1 Lista Op 2

UNIVERSIDADE FEDERAL DE UBERLÂNDIA\nCAMPUS PATOS DE MINAS\nFACULDADE DE ENGENHARIA QUÍMICA\nGRADUAÇÃO EM ENGENHARIA DE ALIMENTOS\nDISCIPLINA: OPERAÇÕES UNITÁRIAS II\nDOCENTE: PROFE. DR. RICARDO CORREIA SALVATERRA\nDOCENTE: LÚCIO GOMES SILVA\nPATOS DE MINAS\n2016 A) TC tubo duplo\n\nm_0 = 0,15 kg/s\nT_fb = 30ºC = 303,15 K\nT_fl = 20ºC = 293,15 K\nU_p = 850 W/(m²·K)\nC_p = 4,18 x 10³ J/(kg·K)\n\nA = 7\n\nQ.U.A. ΔT_f\n Δ = Δ_Q (T_f = 42 K)\n\nΔT: T_f - T_s = 26,5\n\nΔT = 33,66 K\n\n(m_0 m_p C_p (T_f - T_0)) = (0,15 kg)(4,18·b³·3)\n\nA = (6270 W)/(850 W/(374,6 K))\n\n= 0,2193 m² = 219,3 cm² B)\nΔT_f = T_p - T_s = 26,5 K\nΔT = T_s - T_f = 32,5 K\n\nM.D.T = (T_p + T_s)/2 = (326 + 295)/2 = 34,20 K\n\nA = 0,2157 m² = 2157 cm²\n\n2) TC diluição\n-T_f = 27ºC\n-β = ?\n\nρ_f = 961 kg/m³\n= -0,0794 kg/(m³·K)\nK = 0,68 W/(m·°C)\nC_p = 4,211 kg/(kg·°C)\n\nm_w = 0,04 kgf\nl = 2,5 m\n\nc = 1,50 mm\nD_i = 3 mm = 0,003 m\nD_E = 5 mm = 0,005 m Ue = A, , f = ? , ? Ra: ?\n* *\n\n\nh0 = 0.025 m , hi = ? , ho = ?\nD\n- REFINING FOAMING DE FLUIDS\nRef = ? D= , ? , yX? , ? I\nmi = m * (0.03)2 = 7.017 *10-8\n4\n\n\n\n\r\nAvo = ?(0.0256) * (0.001) = (0.001 = 0.0147 m3/s\n\n?* ? mi - 1 = (0.001 = 0.0147 m3/s\nintercambiador : (1.0037/1.06)\n4\n\nPer = ? A i D = 4502,17\n\nD\nNus = 1.96 (A - D/C) (Pr/D3)(Re) \n\nPer = ? mi\n\n\nD = 1\n? = ? 4)\n\nVapor de agua\nmi: 120 Kg/día - 0.08333 kg/s\nTf1: 45°C = 318.15\nTfs: 15°C = 288.15 K\nTf2: 3.98°C = 304.05 K\n\nTC tubo doble: \nsiguiendo\nL: 2.743 m\n\n- tubo de agua\nK: 45 W/mK\nd1: 22.1mm = 0.0221 m\nd0: 25.4mm = 0.0254 m\n\nAio: 71000 W/(m·K)\n\nλ: 2922 K.5/(kg·K) = 2.942×10^5 K·s/kg \nCpa: 4.18 KJ/(kg·K) = 1.4810×10^3 J/(kg·K)\nCpo: 88.91 KJ/(m·K) = 1 KJ/(m·K)\n\nFd0: 20.7% = 76% K/(m·K)\n\nBALANCE DE ENERGÍA\n\nm·λ = mCpt(Tfs-Tfe)\n\nTfs(λ-mCpT)/m·Cpt = 304.05 K\n\nTf1: cons.\nTfs: corr. cont.\n\n0.5\nM.T. = 29733 = 2010\n\n0.5 A = π·D·L\nN·DT = 20.90\nluz para fijamente igual a\npara otra constante\n\nDc: 0.0254\n\nA: = π·Dc / L·(0.0254)·(2×22.1)·(4.18) = 4.1089 m²\n\nQ: = Uc·A·ΔT\n \nUc = Q·A·ΔT\n \nUc = 179733.33/3435.08 W/m²\n\nU0 = Q\nA·ΔTm\n\nUc = 19733.333 = 3453.08 W/m²\n\nb)\nT: = U0 - 1742.95 = 4.61\nUc = 1333.79\n\nc)\nA0 = Q\n \nU0·ΔT\n\nA0 = 79733.333 / (1.0 m²)\nAc: 79733.333 = 1.78\n\n%A0 = AP = 2.1888 = Q·62.20 = 62.20 m\nAc = 2.8603\n 5)\nTC tipo sam/pa, de: 0.075 m\n\nMaterial silicio\nCp: = 3381.0 J/(kg·K)\n\nAgua\nCp: 4187 J/(kg·K)\nmi: 6.98 kg/s\n\nTg1: 60°C = 339.15 K\nTf2: 42°C = 315.15 K\n\nT12 tubo\nUva: 568.8 W/(m·K)\n\nA: 7\nL: 7\n\n(SM. Cp(Tg1-Tgs) = 6.93 kg·- 3381.0 J - 24K = 6.928×10^6 W/K)\n\nQ = m·Cp·(Tf1 - Tfs) & a · mi·Cp·Tf1 = ni·Cp·Tis · (Tf1 - mi·Cp·Tds)\n\noTf1: \n(Tg1·(6.93·4187·285.15)·6.928×10^-7) = 309.17 K\n\nT2: \n\nΔT1 = 54°K\n\nCommo ΔT1 → ΔT2, consider NIDT = 54 K Q= U·A·ΔTm;\n A= m·D2 \n Q= V0·h·ΔTm => l= Q·A = 6.336×10⁶·263.00m² \n U·ΔTm\n568.4·0.0025·54\nA= mD2= π·0.0025·763= 70.66m² \n\nMLDT= AT2-AT1= 50.74K \n\nL= A \n= 6.336×10⁶\nV0·h·ΔTm \n= 568.4·0.0025·5021= \nA+ D2= π·0.0025∙286.72= 22.20m² 6) Aguas a liquido a una presion = P Tns= 10°F \n Tc= 50.74 \n A= 50.74 \n U= 5.5 Btu/h.ft²°F \n Tfs= 20°F \n mf= 7 \n Cpf= 0.293 \n UA= J A F Am \n \n Contro calorente y gas \n AT1= 300 \n AT2= Tgs-70 \n T0= Tgs-370\n \n ln(AT2/AT1) \n Q= h (Tfs-T0) = Tgs- 370 \n ln(ATs-70/AT2) \n (Tgs- 70) (Tgs- 370) \n 300 \n MLDt= AT2-AT1 = \n ln(AT2) \n m*Cp (Tge-Tgs) = mf*Cpf (Tfc-Tfs) = 0 \n m* Cpf(Tgs-T0) = UA F \n m*Cpf (Tge-Tgs)= UAF \n ln(AT2-70) \n \n 3600* 0.293* 500-Tgs= 50.50 (Tgs-370) \n 874*Ro= (Tgs-370=350) \n 2500=(500-Tgs) ln(Tgs-70)/300 \n\n 0.34992= Tgs-370\n 500-70 ln(Tgs-70) \n Tgs= 490°F => 3,14\n Tgs= 300°F => 1,31\n Tgs= 200°F => 0,61\n Tgs= 100°F => 0,29\n Tgs= 50°F => 0,54\n Tgs= 12.5°F => 0,38\n\n (Tgs= 135°F=> 0,349292) \n Tgs= 112°F => 0,3418 \n Tgs= 116°F => 0,3526 \n Tgs= 116°F => 0,3524 \n Relo were 76.0 de tantalio fuse chensian advizar que e aprobaranduento 115°F.\n Q= mi*Cpf (Tgs-Tfs)= 3600*0.293*385-336748.8*m/h.\n Q= mic*Cpf (Tgs-Tfc) + aire = \n= 336769.9- 2500. \ng( Mci )\n 1.20 \n\n 1) Agua\nmi: 6000 b/h\nTiz: 195 °F\nTfg: 210 °F\nCpc: 1\nTc: cara entrada (1-2)\ndo: lin = 1/12\nmi: 30 tubos\n\n\nQ = mi * Cpc * (Teso - Teci) = 60000. * (210 - 185) * 1.5 x 10^6\n\nR = 1.5 * 10^ -2\n\nMOT: 1 = T2 - T1\n = 96 - 71 = 82.81 °F\nLv1 = 1/72 = ln(96/71)\n\n26°\n\nQ = UA * (T1 - T2)\n = A = UD1\n\nR = U * D * L\n = 1 / U * ADTm\n\nL = 1.5 * 10^6\n= 138.10 cm = 462 ml/lub\n\n500.* r.00833. 6397 81 A: 15 Ft\nT3: Cubo-duplo\n\nA0 = T0 = 100 °F\n (liquido): mf = 5000 lb/h\n (vapor): mi = ?\n\nVerificaciones:\n1) Tps < 210 °F\n2) Tgs > 140 °F\n\nU0 = 50 Btu/h ft^2 °F, Cpq = 0.35 Btu/lb °F\n\n Q = mi * Cpq * ΔTm = U * A * ΔTm\nmi * Cpq * ΔTm = U * A * ΔTm\n\nCpt = (50 - 15) / 500 = 1.5\n\nQ = mi * Cpt (Tps - Tqs) = 5000 * 1.5 * 110 = 82500 Btu/h °F\nQ = mi * Cpt (Tgs - Tqs) + mi*q = Qp = 82500 = Cpt (Tgs - Tqs)\n\nmf = 1500 lb/h 9) Tn tubo a corriente 1-2\n\nArea = Tps = 180 °C\n Tgs = 140 °C\n (fluido)\n\nTu (120 °C)/10 °C \n\nQ = UA * F (MILD)CC\n = A = Q / U * F * (MILD)CC\n\nΔTm = D * (MILD)CC\n\nR = T1 - T2 = (376 - 284) = 0.04\n\nT1 + T2 = (302 - 121) = 0.70\n\nProp: Δt = t = (302 - 121) = 0.97\n\nF. Geometrico = F * 0.65\n\nA = 2 x 10^ -5 = 15.65 * t^2\n200.000 \n bl Unidades Cpt=Cpt=A \nQ= m \u00b7 Cp \u00b7 (TgS-Tgi) \u2192 m \u00b7 Cp = Q \nv \u00b7 FLUNIT \u00b7 cc \n\nmg= 2\u00b7105 = 2777.78 \n \n\nmg = 438.89 \n\nmB= Q = 2\u00b7105 = 1.11111\nCpt(TgS-Tgi) \n\nm \u00b7 Cp = 1.3088 \n\n\n\nr2=mg \u00b7 Cpt (Tgef-Tgi) = 1308.88\u00b72= 1\u00b7105 \n\n\nA0) \nAgi \u2212 mg= 1.25 kg/s \n\u2212(Tfin) \u2212 Tef = 35\u00b0C \n(Tssl) \u2212 1=60\u00b0C \nCpt= 4187 J/kg \u00b7 C\n\nU= 250 W/m2\u00b0C \n\nTC= todo esto esicamente como consecuencia \n\nNUT= U \n\nmg= ... \nA \u00b7 NUT (inColums) \nU ? A Q = mg \u00b7 Cp (TgS-Tg) = 9.25 \u00b7 4187 \u00b7 45 = 233519.75 W/m2\n\nA= u\u00b7Q \u00b7 Cpa (TgS-Tg) \u2192mg = 10 \n= 233519.75 \u00b7 1.91 kg/s \nCpg(TgS-TgS) \n2000 \u2216 \n\n(m \u00b7Cpg) = 1.86 \u00b7 2000= 3623.36 min\n\nCpt \u00b7 Cpt= 4.25 \u00b7 4187= 5233.75\n\n\n\u03B1= (mg)/(ml) = 0.6923 \n\nE= [TgS-Tg] - [Tg - 35] = 0.3913 \n[TgT - Tni] \n(150-15)\n\nN= A = 1 ln( e1 \u2212 1) \u2192 A = \n\nA = N (mg Cpa min) \u2192 U = 0.76 \u00b66= 3623.36 \n= 7.5 m2\n\nPara el segundo caso: \n\nmg= 0.9058 \n\n(m \u00b7Cpt) = 0.9058 \u00b7 2000 = 1811.68\n\n(m \u00b7 Cp) = 5233.75 1\u00b0 Construccion : A= 2.5 m2 \u00b710000 m2 = 750000 m2\n\n2\u00b0construccion: a= 1.14 \u00b7120000 m2= 13680 m2 x\n\nA = N \u00b7 ln(e\u2212l) \u2192 0.3461 \nCpt \n\nE = 0.2913 \n\nN= 1 \u2192 1 \n\nN = 1 \u2018 \n\nN = 1 \u2018 ln(0.3913 \u2212 1) \u2192 U = 0.53666\n\n0.5666 \u00b7 1811.68 = =1.94 m2\n\n1\u00b0 Construccion : A= 2.5 m2 = 3000 m2 = 750000 m2 \n\n2\u00b0 Construccion: A= 1.14 \u00b7 12000 m2 = 13680 m2