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

25 0000020 gcm3 therefore the concentration of benzoic

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Unformatted text preview: Use the Chilton-Colburn j-factor method to obtain kc . Combining Eqs. (3-165) and (3-166), 0.023G N Re kc = ρ N Sc N Re = −0 . 2 2/3 0.023ux ρ N Re = ρ N Sc −0.2 2/3 = 0.023ux N Re −0 . 2 N Sc 2/3 (3) DG Dux ρ (5.08)(152)(10) . = = = 86,800 (confirms turbulent flow) (4) µ µ 0.0089 N Sc = 0.0089 µ = = 970 ρDAB (10)(9.18 × 10−6 ) . (5) Exercise 3.32 (continued) Analysis: (continued) Substitution of the results of Eqs. (4) and (5) into (3) gives, k c = 0.023(152) 86,800 970 −0 . 2 2/3 = 0.00367 cm/s From Example 3.15, equilibrium exists at the inside wall of the tube such that the concentration of benzoic acid in the water is at its equilibrium value of cA wall = 0.0034 g/cm3 From Eq. (2), cAout =0.0034 1 − exp − (0.00367)(4,870) (152)(20.25) = 0.000020 g/cm3 Therefore, the concentration of benzoic acid in the exiting water is way below the solubility value. Exercise 3.33 Subject: Sublimation of a long, circular cylinder of naphthalene (A) to air (B) flowing normal to it. Given: Long, circular cylinder of 1-inch (0.0254 m) diameter, D. Air at 100oC (373 K) and 1 atm flows at a Reynolds number, NRe , of 50,000. Physical properties given in Example 3.14. Assumptions: Negligible bulk flow effect. Find: Average sublimation flux in kmol/s-m2. Analysis: Only mass-transfer resistance is in the air. From Eq. (3-105), the flux is, nA / A = N A = k c cA s − cA ∞ (1) Using the Chilton-Colburn analogy, Eq. (3-169) applies, which combined with Eq. (3-165) gives, 0.0266G N Re kc = ρ N Sc −0.195 (2) 2/3 From the Reynolds number, mass velocity, G, is, using µ = 0.0215 cP (2.15 x 10-5 kg/s-m2), G= N Re µ (50,000)(2.15 × 10−5 ) = = 42.3 kg / s - m2 D 0.0254 From Example 3.14, gas density = 0.0327(29) = 0.948 kg/m3 and NSc = 2.41. 0.0266G N Re From Eq. (2), k c = ρ N Sc −0.195 2/3 −0.195 0.0266(42.3) 50,000 = 2/3 0.948 2.41 = 0.080 m/s Using properties in Example 3.14, cA s PAs 10 = yA s c = c= (0.0327) = 4.3 × 10−4 kmol / m3 and cA ∞ = 0 P 760 ( ) From Eq. (1), N A = kc cA s − cA∞ = (0.080)(4.3 × 10−4 − 0) = 3.44 ×10 −5 kmol/s-m2 Exercise 3.34 Subject: Sublimation of a sphere of naphthalene (A) into air (B) flowing past it. Given: Sphere of 1-inch (0.0254 m) diameter, D. Air at 100oC (373 K) and 1 atm flows at a Reynolds number, NRe , of 50,000. Physical properties given in Example 3.14. Assumptions: Negligible bulk flow effect. Find: Initial average sublimation flux in kmol/s-m2 and comparison to that in a bed packed with the same spheres at a void fraction of 0.5. Analysis: Only mass-transfer resistance is in the air. From Eq. (3-105), the flux is, nA / A = N A = k c cA s − cA ∞ (1) Using the Chilton-Colburn analogy, Eq. (3-170) applies, which combined with Eq. (3-165) gives, 0.37G N Re kc = ρ N Sc −0 . 4 2/3 (2) From the Reynolds number, mass velocity, G, is, using µ = 0.0215 cP (2.15 x 10-5 kg/s-m2), G= N Re µ (50,000)(2.15 × 10−5 ) = = 42.3 kg / s - m2 D 0.0254 From Example 3.14, gas density = 0.0327(29) = 0.948 kg/m3 and NSc = 2.41. −0 . 4 0.3...
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