This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ined to be:
Feed, F Extract, D Raffinate, B
(b) Assume the given minimum extract reflux ratio = 3.49 = LR/D in Fig. 8.36a.
Then, LR = 3.49(500) = 1,745 kg/h. Therefore, the flow rate of solvent-free extract in stream VN
= 1,745 + 500 = 2,245 kg/h. But, from above, this stream contains 84 wt% aniline solvent.
Therefore, the flow rate of aniline in stream VN = (84/16)(2,245) = 11,790 kg/h. This is the flow
rate of aniline that must be removed by the separator at the top of the cascade.
(c) From above, the raffinate L1 leaving at the bottom of the cascade is 7.5 wt% aniline.
Therefore, aniline flow rate in B is (7.5/92.5)(500) = 40 kg/h and B = 540 kg/h
The entering solvent rate SB is, therefore, 40 + 11,790 = 11,830 kg/h. This is the amount of fresh
solvent that must be added to the mixer at the bottom of the column.
(a) Referring to Fig. 8.36a, define the raffinate reflux ratio as (L1 - B)/B. Because the cascade is
operating at minimum reflux, assume infinite stages so that VB is in equilibrium with L1. From
the triangular diagram on the previous page, the composition of VB is obtained by the dashed tie
line from L1, which gives 93.5 wt% aniline. A total material balance around the bottom Mixer
SB - B = 11,830 - 540 = 11, 290 = VB - L1 . An aniline balance around the bottom Mixer gives:
11,830 - 40 = 0.935VB - 0.075L1 . Solving these two equations gives L1 = 1,435 kg/h.
Therefore, the raffinate reflux ratio = (1,435 - 540)/540 = 1.66. Exercise 8.21
Subject: Liquid-liquid extraction of hafnium from zirconium in an aqueous nitric acid solution,
using tributyl phosphate (TBP) as the solvent
Given: A flowsheet in Fig. 8.46 for a process comprised of 14 equilibrium stages and a solvent
stripping unit. Feed consisting of 1.0 L/h of 5.10 N HNO3, containing 127 g of dissolved Hf and
Zr oxides per liter, including 22,000 ppm by wt. Hf, enters Stage 5. Stages 5 to 14 constitute an
extraction section. TBP enters Stage 14. Scrubbing water enters Stage 1, with Stages 1 to 4
constituting a scrubbing section. Aqueous raffinate is removed from Stage 14. Organic-rich
extract leaving Stage 1 enters the Stripping Unit, which recovers the TBP for recycle by stripping
with fresh water. The aqueous product from stripping leaves the process. Stagewise data for the
distribution of both Hf and Zr between TBP (containing nitric acid) and aqueous nitric acid.
Assumptions: The 22,000 ppm of Hf refers to a basis of Hf plus Zr, not to the total feed.
(a) and (b) A material balance.
A check on consistency of the given data.
Advantages of the process. Need for all stages. The molecular weights of the components involved are:
Component Molecular weight
From the CRC Handbook, the density of aqueous 5.1 N HNO3 = 1.166 g/cm3
Neglect the effect of dissolved oxides on this density.
1.0 L/h = 1,000 cm3/L. Therefore, in the feed have 1,166...
View Full Document
This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
- Spring '11
- The Land