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

For the continuous drying of beans can use steam tube

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Unformatted text preview: 1 ft 2 Q2 = U 2 ( 273 − T2 ) 350 ( 273 − 197 ) Because the areas are far apart, make another iteration. To do this, adjust the ∆Ts, as follows: Assume the heat duties from above will still hold. That is, Q1 = 18,400(880.6) = 16,200,000 and Q2 = 12,000(1176 – 242) = 11,200,000, both in Btu/h. Then, for equal Areas, A: A(450)(338 – T1) = 16,200,000 and A(350)(T1 – 8 – 197) = 11,200,000 Solving these two equations, T1 = 268oF, compared to 281oF in the first iteration. Exercise 17.40 (continued) Iteration 2: Equations (4) and (5) now become: ms(880.6) = (30,000 – m1)(1172) + m1(210) – 30,000(62) and ( 30, 000 − m1 ) (1172 − 242 ) = ( m1 − 6, 000 ) (1148.6 ) + 6, 000 ( 222 ) - m1 ( 210 ) Solving (4) and (5), m1 = 17,900 lb/h ms = 18,300 lb/h Thus, little change in these flow rates from iteration 1. Now, solve for the heat-transfer areas: 18, 300 ( 880.6 ) Q1 A1 = = = 510 ft 2 U1 ( Ts − T1 ) 450 ( 338.1 − 268 ) A2 = ( 30, 000 − 18, 000 ) (1172 − 242 ) = 510 ft 2 Q2 = U 2 ( 273 − T2 ) 350 ( 260 − 197 ) Thus, the problem is converged. Economy = 24,000/18,300 = 1.31 or 131% compared to 78.2% for one effect (6) (7) Exercise 17.41 Subject: Double-effect evaporation with forward feed. Given: Feed of 16,860 lb/h of 10 wt% aqueous MgSO4 at 14.7 psia and 70oF. Pressure in the second effect = 2.20 psia. Heating medium is saturated steam at 230oF. Estimated heat transfer coefficients in Btu/h-ft2-oF are 400 for Effect 1 and 350 for Effect 2. Concentrated solution from Effect 2 is to be 30 wt% MgSO4. Assumptions: Perfect mixing in the evaporator. No heat losses. Equal heat-transfer areas for the double-effect system. Neglect boiling-point elevations. Find: (a) The pressure in Effect 1 (b) Percent of total evaporation in Effect 1 (c) Heat-transfer area of each effect (d) Economy Analysis: First compute the overall mass balance: MgSO4 in the feed = 0.1(16,860) = 1,686 lb/h Water in the feed = 16,860 – 1,686 = 15,174 lb/h Water in the product solution = (70/30)1,686 = 3,934 lb/h Water evaporated = 15,174 – 3,934 = 11,240 lb/h Product solution = 1,686 + 3,934 = 5,620 lb/h With no boiling-point elevation, from the steam tables, T2 = 130oF at 2.2 psia Therefore, the overall ∆T = ∆T1 + ∆T2 = 230 – 130 = 100oF. Iteration 1: Assume that the ∆T for each effect is inversely proportional to U. Thus, ∆T1 = (U2/U1) ∆T2. Also ∆T1 + ∆T2 = 100oF. Solving these equations simultaneously, ∆T1 = 47oF and ∆T2 = 53oF. Therefore, the exiting temperature in Effect 1 = T1 230 – 47 = 183oF. Enthalpy balances for the two effects are: (1) Q1 = ms ∆H svap = mv1 H v1 + m1 H1 - m f H f Q2 = mv1 ( ∆H vv1ap ) = mv2 H v2 + m2 H 2 - m1 H1 (2) Using Fig. 17.10 and steam tables for enthalpies, these two equations are solved for ms and m1: From (1), ms(958.8) = (16,860 – m1)(1139.3) + m1(82) – 16,860(6) (3) From (2), (4) (16,860 − m1 ) ( 988.5) = ( m1 − 5, 620 ) (1117.7 ) + 5, 620 ( −17 ) - m1 (82 ) Solving (3) and (4), m1 = 11,380 lb/h ms = 7,388 lb/h Exercise 17.41 (continued) Now, solve for the heat...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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