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UNIVERSIDADE FEDERAL DE UBERLÂNDIA\nCAMPUS PATOS DE MINAS\nFACULDADE DE ENGENHARIA QUÍMICA\nGRADUAÇÃO EM ENGENHARIA DE ALIMENTOS\nDisciplina: Operações Unitárias I\nDOCENTE: Prof. Dr. Ricardo Correa Santana\nDiscente: Lúcio Gomes Silva\n2ª LISTA DE EXERCÍCIOS\nPATOS DE MINAS\n2016 A) C, h, b, d\nQ0 = 339,061 L/h, D0 = 0,0466 m, L0 = 1,000 m\nV = 0,0164 m3\n.\nA N = 15000 L/min = 250 lâminas atuam. \n.\nTamanho: d = 4,5 cm e h = 0,075 m\n1) Fluxo na área de alvenaria\ni0 = 0, dimensión dimensiva que pareceria a ser necessária = 4,31*10^-4 m³\n\nρ0 = 1024 kg/m³\ndp = L0 m/s\nμ = 4,10^-3 N.s/m²\n\n.\ndp = [(1.468.193) x dQ0]\n= dp = 1,07 x 10^-6 d\n= v0 = 0,15120\n= dp = 239,167 μm B = (1,733 x 10^-6) (250) (160 - 102.1091)\n= 169,1497 x 10^-3 / (0,03 / 0,1532)\n\n.\n\rho = (167 m/s)\n\n2) Condições iniciais:\nL = 0,020 m - 0,010 m\nQ = 7,5 x 10^-5 m3/s\n\n.\nA = 400 L/min És: 488,318 m3\nρ0 = 1600 kg\nμ = 3 x 10^-3 kg/m\n\n...\n∑ = ?\n∑ v0 L.\nច\ng = 9,81 m/s²\n\nr = z + h0 = R - k0 - 0.05\nr0\n\n∑=· r·L0 (3k h0)² 1.418,979² ∑ = 182,80\n(2,9,8) 3) Calderas Estacionarias : Des 55 cm\nCalor de agua 70ºC Presión atmosferica\nV = 15 ml= 15000 cm3\nDP = 10 mm - 20 x 10-6 m\nVerificando, determinamos la densidad\nm = ?\n(1.30 99 cm3)\nAP = ?\n(1.034, 25 cm3)\n- PROPIEDADES DE FLUIDO:\nμ = ?\n(0.002P)\nμ = ?\n(1.2 x 10-3 g/cm3)\n\nConsiderando gas ideal:\nPV = nRT\n→ P = mM\nV RT\n\nP0 = PMM/RT2 → P0 = T0\nP1 = PMM/T1 → P1 = T1\nP2 = P0T1/T2 → P2 = P1T2/T3\n\nρ = P0T0/R\ng 2 = 3.1.2 x 10-6 x 783.75 = 1.025 x 10-6 g/cm3\n$\n49.25\n\nQ = ? Una familia Estacionaria - D = De = 25 - 11 cm\n5 5\nH1 = 12 - 55 - 274.5 cm = A = K0.B = 11.75 = 302.5 cm2\n\nQ = uA = 1600 300, s = 1.675850\n- VELOCIDAD EN JABON CALORÍFICO\nV0 = A3A62 = 180.99 cm3\nH = .014 ∕ .65 = 14\n\nTAMANO DE CORTE\nQ0 = (20/20) • (200/20) • \n\n(1.51.0) 43.99 x (aplicaciones)\n\nQ0(+ala) = (2/3) =\n\n- CAIDA DE PRESION\nΔP = [(14.58)] = (F0. g) * (P1 - P2)\n(1 - 0)\n - 49 = (1.893)(lA.015\n\n0.5)(l680.73)/2 = 729.96 kg/cm3 5) Calderas Estacionarias : De 49.27 cm\n( 5000 kg/h (30.A1.A2.965710.956 °K - 600°C\nT = min kg (L/T)\nP. = 1.7\n\n0.108 =\n¿6.35? kg\n\nDiseno de установки (volume = 7)\n\nV0 = q T213.07-16\" 7.923 6.1415.00cm2)\n\n(1)•/ = 273(CAPACIDAD DE FLUIDO INTENSIVIDADES)\nPi = ?\n\nP = 1.2 x 10-6\n\ntemperatura.\nValor de presión= 0.028-6.10D-1 Me = 20 g and 100 cm³ = 2000 cm³/\n1 L =\n\nBic = μA + Bi = μe.Bi.Hc\nBc = Bc = 69,491 cm\n\n5 5\nBc = Dc 69.7 28.55 cm\n 1 2\n\nAi = 60.7 (69.491/22.355) Bi = 394/18 cm/\n Ai = 4\n\n.....\n\nD = Lp = 378/18\n\npi = 6.28·r²\n64 / \n\n64 = 754.67 cm/h\n\n1Bp = 5 6 4 30 \n\n1cm³\n\n\n\n\n\n\n\n\n D = 20 - 0.508 nm\n\n51) Zona Infinita = LaTeX\nD = 20 - 0.508 nm\n\nLa capacidad, el componente (D/A): Qr = 3.24·15 g/m³\n\nii) Analiza particula solida de = 15.2 - D: 0.14 cm\n\nMi terminas como.\n\nGral. T: a 210 °C\n\nje 103.0 - Ne (=9nm) = 10-2 Z/m²\n\naut . 6000 cm\n\n\n0.25 26.641·10-5 P/m²\n\nT (°C) | u (m/s)\n 772 18.83\n 220 19.25\n\n y = 27.19\n\n- Densidad de air (tabla)\n\n......\n\nda = 7.13·10-3 kg 1000 cm= 0.013 kg/m³\narc (cm³ = kg (m²)\n slide 50: Muestra nos grafito.\nSus D forma Lf = 0.908 m = 174.40 m/min\n1 km = 1 km 1/4\n\nBc: Dc = 0.508 m = 0.127 m\n 4 4\n\n -> Dimensión de parte ( trayso) 70%.\n\n (....sabe poco acerca de esto.),\n\nai = ( i \n\n ( 120 kg / (kg/m³) )\n\n + .2 v (9j/10-4)\n ( (60 m/30)(b 20.4 - c 2)\n ..........,\n.\n\ny = [10^(-4)]\n\n10.\n\nJD= 0.905 - 0.95 d\n\nD= D² = 19.38.\n\nD' = 4.40 µm\n\nConcluo que para particulas a las diámetros d 4.6 µm.\nson seleccionados un bajo LAPPE con eficiencia de 37%.\nT = 0.057 \n (bDC)\n\n..(a) log-<\n 0.508\n 2.00 x 10^-3 (1.35 x 10^-3) 4.39 x 10^-9 1.35 x 10^-3.\n(2.00 x 10^-3)\n4.51 x 10^-10 x 1.35 x 10^-3 = A 1.57 x 10^-10\n4.51 x 10^-7\nA: 2998.95 m/min. 60 min = 173.297 0157 m/s\n(1)\nP(H2O) (m) AP (m H2O) = (density 1) AP (m H2O) = (0.21) AP (m H2O)\n175.97\n\n\n\n(5)\n\n\nu = B\n( 0.214 )\n\nu = 96.7745 5026 m/s = 246.60 m/s\nh = x no1\nP = (B)\n \nu\n \n(A) = 3.45 (0.345(25 246.50/72) = (0.175 25 373-3, 0.010233 iv = 17 042 km/h 10 3\n0.0\n\nQ: 798.93 m^3/s, 1 min = 49 925 m/s\n60s\n(1) (H) = 39.98 12573683, r P(H) = -\n35 0) Transversando el diáf. α= 10-3 m. K(L) = 2.7797 kg/m3.\nT = 250 °C\nSección A\nl = 2 cm\n d = 50 mm.\nL1 = 80 mm\nL2 = l 60 mm\ns = 0.43\n\rho = 1\n1.00 (altitud)\n0Gg x- exp [ - \n \ndt / \n(tiempo)]\n\n d(ml) = 0.55 = (0.1052 min)\nd(ml) = 0.6 - 0.06 mm.\n\n%\n\n \nL = k(0.97)\n\n(6) to 0\n\nVer con los perfiles, como instrumento lineal= 0.2.\nValidez del paso = 7\nMierzados k tx (g)arg\n\nj = (889.39 ρ 0.3 0.3 0.4)10\n\nk x = (ρ/T) 2\n\n2 = 0.6.0.06 mm\nk- = (0.64)(ml) - ld \n2 = (0.150 (5))\n60 to +120) W = 1024.55\nK = 6.50 48 x 10^-7\nα = 1/3 0.6 - 0.901 k1\nk1 [0.8]\n \n( x |\n k1 8 \n= \n(α x ) \nk3 z\na(\n\nk/ \n2)\n\n\nd1 = 0.55 = (0.105\nA = 2777.3 = 14514 m/s\n A 160.301\n\nCalculando se hace de una y se simplifica usando una en varios pasos.\n\nAP - L < 0 = ∫φ \n φdy = ω \n - g\nρ sistemar\naplicando\nb= \n(A)\n= 0)\n\n- \nAP = - (50)(c)(e) g + M1/ \nN + P \n - ~\nL(A) = [\u200b\n- AP = - 50 (e sin e) -\n\rho \n= mundo.\n \ng\nd + China = Bar. Barina\n1) Pania = 88.06 m/h2\n\nMe resulta: \n- Ap = 800/1200*(4.451)(5.25) - 1.79*(10^-0.04)(1.57) + 1.79*(10^(-0.045))(1.79) = (g+P)(g^2) Bary(g653)\n\n- Ap = 310.97^3,506.1.0m\n\nMe resulta b:\n- A = 1.98*(10)^6, 5273.1m\n\nP(100): 6.001/ H*AP(1.35)\n58.25\n\nQ = 12478.78 m3 = 2.78 x 10^6 m2/h\n1. 100 m2\n\n15. \n12.36 - Ap = 310.97^3,408 + 192.871.503 - 41736.964 - Vt 1.373m= 52874.60\n\nQ(1.07) = (2)^(10^3) * (300g)([L,PB]) = 0.97\n\n36.15 E) Filtros\nAgua a 20º\ngravedad = 9\n\nL(a/h) = 1\n = 2\nL(channel) = 60\n = 30\nE = 0.37\nQ = 0.33\n\n y 0.07\n\n- 12.36\n-20,\n12 \n -70 \n12 \n-78\n\nSolución de variables (b)(m)= 799.99 x 10^4\n= 15\n= 20 = 0.981 m3\n= 28 = 0.1134 m\n= 35 = 0.141m\ng = 780.865 cm/s\n\n- 0.5 + 0.2 = 1\n= 0.6 = 0.42\n= 0.6 = 0.60\n= 0.67\n - 0.078\n\n8.93374 + 4.623\n= 0.973\nDescripción e índices: g= 928. k = 0.000 1m = 9.9822 con m3 Kc = Mg/ 2H^2 - \nKx = g((p1))(e^2)\n363.21(1-e)\n\nKc = (0.0779)(0.0209)\n36.180(1-0.27)\n\nKc = (3.01)(0.04)2\n26.76 10^-0.07\n\nKc = 5.8417 x 10^-6\nKc = 3.2079 x 10^5\n\n= 0.14\n\n= 0.42\n\np1 = (3.10)25\n(2) + (p1 - pb)\n + 1g^2\n\nAplicando el teorema de pruebas es un signo gustos bajo 2:\n(P/fp) + Az = Bt\n\nP/g = 1 10230 = l h u c p g + q g l t f i r e s o n g \n g f k t v e \n\n190 = l m u s t e r g + l a g e + d e s i g n g \n p g = g a t s \n\n480 = m e s u r e g + l a g e + b l a n k 2 \n l g e = f a e g \n\n[ l i t t l e 2 + f t g + t w o v g l 180 = 0 ]\n\n60.155971 + 30.08495 [ g ] 60.1099752 + 30.0020^3 g = 0 \n1.681:0\n1.681:0 \n\n40346837 • 567 152 32539 - 100 = 0 \n\nA b l i t + t a c u = 1.4057 \u00d7 10^8 \n\ny = -105^ 157.90373 \u00d7 1.4057\t10^8 - p + (-105152) + 105.1537 1867\n506.167 886.287 \n\ng = -260.6196 \n\nq = 2.74141 \u00d7 10^{-3} cub \u00d7 \n3000\% / 1 m = 0.07297 m^3 h \n / h = 1hz \n4000m \n\n# FAQ \nG O L F D E E S \nP E T I N E F E E S \nR E S T A R R A M S\n# PAS