Lista Maq Hidraulicas

Alumnos: Mykael Jhenaton Alex Franco\n1º lista ejercicio - Máquinas hidráulicas y neumáticas\n\nDado: ENTRADA:\nD1 = 50 mm\nB1 = 20°\nα1 = 90°\n\nSAIDA:\nD2 = 250 mm\nB2 = 23°\nα2 = 3°\n\nU1 = π.D1.m = π.(50x10^-3).(1720/60)\nU1 = 4,5 m/s\n\nComo entrada, radial, tenemos:\n\nEntrada\nV1m1 = V1m2\nV1t1 = V1t2\ntg(β1) = V1t1/V1m1\n\nV1m = 0\nV1t = V1t1 = 4,5 m/s\n\ntg(β1) = V1t / V1m\n\nV1m1^2 = V1m2^2 + V1t2^2\nV1m1 = √(V1m1^2 + V1t^2)\nV1m1 ≈ 4,99 m/s\n SAIDA\n\nV2t1^2 = π D2 m\nV2t1 = π(250x10^-3).(1720/60)\nV2t1 = 22,51 m/s\n\nV2m1 = V1t2 + W1t2\nV2t2 = V1m2 + V1m2\n\ntg(α2) = V1m2 / W1m2\n\nV2m2 = V1m2.tg(β2) + V1m2.tg(α2) / (tg(β2) + tg(α2))\n\nV1m2 = 22,51.tg(30° + tg(23°))\n\nV1m2 = 1,05 m/s\n\nV1n = 1 m/s\n\nV1t2 = √(20,03^2 + 2,47^2)\n\nV2t2 = 22,51 m/s\n m = 1800 rpm\n\nDADOS:\nQ = 0,6 m³/min\nD1 = 25 mm\nα1 = 90°\nW1 = 6 m/s\n\nENTRADA:\nComo α1 = 90°, tenemos:\n\nV1m = 0\nV1t1 = W1 = 6 m/s\nV1t1 = π.D1.m = π.(25x10^-3).(1800/60)\nV1t1 = 2,36 m/s\nW1t1 = 2,36 m/s\n\nV1m1 = √(6^2 + 2,36^2)\nV1m1 = 5,52 m/s\n\nSAÍDA:\nAgora temos a velocidade dos pontos a 90° para:\nW2m1 = 0\n\nV2m = √(W1m2^2 + 3^2)\nV2 = 9,89 m/s\nα = arctg(W1t / V1m)\nα ≈ 12,66° H'_{t} = \\frac{1}{g} \\left( \\vec{\\underline{v}}_{t_2}^{2} - \\vec{\\underline{v}}_{t_1}^{2} \\right)\nH'_{t} = \\frac{1}{g} \\left( \\vec{\\underline{v}}_{t_2}^{2} \\right)\nH'_{t} = \\frac{1}{g \cdot 9,81} \\left( 9,42^2 - 9,42^2 \\right)\nH'_{t} = 9,04 m \\quad \\text{Altura Teórica} Q = 0,3 m^{3}/s\n\\alpha_{1} = 90^{\\circ} \\rightarrow \\text{entrada radial}\n\\text{Entrada -}\n\\text{Como es entrada i radial, tenemos \\alpha=90^{\\circ}}\n\\vec{\\underline{v}}_{t_{1}}^{2} = \\vec{\\underline{v}}_{mir_{1}}^{2}\n\\vec{\\underline{v}}_{t_{1}}^{1} = \\vec{\\underline{v}}_{t_{1}}^{*} \\rightarrow \\vec{\\underline{v}}_{t_{1}} = 0\n\\vec{\\underline{W}}_{t_{1}} = \\vec{\\underline{W}}_{t_{1}}^{1} = \\dot{m}_{1} '\n\\text{Salida -}\n\\text{Timos a salida radial mas}\\ \\text{legura \\ beta=90^{\\circ}}\n\\text{H'_{t}} = \\frac{1}{g} \\left( \\dot{m}_{2}^{2} + \\dot{m}_{2}^{2} - \\dot{m}_{1} \\dot{v}_{t_{1}} \\right)\nH'_{t} = \\frac{1}{g} \\cdot 33,09^{2} \\rightarrow \\frac{1}{g_{9,81}}\cdot 33,09^2 = 19,35 m^{2}/s^{2}\nH'_{t} = 13,47 m P'_{h} = \\rho Q \\left( \\mu_{2}^{2} - \\mu_{1}^{2} \\right) \\slash v_{t_{1}}\nP'_{h} = 850 kg/m^{3} \\cdot 0,3 m^{3}/s \\cdot (13,09 m/s)^{2}\nP'_{h} = 43,69 kW\\quad \\text{KW}\\rightarrow HP\\quad x 1,341\nP'_{h} = 58,59 HP D1 = 69 mm\nD2 = 316 mm\nm = 1550 rpm\nQ = 5 l/m = 5×10^-3 m^3/s\n\nEntrada:\nV1' = 0\nVi = Vjm = Vwm\n\nW1' = x1 = π·D1·m\nW1' = π·(69×10^-3) · (1550/60)\nW1' = 5.96 m/s\n\nSalida:\nQ = V2m2 · π·D2·b2\n\nx2 = π·D2·m\nx2 = π·(316×10^-3) · (1550/60)\nV2 = 25.64 m/s\nW2' = √(W1' + W2')\nW2' = W2' = 0.44/√(tg 60°)\nW2' = 0.26 m/s\n\nPatiencia W1' = √(V1t^2 + V1m^2)\nV2 = √(V1^2 + V2^2)\nV2 = √(0.44^2 + 25.38^2)\nV2' = 25.38 m/s\n\nH' = 1/g (x2^2 - V2t^2 - x1^2)\nH' = 3/g (x2^2 + V2t^2)\nH' = 3/9.81 (25.64 - 25.38)\nH' = 66.33 m\n\nP1' = ρ Q (x1^2 + V2t^2)\nP1' = 1000 kg/m^3\n\nP1' = 1000 kg/m^3 · 5×10^-3 m^3/s · (25.64 · 25.38)\nP1' = 3.25 kW\n\nP1' = 4.36 HP\nPatiencia Q = 79 m^3/s = 0.022 m^3/s\nm = 1500 rpm\nαj = 90°\nρ = 1000 kg/m^3\nβ2 = 90°\n\nEntrada:\nV1' = 0\nVj = Vjm = Vwm\nW1' = x1\n\nSalida:\nW2't = 0\nV2t = V2m = V2'\nV1' = π·D2·m\nV1' = π·(420×10^-3) · (1500/60)\nV1' = 32.99 m/s\n\nP1' = ρ Q (x1·V2t - x1 · Vjs)\nP1' = 1000 kg/m^3 · 0.022 m^3/s · x1^2\nP1' = 1000·0.022·32.99²\nP1' = 23.94 kW\n\nP1' = 32.33 HP\nPatiencia D2 = 480mm\nm = 1750 rpm\n\nEntrada: \nContinua o mesmo do anterior\nSaída: termos:\n\nvti = xi\n\nvti = πD2 • m\nvti = π • 480 • 10^-3m • 1750/60 rps\nvi2 = 43,98 m/s\n\nPh = ρQ • (vti • vti - vi2 • vi3)\nPh' = 1000kg • 0,022m^3/s • 43,98^2 m/s\n\nPh' = 42,56 kW\nPh' = 57,07 HP\npotência Odão: Q = 0,055 m^3/s\nn = 970 rpm\n\nD1 = 200mm\nD2 = 400mm\nb1 = 44mm\nb2 = 39mm\nb3 = 13°\nα1 = 90°\n\nn'kid = Hman\nHt + Hr = Ht'\nHr = 1/g (vti^2 - vi2^2 - vi3^2)\n\nEntrada:\nvti > 0\nvti = π • D1 • m\nvti = π • 200 • 10^-3 • 970/60\nvti = 30,16 m/s\n\nQ = Vmi • A1 • fe1\nVmi = Q / A1 • fe1\nVm'i = 0,0,55\n\nVm'm = 0,055 Saída:\nvti = πD2 • m\nvi2 = π • 400 • 10^-3 • 970/60\nvi2 = 20,33 m/s\n\nVm2 = Q\nA2 • fe2 (47,91 K2)\nVmi = 0,055\n\nVi2 = V12 + V22\n\nwi2 = √(wi2^2 + wi2^2)\nwi2 = 20,31 m/s\nwi2 = 36,15 m/s\n\nH' = 1/g (vi2^2 - vi1^2)\nH' = 33,44 m\n\nKpf = 1 + 2,01/2\nKpf = J,323 Continuação 06\n\\eta_t = \\eta_{mci} \\eta_{vel} \\eta_{hid}\n\\eta_t = 0,75 \\cdot 0,25 \\cdot 2,18 = 0,409 \\Rightarrow 40,9% \nQual a potência da bomba para os condições?\nP_k = \\rho g QH^+ = 1000 \\cdot 9,81 \\cdot 0,055 \\cdot 25,17 \\div 0,409\nP_h = 33,34 KW \\text{ ou } 44,7 HP\nSe o resto tiver chopa d'ma, k_2=6, qual a quanta de fós para os mesmas condições?\nZ= k_2 \\cdot \\frac{D_2 + D_1}{D_2 - D_1} \\cdot \\frac{\\beta_1 + \\beta_2}{2}\nZ = 6 \\cdot \\left(\\frac{400 \\cdot 10^{-3} + 200 \\cdot 10^{-3}}{400 \\cdot 10^{-3} - 200 \\cdot 10^{-3}}\\right) \\cdot \\left(\\frac{\\sqrt{3} + 30}{2}\\right) \nZ = 6,6 m = Z = 6 m Dados:\nm=3550 npom\nD1=121 mm\nD2=242 mm\n\\alpha_1=90°\nb_2=10 mm\n\\beta_2=40°\nHman=65 m\nQ=80 l/s = 12 l/s\nQ=0,068 m³/s\nPando=0,012 m³/s\nSaída\n\\vec{V}_{1in} \n\\vec{V}_{2out} \n\\vec{V}_{1out} \n\\alpha_2\n\\vec{W}_{1c} \n\\vec{W}_{2c} \n\\vec{W}_{1t} \n\\vec{V}_{2t} \n\\vec{V}_{1t} \n\\vec{V}_{2t} \n\\vec{W}_{1t} \n\\vec{W}_{1t}\n\\vec{V}_{2t} \n\\vec{V}_{1t} \n\\vec{V}_{2t} \n\\vec{W}_{2t}\n\\vec{W}_{2t}\n\\vec{V}_{2t}\n\\vec{V}_{2t}\n\\vec{W}_{2t}\n\\vec{W}_{2t}\n\\vec{V}_{2t}\n\\vec{W}_{2t}\n\\vec{W}_{2t}\n\\vec{W}_{2t}\n\\vec{W}_{2t}\n\\vec{W}_{2t}\n\\vec{W}_{2t}\n\\vec{V}_{2t}\n\\vec{V}_{2t}\n\\vec{W}_{2t}\n\\vec{W}_{2t}\n\\vec{V}_{2t}\n\\vec{W}_{2t}\n\\vec{W}_{2t}\n\\vec{W}_{2t}\n\\vec{V}_{2t} \\vec{W}_{1m} = \\frac{Q}{A_1}\n\\vec{W}_{2m} = 0,068\n\\vec{W}_{m} = 37,89 m/s\n\\vec{W}_{m} = \\vec{W}_{1m}\n\\vec{W}_{2m} = \\sqrt{\\vec{W}_{1m}^2 + \\vec{W}_{1m}^2}\n\\vec{W}_{2m} \\text{=} \\sqrt{28,94 m/s}\n\\vec{W}_{1} = \\sqrt{\\vec{W}_{1m}^2 + \\vec{W}_{2m}^2}\n\\alpha_1 = \\text{arc.tg}(\\frac{\\vec{W}_{1t}}{\\vec{W}_{m1}})\n\\alpha = 38,49^\circ\n\\vec{W}'_{1m} = \\frac{Q}{A_1 F_1}\nF_{2m} = 0,815\n\\vec{W}_m = \\sqrt{\\vec{W}_{1m}^2 + \\vec{W}_{1m}^2}\n\\vec{W}_{m} = 83,44 m/s\n\\vec{W} = 33,16 m/s COMI: mu\u00e7\u00e3o 07-\n\nb-\n\\( n_t = n_{mc} \\cdot n_{id} \\cdot n_{vol} \\)\nn_t = 0.85 \\cdot 0.52 \\cdot 0.85\nn_t = 0.3757 \\rightarrow 37.57\\% \n\nc- \\( P_h = \\rho g Q H_t \\)\n= \\frac{1000 \\cdot 9.81 \\cdot 0.068 \\cdot 125.65}{0.3757}\nP_h = 223.3 kW ou 299.18 HP\n\nd- k_z = 8\n\nZ = k_z \\cdot \\left(\\frac{D_2 + D_1}{D_2 - D_1}\\right)\\ \\ \\text{rem} \\left(\\frac{\\sqrt{b_1 + b_2}}{2}\\right) = 8.\\ (3) \\text{rem} 42.15.\n\nZ = \\sqrt{16.3}\\ \\ %\nE = 16\\ \\ %\n 8 - \\( \\rho = 720 \\text{ kg/m³} \\)\n\\( \\alpha = 90° \\)\n\\( n_1 = 101.6 \\text{ mm} \\)\n\\( b_1 = 36.2 \\text{ mm} \\)\n\\( \\dot{m} = 570 \\text{ kg/s} \\)\n\n\\( \\bar{w}_{mj} = \\bar{v}_{mi} = \\bar{v}_i \\)\n\\( \\bar{u}_i = m\\bar{r} \\cdot D = m \\cdot r \\cdot 0 = 18.62 \\text{ m/s} \\)\n\\( \\bar{V}_{m1} = \\frac{Q}{A} = \\frac{570 \\text{ kg/s}}{\\sqrt{\\rho}} = \\frac{570}{720 \\cdot \\pi \\cdot (12.1)¹²} \\)\n\\( \\bar{V}_{m1} = 36.22 \\text{ m/s} \\)\n\n\\( \\beta_1 = \\arctg \\left(\\frac{\\bar{w}_{mi}}{\\bar{v}_{m1}}\\right) = 43.45° \\) 09 - \\( H_{mon} \\)\nDado:\n\\( \\alpha_2 = 90° \\)\n\\( D_2 = 100 \\text{ mm} \\)\n\\( b_2 = 8 \\text{ mm} \\)\n\\( m = 3500 \\text{ rpm} \\)\n\\( K_{PF} = 1.2 \\)\n\\( \\beta_2 = 30° \\)\n\nSa\u00edda\n\\( \\bar{u}_2 = \\frac{\\pi D_2 m}{60} = 18.33 \\text{ m/s} \\)\n\\( n_H = \\frac{H_{mon}}{4} \\rightarrow H_t = \\frac{12 \\text{ m}}{0.7} = 17.14 \\text{ m} \\)\n\\( H_t = \\frac{\\bar{H'}}{K_{PF}} \\rightarrow H_t = 20.57 \\text{ m} \\)\n\\( \\bar{V}_{2} = 11.01 \\text{ m/s} \\)\n\\( \\bar{w}_{2} = 7.32 \\text{ m/s} \\)\n\\( W_{i1} = W_{1} \\cdot \\cos 60° = 4.22 \\text{ m/s} \\)\n\\( V_{2} = 11.79 \\text{ m/s} \\)\n\\( \\alpha_2 = 20.96° \\)\n\\( Q = \\bar{V}_m \\cdot A = 0.00106 \\text{ m³/s} \\)\n\\( \\rho_f = \\rho g H_t = 2.14 kW \\)