Sol_HW_5_v2_KS_1035_3oct05

# Sol_HW_5_v2_KS_1035_3oct05 - Chemical Engineering 150B Fall...

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Chemical Engineering 150B- Fall 2005 Problem Set #5 Solutions Problem 1. (20 points) In a wetted-wall tower where ammonia, NH 3 , was stripped from an ammonia-water solution into an air stream, the overall gas coefficient, K G , was 3.12 x 10 -9 (kg mole) / (m 2 s Pa). At a plane in the tower, the bulk concentration of the falling aqueous stream was 4 kg mole/m 3 of solution and the partial pressure of ammonia in the rising gas stream was 3.04 x 10 3 Pa. For dilute solutions of ammonia in water at the operating temperature, the equilibrium partial pressure may be evaluated by ,, 3 1360 / Ai Pa p c kg mole m = If the gas phase offered 75% of the total resistance to mass transfer, calculate: (a) the individual gas-film coefficient, k G ; (b) the overall liquid-film coefficient, K L ; 4 pt 4 pt

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(c) the individual liquid-film coefficient, k L ; Note that you can also use both relations 29-15 or 29-17 in W3R to solve for these coefficients where m=Henry’s law constant. (d) the interfacial concentrations, p A,i , and c A,i ; Problem 2. (20 Points) A solid pole, 0.05 m in diameter and 5.0 m tall, is spray painted on the surface parallel to the axis with a very thin coat of paint. The paint primarily contains the 4 pt 2 pt 3 pt 3 pt
solvent benzene. The vapor pressure of the benzene at 298 K is 1.34 x 10 4 Pa. The estimated loading of solvent in the wet paint on the pole is 0.12 g of benzene per cm 2 of cylinder surface. Determine the minimum time to dry the painted pole if dry air at 298 K and 1.013 x 10 5 Pa pressure flows normal to pole axis at a velocity of 1 m/s. This problem is plug and chug and is extremely similar to Example 3 starting on page 615 of the W 3 R text. We go to the book and find the equation for gas phase mass transfer from a single cylinder with flow normal to the axis, which is equation (30-16): () 0.6 0.4 0.281 Re G D M kPS c G = , which is valid for 400 < Re D < 25,000 and 0.6 < Sc < 2.6. We will have to check these bounds as we go through the problem to make sure it is ok to use this correlation. We can calculate the Reynolds number, Re / D vD υ = , because we are given v as 1 m/s and D as 0.05 m. We just need to determine for air at 298 K. If you go to Appendix I, you can determine the viscosity of air at 300 K (1.5689 x 10 -5 m 2 /s) and 280 K (1.3876 x 10 -5 m 2 /s), and then do a linear interpolation to get the viscosity at 298 K, which is 1.5508 x 10 -5 m 2 /s. Using the values and plugging them into the Reynolds number gives Re = 3224. This is good, because it is in the range of allowable Reynolds number values for our equation. Now, we can calculate the Schmidt number, / AB Sc D = . as soon as we determine the diffusivity of benzene in air. We go to Appendix J of our book and find that the diffusivity at P = 1.013 x 10 5 Pa (or 1 atm) is 9.62 x 10 -6 m2/s. Thus, we get a Schmidt number of Sc = 1.568. Looking at the above equation, the only two properties we do not have are G M and k G . G M is the superficial molar velocity of air flowing normal to the cylinder: M v G MW ρ = , where we know v , MW is the molecular weight of air (29 kg-mole / kg), and we can determine the density of air from Appendix I. At 300 K, the density is 1.1769 kg / m 3 , at 280 K the density is 1.2614 kg / m

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Sol_HW_5_v2_KS_1035_3oct05 - Chemical Engineering 150B Fall...

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