Distillation of Hydrocarbon

A feed stock available at the rate of 1000 mol/h and consisting of (all in mol %)

20% Propane (C3) 30% Isobutane (i-C4)20% Isopentane (i-Cs) 30% Normal pentane (Cs)

is to be separated into two fractions by distillation. The distillate is to contain all of the propane fed to the unit and 80% of the isopentane fed to the unit and is to consist of 40% isobutane. The bottoms stream is to contain all the normal pentane fed to the unit.Calculate the complete distillate and bottoms analysis.

Production of Absolute AlcoholThe conventional method for separating ethyl alcohol from an alcohol/water mixture is by distillation. However, this procedure can at best only produce a product 95% by volume alcohol because alcohol and water form a constant boiling mixture called an azeotrope, of that composition. Simple distillation cannot eliminate the azeotrope. Instead, if a pure alcohol product is desired, benzene is added to the feed solution. The benzene itself forms an azeotrope with water but one which has a lower boiling point than the alcohol and thus the alcohol can be purified.

Suppose it is desired to produce 1000 Ib/h pure ethyl alcohol in this manner by distilling a feed mixture containing 60% water and 40% alcohol (in mass %). If the distillate composition (in mass %) is 75% benzene and 24% water, with the rest alcohol, how much benzene must be fed to the column?

Extraction of Benzoic AcidBenzoic acid can be extracted from a dilute water solution by contacting the solution with benzene in a single-stage extraction unit. The mixture will separate into two streams: one consisting of benzoic acid and benzene and another stream consisting of all three constituents as shown in the schematic of Figure. Benzene is slightly soluble in water; thus, stream 4 will contain 0.07 kg benzene/kg water. Benzoic acid will distribute between streams 3 and 4 as follows:

The feed solution, stream 1, contains 2 x 10-2 kg acid/kg water and is fed at the rate of 104 kg/h:Suppose benzoic acid extracted into stream 3 is worth $1/kg and that fresh benzene (stream 2) costs 3¢/kg. Construct a graph of net profit vs. benzene flow and select the optimum benzene flow rate.

Aromatics Separation TrainConsider a separation train consisting of two distillation columns which are designed to separate a three component mixture of benzene, toluene, and xylene into three streams, each rich in one of the species. The system diagram listing the species present in each stream is given in Figure.

Determine how much material will be processed by each unit and how this material will be divided among the outlet streams.

Given a feed rate of 1000 mol/h of a mixture consisting of 20% benzene, 30% toluene, and the rest xylene (all mole %), a bottoms product of 2.5% benzene and 35% toluene is obtained in the first unit and an overhead product of 8% benzene and 72% toluene in the second unit.

Aromatics Separation Train (2)The feed to a two-column fractionating system is 30,000 Ibm/h of a mixture containing 50% benzene (B), 30% toluene (T), and 20% xylene (X). The feed is introduced into column I and results in an overhead consisting of 95% benzene, 3% toluene, and 2% xylene. The bottoms from column I are fed to the second column, resulting in an overhead from column II containing 3% benzene, 95% toluene, and 2% xylene (see Figure P2.19).

Assume that 52% of the feed appears as overhead in the first column and that 75% of the benzene fed-to the second column appears as overhead, calculate the composition and flow of the bottoms stream from the second column.

Aromatics Separation Train (3)The feed to a unit consisting of two columns contains 30% benzene (B), 55% toluene (T), and 15% xylene (X). The overhead stream from the first column is analyzed and contains 94.4% B, 4.54% T, and 1.06% X. The bottoms from the first column is fed to the second column. In this second column, it is planned that 92% of the original T charged to the unit shall be recovered in the overhead stream and that the T shall constitute 94.6% of the stream. It is further planned that 92.6% of the X charged to the unit shall be recovered in the bottoms from this column and that the X shall constitute 77.6% of that stream (see Figure P2.20).

If these conditions are met, calculate:a)Analysis of each stream leaving the unit.b)Percentage recovery of benzene in the overhead stream from the first column.

Multiple Effect EvaporatorEvaporators are typically used to concentrate solutions by boiling off some of the solvent. To economize on the energy input required, evaporation is often carried out in stages with each stage providing some of the evaporation duty. In the multistage evaporation shown in Figure P2.22, a 50% by weight sugar solution is concentrated to 65% by evaporating an equal amount of water in each of the four stages. With a total input of 50,000 Ibm/h, determine the concentrations of the intermediate streams.

2.21 Purification of Titanium DioxideA slurry consisting of Ti02 precipitate in a salt water solution is to be washed in three stages as shown in the flowsheet of Figure P2.21. If the feed slurry consists of 1000 lbm/h of 20% Ti02, 30% salt, and the rest water, calculate the wash water rate to each stage. Assume that:

b) The stages are operated so that the slurry leaving contains one third solids.

c) In each stage, the salt concentration in its waste solution is the same as the salt concentration of the solution entrained with the slurry leaving that stage.

a) 80% of the salt fed to each stage leaves in the waste solution.

2.23 Multiple Stage Process: Alcohol WashingA mixture of alcohol and ketone containing 40 wt % alcohol is washed with successive water washes to remove the alcohol. The wash water used contains 4 wt % alcohol. The water and ketone are mutually insoluble. The distribution of alcohol (A) between the ketone (K) and the water streams (W) is given by:

(See Figure P2.23A) a)If 150 lb of stream 2 is used per 200 lb feed mixture, calculate the composition of stream 4 after one wash. b)If the washing is repeated with 150 lb wash at each stage, how many wash stages are required to remove 98% of the original alcohol (see Figure P2.23B)?

Recycle: Condenser RefluxStaged distillation columns are devices which separate volatile materials by boiling off the more volatile components. In order for a clean separation to take place, these devices require that at least part of the vapor boiled off be condensed and returned to the column. This is necessary to ensure that both liquid and vapor phases coexist throughout the column. Consider the recycle system consisting of a distillation column and its condenser. Suppose the column is used to separate a three-component mixture consisting of 7% acetone, 61.9% acetic acid, and 31.1% acetic anhydride. The column is designed to yield a bottoms stream containing no acetone and a distillate containing 10% acetone and 88% acetic acid. If the column is operated so that 60% of the overhead is returned as reflux, calculate all flows assuming all compositions are in mol % and that 700 mol/h of distillate is to be produced.

DMF Solvent RecoveryA purification system with recycle is used to recover the solvent DMF from a waste gas containing 55% DMF in air. The product is to have only 10% DMF (see Figure P2.24). Calculate the recycle fraction assuming that the purification unit can remove two thirds of the DMF in the combined feed to the unit.

Gas Purification SystemA widely employed method of gas purification is to selectively absorb the undesirable constituents of the gas into a specifically selected liquid medium. The liquid medium is subsequently regenerated by chemical or heat treatment to release the absorbed material.

Removal of Sulfur CompoundIn a particular installation, the purification system for the removal of sulfur compounds, designed to operate at a feed rate of up to 820 mol/h, is temporarily subjected to a feed rate of 1000 mol/h. Since the absorption system simply can accommodate only 82% of this flow, it is proposed that the overload be bypassed and that the exit H2S concentration of the absorption system be reduced far enough so that the mixed exit stream contains only 1% H2S and 0.3% COS on a mole basis (see Figure P2.25). Calculate all flows in the system. The feed stream consists of (mole basis) 15% C02, 5% H2S, and 1.41% COS, with the remainder being CH4.

Removal of Sulfur Compound (2)Suppose the operation of the absorption system of the previous is modified so that all of the COS and one mole of C02 per mole of H2S are absorbed as shown in the schematic in Figure P2.26. The feed stream consists of (mole basis) 15% C02, 5% H2S, and 1.41% COS, with the remainder being CH4. Again, 18% of the feed bypasses the absorption system. Calculate all flows in the system.

By-pass: Concentration of Orange Juice

Fresh orange juice typically consists of about 12% (mass %) dissolved solids, largely sugars, in water. In order to reduce the cost of shipping, the juice is often concentrated prior to shipping and then reconstituted by adding water at the destination.

Concentration must be carried out in specially designed, short residence-time evaporators operated at below atmospheric pressures in order to reduce the loss of volatile and thermally sensitive flavor and aroma components present in trace amounts.

Since some loss of these components is nonetheless unavoidable, a widely accepted approach is to somewhat overconcentrate the juice and then add a small amount of fresh juice (called a cutback) to the concentrate to produce a product of improved aroma and flavor.

Concentration of Orange Juice (Cont’)Suppose that 10% of the feed to such a process is used as cutback and that the evaporator is operated to produce an outlet concentrate containing 80% dissolved solids. If the process feed rate is 10,000 lb/h fresh juice, calculate the rate at which water must be evaporated and the composition of the final product.

Vegetable Oil Extraction

During the extraction step, about 2 lbm solvent, hexane, must be used per 1 lbm clean seeds processed (see Figure P2.29). For each ton of raw seeds to be processed, determine the amount of oil and oil-free meal produced traction unit.

Oilseed protein sources include soybean, cottonseed, peanut, sunflower, copra, rapeseed, sesame, safflower, castor, and flax. Commonly, the separation of the oil from the protein meal is performed by solvent extraction.

The analysis of cottonseed is 4% hull, 10% linter, 37% meal, and 49% oil.

Instant Coffee ProcessInstant coffee is produced via the flowsheet in Figure P2.27. Roasted, ground coffee is charged with hot water to a large percolator in which water-soluble material is extracted. The extract is spray dried to produce the product, and the wet grounds are partially dewatered prior to disposal as land fill or by incineration. For simplicity, the coffee feed is assumed to consist of solubles, insoluble, and no water. The standard charge consists of 1.2 lb water per lb coffee. As a reasonable approximation, it may be assumed that the solubles-to-water ratio in the two streams leaving the percolator is the same. Similarly for the separator and the press, but not for the drier.Calculate the ratio of solubles recovered to solubles lost in the waste stream.

Instant Coffee Process Modification

Calculate the recovery ratio resulting from this modification. Assume that the feed coffee composition is 0% water and 32.7% insolubles and allow the feed water-to-coffee ratio to be variable. The solubles-to-water ratios in both streams leaving the press are the same.

In the plant flowsheet considered in previous problem, recovery of solubles as product is quite low. In an effort to improve the recovery, suppose the waste solution from the press is recycled back to the percolators. However, to reduce the possible release of bitter-tasting material during pressing, the dewatering rate is decreased so that a slurry of 40% insolubles is generated, as shown in the flowsheet in Figure P2.28. To handle the higher water content slurry, the drier operation is also adjusted to produce a 62.5% insolubles coffee grounds.

Thickener in Kraft Paper Mill

As shown in the flow diagram, Figure P2.31, a countercurrent thickener consisting of three stages is used in a Kraft paper mill to wash a "white mud" consisting of 35% solids (CaC03) and 17% NaOH in water. Two wash streams are used, the first (stream 5) contains 4% suspended solids, 6% NaOH, and the rest H20; the second (stream 8) contains no solids, 2% NaOH, and the rest H20. The clear liquids from stage I and stage II contain 0.5% suspended solids; the clear liquid from stage III contains 0.4% solids. The flow of wash liquid to stage II (stream 5) is 1.5 times the flow of the feed sludge (stream 1), and the washed sludge from stage II contains one third solids. The washed sludge from the third stage (stream 9) contains 32.5% solids and 2.5% NaOH, and the flow rates of streams 7 and 9 are equal. All compositions are given on a mass basis. Assume that in each stage the clear solution and the solution contained in the washed sludge have the same concentration. Calculate the concentration of NaOH in stream 4.