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

Now solve for the heat transfer areas 18 300 8806 q1

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Unformatted text preview: f sublimation ρc = density of the crystal layer kc = thermal conductivity of the crystal layer L = length of the tube Tg = bulk temperature of the gas carrying the component that desublimes Tc = bulk temperature of the coolant ALM = log mean area for conduction through the crystal layer ALM = 2πrL − 2πro L 2πrL ln 2πro L (3) Combining (2) and (3), simplifying, separating variables, and putting in integral form, rs ro r ln kc (Tg − Tc ) r dr = ro ρc ∆H s Carrying out the integration and applying the limits gives (1). t 0 dt (4) Exercise 17.39 Subject: Evaporation of aqueous NaOH in a single-effect evaporator Given: Feed of 50,000 lb/h of a 20 wt% aqueous solution of NaOH at 120oF. Concentration to 40 wt% NaOH at a pressure of 3.7 psia. Heating medium is saturated steam at a temperature 40oF higher than the solution temperature in the evaporator. Assumptions: Perfect mixing in the evaporator. No heat losses. Find: (a) Boiling-point elevation of the solution (b) Saturated heating steam temperature and pressure (c) Evaporation rate (d) Heat transfer rate (e) Heating steam flow rate (f) Economy (g) Heat-transfer area if U = 300 Btu/h-ft2-oF Analysis: (a) From the steam tables, at 3.7 psia, saturated steam temperature = 122.3oF. From the During chart of Fig. 17.32, the boiling temperature of 40 wt% NaOH = 166oF. Therefore the boiling-point elevation = 166 – 122 = 44oF. This result is consistent with Fig. 17.33. (b) Saturated heating steam temperature = 166 + 40 = 206oF, with a pressure of 13 psia. Note: Best not to use steam under vacuum. Should change to atmospheric pressure. (c) NaOH flow rate = 0.2(50,000) = 10,000 lb/h. Entering water rate = 50,000 – 10,000 = 40,000 lb/h. Water in concentrated NaOH = (60/40)10,000 = 15,000 lb/h. Therefore evaporation rate = 40,000 – 15,000 = 25,000 lb/h. (d) Calculate heat transfer rate by an enthalpy balance around the evaporator using Fig. 17.35 and the steam tables for enthalpies. Q = mvHv + mpHp - mfHf Note that the vapor produced is superheated steam at 3.7 psia and 166oF. Q = 25,000(1,133.7) + 25,000(145) - 50,000(72) = 28,400,000 Btu/h (e) Heating steam = ms = Q/∆Hs = 28,400,000/974.2 = 29,100 lb/h (f) Economy = 25,000/29,100 = 0.859 = 85.9% (g) From (17-81), A= Q 28, 400, 000 = = 2,370 ft 2 U (Ts − Tp ) 300 ( 206 − 166 ) (1) Exercise 17.40 Subject: Comparison of single-effect to double-effect evaporation. Given: Feed of 30,000 lb/h of 10 wt% NaOH at 100oF. Concentrated to 50 wt% using saturated steam at 115 psia. Assumptions: Perfect mixing in the evaporator. No heat losses. Equal heat-transfer areas for the double-effect system. Find: (a) Heat-transfer area and economy for a single effect, with U = 400 Btu/h-ft2-oF and vapor-space pressure of 4 in Hg. (b) Heat-transfer areas and overall economy for a double-effect system, with forward feed and U = 450 for the first effect and 350 for the second effect. Vapor-space pressure of 4 in Hg for the second effect Analysis: (a) Single-effect First compute the mass balance. NaOH in t...
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