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

# 10 from the initial drop diameter d p1 to the later

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Unformatted text preview: is obtained from the vapor pressure of water at 14.4oC = 12.3 torr, with a total pressure of 1 atm = 760 torr, yA s = PAs 12.3 = = 0.0162 P 760 Substituting into Eq. (8), nA = (5.46)(0.0314)(3.92 × 10 −5 )(0.0162 − 0) = 1.09 × 10 −7 mol/s = 1.96 x 10-6 g/s (d) At any instant of time, the rate of evaporation by mass transfer = rate of loss of drop. Let N = moles of water drop at any time, t. Then, ( ) kc Ac yA s − 0 = − Vdrop = 3 πd p 6 , 2 A = πd p , dVdrop dN ρ = − water dt M water dt Therefore, dVdrop dt From Eq. (7), kc = = (9) 2 3πd p d ( d p ) 6 dt 2 DAB dp Combining these equations and solving for dt, dt = − ρ water d p 4 M water DABcyA s d dp (10) Integrating Eq. (10) from the initial drop diameter, d p1 , to the later drop diameter, d p2 , Exercise 3.31 (continued) Analysis: (continued) 2 2 d p1 − d p2 ρ water (1.0) = t= 4 M water DABcyA s 2 4(18)(0.273)(3.92 ×10 −5 )(0.0162) ( 0.1) − ( 0.02 ) 2 2 2 = 380 s (e) By analogy to mass transfer, the rate of heat transfer from the air to the drop is, hd p 2k k dp o -6 From Perry's Handbook, thermal conductivity of air at 38 C = k = 58 x 10 cal/s-cm2-(oC/cm) h = 2(58 x 10-6)/0.1 = 0.00116 cal/s-cm2-oC. Ts - T∞ = 38 - 14.4 = 23.6oC. A = 0.0314 cm2 Therefore, the initial rate of heat transfer = Q = 0.00116(0.0414)(23.6) = 0.00086 cal/s Q = hA Ts − T∞ , where N Nu = 2 = or h = The rate of heat transfer required to maintain the drop at 14.4oC, based on the initial drop diameter is, Qneeded = ( nevap M water ) ∆H vap + ( CP ) water vapor (Ts − T∞ ) = (1.96 × 10−6 ) [589 + 0.44(38 − 14.4) ] = 0.001176 cal/s since, from Perry's Handbook, the heat of vaporization of water at 14.4oC is 589 cal/g and the specific heat of steam is 0.44 cal/g-oC. Because the heat needed is greater than the rate of heat transfer, the temperature of the drop will decrease with time initially. Exercise 3.32 Subject: Dissolution of benzoic acid when water flows in turbulent motion.through a straight, circular tube of it Given: Tube has an initial inside diameter, D, of 2 inches (5.08 cm), with a length, L, of 10 ft (305 cm). Water at 25oC flows through the tube, with a velocity, ux , of 5 ft/s (152 cm/s). Physical properties in Example 3.15. Assumptions: Fully developed turbulent flow. Find: Average concentration of benzoic acid in the water exiting the tube. Analysis: The rate of increase in benzoic acid (A) in the water (B), by the continuity equation, equals the rate of mass transfer. The only resistance to mass transfer is in the water. Using Eq. (3-105), with a log mean driving force, nA = ux S cA out − cA in = k c A cA wall − cA LM = kc A cA wall − cA in − cA wall − cA out ln (1) cA wall − cA in cA wall − cA out Because cA in = 0, Eq. (1) simplifies to, cA wall kc A = ln or solving, ux S cA wall − cA out cA out = cA wall 1 − exp − kc A ux S (2) Area for mass transfer = A = πDL = 3.14(5.08)(305) = 4,870 cm2 Cross-sectional area for flow = S = πD2/4 = 3.14(5.08)2/4 = 20.25 cm2...
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