{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Find a b c analysis a and b a material balance a

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ined to be: kg/h: Component Feed, F Extract, D Raffinate, B n-Heptane, C 500 25 475 Methylcyclohexane, MCH 500 475 25 Total: 1,000 500 500 (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 gives: 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. Find: (a) (b) (c) Analysis: (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 Hf 178.6 Zr 91.22 HfO2 210.60 123.22 ZrO2 63.01 HNO3 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

{[ snackBarMessage ]}

Ask a homework question - tutors are online