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

# Air is inexpensive disadvantages air can form a

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Unformatted text preview: S=KV/L. The entering gas rate, V, is 946 lbmol/h. Column operating conditions are about 94 psia and a maximum temperature of 126oF. Pertinent properties and factors for the three potential absorbents are: Absorbent C5 s light oil medium oil gpm 115 36 215 ρ, lb/gal 5.24 6.0 6.2 lb/h 36,200 13,000 80,000 MW 72 130 180 lbmol/h 500 100 440 K-value 0.9 0.005 0.0005 S=KV/L 2.3 0.013 0.0013 The light oil can not be used because its flow rate of 100 lbmol/h is much lower than the necessary 368 lbmol/h. The C5s can not be used because their stripping factor is very high. The only possible alternative is the medium oil. Exercise 6.5 Subject: Stripping of VOCs from water effluents by air and water Given: Packed tower for stripping Find: Advantages and disadvantages of air over steam Analysis: Advantages: Air is available anywhere. Air is inexpensive. Disadvantages: Air can form a flammable or explosive mixture with the VOC. With steam, the exit gas can be condensed and the VOC recovered as a liquid. Exercise 6.6 Subject: Preferred operating conditions for absorbers and strippers. Find: Best conditions of temperature and pressure, and the trade-off between number of stages and flow rate of separating agent, using equations. Analysis: Absorbers: For high performance, want a large absorption factor, A = L/KV Therefore, want a small K-value. Assume that: γ iL Pi s Ki = P s Pi = vapor pressure, which increases with increasing temperature Therefore, operate at a high pressure and a low temperature. Strippers: For high performance, want a large stripping factor, S = KV/L Therefore, want a large K-value. Therefore, operate at low pressure and a high temperature. For the tradeoff between number of equilibrium stages, N, and flow rate of mass separating agent, L or V. Consider absorption. The fraction of a component absorbed is given from a modification of Eq. (5-48), A −1 A N +1 − A 1 − φ A = 1 − N +1 = N +1 A −1 A −1 Thus, a large fraction absorbed can be achieved with either N or A = L/KV. The tradeoff is most clearly shown in Fig. 5.9. For low fractions absorbed (i.e. high φA), N has little effect and (1 φA) is approximately equal to A. But for high fractions absorbed, the larger the value of N, the smaller the required value of A, and, thus, the smaller the required value of the flow rate of the liquid separating agent, L. The tradeoff for stripping is similar. The fraction of a component stripped is given by a modification of Eq. (5-51), S −1 S N +1 − S 1 − φ S = 1 − N +1 = N +1 S −1 S −1 Thus, a large fraction stripped can be achieved with either N or S = KV/L. The tradeoff is most clearly shown in Fig. 5.9. For low fractions stripped (i.e. high φS), N has little effect and (1 - φS) is approximately equal to S. But for high fractions absorbed, the larger the value of N, the smaller the required value of S, and, thus, the smaller the required value of the flow rate of the vapor separating agent, V. xercise 6.7 Subject: Absorption of CO2 f...
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