1.0 INTRODUCTION

Design is a creative process whereby an innovative solution to a problem is conceived.

In this modern age of industrial competition, a successful chemical engineer needs more than a

knowledge and understanding of the fundamental sciences and the related engineering

subjects such as thermodynamics, reaction kinetics, and computer technology. The engineer

must also have the ability to apply this knowledge to practical situations for the purpose of

accomplishing something that will be beneficial to society. However, in making these

applications, the chemical engineer must recognize the economic implications which are

involved and proceed accordingly.

All design starts with a perceived need. In the design of a chemical process, the need is the

public need for the product, creating a commercial opportunity, as foreseen by the sales and

marketing organization. Within this overall objective, the designer will recognize sub-objectives,

the requirements of the various units that make up the overall process.

Before starting work, the designer should obtain as complete, and as unambiguous, a

statement of the requirements as possible. If the requirement (need) arises from outside the

design group, from a customer or from another department, then the designer will have to

elucidate the real requirements through discussion. When writing specifications for others,

such as for the mechanical design or purchase of a piece of equipment, the design engineer

should be aware of the restrictions (constraints) that are being placed on other designers. A

well-thought-out, comprehensive specification of the requirements for a piece of equipment

defines the external constraints within which the other designers must work.

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2.0 PROBLEM STATEMENT

Pursuant to instruction from our lecturer we proceeded to come up with a preliminary design

of a process to manufacture 1000kg/h of methyl ethyl ketone from dehydrogenation of 2-

butanol. The design work included coming up with a block diagram, a detailed mass and energy

balance, a flow sheet diagram and a detailed design of a distillation column.

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3.0 LITERATURE REVIEW

3.1 BACKGROUND

3.1.1 Nature of methyl ethyl ketone (product description)

Methyl ethyl ketone, also known as 2-butanone, is a colorless organic liquid with an acetone-

like odor and a low boiling point. It is partially miscible with water and many conventional

organic solvents and forms azeotropes with a number of organic liquids. MEK is distinguished

by its exceptional solvency, which enables it to formulate higher-solids protective coatings.

The molecular formula of methyl ethyl ketone is CH3COCH2CH3; its molecular structure is

represented as:

Some physical and chemical properties of MEK are presented in Table 1 below. Because of

MEK’s high reactivity, it is estimated to have a short atmospheric lifetime of approximately

eleven hours.

Atmospheric lifetime is defined as the time required for the concentration to decay to 1/e

(37percent) of its original value.

3.1.2 Overview of production and use

Generally, Methyl ethyl ketone production is accomplished by one of two processes:

(1) Dehydrogenation of secondary butyl alcohol or

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Fig. 1 2D and 3D dimensional molecular structures of MEK

(2) As a by-product of butane oxidation.

Property Value

Structural formula: CH3COCH2CH3

Synonyms: 2-butanone, ethyl methyl ketone, MEK, methyl acetone

Molecular weight (grams) 72.1

Melting point, °C -86.3

Boiling point, °C 79.6

Density at 20°C, g/L 804.5

Vapor density (air at 101 kPa, 0°C = 1) 2.41

Critical temperature, °C 260

Critical pressure, MPa 4.4

Surface tension at 20°C, dyne/cm 24.6

Dielectic constant at 20°C 15.45

Heat of combustion at 25°C, kJ/mol 2435

Heat of fusion, kJ/(kg*K) 103.3

Heat of formulation at constant pressure, kJ/mol 279.5

Specific heat:

vapor at 137°C, J/(kg*K)liquid at 20°C, J/(kg*K

1732 2084

Latent heat of vaporization at 101.3 kPa, kJ/mol 32.8

Flashpoint (closed cup), °C -6.6

Ignition temperature, °C 515.5

Explosive limits, volume % MEK in air

lowerupper

2 12

Vapor pressure at 20°C, mm Hg 77.5

Viscosity, MPa*s (=cP)

at 0°Cat 20°Cat 40°C

0.54 0.41 0.34

Solubility at 90°C, g/L of water 190

Figure 2 illustrates the production and use of MEK. Major end-users of MEK include protective

coating solvents (61 percent), adhesives (13 percent), and magnetic tapes (10 percent).

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Table 1: Physical and chemical properties of MEK

Vinyls are the primary resins that employ MEK as a solvent. Methyl ethyl ketone is commonly

used as a solvent in rubber cements, as well as in natural and synthetic resins for adhesive use.

It is also the preferred extraction solvent for dewaxing lube oil and is used in printing inks.

Overall, the projected use of MEK is expected to gradually decline. The growing trend towards

water-based, higher-solids, and solvent-less protective coatings, inks and adhesives is reducing

the demand for MEK. The installation of solvent recycling facilities will also reduce

requirements for fresh solvent production. Although MEK is favored as a solvent due to its low

density, low viscosity, and high solvency, its addition on the EPA’s hazardous air pollutants list

will likely cause potential users to consider other comparative solvents such as ethyl acetate.

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PRODUCTION

Dehydrogenation of secondary butyl alcohol

By-product of Butane oxidation

END USE

Protective coating solvent

Adhesive solvent

Magnetic tapes

Lube oil dewaxing

Chemical intermediate

Printing ink

Miscellaneous

Fig. 2 production and uses of MEK

3.2 APPLICATIONS

3.2.1 as a solvent

Butanone is an effective and common solvent and is used in processes involving gums, resins,

cellulose acetate and nitrocellulose coatings and in vinyl films. For this reason it finds use in the

manufacture of plastics, textiles, in the production of paraffin wax, and in household products

such as lacquer, varnishes, paint remover, a denaturing agent for denatured alcohol, glues, and

as a cleaning agent. It has similar solvent properties to acetone but has a significantly slower

evaporation rate. Butanone is also used in dry erase markers as the solvent of the erasable dye.

3.2.2 as a welding agent

As butanone dissolves polystyrene, it is sold as "polystyrene cement" for use in connecting

together parts of scale model kits. Though often considered an adhesive, it is actually

functioning as a welding agent in this context.

3.2.3 Other uses

Butanone is the precursor to methyl ethyl ketone peroxide, a catalyst for some polymerization

reactions. It can also initiate crosslinking of unsaturated polyester resins.

3.3 SAFETY

3.3.1 Flammability

Butanone can react with most oxidizing materials, and can produce fires. It is moderately

explosive; it requires only a small flame or spark to cause a vigorous reaction. Butanone fires

should be extinguished with carbon dioxide, dry chemicals or alcohol foam. Concentrations in

the air high enough to be flammable are also intolerable to humans due to the irritating nature

of the vapor.

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3.3.2 Health effects

Butanone is an irritant, causing irritation to the eyes and nose of humans, but serious health

effects in animals have been seen only at very high levels. When inhaled, these effects included

birth defects.

Butanone is listed as a Table II precursor under the United Nations Convention against Illicit

Traffic in Narcotic Drugs and Psychotropic Substances.

On December 19, 2005, the U. S. Environmental Protection Agency removed butanone from the

list of hazardous air pollutants (HAPs). After technical review and consideration of public

comments, EPA concluded that potential exposures to butanone emitted from industrial

processes may not reasonably be anticipated to cause human health or environmental

problems. Emissions of butanone will continue to be regulated as a volatile organic compound

because of its contribution to the formation of tropospheric (ground-level) ozone.

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Aqueous

OH

H2SO4

OH Zn or Brass

400-550°C

Butene

4.0 METHYL ETHYL KETONE PRODUCTION

This section discusses the methods which are used for production of methyl ethyl ketone.

4.1 SECONDARY-BUTYL ALCOHOL DEHYDROGENATION

The majority of MEK manufactured is produced by dehydrogenation of secondary-butyl alcohol.

This subsection discusses the 2-butanol dehydrogenation process.

4.1.1 Dehydrogenation Process Description

Methyl ethyl ketone manufacture by secondary-butyl alcohol dehydrogenation is a two-step

process where the first step involves the hydration of butenes to produce secondary-butyl

alcohol. The second step consists of the dehydrogenation of secondary-butyl alcohol yielding

MEK and hydrogen gas. These steps are illustrated by the following reactions:

(1) CH3CH=CHCH3 CH3CH2CH3

(2)

CH3CHCH2CH3

Since the first reaction (1) does not involve MEK as a product, this discussion will focus on the

second step of the reaction. Figure 3 illustrates the process of secondary-butyl alcohol

dehydrogenation. Initially, preheated vapours of secondary-butyl alcohol are passed through a

reactor (Step 1) containing a catalytic bed of zinc oxide or brass (zinc-copper alloy) which is

maintained between 400° and 550°C (750° and 1,025°F). A mean residence time of two to eight

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Sec-butyl alcohol

Sec-butyl alcohol

MEK

CH3CCH2CH3 + H2

Hydrogen gas

Solvent Hydrogen

Alcohol to recovery

seconds at normal atmospheric pressures is required for conversion from secondary-butyl

alcohol to MEK.

Product gases from the reaction vessel are then condensed via a brine-cooled condenser (Step

2) and sent to a distillation column for fractioning (Step 3). The main fraction (methyl ethyl

ketone) is typically obtained at an 85 to 90 percent yield based on the mass of secondary butyl

alcohol charged. The uncondensed gas may be scrubbed with water or a non-aqueous solvent

to remove any entrained ketone or alcohol from the hydrogen-containing gas (Step 4).The

hydrogen may then be re-used, burned in a furnace, or flared.

A liquid-phase process for converting secondary-butyl alcohol to methyl ethyl ketone has been

developed and is used sometimes. In this process, secondary-butyl alcohol is mixed with a high-

boiling solvent containing suspended finely divided Raney or copper chromite catalyst. The

reaction occurs at a temperature of 150°C (300°F) and at atmospheric pressure allowing MEK

and hydrogen to be driven off in vapour form and separated as soon as each is formed. The

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Preheater Reactor

Product storage and

loading

Condenser Scru

bber

Colu

mn

Fig. 3 methyl ethyl ketone from secondary butyl alcohol by dehydrogenation

1 24

3

∥ ∥

n-butane Oxygen or air

Acetic acid MEK Water

advantages of this process include a better yield (typically 3 percent better), longer catalyst life,

simpler product separation, and lower energy consumption.

4.2 N-BUTANE OXIDATION

Another method of manufacturing Methyl ethyl ketone is by liquid-phase oxidation of n-

butane. However, MEK has occasionally been commercially available in significant quantities

from the liquid-phase oxidation of butane to acetic acid. Depending on the demand for acetic

acid, this by-product methyl ethyl ketone can be marketed or recycled. This subsection

discusses MEK production via n-butane oxidation.

4.2.1 N-butane oxidation description process

Figure 4 illustrates the liquid-phase oxidation of n-butane. Initially, n-butane and compressed

air or oxygen are fed into a reactor (Step 1) along with a catalyst, typically cobalt, manganese or

chromium acetate to produce acetic acid, MEK and other by-products such as ethanol, ethyl

acetate, formic acid, and propionic acid. This process produces the following chemical reaction:

O O

CH3CH2CH2CH3 + O2 CH3COH + CH3CCH2CH3 + + H2O

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Other by-products

Air is bubbled through the reactant solution at 150° to 225°C (300° to 440°F) with pressures of

about 5.5 MPa (800 psi). Conditions must be carefully controlled to facilitate MEK production

and prevent competing reactions that form acetic acid and other by-products. Process

conditions can be varied producing different ratios of product components through the choice

of raw material, reaction conditions, and recovery methods.

Vapors containing crude acetic acid and the various by-products including MEK are separated

from unreacted n-butane and inert gases (Step 2), then stripped or flashed to remove dissolved

butane and inert gases (Step 3), and sent to the purification section (Step 4). Unreacted

nitrogen leaving the reactor carries various oxidation products (formic, acetic, and propionic

acids; acetone, MEK, methanol, etc.) and some unreacted butane and is sent to a separator

(condenser) for removal/recycling of unreacted hydrocarbons (Step 5).

The purification section of the plant is complex and highly specialized utilizing three phase

distillation in conjunction with straight extraction. The low-boiling organics such as MEK are

separated from the crude acetic acid by conventional distillation. Azeotropic distillation is used

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Fig. 4 Methyl ethyl ketone from n-butane by liquid phase oxidation

to dry and purify the crude acetic acid. Recovery and purification of the various by-products

require several distillation columns and involve extractive distillation or azeotrope breakers or

both. Liquid organic wastes are typically burned in boilers to recover their heat value.

4.3 N-BUTENE OXIDATION

A new one-step process that converts olefins to ketones called OK technology was developed.

Specifically, MEK is produced via direct oxidation of n-butenes at about 85°C (185°F) and 690

kPa (100 psi), using a proprietary, and homogenous non-chloride catalyst. Advantages of this

process are that it is noncorrosive, environmentally clean, and economical because of low

capital investment and low energy needs. The process is currently in lab-scale operation;

however, plans are underway to design a facility for large scale production.

4.4 JUSTIFICATION OF THE PROCESS USED

The justification of the method used was based on the problem statement given to the group

by the supervisor.

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4.5 DISTILLATION

Distillation as a separation process is indispensable in the production of methyl ethyl ketone

from dehydrogenation of 2-butanol.

The separation of liquid mixtures by distillation depends on differences in volatility between the

components. In distillation, the greater the relative volatilities, the easier the separation.

The basic equipment required for continuous distillation consists of column, a re-boiler and a

condenser system.

Vapor flows up the column and liquid counter-currently down the column. The vapor and liquid

are brought into contact on plates, or packing. Part of the condensate from the condenser is

returned to the top of the column to provide liquid flow above the feed point (reflux), and part

of the liquid from the base of the column is vaporized in the re-boiler and returned to provide

the vapor flow.

In the section below the feed, the more volatile components are stripped from the liquid and

this is known as the stripping section. Above the feed, the concentration of the more volatile

components is increased and this is called the enrichment, or more commonly, the rectifying

section.

If the process requirement is to strip a volatile component from a relatively non-volatile

solvent, the rectifying section may be omitted, and the column would then be called a stripping

column.

In some operations, where the top product is required as a vapor, only sufficient liquid is

condensed to provide the reflux flow to the column, and the condenser is referred to as a

partial condenser. When the liquid is totally condensed, the liquid returned to the column will

have the same composition as the top product. In a partial condenser the reflux will be in

equilibrium with the vapor leaving the condenser. Virtually pure top and bottom products can

be obtained in a single column from a binary feed, but where the feed contains more than two

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components; only a single “pure” product can be produced, either from the top or bottom of

the column.

In engineering terms, distillation columns have to be designed with a larger range in capacity

than any other types of processing equipment, with single columns 0.3–10 m in diameter and

3–75 m in height. Designers are required to achieve the desired product quality at minimum

cost and also to provide constant purity of product even though there may be variations in feed

composition.

A distillation unit should be considered together with its associated control system, and it is

often operated in association with several other separate units.

The vertical cylindrical column provides, in a compact form and with the minimum of ground

requirements, a large number of separate stages of vaporization and condensation.

A complete unit will normally consist of a feed tank, a feed heater, a column with boiler, a

condenser, an arrangement for returning part of the condensed liquid as reflux, and coolers to

cool the two products before passing them to storage.

The reflux liquor may be allowed to flow back by gravity to the top plate of the column or, as in

larger units, it is run back to a drum from which it is pumped to the top of the column. The

control of the reflux on very small units is conveniently effected by hand-operated valves and

with the larger units by adjusting the delivery from a pump.

In many cases the reflux is divided by means of an electromagnetically operated device which

diverts the top product either to the product line or to the reflux line for controlled time

intervals.

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2-butanol MEK Hydrogen

5.0 PROCESS DESCRIPTION

5.1 DEHYDROGENATION OF 2-BUTANOL

Methyl ethyl ketone (MEK) is manufactured by the dehydrogenation of 2-butanol. A description

of the processes listing the various units used is given below:

5.1.1 Reactor

A reactor in which the butanol is dehydrated to produce MEK and hydrogen, according to the

reaction:

CH3CH2CH3CHOH CH3CH2CH3CO + H2

The conversion of alcohol to MEK is 88 per cent and the yield is taken as 100 per cent. Initially,

preheated vapours of secondary-butyl alcohol are passed through a reactor (Step 1) containing

a catalytic bed of zinc oxide or brass (zinc-copper alloy) which is maintained between 400°C and

550°C (750°F and 1,025°F). A mean residence time of two to eight seconds at normal

atmospheric pressures is required for conversion from secondary-butyl alcohol to MEK.

5.1.2 Cooler-condenser

In the cooler-condenser the reactor off-gases (i.e. product gases) are cooled and most of the

MEK and unreacted alcohol are condensed. Two exchangers are used but they are modeled as

one unit. Of the MEK entering the unit 84 per cent is condensed, together with 92 per cent of

the alcohol. The hydrogen is non-condensable. The condensate is fed forward to the second

distillation column which is the final purification stage. The MEK is cooled to a temperature of

32 °C. The water is fed to the cooler at a temperature of 24 °C.

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5.1.3 Absorption column

In the absorption column the uncondensed MEK and alcohol are absorbed in water. Around 98

per cent of the MEK and alcohol can be considered to be absorbed in this unit, giving a 10 per

cent w/w solution of MEK. The water feed to the absorber is recycled from the next unit, the

extractor. The vent stream from the absorber, containing mainly hydrogen, is sent to a flare

stack.

5.1.4 Extraction column

In the extraction column the MEK and alcohol in the solution from the absorber are extracted

into trichloroethylane (TCE). The raffinate, water containing around 0.5 per cent w/w MEK, is

recycled to the absorption column. The extract, which contains around 20 per cent w/w MEK,

and a small amount of butanol and water, is fed to the first distillation column.

5.1.5 Distillation column I

In the distillation column the unit separates the MEK and alcohol from the solvent TCE. The

solvent containing a trace of MEK and water is recycled to the extraction column. The recovery

is 99.99%.

5.1.6 Distillation column II

In the second distillation column, also known as the final the purification stage which produces

a 99.9% pure MEK product from the crude product from the first column. The residue from this

column, which contains the bulk of the unreacted 2-butanol, is recycled to the reactor. The

steam generated by the re-boiler in this unit is at a temperature of 140 °C.

The following is the block diagram for the production process of methyl ethyl ketone.

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2-butanol

Unreacted alcohol and MEK

Gaseous products

Uncondensed MEK & alcohol

To flame stack

Water 0.5% w/w MEK

MEK and alcohol

Extract

TCE (trichloroethyl

ane)

Crude product

H2

Pure MEK (99.9%)

Unreacted 2-butanol

Page | 17

Fig. 5 Block diagram for the production of methyl ethyl ketone

Reactor (dehydrogenation)

Cooler- condenser

Absorption column

Extractor

Distillation column 1

Distillation column 2

5.2 MATERIAL BALANCES

Material balances are the basis of process design. A material balance taken over the complete

process will determine the quantities of raw materials required and products produced.

Balances over individual process units set the process stream flows and compositions.

Material balances are also useful tools for the study of plant operation and trouble shooting.

They can be used to check performance against design; to extend the often limited data

available from the plant instrumentation; to check instrument calibrations; and to locate

sources of material loss.

All mass/material balances are based on the principle of conservation of mass that is massr can

neither be created nor destroyed with an exception of nuclear processes according to Einstein’s

equation; E=mc2.

The general conservation equation for any process system can be written as:

Material out−Material∈+Generation−Consumption=Accumulation

For a steady state process the accumulation term is zero and thus for a continuous steady state

process, the general balance equation for any substance involved in the process can be written

as:

Material∈+Generation=Material Out+Consumption

If no chemical reaction takes place, material balance is computed on the basis of chemical

compounds mass basis that are used whereas if a chemical reaction occurs molar units are

used.

Also it is worthwhile to note that when a reaction occurs an overall balance is not appropriate

but a reactant balance (a compound balance) is.

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5.2.1 Choosing a Basis

The correct choice of the basis for a calculation will often determine whether the calculation

proves to be simple or complex.

A time basis was chosen in which the results will be presented. The basis for calculations was

chosen as 1 hour and thus results will be presented in kg/h.

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YieldsCH3CH2CH3CHOH CH3CH2CH3CO + H2

X (kg)

XR

2-butanol XF

X (kg)

5.2.2 MATERIAL BALANCE FOR THE PRODUCTION METHYLETHYLKETONE (MEK) FROM 2-

BUTANOL

Basis used: 1 hour

The material balance was done around the following units:

(1) Reactor

RMM of 2-butanol =74

Moles of 2-butanol =X (kg)

74=0.01335 x

Moles of the2-Butanol that reacted 0.88 × 0.01369 x = 0.01188 x

From the equation:

Mole ratio for the reaction is 1:1

Hence moles of the MEK reacting is 0.01188X

Mass of MEK then is 0.01188x×72=0.8554x

Mass of 2-butanol is 12

100× x=012 x

Mass of then H2 is 0.01188×2=0.0276x

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Reactor

Reactor

MEK

2-butanol

H2

MEK = 0.8554x

2-Butanol=012 x

H2 =0.02376x

MEK¿0.01369 x

2-Butanol¿0.0096 x(Non-condensable)MEK = 0.8554x

2-Butanol=012 x

All the components leaving the reactor are discharged directly into the cooler condenser for the

next operation.

(2) Cooler-condenser

Condensate (which is then directly sent to the final purification column) comprises:

84% MEK= 0.8×0.8554x=0.7185 x

92% 2-Butanol=0.92× 012 x=0.1104 x

Incondensable stream comprises:

H2=0.02376 x

2-Butanol¿0.0096 x

MEK¿0.01369 x

(Condensate) MEK¿0.7185 x

2-butanol¿0.1104 x

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Cooler-condenser

H2=0.02376 xH2=0.02376 x

MEK¿0.5 %=0.005 K

H2OK

2-Butanol¿0.0096 x

JJ

MEK ¿0.98 (0.1369 x )+0.005 K=0.1 J

K

2-Butanol

MEK

(3) MEK balance around the absorption column

0.1369 x+0.005 K=0.02 (0.1369 X )+0.1 J

01369 x+0.005 K=0.002738+0.1 J

J=1.342 x+0.05 K

Overall balance

(0.02376 x+0.0096 x+0.1369 x )+ K=1.342 x+0.05k+(0.02376 x+0.005 k+0.002738 x)

k=1.262033 x

Performing a new balance around the absorption column to express the k -value in terms of x

in the above equations gives the following values:

2

Page | 22

Absorption column

Absorption column

MEK=0.02(0.1369 x)2-Butanol¿0.02(0.0096 x )H2=0.02376 x

(non-condensable)MEK¿0.01369 x

H2=0.02376 x

2-Butanol¿0.98(0.0096 x)H2O=?

MEK

H2O¿1.256 xMEK¿0.1369 x

H2¿0.02376 x

MEK¿0.02 (0.1369 x )Butanol¿0.02(0.0096 x )

H2=0.02376 x

MEK¿0.1342 x+0.00631 x=0.14051 x

H2O¿10256 x and

2−butanol=0.009408x

ϑ

R- Recycle from next operation (TCE)

: MEK ¿0.00631 x H2O¿1.256 x

2-butanol =0.009408 x

Raffinate: MEK¿0.005 K=0.005 (1.262033684 x )=0.00635 x

H2O¿0.995 K=0.995 (1.262033 x )=1.256 x

Stream J: MEK¿0.1 J=0.1 {1.3426 x+0.005 (1.26033684 x ) }=0.1404 x

H2O¿1.262033684 x

2-butanol ¿0.009408 x

(4) Extraction column

Raffinate B

Q

R ¿TC

MEK Balance around the extractor

0.1404 x=0.00631 x+0.2 ϑ

ϑ¿0.6704 x

Overall balance

R+Q=ϑ+B

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Extractor

MEK¿0.1404 x

H2O¿1.256 x

MEK¿0.13409 x

2-butanol¿0.009408 x

TCE¿0.527 x

2-Butanol ¿0.009408 x

1000kg/hr (flow rate as given)

1.399 x+R=1.2623 x+0.67045 x

R=0.533 x

TCE=0.6704 x−(0.13409 x+0.009408 x )

¿0.527 x (Which is approximately =R)

(5) Distillation column 1

For this unit operation, the balances were obtained from the previous unit operation i.e. the

extraction column and are indicated in the block diagram below.

(6) Distillation column 2

The material balance for the second distillation column is given as follows;

Balancing around this gives:

MEK:

0.8526 x=1000

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Distillation column 1

Distillation column 2

TCE¿0.533 x

MEK¿0.13409 x

TCE¿0.527 x

MEK¿0.13409 x2-Butanol¿0.009408 x

MEK¿(0.13409 x+0.71912 x)¿0.8526 x

2-Butanol

¿0.1198 x

2-Butanol (recycled back to the reactor)

x= 10000.8526

=1172.882 kg

2-Butanol:

0.1198 x

x=0.1198 (1172.882)¿140.51 kg (Returning to the reactor)

4.2.3 CALCULATION OF ACTUAL MASS OF THE COMPONENTS IN ALL THE STREAMS

The streams are indicated in the diagrams above.

1) Reactor

From the balances carried out in the previous exercise the value of X was obtained as 1172.883

kg based on the 1 hour basis.

In = out

Entering stream:

XF + XR= X where: XF = feed and XR = feed as recycle

Leaving streams:

MEK = 0.8554 x=0.8554 ×1172.883=1003.24 kg

2-butanol ¿0.12 x=0.12 ×1172.883=140.74 kg

H2¿0.02376 x=0.02376 × 1172.883=27.87 kg

2) Cooler condenser

In = out

MEK ¿0.7185 x=0.7185 ×1172.883=842.716 kg

2-butanol¿0.1104 x=0.1104 ×1172.883=128.486 kg

Non-condensable

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MEK¿0.1369 x=0.1369 ×1172.883=160.568 kg

2-Butanol ¿0.0096 x=0.0096 × 1172.883=11.260kg

H2¿27.87 kg

3) Absorption column

Entering stream:

MEK¿0.1369 x=0.1369 ×1172.883=160.568 kg

2-Butanol ¿0.0096 x=0.0096 × 1172.883=11.260kg

H2¿27.87 kg

Raffinate stream:

MEK 0.00631 x=0.00631 ×1172.883=7.401 kg

H2O:1.256 x=1.256 ×1172.883=1473.141 kg

Leaving stream:

MEK 0.14051 x=0.14051 ×1172.883=164.881 kg

H2O 1.256 x=1.256 ×1172.883=1473.141 kg

2-butanol ¿0.009408 x=0.009488 ×1172.883=11.034 kg

4) Extractor

Entering stream:

MEK 0.14051 x=0.14051 ×1172.883=164.881 kg

H2O 1.256 x=1.256 ×1172.883=1473.141 kg

2-butanol ¿0.009408 x=0.009488 ×1172.883=11.034 kg

Page | 26

Recycle stream = TCE (Tri chloro ethylane)

TCE: 0.527 x=0.527 × 1172.883=618.109 kg

Leaving stream:

MEK:0.13409 x=0.13409 ×1172.883=157.272 kg

2-butanol0.009408 x=0.009408 ×1172.883=11.0334 kg

5) Distillation column 1

Entering stream:

MEK: 0.13409 x=0.13409 ×1172.883=157.272 kg

2-butanol0.009408 x=0.009408 ×1172.883=11.0334 kg

Leaving stream:

MEK:0.13409 x=0.13409 ×1172.883=157.272 kg

2-butanol0.009408 x=0.009408 ×1172.883=11.0334 kg

TCE: 0.527 x=0.527 × 1172.883=618.109 kg (This is recycled back into the extractor)

6) Distillation column 2

In = out

Entering stream:

MEK: 0.13409 x+0.71912 x=0.8526 x

2-butanol: 0.009408 x+0.1104 x=0.1199 x (this is recycled back to the reactor)

Leaving stream

Page | 27

99.99% pure MEK at 1000kg/hr

5.3 ENERGY BALANCES

As with mass, energy can be considered to be separately conserved in all but nuclear processes.

The conservation of energy, however, differs from that of mass in that energy can be generated

(or consumed) in a chemical process. Material can change form, new molecular species can be

formed by chemical reaction, but the total mass flow into a process unit must be equal to the

flow out at the steady state. The same is not true of energy. The total enthalpy of the outlet

streams will not equal that of the inlet streams if energy is generated or consumed in the

processes; such as that due to heat of reaction.

Energy can exist in several forms: heat, mechanical energy, electrical energy, and are the total

energy that is conserved.

In process design, energy balances are made to determine the energy requirements of the

process: the heating, cooling and power required. In plant operation, an energy balance (energy

audit) on the plant will show the pattern of energy usage, and suggest areas for conservation

and savings.

A general equation can be written for the conservation of energy:

Accumulation=Energy∈+Generation−Consumption−Energy out

This is a statement of the first law of thermodynamics. An energy balance can be written for

any process step. Chemical reaction will evolve energy (exothermic) or consume energy

(endothermic). For steady-state processes the accumulation of both mass and energy will be

zero.

The energy balance was carried out around cooler condenser and the second distillation

column. In chemical processes the kinetic and potential energy terms are usually small

compared with heat and work terms, and can normally be neglected.

Page | 28

If the kinetic and potential energy terms are neglected the energy equation reduces to

H 2−H 1=Q−W

For many processes the work term will be zero, or negligibly small, and equation above reduces

to the simple heat balance equation:

Q=H 2−H 1

Where heat is generated in the system; for example in a chemical reactor:

Q=QP+Q S

QS=¿ heat generated in the system. If heat is evolved (exothermic processes) QS is taken as

positive, and if heat is absorbed (endothermic processes) it is taken as negative.

QP=¿ process heat added to the system to maintain required system temperature.

Hence:

QP=H 2−H 1−QS

H 1=¿ enthalpy of the exit stream

H 2=¿ enthalpy of the outlet stream.

For a practical reactor, the heat added (or removed) Qp to maintain the design reactor

temperature will be given by:

QP=¿ H products−Hreactants−Qr¿

Where

H products is the total enthalpy of the product streams, including unreacted materials and by-

products, evaluated from a datum temperature of 25°C;

Page | 29

H reactants is the is the total enthalpy of the feed streams, including excess reagent and inerts,

evaluated from a datum of 25°C;

Qr is the total heat generated by the reactions taking place, evaluated from the standard heats

of reaction at 25°C (298 K).

This equation can be written in the form:

QP=∑ ∫T ref

Tout

ni c pidT−¿∑ ∫T ref

Tout

ni c pidT−¿∑ [−∆ H °rxn ] × mol of product formed ¿¿

C p=A+BT+C T2+D T3

Page | 30

MEK=160.568 kg

2-Butanol¿11.260 kg(Non-condensable)

MEK = 0.8554x

2-Butanol=012 x

MEK =842.716 kg

2-butanol¿128.486 kg

CondensateQR

5.3.1 ENERGY BALANCE FOR THE PRODUCTION METHYLETHYLKETONE (MEK) FROM 2-

BUTANOL

The energy balance was carried around the cooler condenser and the second distillation

column (final purification stage). The balances are as indicated below.

4.3.1.1 Cooler condenser

The temperature at which the products of the reactor leave is 400 °C. The condenser cooler

lowers cools the products to a temperature of 32 °C. The energy balance is given as shown in

the calculations below.

Energy balance for MEK

Sensible heat to lower the temperature of the condensate MEK from 400 °C to 79.6 °C,

¿mC p ∆ T

¿842.716 ×1732 × (400−79.6 )

¿467.65 × 106 J

Sensible heat to lower the temperature of the incondensable MEK from 400 °C to 80 °C,

¿mC p ∆ T

Page | 31

Cooler-condenser

H2=27.87 kgH2=0.02376 x

MEK¿1003.24 kg

2-butanol 140.74 kg H2 27.87 kg

¿160.568 ×1732 × ( 400−80.0 )

¿88.99 ×106 J

Sensible heat to lower the temperature of the condensate MEK from 79.6 °C to 32 °C,

¿mC p ∆ T

¿842.716 × 2084 × (79.6−32 )

¿83.60×106 J

No of moles of MEK condensed

842.716 kg72 kg/kmol

=11.704 kmol

Latent heat of vaporization of MEK,

¿11.704 ×32.8

¿383.89 ×106 J

Total energy required for MEK cooling and condensation,

¿ (467.65+88.99+83.60+383.89 )106

¿1024.13 MJ /h

Energy balance for 2-butanol

Sensible heat to be removed to lower the temperature of 2-butamol from 400 °C to 99 °C is

determined as follows,

Q=mC p ΔT

Q=140.7474

×197.1J

mol . K× ( 400−99 )× 103=112.83× 106 J /h

To condense the 2-butanol,

Page | 32

Q=5.95 × 105 Jkg

×128.486 kg=764.5 ×106 j /hTotal heat to be removed from 2-butanol,

¿Q1+Q2

¿ (112.83+764.49 )× 106 Jh

¿877.32 ×106 J /hTotal heat to be removed from the cooler condenser,

QR=(877.32+1024.13)× 106 j /h

QR=1901.45 MJ /h

5.3.1.2 Distillation column 2

Page | 33

R

QR

QC

D=1000kg/hXD=0.999

F=1140.52kg/hXF=0.88

Taking reflux ratio (R.R) = 1.94

Total energy balance equation is:

HF+QB=QC+HD+HB

QC is obtained by a balance around the condenser

Page | 34

B=140.52kg/hXB=0.0088

QC

DHD

RHR

VHV

An energy balance at steady state is:

HV = QC + HR + HD

Values of enthalpy of product (distillate) and reflux are zero as they are both at the reference

temperature. Both are liquid and the reflux will be at the same temperatures as the distillate.

Enthalpy of vapour:

Hv= latent heat + sensible heat

For methyl ethyl ketone, latent heat is given as:

Ln = 4.56 × 105 J /kg

Latent heat of the vapor stream:

L=mn Ln

L=(2940 ) ( 4.56 ×105 )

L=1340.6 × 106 J /K

Page | 35

Sensible heat

=∑∫ niC pidT

Boiling point of methyl ethyl ketone

=79.6 ℃ (352.6 K)

Sensible heat of MEK,

=0.026362×106 ∫298

352.6

(24.643+33.557¿× 10−2 T−2.057 ×10−4T 2+63.781 ×10−9T 3)dT ¿

¿167.575 ×106 J¿1340.6 ×106+167.575 × 106

H v=1508.18× 106 kJ

H D=(mols of stream)(value of intergral)

¿(0.105488× 106)(2100.98)+(0.026372 ×106)(2203.23)

¿27.9710 6kJ

H R=(molsof stream)(value of intergral)

¿(0.316465× 106)(2100.98)+(0.079116 ×106)(2203.23)

¿111.884 ×106 J

A balance around the condenser yields:

QC=H v−(H R+H D)

¿ [1508.18−(27.97106+111.884 )]× 106 J

Q c=1368.32 ×106 J

The quantity of heat that needs to be extracted from the condenser by the cooling fluid is

obtained as follows.

Page | 36

QR is obtained by an overall energy balance around the column.

QR=QC+H D+HB−H F

H F=41.8 ×103 ×1140.6 J

¿15.884 × 106 J

HB=140.6 × 41.8 ×1000

¿5.852 ×106 J

QR=(1368.32+111.884+5.852−15.884 )×106

QR=1470.2 ×106 J

Page | 37

6.0 DESIGN OF DISTILLATION COLUMN 2

6.1 DISTILLATION PRINCIPLES

Separation of components from a liquid mixture via distillation depends on the differences in

boiling points of the individual components. Also, depending on the concentrations of the

components present, the liquid mixture will have different boiling point characteristics.

Therefore, distillation processes depends on the vapor pressure characteristics of liquid

mixtures.

6.2 VAPOUR PRESSURE AND BOILING

The vapor pressure of a liquid at a particular temperature is the equilibrium pressure exerted

by molecules leaving and entering the liquid surface. Here are some important points

regarding vapor pressure:

energy input raises vapor pressure

vapor pressure is related to boiling

a liquid is said to ‘boil’ when its vapor pressure equals the surrounding

pressure

the ease with which a liquid boils depends on its volatility

liquids with high vapor pressures (volatile liquids) will boil at lower

temperatures

the vapor pressure and hence the boiling point of a liquid mixture depends

on the relative amounts of the components in the mixture

distillation occurs because of the differences in the volatility of the

components in the liquid mixture

6.3 DESIGN OF DISTILLATION COLUMN

Page | 38

Distillation columns are designed using the vapor-liquid equilibrium data for the mixtures to be

separated.

The vapor liquid equilibrium characteristics of the mixture will determine the number of stages

and hence the number of trays required for the separation.

Most distillation columns are designed by use of the McCabe Thiele method.

6.4 McCabe THIELE DESIGN METHOD

The McCabe Thiele approach is a graphical one and use the VLE plot to determine the

theoretical number of stages required to effect the separation of the mixture (binary in our

case).

The method assumes constant molar overflow and this implies that:

Molar heats of vaporization of the components are roughly the same.

Heat effects (heats of solution, heat losses to and from the column etc.) are

negligible.

For every mole of vapor condensed one mole of liquid is vaporized.

The design process is simple. Given the VLE data/relationship for the more volatile component,

operating lines are drawn first.

Operating lines define the mass balance relationships between the liquid and

vapor phases in the column.

There is one operating line for the bottom (stripping) section of the column and

one for the top (rectifying) section of the column.

Use of the constant molar overflow assumption also ensures that the operating

lines are straight.

In the design done for the distillation column 2 the following criteria was followed.

Page | 39

QC

F=1140.52kg/hXF=0.88

1. Specification of degree of separation required

2. Selection of the operating conditions

3. Selection of the type of contacting device e.g. plates , pickings

4. Determining the stage and reflux requirements.

5. Sizing the column e.g diameter and height.

Assumptions made in the design of the distillation column:

Equimolar overflow

Total condenser

Partial reboiler

Density does not vary with temperature

Theoretical plates i.e perfect phase equilibrium exists between both phases

leaving the plate.

1. Degree of separation required

The feed to the distillation column contains 88 mol % of the less volatile component (methyl

ethyl ketone) and 12 mol % of the more volatile component (2-butanol).

An overhead purity of 99.9 mol percent is desired while a bottoms purity of 0.1 mol % is

obtained thus the following mole fraction value relate to the more volatile component:

X F=0.8768

X B=0.0088

X D=0.999

A reflux ratio of 16 was used as calculated based on the minimum reflux ratio.

Page | 40

The following vapour liquid equilibrium data was used to draw the VLE curve.

X 0.088 0.278 0.383 0.467 0.478 0.582 0.702 0.803 0.855 0.900

Y 0.192 0.468 0.583 0.644 0.655 0.737 0.823 0.885 0.905 0.940

Page | 41

0 0.2 0.4 0.6 0.8 1 1.20

0.2

0.4

0.6

0.8

1

1.2

X

Y

y'

xF

The value of y’ is read from the graph as shown above.

Page | 42

0 0.2 0.4 0.6 0.8 1 1.20

0.2

0.4

0.6

0.8

1

1.2

X

Y

y'

2. Determination of stages and reflux requirements

The theoretical number of stages was determined by the McCabe Thiele method. This is a

graphical method for the determination of the ideal number of stages. This was procedure was

carried as follows.

Determining the minimum reflux ratio

The minimum reflux rate can be determined mathematically from the endpoints of the

rectifying line at minimum reflux – the overhead product composition point (xD, yD) and the

point of intersection of the feed line and equilibrium curve(x’, y’).

RDmin=

xD− y '

y '−x '

xD=0.99 y’=0.92 x’=0.8768

RDmin= 0.99−0.92

0.92−0.8768=1.62

Rreal=1.2 RD min=1.2× 1.62=1.94

Rreal=RD

R=R real × D=1.94 ×1000=1940 kg/h

The equation for the rectifying section is given as follows,

yn=RR

1+RRxn+1+

1RR+1

xD

yn=1.94

1+1.94xn+1+

11.94+1

0.999

yn=0.66 xn+1+0.34

The above equation is plotted in the curve as shown below, and the McCabe Thiele method is used to determine the number of stages.

Page | 43

From the above analysis using the McCabe Thiele method, the theoretical number of stages was obtained as 12 stages.

Ideal number of stages obtained= 12

i.e. Rectifying section= 3 stages

Stripping section = 9 stages

Page | 44

0 0.2 0.4 0.6 0.8 1 1.20

0.2

0.4

0.6

0.8

1

1.2

X

Y

q-line

VLE curve

45° line

stripping operat-ing line

rectifying operating line

xDxB

3. Sizing of the column

The sizing of the column was carried out using Carrillo, Martin and Roselle’s correlation (2000).

HETP=P √ ρL

(2712+82.0P )¿¿

Where Fv is defined by the following expression

FV=uGs √ ρL

uGs is the vapor phase superficial velocity

ρL is the liquid phase specific mass

ρGis the vapor phase specific mass

X=¿

y=e−4 x (2)

y=ug

2 a ρg

g ε3 ∆ ρ¿

Where

a=surfacearea

ε=voidage area

V g=velcity of gas∈the column

mL=mass flowrate of the liquid

mg=mass flowrate of the gas

At 760mmHg, data for MEK is as given below

ρg=2.5 ×1=2.5 kgm−3 , ρL=0.806 ×1000=806 kg m−3 ,a=364 m2

m3 , ε=0.63 , g

¿9.80 m / s2

Page | 45

F

VR

VSLS

R=LR

To obtain the mass flow rate of the gas and the liquid the following balance is carried out as below.

Ls=F+LR

¿ F+R

¿1140.5208+1940=3080.52 kg /h

V s=V R=R+D=1940+1000=2940 kg /h

M l=Ls=3080.52 kg/h

M g=V s=2940 kg/h

∆ ρ=(80.6−2.5 )=803.5

μL=0.426 cP , μH 2 O=1.005 cP

x=(3080.522940 )

14 ( 2.5

803.5 )18

x=1.0117 ×0.486=0.492

Using the value of x in equation (1),

y=e−4 (0.492)=0.1399

From equation (3),

Page | 46

υg2= y . g . ε3 Δ ρ

a ρg( μl

μH2 O )0.16

μl

μH 2 O

=0.4261.005

=0.4239

υg=( 0.1399 ×9.81 ×0.633× 803.5364 × 2.5 ×0.42391.6 )

12=( 275.74

793.24 )12=√0.3476=0.5896 m / s

V g=υg . A

volumetric flowrate=flowrate× areaV g=υg × A=υg .π D 2

4

Dc=√ 4 V g

π × υg

V g=υρ=1000 kg /h

2.5 kg /m3=400 m3/hr

¿0.1111m3

s

Dc=√ 4×0.1111π × 0.5896

Dc=0.4898 m

DD=DC

0.75=0.4898

0.75=0.6531 m

Therefore the diameter of the column is,

DD=0.65m

Determining the height of the column using the following procedure,

HETP=P √ ρL

(2712+82.0P )¿¿

FV=uGs √ ρL

¿0.5896 ×√2.5=0.9322 m /s

Page | 47

760 mmHg=101325.024 Pa(N / M 2)

101.325× 103× √ 806 ×0.93220.42

(2712+82.0 P ) ¿¿

Numerator=0.97094 ×28.39 ×101.325 ×103=2793045.141

Denominator=18.365 × 8311.362=15263816.31

¿ 2793045.14115263816.31

=0.18298

HOP=N op× HETP

Nop=number of stages∈theupper section

H op=height of upper part of the column

H ETP=height of equivalent theretical plate

HOL=NOL × HETP

NOL=number of stages∈the lower section

Stages in the upper section= 3

Stages in the lower section = 9

HOP=3 × 0.18298=0.54894

HOL=9 ×0.18298=1.64682

HTotal=HOP+HOL

¿0.54894+1.64684

¿2.19576

HTotal≅ 2.2m

The active part of the distillation column is 2.2 m

4. Selection of the type of contacting device to be used

Raschig rings will be used as the contacting device in the distillation column. They are ceramic

in nature are 1/3 mm in size.

Page | 48

Raschig rings are pieces of tube (approximately equal in length and diameter) used in large numbers as

a packed bed within columns for distillations and other chemical engineering processes. They are usually

ceramic or metal and provide a large surface area within the volume of the column for interaction

between liquid and gas or vapour.

They form what is now known as random packing, and enable distillations of much greater

efficiency than the use of fractional distillation columns with trays.

In a distillation column, the reflux or condensed vapour runs down the column, covering the

surfaces of the rings, while vapour from the re-boiler goes up the column. As the vapour and

liquid pass each other counter-currently in a small space, they tend towards equilibrium. Thus

less volatile material tends to go downwards, more volatile material upwards.

Raschig rings made from borosilicate glass are sometimes employed in the handling of nuclear

materials, where they are used inside vessels and tanks containing solutions of fissile material,

for example solutions of enriched uranyl nitrate, acting as neutron absorbers and preventing a

potential criticality accident.

Page | 49

Fig. 7 Raschig rings used for the operation

7.0 CONCLUSION

Distillation column design requires the selection of the right various packing and tower sizing to

meet the process, hydraulic, efficiency, and mechanical requirements of the service. Process

considerations include operating conditions, flexibility, and solid handling requirements.

Hydraulic and efficiency criteria involve selection of a suitable packing material that allows for

cost-effective optimization of vessel height vs. diameter.

Determining the number of stages required for the desired degree of separation and the

location of the feed tray is merely the first steps in producing an overall distillation column

design.

Other things that need to be considered are tray spacing; column diameter; internal

configurations; heating and cooling duties. All of these can lead to conflicting design

parameters. Thus, distillation column design is often an iterative procedure. If the conflicts are

not resolved at the design stage, then the column will not perform well in practice.

It can be deduced from the previous section on distillation column design that the number of

trays will influence the degree of separation.

As the feed stage is moved lower down the column, the top composition becomes less rich in

the more volatile component while the bottoms contains more of the more volatile component.

However, the changes in top composition are not as marked as the bottoms composition.

Page | 50

8.0 REFERENCES

1. Chemical Engineering Design, 4th Edition by R.K Sinnot.

2. Unit Operations of Chemical Engineering, 5th Edition by McCabe and Smith.

3. Li, Y.L., “Production technology and market analysis of methyl ethyl ketone”, Fine and

Specialty Chemicals, 12(18), 22—25(2004). (in Chinese)

4. Zhang, Y.X., “Production technology and application status of methyl ethyl ketone”,

Journal of Henan Chemical Industry, 11(1), 51—55(2003). (in Chinese)

5. Distillation: An Introduction by M. T Tham.

6. Qi, J., Gao, N., “Market analysis of methyl ethyl ketone”, Petrochemical Industry

Technology, 10(3), 61 — 64(2003). (in Chinese)

7. Ma, Y.S., Su, J., Wang, C.M., “A process of ketone from secondary alcohol by

dehydrogenation”, C.N Pat., 1289753(2001).

8. Perry’s Chemical Engineering Handbook.

9. Coulson and Richardson’s Chemical Engineering, Volume 2, Fifth Edition.

10. Lecture notes from CHP 461 (Chemical Engineering Design I) and CHP 372 (Mass

Transfer I)

11. www.wikipedia.org .

12. www.basf.com

Page | 51