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

# Now solve for the heat transfer areas 7388 9588 q1 a1

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Unformatted text preview: he feed = 0.1(30,000) = 3,000 lb/h Water in the feed = 30,000 – 3,000 = 27,000 lb/h Water in the product solution = (50/50)3,000 = 3,000 lb/h Water evaporated = 27,000 – 3,000 = 24,000 lb/h Now, calculate the product solution temperature. At 4 in Hg, water boils at 125.4oF. From Fig. 17.32, the boiling temperature of 50 wt% NaOH = 197oF, consistent with Fig. 17.33. 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 4 in Hg and 197oF. Q = 24,000(1,148.6) + 6,000(222) - 30,000(62) = 27,000,000 Btu/h Heating steam = ms = Q/∆Hs = 27,000,000/880.6 = 30,700 lb/h Economy = 24,000/30,700 = 0.782 = 78.2% From (17-81), A= Q 27, 000, 000 = = 478 ft 2 U (Ts − Tp ) 400 ( 338.1 − 197 ) (1) Exercise 17.40 (continued) (b) Double-effect system with forward feed and equal heat-transfer areas The overall mass balance is the same as part (a), where the total evaporation rate is 24,000 lb/h. This system is solved iteratively, converging when the two areas are equal. As a first approximation, assume equal evaporation rates, i.e. 12,000 lb/h each. Solution leaving Effect 1 = 30,000 – 12,000 = 18,000 lb/h with a composition of 3,000 lb/h of NaOH and 15,000 lb/h water, or 16.67 wt% NaOH. Boiling-point elevation from Fig. 17.32 is 10oF for Effect 1, while boiling-point elevation for Effect 2 is still 72oF, giving a temperature of 197oF for streams leaving Effect 2. With heating steam at 338oF in Effect 1, the overall ∆T for the two effects is 338 – 197 – the boiling-point elevation of 10oF in Effect 1 (because the vapor produced there and used as heating steam in Effect 2 will condense at a temperature less by the boiling-point elevation) = 131oF. Iteration 1: Assume that the ∆T for each effect is inversely proportional to U. Thus, ∆T1 = (U2/U1) ∆T2. Also ∆T1 + ∆T2 = 131oF. Solving these equations simultaneously, ∆T1 = 57oF and ∆T2 = 74oF. Therefore, the exiting temperature in Effect 1 = 338 – 57 = 281oF. This reduces the boiling-point elevation in Effect 1 to 8oF. Use this. Therefore, the superheated vapor leaving Effect 1 will condense in Effect 2 at 281 – 8 = 273oF, corresponding to a pressure of 44 psia in Effect 1. Enthalpy balances for the two effects are: Q1 = ms ∆H svap = mv1 H v1 + m1 H1 - m f H f Q2 = mv1 ( ∆H v1 − H of saturated condensate of v1 ) = mv2 H v2 + m2 H 2 - m1 H1 (2) (3) Using Fig. 17.35 and steam tables for enthalpies, these two equations are solved for ms and m1: From (2), ms(880.6) = (30,000 – m1)(1176) + m1(222) – 30,000(62) (4) From (3), (5) ( 30, 000 − m1 ) (1176 − 242 ) = ( m1 − 6, 000 ) (1148.6 ) + 6, 000 ( 222 ) - m1 ( 222 ) Solving (4) and (5), m1 = 18,000 lb/h which was the guess. ms = 18,400 lb/h Now, solve for the heat-transfer areas: 18, 400 ( 880.6 ) Q1 A1 = = = 631 ft 2 U1 ( Ts − T1 ) 450 ( 338.1 − 281) A2 = ( 30, 000 − 18, 000 ) (1176 − 242 ) = 42...
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