At 45C, 1,000 kg/h of a mixture of 0.80 mass fraction docosane and 0.20 mass fraction diphenylhexane
Question:
At 45C, 1,000 kg/h of a mixture of 0.80 mass fraction docosane and 0.20 mass fraction diphenylhexane is extracted with pure furfural to remove some diphenylhexane from the feed. Phaseequilibrium data are given in Exercise 8.15. Determine the:
(a) Composition and flow rate of the extract and raffinate from a single equilibrium stage for solvent flow rates of 100, 1,000, and 10,000 kg/h;
(b) Minimum solvent flow rate to form two liquid phases;
(c) Maximum solvent flow rate to form two liquid phases;
(d) Composition and flow rate of the extract and raffinate if a solvent flow rate of 2,000 kg/h and two equilibrium stages are used in a countercurrent-flow system.4
Data From Exercise 8.15
The system docosane–diphenylhexane (DPH)–furfural is representative of complex systems encountered in the solvent refining of lubricating oils. Five hundred kg/h of a 40 wt% mixture of DPH in docosane are to be extracted in a countercurrent system with 500 kg/h of a solvent containing 98 wt% furfural and 2 wt% DPH to produce a raffinate of 5 wt% DPH. Calculate, with a right-triangle diagram, the stages required and the kg/h of DPH in the extract at 45οC and 80οC.
The tie lines in the docosane–diphenylhexane–furfural system are:
Step by Step Answer:
Separation Process Principles Chemical And Biochemical Principles
ISBN: 9780470481837
3rd Edition
Authors: By J. D. Seader, Ernest J. Henley, D. Keith Roper