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Plant design Report on
Production of 140 MTPD Conc. Nitric
Acid (98%) by Recycling and
Rectification
Session
2003(F) – 2007(F)
By
Atiq Ur Rehman
Muhammad Nauman Yousf
Ahmad Saleem
Ali Ayub
Sajjid Hussain
Muhammad Raza Afzal
DEPARTMENT OF CHEMICAL ENGINEERING
NFC INSTITUTE OF ENGINEERING & FERTILIZER RESEARCH
FAISALABAD
Plant Design Report on
Production of 140 MTPD Conc. Nitric Acid
(98%) by Recycling and Rectification
Session
2003(F) – 2007(F)
Project Advisors
Engr. Naveed Asif (Assistant Professor)
Submitted By
Name
UET Registration Number
Sajjid Hussain
2003(F)-UET-IEFR-CHEM-FD-03
Atiq Ur Rehman
2003(F)-UET-IEFR-CHEM-FD-11
Ali Ayub
2003(F)-UET-IEFR-CHEM-FD-18
Ahmad Saleem
2003(F)-UET-IEFR-CHEM-FD-24
Muhammad Raza Afzal
2003(F)-UET-IEFR-CHEM-FD-30
Muhammad Nauman Yousaf
2003(F)-UET-IEFR-CHEM-FD-31
DEPARTMENT OF CHEMICAL ENGINEERING
NFC Institute of Engineering & Fertilizer Research
Faisalabad
Pakistan
Plant Design Report on
Production of 140 MTPD Conc. Nitric Acid (98%)
by Recycling and Rectification
Session 2003(F) – 2007(F)
This project is being submitted to Chemical Engineering
Department, NFC Institute of Engineering & Fertilizer
Research, Faisalabad in the partial fulfillment of
BECHELOR’S DEGREE
IN
CHEMICAL ENGINEERING
Project Advisor:
Engr. Naveed Asif (Assistant Professor)
Approved on:
External Examiner:
Internal Examiner:
---------------------------
----------------------------
Engr. Naveed Asif
(Assistant Professor)
---------------------------Head
Chemical Engineering Department
DEPARTMENT OF CHEMICAL ENGINERING
NFC Institute of Engineering & Fertilizer Research
Faisalabad
All praises to Almighty Allah,
Whose uniqueness oneness &
wholeness is unchangeable. All
respects are for His Holy Prophet,
Muhammad (peace be upon him)
who enabled us to recognize our
Creator.
Our parents who not only supported
us financially throughout our
education but gave us the strength of
character and should always remain
as beacon of light and life for us.
Acknowledgement
All praise to ALMIGHTY ALLAH, who provided us with the
strength to accomplish this main project. All respects are for His
HOLY PROPHET (PBUH), whose teachings are true source of
knowledge & guidance for whole mankind.
Before anybody else we thank our Parents who have always been a
source of moral support, driving force behind whatever we do. We
are indebted to our project advisor Engr. Naveed Asif (Assistant
Professor)for his worthy discussions, encouragement, technical
discussions, inspiring guidance, remarkable suggestions, keen
interest, constructive criticism & friendly discussions which enabled
us to complete this report. He spared a lot of precious time in
advising & helping us in writing this report.
We are sincerely grateful to Engr. Shehla Asif (Assistant Professor)
for their profound gratitude and superb guidance in connection with
the project.
We are also thankful to R & D Department of NFC (IEFR) for their
cooperation in search of design material.
Authors
Table of Contents:
Chapter # 1
Introduction
History
1
Nitric Acid Production in Pakistan
1
Commercial Uses of Nitric Acid
2
Minimum feasible capacity
3
Chemical properties
3
Plant Location & Layout
5
Chapter # 2
Process Description
General scheme for manufacture of concentrated HNO3
11
Commercial manufacturing processes for concentrated HNO3
11
Nitric Acid concentration processes
12
Direct Strong Nitric processes
12
Process Selection
13
Detailed Process Description
13
Chapter # 3
Material Balance
Basis
15
Material Balance around individual equipment
17
Reactor
17
Tail-Gas preheater & Waste Heat Boiler
20
Cooler/Condenser
21
Oxidation Tower
23
Compressor
25
Distillation Column
27
Super Azeotropic Column
28
Absorption Column
30
Bleaching Column
32
Overall Material Balance
33
Chapter # 4
Energy Balance
Energy Balance
36
Chapter # 5
Equipment Design
Equipment design by Atiq-Ur-Rehman: Reactor
49
Equipment Design By Sajid Hussain: Oxidation tower
65
Equipment Design By Ali Ayub: Absorber
74
Equipment Design By Muhammad Nauman Yousaf: Stripper
94
Equipment Design By Ahmad Saleem: Distillation column)
109
Equipment Design By Muhammad Raza Afzal: Partials Condenser
124
Chapter # 6
Mechanical Design
Absorber
134
Oxidation Tower
136
Distillation Column
138
Stripper
140
Chapter # 7
Instrumentation & Process Control
Introduction
142
Process Measurement Instrumentation
142
Types of Instrumentation
144
Process Control Systems
145
Feed Back Control Loop
146
Temperature Measurement & Control
148
Pressure Measurement & Control
148
Flow Measurement & Control
148
Control Schemes of Distillation Column
149
Chapter # 8
Safety
Introduction (Abstract)
152
General goals of safety
152
Major causes of hazards
152
Protection against Hazards
153
Protection Against Mechanical Hazards
153
Personal health Hazards & Protection
153
Chapter # 9
Material Of Construction
Introduction
157
Corrosion Principle
157
High Temperature Ammonia Oxidation
158
NO & NO2 at Intermediate Temperatures
159
Aqueous Nitric Acid Corrosion
160
Chapter # 10
HAZOP Study
Introduction
162
Steps conducted in HAZOP study
163
Chapter # 11
Cost Estimation And Economics Of Plant Location
Plant Cost Estimation
170
Capital Investment
170
Methods of Capital Investment
175
Cost Estimation of our Plant
175
APPENDECES
183207
Chapter: 01
Introduction
Chapter: 01
Introduction
HISTORY
The first reports of nitric acid have been credited to Arab alchemists of the eighth
century. By the middle Ages it was referred to as aqua fortis (strong water) or aqua
valens (powerful water). From that time onward, nitric acid was produced primarily from
saltpeter (potassium nitrate) and sulphuric acid. In the nineteenth century, Chilean
saltpeter (sodium nitrate) from South America largely replaced potassium nitrate.
However, at the beginning of twentieth century newer manufacturing technologies were
introduced. In Norway, where electricity was inexpensive, electric arc furnaces were used
to make nitrogen oxides, and subsequently nitric acid, directly from air. The commercial
life of these furnaces was relatively brief and most were shut down by 1930. At about the
same time, a different production method was being developed. In 1908, at Bochum,
Germany, Ostwald piloted a 3-t per day nitric acid process based on the catalytic
oxidation of ammonia with air. In1913 the synthesis of ammonia from coal,air and water
was successfully demonstrated using the Haber-Bosch process. With a secure and
economical supply of ammonia, ammonia oxidation was firmly established as an
industrial route to nitric acid manufacture.
NITRIC ACID RODUCTION IN PAKISTAN
In Pakistan, there are two main nitric acid producing facilities. These are as under;
1- Pakarab Fertilizers (Pvt.) Ltd. Multan
2- Wah Ordinance Factory
PAKARAB FERTILIZER (PVT.) LTD. MULTAN
The nitric acid plant in this fertilizer complex has an installed capacity of 1200 tons/day.
It is designed in two parallel lines, the capacity of each is 600 metric ton per day,
1
Chapter: 01
Introduction
calculated as 100% and the strength of the product acid is 60% HNO3 by weight. This
plant was designed by a German firm Gmbh UDHE. It is in production since 1977.
WAH ORDINANCE FACTORY
Here nitric acid is used to produce different explosives. These explosives are used further
to make different arms. This unit is currently fulfilling Pakistan Army’s demand for
explosives.
COMMERCIAL USES OF NITRIC ACID
Nitric acid is used in the production of many chemicals (eg, pharmaceuticals, dyes,
synthetic fibers, insecticides, and fungicides) but is used mostly in the production of
ammonium nitrate for the fertilizer industry. Most growth in demand has come from the
production of polyurethanes, fibers, and ammonium nitrate-based explosives. Other uses
of nitric acid are in the manufacture of explosives (trinitrotoluene, nitroglycerin, etc),
metal nitrates, nitrocellulose, and nitrochlorobenzene, the treatment of metals (eg, the
pickling of stainless steels and metal etching ), as rocket propellant, and for nuclear fuel
processing.
For the most part, nitric acid is manufactured and consumed at concentrations of about
60%. But concentrated (90% or more) nitric acid is needed for
The production of chemicals such as;
Isocyanates
Nitrobenzene
which are used as starting materials for diverse commercial chemicals/materials.
Concentrated nitric acid is required for many organic reactions of industrial
importance.
It is used as catalyst for many reactions.
It can promote downstream chemical industry.
2
Chapter: 01
Introduction
MINIMUM FEASIBLE CAPACITY
Presently in Pakistan, there is no plant for the production of concentrated nitric acid. As
an initiative for the futuristic development of chemical industry, initially a minimum
feasible capacity plant can be set up. So our group has selected 140 metric tons/day
capacity plant as design basis.
PHYSICAL PROPERTIES:
1. Crystals of pure nitric acid are colorless
2. The normal boiling point of nitric acid is 83.1˚C
3. The color may range from yellow to red
4. The melting point is -41.6 ˚C
5. Nitric acid is completely miscible with water
6. At high temperature nitric acid is decomposed
7. The V.P and density of acid containing NO2 increase with %age of NO2 present.
8. The density, viscosity and thermal conductivity of HNO3 are given in the table.
CHEMICAL PROPERTIES
ACIDIC PROPERTIES
1. As a typical acid it reacts readily with alkalis, basic oxides and carbonates to
form salts.
2. HNO3 is a strong monobasic acid.
3. It is a powerful oxidizing agent and many organic compounds.
4. An industrial application of HNO3 is the reaction with NH3 to produce
(NH4)NO3.
5. However it is oxidizing in nature HNO3 does not always behave as a typical
acid.
3
Chapter: 01
Introduction
OXIDIZING PROPERTIES
1. HNO3 is a powerful oxidizing agent that reacts violently with many organic
materials e.g. turpentine, charcoal.
2. The concentrated acid may react explosively with ethanol.
3. HNO3 is used with certain organic e.g. furfuryl alcohol and aniline.
4. Depending on acid concentration, temperature and the reducing agent involved
any of the following oxidation may occur.
4 HNO3 + 2e‾
→
2 NO3‾ + 2 H2O + 2 NO2
8 HNO3 + 6e ‾
→
6 NO3‾ + 4 H2O + 2 NO
10 HNO3 + 8e‾
→
8 NO3‾ + 5 H2O + N2O
10 HNO3 + 8e ‾
→
9 NO3 ‾ + 3 H2O + NH4+
16 HNO3 + 12e‾
→
14 NO3‾ + 4 H2O + 2 NH3OH
5. Concentrated HNO3 favors the formation of NO2 while at low strength the
formation of NO.
6. Concentrated HNO3 reacts with all metals except gold, iridium, Pt, Rh,
Tantalum, Titanium and certain alloys.
7. Concentrated HNO3 converts the oxides and sulphides etc. of most elements in a
low oxidation state to a higher level e.g., sulphur dioxide is oxidized to
sulphuric acid.
8.
Chrome, iron, Al readily dissolves in HNO3 but concentrated acid form a metal
oxide.
4
Chapter: 01
Introduction
Plant Location & Layout
As suggested in the process selection that nitric acid plant in Pakistan is not feasible with
out a fertilizer complex. This fertilizer complex will have following main units.
1. Ammonia plant
2. Nitro phosphate plant
3. CAN plant
4. Utilities
ABSTRACT
Following factors should be considered in plant location selection
1. Location w.r.t. the marketing area.
2. Raw materials supply
3. Transport facilities
4. Availability of labor
5. Availability of suitable land
6. Environmental impact and effluent disposal
7. Local community considerations
8. Climate
9. Political and strategic considerations
10. Availability of utilities ,water ,fuel and power
MARKETING AREA
For materials that are produced in bulk quantities such as mineral acids and fertilizers
where the cost of the product per ton is relatively low and the cost of transport a
significant fraction of the sale price the plant should be located close to primary market.
5
Chapter: 01
Introduction
RAW MATERIALS
Main raw material for producing ammonia is natural gas and for plants producing bulk
chemicals are best located close to the source of the major raw material, where it is also
close to the marketing area.
The availability of price of suitable raw materials will often determine the site location.
Plants producing bulk chemicals are best located close to the source of the major raw
material, where this is also close to the marketing area.
TRANSPORT
The transport of materials and products to and from the plant will be an overriding
consideration in site selection. If practicable a site should be selected that is close to
atleast two major forms of transport, road, rail, waterway or a sea port. Road transport is
being increasingly used and is suitable for local distribution from a central warehouse.
Rail transport will be cheaper for the long distance transport of bulk chemicals.
Air transport is convenient and efficient for the movement of personnel and essential
equipment and supplies and the proximity of the site to a major airport should be
considered.
AVAILABILITY OF LABOR
Labor will be needed for the construction of the plant and its operation. Skilled
construction workers will usually be brought in from outside the site area but there should
be an adequate pool of unskilled labor available locally, and labor suitable for training to
operate the plant .Skilled tradesmen will be needed for plant maintenance .Local trade
union customs and restrictive practices will have to be considered when assessing the
availability and suitability of the local labor for recruitment and training.
UTILITIES
Chemical processes invariably require large quantities of water for cooling and general
process use, and the plant must be located near a source of water of suitable quality.
6
Chapter: 01
Introduction
Process water may be drawn from a river from wells or purchased from a local authority.
At some sites the cooling water required can be taken from a river or lake or from sea at
other locations cooling towers will be needed.
Electrical power will be needed at all sites. A competitively priced fuel must be available
on site for steam and power generation.
ENVIRONMENTAL IMPACT AND EFFLUENT DISPOSAL
All industrial processes produce waste products and full consideration must be given to
difficulties and cost of their disposal .The disposal of toxic and harmful effluents will be
covered by local regulations and the appropriate authorities must be consulted during the
initial site survey to determine the standards that must be met.
LOCAL COMMUNITY CONSIDERATIONS
The proposed plant must fit in with and be acceptable to the local community .Full
consideration must be given to the safe location of the plant so that it does not impose a
significant additional risk to the community.
SITE LAYOUT
The process units and ancillary buildings should be laid out to give the most economical
flow of materials and personnel around the site. Hazardous processes must be located at a
safe distance from other buildings. Consideration must also be given to future expansion
of the site. The ancillary buildings and services required on a site in addition to the main
processing units will include
1. Storage for raw materials and products, tank farms and ware houses
2. Maintenance workshops
3. Stores for maintenance and operating supplies
4. Labs for process control
5. Fire stations
7
Chapter: 01
Introduction
6. Utilities, steam boilers, compressed air, power generation, refrigeration,
transformer stations.
7. Effluent disposal plant
8. Offices for general administration
9. Canteens and other amenity buildings.
10. Car parks
When roughing out the preliminary site layout the process units will normally be sited
and arranged to give a smooth flow of materials through the various processing steps,
from raw materials to final product storage .Process units are normally spaced at least
30m apart greater spacing may be needed for hazardous processes.
The location of the principal ancillary buildings should then be decided. They should be
arranged so as to minimize the time spent by personnel in traveling between buildings.
Administration offices and laboratories in which a relatively large number of people will
be working should be located well away from potentially hazardous processes may have
to be sited a safer distance.
The sitting of the main process units will determine the layout of the plant roads, pipe
alleys and drains. Access roads will be needed to each building for construction and for
operation and maintenance .Utility buildings should be sited to give the most economical
run of pipes to and from the process units.
Cooling towers should be sited so that under the prevailing wind the plume of condensate
spray drifts away from the plant area and adjacent properties.
The main storage areas should be placed between the loading and unloading facilities and
the process units they serve .Storage tanks containing hazardous materials should be sited
at least 70 m from the site boundary.
8
Chapter: 01
Introduction
PLANT LAYOUT
The economic construction and efficient operations of a process unit will depend on how
well the plant and equipment specified on the process flow sheet is laid out. The principle
factors are
1. Economic considerations
2. The process requirements
3. Convenience of operation
4. Safety
5. Future expansion
6. Modular construction
COSTS
The cost of construction can be minimized by adopting a layout that gives the shortest
run of connecting pipe b/w equipment and the least amount of structural steel work.
However this will not necessarily be the best arrangement for operation and maintenance.
PROCESS REQUIREMENTS
An example of the need to take into account process considerations is the need to elevate
the base of columns to provide the necessary net positive suction head to a pump.
OPERATION
Equipment that needs to have frequent operator attention should be located convenient to
the control room. Valves, sample points and instruments should be located at convenient
positions and heights. Sufficient working space and headroom must be provided to allow
easy access to equipment.
9
Chapter: 01
Introduction
MAINTENANCE
Heat exchangers need to be sited so that the tube bundles can be easily withdrawn for
cleaning and tube replacement .Vessels that require frequent replacement of catalyst i.e.
ammonia burner should be located on the inside of buildings .Equipment that requires
dismantling for maintenance such as compressors and large pumps should be also placed
under cover.
SAFETY
Blast walls may be needed to isolate potentially hazardous equipment and confine the
effects of an explosion .At least two escape routes for operators must be provided from
each level in process buildings.
PLANT EXPANSION
Equipment should be located so that it can be conveniently tied in with any future
expansion of the process. Space should be left on pipe alleys for future needs and service
pipes over sized to allow for future requirements.
MODULAR CONSTRUCTION
In recent years there has been a move to assemble sections of plant at the plant
manufacturer’s site. These modules will include the equipment structural steel piping and
instrumentation. The modules are then transported to the plant site by road or sea
1. Improved quality
2. Reduced construction cost
3. Less need for skilled labor on site
4. Less need for skilled personnel on overseas sites
Some of the disadvantages are
1. Higher design costs
2. More structural steel work
3. More flanged connection
10
Chapter: 02
Process Description
Chapter: 02
Process Description
GENERAL
SCHEME
FOR
MANUFACTURE
OF
CONCENTRATED HNO3
Standard oxidation of NH3 in the presence of Pt/Rh catalyst to form NO.
Oxidation of NO to NO2 and absorption of NO2 in water to form weak HNO3 (5070 wt%).
Concentration of weak HNO3 (azeotropic i.e., 68 wt%) by distillation (extractive)
or by reaction with nitrogen dioxide.
Since HNO3 forms an azeotrope with water at concentration of about 68%, stronger acid
cannot be produced by simple distillation of weaker acid.
COMMERCIAL MANUFACTURING PROCESSES FOR
CONCENTRATED NITRIC ACID
Almost all commercial quantities of nitric acid are manufactured by the oxidation of
ammonia with air to form nitrogen oxides that are absorbed in water to form nitric acid.
Since nitric acid forms an azeotrope with water at a concentration of about 68.8%,
stronger acid cannot be produced by simple distillation of weaker acid. Two industrial
methods for producing concentrated nitric acid are generally employed
Extractive distillation (nitric acid concentration NAC processes)
Reaction with nitrogen dioxide(direct strong nitric DSN processes)
11
Chapter: 02
Process Description
NITRIC ACID CONCENTRATION PROCESSES
NAC processes use extractive distillation to concentrate weak acid up to 99 wt%. A
dehydrating agent, such as sulphuric acid or magnesium nitrate is used to enhance the
volatility of HNO3 so that distillation methods can surpass the azeotropic concentration of
nitric acid. Weak acid and dehydrating agent are fed to a distillation column. Water
removed from the acid dilutes the dehydrating agent, which is removed as a bottom
stream and later concentrated for reuse in the process. Super-azeotropic acid vapors pass
to the bottom of a rectification section in which the acid is concentrated up to 99 wt%.
The strong nitric vapors are condensed overhead and a portion of the acid is returned to
the column as reflux. The 72% sulphuric acid leaving the bottom of the tower is denitrated by steam stripping in a second tower, and then re-concentrated to 93% by contact
with furnace gases in a drum concentrator.
DIRECT STRONG NITRIC PROCESSES
The second approach to the concentration of nitric acid, of late called the Direct Strong
Nitric process, is an outgrowth of the atmospheric pressure process. At atmospheric
pressure, the rate of oxidation of nitric oxide to the dioxide is extremely slow so that
relatively little dioxide is present when the gas reaches the condenser after the ammonia
burner. Most of the water formed can, therefore, be condensed in an atmospheric plant
forming a solution with only 2-4% by weight nitric acid. Part of this acid is used as
makeup in the process, but most of it has normally been discarded.
The process gas leaving the condenser is contacted by concentrated nitric acid in an
oxidation tower, reversing the absorption reaction by converting nitric oxide plus nitric
acid to nitrogen dioxide and water. The dioxide in the process gas is chemically absorbed
into azeotropic or greater strength acid. Air stripping and rectification of the resulting
stream produces strong acid of 98-99 wt% strength.
12
Chapter: 02
Process Description
PROCESS SELECTION
We have selected concentrated nitric acid (CNA) process which is a direct strong nitric
acid process because of its following advantages over other processes;
Recycle loops and a special rectification step are innovations of this process. They
eliminate the refrigeration, oxygen and chemical dehydrating agents (eg;
sulphuric acid or magnesium nitrate) required by classical manufacturing
techniques, and thus decrease utilities and operating expenses considerably.
In nitric acid concentration processes, drum concentrators are used and
electrostatic precipitators are provided to eliminate acid from the vented gas. They
have operating problems and hence pollution problems are face, whereas in CNA
process, atmospheric emissions have been less than 450ppm in (NOx) without any
abatement units.
This plant can be fully automated – only two operators a shift are required.
This process can also make concentrated and/or weak acid (50-60%) in any
proportion: the product mix can be changed by adjusting the position of two
valves.
This route has a lower investment as compared to other routes.
DETAILED PROCESS DESCRIPTION
Liquid ammonia is vaporized and is mixed with filtered stream of air. The blend reacts
over Pt/Rh catalyst at about 870oC and atmospheric pressure to form nitric oxide and
water. Reaction products pass through tail-gas preheater, waste heat boiler and
cooler/condenser to dissipate the heat generated. At atmospheric pressure, the rate of
oxidation of nitric oxide to dioxide is extremely slow so that relatively little dioxide is
present when the gas reaches the condenser. All the water present is condensed and is
drained off as 2% by wt. nitric acid solution. Operation at atmospheric pressure helps to
keep acid contents down.
13
Chapter: 02
Process Description
The dry gas stream is preheated and sent to bottom of oxidation tower. Here 50-60% acid
from the absorption section of the system enters the top of the column. As the gas and the
liquid flow counter currently, the acid oxidizes NO to NO2
NO + 2HNO3
3NO2 + H2O
Dilute acid from the bottom of the oxidation unit is sent to the absorption section of the
route for concentration. The process gas leaving the top of oxidation tower is combined
with air containing nitrogen oxides and then compressed. The compressed gas is cooled
before entering the absorption section of the process. This section actually consists of two
columns, one produces the superazeotrope and the other makes a weak acid. The gases
first enter the superazeotrope column and are contacted with azeotropic (68%) acid from
the distillation unit. Nitrogen dioxide is absorbed and a superazeotrope of 80% nitric acid
is formed. This superazeotrope, saturated with nitrogen oxides, is stripped with air,
preheated and then fed to distillation column. The normal azeotropic acid mixture used
for absorption is taken off the bottom of distillation column. Concentrated acid, 98% is
obtained at the top.
Since the process gases leaving the top of the superaziotropic absorption tower are still
rich in nitrogen oxides, they are sent to another absorption unit. Here low-concentration
acid from the oxidation tower enters in the middle. HNO3 which is about 50-60% is
removed from the bottom.
The exiting gases have concentration of NOx in the order of 300-400ppm. The tail gas is
preheated to about 450oC by process streams in the ammonia oxidizer and the streams
leaving the compressor.
14
Chapter: 03
Material Balance
Chapter: 03
MATERIAL BALANCE
BASIS
140 metric tons/day of 98% HNO3 on 100% basis
1 hr operation
Establishing the amounts of raw materials for the desired production of
Nitric acid
Acid produced = 140 metric tons/day
= 5833.33kg/hr
(1 metric ton = 1000kg)
Total solution = 5833.33/0.98
= 5952.38 kg/hr
Water produced =5952.38 – 5833.33
= 119.95 kg/hr
(5833.3 kg HNO3 / hr) (I mole of HNO3/63 kg HNO3) (1 Kg mole NH3 required/I Kg
mole HNO3) (17 Kg NH3/I Kg mole NH3) (1 day/130 metric tons)
(24 hr/1 day)
NH3 required = 290.59 kg/metric ton HNO3
(Ref: “Routes to concentrated nitric acid”, L.M.Marzo and S.M.Marzo, Spain 1978.
Fertilizer acids).
15
Chapter: 03
Material Balance
NH3 required = 290.59kg/metric ton x 130 metric tons / day
= 37777.5 kg/day
= 1574.06 kg/hr
= 92.59 kgmol
Overall NH3 to Air ratio = 1:10
(Ref; “Kirk Othmer” encyclopedia for chemical engineering)
Total air required for process = 925.9 kg mol
Primary air entering ammonia oxidation reactor;
For ammonia oxidation, the stoichiometric gas composition is 15% by Vol% of NH3 in
air. However, this composition is in explosive range of NH3 air mixture. So atmospheric
plants are operated at 14%.
Primary air entering reactor = (92.59 / 0.14) – 92.59
= 568.76 kg mol/hr
Oxygen entering reactor = 568.76x 0.21
= 119.44 kg mol/hr
Nitrogen entering reactor = 568.76 – 119.4
= 449.36 kg mol/hr
16
Chapter: 03
Material Balance
MATERIAL
BALANCE
AROUND
INDIVIDUAL
EQUIPMENT
REACTOR (R 01)
3b
NH3=92.54kgmol/hr
O2=119.4kgmol/hr
N2=449.36kgmol/hr
4
NO=92.59 kgmol/hr
H2O=141.57 kgmol/hr
O2=4.6115 kgmol/hr
N2=450.28n kgmol/hr
REACTIONS OCCURING IN REACTOR;
1-
4NH3 + 502
4NO + 6H2O
(98 %)
2-
4NH3 + 3O2
2N2 + 6H2O
(2 %)
According to 1st reaction;
NH3 consumed = 92.59 x 0.98
= 90.73 kgmol/hr
17
Chapter: 03
Material Balance
O2 consumed = (90.73 Kg mole NH3/hr) (5 Kg mole O2 consumed/4 Kg mole NH3)
= 113.4 kg mol/hr
NO produced = 992.59 Kg mole NH3/hr) (4 Kg mole NO produced/4 Kg mole NO /hr)
=92.59 Kg mole NO/hr
H2O produced = (92.59Kg mole NH3/hr) (6 Kgmole H2O/4 Kg mole NH3)
=138.88KgmoleH2O/hr
According to 2nd reaction;
NH3 consumed = 92.59 x 0.02
= 1.85 kg mol/hr
O2 consumed = 1.3888KgmoleO2/hr
N2 produced =(1.85 KgmoleNH3/hr)(2KgmoleN2/4KgmoleNH3)
=0.925KgmoleN2/hr
H2O produced = 1.85KgmoleNH3/hr(6KgmoleH2O/4KgmoleNH3)
= 2.775KgmoleH2O/hr
Total NH3 consumed = 90.73+1.85
= 92.58Kgmole/hr
Total O2 consumed = 113.4+1.3885
= 114.7Kgmole/hr
O2 leaving un-reacted = 119.4-114.7
= 4.6115 kg mol/hr
N2 leaving = N2 entering + N2 produced
= 0.925+449.36
= 450.28 kg mol/hr
NO leaving = 92.59 kg mol/h
18
Chapter: 03
Material Balance
H2O leaving = H2O produced in 1st reaction + H2O produced in 2nd reaction
= 0.925+449.36
= 141.575 kgmol/hr
19
Chapter: 03
Material Balance
TAIL GAS PREHEATER (E02) + WASTE HEAT BOILER (E03);
4
NO=92.59kgmol/hr
H2O=141.575kgmol/hr
O2=4.6115kgmol/hr
N2=450.28kgmol/hr
6
NO2=9.223kgmol/hr
NO=83.36kgmol/hr
H2O=141.575kgmol/hr
N2=450.28kgmol/hr
REACTION;
2NO + O2
2NO2
(100% conversion is assumed)
O2 is limiting reatanct
NO consumed = (4.6115 Kg mole/hrO2)/(2 Kg mole NO./I Kgmole/hrO2)
=9.23 Kgmole/hr(NO)
NO2 produced = (4.6115 kg mol/hr O2) (2KgmoleNO2/1KgmoleO2)
=9.223Kgmole/hr
20
Chapter: 03
Material Balance
COOLER/CONDENSOR (E 05) ;
9
NO2=7.948kgmol/hr
NO=83.785kgmol/hr
N2=450.28kgmol/hr
7
NO2=9.223kgmol/hr
NO=83.36kgmol/hr
H2O=141.57kgmol/hr
N2=450.28kgmol/hr
8
HNO3=0.8kgmol/hr
H2O=141.5kgmol/hr
REACTION;
3NO2 + H2O
2HNO3 + NO
Water is drained off as 2% nitric acid solution.
Let 0.8kg mol of HNO3 produced when water is condensed, then
HNO3 produced = 0.85kgmol
H2O consumed = (0.85 Kgmole/hrHNO3) (1 Kgmole/hrH2O)/(2Kgmole/hrHNO3))
=0.425Kgmole/hrH2O
NO2 consumed =1.275 Kgmole/hrNO
NO produced
= 0.425 Kgmole/hr NO
Check;
0.85 x 63
2% =
x 100%
21
Chapter: 03
Material Balance
(0.8 x 63) + [(141.57 x 18) – (0.425 x 18)]
2% =
2%
Therefore our assumed value is correct.
NO2 leaving = NO2 entering – NO2 consumed
= 9.223-1.275
= 7.948 kgmol/hr
NO leaving = NO entering + NO produced
= 83.36 + 0.425
= 83.785 kgmol/hr
HNO3 produced = 0.85 kgmol/hr
H2O condensed = H2O entering – H2O consumed
= 141.575-0.425
= 141.15 kgmol/hr
22
Chapter: 03
Material Balance
OXIDATION TOWER (D 01) ;
13
11
D
NO=8.375kgmol/hr
NO2=234.18kgmol/hr
N2=450.28kgmol/hr
B (60wt%HNO3)
HNO3=297.27kg/hr
H2O=693.6kg/hr
10
12
A
NO=83.785kg/hr
N2=450.28kg/hr
C (35 wt%HNO3)
HNO3=146.48kg/hr
H2O=769.05kg/hr
REACTION;
NO + 2HNO3
3NO2 + H2O
(90%)
NO consumed = 83.785 x 0.9
= 75.40 kgmol/hr
HNO3 consumed = (75.40 KgmoleNO)(2Kgmole HNO3/1KgmoleNO)
= 150.8 kgmol/hr
NO2 produced =(75.40 KgmoleNO)(3Kgmole of NO2/1Kgmole NO)
= 226.2 kgmol/hr
H2O produced = (75.40NO)/(1KgmoleH2O/1Kgmole NO)
=75.40 kgmole H2O/hr
NO leaving = NO entering – NO consumed
23
Chapter: 03
Material Balance
=83.785-75.40
=8.3785 Kgmole/hr
NO2 leaving =234.18 kgmol/hr
HNO3 BALANCE;
0.60 B – 9500.4 = 0.4 C
H2O BALANCE;
0.4 B + 1375.2 = 0.6 C
Solving these two equations, B & C comes out to be,
B = 31213.65 kg/hr
C =23071.65 kg/hr
24
Chapter: 03
Material Balance
COMPRESSOR (K 01) ;
16
NO2=247.5kgmol/hr
O2=70.81kgmol/hr
N2=732.14kgmol/hr
15
NO=8.37kgmol/hr
NO2=239.181kgmol/hr
O2=74.99kgmol/hr
N2=732.148kgmol/hr
REACTION;
2NO + O2
2NO2
(100% conversion)
NO entering = 8.37 kgmol/hr
O2 consumed = ½ x 8.37 = 4.18 kgmol/hr
NO2 produced = 8.37kgmol/hr = 385.25 kg/hr
Now leaving stream;
NO leaving = 0 kgmol/hr
NO2 leaving = NO2 leaving from oxidation tower + NO2 entering from recycle
stream + NO2 produced
= 10772.28 + X + 385.25 = 11157.57+ X kg/hr
25
Chapter: 03
Material Balance
NO2 entering from recycle stream is 2% of total NO2 entering the compressor
Let NO2 entering from recycle stream is 230 kg/hr;
2%
=
230
11157.57 + 230
2%
So,
= 2.007%
X = 230 kg/hr
NO2 leaving = 11387.53 kg/hr
= 247.5 kgmol/hr
O2 leaving = O2 entering – O2 consumed
= 74.99 – 4.18
= 70.81 kgmol/hr
N2 leaving = 732.14 kgmol/hr
26
Chapter: 03
Material Balance
DISTILLATION COLUMN; (D 05)
24
B
HNO3 (98wt%)=5833.3kg/hr
H2O = 119kg/hr
23
A
HNO3 (80%)=11905.8kg/hr
H2O (20%wt)=2976.4kg/hr
18
C
HNO3 (68%wt)=6072.4kg/hr
H2O (32%wt)=2857.6kg/hr
Pure acid product = 5833.3 kg/hr
Acid solution
= 5833.3/0.98
= 5952.3kg/hr
Overall balance around column;
A=B+C
A = 5952.3 + C
eq.1
HNO3 balance;
0.8 A = 0.68 C + 5833.3
eq.2
Solving equations 1 & 2;
C = 8930 kg/hr;
A = 14882.3kg/hr
27
Chapter: 03
Material Balance
SUPERAZEOTROPIC COLUMN (D02)
NO2=103.6kgmol/hr
NO=46.3kgmol/hr
O2=70.81kgmol/hr
N2=732.14kgmol/hr
HNO3(68%)=6072.4kg/hr
H2O(32%)=2857.6kg/hr
H2O=52.9kgmolhr
NO2=247.5kgmol/hr
O2=70.81kgmol/hr
N2=732.14kgmol/hr
HNO3(80%)=11905.8kg/hr
H2O(20%)=2976.4kg/hr
NO2=10580kg/hr=230kgmol/hr
REACTION;
3NO2 + H2O
2HNO3 + NO
HNO3 balance,
HNO3 entering from distillation column + HNO3 produced = HNO3 leaving
6072.4 + X = 11905.8
X = 5833.4 kg/hr = 92.6 kgmol/hr
NO produced = ½ x 92.6
= 46.3mol/hr
H2O consumed = ½ x 92.6
= 46.3mol/hr = 833.4kg/hr
H2O leaving = H2O entering – H2O consumed
2976.4 = 2857.6 + (water from stram 19) – 833.4
28
Chapter: 03
Material Balance
2976.46 = 2024.2 + (watr from stream 19)
2976.46 = 2024.2 + 952.26
2976.46 = 2976.46
NO2 consumed = 3/2 x 92.6
= 138.9kgmol/hr
NO2 leaving = NO2 entering – NO2 consumed
= 247.9 – 138.9
= 108.6 kgmol/hr
4-5 % of this leaving NO goes into a stream entering bleaching column.
= 108.6 X 0.046 = 4.99 ≈ 5 kgmol/ hr
So in stream 20 NO2 = 108.6 – 5 = 103.6 kgmol/ hr
29
Chapter: 03
Material Balance
ABSORPTION COLUMN; (D 01)
NO2=1.036kgmol/hr
NO=1.389kgmol/hr
O2=11.49kgmol/hr
N2=732.14kgmol/hr
HNO3(40%)=146.48kgmol/hr
H2O(60%)=769.05kgmol/hr
NO2=103.6kgmol/hr
NO=46.3kgmol/hr
O2=70.81kgmol/hr
N2=732.14kgmol/hr
HNO3(60%)=297.27kgmol/hr
H2O(40%) = 693.6kgmol/h
REACTIONS
4NO + 3O2 + 2H2O
4HNO3
eq.1
4NO2 + O2 + 2H2O
4HNO3
eq.2
from eq.1,
NO consumed = 46.3 x .97 = 44.91 kgmol/hr
O2 consumed = ¾ x 44.911
=33.66kgmol/hr
H2O consumed = 2/4 x 44.911
= 22.46kgmol/hr
HNO3 produced = 44.911kgmol/hr
30
Chapter: 03
Material Balance
From eq.2,
NO2 consumed = 103.6 x 0.99 = 102.56 kgmol/hr
O2 consumed = ¼ x 102.56
= 25.64kgmol/hr
H2O consumed = 2/4 x 102.56
= 51.28kgmol/hr
HNO3 produced = 102.56kgmol/hr
O2 leaving = O2 entering – O2 consumed
= 70.8119 – ( 25.64 + 33.68)
= 11.49kgmol/hr
31
Chapter: 03
Material Balance
BLEACHING COLUMN; (D 03)
21
14
HNO3(80%) = 11905.8kg/hr
H20 = 29764kg/hr
NO2 = 5kgmol/hr
22
O2 = 74.9kgmol/hr
N2 = 282.14kgmol/hr
O2 = 74.9kgmol/hr
N2 = 282.14kgmol/hr
NO2= 5kgmol/hr
23
3
HNO3(80%wt)=11905.8kg/hr
H2O (20%wt)=2976.46kg/hr
Total air required for process = 925.9 kgmol/hr
Primary air ysed in Oxidation tower = 568.9kgmol/hr
Remaining secondary air = 925.9 – 568.9 = 357 kgmol/hr
O2 = 357 X 0.21 = 574.9kgmol/hr
N2 = 357 X 0.79 = 282.14kgmol/hr
32
Chapter: 03
Material Balance
OVERALL MATERIAL BALANCE
D
A
H2O=52.9kgmol/hr
A
O2=119.4kgmol/hr
N2=449.36kgmol/hr
B
A
NH3=92.59kgmol/hr
E
A
HNO3=92.59 kgmol/hr
H2O=6.61 kgmol/hr
OVERALL PLANT
BALANCE
F
HNO3=0.85 kgmol/hr
H2O=141.5 kgmol/hr
C
A
O2=74.9 kgmol/hr
N2=282.14kgmol/hr
G
A
NO2=0.1%=1.036 kgmol/hr
NO=0.2%=1.389 kgmol/hr
O2=1.5%=11.49 kgmol/hr
N2=98.12%=732.14 kgmol/hr
A = primary air entering
B = ammonia entering
C = secondary air entering
D = water entering
E = acid product solution leaving
F = water condensate leaving as 2% nitric acid solution
G = tail gases leaving
33
Chapter: 03
Material Balance
CHECK:
Stream A:
O2 = 119.4 kgmol/hr = 3820.8 kg/hr
N2 = 449.36 kgmol/hr = 12582.08kg/hr
= 16402.88kg/hr
Stream B:
NH3 = 92.59 kgmol/hr = 1574.03kg/hr
Stream C:
O2 = 74.9 kgmol/hr = 2396.8 kg/hr
N2 = 282.14 kgmol/hr = 7899.92 kg/hr
= 10296.72kg/hr
Stream D:
H2O = 52.9 kgmol/hr = 952.2kg/hr
Stream E:
HNO3 = 92.59 kgmol/hr = 5833.17 kg/hr
H2O = 6.61 kgmol/hr = 118.98kg/hr
= 5952.15 kg/hr
Stream F:
HNO3 = 0.85 kgmol/hr = 53.55kg/hr
34
Chapter: 03
Material Balance
H2O = 141.15 kgmol/hr = 2540.7 kg/hr
= 2594.25kg/hr
Stream G:
NO2 = 1.036 kgmol/hr = 47.65kg/hr
NO = 1.389 kgmol/hr = 41.67kg/hr
O2 = 11.49 kgmol/hr = 367.68kg/hr
N2 = 732.14 kgmol/hr = 20499.29kg/hr
= 20956.92Kg/hr
INPUT = OUT PUT
A + B + C + D = E + F +G
16402.88 +1574.02 +10296.7 +952.2 = 5952.12 +2594.25 +20956.92
29503.29 = 29225.83
35
PROCESS FLOW SHEET
1
H 01
25
2
M 01
13
D 01
D 03
3a
12
11
E 01
3b
R 01
12
10
9
11
5
E 02
H 02
7
6
4
E 03
E 04
E 05
2a
E 07
G 01
E 06
20
14
19
10
E 11
18
14
24
23
D 02
K 01
N 02
D 04
N 01
16
CW
15
E 09
E 08
17
23
21
CWR
22
E 10
D 05
E 12
18
F 01
H01 Ammonia Filter
H02 Air Filter
M01 Mixer
E01 Heat Exchanger
E02 Tail gas pre heater
E03 Waste Heat boiler
E04 Air Pre heater
E05 Cooler /Condenser
E07 Heat exchanger
E07 Heat exchanger
E08 Heat exchanger
E09 Heat Exchanger
E11 Condnser
E12 Reboiler
D01 Oxidation tower
D02 Super azeotrope tower
D03 Bleaching column
D05 Distillation column
K01 Compressor
F01 Stack
R01 Reactor
E06 Heat exchanger
E10 Acid heat exchanger
D03 Weak acid tower
(absorption tower)
N01 Expander
N02 Motor
Chapter: 04
Energy Balance
Chapter: 04
Energy Balance
ENERGY BALANCE OF REACTOR
MOLAL FLOW RATES OF REACTANTS (kgmol/hr)
NH3 = 92.59 k mol / hr = 14.00 %
O2 = 119.4 kg mol/hr = 18.05%
N2 = 449.36 = 67..94%
661.35
MOLAL FLOW RATES OF PRODUCTS (kg mol/hr)
NO
= 92.59 k mol/ hr = 13.44%
H2 O
=
141.575
O2
=
4.6115
N2
=
= 20.55 %
= 0.67 %
450.28
= 65.347 %
689.05
4NH3 + 5O2
→ 4 NO + 6 H2O
4NH3 +3 O2
→ 2 N2 +6 H2O
ΔH 298 K = -54.1 kcal / mol
ΔH 298 K = -75.7 kcal / mol
Heat of reaction for reaction-1
Heat of formation of reactions and products
NH3 = -10.96 kcal/mol
NO = 21.6 kcal/mol
36
Chapter: 04
Energy Balance
H2O = -57.7979 kcal/mole
Hat of reaction 1 =(ΔHf.p - ΔHfr)
ΔHfp = (92.59 * 1000 * 21.6) + (138.88 * 1000 * -57.7979)
=1999944-8026972.352
=-6027028.35 kcal/hr
ΔHfr= (90.73 * 1000 * -10.96)
=-994400.8 kcal/hr
ΔH r*n 1 = -6027028.35 + 994400.8
=-5032627.55 kcal/hr
For Reaction 2
ΔHf.p = 2.2775 * 1000 * -57.7979
=-160389.17 kcal/hr
ΔHf.r = 1.85 * 1000 *-10.98
=-20276 kcal/hr
ΔH r*n2 =-160389.17 = 20276
=-140113.1725 kcal/hr
Now
Hat Of Reaction = -5032627.55 = (-140113.17)
=-5172740.725 kcal/hr
Calculating Cp of inlet at = 200 ˚C
Cp of NH3 = 9.6299 kcal / k mol ˚C
Cp of O2 = 7.553 kcal / k mol ˚C
Cp of N2 = 6.973 kcal / k mol ˚C
37
Chapter: 04
Energy Balance
Σ Hr = [m Cp) NH3 + (m Cp )O2 + ( m Cp)N2 ] ΔT
= [(92.59*9.6799) + (119.4*7.553) + (449.36*6.973) ] (200 – 25 )
= 863008.475 kcal
Cp of outlet is calculated at 870 ˚C
Cp of NO = 8.19 kcal / kg mol ˚C
Cp of H2O = 10.14 kcal / kg mol ˚C
Cp of O2 = 8.42 kcal / kg mol ˚C
Cp of N2 = 7.643 kcal / kg mol ˚C
Σ Hp = [ (m Cp ) NO + (m Cp) H2O + (m Cp ) O2 + (m Cp ) N2 ] ΔT
= [ (92.59*8.19) + (141.575*10.14 ) + (4.6115*8.42) + (450.28*7.643) ] (870 – 25)
= 4794700.24 kcal
q = Σ Hp + Σ ΔH at 25˚C - Σ Hr
= 4794700.24+(-5172740.725)-863008.4
=--1241048.885 kcal
38
Chapter: 04
Energy Balance
ENERGY BALANCE OF COMPRESSOR
P1 = 1 atm
P2 = 10atm
T1 = 48 C
No. of stages (n) = 2
C.R = (P2/P1)1/n
= 3.16
Cp/Cv = = 1.4
m= - 1
Ec
Ec = efficiency of compressor = 78%
m= 0.366
For stage 1,
T2 = T1 (P2 /P1)0.366
= 321 (3.16)0.366
= 489.09˚K
=216.9 ˚C
For stage 2;
After passing through intercoolers temperature is reduced to 50˚C
Now T1 =50˚C
Again,
T2 = T1 (P2 /P1)0.366
= 323 (3.16)0.366
39
Chapter: 04
Energy Balance
= 492.13˚K
=492.13 ˚C
ENERGY BALANCE OF OXIDATION TOWER
Inlet Temperature of reactants = 50 °C
Moles of Reactants ( kg mol)
NO = 83.784
N2 = 450.28
HNO3 = 297.27
H2O = 693.6
Moles of Products :( kg mol )
NO = 8.375
NO2 = 234.18
N2 = 450.28
HNO3 = 146.48
H2O = 769.05
Specific Heat of Reactants & Products (kcal/k mol °C)
Cp of N O = 7.146
Cp of N2
= 6.95
Cp of HNO3 = 26.29
Cp of H2O = 9.25
Cp of NO2 = 8.9625
Calculating Enthalpy of reactants :(kcal )
H1 of NO = mCp(T2 –T1)
= 83.784 x 7.146 x (50 – 25)
40
Chapter: 04
Energy Balance
= 14968.01
H2 of N2 =m Cp (50 – 25)
= 450.28 x 6.95 x 25
= 78236.15 kcal
H3 of HNO3 = 297.27 x 26.29 x (50 - 25)
=195380.7 kcal
H4 of H2O = 693.6 x 9.25 x25
= 160395 kcal
H of reactants Hr = H1 + H2 + H3 +H4 =450760.66 kcal
Calculating enthalpy of products (kcal)
H1of NO = m Cp (T2 – 25)
= 8.375 x7.146 x (T2 – 25)
= 59.8 (T2 – 25)
H2 of NO2 = 234.18x 8.9625 x (T2 – 25)
= 2098.8 (T2 – 25)
H3 of N2 = 450.28 x 6.9449 x(T2 – 25)
= 2098.8 (T2 – 25)
H4 of H2O = 769.05 x 9.25 x (T2 – 25)
= 7113.7 (T2 – 25)
H5 of HNO3 = 146.48 x 26.29 x (T2 – 25)
=3850.95 (T2-25)
Summing enthalpy of products
H of products Hp = 16250.39 (T2 – 25)
41
Chapter: 04
Energy Balance
For reaction
NO + 2 HNO3 →
3 NO2 + H2O
ΔH rxn at 25 °C = 32.4 kcal /g mol = 2553424.56 kcal (from Kirk Othmer, Vol 13)
Now,
Hp + ΔHrxn at 25oC - Hr = 0
Putting all the above values and calculating T2
16250.39(T2 – 25) + 2553424.56 – 450760.6 = 0
Outlet Temp of products T2 = 48 °C
ENERGY BALANCE OF ABSORPTION COLUMN:
HEAT OF REACTION:
Heat evolved from reaction 1,
No. of moles of NO2 entering = 46.3 kg mol
No. of moles of NO2 leaving = 1.389 kg mol
No. of moles of NO2 consumed=(46.3-1.389) kg mol=44.911 kg mol
Now form reaction1,
Qrxn1 = -263000 kJ/kg mol x 44.911 kg mol
Qrxn1= -1.181x107 kJ/hr
Heat evolved from reaction 2,
No. of moles of NO entering = 103.6 kg mol
No. of moles of NO leaving = 1.036 kg mol
No. of moles of NO consumed=(103..6 – 1.036)kg mol=102.564 kg mol
42
Chapter: 04
Energy Balance
Now form reaction 2,
Qrxn2 = -149000 kJ/kg mol x 102.564 kg mol
Qrxn2= -1.528x107 kJ/hr
Total heat evolved during the reaction,
Qrxn= Qrxn1+ Qrxn2 = -2.709x107 kJ/hr
SENSIBLE HEAT:
At T= 25˚C
CpNO2 = 37.04 kJ/kg mol-˚C
CpNO = 29.7 kJ/kg mol-˚C
CpO2 = 29.39 kJ/kgmol-˚C
CpN2 = 29.1 kJ/kgmol-˚C
CpHNO3 = 110 kJ/kgmol-˚C
CpH2O = 74.88 kJ/kgmol-˚C
Datum temperature = 0C = 273 K
Stream 1:
Streams
Mol%
Cp “kJ/kg mol ˚C”
Cpmix “k J/kg mol˚C”
NO2
10.87
37.04
0.1087*37.04
NO
4.85
29.7
0.0485*29.7
O2
7.43
29.39
0.0743*29.39
N2
76.83
29.1
0.7683*29.1
30.01 kJ/kgmol-˚C
Inlet temperature = 25˚C
43
Chapter: 04
Energy Balance
ń = 1001.22745 kgmol/hr
Q1= ńCpmixΔT
Q1 =7.14x105kJ/hr
Stream 2:
Streams
Wt %
Cp “kJ/kgmol-˚C”
Cp/Mr “kJ/kg˚C”
Cpmix “kJ/kg˚C”
HNO3
35
110
1.746
0.40x1.746
H2 O
60
74.88
4.16
0.60x4.16
3.1944 kJ/kg˚C
Inlet temperature = 25˚C
m = 23071.14 kg/hr
Q2= mCpmixΔT
Q2 =1.84x106kJ/hr
Stream 3:
Streams
Mol %
Cp “kJ/kg mol ˚C”
Cpmix “kJ/kg mol C”
NO2
00.14
37.04
0.0014x37.04
NO
0.18
29.7
0.0018x29.7
O2
01.5
29.39
0.015x29.39
N2
98.13
29.1
0.987x29.1
29.06 kJ/kgmol-˚C
Outlet temperature = 45˚C
ń = 746.05 kgmol/hr
Q3= ńCpΔT
44
Chapter: 04
Energy Balance
Q3 =4.34x105kJ/hr
Stream 4:
Streams
Wt %
Cp “kJ/kgmol-˚C”
Cp/Mr “kJ/kg˚C”
Cpmix “kJ/kg˚C”
HNO3
60
110
1.746
0.60x1.746
H2 O
40
74.88
4.16
0.40x4.16
2.7116 kJ/kg˚C
Outlt temperature = 50˚C
m = 31212.81(50-25) kg/hr
Q4= mCpmixΔT
Q4 =2.11x106kJ/h
Now,
Qin = Q1 + Q2 = 2.55x106 kJ/hr
Qout = Q3 + Q4 = 2.541x106 kJ/hr
ΔQ =Qout-Qin= 10000 kJ/hr
Heat transferred to the cooling coils,
Q= Qrxn- ΔQ = 2.709x107 kJ/hr - 10000 kJ/hr = 2.69x107 kJ/hr
Mass flow rate of water through coils:
Inlet temperature of water in cooling coils = 25˚C
Outlet temperature of water from cooling coils = 80˚C
Change in temperature = ΔT = 55˚C
Heat removed by the cooling medium = Q = 2.69x107 kJ/hr
CpH2O = 4.14 kJ/kg˚C
45
Chapter: 04
Energy Balance
Q= ṁ CpΔT
ṁ=32.8 kg/s;
As there are total 15 coils in absorption column,
So,
Mass flow rate through one coil, ṁ=2.18 kg/s
46
Chapter: 04
Energy Balance
ENERGY BALANCE OF HEAT EXCHANGER (E06)
Cold fluid
Process gases from cooler/condenser
Inlet = 25˚C
Outlet=50˚C
Composition
NO=83.785kgmol/hr = 17.12%
N2 = 450.28kgmol/hr = 82.81%
Cp(mix)=6.94 Kcal/kgmol˚C
Q= n Cp ΔT
= 540.013 x 6.94 x (50-25)
= 94039.25 Kcal/hr
Hot fluid
Gases leaving compressor
Inlet =73˚C
Composition:
N2=23.56 mole%
NO2=69.7 mol%
O2=6.74 mol%
Cp(mix)=7.98 Kcal/kgmol˚C
Q = n Cp ΔT
89100.99 = 1050.45 x 7.98 x (73 – T)
T = 61.78˚C
47
Chapter: 04
Energy Balance
ENERGY BALANCE OF PARTIAL CONDENSER:
Heat duty:
NO=0.1346 4.714 2.2046 3600=5035.78 lbm/hr
NO2=67.343 lbm/hr
N2=24490.513lbm/hr
H2O=7815.56 lbm/hr
Refrence temperature=77oF
Enthalpies or heat rates carried:
Q1 for NO=380705.1373 Btu/hr
Q2 for NO2 =6151.8095 Btu/hr
Q3 for N2=2005773.01 Btu/hr
Q4 for H2O =6510364.77 Btu/hr
Total heat load without condensation load
∑Q=Q1+Q2+Q3=2392629.962 Btu/hr
If the mixture is cooled up to 50oF below the saturation temperature of steam then
Q5=∑mCp∆T=6008.159 50 oF
=300407.9847Btu/hr
Total duty of condenser
Qc=Q4+Q5
=6810772.75 Btu/hr
Amount of coolant Water (treated)
Temperature limit 77 oF to 203 oF
Qc = ∑mCp∆T
m=54053.75 lbm/hr
48
Chapter: 05
Equipment Design
Chapter: 05
Equipment Design
REACTOR
Reactor is a container for chemical reaction and is the heart of reaction process.
CHOICE OF REACTOR TYPE
The choice of reactor is dictated by the process conditions, the type of reaction and the
mode of catalyst exposition. The oxidation of ammonia is a gas-solid reaction of catalytic
type. Since the catalyst is in the form of wire gauze, so a fixed bed reactor is best suited.
The reaction is exothermic and the heat of reaction is to be conserved as it will be used to
supply power required for compression, to generate steam and to preheat air, hence the
mode of operation will be adiabatic.
The reactor is classified as a heterogeneous, catalytic, shallow fixed bed, adiabatic, down
flow reactor. Most of the reactors employed in nitric acid industry are tapered along with
700 angles.
CHOICE OF CATALYST
Catalysts which increase the rate of oxidation of ammonia to nitrogen oxide (NO) include
platinum, its alloys with metals of platinum group, oxides of iron, manganese, cobalt etc.
The activity of platinum and platinum alloy catalysts is higher than that of others. Nonplatinum catalysts are cheaper but they are less active and unstable. For these reasons at
most of plants where HNO3 is made from ammonia, platinum catalysts are used.
90% Pt with 10% Rh is selected as catalyst for this reaction because it can operate at
high temperatures, maximum yield is obtained and the contact time can be accurately
controlled.
49
Chapter: 05
Equipment Design
SHAPE OF CATALYST
The platinum-rhodium catalyst is employed in the form of gauze made of fine wire. For
oxidation at atmospheric pressure, 2-4 layers of 80 mesh gauze are used. At higher
pressure, 20-30 layers are used.
Since we are operating at atmospheric pressure, so we use 2-3 layers of 80 mesh gauze,
0.06mm wire diameter with 1024 holes per cm2 .The free space is 69.8%
The gauzes are stacked in a fixed bed assembly and operate under bulk mass transfer
control.
CATALYST POISONING
Even minute traces of some materials are such serious poisons for platinum catalysts that
extreme care must be taken to eliminate them. Dusty air and pipe line scale may poison
or damage the catalyst and must be avoided. The effect of various poisons is shown in
table.
POISON
CONC. BY VOLUME
DECREASE IN EFFICIENCY
SiH4
0.0016%
50%
Fi(CO4)
0.02%
40%
Arsine
0.0024%
10%
Lead
0.2%
Soaps
5%
0.0045%
50
Chapter: 05
Equipment Design
DESIGN OF REACTOR (salient features)
The reaction is so rapid that the amount of catalyst required is very small and heat
transfer is not feasible. The reaction is limited by mass transfer and the design of screen
packs has been based on pilot plant and plant experience.
A major issue in design is arrangement of layers of screens and supports in a manner that
will assure good distribution of flow. The L/D ratio of the bed is very small i.e(6.25x103
). Poor distribution not only reduces capacity but also causes hot spots in exothermic
reactions that can volatize metal catalyst. Excessive loss of a precious metal catalyst in
this manner is, of course, intolerable. The distributor installed after the Pt/Rh gauze
restricts the flow enough so that the fluid is spread evenly over the Pt/Rh gauze. The
porous pack is made of nichrome and has randomly oriented pores. Bt using an efficient
distributor of this type, savings in expensive precious metal are realized. This not only
reduced Pt inventory (from 2 troy oz per daily ton HNO3 to 0.8) but also reduced the
amount of Pt subject to volatization.
The loss of platinum has a significant impact on the cost of producing HNO3.
Thus its recovery was an incentive for the development of Pt recovery process. The most
efficient recovery process is the use of a Pd-rich alloy gauze located immediately below
the oxidation gauze which captures the Pt species. The overall reaction may be
represented by
PtO2 + Pd
PtPd + O2
Recovery of Pt involves the dissociation of PtO2 to its respective elements and formation
of an alloy between Pt & Pd. Early recovery gauzes contained 80%Pd and 20%Au and
were capable of recovering about 35-40% of the Pt involved. More recently, the Pd
content has been increased and it has become recognized that gauze geometry
optimization leads to increased recovery efficiencies. Current recovery systems can be
designed specifically for each plant. These operate in the mass transfer limited regime
providing recoveries up to 80%.
51
Chapter: 05
Equipment Design
REACTIONS
NH3 + 5O2
4NO + 6H2O
The efficiency of reaction is 98%.The major competing reaction yields nitrogen and is
represented as;
4NH3 + 3O2
2N2 + 6H2O
INLET STREAM;
Kgmol/hr
%
NH3=92.59
02
14.0
=119.4108
18.05
N2 =449.3680
67.95
INLET TEMPERATURE;
T=2000C = 473K
INLET PRESSURE;
P = 1atm =1.01 bar
INLET MOLAR FLOW RATE;
n = 661.35 kgmol/hr
VOLUMETRIC FLOW RATE AT 1atm & 150C
PV = znRT
Mole fraction
NH3
O2
N2
0.14
0.1805
0.679
Tc
405.16K
154.2K
125.8K
Pc
112.8bar
50.43bar
34.00bar
(Ref:
)
52
Chapter: 05
Equipment Design
Tc (mix) = (0.14x405.16)+(0.1805x154.2)+(0.6795x125.8)
=56.722+27.83+85.48
=170.033K
Pc (mix) = (0.14x112.8)+(0.1805x50.43)+(0.6795x34)
=15.792+9.1025+23.103
=47.9975 bar
Reduced temperature =
T/Tc
= 473/170.033
= 2.7818
Reduced pressure
=
P/Pc
= 1.01/47.9975
=0.02104
z =1
(from graph)
V = znRT/P
R=0.08314 m3 bar/ Kmol K
V = 1x661.35 x.0.08314 x288 /1.01
= 15.68x103 m3/hr
Contact time= Volume of catalyst/Volumetric flow rate at 1atm & 150C
Contact time= 0.03 sec
= 8.3x10-6hr (from Pak-Arab fertilizer)
Volume of catalyst = contact time x Volumetric flow rate
53
Chapter: 05
Equipment Design
=8.3 x10-6 x 15.68 x10-3
= 1.3014 x 10-1 m3
Density of 10%Rh, 90%Pt alloy = 20150 kg/m3
% free space in the gauze = 69.8%
Density of gauze = 20150(1-0.698) = 6085.3 kg/m3
Weight of gauze = 6085.3 kg/m3 x 1.3014x 10 -1 m3
= 7.919 kg
After applying safety factor = 1.2 x 7.919
= 9.5 kg
Losses;
Operating days = 330
Losses per ton = 0.05g/ton HNO3 (100% basis)
Catalyst losses = 0.05g/ton HNO3 x 140 ton HNO3 x 330
= 2310.5 g
= 2.310 kg
Total weight of catalyst = 9.5 + 2.310
= 11.81 kg
Inlet volumetric flow rate to convertor;
V= znRT/P
n = 661.35 kg mol/hr
R = 8.314 KPa m3 /kgmol K
T = 473 K
P = 101.325 KPa
Z =1 (from graph)
54
Chapter: 05
Equipment Design
V = 1x 661.35 8.314x 473 / 101.325
=25667.63 m3 /hr
= 7.1298 m3 /sec
INLET PIPE DIAMETER OF CONVERTOR;
D
i,opt
=0.363 mv
0.45
p
0.13
K.D;”Plant design & Economics for Chemical
(ref; Peters,M,S. & Timmerhaus,
Engineers ;” 5th Ed.,Mc. Graw Hill.)
Where,
D i,opt = optimum pipe dia in meter
mv
p
= volumetric flow rate in m3/sec =7.1298 m3/sec
= fluid density in kg/m3
R=8.314 KPa m3/kgmol K
P=101.3 KPa
Mav=(17 x 0.144) + (32 x 0.1805) + (28 x 0.6795)
= 27.25
Z=1
p = PM/zRT
=101.3 x 27.25/1 x 8.314 x 473
=0.7019 kg/m3
D i,opt = 0.363 x (7.1298)0.45x (0.7019)0.13
= 0.839m
55
Chapter: 05
Equipment Design
Composition of exit gas;
Component
kgmol/hr
%
NO
95.59
13.43
O2
4.6115
0.66
N2
450.28
65.34
H2 O
141.575
20.54
Component
Mole fraction
Critical Temperature (K)
Critical pressure(Bar)
N2
0.6534
126.2
34
O2
0.0066
154.6
50.43
NO
0.1343
180.2
64.8
H2 O
0.2054
647.1
220.55
Mav =(28 x 0.6534) + (32 x 0.0066) + (30 x 0.1343) + (18 x 0.2054)
= 26.2326
Tc(mix) = (0.6534 x 126.2) + (0.0066 x 154.6) + (0.1343 x 180.2) + (0.2054 x 647.1)
= 240.59 K
Pc(mix) = (0.6534 x 34) + (0.0066 x 50.34) + (o.1343 x 64.8)+ (0.2054 x 220.55)
= 75.52 bar
Pr = 1.01/75.52 = 0.01337
56
Chapter: 05
Equipment Design
Tr = 1143/240.59 = 4.75
Z=1 (from graph)
n = 661.35 kgmol/hr
R = 0.08314 m3 bar/kgmolK
V = znRT/P
= 1 x 661.35 x 0.08314 x 1143 / 1.01
= 62225.19 m3/hr =17.28
m3/sec
OUTLET PIPE DIAMETER OF CONVERTOR;
D i,opt =0.363 mv 0.45 p 0.13
P = 101.3 kPa
T = 1143 K
R= 8.314 KPam3/kgmolK
p =PM/Zrt
= 101.3 x 26.23 / 1 x 8.314 x 1143
= 0.2796 kg/m3
mv = 17.28 m3/sec
D i,opt =0.363 (17.28) 0.45 (0.2796) 0.13
= 1.1087 m
57
Chapter: 05
Equipment Design
DIAMETER OF REGION WHERE GAUZE IS PRESENT;
D i,opt = 1.1087 m is based on empty pipe
As free space in gauze is 69.8% , so diameter of convertor is given as;
D = 1.1087/0.698
= 1.588 m
L/D for catalyst bed = 6.25 x 10-3 ( Ref; Howard F.Rase,”Chemical Reactor design for
Process plants Vol.1)
Catalyst screen depth = L = 6.25 x 10-3 x 1.588
=9.83 x 10-3 m
SKETCH
NH3+air mixture
D1
D2
Random pack
Pt/Rh gauze
Pd/Au getter
Support screen
NO gas+ H2O
58
Chapter: 05
Equipment Design
HEIGHT OF CONVERTOR;
As top and bottom of convertor is cone shaped ,
As
D1=0.818m
D2=1.574m
D2 - D1=1.574 - 0.818
=0.756
a1 = 0.756/2 = 0.378
NOW
As θ = 700
Tan θ =h1/a1
h1
= a1 tan θ
= 0.378 tan 70
= 1.0385 m
NOW
D2 = 1.574m
59
Chapter: 05
Equipment Design
D3 = 1.0985m
D2 – D3 = 1.574 – 1.0985
= 0.4755
a2 = 0.477/2 = 0.23775
Now
tan θ = h3 / a2
h3
= a2 tan θ
= 0.23775 tan70
= 0.6532 m
As catalyst screen depth = L = 9.83*10-3 m
60
Chapter: 05
Equipment Design
SO
Total height of convertor =
h1 +h2 + L
= 1.0385 + 0.6532 + 9.83 x 10-3
=
1.7015 m
VOLUME OF CONVERTOR;
AS
h1 = 1.0386m
h3 = 1.7015m
h2 = R2 tan θ
= (1.574/2) tan 70
= 2.162 m
V1 = ∏ R12 h1
= ∏ (0.81/2)2 x 1.0385
= 0.5351 m3
V2 = ∏ R22 h2
= ∏ (1.574/2)2 x 2.162
= 4.206 m3
∆ V = V2 –V1
= 4.206 – 0.535
= 3.67 m3
61
Chapter: 05
Equipment Design
V3 = ∏ R32 h3
= ∏ (1.0985/2)2 x 1.7015
= 1.6211 m3
∆ V’= V2 – V3
= 4.206 – 1.6211
= 2.584 m3
As volume of catalyst = 1.2307 * 10 -3 m3
So
Total volume of convertor = V! + V3 + ∆ V/2 + ∆ V’/2 + volume of catalyst
= 0.5351 + 1.6211 + 3.67/2 + 2.58/2 + 1.2307 x 10-3
= 5.3 m3
PRESSURE DROP CALCULATIONS
P = bs ρ u2/g c ε2 ds
where,
= 8.61 ( awR2 ds / ρ u1)
= tortuosity factor = 1
u1= upstream velocity
awR = surface area / volume of screen wire
bs = screen thickness
ds = screen pore diameter
62
Chapter: 05
Equipment Design
Wire diameter = 0.0056 inch
bs = 2 x 0.0056 = 0.0112 inch
N = mesh size = 80
d= wire diameter
L = [ 1/N2 + d2 ]
= 0.01357 inch
awR = π L N2
= 272.84 inch2 /inch3
ds= 0.0069 inch
ε = 1 – π L N2d = 0.618
4
μmix= 0.0048 /3600 lbm/in sec
ρmix = 0.04819/ (12)3 lbm/in3
u1 = 2491.97 in/sec
= 0.0867
u2 = 1326.37 in/sec
u =1909.17 in/sec
∆ P = 0.969 psi
63
Chapter: 05
Equipment Design
SPECFICATION SHEET
Identification:
Item
Reactor
Item No.
Function:
Date: 29-10-2005
R-01
To convert NH3 – air mixture to nitric oxide mixture.
Operation : Continuous
Materials handled:
NH3-air mixture
NO gas
Quantity
25.667x10 m /hr
62.225x103m3/hr
Pressure
1.01 bar
1.01 bar
200oC
870oC
Temperature
Design Data:
3
3
Catalyst
90% Pt, 10% Rh
Weight of catalyst
11.81 kg
Diameter of converter
1.146m
Height of converter
2.135m
Material of construction
Inconel 601
Inlet pipe diameter
0.8m
Controls: Ammonia air ratio control
64
Chapter: 05
Equipment Design
Oxidation Tower
In absorption gas mixture is contacted with liquid solvent which absorbs one or more
components from the gas stream. The absorbed gas and solvent leave at the bottom and
unabsorbed components leave as gas from the top.
Chemical reaction:
The gas NO and HNO3 flow counter currently ,the acid oxidizes NO to NO2 .this
reaction is endothermic .
NO + 2 HNO3 →
3 NO2 + H2O
WHY A PACKED COLUMN ?
•
PACKED COLUMNS
1.
Suitable for high liquid flow rate
2. Pressure drop per equilibrium stage is
lesser than plate towers
3. Replacement of packing is easier and
cheaper
4. Suitable for small diameters
5. Suitable when no inter-cooling and
withdrawal of side-streams is required
6. Should be considered if separation
process involves corrosive fluids.
7. Packed height of less than 6m.
• TRAY COLUMNS
1. Suitable for Larger Diameter
Towers
2. Liquid Distribution is good
3. Repair and Maintenance is difficult
for smaller towers
4. Easier to make provision for intercooling and withdrawal of sidestreams
5. Not advisable to use for corrosive
and toxic liquids
65
Chapter: 05
Equipment Design
Packing Selected : Ceramic Berl Saddles
Comparison of Common Packing Types
RASCHIG RINGS
BERL SADDLES
INTALOX SADDLES
•
Considerable
Side Thrust
•
Lesser Side
Thrust
•
A Fairly
Uniform Bed
•
Less Efficient
•
•
•
Channeling is
favored
•
More efficient
than Rings
Lesser
Channeling than
rings
Very Little
Tendency for
Channeling
•
Smaller HTU
•
Higher HTU
•
Smaller HTU
•
•
Lower Flooding
Point
Higher Flooding
Point
•
High Flooding
Point
•
More Costly
than Berl saddles.
•
Cheap
•
More Costly than
Raschig rings.
OPERATING CONDITIONS
•
Leaving Gas(23631.28kg/hr)
•
•
•
NO =251.85
NO2=10772.28
N2=12607.84
•
Entering Gas(15486.998kg/hr)
•
•
NO= 2513.55
N2=12607.84
Entering Solvent(31510.08kg/hr)
HNO3=60%=19025.08
H2O=40%=12484.8
Leaving Solvent(23217.624kg/hr)
HNO3=35%=9374.72
H2O=13842.9
66
Chapter: 05
Equipment Design
Operating temperature
Inlet temperature = 50 °C
Outlet temperature = 48°C
Design Calculations of Absorption Column
Temperature = 50 °C = 323 K
Pressure
= 1 atm = 101 kPa
Density of 60% HNO3 soln , ρL = 1338.35 kg /m³ (Kirk Othmer Vol 13)
μ of 60% HNO3 soln at 50 °C = 1.5 cp
μ of NO at 50°C = 0.02 cp
Volume of NO ,V= nRT/ P
= 87.5666 x 8.314 x 323
101
=
2227.70 m³/ hr
Density of NO , ρv = 2513.55 kg / hr
2227.70 m³/ hr
= 1.1283 kg / m³
ESTIMATION OF NUMBER OF TRANSFER UNITS ( NOG)
If the reaction is essentially irreversible at absorption conditions ,the equilibrium partial
pressure is zero and NOG can be calculated just from the change in gas composition .
b
NOG = ∫dy/y = ln yb / ya
a
(Unit operations of chemical engr,Mccabe&Smith , 5th ed.)
NOG = 3
ya, mol fraction of NO in leaving gas = 0.012
yb, mol fraction of NO in entering gas = 0.154
67
Chapter: 05
Equipment Design
Using CORNELLS METHOD
Calculating % age flooding
The liquid-vapor flow factor is given by
FLV = (Lw / Vw) x (ρV/ρL)0.5
(Coulson & Richardsons Chemical Engineering,Vol 6)
Where,
Liquid mass flow rate, Lw = 8.75 kg/s
Vapor mass flow rate, Vw = 4.30 kg/s
FLV = 7.723 х (1.1283/ 1338.35 )^0.5
4.026
FLV = 0.056
From graph
K4 at flooding line = 4
From Coulson & Richardsons Chemical Engr, Vol 6,fig 11.44
Recommended design values, 15 ~ 50 mm H2O / m of packing
Design for a pressure of 42mm of H2O / m of packing K4 = 1.5
%age flooding = (K4 at 42mm H2O per m of packing /K4 at flooding line)0.5 x 100
= (1.5 / 4 ) 0.5x 100
= 61%
Estimation of COLUMN DIAMETER:
Calculating gas mass flow rate per unit column cross-sectional area
Vw*= [k4 ρV (ρL – ρV) /13.1FP(μL/ ρL)0.1]0.5
Where,
Fp = Packing Factor
Using Ceramic Berl saddles 51mm = 2in
68
Chapter: 05
Equipment Design
Fp = 35 mֿ¹ for Ceramic Berl Saddles (from unit operations of chemical engineering, 5th
ed, pg 689)
μL = 0.0015 Ns / m² at 50°C
Vw* =[ 1.5x 1.1283x(1338.35 – 1.1283) / 13.1 x 35 x (0.0015/1338.35)0.1]0.5
Vw* = 4.4075 kg / m²s
Column area required = flow rate of NO entering / Vw*
= 4.30/ 4.407
=
Column Diameter
0.975 m²
= (4/π x column area req)0.5
= ( 4 /π x 0.9135 )0.5
= 1.11 m
Column area
= π / 4 x d2
= π / 4 x 1.072
= 0.975 m²
Calculating height of liquid phase transfer unit HL & height of gas phase transfer unit
HG:
Diffusivity of vapor =Dv = 3.17 x 10 -5 m²/sec
Diffusivity of liquid = DL = 1.456 x 10-9 m²/sec
µ of vapor = 0.00002 Ns/m²
μ of liquid = 0.0015 Ns /m²
Gas Schmidt number (Sc)v
69
Chapter: 05
Equipment Design
(Sc)v = μv / ρv x Dv
= 0.00002
1.1283 x 0.0000317
= 0.559
(Sc)L = μL / ρL x DL
= 0.0015
1338.35 x 1.456 x 10 ^-9
= 770
Liquid mass flow rate per unit area = Lw / column area
= 8.75/ 975
= 8.97kg /m² s
From fig at 62 % flooding (fig.11.41)
%age flooding factor K3 = 0.85
From fig at 62 % flooding (fig=11.42)
HG factor Ψh = 82
From fig at Lw*= 8.59
HL factor φh = 0.08
(Figures from Coulson &Richardsons Chem Engr,3rd ed ,Vol 6)
HOG can be expected to be around 1m so as a first estimate Z can be taken as 3m .
Cornell recommended that for design purposes the dia correction term should be taken as
a fixed value of 2.3 for columns above 0.6 m dia .
Using CORNELLS Eqs :
Height of gas phase transher unit HG
HG = 0.011 Ψh x (Sc)v 0.5 x (Dc/0.305)1.11 x (Z / 3.05)0.33 /Lw* 0.5
70
Chapter: 05
Equipment Design
=0.011 x 82 x 0.5590.5 x 2.3 x (3/3.05)0.33
8.59 0.5
= 0.5448m
Height of liquid phase transfer unit HL
HL = 0.305 x Φh x (Sc)L0.5 x K3 x (Z/3.05)0.15
= 0.305x 0.09x 770 0.5 x 0.85 x (3/3.05)0.15
= 0.646 m
ESTIMATION OF HOG:
Height of overall gas phase transfer unit :
HOG = HG +( m Gm/ L ) x HL
mGm / L optimum value of this term as suggested by Colburn will lie b/w 0.7 – 0.8
using average 0.75
HOG = 0.5448 + 0.75 x 0.646
HOG = 1.03 m
Height of column Zt
Height of packing Z
Z = NOG x HOG
= 3 x 1.03
Z = 3.09 m
Giving additional 0.5m height on top and bottom of packed bed.
TOTAL HEIGHT OF COLUMN = Zt = 4.09 m = 13 ft
PRESSURE DROP CALCULATION :
71
Chapter: 05
Equipment Design
Pressure drop can be calculated from the correlation;
ΔP = α x 10βLx (G2/ρg)
2
Where α and β are the factors related to the packing material ceramic berl saddle of
51mm.
α = 0.16
β = 0.12
L = liquid mass velocity = 1.76 lbs /ft 2 s
G = gas mass velocity = 0.917 lbs / ft sec
ΔP = 3 in of water = 0.008 atm
So, total pressure drop due to packing is = 0.008 x 3.09 = 0.025 atm
ΔP due to packing support and distributor= 2x0.00073atm
Total pressure drop across column = 0.025 + 0.00073x2
= 0.026 atm
LIQUID HOLD UP :
The amount of liquid holdup is given by the following formula,
Hw= 0.1434 x ( Lw* / d) 0.6
Lw*= 8.6 kg/m²s
d = equivalent diameter of packing = 51mm
Hw = 0.1434 x (8.6 /51)0.6 = 0.05 m³liq / m³bed
72
Chapter: 05
Equipment Design
SPECIFICATION SHEET :
Unit:
Function:
Operating pressure:
Diameter of the column :
Number of Transfer Units:
Height of Transfer Unit :
Height of Packing :
Height of Column :
Packing :
Total Pressure Drop :
Distributor:
Material of construction
Packing Support
Absorber
Absorption of NO in HNO3
1.0 atm
1.11
3
1.03m
3.09m
4.09 m
0.051 m Ceramic Berl Saddles
0.026 atm
Orifice type
Stainless Steel
Ceramic grid plate
73
Chapter: 05
Equipment Design
Absorber
This Section covers the following topics:
Introduction
Choice between Plate and Packed Column.
Choice of Plate Type.
Design Steps of Absorption Column
Design Calculations of Absorption Column
Specification sheet of Plate Column
References.
74
Chapter: 05
Equipment Design
INTRODUCTION:
TYPES OF GAS ABSORPTION:
Physical Absorption
Chemical Absorption (Gas Absorption With Chemical Reaction)
GAS ABSORPTION WITH CAHEMICAL REACTION:
In gas absorption with chemical reaction a soluble vapor is absorbed chemically
from its mixture with an inert gas by means of a liquid in which the solute gas is more or
less soluble.
Advantages:
Equilibrium partial pressure of the solute over the solution reduces which greatly
increase the driving force for mass transfer.
Increase in mass transfer coefficient.
Height of absorption tower reduces.
More economical.
FUNCTION OF THIS ABSORPTION COLUMN:
The main function of this column is to reduce the concentration of NOx leaving
from the superaziotropic column from 2000 ppm (which is harmful for the environment)
to 300ppm and to improve the economics by concentrating the acid which is coming from
oxidation tower from 35% to 60% which is then recycled to the oxidation tower.
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Chapter: 05
Equipment Design
TYPES OF ABSORPTION COLUMN:
Tray Column
Packed Column
CHOICE BETWEEN PACKED COLUMN PLATE AND:
Vapor liquid mass transfer operation may be carried either in plate column or packed
column. These two types of operations are quite different. A selection scheme
considering the factors under four headings.
i)
Factors that depend on the system i.e. scale, foaming, fouling factors, corrosive
systems, heat evolution, pressure drop, liquid holdup.
ii)
Factors that depend on the fluid flow moment.
iii) Factors that depends upon the physical characteristics of the column and its
internals i.e. maintenance, weight, side stream, size and cost.
iv) Factors that depend upon mode of operation i.e. batch distillation, continuous
distillation, turndown, and intermittent distillation.
The relative merits of plate over packed column are as follows:
i)
Plate column are designed to handle wide range of liquid flow rates without
ii)
For large column heights, weig flooding.
iii)
If a system contains solid contents, it will be handled in plate column, because
solid will accumulate in the voids, coating the packing materials and making it
ineffective.
iv)
Dispersion difficulties are handled in plate column when flow rate of liquid
are must be removed before cleaning.
v)
For non-foaming systems the plate column is preferred.
76
Chapter: 05
vi)
Equipment Design
Design information for plate column is more readily available and more
reliable than that for packed column low as compared to gases. ht of the
packed column is more than plate column.
vii)
If periodic cleaning is required, man holes will be provided for cleaning. In
packed columns packing.
viii)
Inter stage cooling can be provide to remove heat of reaction or solution in
plate column.
ix)
When large temperature changes are involved as in the distillation operations
tray column are often preferred because thermal expansion or contraction of
the equipment components may crush the packing.
x)
Random-Packed Column generally not designed with the diameter larger than
1.5 m and diameters of commercial tray column is seldom less than 0.67m.
Type of Column Selected:
On the basis of merits and demerits of both types of columns, I have selected the
“Tray Column” for my system because:
System is non-foaming
Diameter is greater than 0.67 meter.
Height of the Column is very large.
77
Chapter: 05
Equipment Design
TYPES OF TRAYS:
Sieve trays
Bubble cap trays
Valve trays
Downcomerless trays
SELECTION CRITERIA ON THE BASIS OF TYPE OF TRAYS:
Type of tray
Capacity
Efficiency
Cost/area
Flexibility
Remarks
(Turndown ratio)
1. Sieve
2.Bubble Cap
Med – High
High
Better than
bubble cap
As good or
better than
bubble cap
Low – Med.
Med. – High
3.Valve
Ballast
Flexi tray
Float valve
Med. – High
High
As good as
sieve trays
As good as
sieve trays
Lowest of all
trays with
downcomers
Med. or Good
2/1 (3/1 can usually
be achieved)
High
Med. – Good
About twice the
cost of sieve trays
3/1 or slightly
higher
Will be used in most
application
Use for higher
flexibility or dirty
service
Esp. recmd. when high
flexibility is required
Med.
Good
but some care must be
10 – 20% greater Possibly up to 5/1
taken about
than sieve
maintenance problems
Hi-contact valve
4.Downcomerless
Dual flow
Ripple
Highest in
Med. – Good
some instances
Lowest – Med.
Low efficiency
Of interest for
bottleneck removal if
poor flexibility
tolerable
Turbo grid
78
Chapter: 05
Equipment Design
Type of Tray Selected:
I have selected sieve tray because:
They are lighter in weight and less expensive.
It is easier and cheaper to install.
Pressure drop is low as compared to bubble cap trays.(1/3 of bubble cap)
Maintenance cost is reduced due to the ease of cleaning.
SCHEMATIC DIAGRAM:
79
Chapter: 05
Equipment Design
DESIGN STEPS OF SIEVE TRAY ABSORPTION COLUMN
FROM RICHARDSON& COULSON VOL-6 (3rd Ed) P=458
Calculation of theoretical no. of stages.
Calculation of actual no. of stages.
Calculation of diameter.
Liquid Flow Arrangement.
Trial Plate Layout.
Calculation of weeping rate.
Calculation of plate pressure drop.
Calculation of downcomer back-up.
Calculation of downcomer residence time.
Calculation of no. of holes.
Checking of entrainment.
Calculation of height.
80
Chapter: 05
Equipment Design
Operating Conditions
T3=45oC, P3=8.62 bar
ρv=9.8 kg/m3
Q3=5.4x105kJ/hr
mol% kgmol/hr kg/hr
3
NO2 O.14 1.036
47.656
NO 0.186 1.389
41.67
2
O2
1.54 11.49
367.68
N2 98.13 20928.93732.14 20471.92
746.06
T1=25oC, P1=9 bar
ρv=11.02 kg/m3
Q1=7.06x105kJ/hr
mol% kgmol/hr
kg/hr
NO2 10.87 103.6
4765.6
NO 4.85 46.3
1389
O2
7.45 70.82
2266.24
N2 76.84 732.84
20499.92
952.86
28920.76
2
1
4
T2=25oC, P2=10 bar,
ρL=1209.8 kg/m3
Q2=1.8x106kJ/hr
wt%
kg/hr
HNO3 40 9228.24
H2O 60 13842.9
23071.14
T4=50oC, P4=9.7 bar
ρL=1360 kg/m3
Q4=2.12x106kJ/hr
wt%
kg/hr
HNO3 60 16634.5086
H2O
40 11128.5130
27763.0216
CHEMICAL REACTIONS:
1. 4NO 2 O 2 2H 2 O 4HNO 3
ΔH r 25C 149000kJ/kgmol
2. 4NO 3O 2 2H 2 O 4HNO 3
ΔH r 25C 263000kJ/kgmol
The overall reaction,
3. 3NO 2 H 2 O 2HNO 3 NO
ΔH r 25C 46000kJ/kgmol
81
Chapter: 05
Equipment Design
Number of plates calculation:
The equilibrium data for the solution is determined by the following relation:
Ye = .6X
(1)
X .02 .04 .06 .08 .10 .12 .14
Ye .012 .026 .04 .048 .06 .07 .081
This data is for equilibrium curve. Now applies the overall balance on the absorber:
Lm(X1-X2) = Gm (Y1-Y2)
As there is no concentration of NO2 in inlet liquid stream, so X2 = 0
Now the above equation becomes:
Lm(X1-0) = Gm (Y1-Y2)
LmX1 = Gm( Y1-Y2 )
Y1= LmX1/Gm +Y2
This is Eq. for operating line
Lm = Liquid flow rate/area
Lm = 136.54 Kg mol/hr m2
Gm = Gas flow rate/area
Gm = 148.85 Kg mol/hr m2
Now equation (2) becomes;
Y1 = .92X1 + .001
(3)
Data for the operating line can be obtained by letting the suppose value of X1 as given
below:
X1 .02 .04 .06
.08 .10 .12 .14
Y1 .019 .038 .056 .075 .093 .11 .129
Now draw the operating line and equilibrium curve on the graph in order to find the
number of stages as given in figure.
Number of theoretical trays find from the graph are 6
82
Chapter: 05
Equipment Design
83
Chapter: 05
Equipment Design
Now calculate actual No. of trays:
Nactual = NO. Of Theoretical trays/ E
Efficiency of column is finding by the following relation:
By drichmer and bradform relation:
E = 17 – 61.1Log (μFavg)
μFavg = Viscosity of feed at average temperature and pressure
E = 17– 61.1Log (0.23)
= 55%
E = 0.55
Now, actual number of trays (Nactual) is:
N actual = Theoretical trays/E
= 6/0.55
N actual = 10
CALCULATION OF COLUMN DIAMETER:
On the basis of flow rates by following the method given in Chemical
Engineering, Coulson & Richardson, Vol.6 (3rd edition)
At bottom:
Maximum liquid flow rate at bottom:
Liquid mass flow rate = Lw = 31212.81 kg/hr = 8.67 kg/s
ρL at 50˚C = 1.312 g/cm3 = 1312 kg/m3
Maximum vapor flow rate at bottom:
Vapor mass flow rate = Vw = 28920.76 kg/hr
Vapor molar flow rate = Vm = 952.86 kgmol/hr
Average molecular weight of inlet gaseous stream:
Mr(avg) = 28920.76Kg /952.86 =30.35
P = 9 bar = 900 kpa
T= 25˚C = 298 K
Pv = PMr/RT = 900 *30.35 /8.314 * 298 = 11.02
Now from Coulson & Richardson vol 6 the vap-liquid factor is calculated as:
84
Chapter: 05
Equipment Design
FLV = Lw/Vw (Pv/PL)1/2 = 31212.81/28920.76 (11.02/1312)1/2 = 9.7 * 10-2
By selecting plate spacing=0.6m
Now from figure 11.27 pg#567, Coulson & Richardson vol. 6 (3rd edition), value of
K1=0.11
Flooding velocity;
Uf K 1
ρL ρ V
1360 11.08
0.11
1.22m/s
ρV
11.08
Superficial velocity:
Design for 85% flooding at maximum flow rate,
Û v U f 0.85 1.037 m/s
Maximum volumetric flow rate=Vv = Vw/(ρvx3600)=0.73m3/s
Net area required=(volumetric flow rate/superficial velocity) = Vv/Ǔv
An=0.70m2
By taking downcomer area as 12% of the total column cross sectional area.
Column cross sectional area:
Ac= 0.70/0.88 = 0.80m2
Column Diameter:
Dc
Ac 4
1m
Π
At top:
Maximum liquid flow rate at top:
Liquid mass flow rate = Lw = 23071.14 kg/hr = 5.359 kg/s
ρL at 25˚C = 1.2098 g/cm3 = 1209.8 kg/m3
Maximum vapor flow rate at bottom.
85
Chapter: 05
Equipment Design
Vapor mass flow rate = Vw = 20928.93 kg/hr
Vapor molar flow rate = Vm = 746.06 kgmol/hr
Average molecular weight of inlet gaseous stream =
Mr(avg) = 20928.93/746.06 = 28.05 Kg/Kgmol
T= 40˚C = 313 K
Pv = PM/RT= 864.2 * 28.51 /8.314 *298 =9.8 Kg/m3
Vapor liquid flow factor,
FLv = Lw/Vw (Pv/PL) = 23071.14/20298.93 (9.3/1209.8)1/2 = 9.6 * 10-2
By selecting plate spacing = 0.6m
Now from figure 11.27 pg#567, Coulson & Richardson vol. 6 (third edition), value of
K1=0.1
Flooding velocity:
Uf K 1
ρL ρ V
1209.8 9.8
0.1
1.106m/s
ρV
9.8
Superficial velocity:
Design for 85% flooding at maximum flow rate,
ˆ V 0.85 1.106m/s 0.85 0.9401m/s
U
v
v
Maximum volumetric flow rate=Vv = Vw/(ρvx3600) = 0.59 m3/s
Net area required = (volumetric flow rate/superficial velocity) = Vv/Ǔv
An = 0.63m2
By taking downcomer area as 12% of the total column cross sectional area.
Column cross sectional area:
Ac= 0.67/0.88 = 0.72m2
86
Chapter: 05
Equipment Design
Column Diameter:
Dc
Ac 4
1m
Π
So it can be concluded from the calculation at the top that the diameter of the
column at bottom and at the top is same and is found out to be 1 m.
LIQUID FLOW ARRANGEMENT:
Maximum volumetric flow rate at bottom of the column = Lw/(ρLx3600)
= 6.38x10-3 m3/sec
Maximum volumetric flow rate at top of the column = Lw/(ρLx3600)
= 5.30x10-3 m3/sec
Now from the figure 11.28 pg#568, Coulson & Richardson vol. 6 (3rd edition), for
1m dia. column at the given flow rates, required pattern is “Cross Flow (single pass)”
PROVISIONAL PLATE DESIGN:
Diameter of column 1m
D2
Column cross sectional area A c Π
0.785m 2
4
Downcomer Area:
Downcomer area is 12% of cross sectional area;
87
Chapter: 05
Equipment Design
Downcomer area A d 0.12 A c 0.12 0.785 0.0942m 2
Net Area:
Net area A n A c A d 0.785 0.0942 0.6908m 2
Active Area:
Active or bubbling area A a A c 2A d 0.785 2(0.0942) 0.5966m 2
Hole Area:
Let take hole area 10% of Aa as first trial ;
Hole area A h 0.1 A a 0.1 0.5966 0.05966m 2
Weir height:
Weir Height h w Take weir height 50mm
Weir length:
Ad
100 12%
Ac
From Graph b/w (Ad/Ac)*100
vs. lw / Dc on page # 572 by “Coulson and
Richardson’s”, volume 6 (3rd edition)
lw
0.77
Dc
l w 0.77 Dc 0.77 1m 0.77m
Hole size:
Take hole diameter = 5mm
Plate thickness:
Plate thickness = 3mm (for austentic stainless steel 304L)
88
Chapter: 05
Equipment Design
WEEP POINT:
Weir Liquid Crest:
Maximum liquid rate=Lw(max) = 7.712 kg/sec
Take minimum feed rate as 70% of maximum fee rate of turn down ratio 70%
Minimum liquid rate= Lw(min) = 7.712x0.70 = 5.3984 kg/s
how max
Lw
750 max
ρL w
how min
Lw
750 min
ρL w
2/3
2/3
7.712
750
1360 0.77
5.3984
750
1360 0.77
2/3
28.4mm
2/3
22.4mm
At minimum liquid rate,
hw + how(min) = (50 + 22.4) mm = 72.4 mm
From graph 11.30, page # 571,”Coulson and Richardson” Vol. 6. At, hw + how = 72.4 mm,
the value of,
K2 = 30.6
The minimum design vapor velocity is given by;
[K 0.9(25.4 dh )] [30.6 0.9(25.4 5)]
Uh 2
3.7 m/s
(ρ v )1/2
11.08
Actual minimum vapor velocity:
Actual minimu, vapour velocity = minimum vapour rate /Ah
= .7 * 28920.76 kg/hr /3600 s/hr 8 11.02 kg/m3 /
.0596 m2 = 8.52m/s
As, actual minimum velocity is greater than design vapor velocity, so the
minimum operating rate will be well above the weep point,
89
Chapter: 05
Equipment Design
PLATE PRESSURE DROP:
Dry Plate Pressure Drop:
Maximum vapor velocity through holes
Maximum vapour velocity = maximum vapour rate /Ah
28920.76 Kg/hr /3600 s/hr * 11.02 kg/m3/.0596m2
=12.84m/s
Plate thickness 3mm
0.6
Hole diameter
5mm
Ah
A
100 h 100 0.1 100 10
Ap
Aa
Now,
From figure 11.34, 6th Ed. “Coulson and Richardson’s” ,At (Ah/Aa)*100=10,
When Plate thickness to plate diameter ratio is 0.6;
Then, Co = 0.74
Dry plate drop:
hd =51(Uhmax /Co)2 Pv/PL =51(12016/.74)2 *11.02/1360 =111.59
Reisdual Drop:
12.5 10 3 12.5 10 3
hr
9 mmliq
ρL
1360
Total Plate Pressure Drop:
ht = hd + (hw + how max ) + hr = (111.5 + (50 + 2804 ) + 9 = 198.2 mm
Note:
200mm liquid was assumed to calculate the top pressure 212.5 mm is considered
acceptable.
90
Chapter: 05
Equipment Design
DOWNCOMER LIQUID BACKUP/LIQUID HEIGHT IN DOWNCOMER:
Let,
hap= hw-10mm = 50-10mm = 40 mm = 0.04m
Clearance area under the downcomer;
Area under apron = hap*lw
Aap = (0.04*0.77)m2 = 0.0308m2
As, Aap is less than “Ad = 0.0942 m2”
Am = Aap = 0.0308 m2
Head loss in the downcomer:
2
L
7.712
hd 166 wd 166
5.6mmliq
1360 0.0308
ρL A m
2
In term of clear liquid down comer backup
hb = ht + (hw + how max ) + hdc = 198.5 +(50 +28.4 ) 5.6 =282.5 mmliq
Now,
1/2 (plate spacing + wier height ) > ho
½(.6m + .05m ) > .282m
.325m > .282m
So, tray spacing is acceptable.
RESIDENCE TIME:
Tr = AdhbcPL /Lwd = .0942m2 * .282m * 1360Kg/m3 /7.712Kg/s =4.7s > 3s
(satisfactory)
As residence time is greater than 3 sec, so satisfactory
91
Chapter: 05
Equipment Design
NO. OF HOLES:
ΠD2
1.9634 10 5 m2
4
Total hole area
0.05966
Total no. of holes
Area of one hole 1.9634 10 5
N 3038 no. of holes
Area of one Hole
ENTRAINMENT:
Percentage flooding
Uv
100 (Uv based on net area)
Uf
Volumetric flow rate Vv 0.7659m 3 /s
Uv
Vv 0.7659m 3 /s
1.11m/s
An
0.6908m 2
Uf 1.22m/s
Percentage flooding
1.11
100 90%
1.22
FLV 8.2 10 2
Now, from figure#11. 29 at page#570 from C & R vol 6, the value of fraxtional entrainmen t is,
ψ 5.1 10 2 0.051 (well below 0.1)
As, entrainment is less than 0.1, process is satisfactory.
HEIGHT OF THE COLUMN:
No. of plates = 10
Tray spacing = 0.6 m
Distance between 15 plates = 0.6 m 10 = 6 m
Top clearance = 1.2 m
Bottom clearance = 1.2 m
Tray thickness = 3 mm/plate
Total thickness of trays = 0.003 m 10= 0.03 m
Total height of column
= (6 + 1.2 + 1.2+ 0.03) m = Ht = 8.43 m
92
Chapter: 05
Equipment Design
PLATE SPECIFICATION SHEET:
Plate No.
1
Plate I.D.
1m
Hole Size
5mm
Hole Pitch lp
12.5mm Δ
Active Holes
3038
Turn Down
70% of max rate
Plate Material
Austentic stainless steel 304L
Downcomer Material
Austentic stainless steel 304L
Plate spacing
0.6m
Plate Thickness
3mm
Plate pressure Drop
2.34 kpa
93
Chapter: 05
Equipment Design
STRIPPING COLUMN DESIGN
STRIPPING
A process in which mass transfer occur from liquid phase to gas phase is called stripping.
It is a unit operation in which solute is separated from solvent on the basis of difference
in volatility in the presence of suitable solvent or passing through a reboiler.The stripping
column contains normal fractionation trays and is designed to strip the gases from the
rich solvent. Feed is entered from the top to the stripping column. The stripping reboiler
heating medium is usually steam from the pressure steam system. The heat input is set to
ensure essentially complete stripping of gases from the rich solvent. The amount of lean
solvent diverted to the stripper feed is determined by stripper selectivity. High
hydrocarbon concentrations in the stripper feed stream make the selective stripping of
gases more difficult. Lean solvent for this purpose may be added from the lean solvent
line. The overhead vapors which contain hydrogen chloride, methyl chloride, and some higher
boiling chloromethane, are sent to scrubber and bottoms, containing methylene chloride,
chloroform, and carbon tetrachloride are to heat exchanger
stream. Component to be
removed from an entering liquid is called solute.
Absorbers and strippers are often used in conjunction with each other. Absorbers are
often employed to remove trace components from gas streams. Strippers are often applied
to remove the trace components from the liquid in a more concentrated form. Absorption
and stripping operations are carried out in vertical, cylindrical columns or towers
containing plates or packing elements. The plates and packing provide a surface area for
the liquid and gas to come into contact facilitating mass transfer between the two streams.
The gas and liquid streams for both operations are commonly counter-current for a more
effective mass transfer. The columns are simpler than those for distillation are because
they commonly do not include a condenser.
The solute in the solvent is removed in the stripping column by a stripping gas that enters
at the bottom of the column. The solvent now exits from the bottom of the column and is
condensed before it is recycled back to the absorption column. The gas exiting the
stripping column can now be stored or processed easier. Absorption and stripping are
conducted in tray towers, packed columns, spray towers, bubble columns, and centrifugal
contactors. For the additional design considerations, one needs to specify an isothermal or
non isothermal absorber. Each case will be dealt with separately, starting with the
isothermal, and followed by the non-isothermal with an example problem. The nonisothermal case is of more importance as it more closely models current topics in our
design of the anhydride plant and is presented in more detail than the isothermal case.
Reboiled strippers typically have forced reboilers and employ a reboiling medium at a
temperature as low as possible to minimize solvent decomposition. The reboiler outlet
temperature arises as the bottom heavies concentration increases. Waste is withdrawn
when reboiler capacity is expended or when temperature target is reached .Because feed
vaporizations Occurs primarily in reboiler , the tray liquid flow rates across the column
are fairly constant , simplifying the tray design .Rebolied stripper operating pressure are
lower than any other part of extraction unit .
94
Chapter: 05
Equipment Design
CHOICE BETWEEN PLATE AND
PACKED COLUMN
If the diameter of column is greater than 0.762m then a plate column is
selected.
Plate columns are designed to handle wide range of liquid flow rates without
flooding
For large column heights, weight of the packed column is more than plate
column.
Periodic cleaning is easy in plate columns as compared to packed columns.
When temperature change is involved, packing may be damaged.
If a system contains solid contents, it will be handled in plate column, because
solid will accumulate in the voids, coating the packing materials and making it
ineffective.
Inter stage cooling can be provide to remove heat of reaction or solution in plate
column.
SELECTION OF TRAYS
The terms "trays" and "plates" are used interchangeably. There are many types of tray
designs, but the most common ones are:
SIEVE TRAYS
Sieve trays are simply metal plates with holes in them. Vapour passes straight upward
through the liquid on the plate. The arrangement, number and size of the holes are design
parameters
95
Chapter: 05
Equipment Design
VALVE TRAYS
In valve trays, perforations are covered by lift able caps. Vapour flows lifts the caps, thus
self creating a flow area for the passage of vapour. The lifting cap directs the vapour to
flow horizontally into the liquid, thus providing better mixing than is possible in sieve
trays.
Bubble cap trays
A bubble cap tray has riser or chimney fitted over each hole, and a cap that covers the
riser. The cap is mounted so that there is a space between riser and cap to allow the
passage of vapors. Vapors rises through the chimney and is directed downward by the
cap, finally discharging through slots in the cap, and finally bubbling through the liquid
on the tray.
Comparison of sieve tray and other trays
Sieve tray
1.
2.
3.
4.
5.
6.
7.
8.
9.
Their maintenance cost is low.
Their fouling tendency is low.
They have large capacity.
They are highly efficient (60-85%).
Their design procedures are well known.
Their operating range is satisfactory (50-120% of design capacity).
Sieve trays give the lowest pressure drop.
Minimum entrainment as compared to bubble cap and valve tray.
Simple to construct.
96
Chapter: 05
Equipment Design
Bubble cap tray
1.
2.
3.
4.
5.
They cause entrainment.
They are appreciably more expensive.
Their capacity rating is low.
They give highest pressure drop.
They are used when low vapors (gas) rates have to be handled and a positive
liquid seal is essential at all flow rates.
Valve tray
1. They are essentially sieve trays with large diameter holes covered with by
movable flaps
2. Their operating capacity is low as compared to sieve trays but higher than bubble
caps trays.
3. They give higher pressure drop than sieve trays.
CALCULATION OF EQUILIBRIUM STAGES
Si =Ki V / L
Where
Si = Stripping factor for component i
Ki = vapour / liquid equilibrium ratio for component i
V= entering vapour flow rate into the column
L = entering liquid flow rate into the column
Feed is entering at 37.78oC into the stripper .
Top remperature = 13oC
Bottom temperature Ts =328 K.
Ki at 35oC = 1.45
So Si = 1.45 * 422.1768 /334
Si = 1.8328
97
Chapter: 05
Equipment Design
Calculation of theoretical number of stages
To find the theoretical number of stages use Kremser Edmister method
According to this method
Solute fraction stripped = Si
N+1
- Si / Si N+1 -1
Where
Si = stripping factor fpr component i
N = number of equilibrium stages required for specified solute fraction stripped.
Now calculate the % age stripped of the key component in each stage .
KEY COMPONENT
STAGES
% STRIPPED
1
2
3
4
5
6
7
8
9
82.91
84.23
88.76
96.59
99.30
99.86
99.97
99.98
99.99
So N=9 theoretical stages should be sufficient to achive 99.99 % recovery .
DETERMINATION OF ACTUAL # OF STAGES
N act = N / E
= 9/0.5
= 18
L
( ρV / ρL ) 1/ 2
V
=
334
( 1.362/1425 )
427.64
1/ 2
= 0.024
From fig
Csb =0.05 m/sec
98
Chapter: 05
Equipment Design
Then vapour velocity at flood conditions Vnf in the column is given by
Vnf = Csb (σ/20) 0.2 (ρL -ρV / ρV )0.5
Csb = souders and brown factor at flood conditions in m/sec
σ = Tension Surface (σ)
ρL = density of liquid in the column
ρV = density of vapour sreams in the column
Vnf = 0.05 (155 / 20) 0.2 (1425 – 1.362/1.362) 0.5
=1.3428 m/sec
The actual vapour velocity Vn is selected by assuming atat it is 50 to 90 percent of the
net vapour velovity at flood conditions
So assume V n is 80 % of Vnf .
Vn = 1.3428* 0.8
Vn = 1.07 m/sec
The net column area area used in the sepration process = An
An = m . v / V n
Where
m . v = volumetric flow rate of vapour
An = 29.04*427.64 / 3600*1.07 * 1.362
An = 0.777 m2
Assume the downcomer occupies 15 % of the cross section area of the column,thus
Ac = An / 0.85
Ac = 0.777/ 0.85
Ac = 0.914
The column diameter is calculated as
D = (4 Ac / π )1/ 2
D = 1.078 m
HEIGHT EVALUATIN OF STRIPPER
Height of the stripper is given by the relation
Hc = ( Nact -1) * Hs + H
Where
Hc =Actual column height
Nact = Actual # of trays
Hs = plate spacing
H = Additional height required for column operation
Hc = (18-1)*0.46 + 0.92
Hc = 8.74 m
99
Chapter: 05
Equipment Design
Sieve Plate Design
Column dia =1.078m
AC (cross flow area ) =0.777m2
Down comer area Ad= 0.15*Ac = 0.914 m2
Net area An = Ac – Ad =1.55-0.233
= 1.105 m2
Active area Aa = Ac -2*Ad
Aa =0..91 m2
Hole area An, Take 10% of Aa as 1st =0.1*0.91
Hole area = .091 m2
Weir length
Ad / Ac = 0.195/ 1.3 = 0.15
(from figure 11.31 vol.6)
Lw / dc
=
0.81
Lw
1 . 28 0 . 81
=
1.0368 m
Take weir height , hw
=
50 mm
Hole diameter, dh
=
5 mm
Plate thickness
=
5 mm
100
Chapter: 05
Equipment Design
Check Weeping
Maximum liquid rate Lm’
=
Minimum liquid rate at 70% turn down
329 × 63 / 3600 = 5.7575 kg/sec
0.7 5.7575
= 4.03 kg/sec
how
= weir crust
Maximum how
5.7575
750
1402 1.0368
2/3
= 19.48 mm liquid
Minimum how
4.03
750
1402 1.0368
2/3
= 15.4 mm liquid
at minimum hw + how = 50 + 15.4
=65.4mm liquid
from fig 11.30, Coulson and Richardson Vol.6
K2 = 30.6
The design vapour velocity
K 0.925.4 d h
U min 2
v 1 / 2
101
Chapter: 05
Equipment Design
30.6 0.925.4 5
U min
1.362 1 / 2
= 10.488 m/s
Vapour velocity at Base of column =422.62*63/3600*1.284
Actual minimum vapour velocity
= 0.98m/sec
minimum vapour rate
Ah
0.98 0.7
0.091
= 7.54 m/s
So minimum vapor rate will be well above the weep point.
Plate Pressure Drop
Dry Plate Drop
Max. Vapour velocity through holes
Û h = Volumetric Flow Rate / Hole Area
0.98
Uˆ h
10.77 m/s
0.091
from fig. 11.43 for plate thickness/hole dia = 5/5 = 1
and
Ah Ah 0.091
0 .1
A p Aa
0.91
Co
= 0.84
From Eq.11.88 Coulson vol.6
102
Chapter: 05
Equipment Design
2
Uˆ
hd 51 h V
C o L
2
10.77 1.284
=7.54mm liquid
hd 51
0.84 1425
Residual Head
hr
12.5 10 3
8.77 mm liquid
1425
Total Pressure Drop
ht = hd + (hw + how) + hr
Total pressure drop = 7.54 + (50 +19.48) + 8.77
ht = 85.8 mm liquid
Down comer Liquid Backup
Take hap = hw – 10 = 40 mm
Area under apron “Aap” 1.0368 40 10 3
= 0.04147 m2
As this is less than Ad use Aap in eq. 11.92 coulson vol.6 i.e,
103
Chapter: 05
Equipment Design
l
hdc 166 wd
L Aap
2
5.757
hdc 166
1425 0.041472
2
= 1.57529 mm liq.
Backup in down comer
hb
= (hw + how) + ht + hdc
hb
=(50 + 19.48) + 85.804 + 1.5753
= 156.86 mm liq.
hb < ½ (Tray spacing + weir height)
0.157<1/2(0.46+0.81)
So tray spacing is acceptable
Check Residence Time
tr
Ad hbc L
Lwd
tr
0.195 0.157 1425
5.757
= 7.58 sec
≡ 8 sec
3 sec. so, result is satisfactory
104
Chapter: 05
Equipment Design
Check Entrainment
Uv
= Maximum Volumetric Flow Rate of vapors/Net Area
UV
=
0.98/1.105
=
0.89 m/s
Flv= Lw / Vm (ρv/ρL) 1/2
=0.1
= K1 = 0.075
Uf =k1{( ρL- ρv ) / ρv}1/2
=2.5 m/sec
Percent flooding = UV actual velocity (based on net area ) / Uf
Percent flooding
=
0.89/2.5
=
35.6 %
FLV(base) = .1
from fig. 11.29 coulson vol.6
Fractional Entrainment = 0.003 well below 0.1 Satisfactory
Trial Lay Out
Use cartridge type construction. Allow 50 mm imperforated strip round plate edge;
50 mm wide calming zone.
From fig. 11.32
Lw/Dc =
0.825
Q
107o
=
105
Chapter: 05
Equipment Design
Angle subtended at plate edge by imperforated strip = 180 – 107 = 73o
Max length, unpeeforoted edge strip:
t r 1.41 50 10 3 73
1.73
180
Area of imperforated edge strip Ap/ 50 10
3
1 . 73 ) 0 . 0865 m2
Mean length of calming zone 1.41 50 10 3 sin 992 1.03 m
Area of calming zone Acal 2 1.03 50 10 3 0.1 m2
Total area of perforations, Ap = Aa – Ap/ - Acal = 0.89 – 0.0865 – 0.1 =0. 70m2
Ah 0.089
0.127
Ap
0.70
From fig. 11.33 Coulson vol.6
lp/dh = 2.8
Satisfactory within 2.5 - 4.0
No of Holes
Area of one hole 1.964 10
5
Number of Holes = Hole Area / Area of one hole
No. of holes
0.91
1.964 10 5
106
Chapter: 05
Equipment Design
= 4531.57
≡ 4634
107
Chapter: 05
Equipment Design
Specification Sheet
Identification
Item
Main Stripper
Item #
D – 102
Tray type:
Sieve tray
Function :
To strip NO2
Operation:
Continuous
Top
Bottom
286 K
328 K
Operating Pressure = 1 atm
No of actual Trays = 18
Diameter of Column = 1.28 m
Tray spacing = 0.46 m
Height of Column = 8.74 m
Fractional Entrainment = 0.003
Plate pressure drop = 85.804 mm liquid
Weir Height = 50mm
Weir length = 1.0368 m
Hole Area = .000964 m2
Hole Diameter = 5 mm
No of Holes =4634
Overall efficiency = 50 %
Temperature
Design Data :
108
Chapter: 05
Equipment Design
DESIGN OF DISTILLATION COLUMN
In industry it is common practice to separate a liquid mixture by distilling the
components, which have lower boiling points when they are in pure condition
from those having higher boiling points. This process is accomplished by partial
vaporization and subsequent condensation.
Definition
“Process in which a liquid or vapor mixture of two or more substances is separated
into its component fractions of desired purity, by the application and removal of
heat”.
109
Chapter: 05
Equipment Design
TYPES OF DISTILLATION COLUMNS
There are many types of distillation columns, each designed to perform specific
types of separations, and each design differs in terms of complexity.
Batch columns
Continuous columns
Batch Columns
In batch operation, the feed to the column is introduced batch-wise. That is, the
column is charged with a 'batch' and then the distillation process is carried out.
When the desired task is achieved, a next batch of feed is introduced.
Continuous Columns
In contrast, continuous columns process a continuous feed stream. No
interruptions occur unless there is a problem with the column or surrounding
process units. They are capable of handling high throughputs and are the more
common of the two types. We shall concentrate only on this class of columns.
CHOICE BETWEEN PLATE AND PACKED COLUMN
Vapor liquid mass transfer operation may be carried either in plate or packed
column. These two types of operation are quite different. The relative merits of
plate over packed column are as follows:
1-Plate column are designed to handle wide range of liquid flow rates without
flooding.
110
Chapter: 05
Equipment Design
2-If a system contains solid contents; it will be handled in plate column, because
solid will accumulate in the voids, coating the packing materials and making it
ineffective.
3-Dispersion difficulties are handled in plate column when flow rate of liquid are
low as compared to gases.
4-For large column heights, weight of the packed column is more than plate
column.
5-If periodic cleaning is required, man holes will be provided for cleaning. In
packed columns packing must be removed before cleaning.
6-For non-foaming systems the plate column is preferred.
7-Design information for plate column is more readily available and more reliable
than that for packed column.
8-Inter stage cooling can be provided to remove heat of reaction or solution in
plate column.
9-When temperature change is involved, packing may be damaged.
Our mixture which is to be processed is “Water, Isopropyl alcohol, Acetone”. I’ve
selected plate column because:
1-System is non-foaming.
2-Temperature is high i.e. 90C.
CHOCE OF PLATE TYPE
There are three main types, sieve tray, bubble cap, valve tray. I’ve selected sieve
tray because:
1-They are lighter in weight and less expensive. It is easier and cheaper to install.
2-Pressure drop is low as compared to bubble cap trays.
3-Peak efficiency is generally high.
4-Maintenance cost is reduced due to the ease of cleaning.
111
Chapter: 05
Equipment Design
Sieve trays
Sieve trays are simply metal plates with holes in them. Vapor passes straight
upward through the liquid on the plate. The arrangement, number and size of the
holes are design parameters.
Because of their efficiency, wide operating range, ease of maintenance and cost
factors, sieve and valve trays have replaced the once highly thought of bubble cap
trays in many applications.
Main Components of Distillation Columns
Column internals such as trays/plates and/or packing which are used to enhance
component separations.
A reboiler to provide the necessary vaporization for the distillation process.
The liquid removed from the reboiler is known as the bottoms product or
simply, bottoms.
A condenser to cool and condense the vapor leaving the top of the column.
The condensed liquid that is removed from the system is known as the
distillate or top product.
A reflux drums to hold the condensed vapor from the top of the column so
that liquid (reflux) can be recycled back to the column. The condensed
liquid is stored in a holding vessel known as the reflux drum. Some of this
liquid is recycled back to the top of the column and this is called the reflux.
112
Chapter: 05
Equipment Design
A schematic of a typical distillation unit with a single feed and two product
streams is shown above.
113
Chapter: 05
Equipment Design
FACTORS AFFECTING DISTILLATION COLUMN OPERATION
Vapor Flow Conditions
Adverse vapor flow conditions can cause:
Foaming
Entrainment
Weeping/dumping
Flooding
Foaming
Foaming refers to the expansion of liquid due to passage of vapor or gas. Although
it provides high interfacial liquid-vapor contact, excessive foaming often leads to
liquid buildup on trays. In some cases, foaming may be so bad that the foam mixes
with liquid on the tray above. Whether foaming will occur depends primarily on
physical properties of the liquid mixtures, but is sometimes due to tray designs and
condition. Whatever the cause, separation efficiency is always reduced.
Entrainment
Entrainment refers to the liquid carried by vapor up to the tray above and is again
caused by high vapor flow rates. It is detrimental because tray efficiency is
reduced: lower volatile material is carried to a plate holding liquid of higher
volatility. It could also contaminate high purity distillate. Excessive entrainment
can lead to flooding.
114
Chapter: 05
Equipment Design
Weeping/Dumping
This phenomenon is caused by low vapor flow. The pressure exerted by the vapor
is insufficient to hold up the liquid on the tray. Therefore, liquid starts to leak
through perforations. Excessive weeping will lead to dumping. That is the liquid
on all trays will crash (dump) through to the base of the column (via a domino
effect) and the column will have to be re-started. Weeping is indicated by a sharp
pressure drop in the column and reduced separation efficiency.
Flooding
Flooding is brought about by excessive vapor flow, causing liquid to be entrained
in the vapor up the column. The increased pressure from excessive vapor also
backs up the liquid in the down comer, causing an increase in liquid holdup on the
plate above. Depending on the degree of flooding, the maximum capacity of the
column may be severely reduced. Flooding is detected by sharp increases in
column differential pressure and significant decrease in separation efficiency.
Reflux Conditions
Minimum trays are required under total reflux conditions, i.e. there is no
withdrawal of distillate. On the other hand, as reflux is decreased, more and more
trays are required.
115
Chapter: 05
Equipment Design
Feed Conditions
The state of the feed mixture and feed composition affects the operating lines and
hence the number of stages required for separation. It also affects the location of
feed tray.
State of Trays and Packings
Remember that the actual number of trays required for a particular separation duty
is determined by the efficiency of the plate. Thus, any factors that cause a decrease
in tray efficiency will also change the performance of the column. Tray
efficiencies are affected by fouling, wear and tear and corrosion, and the rates at
which these occur depends on the properties of the liquids being processed. Thus
appropriate materials should be specified for tray construction.
Column Diameter
Vapor flow velocity is dependent on column diameter. Weeping determines the
minimum vapor flow required while flooding determines the maximum vapor flow
allowed, hence column capacity. Thus, if the column diameter is not sized
properly, the column will not perform well.
DESIGNING STEPS OF DISTILLATION COLUMN
a) Calculation of minimum reflux ratio.
b) Calculation of optimum reflux ratio.
c) Calculation of theoretical no. of stages.
d) Calculation of actual no. of stages.
e) Calculation of diameter.
f) Calculation of height.
116
Chapter: 05
Equipment Design
g) Calculation of pressure drop.
h) Calculation of no. of holes.
i) Calculation of weeping and flooding.
Design of Regenerator
Diameter of the Column:
Top
Product
Feed
Bottom
Product
At Top:
T
= 185 0F = 185 + 460 = 6450R
P
= 14.7 Psig =29.4 Psia
Avg. M.Wt. = 62.1Kg/Kgmol
ρL
= 60.47 lbm/ft3 (From Plant Design book)
117
Chapter: 05
Equipment Design
R
= 10.73 lbf.ft3/in2 lbmol 0R
ρg
= PM/RT
=29.4x62.1/10.73x645= 0.26 lbm/ft3
From Plant Design by Peters & Timmerhaus; (Page No.656,Edition 4rth)
Vm
= Kv √ [(ρL - ρg) / ρg]
Vm
= max. allowable superficial vapour velocity (ft/sec)
Kv
= an empirical constant (ft/sec)
(Souders $ Browns Eq)
Where,
For the tray spacing of “24 inch” (From Plant Design book ,Page No.684 ,4rth Edition)
Vm= 0.3 √ [(60.47-0.26) / 0.26]
Here Kv=0.3(From Graph,Plant
Design Book,4rth Edition,Page No.657)
Vm
= 4.56 ft/sec
For the diameter of the column;
Mass flow rate / Area
= Vm x ρg
Area
= Mass flow rate(G) / Vm x ρg
(Л / 4)D2
= G / Vm x ρg
D
= √ [4 G / (Л x Vm x ρg x 3600)]
D
= √ [(4 x 13095.06) / (Л x 4.56x0.26 x 3600)]
D
= 2 ft
At Bottom:
T =248 0F=248+460=7080R
P
= 14.7 Psig = 29.4Psia
Avg.M. Wt.
= 48.6 Kg/Kgmol
ρL
= 58 lbm/ft3 (From Plant Design book)
R
= 10.73 lbf.ft3/in2 lbmol 0R
ρg
= PM/RT
= 29.4x48.6/10.78x70 =0.18 lbm/ft3
For the tray spacing of “24 inch” (From Plant Design book ,Page No.684 ,4rth Edition)
= Kv √ [(ρL - ρg) / ρg]
Vm
Here Kv=0.3(From Graph,Plant Design
Book,4rth Edition,Page No.657)
Vm
= 0.3 √ [(58-0.18)/0.18]
118
Chapter: 05
Equipment Design
Vm
=5.38 ft/sec
D
= √ [4 G / Л x Vm x ρg x 3600]
D
= √ [(4x19646) / Л x5.38x 0.18 x 3600]
Now;
D= 2.68 ft
No. of Plates Required:
D = 99.203 Kgmol/hr
Xd = 0.933
F = 354.34 Kgmol/
hr
Xf = 0.533
W = 255.143 Kgmol/hr
Xw = 0.378
At 216.50F,=1000C approx.(From Perry)
Vapour pressure of HNO3
= 545 psi
Vapour pressure of H2O
= 185psi
Relative Volatility;
= 545/185=2.946
α
For the calculation of minimum reflux ratio;
Rmin.
= (1/ α - 1) [Xd / Xf – α ( 1 – Xd / 1 – Xf )]
Rmin.
= (1/ 2.946 – 1) [0.933 / 0.533 –2.946 ( 1 – 0.933 / 1 –0.533 )]
Rmin.
= 0.514x(1.75-0.421)
Rmin.
= 0.683
(Underwood and Fenske equation)
As the optimum reflux ratio is (1.1 to 1.5) times the minimum reflux ratio;(In Richardson
$ Coulson,Vol-2)
119
Chapter: 05
Equipment Design
So,
R
= 1.2 (0.683 )
R
= 0.82
Equation of the Top operating line;
Yn
= ( Ln / Vn )Xn+1 + ( D / Vn )Xd
Ln / D = R
=0.82
=>
Ln
= 0.82 (99.203)
Vn
= Ln + D = 81.346+99.203
Ln =0.82 D
= 81.346 kgmol/hr
= 180.549 kgmol/hr
Therefore,
Yn
= ( 81.346/180.549 )Xn+1 + (99.203/180.549)x0.933
Yn
= 0.45 Xn+1 + 0.512
Equation of the Bottom operating line;
Ym
= ( Lm / Vm )Xm+1 - ( W / Vm )Xw
Vm
= Vn
Lm
= Ln + F
=180.549 kgmol/hr
= 81.346+354.34
=435.686kgmol/hr
Therefore;
Ym
= (435.686/180.549 )Xm+1 - ( 255.143/180.549 )x0.378
Ym
=2.413 Xm+1 – 0.534
Equilibrium Data; We have eq.
ya
= α xa / [1 + (α – 1)xa]
xa
0
0.1
0.2
0.3
0.4
0.5
0.6
ya
0
0.25
0.42
0.56
0.66
0.75
0.82
By “McCabe-Thiele Method”:
Using “top and bottom operating lines” and the “equilibrium curve” from the
graph,
The theoretical no. of stages required = 08
120
Chapter: 05
Equipment Design
Efficiency: (Drickamer $ Bradford Eq.)
E
Now,
= 17 – 61.1 log ( µ Favg. )
(From Plant Design book ,Page No.665 ,4rth Edition)
At 216.5 0F
µ Favg. = [(0.45x0.533)+(0.028x0.467)]=0.253Cp , U,HNO3 is taken from
Perrys,Transport Properties. (2-322,2-323) that is 0.45Cp $ U,H2O is taken from Plant
Design Book,Page No.(872,873) that is 0.028Cp
E
= 17 – 61.1 log ( 0.253 )
E
= 17 – 61.1 x(-0.597)
E
= 53.5 %
As,
Efficiency
= Theoretical no. of stages / Actual no. of stages
Actual no. of stages
= Theoretical no. of stages / Efficiency
Actual no. of stages = 8 / 0.535
= 14.95 ~ 15
Height of the Column:
Height of column occupied by trays = No. of trays x Tray spacing
No. of trays
=15
Tray spacing
= 24 in
= 2 ft
Height of column occupied by trays = 15 x 2
=30 ft
Height occupied by one distributor
= 1.64 ft (By Rule of
= 0.5 m
Thumb, Height occupied by one distributor is ( 0.3 m to 0.5m)
No. of distributors used in column
=3
Height occupied by distributors
= 3 x 1.64
= 4.92 ft
By Rule of Thumb,We have:
Height at the bottom of the column to maintain the liquid level
= 4 ft
Height at the bottom for security purpose
= 4 ft
Height at the top of the column above the top most tray
= 4 ft
So,
Total Height of the column
= 30 + 4.92 + 4 + 4 + 4
121
Chapter: 05
Equipment Design
=46.92 ft
Pressure Drop in the Sieve Tray column:
Diameter of the tray (Avg.)
=2.34 ft
As (Dia of tray = Dia of column)
Total column area
= (Л / 4) D2
= (Л / 4) 2.342
= 4.3 ft2
Taking “15%” Down comer area,
Down comer area
= 4.3 x 0.15
= 0.645 ft2
Taking “85%” active area of the column,
Active area of the column
= 4.3 x 0.85
= 3.6 ft2
Taking hole area as “10%” of the active area,
Sieve hole area
= 3.6 x 0.10
= 0.36 ft
Taking hole diameter
= 3/16 in
Taking tray thickness
= 3/16 in
Pressure drop
= ∆PT = (ρL x g x ht) / 144 gc
ht
= hc + ho + hw + 0.5hg
hc
= Head equivalent to gas pressure drop
As,
Where,
through the holes in sieve tray
ho
= Height of liquid crest
hw
= Weir height
hg
= Average liquid gradient
Now,
122
Chapter: 05
Equipment Design
hc
= K.H [(Vc2 x ρg) / 2gc x ρL)]
Vc
= Linear velocity through the sieve hole
Vc
= Superficial velocity (Total column area / Total sieve hole area)
Where,
= 4.97 (4.3 / 0.36 )
= 59.36 ft/sec
To calculate the Kinetic head,
Hole area / Active area
= 0.36 / 3.6
= 0.1
Tray thickness / Hole diameter
= (3/16) / (3/16)
=1
From the fig. 16.12 by Peters & Timmerhaus,Page No.670
K.H
= 1.4
So,
hc
= 1.4 [(1.4x59.36^2x0.22 / 2 x 32.2 x 59.24)]
hc
= 0.28 ft
ho
= [ 1.7 QL /lwg ]2/3
QL
= Vol. flow rate of liquid (ft3/sec)
lw
= Weir length (ft)
QL
= 19646 lbm
hr
Now,
Where;
So,
ft^3
59.24lbm
1 hr
3600 sec
= 0.092 ft3 / sec
lw
= 75% of tray diameter(By Rule of Thumb)
= 0.75 x2.34
= 1.8ft
Therefore
ho
= [ 1.7 QL / lwg ]2/3
= [(1.7 x 0.092) /1.8x32.2]2/3
= 0.092 ft
123
Chapter: 05
Equipment Design
How,
Weir height = hw
= 2 in
= 0.167 ft
(R & Coulson vol. 6 or Rule of
Thumb)
Let liquid gradient is negligible or usually ignored in case of sieve trays,
(From Plant Design
book ,Page No.672 ,4rth Edition)
hg
=0
Therefore,
ht
= 0.28+0.167+0.06
= 0.507 ft
∆PT
= (59.24.x 32.2 x 0.507) / (144 x 32.2)
∆PT
= 0.208 Psig
%Pressure Drop due to liquid head above the sieve holes:
% ∆P
= [( ho + hw ) / ht] x 100
= [( 0.06 + 0.0.167 ) /0.507] x 100
% ∆P
= 44.8 %
124
Chapter: 05
Equipment Design
DESIGN OF PARTIAL CONDENSER
Sometimes it is desirable to condense only a portion of the vapor in a condenser such as
may be needed only for reflux. Such a condenser is a partial condenser although the term
dephlagmator was formally used. The calculation of a partial condenser does not alter the
method of computing the condensing film co-efficient. The calculation of a pressure drop
for a partial condenser is obtained with sufficient accuracy from the average of the
pressure drops based on the inlet and the outlet conditions.
So we say that partial condenser condenses the vapors at a point high enough to provide a
temperature difference sufficient to preheat a cold stream of process fluid . this save heat
and eliminates the need for providing a separate preheater .
DESIGN CALCULATIONS:
Heat duty:
NO=0.1346 4.714 2.2046 3600=5035.78 lbm/hr
NO2=67.343 lbm/hr
N2=24490.513lbm/hr
H2O=7815.56 lbm/hr
Reference temperature=77oF
Enthalpies or heat rates carried
Q1 for NO=380705.1373 Btu/hr
Q2 for NO2 =6151.8095 Btu/hr
Q3 for N2=2005773.01 Btu/hr
Q4 for H2O =6510364.77 Btu/hr
Total heat load without condensation load
∑Q=Q1+Q2+Q3=2392629.962 Btu/hr
If the mixture is cooled up to 50oF below the saturation temperature of steam then
125
Chapter: 05
Equipment Design
Q5=∑mCp∆T=6008.159 50 oF
=300407.9847Btu/hr
condenser duty total
Qc=Q4+Q5
=6810772.75 Btu/hr
Amount of coolent Water (treated)
Temperature limit
77 oF to 203 oF
Qc = ∑mCp∆T
m=54053.75 lbm/hr
Design of partial condenser
Film co-efficient of steam can be taken as
T1=392 oF
, T2=342 oF
t1=77 oF
, t2 =203 oF
R= T1 - T2/ t2- t1
=0.4
S= t2- t1 / T1 - t2 =0.4
FT=0.98
1 shell pass
2 tube passes
LMTD= 224.864 oF
Corrected T = FT LMTD=0.98 224.864=220.366oF
Suppose overall heat transfer coefficient= UD = 200 Btu/lb.ft2. oF
Heat transfer area ,
126
Chapter: 05
Equipment Design
Q
= UDA(t)w
A
= 154.53 ft2
Exchange Layout
1-2, shell & tube heat exchanger
1OD, 5/4 sq. pitch,
A = at Nt Lt
Lt = 18 ft.
a't = 0.2618 ft2/ft
A = 0.2618 18 Nt
Nt = 32.792
Nt = 32
(Nearest count)
So
A = 150.7968ft2
UD = 204.955 Btu/hr. ft2. oF
Shell side Tav=392+342/2=367 oF
Tube side Tav =203+77/2=140 oF
Assume ho=150 Btu/hr ft2 oF
Then
(T-tw)ho=(T-t)UD
(367-tw)150=(367-140)205
tw=57 oF
average condensate temperature =(57+367)/2
=211.88 oF
Tube bundle dia
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Chapter: 05
Equipment Design
Db=2504(32/0.244)1/2.207
=0.752 ft
Hc=0.95KL[ L ( L - V )g/µTl]Nr-1/6
Nr = Db /Pt
=35.305
=2/335.305=23.536
For,
ρv weighted must be used
ρv = ρVNO2yNO2+ ρVNOyNO+ ρVN2yN2+ ρVH2OyH2O
=1.7450.0018+0.765990.1346+0.71490.6546+0.50780.2089
=0.6923 lbm/ft3
So,
Tl =ml/Lnt
=93.84
By putting the values
hc=1777.88 Btu/hr. ft2. oF
Assume, ho =1500 Btu/hr. ft2. oF
tw=366 oF
average temperature of condensate =366.89 oF
KL =0.436
µ L =0.0217 lbm/hr ft
L =50.05 lbm/hr ft
V =0.55 lbm/ft3
Now
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Chapter: 05
Equipment Design
hc =0.950.43[50.05(50.05-0.55) 4.16108/0.021793.84]1/3 23.536 -1/6
=19050 Btu/hr. ft2. oF
TUBE SIDE CO-EFFICIENT (WATER)
at = 0.016136 m2
Velocity
= 0.17359 m/ sec
hi
= 4200(1.35+0.0265)0.173590.8 /22.0980.2
=266 Btu/hr. ft2. oF
hio
= 231 Btu/hr. ft2. oF
UC
=207 Btu/hr. ft2. oF
RD
=0.0001 hr. ft2. oF/ Btu
Pressure drop for shell side is negligible
=0.00412 ft2
at
flow area for 32 tubes
at’=at32/2=0.06605 ft2
Gs = m/at = 818351.629 lb/hr ft2
Res = De Gs/
=2045879.07
ρ = 58 lbm/ft3
s.gr=0.929
f=0.00001
Pt
f G 2t L n
10
5.22 10 D s
=0.6857 psia
P2
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Chapter: 05
Equipment Design
V2/2g’= from fig. 27=0.098
PL =4n N2/2g’ ρ
=4*2*0.098/1=0.784
Pt =0.784+0.6857
=1.4697 psi
Which is satisfactory
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Chapter: 05
Equipment Design
Specification Sheet
Identification: Partial condenser
No. Required = 1
Function: Condense steam by removing the latent heat of vaporization
Operation: Continuous
Type: 1-2 Horizontal Condenser
Shell side condensation
Heat Duty = 6810772.75 Btu/hr = 7.18 * 10 6 KJ/hr
Tube Side:
Tubes: 1 in. dia. 18 BWG
Fluid handled cold water
33 tubes
Flow rate = 54053.75 lbm/hr = 24569.5
L = 18 ft = 5.48 m
Kg/hr
2 passes
Pressure = 1 atm= 101.325 Kpa
o
5/4 in triangular pitch
o
Temperature =77 F to 203 F
25 0C to 95 0C
Shell side:
Flow rate= 37409 lbm/hr =
Temperature 392oF to 342 oF`
200 0C to 172.22 0C
17004 kg/hr
Pressure 1 atm = 101.325 Kpa
Utilities: Cold water
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Chapter: 05
Equipment Design
Ud assumed = 200 Btu/hr.ft2. oF =
UDcalculated=UD= 204.955Btu/hr.ft2. oF=
4088.133 KJ/hr m2 0C
4169.8 KJ/hr m2 0C
UC
=207 Btu/hr. ft2. oF = 4231 KJ/hr RD =0.0001 hr. ft2. oF/ Btu = 4.89*10-6 hr
m2 0C
m2 0C/KJ
NOMENCLATURE
as = Area of shell
a//t = Flow area
at = Flow area per tube
B = Baffle Spacing
C = Clearance b/w tubes
Cp = Specific Heat Capacity
D = Inside diameter of tubes
De = Equivalent diameter of shell
Ds = Inside diameter of shell
f= Friction Factor
Gt = Tube side mass velocity
Gs = Shell side mass velocity
hi , ho = Inside and Outside film coefficient
hio = Value of hi when referred to the tube ,OD,Btu/hr.ft2.oF
Hv = Heat of vaporization
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Chapter: 05
Equipment Design
ID = Internal Diameter
L = Tube Length,ft
LMTD = Log Mean Temperature Difference
m= Mass flow rate of vapors
mm,mw,mDME = Mass Flow rate of vapors of methanol,water,DME
n = Number of passes
OD = Outer Diameter
PT = Tube Pitch
ΔP, ΔPt, ΔPr = Total, tube, return pressure drop
ΔPs = Pressure drop of shell
Q = Heat Flow
Rd = Combined dirt factor
Re = Reynolds number, dimensionless
s= Specific Gravity
UC, UD = Clean overall coefficient for condensation, desuperheating
w = Mass Flow Rate of water
= Viscosity ;
= Viscosity ratio, (/w) 0.14
133
Chapter: 06
Mechanical Design
Chapter: 6
Mechanical Design
Absorber
Mechanical design:
For stainless steel vessel:
So, max allowable stress f is;
f = 165 N/mm²
Operating pressure =
130.9psi
Design pressure is 10% of operating pressure
So, design pressure,
Pi = ((130.9×10/100)) + 130.9
Pi = (13.09+130.9)
Pi =
143.99psi
Thickness of shell:
Taking joint efficiency 80%
J
=
0.80
t
=
Pi Di/2fJ−Pi
=
((143.99×3.2808)/ (2× 479042.95 ×.8) − (143.99)
=
0.0006ft
t
= .0073 in
As less corrosive service,
So, taking corrosion allowance is taken as 0.118 inch /year
As plant life is 10 years, so for 10 years corrosion allowance is 1.18 inch
So, total thickness is;
T =
0.0073 + 1.18
T =
1.187 inch
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Chapter: 06
Mechanical Design
Thickness of head:
We choose torispherical head. So, the equation for thickness of tori spherical head is;
Cs = 1/4[3+(Rc/Rk)1/2]
= ¼[3+1.29] = 1.07
t = PiRcC s / 2fj + Pi(Cs-.2)
= .00065ft
= .008in
. So, taking corrosion allowance is taken as 0.118 inch /year
As plant life is 10 years, so for 10 years corrosion allowance is 1.18 inch
So, total thickness is;
T/=
.008 + 1.18
=
1.188 inch
135
Chapter: 06
Mechanical Design
Oxidation Tower
Mechanical design:
For stainless steel vessel:
So, max allowable stress f is;
f = 165 N/mm²
Operating pressure =
14.7 psi
Design pressure is 10% of operating pressure
So, design pressure,
Pi = ((14.7×10/100)) + 14.7
Pi = (14.7 + 1.47)
Pi =
16.17 psi
Thickness of shell:
Taking joint efficiency 85%
J
=
0.85
t
=
Pi Di/2fJ−Pi
=
((16.17 × 3.64)/ (2× 479042.95 ×.85) − (16.17)
=
0.00007ft
t
= .0009 in
As less corrosive service,
So, taking corrosion allowance is taken as 0.11 inch /year
As plant life is 10 years, so for 10 years corrosion allowance is 1.18 inch
So, total thickness is;
T =
0.0009 + 1.18
T =
1.1809 inch
136
Chapter: 06
Mechanical Design
Thickness of head:
We choose torispherical head. So, the equation for thickness of tori spherical head is;
Cs = 1/4 [3+ (Rc/Rk)1/2]
= ¼[3+1.29] = 1.02
t = PiRcCs / [2fj + Pi (Cs-.2)]
= .00007ft
= .00084in
So, taking corrosion allowance is taken as 0.11 inch /year
As plant life is 10 years, so for 10 years corrosion allowance is 1.18 inch
Total thickness of head
T/=
=
.00084 + 1.18
1.1808 inch
137
Chapter: 06
Mechanical Design
Distillation Column
Mechanical Design:
Carbon steel is used as the material of construction.
Diameter of the column
= 2.34 ft =28.08 in
Pressure of the column
= 14.7 psig
Max. allowable pressure or Design pressure = 14.7 (1.1 )= 16.7 Psig
Pressure = 10 to 15%
As (Design
of Working Pressure)
psigMax. allowable design stress
= 13700 psig
Joint efficiency(For fully Radiographed) = 1.0 ,
(From Plant Design book ,Page No.(537,538) ,4rth
Edition)
Now wall thickness;
t
= PiDi / (2Jf – Pi)
= (16.17 x 28.08) / (2 x 1 x 13700 – 16.7)
= 0.016 in
During the normal operation;
Corrosion allowance for carbon steel
= 1.0 mm
= 0.04 in (From R $ C, Vol- )
Therefore,
Total thickness
= 0.016 + 0.04
= 0.056in
As
The design pressure
= 14.7 psig
= 1.114 bar < 15 bar
So the head is “Torispherical head”.
For Torispherical head,
t
= PiRcCs / [2Jf + Pi(Cs – 0.2)]
Cs
= 1/4 [3 + √(Rc/Rk)]
Where,
Where
Cs
=Stress Conc. factor
Rc
= Crown Radius
= Di
Rk
= Knuckle Radius
= 6 % of Di
= 0.06 x 28.08
= 1.68
Cs
= 28.08 in
= 1/4 [3 + √(28.08/1.68)]
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Chapter: 06
Mechanical Design
= 1.77
Now,
t
= (16.17 x 28.08x 1.77) / [(2 x 1 x 13700) + 16.17x(1.77 – 0.2)]
= 0.029 in
For Carbon steel,
Corrosion allowance
Total Head thickness
= 1.0 mm
= 0.04 in
= 0.029 + 0.04
= 0.069 in
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Chapter: 06
Mechanical Design
Stripper
Mechanical design:
For stainless steel vessel:
Design stress = 145 N/mm2
Operating pressure =
14.7psi
Design pressure is 10% of operating pressure
So, design pressure,
Pi = ((14.7×10/100)) + 14.7
Pi = (1.47 + 14.7)
Pi =
16.17 psi
Thickness of shell:
Taking joint efficiency 85%
J
=
0.85
t
=
Pi Di/2fJ−Pi
=
(16.17× 4.199)/ (2× 420977.13×.85) − (16.17)
=
0.00009ft
t
= .0011 in
As less corrosive service,
. So, taking corrosion allowance is taken as 0.118 inch /year
As plant life is 10 years, so for 10 years corrosion allowance is 1.18 inch
So, total thickness is;
T =
0.0011 + 1.18
T =
1.181 inch
Thickness of head:
We choose torispherical head. So, the equation for thickness of tori spherical head is;
Cs = 1/4[3 + (Rc/Rk)1/2]
= .1/4[3 + 4.09] = 1.77
t = PiRcC s / 2fj + Pi(Cs-.2)
= .00007ft
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Chapter: 06
Mechanical Design
= .0008in
So, taking corrosion allowance is taken as 0.118 inch /year
As plant life is 10 years, so for 10 years corrosion allowance is 1.18 inch
So, total thickness is;
Total thickness of head
T/=
=
.0008 + 1.18
1.18 inch
141
Chapter:07
Instrumentation & Plant Control
Chapter: 07
INSTRUMENTATION & CONTROL
INRODUCTION:
Measurement is a fundamental requisite to process control. Either the control can be
affected automatically, semi-automatically or manually. The quality of control obtainable
also bears a relationship to the accuracy, re-product ability and reliability of the measurement
methods, which are employed. Therefore, selection of the most effective means of
measurements is an important first step in the design and formulation of any process control
system.
Objectives of Process Control
Safety.
Production Specifications.
Operating Constraints.
Economics.
Environmental regulations
PROCESS MESUREMENT INSTRUMENTATION:
The importance of process measurement in a process plant cannot be neglected. Current
process control technology relies on accurate and repeatable measurements in order to
produce a return on investment consistent with the design of the process plant.
The measurement instruments cannot measure the process parameter directly. Rather
measures some other parameter, and infers that in the desired parameters. Consider the
following examples.
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Instrumentation & Plant Control
I. Suppose it is required to read the temperature of a room using alcohol in a glass
thermometer. It is clear that, in reality the expansion and contraction of alcohol is
what is being measured relative to the room temperature. The height of the alcohol
column is, intern, compared to a calibrated scale and a temperature measurement is
inferred.
II. If a steam pressure in a boiler drum is required by using a common pressure gauge, in
reality the mechanical deflection of the measuring element inside the gauge is what is
being measured relative to the steam pressure. This deflection of the measuring
clement is converted to angular motion; a pointer is linked to the measuring element
and compared to a calibrated scale as an inferred pressure measurement.
The phenomenon of inferred measurement causes the instrument to carry out a.c. conversion
from one energy to another, i.e. in case of room thermometer, from heat energy to
mechanical energy. However, this conversion process increases the complexity of the
instrument and consequently increases the potential sources of error.
The most common parameters measured on a process plant are Sometimes referred to as Big
Four” i.e. pressure, temperature, level and flow. These accounts for about 80% of all
measurements on a typical process plant.
The remaining 20% include analytical measurements cf density, humidity and speed etc.
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Chapter:07
Instrumentation & Plant Control
TYPES OF INSTRUMENTATION
Following are some of the main types of instrumentation
RECORDING INSTRUMENTATION
In Practical work process parameters such as temperature, pressure, and flow require
continuous measurements. If a review of the measurements is required provision must be
made to note the parameters with respect to time. The recorder is a device to accomplish this
task and may take many forms depending upon the application. The usual method is to
inscribe the measurement of the parameter on a chat with respect to time. These charts can be
circular or linear, and may be driven by a clock mechanism. The process parameter is record
by a pen which leaves a trace on the chart. The duration of the record is a Function of chart
speed (time base) and length of chart paper.
INDICATION INSTRUMENTATION
To provide visual indication of process parameters to the plant operators, instrument
indicators are used extensively in process. The indicators may be traditional circular dial type
digital display. Indicator may be field mounted, or panel mounted in a control room.
The indicator mechanism is selected to match the parameter signal and desired range of
measurement, i.e. it may use a pointer on a calibrated scale to, indicate. In recent years
circular gauges have been superseded by linear indicator displaying the process parameter in
the form of a pointer against a calibrated scale. I)igital display unit, which give a numerical
indication of the process parameter are becoming increasingly common.
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Instrumentation & Plant Control
ALARM INSTRUMENTATION
The purpose of alarm instrumentation is to detect process parameters that deviate outside the
predetermined limits. An alarm instrument loop usually consists of a switch or contact
closure device as a field device, and on end device such as horn, light or bell. The alarm loop
is normally digital having two conditions corresponding to an alarm state and a non alarm
state.
PROCESS CONTROL SYSTEM
The three main objects of control systems are.
I. To suppress the influence of external disturbances on a process.
II. To ensure the stability and safety of a process plant.
III. To optimize the productivity of a process plant.
CANCELING THE EFFECT OF DISTURBANCES
To suppress the influence of disturbances is the most common objective of’ a control system.
This control objective attempts to return the process parameter back to the set point as
effectively as possible. The control systems made changes to the process to cancel the effects
of any external disturbances.
PROCESS STABILITY
Ensuring the stability of a process is an important aspect of control. With out stable control
the behavior of control process can range form virtually inactive, to very little control action
taking place. This may lead to unstable process.
An unstable condition affects the safety of plant and personnel, in addition causing poor
productivity.
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Chapter:07
Instrumentation & Plant Control
OPTIMIZING THE PRODUCTIVITY OF THE PROCESS
With proper control of process variables and stability of the process it is possible to run the
plant at the optimum production rate.
THE FEED BACK CONTROL LOOP
A fundamental part of any industrial control system is the feed back control loop. It consists
of following parts.
I. Controller
II. Measurement Element
III. Final Control Element
IV. Process
The process variable may be energy or material which is to be adjusted to a definite value
Examples of process variables are pressure, level, flow, temperature and composition.
SET POINT
The set point is the desired value of the process variable.
ERROR
It is the difference between the actual value of the controlled variable and the set point.
OR
It is the margin by which an automatic controller misses the desired value.
FEED BACK
Feed back is the form of information gained by monitoring the controlled variable, and
comparing the controlled variable signal to the set point. If the feed back causes the
difference between the set point and the controlled variable to increase, then the feed back is
positive. This situation is dangerous and undesirable.
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Instrumentation & Plant Control
However under negative fed back control, the set point (usually a fixed value) and the
controlled variable are continually compared and the error between the two diminishes.
CLOSE FEED BACK CONTROL LOOP
If all the components of the loop are interconnected so that information can be passed
continuously around the loop, the loop is called closed fed back control loop.
OPEN FEED BACK CONTROL LOOP
If the flow of information around the loop is interrupted in any way (as, for example when
the controller is placed on manual control) then the loop is said to be open and automatic
feed back cases.
Following are the main parts of control loop.
I. CONTROLLER
The principle function of’ a controller is to generate an output signal that will control the
final control clement in order to remove error. The controller is normally the only component
in the loop that is capable of counteracting the effect of disturbances on the process.
II. SENSING OR MEASUREMENT ELEMENT
This measures the value of a controlled variable. Sensing elements are used to convert a
measurement. such as pressure, temperature, or flow, into a movement or signal that can be
used for transmission to a controller, recorder or an indicator. They are also called detectors
or sensors.
III. FINAL CONTROL ELEMENT
This is the component of a control system that adjust the value of the controlled variable e.g.
valve, variable transformation.
IV. PROCESS
Process may be defined as manufacturing operation which uses energy to produce a change
in a material, or to produce another energy conversion. The process may take many forms. It
may be the maintenance of water level in a boiler tank, the control of flow rate of various
liquids and gases, or the maintenance of pressure in a vessel.
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Instrumentation & Plant Control
TEMPERATURE MEASUREMENT AND CONTROL
Temperature measurement is used to control the temperature of outlet and inlet
streams in heat exchangers, reactors, etc.
Most temperature measurements in the industry are made by means of thermocouples to facilitate bringing the measurements to centralized location. For local
measurements at the equipment bi-metallic or filled system thermometers are used to a lesser
extent. Usually, for high measurement accuracy, resistance thermometers are used.
All these meters are installed with thermo-wells when used locally. This provides
protection against atmosphere and other physical elements.
PRESSURE MEASUREMENT & CONTROL
Like temperature pressure is a valuable indication of material state and composition.
In fact, these two measurement considered together is the primary evaluating devices of
industrial materials.
Pumps, compressor and other process equipment associated with pressure changes in
the process material are furnished with pressure measuring devices.
Thus pressure
measurement becomes an indication of energy increase or decrease.
Most pressure measurement in industry are elastic element devices, either directly
connected for local use or transmission type to centralized location. Most extensively used
industrial pressure element is the Bourdon Tube or a Diaphragm or Bellows gauges.
FLOW MEASUREMENT AND CONTROL
Flow-indicator-controllers are used to control the amount of liquid. Also all manually
set streams require some flow indication or some easy means for occasional sample
measurement. For accounting purposes, feed and product stream are metered. In addition
utilities to individual and grouped equipment are also metered.
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Instrumentation & Plant Control
Most flow measures in the industry are/ by Variable Head devices. To a lesser extent
Variable Area is used, as are the many available types as special metering situations arise.
.
CONTROL SCHEMES OF DISTILLATION
COLUMN
GENERAL CONSIDERATION
OBJECTIVES
In distillation column control any of following may be the goals to achieve
1.
Over head composition.
2.
Bottom composition
3.
Constant over head product rate.
4.
Constant bottom product rate.
.
MANIPULATED VARIABLES
Any one or any combination of following may be the manipulated variables
1. Steam flow rate to reboiler.
2. Reflux rate.
3. Overhead product withdrawn rate.
4. Bottom product withdrawn rate
5. Water flow rate to condenser.
LOADS OR DISTURBANCES
Following are typical disturbances
1.
Flow rate of feed
2.
Composition of feed.
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Chapter:07
Instrumentation & Plant Control
3.
Temperature of feed.
4.
Pressure drop of steam across reboiler
5.
Inlet temperature of water for condenser.
CONTROL SCHEME
Overhead product rate is fixed and any change in feed rate must be absorbed by
changing bottom product rate. The change in product rate is accomplished by direct level
control of the reboiler if the stream rate is fixed feed rate increases then vapor rate is
approximately constant & the internal reflux flows must increase.
ADVANTAGE
Since an increase in feed rate increase reflux rate with vapor rate being approximately
constant, then purity of top product increases.
DISADVANTAGE
The overhead reflux change depends on the dynamics of level control system that adjusts it.
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Instrumentation & Plant Control
Control scheme of distillation column
151
Chapter: 08
Safety
Chapter: 08
Safety
INTRODUCTION (ABSTRACT):
Every organization has a legal and moral obligation to safe-guard the health and welfare of
its employees and general public. Safety also leads to good business. So by good
management practice and ensuring safe operations personal safety, plant safety and efficient
operations can he achieved.
GENERAL GOALS OF SAFETY
The ultimate goal of safety is the complete protection of
i.
Personnel injury and loss of life.
ii.
Destruction of plant and property as a result of:
a. Accidents
b. Fires
c. Explosions or
d. Other hazards situation
MAJOR CAUSES OF HAZARDS:
Process industries have a wide range of hazards. These hazards are the result of the presence
of:
i.
Sizable quantities of flammable and unstable materials.
ii.
I ugh temperatures which promote ignition and decomposition.
iii.
High pressure which may release in the case of structural failure or explosion.
iv.
Improper handling of material by employees.
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Chapter: 08
Safety
PROTECTION AGAINST HAZARDS:
The Hazards and their prevention may be divided into two major groups.
1. Mechanical Hazards and protection.
2. Personal health Hazards and protection
PROTECTION AGAINST MECHANICAL HAZARDS
MAJOR MECHANICAL HAZARDS
Major Hazards are accidents which may happen due to mechanical failure of equipment
under sever operating conditions of temperature and pressure.
PROTECTION FROM MECHANICAL HAZARDS
The mechanical failure of equipment can be prevented by taking following steps into
account.
i.
To prevent mechanical failure, the equipment should he designed according to
standards recommended by authorities. For example, pressure vessels and storage
tanks should be designed according to American Petroleum Institute (A.P.I.) code,
and they should be tested for 2 or more times at the designed pressure.
ii.
While designing the equipment, it is necessary to select proper instruments and to
control the process at the safe operating conditions. Safety acts a guide line in the
design of control systems along with other factors.
iii.
Clear and effective procedures play an important role in the safe operations of a
chemical plant. The equipment manufactures provide operating instructions. But in
the plant where hundreds of small units are looked, it is necessary to lay down
standard operating procedures (S.O.P’s) to ensure safe start up, shut down and normal
operations.
PERSONAL HEALTH HAZARDS AND PROTECTION
Accidents may also result during handling and storage of 1-lazardous material, Injury to plant
personal may also result due to toxicity of chemicals being handled during operations. It is
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Chapter: 08
Safety
therefore, necessary to have a MI understanding of chemical and physical properties of
material being handled,
MAJOR HEALTH HAZARDS AT NITRIC ACID PLANT
At a typical nitric acid plant, large amount of nitric acid and nitrogen oxides are handled. So
major HAZARDS of nitric acid and nitrogen oxides are given below.
Nitric acid, its vapors and nitrogen oxides are
i.
Highly toxic
ii.
Fast in action
iii.
Capable of producing sever injuries or death if improperly handled.
TYPICAL HAZARDS OF NITRIC ACID AND NITROGEN OXIDES
Typical Hazards of these chemicals can be divided into two groups.
i.
Contact Hazards
ii.
Respiratory Hazards
i.
CONTACT HAZARDS
I. An injury may result by contact of skin with acid.
II. The symptoms resulting from skin contact vary from moderate irritation to sever
burns, depending on
a. Contact time and
b. Strength of acid
III. Signs of contact may include a yellow discoloration of skin.
IV. Sever burns may penetrate deeply causing ulceration.
ii.
RESPIRATION HAZARDS
1. Due to highly toxic action of nitrogen oxides these gases are taken as the most toxic
gases.
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Chapter: 08
Safety
2. The major Hazards of these gases is the sever pulmonary problems.
3. The detail of problem is as follows
i.
Sever pulmonary symptoms may set with in 5-8 hours after breathing as little
as 25 PPM through out an 8-hour period.
ii.
Breathing 100-150 PPM for 0.5-1 hour may produce pulmonary edema.
iii.
A few breaths of 200-700 PPM may produce sever pulmonary edema with in
5-8 hours.
4. Most cases of nitrous fume poisoning can be classified into the following categories.
i.
Slight symptoms
ii.
Mild immediate effects
iii.
Sever effects
i.
Slight symptoms
a. Slight symptoms may occur with in 48 hours breathing light nitrous oxide
fumes.
b. This form of poisoning occurs in industry.
ii.
Mild Immediate Effects
From mild effects the recovery is apparently complete but after words the sever attacks
eventually continue. In typical case of NO poisoning the sequence of attacks may be
a. A few breaths of apparently harmless gas.
b. Only slight discomfort may be felt with worker continuing his job.
c. 5-8 hours after exposure, the victim may face very sever attack?
iii.
Sever Effects
In case of sever attacks, attacks ma be followed by
a. Shocking
b. Dizziness and
c. Irregular respiration
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Safety
PROTECTION AGAINST HEALTH HAZARDS
Protection against health Hazards may he classified as:
I. Protection against contact Hazards.
II. Protection against respiratory Hazards.
I. PROTECTION AGAINST CONTACT HAZARDS
To protect contact Hazards following safety equipment should be provided to the workers.
a. Acid proof safety shoes
b. Acid proof over-all clothes
c. Rubber gloves
d. Safety goggles
e. Gas masks
f. Fire extinguisher
g. Some of employees should be given proper first aid training.
TREATMENT OF ACID CONTACT INJURY
Generally contact injury occur on
i.
Skin
ii.
Eye
i.
SKIN INJURY TREATMENT:
a. II skin is a Heeled by the acid the affected part should he washed with
Na2CO3 solution or weak alkali solution.
b. Then the wound should he washed with water and dried.
c. Moist picric acid gauze is applied and bandaged.
d. If acid is swallowed, the victim should drink warm water soap-water or milk
in large quantities.
e. Vomiting should be introduced if it does not occur.
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ii.
Safety
EYE INJURY TREATM ENT
a. If eye is affected immediately wash the eye with saturated solution of
Na2CO3.
b. Cover the eye and bandage.
c. Report the case at once so that a doctor may see it as soon as possible.
II. PROTECTION AGAINST RESPIRATORY HAZARDS
To protect respiratory Hazards the workers should he provided the safety equipment
including
a. Safety
b. Acid proof safety shoes
c. Acid prod over all clothes
d. Rubber gloves
e. Safety goggles
f. Gas masks
g. Fire extinguisher
TREATMENT OF RESPIRATORY HAZARDS
i.
In the case of nitrogen oxides poisoning first acid should be given immediately.
ii.
In such situation patient should he moved to open and un-contaminated atmosphere.
iii.
No exertion should he allowed to the patient. Patient should breath pure
iv.
Oxygen for 30 minutes to 1 hr for 6 hours.
v.
If breathing is normal O2 inhaling should be discontinued.
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Material of Construction
Chapter: 09
Material of Construction
INTRODUCTION
The corrosion problem affects many industries. In chemical industries it is one of the most
serious, and perhaps the most costly, of the problems confronting the chemical engineer, not
only because of the added cost to production due to replacement of apparatus, shutdowns for
repair, etc., but also because of possibility of contamination of the product through catalytic
action or other wise. Therefore losses due to Corrosion Can run into hundreds of millions of
rupees. Any development, therefore, that will have for its object the control or reduction of
this heavy loll must be welcomed by the chemist and the engineer.
The ammonia may be converted into nitric acid by an oxidation process was known as a
theoretical possibility for many years. The success of this process should not be considered
as due solely to the skill and ingenuity of the chemist. It is due as well to the availability of
suitable materials which satisfactorily resisting the corrosive action of the acid and capable of
fabrication into required forms, have made possible the construction of the necessary plants
and equipment. These materials are the iron chromium alloys. Therefore early nitric acid
plants were severely restricted in design by the available materials of construction until high
chromium content iron was developed in the 1920, commonly used construction materials
were acid brick, earthen ware and glass. As better materials of construction came along,
nitric acid process design could he improved. Higher operating pressures resulted in lower
plant capital cost.
CORROSION PRINCIPLES
In case of pure metals, what may be called the affinity of its atoms for the stranger atoms of
some other substance determines the readiness with which it will combine with this
substance, generally displacing some other element with which the stranger atoms have
previously been in combination. This is the basis of the process called “corrosion”. IF this
157
Chapter: 09
Material of Construction
affinity can be reduced by the atoms of some other metal, as is the case in solid solutions (a
solid solution is formed when two or more elements or compounds share a common lattice.
The composition may vary with in very wide limits), susceptibility to attack will be
considerably reduced. If this energy of affinity is still further reduced by the addition of a
third metal in the solid solution, there will result still greater resistance to attack. This is well
illustrated by numerous alloys. Iron is readily attacked by oxygen even at normal
temperatures. The addition of chromium in suitable proportions, which forms a solid
solution, produces practically complete resistance at ordinary temperatures and even at
moderately elevated temperatures. If a third metal, as nickel or silicon in proper is added,
which also enters solution with the iron and chromium resistance to oxidation is extended to
still higher temperatures, and the alloy becomes resistant to a number of materials which
neither the plain iron, nor even the iron chromium solid solution, will resist. The solid
solution class of alloys is inherently better, not only From the chemical resistance, but also
From a mechanical stand point, for such alloys almost always possess the best physical
properties as regards both strength and toughness of any alloy of the series.
Equipment for an ammonia oxidation plants includes.
I. Converters.
II. Absorption towers.
III. Heat exchangers.
IV. Valves.
V. Pipe lines, etc.
HIGH TEMPERATURE NH3 OXIDATION:
Alloys that will with stand high temperature and have good strengths at 1000- 1900oF are
available hut are usually very expensive. These alloys, for use in the ammonia burner, should
resist oxidation and nitriding and have good sealing resistance. Also, good thermal shock
properties are desirable, particularly for internals and un-cooled items that undergo high
stresses.
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Material of Construction
The Fe-Ni-Cr alloys such as Inconel 601, Hastelloy X, Nichrorme, RA-333 etc, are suitable
alloys to consider for use in these high heat environment.
The major concern with equipment in the hot areas of the nitric acid is not strength, although
the alloys used must have some hot strength, but is whether the material can take the
environment without breaking down. In general ideally, the alloy should contain high nickel,
chrome and silicon. High nickel means over 35% Ni, high Cr means over 20% Cr, and high
Si means over 0.5% Si. Many wrought and even a few cast alloys ft into this analysis, and
any one of them can do the
NO & NO2 AT INTERMEDIATE TEMPERATURE:
As temperatures fall to 1000°F and below, and if the gases are dry, the standard austenitic
stainless steel arc used, providing they are not subjected to attack from cooling media that
contain high chloride levels. Alloys such as 304L, 321, 430, 329, 310 and Incoloy 800, may
be used in this area. Most of this equipment involves tubular components, although the dry
gases being cooled are not trouble some, the cooling media on the water side are often the
cause of failures. A careful look at the water side may be deciding factor in the choice of
materials.
Examples of trouble areas are in items of equipment. Such as tail-gas heaters, cooler
condensers and tail-gas pre-heaters. Type 304L and type 430 are the usual materials of
construction, but in some cases are being replaced by type 329, which has better corrosion
resistance. Some plants attributed their problems to acid condensate in the tail gas impinging
on the O.D. others felt the attack was on the tube I.D. as a result of the relatively cool tail gas
(from tail-gas heater) impinging on the tubes containing hot process gas and causing
localized cooling and re-boiling of weak acid.
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Material of Construction
AQUEOUS NITRIC ACID CORROSION
The aqueous acid is strongly oxidizing and extremely corrosive to most materials such as
steel, copper alloys, rubber products, aluminum (low concentrations) and many more.
Materials for weak-acid condensers and absorption towers must be carefully selected since
these are the areas of most sever aqueous corrosion.
Choice of materials for wet acid service is the normal stainless alloys, special purpose
stainless alloys or high silicon irons (seldom used in modern construction except for pumps
and valves).
Type 304L cooling coils are used in absorption towers but the alloy is just barely adequate,
and alloys with better corrosion resistance are being looked at, particularly as higher
absorption pressures are being use. Most of’ the other standard grades of’ austenitic stainless
steels would have similar corrosion rates. Type 316 is not selected specifically for nitric acid
service because it has slightly less resistance than 304L and is more costly. Type 304L shows
poor corrosion resistance to hot, very strong acid, including fuming nitric acid (but has good
resistance to red and white fuming at room temperatures).
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Material of Construction
Aluminum alloys are only good for very for very high acid concentrations (80-100%) at
ambient temperatures. Any dilution of the acid will cause accelerated attack.
High nickel alloys such as Inconel 600, Hstelloy B and C, Chlorimet 3 etc. do not perform as
well as regular type 304L, and so are not considered for use in aqueous acid conditions.
Gold, tantalum and platinum exhibit excellent resistance but are expensive.
Non metallic, and especially teflon and kynar, are used for nitric acid service when the acid is
weak and cold. Both teflon and kynar are good in the weak acids (to 20%) at high
temperatures, but at higher concentrations these plastics are only good at ambient
temperatures. Glass has excellent resistances to 400°F at all concentrations Viton linings are
the one rubber product that is popular. It is good to 100°F at all concentrations.
161
Chapter: 09
Material of Construction
Table 1 lists the concentrations and limiting temperatures for many materials that should or
should not be used in nitric acid service.
162
Chapter:10
Hazop Study & Envoirmental Impact Assessment Report
Chapter: 10
Hazop Study & Envoirmental Impact Assessment
Report
INTRODUCTION
A HAZOP survey is one of the most common and widely accepted methods of
systematic qualitative hazard analysis. It is used for both new or existing facilities and
can be applied to a whole plant, a production unit, or a piece of equipment It uses as its
database the usual sort of plant and process information and relies on the judgment of
engineering and safety experts in the areas with which they are most familiar. The end
result is, therefore reliable in terms of engineering and operational expectations, but it is
not quantitative and may not consider the consequences of complex sequences of human
errors. The objectives of a HAZOP study can be summarized as follows:
1)
To identify (areas of the design that may possess a significant hazard
potential.
2)
To identify and study features of the design that influence the probability of a
hazardous incident occurring.
3)
To familiarize the study team with the design information available.
4)
To ensure that a systematic study is made of the areas of significant hazard
potential.
5)
To identify pertinent design information not currently available to the team.
6)
To provide a mechanism for feedback to the client of the study team's detailed
comments.
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Hazop Study & Envoirmental Impact Assessment Report
STEPS CONDUCTED IN HAZOP STUDY
1)
Specify the purpose, objective, and scope of the study. The purpose may be
the analysis of a yet to be built plant or a review of the risk of unexisting unit.
Given the purpose and the circumstances of the study, the objectives listed
above can he made more specific. The scope of the study is the boundaries of
the physical unit, and also the range of events and variables considered. For
example, at one time HAZOP's were mainly focused on fire and explosion
endpoints, while now the scope usually includes toxic release, offensive odor,
and environmental end-points. The initial establishment of purpose,
objectives, and scope is very important and should be precisely set down so
that it will be clear, now and in the future, what was and was not included in
the study. These decisions need to be made by an appropriate level of
responsible management.
2)
Select the HAZOP study team. The team leader should be skilled in HAZOP
and in interpersonal techniques to facilitate successful group interaction. As
many other experts should be included in the team to cover all aspects of
design, operation, process chemistry, and safety. The team leader should
instruct the team in the HAZOP procedure and should emphasize that the end
objective of a HAZOP survey is hazard identification; solutions to problems
are a separate effort.
3)
Collect data. Theodore16 has listed the following materials that are usually
needed.
Process description.
Process flow sheets.
Data on the chemical, physical and toxicological properties of all raw
materials,, intermediates, and products.
Piping and instrument diagrams (P&IDs).
Equipment, piping, and instrument specifications.
Process control logic diagrams.
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Chapter:10
Hazop Study & Envoirmental Impact Assessment Report
Layout drawings.
Operating procedures.
Maintenance procedures.
Emergency response procedures.
Safety and training manuals.
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Hazop Study & Envoirmental Impact Assessment Report
HAZOP Guide Words and Meanings
Guide Words
4)
Meaning
No
Negation of design intent
Less
Quantitative decrease
More
Quantitative increase
Part of
Qualitative decrease
As well as
Qualitative Increase
Reverse
Logical opposite of the intent
Other than
Complete substitution
Conduct the study. Using the information collected, the unit is divided into
study "nodes" and the sequence diagrammed in Figure , is followed for each
node. Nodes are points in the process where process parameters (pressure,
temperature change between nodes as a result of the operation of various
pieces of equipment' such as distillation columns, heat exchanges, or pumps.
Various forms and work sheets have been developed to help organize the node
process parameters and control logic information.
When the nodes are identified and the parameters are identified, each node is
studied by applying the specialized guide words to each parameter. These guide words
and their meanings are key elements of the HAZOP procedure. They are listed in Table.
Repeated cycling through this process, which considers how and why each
parameter might vary from the intended and the consequence, is the substance of the
HAZOP study.
5)
Write the report. As much detail about events and their consequence as is
uncovered by the study should be recorded. Obviously, if the HAZOP
identifies a not improbable sequence of events that would result in a disaster,
appropriate follow-up action is needed. Thus, although risk reduction action is
not a part of the HAZOP, the HAZOP may trigger the need for such action.
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The HAZOP studies are time consuming and expensive. Just getting the P & ID's
up to date on an older plant may be a major engineering effort. Still, for processes with
significant risk, they are cost effective when balanced against the potential loss of life,
property, business, and even the future of the enterprise that may result from a major
release.
HAZOP Study of Ammonia Air mixing tee:
Intention: Transfer vapor to mixing tee.
Guide Word
Deviation
Cause
Consequences and Action
LESS
As no flow
Flow
CV plug gage
FRC failure
MORE
Flow
FR/ ratio control
Danger of high ammonia
Miss-operation
concentration: fit alarm, fit
analyzers (duplicate) with
high alarm 14 percent NH3
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Chapter:10
Hazop Study & Envoirmental Impact Assessment Report
Environmental Impact Assessmeent (EIA) Report
NOx: What is it? Where does it come from?
Nitrogen oxides, or NOx, is the generic term for a group of highly reactive gases, all of
which contain nitrogen and oxygen in varying amounts. Many of the nitrogen oxides are
colorless and odorless. However, one common pollutant, nitrogen dioxide (NO2) along
with particles in the air can often be seen as a reddish-brown layer over many urban
areas.
Nitrogen oxides form when fuel is burned at high temperatures, as in a combustion
process. The primary manmade sources of NOx are motor vehicles, electric utilities, and
other industrial, commercial, and residential sources that burn fuels. NOx can also be
formed naturally.
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Health and Environmental Impacts of NOx
NOx causes a wide variety of health and environmental impacts because of various
compounds and derivatives in the family of nitrogen oxides, including nitrogen dioxide,
nitric acid, nitrous oxide, nitrates, and nitric oxide.
Ground-level Ozone (Smog) –
is formed when NOx and volatile organic compounds (VOCs) react in the presence of
sunlight. Children, people with lung diseases such as asthma, and people who work or
exercise outside are susceptible to adverse effects such as damage to lung tissue and
reduction in lung function. Ozone can be transported by wind currents and cause health
impacts far from original sources. Millions of Americans live in areas that do not meet
the health standards for ozone. Other impacts from ozone include damaged vegetation
and reduced crop yields
Acid Rain
- NOx and sulfur dioxide react with other substances in the air to form acids which fall to
earth as rain, fog, snow or dry particles. Some may be carried by wind for hundreds of
miles. Acid rain damages; causes deterioration of cars, buildings and historical
monuments; and causes lakes and streams to become acidic and unsuitable for many fish.
Particles
- NOx reacts with ammonia, moisture, and other compounds to form nitric acid and
related particles. Human health concerns include effects on breathing and the respiratory
system, damage to lung tissue, and premature death. Small particles penetrate deeply into
sensitive parts of the lungs and can cause or worsen respiratory disease such as
emphysema and bronchitis, and aggravate existing heart disease.
Water Quality Deterioration
- Increased nitrogen loading in water bodies, particularly coastal estuaries, upsets the
chemical balance of nutrients used by aquatic plants and animals. Additional nitrogen
accelerates "eutrophication," which leads to oxygen depletion and reduces fish and
shellfish populations. NOx emissions in the air are one of the largest sources of nitrogen
pollution in the Chesapeake Bay.
Climate Change
- One member of the NOx, nitrous oxide or N2O, is a greenhouse gas. It accumulates in
the atmosphere with other greenhouse gasses causing a gradual rise in the earth's
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Hazop Study & Envoirmental Impact Assessment Report
temperature. This will lead to increased risks to human health, a rise in the sea level, and
other adverse changes to plant and animal habitat.
Toxic Chemicals
- In the air, NOx reacts readily with common organic chemicals and even ozone, to form
a wide variety of toxic products, some of which may cause biological mutations.
Examples of these chemicals include the nitrate radical, nitroarenes, and nitrosamines.
Visibility Impairment
- Nitrate particles and nitrogen dioxide can block the transmission of light, reducing
visibility in urban areas and on a regional scale in our national parks.
National Environmental Quality Standards For
Industrial Gaseous Emissions(mg/Nm3,Unless
Otherwise Defined)
Parameter
Oxides of Nitrogen
(NOx)
Source of emission
1) Nitric Acid manufacturing unit
2) Gas fired
3) Oil fired
4) Coal fired
Standards
400
400
_
_
Revised
Standards
3000
400
600
1200
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Chapter: 11
Cost Estimation
CHAPTER: 11
COST ESTIMATION AND ECONOMICS OF PLANT
LOCATION
PLANT COST ESTIMATION
As the final process-design stage is Complete, it becomes possible to make accurate cost
estimation because detailed equipment specification and definite plant facility
information are available. Direct price quotation based on detailed specification can then
be obtained from various manufacturers. However o design project should proceed to the
final stages before costs are considered and cost estimate should be made through out all
the early stages of the design when complete specifications are not available. Evaluation
of costs in the preliminary design is said predesign cost estimation. Such estimation
should be capable of providing a basis for company management to decide if further
capital should be invested in the project.
Evaluation of costs in the preliminary design phase is some time called guess estimations.
A plant design obviously must present a process that is capable of operating under
condition which will yield a profit.
A capital investments is required to any industrial process, and determination of the
necessary investment is an important part of a plant design project. The total investment
for any process consists of the physical . equipment and facilities in the plant plus the
working capital for money which must be available to pay salaries keep raw materials
and products on hand and handle other special items requiring a direct cast out lay.
CAPITAL INVESTMENTS
Before an industrial plant can be put into operation, large amount of
-money must be
supplied to purchase and install the necessary machinery and equipment, land and service
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Chapter: 11
Cost Estimation
facilities must be obtained and the plant-must be erected. Complete with all pipe controls
inn services. In addition it is necessary to have money available for payment of expenses
involved in the plant operation.
The capital needed to supply the necessary manufacturing and plant facilities is called the
fixed capital investment while the necessary for the operation of the plant is termed as the
working capital investment.
1.
Working Capital Investment
The capital which is necessary lor the operation of the plant is called working capital
investment.
2.
Fixed Capital Investment
The capital needed to supply flu- necessary maMiif'acttirini1 and plant facilities is called
fixed capital investment.
The fixed capital investment classified in to two sub divisions,
i.
Direct Cost
ii.
Indirect Cost
DIRECT COST
The direct cost items arc incurred in the construction of the plant in addition to the cost of
equipment.
1.
Purchased Equipment
2.
Purchased Equipment Installation
3.
Instrumentation and Control
4.
Piping
5.
Electrical Equipment and Materials
6.
Building (Including Services)
7.
Yard Improvement
8.
Services Facilities
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Chapter: 11
9.
Cost Estimation
Land
INDIRECT COST
1.
Design and Engineering
2.
Contractor's Expenses
3.
Contractor's Fee
4.
Contingency
METHODS OF CAPITAL INVESTMENT
Various methods are employed for estimating capital investment. The choice of any
method depends on the foil owing-factors,
a.
Amount of detailed information available
b.
Accuracy Desired
Seven methods of estimating capital investments are outlined, estimate
1.
Detailed item estimate
2.
Unit estimate
3.
Percentage of delivered equipment cost
4.
“Lang” factor approximation of capacity ratio
5.
Investment cost per capacity
The accuracy of an estimate depends on the amount of design detail available; and the
accuracy of the cost data available; and the time spent on preparing the estimate. In the
early stages of a project only an approximate estimate will be required an justified by the
amount of information by then developed.
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Chapter: 11
Cost Estimation
PERCENTAGE DELIVERED EQUIPMENT
This
method
for
estimating
total
investment
requires
the
determination
of the delivered equipment cost. The cost of purchased equipment is the basis of several
pre design methods for estimating capital investment.The most accurate methods for
determining process equipment costs is to obtain firm bids from fabricators or suppliers.
Percentage of delivered equipment cost is the method used for estimating the fixed or
total capital investment requires determination of the delivered equipment cost. The other
items included in the total direct plant cost are then estimated as percentage of the
delivered equipment The addition components of the capital investment are based on
average percentage of total direct plant cost total direct and indirect plant costs or total
capital investment.
Estimating by percentage of delivered equipment cost is commonly used for preliminary
and study estimates. It yield most accurate results when applied to a project similar in
configuration to recently constructed plants.
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Chapter: 11
Cost Estimation
DIRECT COST
PURCHASED Equipment Cost =E
COMPONENTS
% AGES OF E
COST ($)
Purchased equipment installation
47% E
a
Instrumentation (installed)
12%E
b
Piping (installed)
66% E
c
Electrical (installed)
11 % E
d
Building (including Service)
18% E
e
Yard improvement
10% E
f
Service facilities
70%. E
g
Land
6% E
h
Total direct cost
Total direct cost
D
=
D
INDIRECT COST
Engineering and supervision
33%E
Construction Expenses
41%E
Total indirect Cost
I
Total direct and indirect cost
D+I
Contractor's fee
5%(D+I)=y
Contigency
10%(D+I)= x
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Chapter: 11
Cost Estimation
E/fixed Capital investment
D+I+x+y
Working Capital investment
W.C.I
W.C.I
15% total capital
COST ESTIMATION OF OUR PLANT
EQUIPMENT PURCHASE COST:
1. ABSORPTION COLUMN
Cost of trays:
*Cost of 1 tray = 360 $
Material factor = 1.1 (for cast iron)
Corrected cost/tray = 360 X 1.1 = 399.6$
Cost of 10 trays = 10 X 399.6 = 3996$
Cost of column:
For D = 1m
*Cost = 18000$
Material factor = 1.5 (for SS)
Pressure = 1.1
Purchased cost for column = 1.1 x 1.5 x 18000= 29700$
Total column cost = 29700 + 3996 = 33696$
2. SUPERAZEOTROPIC COLUMN:
Cost of trays:
Cost of one tray = 360$
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Chapter: 11
Cost Estimation
Material factor = 1.11 (For SS)
Corrected cost per tray = 360x1.11 = 399.6
Cost of 10 trays = 10x399.6= 3996$
Cost of column:
For D = 1m
Cost = 18000$
Material correction factor = 1.5 (for SS)
Pressure factor = 1.1
Purchased cost for column = 1.1x1.5x18000= 29700$
Total column cost = 29700+3996 = 33696$
3. DISTILLATION COLUMN:
Cost of trays:
Cost of one tray = 400$
Material factor = 1.065 (for Carbon Steel)
Corrected cost per tray = 400x1.065x426 =
Cost of 15 trays = 15x426x426 = 6390$
Cost of column:
For D = 1m
Cost = 18000$
Material correction factor = 1.5 (for SS)
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Chapter: 11
Cost Estimation
Pressure factor = 1.1
Purchased cost for column = 1.1x1.5x18000= 29700$
Total column cost = 29700+3690 = 36090$
4. STRIPPER
Cost of trays:
Cost of one tray = 360$
Material factor = 1.0
Corrected cost per tray = 360x1.0 = 360$
Cost of 18trays = 18x360 = 6480$
Cost of column:
For D = 1.25m
Cost = 20000$
Material correction factor = 1.5
Pressure factor = 1.1
Purchased cost for column = 1.1x1.5x20000= 33000$
Total column cost = 6480+33000 = 39480$
5. Reactor
Cost of catalyst:
Cost of 90% Pt, 10% Rh = 11257 x 0.9 + 29907 x 0.1 = 13122 $ / kg
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Chapter: 11
Cost Estimation
Weight of catalyst = 11.8 kg
Cost = 13122 x 11.8 = 154839.6$
Reactor Head Cost:
Cost in 1977 = 7777.77$
Cost in 2005 = 89720.96$
Total Cost = 154839.6+89720.96 = 244560.56$
6. BOILER COST
Cost in 1977 = 34839.68$
Cost in 2006 = 374958.54$
7. NH3 EVAPORATOR COST:
Cost in 1977 = 22222.22 $
Cost in 2006 = 238805.96$
8. TAIL GAS TURBINE
Cost in 1977 = 236185.73$
Cost in 2006 = 2542951.9$
9. COMPRESSOR
Cost in 1977 = 409618.57$
Cost in 2006 = 4409618.56$
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Chapter: 11
Cost Estimation
10. COOLER CONDENSOR
Heat transfer area = 30.23 m2
Pressure factor = 1.0
Type factor = 0.8
Material of construction:
Shell and tubes = stainless steel
Base cost = 30000$
Actual cost = 30000x 0.8 x 1.0 = 24000$
Its the cost in 1992
Cost is 2006 = 27443.28$
Total Equipment Cost =7981300.8$
DIRECT COST
Purchased equipment cost
= 7981300.8$
Purchased equipment installation = 0.47 7981300.8
= 3751211.3$
Instrumentation & Process Control = 0.12 7981300.8
= 957756.0$
Piping (installed) = 0.66 7981300.8
=5267658.5$
Building (Including Services) = 0.18 7981300.8
= 1436634.1$
Yard improvements = 0.1 7981300.8
= 798130.0$
Service facilities (installed) = 0.7 7981300.8
= 5586910.5$
Land = 0.06 x 7981300.8
= 478878.0$
Total direct plant cost
=26258479.24$
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Chapter: 11
Cost Estimation
INDIRECT COST
Engg & Supervision = 0.33 7981300.8
= 2633829.2$
Construction expenses = 0.41 7981300.8 = 3272333.3$
Total Indirect Cost
= 5906162.5$
Total Direct & Indirect Cost
= 32164641.77$
Contractor’s fee = 0.05 32164641.77
= 1608232.08$
Contingency = 0.1 32164641.77
= 3216464.17$
FIXED CAPITAL INVESTMENT
Fixed Capital Investment = Total direct + indirect cost + contingency + Contractor’s fee
= 36989338.02$
Total capital investment
=
F.C.I + W.C
Now
W.C
=
0.15 (T.C.I)
= 0.15(36989338.02 + W.C)
=(5548400.703/0.85)
= 65275530.239 $
Total capital investment = 43516868.26$
180
Chapter: 11
Cost Estimation
PRODUCT COST.
Depreciation Value.
d=(V – Vs)/n
Where
V=36989338.02$
Vs=5%V
n=10 years
d= 3513987.112$
Raw Material Cost.
For 1 M.T NH3 price=2366.7 Rs (Pak Saudi Fertilizer limited)
For 1 hr operation NH3 required = 92.59 Kgmole
For one day operation NH3 required
(92.59Kgmol NH3/hr)*(17 Kg NH3/1 Kgmole NH3)*(1 M.T/1000Kg)*(24hr/1 day)
=37.77 M.T/day
For one year NH3 required
=(37.77 M.T NH3/day)*(330 day/1 year)
=12466.317 M.T NH3/year
Price of NH3 for one year=12466.317*2366.7
=29504033.86 Rs
=491733.89$
W.C.I.
W.C=65275530.239 $
Total Amount=W.C+Raw material cost+Depreciation Value
Total Amount=69281251.23$
181
Chapter: 11
Cost Estimation
Cost of product= Total Amount/M.T of HNO3 produced in one year
=69281251.23$/140*330
= 500.6$/M,T
=0.523$/Kg
=31.38 Rs/Kg
182
Appendices
APPENDECES
183
Appendices
184
Appendices
185
Appendices
186
Appendices
187
Appendices
188
Appendices
189
Appendices
190
Appendices
191
Appendices
192
Appendices
193
Appendices
194
Appendices
195
Appendices
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Appendices
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REFERENCES:
1. Coulson & Richardson, “Chemical Engineering”, Volume 6, 3rd edition
2. Coulson & Richardson, “Chemical Engineering”, Volume 2, 2nd edition
3. Peters, M, S. & Timmerhaus, K.D., “Plant Design & Economics for Chemical
Engineers”, 5th Edition, McGraw Hill.
4. Howard F. Rase, “Chemical Reactor Design For Process Plants”, Volume 1.
5. Smith, J. H., “Chemical Engineering Kinetics”, 2nd Edition, McGraw Hill.
6. Ludwig, E. “Applied Process Design for Chemical & Petrochemical Plant”,
Volume 3, Gulf Publishing Co., Houston 1961.
7. Kirk Othmer, “Encyclopedia of Chemical Engineering”.
8. Perry’s Chemical Engineering Handbook, 6th edition.
9. Design manuals of Pak Arab Fertilizers.
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