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

# Now solve for the heat transfer areas 18 300 8806 q1

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online