Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# From fig 28 k values and relative volatilities are

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Unformatted text preview: 15.9(350.99) = 5,580 lbmol/h. V = L + D = 351 + 5,580 = 5,931 lbmol/h. From Perry's Handbook, at 116oF = 576oR = 320 K, vapor density = ρV = 2.64 lb/ft3 and liquid density = ρL = 29.0 lb/ft3. FLV 1/ 2 LM L ρV = VMV ρ L (5,580)(42.1) 2.64 = (5,931)(42.1) 29.0 0.5 = 0.284 Assume 24-inch tray spacing. From Fig. 6.24, CF = 0.25 From below Eq. (6-44), Ad/A = 0.1+ (0.284 - 0.1)/9 = 0.120 Assume surface tension, σ, = 10 dyne/cm. From below Eq. (6-42), FST = (10/20)0.2 = 0.87 Assume that the foaming factor, FF = 1.0 and take FHA = 1.0. From Eq. (6-42), C = 0.87(1.0)(1.0)(0.25) = 0.22 ft/s ρ − ρV From Eq. (6-40), U f = C L ρV 1/ 2 = 0.22 29.0 − 2.64 2.64 0.5 = 0.70 ft / s From Eq. (6-44), 4VM V DT = fU f π(1 − Ad / A)ρV 1/ 2 4(5931/ 3600)(42.1) = 0.85(0.70)(3.14)(1 − 0.12)(2.64) 0.5 = 8.0 ft Bottom tray of first column: Use the properties of pure propane. L in first column = L in second column + feed = 5,580 + 360 + 240 = 6,180 lbmol/h. V in first column = L in first column - B = 6,180 - (600 - 351) = 5,931 lbmol/h From Perry's Handbook, at 136oF = 596oR = 331 K, vapor density = ρV = 2.93 lb/ft3 and liquid density = ρL = 27.0 lb/ft3. Analysis: (continued) FLV Exercise 7.44 (continued) 1/ 2 LM L ρV = VMV ρ L = (6,180)(44.1) 2.93 (5,931)(44.1) 27.0 0.5 = 0.343 Assume 24-inch tray spacing. From Fig. 6.24, CF = 0.22 From below Eq. (6-44), Ad/A = 0.1+ (0.343 - 0.1)/9 = 0.127 Assume surface tension, σ, = 10 dyne/cm. From below Eq. (6-42), FST = (10/20)0.2 = 0.87 Assume that the foaming factor, FF = 1.0 and take FHA = 1.0. From Eq. (6-42), C = 0.87(1.0)(1.0)(0.22) = 0.19 ft/s ρ − ρV From Eq. (6-40), U f = C L ρV 1/ 2 27.0 − 2.93 = 019 . 2.93 0.5 = 0.55 ft / s From Eq. (6-44), 4VM V DT = fU f π(1 − Ad / A)ρV 1/ 2 4(5, 931/ 3600)(44.1) = 0.85(0.55)(3.14)(1 − 0.127)(2.93) 0.5 = 8.8 ft Tray Efficiencies: Use Fig. 7.32. Need relative volatility and average liquid viscosity. From Fig. 2.8, K-values and relative volatilities are: Component K-value at 116oF, 280 psia K-value at 136oF, 300 psia Propylene 1.00 1.14 Propane 0.87 1.00 At the top, α = 1.00/0.87 = 1.15. At the bottom, α = 1.14/1.00 = 1.14 At the top, propylene viscosity = 0.45 cP. At the bottom, propane viscosity = 0.44 cP. At the top, αµ = 1.15(0.45) = 0.52. From Fig. 7.32, Eo = 60%. At the bottom, αµ = 1.14(0.44) = 0.50. From Fig. 7.32, Eo = 62%. Actual trays and column heights: Actual tray requirements are 90/0.62 = 145 trays for the bottom column and 90/0.60 = 150 trays for top column. Column heights are: Bottom column = 144(2) + 4 + 10 = 302 ft Top column = 149(2) + 4 + 10 = 312 ft Exercise 7.45 Subject: Sizing of a vertical flash drum. Given: Flash temperature and pressure, with product flow rates in Fig. 7.48. Find: Drum height and diameter. Analysis: To determine minimum drum diameter, use Eq. (6-44) with f = 0.85 and Ad = 0. The flooding velocity is obtained from Eqs. (6-40) and (6-42), using FST = 1.0, FHA = 1.0 and FF = 1.0, with CF from the 24-inch plate spacing curve in Fig. 6.24...
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