) R^4 dy)\n Ro = ( k y^7 ) / m^7)\nB = 2 × 60 × 4 = R^1 1.02 ∴ B = (1.02)\nD = 150 mm\nD = 75 m\nPa = 103 kN/m^2 = 3.6 m/h\n= \n∑ Q = Q1 = V 1 ∫ A R\n( 0 )\n... \nV = 3.6 × (150) / (yR) = [ 1000 ]\n= 1000 m^2 / m^3 ∴ p1 = 3. = 103 0.6 2p \n= 9.398 kN / m^2 P B = Sem condição 2\nP B = 35,83 kN/m²\n\nExample 13\nN = 9,75; Q n = 15 l/s\n\n200 m\nΔHm = 0,56 m; ΔHj = 17,92 m\n\nH\n\n1,950 m\n\nP + z + V²\n z = H e - ΔH m\n z M = H n - ΔH j \nH e = 150 + 0,56 = 150,56 m\nH j = 200 + 17,92 = 217,92 m\nHe = H n + z\n\nb) V b = Q/A = 15,10 m³/s = 1,97 m/s\n\nQ s = Q m => V q = 0,85 m/s\n\nP x = χ (H z - z e - V² / 2g) = 9,81 * 10 m²/s² [197,92 - 151,50 - (1,67)² / 2,91]\n\nP x = - 2,097 m\nH s = P n + z n + V² / 2g\n\nP 1 = 217,92 - 151,50 - 1,67² / 2,91 = 66,23 m\n\nTERRA DO SOL 2 2 du = 2 arcdanh ( u ) \n √c 1 - u²\n \n = 2 arctanh ( √v ) = 1 \n √c \n \n arctanh ( √v ) = -√c => V2 = √c . tanh ( t . √c 2 ) \n c = 2y ( z1 - z2 ) \n \n V2 = √2g ( z1 - z2 ) tanh ( t . √g ( z1 - z2 ) ) \n 2 \n \n = [ 7.67 . tanh ( 0. 6343 t ) ]\n = 7.67. tanh ( 0.6343 t ) \n \n \n Scanned with CamScanner