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

# The results are zl 0 1 2 3 4 5 6 7 8 9 ws 00058 00094

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Unformatted text preview: rom (17-66), crystal-layer thickness = (ri – rs) = 2kc (Tm − Tc ) t 1/ 2 (1) ρc ∆H f From Fig. 17.24, the temperature of the saturated solution at 60 wt% naphthalene = Tm = 44oC. It is clear from Fig. 17.24 that the crystals will be naphthalene. Assume that the coolant temperature will not change much, so that Tc = 10oC. From Example 17.13, ρc = 71.4 lb/ft3 kc = 0.17 Btu/h-ft-oF ∆H f = 63.9 Btu/lb From (1), (ri – rs) in feet = 2 ( 0.17 ) ( 44 − 10 )1.8 t 71.4 ( 63.9 ) 1/ 2 = 0.0675t 1/ 2 with t in h. Results of calculations: (ri – rs), cm 0.5 1.0 1.5 2.0 (ri – rs), ft 0.0164 0.0328 0.0492 0.0656 t, h 0.059 0.236 0.531 0.944 t, min 3.5 14.2 31.9 56.6 Exercise 17.31 Subject: Melt crystallization of paradichlorobenzene (PDCB) from a mixture with orthodichlorobenzene (ODCB) in a falling-film crystallizer. Given: A mixture of 80 wt% PDCB (melts at 53oC) and 20 wt% ODCB (melts at -17.6oC) at the saturation temperature of 43oC. A eutectic forms at 87.5 wt% ODCB and -23oC. Coolant enters at 15oC. Tubes are 8 cm inside diameter. Properties in Perry ‘s Handbook, except for kc = 0.15 Btu/h-ft-oF. Assumptions: Equilibrium. Cylindrical crystallizer with crystallization inside the tube. Find: Which isomer will crystallizer? Find the time for crystal layer thickness to reach 2 cm. Analysis: Assume the major resistance to heat transfer is the crystal layer building up on the inside of the cylindrical tube. Let ri = inside radius of the tube, and rs = distance from the center of the tube to the surface of the crystal layer, which starts at ri and decreases with time. From (17-67), k T −T t 1 2 2 rs2 r ( ri − rs ) − 2 ln ri = c (ρ m H c ) 4 s c∆ f (1) It is clear from the given melting and eutectic temperatures that the crystals will be PDCB. Assume that the coolant temperature will not change much, so that Tc = 15oC. From Perry’s Handbook, ρc = 1.458 g/cm2 ∆H f = 29.67 cal/g Also, ri = 4 cm rs = 2 cm (Tm − Tc ) = 43 – 15 = 28oC kc = 0.15 Btu/h-ft-oF or 0.00062 cal/s-cm-oC Use cgs units in (1), 2 0.00062 ( 28 t 12 ( 4 − 22 ) − 22 ln 4 = 1.614 = 1.458 ( 29.67)) 4 2 Solving, t = 4,021 s or 1.17 h Exercise 17.32 Subject: Melt crystallization in a cylindrical tube. Given: Cylindrical tube with inside radius ri, inside of which crystallization is occurring due to passage of a coolant on the other side of the tube wall. Assumptions: The only resistance to heat transfer is the crystal layer. Find: Derive (17-67): k T −T t 1 2 2 rs2 r ( ri − rs ) − 2 ln ri = c (ρ m H c ) 4 s c∆ f (1) Analysis: Let rs = radial distance from tube centerline to the inside surface of the crystal layer. Tm – Tc = overall temperature driving force (melt to coolant) For a cylindrical tube, the rate of heat conduction through the crystal layer = heat released by crystallization, based on a shell balance is: ∆H f dm dr kc (Tm − Tc ) = −2πrLρc ∆H f = ALM dt dt ri − r where, ALM = 2πL ( ri − r ) r ln i r (2) (3) Combining (2) and (3)...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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