{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

# Let yi mass fraction of oil in the overflow from

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: ge have the same composition. Find: The percent recovery of NaOH in the final overflow (extract) and the wt% NaOH in the dried CaCO3. Also, is it worthwhile to add a third washing stage? Analysis: Because the slurry composition and flow rate are given, it is not necessary to consider the precipitation reaction, which is: CaO (s) + Na2CO3 (aq) + H2O (l) = CaCO3 (s) + 2 NaOH (aq) First compute the component flow rates in the slurry leaving the precipitation vessel and entering washing Stage 1. Component Wt% Flow rate, lb/h CaCO3 5.0 5,000 NaOH 0.1 100 Water 94.9 94,900 Now consider the underflows. Each underflow will contain all of the solid CaCO3 or 5,000 lb/h of solids. Therefore, the liquid in each underflow of 20% solids will be 20,000 lb/h of aqueous NaOH. Because the underflow basis is on total solution (not just the water), use mass fractions to express the NaOH composition in the overflow and the liquid in the underflow. Let: yi = mass fraction of NaOH in the overflow from Stage i xi = mass fraction of NaOH in the underflow from Stage i Vi = lb/h of liquid in the overflow from Stage i. Li = lb/h of liquid in the underflow from Stage i. S = lb/h of fresh wash water entering Stage 2. Then: xi = yi Li = 20,000 lb/h Overall mass balance for the two washing stages gives: 100,000 + 20,000 = 5,000 + 20,000 + V1 Solving, V1 = 95,000 lb/h Total mass balance around Stage 2 gives: 20,000 + 5,000 + 20,000 = 5,000 + 20,000 + V2 Solving V2 = 20,000 lb/h Exercise 16.3 (continued) Mass balances for NaOH for each stage: Stage 1 (to which the slurry from precipitation enters): 100 + y2V2 = y1V1 + x1L1 or, 100 + y2(20,000) = y1(95,000) + y1(20,000) (1) Stage 2 (to which the fresh wash water is sent): 0 + x1L1 = y2V2 + x2L2 or, y1(20,000) = y2(20,000) + y2(20,000) (2) We have two linear equations in two variables: y1 and y2. Solving (1) and (2), gives: y1 = x1 = 0.000952 and y2 = x2 = 0.000476 Flow rate of NaOH in final overflow = y1V1 = 0.000952(95,000) = 90.44 lb/h Therefore, the recovery of NaOH in the extract = 90.44/100 = 0.9044 or 90.44% Flow rate of NaOH in the dried solids = 100 – 90.44 = 9.56 lb/h Therefore, the wt% NaOH in the dried solids = 9.56/(9.56 + 5,000) = 0.00191 or 0.191 wt%. If a third washing stage is added, the three mass balance equations for NaOH are: Stage 1: 20,000 y2 + 100 = 20,000 y1 + 95,000 y1 Stage 2: 20,000 y3 + 20,000 y1 = 20,000 y2 + 20,000 y2 Stage 3: 0 + 20,000 y2 = 20,000 y3 + 20,000 y3 Solving these three linear equations, y1 = x1 = 0.000984 y2 = x2 = 0.000656 y3 = x3 = 0.000328 The % recovery of NaOH in the extract = (95,000)(0.000984)/100 = 0.9348 or 93.48% The lb/h NaOH in the dry solids = 100 – 93.48 = 6.52 lb/h The wt% NaOH in the dry solids = 6.52/(6.52 + 5,000) = 0.00130 or 0.130 wt% The use of three stages does give a significantly increased recovery of NaOH or a significantly increased purity of CaCO3. Exercise 16.4 Subject: Recovery of zinc from a ZnS ore by oxidizing the ZnS to ZnO, followed by leaching with aq. H2SO4 to produce a decanted sludge of water soluble ZnSO...
View Full Document

{[ snackBarMessage ]}