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

32 psia the equations that apply at the secondary dew

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Unformatted text preview: ontinued) Analysis: (c) (continued) For this condition, the partial pressure of water is 0.2(3.32) = 0.664 psia (less than 1.79 psia). The secondary dew point is much more difficult to determine. We now know that as the pressure is increased, the hydrocarbons condense first. At the primary dew point, the partial pressure of W is 0.664 psia. W will not condense until its partial pressure increases to its vapor pressure of 1.79 psia. As the pressure is increased above the primary dew point pressure of 3.32 psia, the hydrocarbons will condense, reducing their mole fractions in the vapor phase and causing the mole fraction of water to increase in the vapor phase. This plus the increased total pressure will eventually cause the water partial pressure to reach its vapor pressure and begin condensing. We know the secondary dew point pressure must lie between the bubble point pressure of 5.29 psia and the primary dew point pressure of 3.32 psia. The equations that apply at the secondary dew point, where all pressures are in psia, are: Dalton's law: P = pB + pT + 1.79 = yBP + yTP + 1.79 Raoult's law: pB = xB PBs = xB (5.22) (2) (3) pT = xT P = xT (178) . (4) Substituting Eqs. (3) and (4) into (2), P = 5.22xB +1.78xT + 1.79 (5) But in the hydrocarbon-rich liquid phase, xT = 1 - xB . Therefore, Eq. (5) becomes: (6) P = 5.22xB +1.78(1 - xB) + 1.79 = 3.44 xB + 3.57 s T Material balance equations for a basis of 100 moles, with 40 moles of B, 40 moles of T, and 20 moles of W: At the secondary dew point, only HCs have condensed. Therefore, the total moles of vapor is: V = 20P/1.79 = 11.17P (7) The moles of liquid is: L = 100 - V (8) A benzene material balance is: 40 = yBV + xBL (9) Substituting Eqs. (7) and (8) into (9), We can also write from Eq. (3), 40 = 11.17 yBP + 100xB -11.17xBP yB = 5.22xB/P (10) (11) Combining Eqs. (6), (10), and (11) to eliminate xB and yB , 3.247P2 -57.61P + 204.29 = 0 (12) Solving for the root between the bubble point and the primary dew point, the secondary dew point pressure = P = 4.90 psia. Exercise 4.72 Subject: Bubble and dew points of a mixture of benzene, toluene, and water. Given: Pressure of 2 atm. (a) 50 mol% benzene and 50 mol% water. (b) 50 mol% toluene and 50 mol% water. (c) 40 mol% benzene, 40 mol% toluene, and 20 mol% water. Assumptions: hydrocarbons and water immiscible. Find: For each mixture, bubble-point, primary dew-point, and secondary dew-point temperatures. Analysis: Use the CHEMCAD simulator with API-SRK for K-values, with the immiscible option and the three-phase flash model. The following results are obtained: Bubble point Secondary dew point Primary dew point T for Case (a), oC 89.4 89.4 100.0 T for Case (b), oC 104.3 104.3 110.3 T for Case (c), oC 93.9 98.0 99.7 Exercise 4.73 Subject: Bubble-point of a mixture of toluene, ethyl benzene, and water. Given: Mixture of 30 mol% toluene, 40 mol% ethylbenzene, and 30 mol% water at 0.5 atm. Assumptions: Hydrocarbons and water are immiscible. Find: Bubble-point temperature and vapor composition. Analysis: : Use the CHEMCAD simulato...
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