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 (1781), A= Q
27, 000, 000
=
= 478 ft 2
U (Ts − Tp ) 400 ( 338.1 − 197 ) (1) Exercise 17.40 (continued)
(b) Doubleeffect system with forward feed and equal heattransfer 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. Boilingpoint elevation from Fig. 17.32 is
10oF for Effect 1, while boilingpoint 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 boilingpoint 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 boilingpoint 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 boilingpoint
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 heattransfer 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...
View
Full Document
 Spring '11
 Levicky
 The Land

Click to edit the document details