1 Lista de Op1

4) M= 70g, de al = 1, g= 15g/cm3\nTipo \u0394xi (mili) Anterior (cm) Dif.\n - 35 0 0.420 0.120 0.020 \n - 40 0 0.297 0.094 0.035 \n - 65 0.36 0.240 0.210 0.025 \n - 100 0.36 0.349 0.249 0.117 \n - 200 0.40 0.074 0.014 0.041 \n m 0.204 - - - \n \nDi = ... + 0.420 - 0.021cm Di35 = 0.041+\u00b1 0.02791 - 0.03585cm\n 2\nPi65 = 0.297 + 0.0210 - 0.02559cm Pi65 = 0.0210 - 0.0149 - 0.02795\n 2\nDi = 0.0149 + 0.0274 = 0.01175\n\nV= ... C _i = r^2 * Ni = Maxi = Maxi\n \u03b6 = 6 \u03b3 = pi Di^3 \n 6 \n \nNi = (20) (1) = 0.000 Ni = (20) (1) = 0.000\n(5.1) (0.1) (0.0210)(0.0203) \n\nNi = (20)(0.56) - 8.75 x 10^-6m Ni = (200)(0.32) - 1.932 x 10^6\n\nNi = (60)(0.10) = 1.941 x 10^-6 mol xx%\nDistribuição acumulativa\n\nb) D = 1\n∑Δxi = 0.028 + 0.081 + 0.142 + 0.106 + 0.135 + 0.065 + 0.076\nD_i 0.205 0.08205 0.0074 0.051 0.105 0.060 0.0076\n\n0.073 0.029 0.047 + 0.044 + 0.068\n0.021735 0.023 0.01935 0.023\n\n\n D = 0.0448 cm = 44.8 μm\nn = 2.2 = D' = 616.28 μm\nRBF = R(1 - (1/n))x n = (1 - 1/n)^n \n\n(0.9657) = Γ(0.5451) = Γ[0.5457 + 1] = Γ(0.5454)\n0.38887 = 1.62\n\nD' = D'/(Γ(1/n) = 646.26 = 300.47\n\n m = 258\nD = 0.8 cm²/s\nv = 0.05 m/s;\n\nρ = 0.99534 kg/m³\n\nμ = 0.8418 cp = 0.81810⁻³ p = g = 980 cm/s²\n\nEstimación de Ax = πD²/4\n\nD = k; D = C k' F\n\n1 3.0 37.71 7\n2 4.0 12.52 3.98x10⁻²\n3 6.0 28.27 1.75\n4 12.0 143.10 4.62\n\nD_x = 4.1947 * 0.2957A = 911.468/8.71890 = 12.45 – 10 x 10⁵\nke_1 = 3 · 3.53x10⁶ 10 x 10⁶\n\nxel = 78 2R + 0.97V_m = 0.95\n\n Como es claro, la validez de la presentación cuantitativa de la misma a un valor de 4.23 x 10^{-10} calculaciones así exige: C_{0} = 1/(1.63 x 10^{3})J_{m}^{-1} = (1.6-0.99-0.34)(1.66 x 10^{3})(10^{8}) = (1.472619.69)\n\n\n\nVe_{f} = 3.14 x 10^{-3}\n\n\n\n(3.14/3).(4.03 x 10^{-4}) = 4.03 x 10^{-4}\n\n(1.24 k )}_{C_{0}} [ ( 0.83 m/ C_{0})\ }\frac{(2)(2)}{C_{0}}\ [ 8.64 x 10^{-5} cm.\ = 6.64 \mu m\n\n\n\nd_{p} = C_{O}\cdot k, 1.67 x 10^{-3} = (1.4)(1.98)(10^{8})(10^{-6}) = 57.4x10^{-3} = 5.74 μm\n\nd_{p} = (1.8-0.9*0.9)( 0.1) = 0.82μm\n\n\n\nd_{pt} = [(4.39 x 10^{-3}) (1.888)(0.616 x 10^{-2})]^{1/(4.8-0.95231) (830)}=2.50 x 10^{-1} = 25.0μm\n ρ_{g} = \frac{m_{g}}{V_{g}} = (0.75-10^{-6})\ (0.22-17cm)\ \frac{5345}{1.3327}\ (g/cm^{3})\ T = 25°C\n\n\n\nT= 25°C \; \rho= 0.9978 g/cm^{3} \;\mu = 8.9 g/cm^{3}\n\n\nC (calories)\ 0.056\ 0.833\ \ 0.947\ 0.493\ 0.236\ 0.276\nN (cals):\ 4.22 x 10^{-6}\ 3.376\ \ 2.974\ 1.894\ 1.53\ 2.84\ ( 1.60)\n\nX (log(1))\ 0.625\ 0.578\ 0.736\ 0.765\ 0.490\ 0.146\ 0.725\n\nd_{m} = 0.14970\ a=0.285\ y_{O}=\ 0.373\ \)\n\n\n\nlog(10)= 0.285\ @log(Φ) = Vap\ = (10^{60}) lev. 60s.\n\nb) Supongamos que esos potenciales son estirados a pieked\n\n u_{k^{2}}(6^{5}-p_4\C) = 16μm:Kid^{2}/(pos).pu d_{p}^{*} = 18μm\ \Rightarrow d_{p} = \sqrt{\frac{18μm}{K_{lg}\cdot g}}\n\nd_{p \; cm} = (1.291\cdot g\cdot(10^{-3})) \cdots(W)\ = {3.297 x 10^{-3} cm = 32.7 \mu m\n\nd_{p^{*}} = 4.71x10^{-10} cm = 8.47 \mu m \n\nd_{p^{*}} = 11.57 x 10^{-5} cm = 7.57 \mu m\n\nd_{p^{*}} = 6.24 x 10^{-2} cm = 6.21 μm\n\nd_{p^{*}} = 5.51 x 10^{-2} cm = 5.59 μm\n\nd_{p^{*}} = 5.13 x 10^{-2} cm = 5.13 μm\n\nd_{p^{*}} = 4.93 x 10^{-2} cm = 4.986μm\n\nj = \frac{t}{(p_{g}-p_{1})}\cdot K\nD_{k} = (5.927 cm = 1 cm / 5.937 cm (capilar.); g-980 cm/s²\n10 μm\nH:100 mm = 1 cm = 10 cm\nD_{g.} = 1.69 m\np_{g} \approx max. K= 1.85\n\n\np = 0.74 Nuco2: K1 (j-s-p) dP2 = 1/(h(s-p) (1.4215)(-98) = 61.081(das/p) / (g/µ) \nMez-stro: KWW \nH1 = m/ t \nK1 = 1/(7.056)(s(p-r)ktv) \nHk = 6.991 kW \nK = (67.081 kW)/(H/A) \nKv x 1.75 = 1.75 - 0.05099 \nK = 67.081/0.05099 \n K = 0.340 kcal/(s*m²) \n0 - 3.5 µg/cm³, β = 0, μ = 0, 0 - 1 µm | 1 - 2 µm | 2 - 3 µm | \nÁgua a 20°C - \nA = 0.9482 g/cm³ \n \nTendo que q = 0 \n \nSuponha (data -)\nA: dp = 1µm - 1x10^4 am \nC³ = 4/3 \n \n\ndp(s-η) φ = + (1.10^5)(0.989982)/(3)(σ.09821)(k) = 3.14 + \nd=|S| \n Kx = 0.4632, log(1/0.065) = 0.933 log(0.0710/0.065) = 0.447\n\nKx = 5.7 (1.968-0.05) + 5.34 (1.88-0.11) = 1.584\n\nRe = 3.76 * 10^6 \nReje de Stokes K ≤ 0.5\n\nTodo processo em que ocorre uma fluição de Stokes, os consideramos mais de superfície - \n\nmédio -\n\n e dp. ~ 3.6 µm podemos condizer \n\nmédio da partícula não flui. um \n\nvalor fixado \ndp = 3.6 µm 9/9/\ng5 \ndp= 3.6 \n 1. Seguir cálculos da experiência: \nConsiderando sobre a manutenção de sí. dos impermeáveis etc. e \n\nc = \ntrabalho volumetria e eficiência em separação. \n\nφ1-seção adensição fluido do Stokes. \n\nAdmitido como a menor de Stokes: \n\nA0= 10 cm³ (s = 7.83 x 10⁻⁴) \nK = 92 cm 100 cm = 300 mm \nK = 5 % p = 5% para condição inicial: \n\nA = 1.2 m³ 100 cm = 1000 cm = 1 m³\n\nA1 = L.h = 30 cm³(1000 = 6000 cm²) \n\nv = Q = 2.783 x 10³ cm³/h + A 6000 cm² = 0.467296 cm/s \n\nSólido tem água ? Considerando que a temperatura é 20 °C (sal) \n\nρ = 0.9982 g/cm³ \n\nm = 10.05 g/cm³ \n\nPese 100 g. \n\nC: valor líquido sólido - m/s²/s = 517,85 = 0.01080 \n\nvolume líquido + va. liquido = 5.125.642 + 5(10^4) K= 0.999 \nK2 Re = 0,0.9758 = 0.799 \nK2 = 531 + 688.77 + 406 \n\n!! REGIMEN INTERMEDIARIO !! \n\nCp/Ct, 4/3 18.85175,110951(036) = 7.96 x 10^3\n5. (0.97821)(0.82676) = 0.10 \nHe= g/ρ μ = g/b \n\nD = Kμ = (0.010)(1.00)/(0.0675) = 0.42111 = 22.91 / m2\n\nInterpolando: \nest. 22 x (1.49)< = 11.4 Esto menor que 29.91 eso considerado \n . \n21.91gm 12) Ecuación de Bernoulli en O-1: P1 + 0.5 * ρ * v1^2 + ρ * g * z1 = P2 + 0.5 * ρ * v2^2 + ρ * g * z2 \n¿? = (P1 + z1) \n¿? = (P2 + z2) \nD = 70.30 N/m2 \nH = 9 m \nQ1 = P1 * H / 9 \nR = (70.30 N/m2 * 9m) / (9.81 * 9.81) - 155.800 N/m2 \nR = (P1 + z1) - (P2 + z2) = (91600.985 - 155800.122 = 40m \n\n\n\nEsto significa que si se sube un par de cañas. \n\nMs. 1. 10^-3,; f: 7 ; Deg: 7; ∇: 7\nP= 75* 2. 0.048*3in Tubos/Accesorios Log \n2 tramos rectos 3.5 m \n1 tramo intermedio 2.0 m \n4 codos de 90° Min \nMlacha 10 m \nTramos rectos 1.7m \nTotal 25.5m \n\nAc. = A _ B = 0.09555 * 0.07264 == 0.072248 / A # 1.0 10 ^-4 = 4 \nF(1) - F \nRe= ρ * D = Re = 0.31, 0.746 = 414.84 v\nμ = 5*10^-6 \nF = 6.4 / 0.015493\n\nM. 2, codo: = (0.154)(2155)(107.2)*2 = 2.4/ d2g\n0.34 * 2.46\n\nComo los sistemas pueden disminuir directo para eliminar o costo de operación.