Contents
1.0 Project Objectives.......................................................................................................................................... 4
2.0 Project Objectives.......................................................................................................................................... 4
2.1 Technical Objectives .................................................................................................................................. 4
2.2 Personal Objectives ................................................................................................................................... 5
3.0 Introduction ................................................................................................................................................... 7
3.1 Components of Distillation Columns ......................................................................................................... 8
3.1.1Reboiler [5][6] ......................................................................................................................................... 8
3.1.2 Condenser [7] ....................................................................................................................................... 9
4.0 Chemical Engineering Design ...................................................................................................................... 10
4.1 Process Description [8].............................................................................................................................. 10
4.2 Distillation Column Design [9] ................................................................................................................... 10
4.3 Key Components...................................................................................................................................... 11
4.4 Operating Conditions............................................................................................................................... 11
4.6 Packed Columns [10] ................................................................................................................................. 12
4.6.1 Types of Packing [10] .......................................................................................................................... 12
4.7 Plate Columns [11] ..................................................................................................................................... 13
4.7.1 Types of Plate [12] [13].............................................................................................................................. 13
4.8 Mass Balance ........................................................................................................................................... 16
4.8.1 Adjusted Mass Balance ..................................................................................................................... 17
4.9 Vapour Pressure Calculation ................................................................................................................... 18
4.9.1 Relative Volatility .............................................................................................................................. 19
4.10 Reflux Ratio [17]....................................................................................................................................... 19
4.11 Fenske Equation and Gilliland Correlation [19] ....................................................................................... 21
4.12 Erbar–Maddox Correlation [21] ............................................................................................................... 23
4.13 Column Diameter .................................................................................................................................. 25
4.13.1 Internal Traffic [24] ........................................................................................................................... 25
4.13.2 Vapour Density ............................................................................................................................... 26
4.13.3 Liquid Density ................................................................................................................................. 27
4.13.4 Flooding .......................................................................................................................................... 29
4.13.5 Maximum Volumetric Flowrate...................................................................................................... 32
4.13.6 Calculation of Column Diameter .................................................................................................... 34
4.14 Plate Design ........................................................................................................................................... 35
1
4.14.1 Provisional Plate Design
[30]
........................................................................................................... 35
4.14.2 Weir Height (hw) ............................................................................................................................. 36
4.14.3 Weeping.......................................................................................................................................... 36
4.14.4 Plate Pressure Drop ........................................................................................................................ 39
4.14.5 Downcomer Design [40] ................................................................................................................... 42
4.14.6 Residence Time [41].......................................................................................................................... 44
4.14.7 Entrainment [42] ............................................................................................................................... 44
4.14.8 Trial Layout ..................................................................................................................................... 46
4.14.9 Liquid Flow Arrangement ............................................................................................................... 48
4.14.10 Hydraulic Gradient........................................................................................................................ 49
4.14.11 Actual Plate Efficiency [47] ............................................................................................................. 49
4.14.12 Actual Number of Plates............................................................................................................... 51
4.14.13 Height of Column.......................................................................................................................... 51
5.0 Energy Balance ............................................................................................................................................ 52
5.1.1 Condenser Heat Load (QC) .................................................................................................................... 52
5.1.2 Condenser Design ................................................................................................................................. 55
5.1.2.1 Heat Transfer Area and Number of Tubes (NT)............................................................................. 55
5.1.2.2 Tube Bundle Diameter (Db) [ ......................................................................................................... 56
5.1.2.3 Length of Condenser ..................................................................................................................... 57
5.2 Tube Side Coefficient [55] .......................................................................................................................... 58
5.3 Shell Side Coefficient [58] .......................................................................................................................... 59
5.4 The Overall Heat Transfer Coefficient ..................................................................................................... 62
5.5 Reboiler Design ........................................................................................................................................ 63
5.5.1 Reboiler Duty .................................................................................................................................... 63
5.5.2 Kettle Reboiler Design ...................................................................................................................... 64
6.0 Mechanical Design....................................................................................................................................... 67
6.1 Shell Thickness ......................................................................................................................................... 67
6.2 Skirt Design [66] ..................................................................................................................................... 68
6.3 Heads and Closures ............................................................................................................................. 72
6.4 Material Selection.................................................................................................................................... 72
6.5 Pipe Sizing ............................................................................................................................................ 73
6.5.1 Pipe Sizing at the Feed...................................................................................................................... 74
6.5.2 Pipe Sizing at the Distillate ............................................................................................................... 76
6.5.3 Pipe sizing at the bottoms ................................................................................................................ 77
2
6.6 Concept Drawing ..................................................................................................................................... 80
7.0 Control and Instrumentation [74].................................................................................................................. 80
7.1.1 Quality Control ..................................................................................................................................... 81
7.1.2 Pressure Control ................................................................................................................................... 82
7.1.3 Temperature Control ............................................................................................................................ 83
7.1.4 Flow Control ......................................................................................................................................... 84
7.1.5 Level Control......................................................................................................................................... 85
7.1.6 Alarms and Safety Trips [74] ................................................................................................................... 85
7.1.7 Valve Selection [76] ................................................................................................................................ 87
7.1.8 Instruments and Controllers................................................................................................................. 88
7.1.9 Pipe Labelling........................................................................................................................................ 89
8.0 Hazard and Operability Study (HAZOP) ....................................................................................................... 92
9.0 Economic Appraisal ..................................................................................................................................... 93
9.1.1 Estimation of Capital Costs ................................................................................................................... 93
9.1.1.1 Wilson’s Method [81] ...................................................................................................................... 94
9.1.1.2 Zevnik and Buchanan Method [81] .................................................................................................. 96
9.1.1.3 Bridgewater‘s Method [81].............................................................................................................. 97
9.1.2 Cost of Raw Materials......................................................................................................................... 100
9.1.3 Operating Costs .................................................................................................................................. 101
9.1.4 Plant Income ....................................................................................................................................... 102
9.1.5 Cash Flow ........................................................................................................................................ 102
9.2 Breakeven and Profitability Analysis ..................................................................................................... 105
9.2.1 Amortisation ................................................................................................................................... 105
9.2.3 Discounted Cash Flow Rate of Return ............................................................................................ 107
9.2.4 Return on Investment..................................................................................................................... 109
9.2.5 Analysis of Profitability and Improvements to Profitability ........................................................... 109
10.0 Economic Optimisation ........................................................................................................................... 110
10.1.1 Potential Changes to the Existing Plant.................................................. Error! Bookmark not defined.
10.1.1.1 Re-define Product Purity ................................................................. Error! Bookmark not defined.
10.1.1.2 Improved Heat Integration ........................................................................................................ 111
10.1.1.3 Control System Upgrade............................................................................................................ 112
10.1.2 Divided Wall Column .............................................................................. Error! Bookmark not defined.
References: ...................................................................................................................................................... 115
3
1.0 Project Objectives
This section of the advanced process design module involves designing a product distillation column. It is a
continuation from the group project which involved the production of 100000 tonnes per year of Methyl
Ethyl Ketone to be located in China.
This report will focus on certain chemical engineering design features including calculations of length and
diameter. Also, it will be important to consider materials of construction and other characteristics associated
with designing a distillation column. Within this part of the report cost and economical appraisal of designing
the distillation column will be included.
The type of alarms, trips and relief devices which will be used will be included in the control and
instrumentation section of the report. When selecting the types of control and instrumentation devices for
the distillation column it will be important to consider temperatures and pressures.
A piping and instrumentation drawing will be constructed for the product distillation column. This will
include all alarms, control valves, pressure relief devices and utilities.
This report will also include a Hazard and Operability study (HAZOP). This will be carried out as a group with
the group supervisor. Each member will select a process line and a HAZOP will be carried out.
Within this report a costing and economic optimisation of the product distillation column design will also be
carried out. The last section of the report includes the economic appraisal for the process. Within this
section the capital costs of the process will be determined. Standard correlations will be used to determine if
the project is economically feasible. It will identify whether the project should be carried out in the real
world.
2.0 Project Objectives
This is a continuation from the group project and the main aim is to design a product distillation column. The
report will include detailed and constructive explanation of how the distillation column should be designed.
The following objectives have been set:
2.1 Technical Objectives
The report is expected to complete by week 22. In order to meet this deadline technical objectives will be
set. From the Gantt Chart start dates and end dates for each task can be seen. The following are the
technical objectives which have been set:

Submit the project plan and Gantt Chart by 20/01/2012.
4

Research in detail and evaluate the design of a distillation column. Including characteristics such as
number of trays, column height, column diameter, efficiency and flooding parameters.

Research the various types of control systems, instrumentation and monitoring systems for the safe
operation of the distillation column.

Construct a P +ID indicating clearly all equipment items, piping, process and utility lines, control
loops, valves, instruments including alarms and trips, pressure relief devices. Include a key of
symbols and ensure that the diagram is correctly numbered.

Carry out the HAZOP with the group. This will include systems which are required for the safe
operation of the distillation column and reduce the chance of any accidents occurring.

An economic appraisal should be carried out. This will include an estimate of the overall capital costs
and annual operating costs. Parameters including product price, construction time and interest rate
should also be justified.

References used throughout the project should be listed.

All appropriate documents including the Gantt Chart, lengthy calculations and any large drawings
should be placed in the appendices.
2.2 Personal Objectives

It is important that there is efficient communication within group members.

Important that the group works together as a team to put across ideas to ensure that the tasks are
completed on time.

Enhance communication, individual and team-working skills through the project timeline.

Remain focussed and motivated to ensure that work is produced to a high grade.

Organisation is a very important factor for the competition of this report.

It will be important that the Gantt Chart is followed closely as time as been allocated to each task
equally to ensure that the deadline is met.

Allocate sufficient number of hours during the week to work on the project.

Regular meetings should be held with the supervisor this will ensure that everything is on task and
any changes that need to be made quickly and efficiently.

Research is a major section of this report. It is therefore very important that enough time is allocated
to carry out adequate research. Extensive research will be required to gain sufficient knowledge. It is
important that various methods of research are utilised to ensure that a wide range of information is
available.

Build vital organisational and time-management skills whilst working on the project.
5

Have complete and clear understanding of how a distillation column operates on completion of the
report.

Improve technical skills by carrying out a HAZOP and constructing a P +ID which may be useful in
future projects.

Proof read all work to correct any errors which may occur and to ensure that work is of a high
standard.
6
3.0 Introduction
One of the most important operation in the chemical and petroleum industry is the separation of liquid
mixtures into several components. It is sometimes referred to as fractional distillation. It is one of the oldest
unit operation processes. The technical publication of distillation was first developed in 1957, however,
distillation had been practiced for many centuries prior to this. Distillation is one of the most common and
widely used separation processes in the chemical industry. However, it is also an extremely energy intensive
process. [1] It requires large amounts of energy for both cooling and heating. 50% of plants operating costs
are produced from distillation. At present distillation is commonly used in the petroleum, chemical,
petrochemical, beverage and pharmaceutical industries. Distillation is a process which is important in the
development of new products and for the recovery and reuse of volatile liquids.[2] A great deal of research
has been carried out into techniques of distillation due to the demand for purer products and a persistent
requirement of greater efficiency.
When designing a distillation column it is essential to consider process control. Many distillation columns
usually operate with the combination of many other separate units.
[3]
The correct design of a distillation
column is not always a simple procedure as it is regarded as a specialised technology. [2] Compared to other
types of processing equipment distillation columns have to be designed with a larger range in capacity with
single columns varying from 0.3 to 10m in diameter and 3m to 75m in height. It is important that designers
are able to provide the desired product quality at a minimum cost but also at a constant purity.
Distillation is usually used to separate liquid mixtures into two or more vapour or liquid products which have
different compositions. [1] The separation of liquid mixtures is dependent on the differences in the volatility
between the components. Separation is easier if the relative volatilities are larger. [3] There are two major types
of distillation, this includes continuous distillation and batch distillation.[4] In continuous distillation the feed
is supplied continuously. There are usually no interruptions however, problems may occur with the column
or surrounding units. This type is the more common of the two types of distillation. However, in batch
distillation the feed is supplied to the column batch-wise. The column is charged with a batch and the
distillation process is then carried out. Once the desired task has been achieved the next batch of the feed is
then introduced. [5]
7
3.1 Components of Distillation Columns
The following section is a description of the components required for the operation of a distillation column.
This includes a reboiler and a condenser.
3.1.1Reboiler [5][6]
The main objective of a reboiler in distillation columns is to vaporise a fraction of the bottom product. They
are used to provide the necessary vaporisation required for the distillation process.[5] There are three
principal types of reboilers used in distillation columns. They are as follows:
1. Forced Circulation Reboiler: the fluid is pumped through the exchanger and the vapour which is
formed is separated at the base of the column. Figure 1 shows a diagram of a forced circulation
reboiler.
Figure 1: Forced Circulation Reboiler [6]
2. Thermosyphon Natural Circulation Reboiler: it is a vertical exchanger with vaporisation in the tubes
or the shell. The difference in density between the two-phase mixture of vapour and liquid and the
single phase liquid in the base of the column helps maintain the liquid circulation through the
exchanger. The most frequently used reboiler is the shell and tube thermosyphon reboiler this is
because it is the most economical type of reboiler for most applications. Figure 2 shows a diagram
of a thermosyphon reboiler.
Figure 2: Thermosyphon Reboiler [6]
8
3. Kettle type: in which boiling takes place on tubes which are immersed in a pool of liquid. In this type
of reboiler there is no circulation of liquid. This type of reboiler is sometimes also called a submerged
bundle reboiler. The bundle may also be stored in the base of the column in some applications. This
helps save the cost of the exchanger shell. Figure 3 shows a diagram of a kettle type reboiler.
Figure 3: Kettle Type Reboiler [6]
The choice of the reboiler to be used for a given duty will depend on certain factors including [6]:
1. The nature of the process fluid i.e. the viscosity and propensity to fouling.
2. The operating pressure i.e. vacuum or pressure.
3. The equipment layout.
3.1.2 Condenser [7]
A condenser is used in a distillation column to cool and condense the vapour leaving the top of the column.
The vapour is cooled and condensed to its liquid state. The most common type of condenser used is the
horizontal shell-side and vertical tube side. This means the processor has the option of condensing on either
the shell side or the tube side. In condensers the use of cooling water as a medium to cool the substance is
of vital importance.
Condensers are available in a range of designs and in many different sizes. Capital costs of condensers can be
reduced by using a carbon steel shell. There are four possible condenser configurations which may occur as
they. They are as follows:
1. Horizontal in design with condensation occurring in the shell and the cooling medium in tubes.
2. Horizontal in design with condensation in the tubes.
3. Vertical in design with condensation in the shell.
4. Vertical in design with condensation in the tubes.
9
4.0 Chemical Engineering Design
4.1 Process Description [8]
A mixture containing Methyl Ethyl Ketone and Secondary Butyl Alcohol is obtained from the solvent recovery
column and this is fed to the product distillation column.
Methyl Ethyl
Ketone
Methyl Ethyl Ketone and
Secondary Butyl Alcohol
Mixture from Solvent
Recovery Column
CO4
Secondary Butyl
Alcohol to the recycle
stream
Figure 4: Process Description Diagram
The product distillation unit will be fed with a Methyl Ethyl Ketone (MEK)/ Secondary Butyl Alcohol (SBA)
mixture. The mixture fed to the product distillation column is obtained from the distillate of the solvent
recovery column. Before the mixture is fed to the product distillation column, the two streams are firstly
mixed in an intermediate storage holding tank. Within the product distillation column cooling water is
utilised, which enters the condenser at 24°C and leaves at 40°C. In the reboiler steam enters at 140°C and at
a pressure of 2.0 bar.
4.2 Distillation Column Design [9]
There are various stages in the design of a distillation column. They can be divided into the following steps:

Firstly the degree of separation required is specified. The product specifications are set.

The operating conditions are selected; i.e. batch or continuous, temperature and pressure.

Select the type of contacting device which is required; i.e. tray design or packed column.

Determine the stage and reflux requirements; i.e. the number of stages required for distillation
using various correlations, calculating the minimum reflux ratio and the reflux ratio.
10

Size the column; i.e. diameter, height and number of real stages.

Design the column internals; i.e. plates and packing supports.

Consider the mechanical design.

Design the condenser and reboiler.
4.3 Key Components
Before the design stage of a distillation column the designer must select the key components which are to
be separated. The light key is described as the component that is desired to be kept out of the bottom
product. The heavy key is described as the component that is desired to be kept out of the top product.
Usually it is relatively easy to determine which the key components are. However, there may be situations in
which close boiling isomers are present so judgement must be used in their selection. The light key is
described as the most volatile component in the bottom product and the heavy key is described as the least
volatile component in the top product. In this case the light and heavy keys are as follows:

Light Key: Methyl Ethyl Ketone (A)

Heavy Key: Secondary Butyl Alcohol (B)
4.4 Operating Conditions
It is assumed that the process operates at a steady state and the system is ideal. It is assumed that the inlet
temperature of the feed to the column will be at the boiling point i.e. ‘boiling liquid feed’. At this point q=1,
assuming that all of the feed to the column is in liquid phase.
It is assumed that the operating temperature of the column will be at an average temperature of 89°C
(362K). This is because Methyl Ethyl Ketone has a boiling point of 79.79°C and Secondary Butyl Alcohol has a
boiling point of 99°C. So the feed has been set to operate at a temperature in between the boiling points of
the two components.
The column will be operated on a continuous basis and will be operated at atmospheric pressure at 89°C
(362K).
11
4.6 Packed Columns [10]
Packed columns are used for a variety of processes including distillation, gas absorption and liquid-liquid
extraction. In packed bed columns the gas liquid contact is continuous, however, in plate columns it occurs
stage-wise. In packed columns the liquid flows down the column and over the packed surface and the
vapour flows counter-currently up the column. The adequate operation and performance of a packed
column relies greatly on the maintenance of good liquid and gas distribution throughout the packed bed.
Figure 5 shows a diagram of a packed column. Packed distillation columns and plate columns are similar.
However, the difference being that in packed columns the plates are replaced with packed sections.
Figure 5: Schematic Diagram of a Packed Column[11]
4.6.1 Types of Packing [10]
Packing is required for certain requirements. They are as follows:




They provide a large surface area i.e. to provide a high interfacial area between the vapour and
liquid.
They should have an open structure.
They should promote the uniform distribution of liquid on the packing surface.
They should promote uniform vapour gas flow across the column cross-section.
Various types of packing have been developed with many shapes and sizes to satisfy the requirements. They
are usually divided into two categories:
1. Random Packing
2. Structured Packing
The various types of random and structured packing can be seen in appendix ##.
12
4.7 Plate Columns [11]
In distillation columns cross-flow plates are the most common type used. In this type design the liquid flows
across the plate and the vapour flows up through the plate. The liquid is passed from one plate to the next
through vertical channels which are known as downcomers. Figure 6 shows a diagram of a cross-flow plate.
In certain occasions plates may be used which do not have any downcomers. They are known as non-crossflow plates. This type of plate may be utilised when a particularly low pressure drop is required.
Figure 6: Diagram of a Cross-Flow Plate [13]
4.7.1 Types of Plate [12] [13]
There are three principal cross-flow plate types which are used in plate columns. They are classified
according to the method used to contact the vapour and liquid. They are as follows:
1. Sieve Plate also sometimes called perforated plate: This type of plate is the simplest type of
cross-flow plate. The vapour passes the holes in the plate and the liquid is retained on the plate
due to the vapour flow. In occasions when flow rates are low liquid weeps through the holes and
this reduces plate efficiency. Usually the perforations are small holes however, in some cases
larger holes and slots are also made use of.
13
Figure 7: Diagram of a Sieve Plate [13]
2. Bubble-cap Plate: This type of plate is the most traditional and oldest type of cross flow plate.
Various designs have been developed. For most applications the standard cap design would be
specified. In this type of plate the vapour passes up pipes which are known as risers. The risers
are enclosed by a cap with a jagged edge or slots. Risers ensure that a level of liquid is
maintained on the tray at all vapour flow-rates.
Figure 8: Diagram of a Bubble-cap Plate [13]
3. Valve Plate also sometimes called floating cap plate: This type of plate is very much similar to
sieve plates however, the only difference being that they have large diameter holes which are
covered by movable flaps. When the vapour flow increases the movable flaps lift. Valves plates
are able to operate more efficiently at lower flow rates in comparison to sieve plates. At low
flow rates the valves in the valve plate’s close.
14
Figure 9: Diagram of a Valve Plate [13]
It can be observed from the mass balance in section 4.8 that a plate column will be more suitable. This is
because the flow rates in this process are large and these will require a large diameter. In cases when the
diameter is large it is possible to have plates or trays.
When selecting the plate type many factors are considered including cost, capacity, operating range,
efficiency and pressure drop. Of the three types sieve plates are the cheapest and are satisfactory for most
applications. The operating costs of sieve plates are the pressure drop is lower compared to the other types
of plates. For these reasons the selected type of plate for the distillation column is sieve plates.
15
4.8 Mass Balance
18
Product Distillation
Column
17
19
Figure 10: Block Diagram of the Product Distillation Column
Previously in the group report it was assumed that hundred percent separation i.e. complete separation was
achieved. However, this is not likely to occur to reality. The temperature difference between the
components is only 20°C as Methyl Ethyl Ketone has a boiling point of 79.79°C and Secondary Butyl Alcohol
(SBA) has a boiling point of 99°C. The distillate temperature is assumed to be the boiling point of Methyl
Ethyl Ketone i.e. the desired product. A small amount of SBA is assumed to be present in the distillate. Also,
a small quantity of MEK is also present in the bottom product. This means that the mass balance calculated
in the previous report has to be adjusted.
The following table shows the summary of the compositions entering and leaving the product distillation
column before recalculation.
Inlet
17
kmol/hr
170.2744204
19.11048489
Stream:
Components
kg/hr
wt%
MEK
12276.78571
89.66
SBA
1416.08693
10.34
Hydrogen
Water
TCE
Total
13692.87264 189.3849053
100
Table 1a: Inlet Flowrates of the product Distillation Column [8]
16
Outlet
Stream:
Components
MEK
SBA
Hydrogen
Water
TCE
Total
kg/hr
19
kmol/hr
wt%
1292.07899
17.43696343
100
12400.79365 171.9479419
100
1292.07899 17.43696343
Total Outlet Mass flowrate = 13692.87264 kg/hr
Table 1b: Outlet Flowrates of the Product Distillation Column [8]
100
kg/hr
12276.78571
124.00794
18
kmol/hr
170.2744204
1.673521457
wt%
99.0
1.00
4.8.1 Adjusted Mass Balance
It was previously assumed in the group project that all the MEK is recovered in the distillate i.e. meaning that
there would be no MEK in the bottom product stream. This does not usually occur. So it is assumed that
there is a 99% separation. This means that 99% of the MEK from the feed is recovered in the distillate and
therefore the remaining 1% is recovered in the bottom product. This will mean the previous mass balance
will have to be adjusted. This is calculated in this section of the report.
The inlet stream to the distillation column does not change. This can be seen from table 2a below:
Inlet
Stream:
17
Components
kg/hr
kmol/hr
wt%
MEK
12276.786
170.274
90
SBA
1416.0867
19.110
10
Total
13692.873
189.385
100
Table 2a: Inlet Flowrate of the Product Distillation Column [8]
Outlet
Stream:
Components
MEK
SBA
Total
18 (Distillate)
19 (Bottom Product)
kg/hr
wt%
kmol/hr
wt%
kg/hr
wt%
Kmol/hr
12140.239
99.0
168.381
99.0
13.92
1.00
0.19
126.030
1.00
1.701
1.00
1416.09
99.0
19.11
12266.269
100
170.082
100
1430.00
100
19.30
Table 2b: Outlet Flowrates of the Product Distillation Column
wt%
1.00
99.0
100
17
4.9 Vapour Pressure Calculation
To determine the vapour pressure of the components the Antoine equation must be used. The relative
volatility can then be calculated using this information.
Equation 1: Antoine Equation[14]
Where:
P = Vapour Pressure (bar)
T = Temperature (K)
A, B and C are constants
The following table shows the Antoine constants, the boiling point and latent heat of vaporisation for both
MEK and SBA.
Component
Methyl Ethyl Ketone
(MEK)
Secondary Butyl Alcohol
(SBA)
A
3.9894
B
1150.207
C
-63.904
B.P (°C)
79.6
ΔHvap (kJkmol-1)
31.3
4.32943
1158.672
-104.683
99
40.75
Table 3: Anotine Constants for MEK and SBA [15][16]
Methyl Ethyl Ketone (MEK)
= 0.131
P=
P = 1.35bar
Secondary Butyl Alcohol (SBA)
= -0.1735
P=
P = 0.67 bar
18
4.9.1 Relative Volatility
From the calculated vapour pressure for each of the components the relative volatility can be determined.
Where:
A = Light Key
B = Heavy Key
Therefore Relative Volatility:
αAB = 2.015
4.10 Reflux Ratio [17]
The reflux ratio (R) is defined as:
The reflux ratio is a very important factor in the determination of the number of stages required for
separation. An increase in the reflux ratio reduces the number of stages required for separation. This leads
to a decrease in capital costs, however, operating costs and service requirements such as steam and water
increases. The optimum reflux ratio will be the ratio at which the annual operating costs are its lowest.
The minimum reflux ratio Rmin is calculated using the Underwood equation. It is assumed that the feed
enters at its boiling point. Therefore q=1.
[
](
)
Equation 2: Underwood Equation to calculate the minimum reflux ratio
Where:
Rmin : The minimum reflux ratio
: The relative volatility = 2.015
19
: The mole fraction of MEK in the Distillate = 0.99
: The mole fraction of MEK in the feed =0.90
Substitute in values:

R
min
 0.99
1
(1  0.99) 
 2.015

2.015  1  0.90
(1  0.90) 
Rmin = 0.88
It is suggested that for many systems the optimum reflux ratio lies between 1.2 to 1.5 times the minimum
reflux ratio.
Therefore:
R= Rmin x 1.5
R= 0.88 x 1.5
R= 1.323
20
4.11 Fenske Equation and Gilliland Correlation [19]
The Fenske Equation and Gilliland correlation are used together to calculate the number of theoretical
stages. The minimum number of stages required at the total reflux can be calculated using the Fenske
equation. The number of theoretical plates required can then be estimated using the Gilliland correlation,
figure 11. The Gilliland correlation is a relationship between the reflux ratio, the minimum reflux ratio and
the minimum number of stages. [18]
[(
) (
) ]
Equation 3: Fenske Equation [20]
Where:
: The relative volatility = 2.015
: The mole fraction of MEK in the distillate = 0.99
: The mole fraction of SBA in the distillate = 0.01
: The mole fraction of SBA in the waste (bottoms) = 0.99
: The mole fraction of MEK in the waste (bottoms) = 0.01
Substitute in values:
*(
) (
) +
= 13.099
The Gilliland correlation can now be used to calculate the number of theoretical stages.
21
Figure 11: Gilliland Correlation
Where:
R: The Reflux Ratio = 1.32
Rm: The Minimum Reflux Ratio = 0.88
Substitute in values:
From figure 11 it can be determined that the curve is intersected at 0.47 when the x-axis is 0.189.
Substitute in the values and rearrange to calculate n (theoretical number of stages).
Therefore,
22
4.12 Erbar–Maddox Correlation [21]
The Erbar–Maddox correlation is a different method which can also be used in the determination of the
number of theoretical plates. This method gives the ratio of number of stages required to the number at
total reflux. It is given as a function of the reflux ratio with the minimum reflux ratio. Figure 12 shows the
Erbar-Maddox correlation.
Figure 12: Erbar-Maddox Correlation [22]
The following calculations have to be carried out in order to determine the number of stages:
Where:
R: The Reflux Ratio = 1.32
Substitute in values:
Where:
Rm: The Minimum Reflux Ratio = 0.88
23
Substitute in values:
The Nm /N value can be obtained from the graph in figure 12.
Where Nm = 13.099
Therefore, substitute in the values and rearrange to obtain N. Where N is the number of theoretical plates.
The values obtained for the number of plates are only preliminary value. The actual number of stages cannot
be determined at this point as the plate efficiency is not yet known. The plate efficiency will be calculated at
a later stage of the report and that will be used to calculate the actual number of stages.
24
4.13 Column Diameter
The number of stages is not required to calculate the diameter of the column. However, the liquid and
vapour flowrates are required in order to calculate the diameter. The flowrate is the principal factor in
determining the column diameter. It is important that the vapour velocity is lower than that velocity which
would cause entrainment.[23] There are various steps which will be needed to calculate the column diameter
for the distillation column. A number of calculations will be carried out in this section to determine the
column diameter.
The chosen plate is a sieve plate. Certain specifications have to be put in place in order to calculate the
diameter of the column. The following specifications were made:

Hole Diameter – 5mm

Tray Spacing – 600mm

Plate Thickness – 5mm

Hole Pitch – 15mm
4.13.1 Internal Traffic [24]
The liquid and vapour flowrates will be calculated using the four following equations:
1. Lo = RD
2. V = Lo + D
3. L’ = Lo + qF
4. V = V’ + (1 – q)
Where:
R: The reflux ratio = 1.323
D: The total distillate flowrate = 170.081 kmol/hr which is equal to 12266.269 kg/hr
F: The total feed flowrate = 189.385 kmol/hr which is equal to 13692.873kg/hr
q: The feed enters as liquid at its boiling point = 1
1. Lo = RD
A) Lo = 1.323 x 170.08 = 225.028 kmol/hr
B) Lo = 1.323 x 12266.269 = 16228.274 kg/hr
2. V = Lo + D
A) V = 225.028 + 170.081 = 395.109 kmol/hr
B) V = 16228.274 + 12266.269 =28495.543 kg/hr
3. L’ = Lo + qF
25
A) L’ = 225.028 + (1 x 189.385) = 414.413kmol/hr
B) L’ = 16228.274 + (1 x 13692.873) =29921.147 kg/hr
4. V = V’ + (1 – q)
A) V =395.109 + (1 – 1) = 395.109 kmol/hr
B) V = 28495.543 + (1 – 1) = 28495.543 kg/hr
4.13.2 Vapour Density
The ideal gas equation will be used to calculate the vapour density for both Methyl Ethyl Ketone and
Secondary Butyl Alcohol.
PV=nRT [25]
Where:
P: is the Pressure in Pa
V: is the Volume
n: is the number of Moles
R: is the gas constant
T: is the temperature in Kelvin
PV=nRT
Where M is the mass of the feed.
Rearrange the above equation to make density (ρ) the subject:
The column pressure is assumed to be operating at atmospheric pressure. So the pressure will be taken as
101325Pa.
26
Methyl Ethyl Ketone Vapour Density
P = 101325 Pa
Mr = 72.1
R = 8314 m3PaK-1kmol-1
T = 362 K
Substitute the values into the above equation.
Secondary Butyl Alcohol Vapour Density
P = 101325 Pa
Mr = 74.1
R = 8314 m3PaK-1kmol-1
T = 362 K
Substitute the values into the above equation.
4.13.3 Liquid Density
To calculate the liquid density for both Methyl Ethyl Ketone and Secondary Butyl Alcohol the following
equation will be used:
L  A* B
 1T
 
 Tc




n
Equation 4: Liquid Density [25]
Where:
A, B and n: are regression coefficients (shown in table 4)
T: The operating temperature in Kelvin = 362k
Tc: The critical Temperature in Kelvin (shown in table 4)
27
A
B
n
0.2857
Critical
Temperature,
Tc (K)
535.5
Operating
Temperature
(K)
362
0.2676
0.2514
0.2734
0.2635
0.2604
536.01
362
Component
Methyl Ethyl Ketone
(MEK)
Secondary Butyl Alcohol
(SBA)
Table 4: Regression Coefficients for MEK and SBA
[26]
Methyl Ethyl Ketone Liquid Density
Substitute the values into the above equation:
  0.2676  0.2514
 1362 


 535.5 
0.2857
  0.2676  0.2514 0.893
Convert to kg/m3
Secondary Butyl Alcohol
Substitute the values into the above equation:
  0.2734  0.2635
 1362 


 536.01 
0.2604
  0.2734  0.2635 0.902
Convert to kg/m3
28
4.13.4 Flooding
Flooding occurs when the vapour flow is excessive and this causes liquid to be entrained in the vapour up
the column. The excessive vapour flowrate also cause an increase in pressure and this backs up the liquid in
the downcomer. This causes an increase in liquid holdup on the plate above. The maximum capacity of the
column can be reduced severely depending on the degree of flooding. Flooding can be detected by a sharp
increase in the column differential pressure and a significant decrease in the separation efficiency.
[27]
The
following equation is used to calculate the Liquid-Vapour Flow (FLV) factor.
FLV
L 
   *  V
V   L



0.5
Equation 5: Liquid Vapour Flow Factor [28]
Where:
FLV: Liquid-Vapour Flow Factor
L: Liquid molar flowrate
V: Vapour molar flowrate
ρv: Vapour Density
ρL: Liquid Density
The Liquid-Vapour Flow Factor will be calculated for both the enriching (Top) section and the stripping
(Bottom) section.
Enriching Section (FLVTOP)
FLV
L 
   *  V
 V   L



0.5
FLV: Liquid-Vapour Flow Factor
L: Liquid molar flowrate = 225.028 kmol/hr
V: Vapour molar flowrate = 395.109 kmol/hr
ρv: Vapour Density = 2.427 kg/m3
ρL: Liquid Density = 780 kg/m3
Substitute the values into equation5:
(
)
(
)
0.0312
29
Assuming a tray spacing of 600m (0.6m) which is suitable for maintenance, inspection and cleaning. From
the calculated FLV the corresponding K1 TOP value can be obtained. From figure13 it can be seen that the
corresponding K1TOP value is 0.11.
Figure13: Flooding Velocity [28]
Stripping Section (FLVBOTTOM)
FLV
 L'   
   *  V
V '   L



0.5
Equation 6: Liquid Flow Factor [28]
FLV: Liquid-Vapour Flow Factor
L’: Liquid molar flowrate = 414.413 kmol/hr
V’’: Vapour molar flowrate = 395.109 kmol/hr
ρv: Vapour Density = 2.495 kg/m3
ρL: Liquid Density =820 kg/m3
Substitute the values into equation 6
(
)
(
)
1.048856138 × 0.055155504
0.058
30
Assuming a tray spacing of 600m (0.6m) which is suitable for maintenance, inspection and cleaning. From
the calculated FLV the corresponding K1BOTTOM value can be obtained. From figure13 it can be seen that the
corresponding K1BOTTOM value is 0.1.
The flooding velocity is then calculated using the following equation:
   V 
U F  K1  L

 V 
0.5
Equation 7: Flooding Velocity Correlation [28]
Where:
UF: Flooding Vapour Velocity
K1: A Constant
ρL: Liquid Density
ρV: Vapour Density
Enriching Section Flooding Velocity
   V 
U F  K1  L

 V 
0.5
UF TOP: Flooding Vapour Velocity
K1: A Constant = 0.11
ρL: Liquid Density = 780 kg/m3
ρV: Vapour Density = 2.427 kg/m3
Substitute the values into equation 7
[
]
UF TOP = 1.969 m/s
Stripping Section Flooding Velocity
   V 
U F  K1  L

 V 
0.5
31
UF TOP: Flooding Vapour Velocity
K1: A Constant = 0.1
ρL: Liquid Density = 820 kg/m3
ρV: Vapour Density = 2.495 kg/m3
Substitute the values into equation 8
[
]
UF BOTTOM = 1.81 m/s
The flooding condition fixes the upper limit of vapour velocity. For high plate efficiencies a higher vapour
velocity is required, the velocity will be normally 79-90% of that which could flooding. In chemical
engineering design a flooding velocity between 80-85% would be effective. [29]
4.13.5 Maximum Volumetric Flowrate
Taking 80% flooding the velocity at flooding is as follows:
Flowrate in the enriching section (UV,TOP) = 0.8 × 1.969 = 1.575 m/s
Flowrate in the stripping section (UV,BOTTOM) = 0.8 × 1.81 =1.448 m/s
To calculate the area for the top and the bottom of the column the maximum volumetric flowrates are
used. The maximum volumetric flowrate is calculated using the following equation:
Where:
Umax: Maximum Volumetric Flowrate
V: Molar Vapour Flowrate
ρ: Vapour Density
Mr: Relative Molecular Mass
32
Maximum Volumetric Flowrate in the Enriching Section
Where:
Umax: Maximum Volumetric Flowrate
V: Molar Vapour Flowrate = 395.109 kmol/hr
ρ: Vapour Density = 2.427 kg/m3
Mr: Relative Molecular Mass = 72.1
Substitute values into the above equation:
Umax = 3.260 m3/s
Maximum Volumetric Flowrate in the Stripping Section
Where:
Umax: Maximum Volumetric Flowrate
V: Molar Vapour Flowrate = 395.109 kmol/hr
ρv: Vapour Density = 2.495 kg/m3
Mr: Relative Molecular Mass = 74.1
Substitute values into the above equation:
Umax = 3.260 m3/s
33
4.13.6 Calculation of Column Diameter
The following equation is used to calculate the column diameter of the distillation column:
√
Equation 9: Column Diameter
Where:
d: The diameter
A: Area
Enriching Section (Top) Diameter
Area:
So the diameter is therefore calculated to be:
√
d = 1.623 m
Striping Section (Bottom) Diameter
Area:
So the diameter is therefore calculated to be:
√
d = 1.693 m
It can be seen that the both the top and bottom diameter are very similar. However, the largest diameter is
selected to be the column diameter of the entire distillation column, i.e. Dc = 1.693 m.
34
From the diameter calculated it can be seen that the diameter is greater than 0.6m. Therefore, it would be
suitable for the column to be designed as a tray design.
4.14 Plate Design
4.14.1 Provisional Plate Design
[30]
In section 4.13.6 the column diameter was determined.
The area of the column is calculated as follows:
Equation 10 Area of the Column
Ac = 2.251 m2
The Downcomer area (Ad) is taken at 12% of the column area (Ac):
Ad = 0.12 × 2.251 = 0.270 m2
Net Area (An) = Ac – Ad
(An) = 2.251 – 0.270 = 1.981 m2
Active Area (Aa) = Ac – 2Ad
(Aa) = 2.251 – (2 × 0.270) = 1.711m2
The Hole area (Ah) is taken at 10% of the Active area (Aa):
Ah = 0.1 × 1.711 = 0.171m2
The Weir length (lw) is calculated using figure14.
Where:
Ad: Downcomer area = 0.270 m
Ac: Column Area = 2.251 m
Substitute in the values:
The corresponding value for
can now be obtained from figure14. The value determined from the graph is
0.76.
35
Weir length (lw) = 0.755 × 1.693 = 1.278m.
Figure14: Correlation between downcomer area and weir length [31]
4.14.2 Weir Height (hw)
The volume of liquid on the plate is determined by the height of the weir. It is also an important factor in the
determination of plate efficiency. Plate efficiency increases as the weir height increases. However, this is at
the expense of a higher plate pressure drop. For distillation columns which require a vacuum lower weir
heights are suggested as this reduces the pressure drop. Recommended weir heights are typically in the
range of 6 to 12mm for vacuum operation. For columns which operate above atmospheric pressure weir
heights are generally between 40mm and 90mm. It is recommended that weir heights of 40 to 50mm are
used. In this case the selected weir height (hw) is 50mm. [31]
4.14.3 Weeping
Weeping occurs when the flowrate in the distillation column is low. Due to this the pressure exerted by the
vapour is insufficient to hold up the liquid on the tray. As a result of this, liquid starts to leak through the
perforations (holes). Dumping occurs as a result of excessive weeping. This will mean that the liquid on all
the trays will crash through to the base of the column. This in turn results in a domino effect and the
therefore the column will have to be re-started. Significant pressure drops and reduced separation efficiency
indicate the presence of weeping.
[27]
The weep point occurs when liquid leakage through the plate holes
36
becomes excessive. The vapour velocity at the weep point is the minimum velocity required for stable
operation. The vapour flow velocity at the lowest operating rate must be well above the weep point when
specifying the hole area. The minimum design vapour velocity is given in equation 11. [32]
[
]
[
]
[
]
Equation11: Minimum Design Vapour Velocity [32]
Where:
uh: The minimum vapour velocity through the holes
dh: The hole diameter
K2: A constant, which is dependent on the depth of clear liquid on the plate (obtained from figure 15)
Weeping Check [33]
K2 in equation 11 is a constant value. In order to determine K2 the depth of the crest of liquid over the weir
(how) must first be calculated. The Francis weir formula can be utilised to find the height of the liquid crest
over the weir. The Francis weir formula is given in equation 12.
[
]
Equation12: Francis weir equation [34]
Where:
how: Weir Crest
Lw: Weir Length
Lw: Liquid flow-rate
: Liquid Density
Firstly, the values required to determine the weir crest must be calculated.
The maximum liquid rate ( L’):
The minimum liquid rate at 70% turn down:
= 0.7 x 8.312
= 5.818 kg/s
37
Maximum how:
[
]
Minimum how:
[
]
At the minimum liquid rate: hw + how = 50 + 24.312 = 74.312
The K2 value can now be determined from figure15.
Figure15: Weep point correlation [35]
The corresponding K2 value at 74.312mm is 30.7.
Now the minimum design vapour velocity can be calculated from equation 11.
Where:
dh: The hole diameter = 5 mm
K2: A constant = 30.7
ρv: Vapour Density = 2.427 kg/m3
Substitute values into equation 11:
[
]
[
]
[
]
38
It is important to calculate the actual minimum vapour velocity. This is calculated as follows:
It can be seen that the minimum operating rate is well above the weep point.
4.14.4 Plate Pressure Drop
In the design of distillation columns the pressure drop is an important aspect. There are two causes of
pressure drop: as a result of vapour flow through the holes and due to the static head of liquid on the plate.
The total pressure drop is the sum of the dry plate pressure drop (hd), the head of the clear liquid on the
plate (hw + how) and residual losses (hr). Residual losses account for other minor sources of pressure losses
which may occur. The residual loss is the difference between the experimental pressure drop and the sum of
the dry plate drop and the clear-liquid height. [36]
4.14.4.1 Dry Plate Drop [37]
The pressure drop through the dry plate is calculated using equation 13.
[
]
Equation 13: Pressure drop through the dry plate [38]
Where:
Uh: Maximum vapour velocity through the holes
Co: Orifice coefficient. It is a function of plate thickness, hole diameter and the hole to perforated area ratio
(obtained from figure16).
ρv: Vapour Density
ρL: Liquid Density
39
Figure16: Discharge Coefficient [38]
The maximum vapour velocity through the holes (Uh) is calculated as follows:
To determine Co:
Assuming from the graph:
The corresponding Co value can be obtained 0.84
Now the dry plate drop can be determined by substituting into equation 13.
[
]
40
4.14.4.2 Residual Head [39]
Many methods have been developed to estimate the residual head which have been a function of liquid
surface tension, froth density and height. As a result of the correction term being small the estimation is not
justified. However, an equation by Hunt et al (equation 14) has been proposed to find the residual head.
Equation 14: Residual Head [39]
Where:
hr: The Residual Head
ρL: Liquid Density
4.14.4.3 Total Drop
The total pressure drop can now be calculated. The total plate drop is given in equation 15.
Equation 15: Total Plate Drop [39]
Where:
ht: Total Drop
hd: Dry Plate Drop = 81.673 mm
hw+how: Head of clear liquid on the plate = (50 + 30.840mm)
hr: Residual Head = 16.026 mm
Substitute in the values:
ht = 81.673 + 50 + 30.840 + 16.026 = 178.539 mm liquid
The total plate drop is expressed in terms of millimetres however it can also be given in pressure units. This
is given as follows:
Equation 16: Total Pressure Drop [36]
41
Where:
ΔPt: Total Plate Pressure Drop (Pa(N/m2))
ht: Total Plate Pressure Drop (mm liquid)
ρL: Liquid Density
Substitute in the values:
ΔPt = 9.81 x 10-3 x 178.539 x780 = 1366.149 Pa
= 1.366 kPa
4.14.5 Downcomer Design [40]
When designing the downcomer it is important to ensure that the level of the liquid and the froth in the
downcomer is considerably below the top of the outlet weir on the plate above it. The column is likely to
flood if the level rises above the outlet weir. The pressure drop over the plate and the resistance to flow in
the downcomer may cause a backup of liquid in the downcomer. A diagram of the downcomer backup is
shown in figure17.
Figure17: Downcomer Back-up [40]
The head loss in the downcomer can be estimated using the equation is equation 17.
[
]
Equation17: Head loss in downcomer [40]
Where:
Lwd: Liquid flow rate in downcomer
Am: is either the downcomer area (Ad) or the clearance area under the downcomer (Aap). The smallest value
42
is used.
ρL: Liquid Density
The clearance area under the downcomer (Aap):
(Aap) = hapIw
hap is the height of the bottom edge of the apron above the plate. The height is usually set at 5 to 10mm. In
this case it has been set to be 10mm. So:
hap = hw – 10 = 50 – 10 = 40mm
= 0.04m
The clearance area under the downcomer (Aap):
=0.04 x 1.278 = 0.051 m2
It can be seen that Ad = 0.270m2. It can therefore be concluded that the smallest value for Am in this case is
equal to Aap. i.e. Am = Aap
Substitute the values into equation 17:
[
]
The downcomer backup can now be calculated:
hb = 50 + 30.840 + 178.539 + 7.248 = 266.627 mm
The downcomer back up height should be less than 0.5 times the plate spacing and weir height for a safer
plate design. This is shown below:
[
]
[
]
0.266m < 0.325m
This shows that the plate spacing of 600m is acceptable as the downcomer backup is less.
43
4.14.6 Residence Time [41]
It is important to ensure that enough time is allowed in the downcomer for any entrained vapour to
disengage from the liquid stream and prevent the liquid being carried under the downcomer. A time of at
least 3 seconds is recommended.
Equation 18: Residence Time for the downcomer [41]
Where:
tr: Residence Time
Ad: Downcomer Area = 0.270 m
hbc: Clear Liquid back-up = 0.266 m
ρL: Liquid Density = 780 kg/m3
Lwd: Liquid flow rate in downcomer = 8.312 kg/s
Substitute in the values:
It can be seen that the calculated residence time is greater than the recommended time of least 3 seconds.
4.14.7 Entrainment [42]
Entrainment is a result of high vapour flow rates and refers to the liquid carried up by vapour to the tray
above. It is unfavourable as tray efficiency is reduced. The lower volatile material is carried to a plate holding
liquid of a higher volatility. High purity distillates can also become contaminated. In the event of excessive
entrainment flooding can occur.
[27]
The correlation developed by Fair (figure18) can be used to estimate
entrainment. It shows the fractional entrainment (ψ) as a function of the liquid-vapour factor (FLV), with the
percentage of flooding as a parameter.
It can be seen from section 4.13.5 that the percentage flooding is taken to be 80%.
FLV = 0.058
The corresponding fractional entrainment (ψ) can be obtained from figure18 below. The (ψ) is found to be
0.041.
44
It is important to ensure that the fraction entrainment is lower than 0.1. It can be seen that the value
obtained is significantly lower than 0.1 and is therefore within a safe operating range.
Figure18: Correlation to find Fractional Entrainment [42]
45
4.14.8 Trial Layout
4.14.8.1 Perforated Area [43]
Figure19: Relation between angle subtended by chord, chord height and chord length [44]
Obstruction caused by structural members such as support rings and beams and by the use of calming zones
reduces the area available for perforation. Calming zones are referred to unperforated strips of plate at the
inlet and outlet sides of the plate. The widths of each zone are usually made the same and have
recommended values of below 1.5m, 75mm; above 100mm. For sectional plates the width of the support
ring is usually between 50 to 75mm. It is important to ensure that the support rings do not enter into the
downcomer area. Using figure19 the unperforated area can be calculated from the plate geometry. [43]
From figure19:
From the y axis the corresponding value determined = 99°
The angle subtended at the plate edge by unperforated strip = 180 – 99 = 81°
Mean Length of unperforated edge strips =
Area of unperforated edge strips =
46
Area of calming zone = 2(1.328 x 50 x 10-3) = 0.133 m2
Total Area for perforations (Ap) = Active Area - Area of unperforated edge strips - area of the calming zone
Ap = 1.711 – 0.116 – 1.328 = 0.267 m2
The distance between the hole centres i.e. the hole pitch should not be less than 2.0 hole diameters. The
normal range falls between 2.5 to 4.0 diameters. From the range the pitch can be selected to give the
number of active holes required for the total hole area. Usually square and equilateral triangular patterns
are used. Of these two the equilateral triangular pattern is preferred. The total hole area as fraction of the
perforated area Ap is expressed in the following equation.
[
]
Equation 19: The total hole area as a fraction of the perforated area [43]
From figure 20 below the
can be determined using the value calculated for
above.
Figure 20: Correlation to show the relationship between hole area and pitch [45]
The value obtained for
does not fall within the range. This means that the hole area is too large. Within
the provisional plate design it was originally assumed that the hole area will be taken as 10% of the active
area. However, now we shall assume a hole area as 3% of the active area:
47
Ah =0.03 x 1.711 m2 = 0.051 m2
Therefore,
can now be recalculated:
The corresponding
can now be obtained figure20:
4.14.8.2 Number of Holes
The diameter of hole = 5mm
Therefore,
4.14.9 Liquid Flow Arrangement
Figure21: Graph for selection of liquid-flow arrangement [46]
48
The liquid flow-rate and column diameter are the factors which determine the choice of the plate type i.e.
reverse, single pass or multiple pass. Figure21 can be used to find the liquid-flow arrangement. [46]
From figure21it can be seen that a cross –flow (single pass) can be used.
4.14.10 Hydraulic Gradient
The difference in liquid level which is needed to drive the liquid flow across the plates is referred to as the
hydraulic gradient. On sieve plates no hydraulic gradient occurs because of small resistance to liquid flow. In
sieve plate designs the hydraulic gradient is usually ignored. [43]
4.14.11 Actual Plate Efficiency [47]
Van Winkle et al (1972) published a correlation for the determination of plate efficiency which can be used
for binary systems. Dimensionless groups which affect plate efficiency are included within the correlation.
The equation is as follows:
Emv  0.07 Dg 0.14 Sc0.25 Re 0.08
Equation 20: Van Winkle Correlation for plate efficiency [47]
Where:
μL : Liquid Viscosity
(calculation shown in appendix ##)
49
σL: Liquid surface tension = 0.0213 N/m (calculation shown in appendix ##)
DLK: Liquid Diffsivity =
(calculation shown in appendix ##)
hw: Weir height = 50 mm
ρL: Liquid Density = 780 kg/m3
ρv: Vapour Density = 2.427 kg/m3
Calculating the unknowns from above:
Substitute all the values into the Van Winkles equation:
Emv = 0.07 × (67.046)0.14 × (74.139)0.25 ×(10537.998)0.08
Emv = 0.776
Therefore, using Van Winkle’s equation and actual plate efficiency of 77.6% is obtained.
50
4.14.12 Actual Number of Plates
Previously two methods were used to determine the theoretical number of plates. According to Gilliland and
Fenske’s correlation 24 plates would be required. According to the Erbar-Maddox correlation 23 plates
would be required. We must select the worst case scenario i.e. pick 24 plates.
Therefore, the actual number of plates required taking account of the calculated plate efficiency would be as
follows:
In order to achieve effective distillation 31 stages would be required.
4.14.13 Height of Column
The equation below can be used to predict the height of the distillation column:
Hc = (N+1)*Hs+∆H+plate thickness
Equation 21: Determination of Column Height
Where:
N: The actual number of plates (stages) required = 31
Hs: The tray spacing = 600mm (0.6m)
ΔH: The distance of liquid holdup and vapour disengagement = 1m
Plate Thickness = 5mm (0.005m)
Substitution of all values gives a column height of: [(31 + 1) × 0.6] +1 + 0.005 = 20.205m ~ 21m
The height of the column is therefore predicted to be approximately 21m tall.
51
5.0 Energy Balance
[48]
Process streams have kinetic and potential energies; however, they are neglected as they are small. In all
systems a transfer of heat occurs between the inlet and outlet streams. This is shown in figure22.
Figure22: Transfer of Energy within a system [48]
Input of energy is provided by two means: from the feed (Hf) and the reboiler (Qb).
So: Qb + feed sensible heat Hf
Output of energy occurs from the top (HD) and bottom (HW) products and from the condenser (QC).
So: QC + top and bottom products sensible heat HD + HW
In order to calculate the condenser and reboiler heat load the following have to be calculated:

Latent Heat of Vaporisation of the Components ( Calculations found in appendix ##)

Specific Heat Capacity ( Calculations found in appendix ##)
5.1.1 Condenser Heat Load (QC)
In order to determine QC the heat balance must be calculated around the condenser. The temperature of
the column is 89°C (362 K), however the temperature at the top of the column must be lower than this. The
inlet temperature to the condenser is assumed to be 79.98°C (352.98K) and the outlet temperature is
assumed to be 60°C (333 K). From the mass balance in section 4.8 it can be seen that the distillate contains
99% MEK so the calculation will be based on the capacities of only MEK.
52
12266.269 kg/hr
Figure23: Top of column and condenser diagram [49]
At a steady state as shown in figure 23 the following are true:
INPUT = OUTPUT
And this can be rearranged to determine QC:
It is assumed that complete condensation occurs so:
Enthalpy of vapour HV = Latent + Sensible
189.809 J/molK
1.591 J/molK
[
]
Where:
189.809 J/molK
1.591 J/molK
V’= 28495.543 kg/hr
Substitute in all the values:
53
[
]
QC = 6317039.579 kj/hr
= 1754.733 KW
The top product and reflux will be at the same temperature so QC = HV
Cooling water required
The amount of cooling water required in the condenser will be calculated in this section.
It is assumed that:
It is know that:
The above equation can be rearranged to calculate the amount of cooling water required. This is as follows:
Where:
Cp = 4.187 kj/kg.K
ΔT = From the process description in section 4.1 it can be seen that the cooling water enters the condenser
at 24°C (297K) and leaves at 40°C (313 K).
Substitute the values in:
54
5.1.2 Condenser Design
This next focuses on the detailed design of the condenser.
There are certain specifications which are put in place in order to carry out the appropriate calculations.
They are as follows:

Outside Diameter (OD) of 25mm

Internal Diameter (ID) of 20mm

Tube length of 5m

Baffle cut of 25% is optimum as this gives good heat transfer rates [50]

The condensing material MEK will be located on the shell side as the shell side copes better with
changes in density. The cooling water will flow on the tube side.

Pipes are assumed to be arranged in a square arrangement. This is due to the ease of maintenance
such as cleaning. [51]

The recommended minimum clearance between the tubes is 25 inches (6.4mm) when using a square
pattern. [51]

The overall heat transfer coefficient (U) is assumed to be 850 W/m2 °C

Fouling Coefficient of 5000 W/m2 °K
[52]
5.1.2.1 Heat Transfer Area and Number of Tubes (NT)
Calculating the provisional Area:
Firstly the log mean temperature needs to be calculated using the following equation:
[
(
[
]
)
]
Equation 22: Calculation of log mean temperature [53]
Where:
ΔTLM: Log mean temperature difference
T1: Inlet shell side fluid temperature = 79.98°C
T2: Outlet shell side fluid temperature = 60°C
t1: Inlet tube side temperature = 24°C
t2: Outlet tube side temperature = 40°C
55
Substitute values into equation 22:
[
(
[
]
]
)
We are aware that Q=UAΔT. This can be rearranged to obtain the area. This is shown below:
From the above assumptions and specification it can be seen that U is assumed to be 850 W/m2 °C.
Therefore;
(
)
From this information the number of tubes can now be calculated:
Square pitch of 1.25do
Therefore pitch (Pt) = 1.25 x 25 = 31.25 mm
5.1.2.2 Tube Bundle Diameter (Db) [54]
The diameter of the tube bundle depends on both the number of tubes and the number of passes. The tube
bundle diameter can be estimated from equation 23. The constants are shown in table 5.
(
)
Equation 23: Tube Bundle diameter page [54]
Where:
Db: Bundle Diameter
NT: The number of tubes = 139
56
do: Outside diameter = 25mm
K1 and n1: Constants = assumed a square pitch with 2 passes (The constants are shown in table 5)
Table 5: Constants required for tube bundle equation[54]
Substitute the values into equation 23:
(
)
Where:
Db: Bundle Diameter = 484.774 mm
Pt: Tube Pitch = 31.25 mm
Therefore:
5.1.2.3 Length of Condenser
The length of the condenser can be calculated as follows:
Equation 24: Length of Condenser
Where:
57
L: The length of the condenser
A: The heat transfer area = 54.391 m2
do: Outside diameter = 25 mm
NT: The number of tubes
Substitute in the values into Equation 24:
5.2 Tube Side Coefficient [55]
The tube side coefficient is calculated using equation 25 below. The equation has been adapted from data
provided Eagle and Ferguson (1930).
Equation 25: Tube Side Coefficient [56]
Where:
hi: Inside coefficient for water
t: water temperature = 32°C - calculated below
ut: Water Velocity - calculated below
di: Tube inside diameter = 20mm
Mean temperature of water:
Tube cross-sectional area:
Tube side water velocity:
(
)
58
Density of water = 998 kg/m3 [57]
Substitute into Equation 25:
5.3 Shell Side Coefficient [58]
Firstly the shell diameter has to be determined. The bundle diametrical clearance is found from figure24
below. Then this is used with the bundle diameter to calculate the shell diameter.
Choosing the pull through floating head the bundle diametrical clearance is found to be 90.5mm.
Shell Diameter (DS) = 484.774 + 90.5 = 575.274 mm
Figure24: Shell-bundle Clearance [59]
In the shell’s baffles are used to direct the fluid stream across the tubes. They help to increase fluid velocity
and therefore improve the rate of transfer. Single segmental baffles are the most common type of baffle
used. This type of baffle is show in figure25.
59
Figure25: Segmental Baffle Design [60]
Tube Pitch (Pt) = 1.25 x 25 = 31.25 mm
(
)
The density of MEK has to be calculated at 69.99°C. This is shown in appendix ##.
Density of MEK at 69.99°C (342.99 K) = 795 kg/m3
The viscosity of MEK has to be calculated at 69.99°C. This is shown in appendix ##.
Viscosity of MEK at 69.99°C (342.99 K) = 0.2548 MNs/m2
One is also required to calculate the thermal conductivity (K) The calculation is shown in appendix ##.
The thermal conductivity (K) = 0.188 w/m◦C
60
From the assumptions it can be seen that a baffle cut of 25% has been specified. From figure26 the
corresponding heat transfer factor (jh) can be determined from the Reynolds number calculated.
Figure26: Shell-Side Heat Transfer Factor [61]
Jh = 2.7 ×10-3
The shell side heat transfer coefficient without the viscosity correction term is as follows:
61
5.4 The Overall Heat Transfer Coefficient [62]
The overall heat transfer coefficient is given by the following equation:
d
d o ln  o
1
1
1
 di



U ho hod
2k w


  do * 1  d o * 1
d i hid d i hi
Equation 26: Overall heat transfer coefficient [62]
Where:
U: Overall Coefficient
ho: Outside fluid film coefficient = 1906.365
hod: Outside dirt coefficient = 5000 W/m2 °K
hi: Inside fluid film coefficient =
hid: Inside dirt coefficient = 5000 W/m2 °K
kw: Thermal conductivity of the tube wall = 50
di: Inside diameter = 20 mm
do: Outside diameter = 25mm
Substitute values into equation 26:
 0.025 
0.025ln

1
1
1
1
 0.020   0.025  1  0.025 



 789.407
U 1906.365 5000
2  50
0.020 5000 0.020 5286.993
U = 789.407
It can be seen that the calculated value for the overall heat transfer coefficient is different from the assumed
value of 850 W/m2 °C. So in this case it would be suitable to assume an overall heat transfer coefficient of
700 W/m2 °C and carry out the calculations again.
62
5.5 Reboiler Design
This section will involve calculating the heat load for the condenser and also a brief design of the reboiler
required for the distillation column.
5.5.1 Reboiler Duty
The heat needed in the reboiler can be determined from the overall heat balance over the distillation
column. This is given below in equation 27.
Equation 27: Overall heat balance
Where:
HD and Hb are the sensible heats of the distillate and bottom product.
In order to determine the reboiler duty the other parameters have to be calculated. Then the above
equation can be rearranged to find the heat required in the reboiler. They are as follows:
Firstly the specific heat capacity (Cp) has to be calculated for Methyl Ethyl Ketone (MEK) and Secondary Butyl
Alcohol (SBA). This is shown in appendix##. The feed inlet to the distillation column is assumed to be 70 °C.
The sensible heat of the feed at a basis of 25 °C is calculated below.
Now HF can be calculated as follows:
The bottom product is assumed to be at a temperature of 98.807°C. The sensible heat of the bottom product
is calculated as follows:
HD = 0
Qc was previously calculated in section 4.15.1.
Qc = 6317039.579 kj/hr
Now all the parameters have been found the heat required can now be calculated.
63
QB = QC + HD + HB – HF
QB = 6317039.579 + 0 + 355253.736 – 2091312.493 = 4580980.822 kj/hr
= 1272.495 kw
Heat losses may occur. Therefore add 10% for heat losses:
QB =1.10 × 4580980.822 = 5039078.904 kj/hr
= 1399.744 kw
It is assumed that the steam which is supplied to the reboiler is at 2 bar (200 kN/m2) at a temperature of
approximately 140°C. The latent heat of steam is therefore 2144.4 kj/kg. [63]
The steam required is calculated using the following equation:
Where:
QB: Reboiler heat load = 5039078.904 kj/hr
S: Steam Flowrate
hfg: Latent heat of vaporisation = 2144.4 kj/kg
The above equation can be rearranged to calculate the steam flowrate:
5.5.2 Kettle Reboiler Design
Section 3.1.1 describes the three types of reboilers which are usually used. In this design a kettle reboiler
will be used. In kettle reboilers boiling takes place on tubes which are immersed in a pool of liquid.
In order to carry out the appropriate calculations certain specifications and assumptions will have to be set.
They are as follows:

Inside Diameter (ID) of 25 mm

Outside Diameter (OD) of 30 mm

Nominal tube length of 5m

Steam at 2 bar

Overall heat transfer coefficient 900 W/m2 °C
64
The temperature of the bottom product is assumed to be 98.807 °C (371.807 K). From the steam table it can
be determined that the temperature of the steam at 2 bar (200 kN/m2) is 120.23 °C (393.23 K).
The log mean temperature (ΔT):
We are aware that Q=UAΔT. This can be rearranged to obtain the area. This is shown below:
From the above assumptions and specification it can be seen that U is assumed to be 900 W/m2 °C.
Therefore;
The number of tubes required for the reboiler is calculated as follows:
Equation 28: Number of tubes
Where:
NT: Number of tubes
A: Heat transfer Area = 72.598 m2
L: Nominal length = 5m
do: Outside diameter = 30mm
Substitute the values in the above equation:
(
)
Square pitch of 1.25do
Therefore pitch (Pt) = 1.5 x 30 = 45 mm
The bundle diameter can then be calculated from equation23.
Where:
Db: Bundle Diameter
NT: The number of tubes = 155
65
do: Outside diameter = 30mm
K1 and n1: Constants = assumed a square pitch with 2 passes (The constants are shown in table 5)
Substitute the values into equation23:
(
)
The shell diameter is twice the bundle diameter. Therefore:
To calculate the freeboard assume the liquid level as 500mm from the base:
Freeboard = 1220.124 – 500 = 720.124 mm
720.124
720...
mm
.....12
4m
m
5001.5
mm
610.062
1.3 m
mm
Figure27: Shell and Bundle design
66
6.0 Mechanical Design
[64]
This section will focus on the mechanical design of the distillation column. The detailed mechanical design of
processing equipment will not be usually carried out by chemical engineers. It is usually mechanical
engineers who are responsible for this. However, in this section of the report certain aspects of the
mechanical design will be investigated including; shell thickness, skirt design and weight of plates.
In order to carry out the mechanical design certain assumptions and specifications have to be stated. They
are as follows:

The Diameter of the column (Dc) = 1.693 m

The material of construction for the shell is Stainless Steel (18Cr/8Ni, Ti stabilised)[65]

The column is operating at atmospheric pressure i.e. 1 atm = 1.01325 bar

It is assumed that the column is designed for 10% above the normal operating pressure.
[65]
So the
design pressure will be 1.1 × 1.01325 = 1.114575 bar. This is equal to 0.114575 N/mm2

The shell will be insulated with 75 mm of mineral wool
6.1 Shell Thickness
In order to calculate the shell thickness the following equation has to be used:
Equation 29: Minimum Shell Thickness [65]
Where:
67
e: cylindrical column minimum shell thickness
Pi: The design pressure = 0.1114575 N/mm2
Di: The column diameter: 1.693 m
f: The design stress = from table 8 at 100⁰C the design stress is 150 N/mm2
Substitute the values into equation 29:
According to Coulson and Richardson Volume 6 a corrosion allowance of 2mm should be used. Therefore,
the column thickness will be:
From the calculated wall thickness it can be seen that thickness is too small to withstand the pressure of the
column. It is important that the thickness of any vessel is strong enough to withstand its own weight and any
incidental loads. The table below gives an indication of wall thicknesses of any vessel. The thickness should
not be less than the values stated in the table.
Vessel Diameter (m)
1
1 to 2
2 to 2.5
2.5 to 3.0
Minimum Thickness (mm)
5
7
9
10
3.0 to 3.5
12
Table 6: Optimum wall thickness [65]
From table 6 it can be seen that with a vessel diameter of 1 to 2m the minimum thickness should be 7mm.
The diameter of this distillation column been calculated and found to be 1.693m so it is important that the
minimum thickness is 7mm.
6.2 Skirt Design [66]
Certain factors such as size, shape and weight of the vessel will determine the type of support a vessel uses.
Skirt supports are usually used for tall vertical columns. It is support that the supports are designed to
withstand the weight of both the column and the contents within it. Also, it must be able to withstand other
loads such as wind loads. When designing supports it has to be taken into that there should be ease of
access to the vessel for inspection and maintenance. [66] The skirt support will be made of a cylindrical shell
that is welded to the base of the distillation column. A flange at the bottom of the skirt will transmit the load
to the foundations. When designing the skirt support it is important that gaps are provided for access and
68
for connecting pipes. The thickness of the skirt must be sufficient enough to withstand bending moments
and the dead weight of the column. [66] Figure 28 below shows the two types of skirt designs.
Figure 28: Skirt –support designs [66]
The resultant stresses occurring in the skirt will be:

σs(compressive) = (σbs) + (σws)

σs(tensile) = (σbs) + (σws)
Where:
σbs: Bending stresses in the skirt
σws: The dead weight stresses in the skirt
Bending Stresses
The following equation is used to determine the bending stresses in the skirt support:
Equation 30: Bending stresses [66]
Where:
Ms: The maximum bending moment elevated at the bottom of the skirt, due to wind, seismic and eccentric
loads
Ds: The diameter of the column , including shell thickness and insulation (calculated below)
ts: is the shell thickness = 7 mm
The dynamic wind pressure is taken to be 1280 N/m2 [67]
69
The diameter of the column including the shell thickness and insulation has to be calculated. This is
calculated as follows: 1.693 + 2 x (7 + 75) x 10-3 = 1.857 m
The loading Fw = 1280 x 1.857 = 2376.96 N/m
From section 4.14.13 it can be seen that height of the column is 21m.
So,
Now the bending stresses in equation 30 can be calculated:
Dead Weight Stresses
The following equation is used to determine the dead weight stresses in the skirt support:
Equation 31: Dead weight stresses [68]
Where:
Ms: The maximum bending moment elevated at the bottom of the skirt, due to wind, seismic and eccentric
loads
Ds: The diameter of the column , including shell thickness and insulation (calculated below)
ts: is the shell thickness = 7 mm
W: Total weight of the vessel and contents (calculated below)
The total weight of column (W)= weight of the shell (w1) + weight of the plates (w2) + weight of the insulation
(w3)
Weight of the Shell (w1):
The following equation is used to find the weight of the shell:
Equation 32: Weight of Vessel [69]
70
Where:
Cv: Factor accounts for the weight of the nozzle, for distillation columns = 1.15
Hv: The height of the column = 21m
t: The shell thickness = 7mm
Dm: The diameter of the column including the shell thickness (calculated below)
The diameter of the column including the shell thickness:
(
)
Substitute the values into equation 32:
Weight of Plates (w2):
The plate area if calculated as follows:
Weight of one plate = 1.2 × 2.251 = 2.701 kN
In the distillation column there are 31 plates in total therefore the total weight of the 31 plates is as follows:
= 2.701 × 31 = 83.731 kN
Weight of Insulation (w3):
The volume of insulation has to be calculated. It is specified that the column will be insulated with 75 mm
mineral wool. Therefore, the insulation volume:
Weight of insulation: 8.377 x 130 x 9.81 = 10683.188 N
The value for the weight of insulation has to be doubled in order to allow for fittings:
= 2 × 10683.188 = 21366.376 N = 21.366 kN
Therefore, the total weight of the column can be determined:
w1 + w2 + w3 = 73.435 + 83.731 + 21.366 = 178.532 kN
The dead weight stresses can now be calculated by substitution into equation 31.
71
The resultant stresses occurring in the skirt will be:

σs(compressive) = (σbs) + (σws) = 2.754 + 4.756 = 7.510 N

σs(tensile) = (σbs) - (σws) = 2.754 – 4.756 = -2.00 N
ADD IN THE EXPLANATION
6.3 Heads and Closures
There are four typical types of heads of various shapes which are used for the end of a cylindrical vessel
including:

Flat plates and formed flat heads

Hemispherical heads

Ellipsoidal heads

Torispherical heads
Flat plates are usually used as covers for manways and are used as channel covers for heat exchangers.
Formed flat ends are sometimes known as ‘flange-only’ heads. This type of head is the cheapest type to
manufacture as fabrication costs are low, however, they are limited to low pressures and small diameter
vessels. Torispherical head are the most commonly used type for vessels with operating pressures up to 15
bar. They can be used with vessels with higher pressures however; the costs have to be compared with that
of an ellipsoidal head. For pressures over 15 bar ellipsoidal heads would be the most economical type. The
hemispherical head type has the strongest shape and is able to withstand double the pressure of a
torispherical head which is of the same thickness. For the product distillation column the flat head closure
would be suitable as the column is assumed to operate at atmospheric pressure. [65]
6.4 Material Selection
In the selection of engineering materials many factors have to be considered. The most important factor is
the ability of the material to resist corrosion. The designer has a role to recommend the best possible
material that will be suitable for the process conditions. Not only should the material satisfy the chemical
properties but the mechanical properties should also be considered. The material that gives the lowest cost
over the working plant life, allowing for maintenance and replacement, is also another factor in the selection
of materials. Product contamination and process safety are also other factors that must be considered. Also,
in the selection of a suitable material the suitability of the material for fabrication should be considered.
72
Stainless steel is one of the most commonly used material in the chemical industry, as It is highly corrosion
resistant. [70]
6.5 Pipe Sizing
The selection of pipe size depends on pressure differences which occur in the pipe. If the fluid has to be
pumped through the pipe the size of the pipe selected should give the least annual operating costs.
However, if there is enough head for gravity flow, the smallest pipe diameter is selected which gives the
required flowrate. Table 7 shows the typical pipe velocities along with the allowable pressure drops. This
information can be used to estimate pipe sizes. [71]
Velocity m/s
ΔP kPa/m
Liquids, pumped (not viscous)
1-3
0.5
Liquid, gravity flow
-
0.05
Gases and vapour
15 - 30
0.02 % of line pressure
High pressure steam, > 8 bar
30 - 60
-
Table 7: Typical pipe velocities and allowable pressure drops [71]
From section 4.16.4 it can be seen that stainless steel has been chosen as the material to build the pipes.
Equation 31 has been developed to calculate the optimum diameter for turbulent flow. The following
equation can be used to give an estimation of the economic pipe diameter for normal pipes.
Equation 33: optimum diameter for stainless steel pipes [71]
Where:
G: The flowrate
ρ: The liquid density
The optimum pipe diameter will be calculated for the pipes at the feed, the distillate and the bottom
product. Also, the wall thickness will be calculated for the feed pipe, the distillate pipe and the bottom
product pipe. In order to resist internal pressure the pipe wall thickness is also calculated which has an
allowance for corrosion. Process pipes are usually considered as thin cylinders. However, in cases such as
73
high-pressure streams they are considered as thick cylinders. The wall thickness is calculated from equation
34.
[71]
Equation 34: Formula for wall thickness [71]
Where:
P: Internal Pressure
d: Pipe outer diameter (will be calculated for each pipe and calculations are shown below)
σ: Design stress at working temperature (obtained from table 8)
Materials Tensile
Stainless strength
(N/mm2)
Steel
(18Cr/8Ni,
540
Ti
stabilised)
0 to
50
165
Design stress at temperature °C (N/mm2)
100 150 200 250 300 350 400 450
500
150
115
140
135
130
130
125
120
120
Table 8: Typical Design stresses [72]
6.5.1 Pipe Sizing at the Feed
The temperature of the feed inlet to the column is assumed to be 70°C (343 K). We are aware that the feed
contains both Methyl Ethyl Ketone and Secondary Butyl Alcohol. Therefore, the density of the mixture has to
be calculated at 343 K. This is shown as follows:
Using equation 4 and table 8:
Methyl Ethyl Ketone:
L  A* B
 1T
 
 Tc




n
 1343 


 535.5 
0.2857
  0.2676  0.2514
  0.2676  0.2514 0.8798
74
Convert to kg/m3
Secondary Butyl Alcohol:
L  A* B
 1T
 
 Tc




n
 1343 


 536.01 
0.2604
  0.2734  0.2635
  0.2734  0.2635 0.8896
Convert to kg/m3
ρ
ρ
Therefore, the density of the mixture is:
ρ
Now the optimum can be calculated using equation 33:
(
)
75
An optimum diameter of 50 mm should be used. However, the diameter has to be converted to inches.
According to a reference a nominal pipe size of 2 inches should be used. [73]
Using this information the wall thickness can be calculated using equation 34. However, firstly the design
stress has to be calculated using the information provided in table 8 interpolation as follows:
It is assumed that the feed to the column is not viscous it is a pumped liquid. Therefore, from table 7 it can
be seen that the velocity falls between the 1-3.0 m/s range. So the pressure will be:
The wall thickness can now be determined using equation 34:
6.5.2 Pipe Sizing at the Distillate
The temperature of the product stream from the column is assumed to be 79.982°C (352.982 K). The
distillate contains 99 per cent MEK so the density will only be calculated for MEK in this instance. This is
shown as follows:
Using equation 4 and table 8:
Methyl Ethyl Ketone:
L  A* B
 1T
 
 Tc




n
 1 352.982 


 535.5 
0.2857
  0.2676  0.2514
  0.2676  0.2514 0.8870
Convert to kg/m3
76
Now the optimum can be calculated using equation 33:
(
)
An optimum diameter of 50 mm should be used. However, the diameter has to be converted to inches.
According to a reference nominal pipe size of 2 inches should be used. [73]
Using this information the wall thickness can be calculated using equation 34. However, firstly the design
stress has to be calculated using the information provided in table 8 interpolation as follows:
It is assumed that the feed to the column is not viscous it is a pumped liquid. Therefore, from table 7 it can
be seen that the velocity falls between the 1-3.0 m/s range. So the pressure will be:
The wall thickness can now be determined using equation 34:
6.5.3 Pipe sizing at the bottoms
The temperature of the bottom product is assumed to be 98.8 °C (371.81 K). The bottom product contains
mainly SBA. So the density will only be calculated for SBA in this instance. This is shown as follows:
Using equation 4 and table 8:
Secondary Butyl Alcohol:
L  A* B
 1T
 
 Tc




n
77
 1 371.81 


 536.01 
0.2604
  0.2734  0.2635
  0.2734  0.2635 0.9086
Convert to kg/m3
Now the optimum can be calculated using equation 33:
(
)
An optimum diameter of 20 mm should be used. However, the diameter has to be converted to inches.
According to a reference a nominal pipe size of 3/4 inches should be used. [73]
Using this information the wall thickness can be calculated using equation 34. However, firstly the design
stress has to be calculated using the information provided in table 8 interpolation as follows:
It is assumed that the feed to the column is not viscous it is a pumped liquid. Therefore, from table 7 it can
be seen that the velocity falls between the 1-3.0 m/s range. So the pressure will be:
The wall thickness can now be determined using equation 34:
Using all the information a data sheet has been constructed for the column, the condenser and the reboiler.
This can be found in appendix##.
78
79
6.6 Concept Drawing
Distillate Pipe Size 50mm
Feed Pipe Size 50mm
E-1
Column Height 21 m
Shell thickness 7mm
Bottoms Pipe Size 20mm
Colum Diameter 1.693m
1.693 m
Cross-Sectional View of Column
Number of holes = 2599
The diameter of hole = 5mm
Active Area = 1.711m
Downcomer Area = 0.270 m
80
7.0 Control and Instrumentation
[74]
One of the main priorities of a chemical plant is the safe operation of the process as a potential disaster
could possibly lead to major negative consequences. Therefore, control and Instrumentation is an essential
requirement for all chemical plants. The design intention is unachievable if the chemical plant lacks control.
Control can only be met when the design intention is safe and any disturbances which may occur are within
bounds which can be predicted. Safe plant operation occurs when the process variable are within the safe
operating limits. Using control and recording instruments information can be obtained about the process.
Measuring and indicating instruments such as flow meters and pressure indicators are used to measure the
process variables. Process instruments are sometimes linked together using control valves to form control
loops which provide more efficient control and understanding of the process. The readings are taken to
ensure that pressure, temperature and other variables such as flow are within safe limits. Instruments are
defined as devices or control systems which are designed in a way that maintains a functional relationship
between a prescribed property of a substance and a physical variable. [74][75]
The designer has certain primary objectives such as:

Safe plant operation – To ensure that process variables are kept within known safe operating
limits. Also, any deviations which may occur are brought to the attention of operators by the use
of alarms. Finally, in dangerous situations automatic shut-down systems should be put in place.

Production rate – The desired product should be produced, i.e. to ensure the product meet the
desired specifications.

Cost – it should be aimed to operate at the lowest production cost.
The piping and instrumentation diagram can be seen in figure 42 and in a larger size in appendix##.
7.1.1 Quality Control
Quality control is an important aspect as the cost at which the product is sold depends on the quality.
Therefore it is extremely important that the desired specification is achieved in the distillation column. A loss
of investment and could occur if the specifications are not met accurately. Higher quality products are
valued to sell at higher prices. However, to achieve higher quality products higher operating costs are
incurred so a compromise is required.
For quality control a mixture of both automatic and manual control systems can be used to achieve the most
effective results. For example, on-stream composition analysers can be installed coming off the distillation
column. However, they are rarely used as they are expensive and they require a great deal of maintenance.
81
It is important that the analysers are placed in a protected location they cannot simply be mounted to the
distillation column.
Various disturbances such as environmental disturbances i.e. weather conditions can occur which can affect
the quality of the product exiting the distillation column. Environmental disturbances are an example of
uncontrolled disturbances. To overcome the effects of uncontrolled disturbances a controller can be used
and this will help to restore the quality of the desired product.
The compositions of the distillate and bottom product must be determined externally so the product purity
can be measured continuously. To overcome this, a sample could be sent off to be tested in the onsite
laboratory for analysis. Adjustments can be made once the results have been obtained from analysis.
7.1.2 Pressure Control
Pressure control is essential in systems where vapours/gases are present. [74] Therefore, it is vital to control
the pressure drop across the product distillation column. Deviations which occur can affect the temperature
profile and this therefore will affect the separation of MEK and SBA. A pressure drop controller can be used
to monitor the pressure across the distillation column. This can be seen from figure 29 below. The operating
pressure of the distillation column is 1atm. The pressure drop controller (PDC) will detect any pressure
differences which may occur in the column. This is linked to the steam control valve of the reboiler and this
causes the control valve to prevent the flowrate from entering the column from the reboiler. Once the
pressure in the column returns to its set point, the PDC will regulate the control valve to open again allowing
the flowrate to pass from the reboiler to the column.
Figure 29: Pressure Drop Indicator and Controller
82
Also from figure 30 it can be seen that pressure indicators are located next to each pump. This measures the
pressure in the delivery line. From the indicator operators will be able to deduce if the pump is operating
normally and adequately as the operator as the pressure in the delivery line can be analysed.
Figure 30: Pressure indicator and Transmitter for the pump delivery line
7.1.3 Temperature Control
Usually in distillation columns the feed temperature is not controlled. It is only controlled if a feed preheater
is utilised. Temperature is used as an indication for the composition of the streams. For monitoring the
performance of the column it is recommended that additional temperature indicating or recording points
should be included up the column. It is advised that temperature sensors are located at the position in the
distillation column where the rate of change of temperature with the change in composition of the key
component is a maximum.
From figure 31 below it can be seen that in order to maintain and achieve the required temperature at the
bottom of the column a temperature control device has been put in place. The steam temperature is
adjusted to control the temperature of the vapour that returns from the reboiler. The temperature indicator
will detect a temperature rise over the set point and this will in turn be controlled by the temperature
controller. The steam supply to the reboiler will be either reduced or shut off until the temperature has
returned to the set point. In the case where a drop in temperature is detected then a signal will be sent to
increase the steam supply. We are aware that temperature is used as an indication of composition.
Therefore, in order to obtain the specified purity it is extremely important to monitor this section of the
distillation column.
83
Figure 31: Temperature control for the reboiler
The other point at where a temperature control loop is used is to measure the temperature of the cooling
water in coming into the condenser. This is used to control the temperature of the product stream which
leaves the condenser. From figure ## it can be seen that a signal is transmitted along the line to vary the
flowrate of the cooling water into the condenser. The reflux ratio can also be varied to control the
temperature at the top of the column.
Figure 32: Temperature control for the condenser
7.1.4 Flow Control
From figure 33 it can be seen that flow control loop is used in the feed stream to the distillation column. FIC1 measures the flow rate of the stream entering the column. Deviations which may occur in the feed line are
detected and a signal is sent to the control value to adjust the flow. The flow rate is regulated either by
increasing or decreasing the flowrate depending on the deviation.
84
Figure 33: Feed Stream Flow Control Loop
7.1.5 Level Control
From the figure 34 below it can be seen that a level indicator is attached to the reflux drum as it is important
to maintain the ratio of the distillate to the reflux liquid. In the event of an increase in the level within the
reflux drum it is detected by the level indicator and a signal is transmitted to the control value (V-16). This
will result in the opening of the valve in order to reduce the flow in the drum until the level return to the set
point.
Figure 34: Level control in the reflux drum
7.1.6 Alarms and Safety Trips [74]
Operators are alerted of serious and potentially hazardous deviations in the process conditions by the use of
alarms. Alarms will help alert when any deviations occur in the system which are different from the normal
set point. Therefore, this will help to prevent any potential hazards from occurring. In cases when the
response of the operator may be delayed which potentially could lead to the development of a hazardous
situation, instruments can be fitted with trip systems which automatically prevent the hazard. For example,
shut down procedures on pumps and valves to automatically close. In the event of a deviation a visual or
audible alarm can be used to alert the operator to carry out the appropriate action.
85
There are three components of an automatic trip system they are as follows:

A sensor monitors the control variable and provides an output signal when the set value is exceeded.

A link transfers the signal to the actuator and this consists of pneumatic or electric relays.

The actuator carries out the required action. For example, to close or open a valve.
Interlock systems are utilised when a fixed sequence of operations has to be followed. Interlock systems are
included in these situations to prevent operators from departing from the required sequence. For example
these types of systems are usually applied in start-up and shut down procedures. It is possible to incorporate
them into the control system design or mechanical interlocks can also be used. [74]
Alarms have been included in the P&ID this can be seen in figures 35 and 36 below. It can be seen that the
distillation column has a level indicator attached to it with a level indicator alarm. The alarm has a high level
and low level alarm fixed to it. The main objective of the alarm is to alert operators and to ensure that the
level within the distillation column is not too high or too low.
If the level increases above the set point the alarm sends a warning to the control and this helps to alert
operators. A message is sent to the control room and then the correct regulatory procedure is carried out to
return the level back to the set point. The same procedure would occur if the level fell below the set point.
The procedure may be automatic or may have to be done manually by the operator.
In the case where the level deviates significantly i.e. high-high limit or low-low limit operators would be
alerted by the alarm. Also, at the same time safety trips would be used which in turn would cause the
automatic shut down of the process.
Figure 35: Level Alarm on Distillation Column
It is possible to incorporate a safety trip into a control loop. In some situations it may be more suitable to
separate the trip system from the control loop to increase the reliability of the control equipment. The
control loop and the trip system may be separated because a fault in the control loop can potentially affect
the trip system and may increase the risk even further.
86
From figure 36 it can be seen that safety trips have been fitted to the active pumps. In the event that a
deviation occurs from the set point in the column the pump will shut down and this prevents the flow from
entering into the distillation column. Once the deviation returns to the set point the pump will be started up
again so the flow can enter the column.
Figure 36: Safety Trips on Pumps
7.1.7 Valve Selection [76]
Valves used in chemical plants can be divided into two categories depending on their function:
1. Shut off valves – The main purpose of shut off valves is to close off the flow.
2. Control valves – They can be either automatic or manual and their objective is used to regulate the
flow.
Control Valves
Selection of control valves is an important factor. It is important that good flow control is achieved whilst the
pressure drop is kept low as possible. Control valves may fail open this is the position the valves take when
power supply failure occurs. Diaphragm valves are commonly used and this is the type used in this case. This
type of valve can be seen from figure 37 below.
Figure 37: Diaphragm valve
Flanged Valves
Flanged valves can be used for drainage. An example of the type of flanged valve used in this case is shown
in figure 38below. These are generally closed and are in operation usually during site or unit maintenance.
Figure 38: Flanged valve
Non- Return Valves
87
This type of valve is used to prevent the back-flow of the fluid in the process lie. It is important that nonreturn valves have been correctly installed to ensure they are working adequately, i.e. they should be fitted
in the correct orientation. In this case non-return valves have also been utilised an example of this type of
valve is shown in figure 39 below.
Figure 39: Non-return valve
Gated Valves
Gated valves are frequently used for shut –off purposes. It is important that a valve selected for this purpose
gives a positive seal in the closed position and minimum resistance to the flow when the valves are open.
Gate valves exist in a wide range of sizes and it is possible to operate them manually or atomically by the use
of a motor. When gate valves are fully open they have a low pressure drop. When operating gate valves
attention must be paid to ensure they are not operated partially open. This is because the valve seal can
become deformed, so as a result the valve will not seal properly. Below figure 40 shows a diagram of a gate
valve which has been frequently used in the piping and instrumentation diagram.
Figure 40: Gate type valve
7.1.8 Instruments and Controllers
There are various locations of instruments in a chemical plant, for example local, mounted on a central
control panel. Within the P&ID there are many different shapes on instruments which are used. A circle
symbol represents a device which is field mounted. A circle which has a line through the middle indicates a
device which can be easily accessible by the operators.
88
Figure 41: Diagram to show Instrument type and location [77]
7.1.9 Pipe Labelling
Within the P&ID it is important that the process lines are labelled. The line label will show the required
information which is associated with that particular stream. No fixed standard exists on line piping. In this
case only the three main process lines i.e. the feed line, the distillate line and the waste products line will be
labelled. This is because the diameters have only been calculated for the three streams mentioned. The
following information is required to label each of the streams:

The pipe size in inches

The pipe material of construction

The line identification number
For example, the feed line which 2 inches made out of stainless steel is labelled as follows:
2 – SS – 01
89
Figure 17: Piping and Instrumentation Diagram for Product Distillation Column
90
91
8.0 Hazard and Operability Study (HAZOP)
Hazard and operability studies are also commonly known as HAZOP. They have been used and developed for
identifying potential hazards and operability problems which are caused by deviations from the design intent
either due to plant design or human error. [78]
It is a detailed and systematic application of prescribed guidewords to a design drawing.
[79]
Conducting a
HAZOP is an important stage of any chemical plant for the prevention of major accidents from arising. Not
only does a HAZOP prevent major accidents but they also provide major financial benefits to the plant
owner.
[80]
A HAZOP is usually carried out for the whole process. However, this is very time consuming
therefore for the purpose of this report it will only be carried out for one process line.
The method of a HAZOP involves using guidewords such as ‘pressure’, ‘flow’, ‘temperature’ and ‘level’. For
each of the guidewords all deviations that could potentially occur are identified, for example ‘no flow’. Then
the possible causes of the deviation are identified and finally the consequences and actions required are
stated before moving on to the next deviation. Once all deviations have been covered move onto the next
guideword. [79]
The line chosen to conduct the HAZOP was the line connecting the feed to the solvent recovery distillation
column. The HAZOP can be seen below in section.......
92
9.0 Economic Appraisal
In this section of the report an economic appraisal will be carried over the Methyl Ethyl Ketone plant by
estimating and calculating the relevant economical factors of the process plant.
There are many factors which have to be considered to determine whether the proposal of a plant is
worthwhile. A very important factor is the cost required to build the plant and to determine whether the
investment is financially viable.
To determine the feasibility of the plant a thorough economical evaluation will be carried out in which all the
costs involved must be accounted for. Costs such as services and equipment required to operate the plant
must be considered. Other factors such as capital costs, operating costs and raw materials must also be
considered to determine the feasibility. There are many methods which have been developed to estimate
the cost to set up and run a chemical plant over a number of years. From the various methods the capital
costs of the plant can be calculated and these are analysed carefully before deciding whether investing in
the plant is viable or not.
Using various methods the capital costs of the chemical plant will be calculated. Also, a cash flow table will
be constructed and from this the cost going in and out of the project during the plant life will be analysed.
From the cash flow table the expected targets on an annual can also be predicted. The breakeven point i.e.
the point at which the initial investment is recovered will then determined.
9.1.1 Estimation of Capital Costs
As mentioned above there are a number of methods to determine the capital costs associated with the start
up of a chemical plant. The following methods will be used to estimate the capital costs:

Wilson Method

Zevnik-Buchanan Method

Bridgewater Method
The three methods mentioned above are known as step counting methods as the capital cost is determined
by the number of significant process steps in the overall process.
93
9.1.1.1 Wilson’s Method [81]
According to the Wilson Method the principle is that the average cost of one item of equipment is a function
of the process size and complexity, i.e. the capital cost is calculated based upon the number of significant
plant items in the overall process. Firstly the Average unit cost of main plant items has to be calculated. The
following equation is used to estimate this:
AUC = 21 X V0.675
Where:
V: Is the capacity of the plant, t/y of product = 100000 t/y
So the AUC can be calculated:
AUC = 21 x 1000000.675 = £49798.85 = £49799
This is then used to calculate the capital cost from the equation below:
Equation 35: Wilson’s Method Equation [81]
Where:
C= fixed capital cost
F= installation factor obtained graphically and is based on AUC
n: number of plant items (including duplicates however excluding pumps)
FM= factor for materials of construction.
FP= factor for design pressure, obtained graphically
FT= factor for design temperature, obtained graphically
From the AUC calculated above the installation factor (f) can be determined from figure 43below.
94
Figure 43: Installation Factor [81]
From Figure 43 it can be seen that the installation factor F is approximately 1.9.
Assuming that the material of construction is stainless steel the materials factor to be used is 1.3.
The number of significant process steps in the process is found to be 7.
From figure 44 below the pressure factor is found to be 1.0 as the operating pressure in the reactor is
assumed to be 2.5 bar. A pressure of 2.5 bar is the maximum operating pressure in the process.
Figure 44: Pressure Factor used for Wilson’s Method [81]
95
Next the temperature factor has to be determined. From figure 45 below the temperature factor is found to
be 1.1 as the operating temperature in the reactor is assumed to be 800K (527°C). A temperature of 800K
(527°C) is the maximum operating temperature in the process.
Figure 45: Temperature Factor used for Wilson’s Method [81]
Using all of the above information the cost can be calculated as follows:
9.1.1.2 Zevnik and Buchanan Method [81]
This method uses the number of process steps rather than the number of plant items. A process step is
defined as a unit operation, unit process or separation unit which has energy transfer. For plant capacities
above 4500 t/y the following equation should be utilised to calculate the capital costs:
Equation 36: Zevnik and Buchanan’s Method [81]
Where:
C: Fixed capital costs
Q: Plant capacity, t/y = 100000 t/y
N: The number of process steps = 7
In order to calculate the costs using the above firstly X has to be calculated. This is shown as follows:
[
]
Where:
Pmax: Maximum Process Pressure = 2.5 bar
96
Tmax: Maximum Process Temperature = 800 K
Fm: Materials of Construction Factor = 0.1
So, X is calculated below as follows:
[
]
Now the total costs can be estimated as follows:
9.1.1.3 Bridgewater‘s Method [81]
This method is also based on the step counting method. For plant capacities above 60,000 t/y the following
equation should be utilised to calculate the capital costs:
( )
Equation37 Bridgewater’s Method [81]
Where:
C: Fixed Capital Cost
N: The number of process steps = 7
Q: Plant Capacity, t/y = 100000 t/y
s: Reactor Conversion = 87.57% [8]
Therefore:
(
)
Table 9 below summarises the calculated capital costs:
Method
Cost (million £2000)
Wilson
50.54
Zevnik & Buchanan
88.76
Bridgewater
35.04
Table 9: Estimated Capital Costs
The Wilson method will be used as the initial capital costs. This method has been selected as it is the middle
value of the three calculated capital cost. Bridgewater’s method may be deemed as not reliable as it is too
small and Zevnik and Buchanan method may be too large. So in this case Wilson’s method has been selected.
Another justification as to why Wilson’s method may be selected is because it accounts for all the plant
97
items. So this provides a more clear indication on how much the initial cost will be taking account for all
plant items not just the ‘main process steps’.
The calculated capital cost is that for 2000 so it is important to find the capital cost for 2012. Also, the
proposed plant is to be located in China therefore location factors and exchange rates have to be taken into
account. To determine the capital cost for 2012 inflation rates of both China and Great Britain have to be
considered. This is done as follows:
Table 10 below shows the inflation rate for Great Britain from 2000 to 2012.
Year
Inflation
Inflation Factor
2000
+0.79%
1.0079
2001
+1.24%
1.0124
2002
+1.26%
1.0126
2003
+1.36%
1.0136
2004
+1.34%
1.0134
2005
+2.05%
1.0205
2006
+2.33%
1.0233
2007
+2.32%
1.0232
2008
+3.61%
1.0361
2009
+2.18%
1.0218
2010
+3.29%
1.0329
2011
+4.47%
1.0447
2012
+3.59%
1.0359
Table 10: Inflation rate for Great Britain from 2000 to 2012 [82]
Location factors are only available for the year 2004 so the plant has to be moved in 2004. However, the
initial capital cost calculated above is for the year 2000 so this is where the UK inflation is taken into
consideration. So the cost in the UK in the year 2004 will be:
C2004 = C2000 × 1.0079 ×1.0124 × 1.0126 × 1.0136 × 1.0134
C2004 = 50.54x106× 1.0079 ×1.0124 × 1.0126 × 1.0136 × 1.0134
C2004 = £53.64 million
98
As mentioned above the location factors are only available for 2004. So using the location factors below the
plant has to be moved from UK to China. So the capital cost in 2004 taking account of location factors is as
follows:
Figure 46: Location Factors for 2004 [81]
The location factor for China in 2004 is 0.9. So the capital cost of the proposed plant in China in 2004 is:
£53.64 million × 0.9 = £48.276 million
It is also important to take into account the exchange rate as the plant is to be built in China. The exchange
rate of Great British Pounds to Chinese Yuan in 2004 was ¥14.8173. [83] Using the exchange rate the capital
cost in 2004 would be as follows:
£48.276 million × ¥14.8173 = ¥715.32 million
The next step is to determine the capital cost of the plant for 2012. China inflation rates from 2004 to 2012
have to be taken into account. This is shown below:
Table 11 below shows the inflation rate for China from 2004 to 2012.
Year
Inflation
Inflation Factor
2002
-0.73%
0.9927
2003
1.13%
1.0113
2004
+3.84%
1.0384
2005
+1.78%
1.0178
2006
+1.65%
1.0165
2007
+4.82%
1.0482
99
2008
+5.97%
1.0597
2009
-0.72%
0.9928
2010
+3.17%
1.0317
2011
+5.53%
1.0553
2012
+4.58%
1.0458
Table 11: Inflation rate for China from 2004 to 2012 [84]
So, therefore the capital cost in 2012 for China is calculated as follows:
C2012 = C2004 × 1.0384 × 1.0178 × 1.0165 ×1.0482 × 1.0597 × 0.9928 × 1.0317× 1.0553× 1.0458
C2012 = ¥715.32 million × 1.0384 × 1.0178 × 1.0165 ×1.0482 × 1.0597 × 0.9928 × 1.0317× 1.0553× 1.0458
C2012 = ¥964.94 million
9.1.2 Cost of Raw Materials
The next step to be determined is the cost of the raw materials which are required for the process. There are
only two raw materials which are required for this process. In order to determine the total cost of the raw
materials the amount required in tonnes per year will have to be calculated. It is important to remember
that in this process 1 year of production is equal to 8064 hours of operation.
Also, it has to be taken into account that the prices of the raw materials obtained are in dollar per tonne.
However, these have to be converted into the Chinese currency i.e. Yuan per tonne. For this economic
appraisal an exchange rate of $1: ¥6.30298 [85] be used. The table below shows the cost of the raw materials:
Raw Material
Cost ($/t)
Cost (¥/t)
Flowrate
Flowrate
Flowrate
Total Cost
(kg/hr)
(kg/yr)
(tons/yr)
(¥)
SBA
1080
6807.22
11576.71
93354589.44
93354.58944
635,485,228.3
TCE
1180
7437.52
2232.14
17999976.96
17999.98
133,875,211.2
Total
769,360,436.5
Table 12: Cost of Raw Materials [86][87]
From table 12 above it can be seen that in order to produce 100000 t/y of Methyl Ethyl Ketone the total cost
of raw materials will be ¥769.36 million. However, from the mass balance produced in the previous report [8]
it can be seen that the amount of TCE entering stream 14 is completely recycled and is used again. So, the
cost of TCE can be ignored and the cost can be included in the start up cost. Therefore, TCE is not considered
100
as an operating cost. Also, It can be seen from the group project mass balance there is a recycle of 1292.08
kg/hr of SBA [8]. This amount can therefore be deducted from the amount of SBA required. This will help to
decrease the cost of raw materials.
This therefore means that total cost of raw materials is ¥635,485,228.3.
9.1.3 Operating Costs
Operating costs compromise of fixed and variable costs. They are the costs which occur during construction
and production. Fixed operating costs do not vary with the level of production i.e. they don’t vary in
accordance to that produced. Examples include insurance and taxes. On the other hand variable operating
costs do vary with the level of production in a plant for example raw materials which may be required.
An estimate of the operating of the costs associated with the Methyl Ethyl Ketone can be found from the
cost of raw materials. From table 13 it can be seen that raw materials cost represent 31% of the total
operating costs. Therefore, the total operating costs are shown below:
% of Total
Operating Cost
Cost (¥/Y)
% of Total
Operating Cost
Cost (¥/Y)
Raw Materials
31
635,485,228.3
Packaging
1
20,499,523.49
Labour
11
225,494,758.4
Shipping
1
20,499,523.49
Supervision
2
40,999,046.98
Depreciation
4
81,998,093.96
Maintenance
2
40,999,046.98
Property Taxes
1
20,499,523.49
Plant Supplies
1
20,499,523.49
Insurance
1
20,499,523.49
Royalties and
Patents
2
40,999,046.98
Administration
4
81,998,093.96
Utilities
8
163,996,187.9
Sales
11
225,494,758.4
Payroll Overhead
2
40,999,046.98
Research
5
102,497,617.5
Laboratory
2
40,999,046.98
Finance
2
40,999,046.98
Plant Overhead
9
184,495,711.4
Operating Cost
Operating Cost
Table 13: Typical Operating Costs [81]
From the above the total operating cost therefore equates to: ¥2,049,952,349
101
9.1.4 Plant Income
The plant makes an income by selling the product which is Methyl Ethyl Ketone. Currently MEK is being sold
for $1800 /t. Using the conversion above i.e. $1: ¥6.30298, the selling price of MEK is therefore calculated to
be ¥11345.364. Therefore, this gives an annual amount of:
100000 × ¥11345.364 = ¥1,134,536,400
9.1.5 Cash Flow
The cash flow table shows all the expenditure which includes capital costs, working capital and the operating
costs. All expenditure is given a negative sign as this shows money being spent. On the other hand all income
i.e. sales of the product are positive.
The plant has been assumed to have a plant of 15 years. Also, it is assumed that construction is split equally
over two years with 50% construction in year 0 and 50% construction in year 1. This therefore, means that
production will begin in year 2.
The working capital is paid in the year before operation and is recovered in the final year of operation which
is an income to the plant. The working capital is usually required in the year before production to purchase
the initial raw materials to get the plant running. In this cash flow the working capital will be 10% of the
capital cost. The final income to account for is the scrap value. The scrap value will again be 10% of the
capital cost.
All of the data which has been previously found is bought together in order to construct a cash flow table.
The cash flow table can then be assessed to determine the profitability of the process.
The following diagram can be produced using the data from the cash flow in Table 14.
102
Cash Flow Diagram
0
Cumulatine Net Cash Flow
-2,000,000,000
0
5
10
15
20
-4,000,000,000
-6,000,000,000
-8,000,000,000
-10,000,000,000
-12,000,000,000
-14,000,000,000
-16,000,000,000
Year
Figure 47: Cash Flow Diagram
9.1.6 Payback Time
From figure 47 and table 14 the payback time can be determined. The payback time is the time required to
recover the initial costs from the income which is produced by selling the product. The payback time is also
be said to be a measure of how long the initial investment is at risk of not making a profit. However, in this
instance from figure 47 it can be seen that the payback time cannot be determined. There is not a point
where the graph crosses the axis. From the graph it can be seen that not there is not a single year where the
plant is making a profit. The payback time cannot be achieved in this case. Therefore, it can be said that the
plant is not profitable as the initial investment is not recovered over the lifetime of the plant.
103
Year
Capital Cost
0
-482471590.3
1
-482471590.3
Working Capital
Scrap Value
Operating Cost
Income
-96494318.05
Net Cash Flow
Cumulative NCF
-482471590.3
-482471590.3
-578965908.3
-1061437499
2
-2049952349
1134536400
-915415949
-1976853448
3
-2049952349
1134536400
-915415949
-2892269397
4
-2049952349
1134536400
-915415949
-3807685346
5
-2049952349
1134536400
-915415949
-4723101295
6
-2049952349
1134536400
-915415949
-5638517244
7
-2049952349
1134536400
-915415949
-6553933193
8
-2049952349
1134536400
-915415949
-7469349142
9
-2049952349
1134536400
-915415949
-8384765091
10
-2049952349
1134536400
-915415949
-9300181040
11
-2049952349
1134536400
-915415949
-10215596989
12
-2049952349
1134536400
-915415949
-11131012938
13
-2049952349
1134536400
-915415949
-12046428887
14
-2049952349
1134536400
-915415949
-12961844836
15
-2049952349
1134536400
-915415949
-13877260785
-2049952349
1134536400
-818921631
-14696182416
96494318.05
-14599688097
16
17
96494318.05
96494318.05
Table 14: Cash Flow Table for the Methyl Ethyl Ketone Plant
104
9.2 Breakeven and Profitability Analysis
Al the relevant costs and income have now been calculated for 1 year’s production of MEK. Now the
profitability of the proposed plant has to be determined. There are various methods which can be used to do
this they are as follows:

Amortisation

Discounted Cash Flow Rate of Return

Return on Investment
9.2.1 Amortisation
The initial capital which is necessary to purchase all required equipment is usually borrowed from the bank.
As the capital is repaid the outstanding loan decreases and the interest on the initial loan is reduced. The
borrowed capital is often repaid on an amortisation basis. This means that over the life time of the plant a
constant annual repayment is made. The following equation is used to calculate the amortisation rate:
Equation 38: Calculation of Amortisation Rate [81]
Where:
A: The annual repayment which includes capital and interest. It is constant over the life of the loan.
I: The initial capital cost = ¥964,943,180.5
r: The interest rate = 10%
n: The number of years the loan is repaid over. It is usually the lifetime of the plant = 10 years
So substitution of all the values into Equation 38 gives:
The following table show the effect of amortisation. Table 15 shows the effect if a loan was taken out based
on the full capital cost.
105
Year
Working Capital (¥)
Scrap Value (¥)
Amortisation (¥)
Operating cost (¥)
Income (¥)
Net Cash Flow (¥)
Cumulative NCF (¥)
0
1
-96494318.05
-126864724.4
-223359042.5
-223359042.5
2
-126864724.4
-2049952349
1134536400
-1042280673
-1265639716
3
-126864724.4
-2049952349
1134536400
-1042280673
-2307920389
4
-126864724.4
-2049952349
1134536400
-1042280673
-3350201063
5
-126864724.4
-2049952349
1134536400
-1042280673
-4392481736
6
-126864724.4
-2049952349
1134536400
-1042280673
-5434762409
7
-126864724.4
-2049952349
1134536400
-1042280673
-6477043083
8
-126864724.4
-2049952349
1134536400
-1042280673
-7519323756
9
-126864724.4
-2049952349
1134536400
-1042280673
-8561604430
10
-126864724.4
-2049952349
1134536400
-1042280673
-9603885103
11
-126864724.4
-2049952349
1134536400
-1042280673
-10646165776
12
-126864724.4
-2049952349
1134536400
-1042280673
-11688446450
13
-126864724.4
-2049952349
1134536400
-1042280673
-12730727123
14
-126864724.4
-2049952349
1134536400
-1042280673
-13773007797
15
-126864724.4
-2049952349
1134536400
-1042280673
-14815288470
-126864724.4
-2049952349
1134536400
-945786355.4
-15761074825
96494318.05
96494318.05
Table 15: Cash Flow Table for the Methyl Ethyl Ketone Plant taking account for amortisation
-15664580507
16
17
96494318.05
106
9.2.3 Discounted Cash Flow Rate of Return
The discounted cash flow rate of return also known as DCFrr is another method which is used to assess
profitability. This method uses the net present value to determine the profitability. Money which is invested
into a project will earn interest. This means that any sum of money which is spent or any money which arises
in the future has an equivalent value today. [81] This depends on the interest rate assumed and the number of
years. The following equation is used to calculate the present value:
Equation 39: Present Value Calculation [81]
Where:
P: The present value
F: The future value
I: The assumed interest rate
n: The number of years
The net present value is the cumulative present value over the life time of the plant. The net present value
will be calculated for two interest rates; 10% and 20%. The number of years of the MEK is 15 years. The net
present value can then be used to determine the DCFrr. This is the discount rate which is required to make
the net present value equal zero.
DCFrr
Cumulative Net Present Value
0
-2,000,000,000
0
5
10
15
20
25
-4,000,000,000
-6,000,000,000
-8,000,000,000
-10,000,000,000
-12,000,000,000
-14,000,000,000
-16,000,000,000
Discount Rate
Figure 48: DCFrr for the Methyl Ethyl Ketone Plant
From figure 48 it can be seen that a DCFrr cannot be determined as the graph does not cross the axis.
107
Net Cash Flow (¥)
Cumulative NCF(¥)
NPV at 10% (¥)
CUM NPV at 10% (¥)
NPV at 20% (¥)
CUM NPV at 20% (¥)
-482471590.3
-482471590.3
-482471590.3
-482471590.3
-482471590.3
-482471590.3
-578965908.3
-1061437499
-526332643.9
-1008804234
-482471590.3
-964943180.5
-915415949
-1976853448
-756542106.6
-1765346341
-635705520.1
-1600648701
-915415949
-2892269397
-687765551.5
-2453111892
-529754600.1
-2130403301
-915415949
-3807685346
-625241410.4
-3078353303
-441462166.8
-2571865468
-915415949
-4723101295
-568401282.2
-3646754585
-367885139
-2939750606
-915415949
-5638517244
-516728438.4
-4163483023
-306570949.1
-3246321556
-915415949
-6553933193
-469753125.8
-4633236149
-255475791
-3501797347
-915415949
-7469349142
-427048296.2
-5060284445
-212896492.5
-3714693839
-915415949
-8384765091
-388225723.8
-5448510169
-177413743.7
-3892107583
-915415949
-9300181040
-352932476.2
-5801442645
-147844786.4
-4039952369
-915415949
-10215596989
-320847705.6
-6122290351
-123203988.7
-4163156358
-915415949
-11131012938
-291679732.4
-6413970083
-102669990.6
-4265826348
-915415949
-12046428887
-265163393.1
-6679133476
-85558325.48
-4351384674
-915415949
-12961844836
-241057630.1
-6920191106
-71298604.57
-4422683278
-915415949
-13877260785
-219143300.1
-7139334406
-59415503.81
-4482098782
-818921631
-14696182416
-178221206.8
-7317555613
-44293745.49
-4526392528
96494318.05
-14599688097
19090886.41
-7298464727
Table 16: Net Present Value at 10% and 20%
4349311.953
-4522043216
108
9.2.4 Return on Investment
Return on Investment (ROI) is another method which is used to analysis profitability. ROI is the percentage
return of the initial investment over the plant lifetime. This method takes into account the capital cost, the
cumulative net cash flow and the plant operating life. The following equation is used in order to calculate the
ROI.
⁄
Equation 40: Return on Investment [81]
The return on investment is calculated for the return on the initial capital cost is shown below:
⁄
It can be seen from the calculation that the ROI is negative. It again shows that the plant is suffering a major
loss. It would be classed as not being ‘worthwhile’. It shows that the plant is not profitable at all. From the
ROI achieved it can be said it would be better to invest in a bank rather than to build the plant. It would be
less of a risk to put the money in a bank rather than to build the plant.
9.2.5 Analysis of Profitability and Improvements to Profitability
From the cash flow tables above it can be seen that a company is making a very large loss. It can be observed
that the capital cost will not be paid over the operation of the plant. From table 15 it can be seen that if the
money required for the capital cost was loaned from the bank the company would still suffer a large loss. A
larger debt would occur if the money required for the capital cost was loaned. This is because the company
is not generating a profit in the first place. So, if a loan was taken out this would just add more to the debt as
interest would also have to be paid. It may be advisable to invest money from elsewhere in the company
rather than borrowing the initial capital cost.
It can be seen that the plant is not generating any profit as the operating costs are vastly greater than the
income into the plant. It can therefore be assumed that the project is not feasible any may not worth be
constructing.
Profitability to the plant may be increased if the cost of raw materials was decreased. The cost of raw
materials could be reduced by locating the plant closer to raw materials. This would mean that costs of
transporting the raw materials would also be decreased.
109
Also, the plant could negotiate a contract with the supplier in which a discount could be obtained if the plant
was to buy the raw materials in bulk. Therefore, this would decrease the price of raw materials which would
in turn reduce the operating cost.
Another suggestion to improve the profitability of the plant is to increase the selling price of the product
Methyl Ethyl Ketone. If the selling price increases this would in turn increase the revenue into the project
and would slowly decrease the debt. From the first term project it was seen that there is a large market for
MEK in China so it may be viable to increase the selling of MEK. However, if the selling price is set to high
then there is a threat of competition from rival companies. Also, an increase in income will also cause an
increase in the operating cost therefore; the plant again would still suffer a major loss.
10.0 Economic Optimisation
Optimisation is an important strategy used for decision making. As mentioned in the introduction it is known
that distillation is a very energy intensive process. It is possible that distillation columns can be fitted in such
a way that provides greater attractive economic benefits.
The cost of the major parts of equipment in a distillation column are the vessel, the condenser and reboiler.
In section 10.1 the costs of the three parts of equipment are estimated.
10.1 Cost of Major Equipment
10.1.1Cost of Distillation Column
It is important to calculate the cost of the product distillation. The price of a distillation column for the year
2002 was found to approximately $96,000. [88] However, the selling price has to be determined for 2012. To
calculate the price of the column in 2012 inflation has to be accounted for. Also, the price has to be
converted to Chinese Yuan. The conversion rate in 2002 from dollars to Yuan was ¥8.2766.[85] Therefore, the
cost in 2002:
$96,000 × ¥8.2766 = ¥794,553
Using the inflation rates determined in section 9 the 2012 cost can be calculated. This is show as follows:
= ¥794,553 × 0.9927 × 1.0113 × 1.0384 × 1.0178 × 1.0165 ×1.0482 × 1.0597 × 0.9928 × 1.0317× 1.0553×
1.0458
=¥1,076,024.7
110
10.1.2 Cost of Condenser
A shell and Tube type condenser is used for the distillation column. Again the price of the condenser
obtained is from 2002 and will therefore have to be inflated and converted into Chinese Yuan. The price of
the exchanger is found to be approximately $16,600.
$16,600 × ¥8.2766 = ¥137,391 [88]
Using the inflation rates determined in section 9 the 2012 cost can be calculated. This is show as follows:
= ¥13,7391 × 0.9927 × 1.0113 × 1.0384 × 1.0178 × 1.0165 ×1.0482 × 1.0597 × 0.9928 × 1.0317× 1.0553×
1.0458
=¥186,062
10.1.3 Cost of Reboiler
10.2 Potential Changes to existing plant
10.2.1 Number of Trays
An optimisation technique that could be considered is setting the total number of trays i.e. actual number of
trays to twice the minimum number of trays and adding two stages for the reflux drum and the reboiler. The
Fenske equation is used to calculate the minimum number of trays. An increase in the minimum number of
trays will in turn lead to a higher number of theoretical stages. However, an increase in the number of trays
leads to an increase in the height of the column. As the diameter decreases the reboiler heat input also
decreases. This therefore means that heat exchanger costs and energy costs also decreases. As a result of a
taller column the cost of the vessel increases. The number of trays in the column can be increased by
selecting smaller tray-spacing or installing more efficient contacting devices.[89]
10.2.2 Re-define Product Purity
In most occasions it is acceptable that many chemical plants produce extremely pure products. However, it is
not worthwhile to purify a product to 99% when 95% may be an acceptable purity. In this type of situation it
would be advised to decrease the reflux rate slowly until the desired product is obtained.
10.2.3 Improved Heat Integration
In order to reduce the duty on both the condenser and reboiler it may be possible to use the heat of the
MEK distillate can be recovered to preheat the feed stream into the product distillation column. The MEK
distillate stream leaves the condenser at 60°C and this then enters a product cooler. This heat can be
recovered and can then be used to preheat the inlet stream to the distillation column.
111
10.2.4 Control System Upgrade
Under a poor control system a perfectly designed column will be unable to utilise energy inputs adequately.
Operators will be responsible to make more decisions in situations where control systems are less advanced.
By upgrading the control system optimal operating points can be achieved. Therefore, it is important to use
up to date control systems.
11.0 Conclusion
This project has been a indication of the level of detail that is involved in the design of a distillation column.
The project has been a major learning experience and has helped to further develop skills including time
management and analytical skills. The report has expanded my perspective and conception of chemical
engineering.
During the design phase it was found that many educated assumptions had to be made using the data
available and in cases were data was not available chemical engineering knowledge had to be used.
112
12.0 Nomenclature
Cp
Specific heat at constant pressure
J/kgK
Dc
Db
Diameter
Bundle diameter
m
m
di
do
Internal diameter
Outer diameter
m
m
Ds
Shell diameter
m
e
Column shell thickness
F
FLV
Feed mass flow rate
Liquid-vapour flow factor
kg/s
_
G
g
H
jf
Vapour mass flow rate
Acceleration due to gravity
Height
Friction factor
kg/s
m/s2
m
_
jh
Heat transfer factor
_
k
KL
Thermal conductivity
Thermal conductivity of liquid
W/mK
W/mK
l
Lo
L’
N
Np
Length of tube
Liquid mass flow in rectifying region
Liquid mass flow in stripping region
Number of theoretical stages
Number of passes in a tube
m
kg/s
kg/s
_
_
NT
Number of tubes
_
p
p
P max
Pressure
Pitch
Maximum process pressure
N/m2
_
N/m2
P°
q
Q
R
Rmin
Saturated vapour pressure
Feed condition
Heat transferred per unit time
Reflux ratio
Minimum reflux ratio
N/m2
_
W
_
_
S
t
T max
TIN, TOUT
Allowable design stress
Thickness of vessel
Maximum process temperature
Shell side temperature of process fluid IN,OUT
N/m2
m
K
K
tIN, tOUT
Tube side temperature of service fluid IN,OUT
K
u
uf
Velocity
Flooding velocity
m/s
m/s
Uo
Overall heat transfer coefficient
W/m2K
113
us
Shell side velocity
m/s
uv
Vapour velocity
m/s
V
V*
Vapour mass flow in the rectifying region
Vapour mass flow rate per unit area
kg/s
kg/m2s
V’
W
xf
Vapour mass flow in the stripping region
Bottom mass flow rate
Mole fraction of component in feed
kg/s
kg/s
_
xHK
Mole fraction of heavy key
_
xLK
Mole fraction of light key
_
αa, αb
Relative volatilities of component a,b..
_
μL
Liquid viscosity
kg/ms
ρL
Liquid density
kg/m3
ρV
Vapour density
kg/m3
∆Ps
Pressure drop in shell
N/m2
∆Pt
Pressure drop in tube
N/m2
∆T
∆TLM
Temperature difference
Log mean temperature difference
K
K
Nu
Pr
Nusselt number
Prandtl number
Re
Reynolds number
114
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[6]
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[8]
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[9]
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