CHAPTER 1 INTRODUCTION
1
Urea is an oraganic compound with the chemical formula (NH2)2CO. Urea is also
known by the International Nonproprietary Name (INN) carbamide, as established by
the World Health Organization. Other names include carbamide resin, isourea,
carbonyl diamide, and carbonyldiamine.
Synthetic urea It was the first organic compound to be artificially synthesized from
inorganic starting materials, in 1828 by Friedrich Wöhler, who prepared it by the
reaction of potassium cyanate with ammonium sulfate. Although Wöhler was
attempting to prepare ammonium cyanate, by forming urea, he inadvertently
discredited vitalism, the theory that the chemicals of living organisms are
fundamentally different from inanimate matter, thus starting the discipline of organic
chemistry.
This artificial urea synthesis was mainly relevant to human health
because of urea cycle in human beings. Urea was discovered; synthesis in human
liver in order to expel excess nitrogen from the body. So in past urea was not
considered as a chemical for agricultural and industrial use. Within the 20th century it
was found to be a by far the best nitrogenic fertilizer for the plants and became
widely used as a fertilizer. Urea was the leading nitrogen fertilizer worldwide in
the 1990s.Apart from that urea is being utilized in many other industries.
Urea is produced on a scale of some 100,000,000 tons per year worldwide.
For use in industry, urea is produced from synthetic ammonia and carbon dioxide.
Urea can be produced as prills, granules, flakes, pellets, crystals, and solutions.More
than 90% of world production is destined for use as a fertilizer. Urea has the
highest nitrogen content of all solid nitrogenous fertilizers in common use (46.7%).
Therefore, it has the lowest transportation costs per unit of nitrogen nutrient. Urea is
highly soluble in water and is, therefore, also very suitable for use in fertilizer
solutions (in combination with ammonium nitrate).
Commercial production of urea Urea is commercially produced from two raw materials, ammonia,
and carbon dioxide. Large quantities of carbon dioxide are produced during the
manufacture of ammonia from coal or from hydrocarbons such as natural gas and
2
petroleum-derived raw materials. This allows direct synthesis of urea from these raw
materials. The production of urea from ammonia and carbon dioxide takes place in an
equilibrium reaction, with incomplete conversion of the reactants. The various urea
processes are characterized by the conditions under which urea formation takes
place and the way in which unconverted reactants are further processed.
Unconverted reactants can be used for the manufacture of other products,
for example ammonium nitrate or sulfate, or they can be recycled for complete
conversion to urea in a totalrecycle process. Two principal reactions take place in
the formation of urea from ammonia and carbon dioxide. The first reaction is
exothermic:
2 NH3 + CO2 ↔ H2N-COONH4 (ammonium carbamate)
Whereas the second reaction is endothermic:
H2N-COONH4 ↔ (NH2)2CO + H2O
Both reactions combined are exothermic.
Properties of urea
Mol. Formula - NH2CO NH2
Melting Point - 132.60C
Sp. Gravity - 1.335@20 0 C
Heat of Fusion - 57.08 Cal/gm
Heat of Combustion - 2531 Cal/gm
Crystal Form - Tetragonal
Nitrogen Content - 46.6%
Boiling Point - Decompose on boiling
Colour - White
Bulk Density - 740 to 750 kg/m3
Angle of Repose - 23 to 300C
Viscosity - [email protected] 0 C
Triple point -1020C
3
Dielectric Constant -3.52 to 0.2
Sp. Heat -0.42Kcal/cm2
Raw materials of urea manufacturing 1. Ammonia
Ammonia, NH3, is a comparatively stable, colourless gas at ordinary
temperatures, with a boiling point of -33 C. Ammonia gas is lighter than air, with a
density of approximately 0.6 times that of air at the same temperature. Ammonia is
highly soluble in water, although solubility decreases rapidly with increased
temperature. Ammonia reacts with water in a reversible reaction to produce
ammonium (NH4)+ and hydroxide (OH)- ions, as shown in
equation. Ammonia is a weak base, and at room temperature only about 1 in 200
molecules are present in the ammonium form (NH4)+. The formation of
hydroxide ions in this reaction increases the pH of the water, forming an alkaline
solution.
NH3 + H2O (NH4)+ + OH-
.
Ammonia Production Essentially all the processes employed for ammonia synthesis are
variations of the Haber-Bosch process, developed in Germany from 1904-1913.
This process involves the reaction of hydrogen and nitrogen under high
temperatures and pressures with an iron based
catalyst.
N2 +3 H2 2NH3
The source of nitrogen is always air. Hydrogen can be derived from a
number of raw materials including water, hydrocarbons from crude oil refining,
coal, and most commonly natural gas. Hydrogen rich reformer off-gases from oil
refineries have also been used as a source of hydrogen. Steam reforming is generally
employed for the production of hydrogen from these raw materials. This process
also generates carbon dioxide, which can then be used as a raw material in the
production of urea.
Ammonia storage
4
Anhydrous ammonia is usually stored as a liquid in refrigerated tanks at -
33.3 C and atmospheric pressure, often in doubled-walled tanks with the capacity
for hundreds or thousands of tonnes. The low temperature is usually maintained by
the venting of ammonia gas.
2. Carbon Dioxide CO2 is a odourless and colourless gas which contain 0.03% in the
atmosphere. It is emitted as a pollutant from number of industries. CO2 can be
obtained from ammonia production process as a by product.
Applications of urea 1. Agricultural use
More than 90% of world production is destined for use as a fertilizer. Urea
is used as a nitrogen-release fertilizer, as it hydrolyses back to ammonia and carbon
dioxide, but its most common impurity, biuret, must be present at less than 2%, as it
impairs plant growth. Urea has the highest nitrogen content of all solid
nitrogeneous fertilizers in common use (46.4%N.) It therefore has the lowest
transportation costs per unit of nitrogen nutrient. In the past decade urea has surpassed
and nearly replaced ammonium nitrate as a fertilizer
In the soil, urea is converted into the ammonium ion form of nitrogen.
For most floras, the ammonium form of nitrogen is just as effective as the nitrate
form. The ammonium form is better retained in the soil by the clay materials than the
nitrate form and is therefore less subject to leaching. Urea is highly soluble in water
and is therefore also very suitable for use in fertilizer solutions, e.g. in “foliar feed‟
fertilizers.
2. Industrial use Urea has the ability to form 'loose compounds', called clathrates, with
many organic compounds. The organic compounds are held in channels formed
by interpenetrating helices comprising of hydrogen-bonded urea molecules.
This behaviour can be used to separate mixtures, and has been used in the
production of aviation fuel and lubricating oils. As the helices are interconnected, all
helices in a crystal must have the same 'handedness'. This is determined when the 5
crystal is nucleated and can thus be forced by seeding. This property has been used
to separate racemic mixtures.
3. Further commercial uses
A stabilizer in nitrocellulose explosives
A component of fertilizer and animal feed, providing a relatively
cheap source of nitrogen to promote growth
A raw material for the manufacture of plastics, to be specific, urea-
formaldehyde resin
A raw material for the manufacture of various glues (urea-formaldehyde or
urea-melamine-formaldehyde); the latter is waterproof and is used for marine
plywood
An additive ingredient in cigarettes,designed to
enhance flavor
An ingredient in some hair conditioners, facial cleansers, bath oils, and lotions
A flame-proofing agent (commonly used in dry chemical fire extinguishers as Urea-potassium bicarbonate) An ingredient in many tooth whitening products
A cream to soften the skin, especially cracked skin on the bottom
of one's feet
An ingredient in dish soap.
4. Medical use
Urea is used in topical dermatological products to promote rehydration of
the skin. If covered by an occlusive dressing, 40% urea preparations may also
be used for nonsurgical debridement of nails. This drug is also used as an earwax
removal aid. Like saline, urea injection is used to perform abortions. It is also the
main component of an alternative medicinal treatment referred to as urine therapy.
6
5. Textile use
Urea is a raw material for urea-formaldehyde resins production in the adhesives
and textile industries. A significant portion of urea production is used in the
preparation of urea- formaldehyde resins. These synthetic resins are used in the
manufacture of adhesives, moulding powders, varnishes and foams. They are also
used for impregnating paper, textiles and leather. In textile laboratories they are
frequently used both in dyeing and printing as an important auxiliary, which
provides solubility to the bath and retains some moisture required for the
dyeing or printing process.
7
CHAPTER 2 PROCESS SELECTION
8
Process Selection Several processes are used to urea manufacturing. Some of them are used
conventional technologies and others use modern technologies to achieve high efficiency. These
processes have several comparable advantages and disadvantages based on capital cost,
maintenance cost, energy cost, efficiency and product quality. Some of the widely used urea
production processes are
1. Conventional processes
2. Stamicarbon CO2 - stripping process
3. Snamprogetti Ammonia and self stripping processes
4. Isobaric double recycle process
5. ACES process
Snamprogetti Ammonia and self stripping processes In the first generation of NH3 and self strip ping processes, ammonia was used as
stripping agent. Because of the extreme solubility of ammonia in the urea containing synthesis
fluid, the stripper effluent contained rather large amount s of dissolved ammonia, causing
ammonia overload in down stream section of the plant. Later versions of the process abandoned
the idea of using ammonia as stripping agent; stripping was achieved only by supply of heat.
Even without using ammonia as a stripping agent, the NH3:CO2 ratio in the stripper effluent is
relatively high. So the recirculation section of the plant requires an ammonia-carbomate
separation section
The process uses a vertical layout in the synthesis section. Recycle within the
synthesis section, from the stripper via the high pressure carbamate condenser, through the
carbamate separator back to the reactor, is maintained by using an ammonia-driven liquid-liquid
ejector. In the reactor, which is operated at 150 bars, NH3:CO2 molar feed ratio of 3.5 is applied.
The stripper is of the falling film type. Since stripping is achieved thermally, relatively high
temperatures (200-210 0C) are required to obtain a reasonable stripping efficiency. Because of
this high temperature, stainless steel is not suitable as a construction material for the stripper
from a corrosion point of view; titanium and bimetallic zircornium - stainless steel tubes have
been used Off gas from the stripper is condensed in a kettle type boiler. At the tube side of this
condenser the off gas is absorbed in recycled liquid carbamate from the medium pressure
recovery section. The heat of absorption is removed through the tubes, which are cooled by the
9
production of low pressure steam at the shell side. The steam produced is used
effectively in the back end of the process.In the medium pressure decomposition and recirculation
section , typically operated at 18 bar, the urea solution from the high pressure stripper is subjected
to the decomposition of carbamate and evaporation of ammonia. The off gas from this medium
pressure decomposer is
rectified. Liquid ammonia reflux is applied to the top of this rectifier; in this way a top product
consisting of pure gaseous ammonia and a bottom product of liquid ammonium carbamate are
obtained. The pure ammonia off gas is condensed and recycled to the synthesis section. To
prevent solidification of ammonium carbamate in the rectifier, some water is added to the bottom
section of the column to dilute the ammonium carbamate below its crystallization point. The
liquid ammonium carbamate-water mixture obtained in this way is also recycled to the synthesis
section. The purge gas of the ammonia condenser is treated in a scrubber prior to being purged to
the atmosphere.
The urea solution from the medium pressure decomposer is subjected to a second
low pressure decomposition step. Here further decomposition of ammonium carbamate is
achieved, so that a substantially carbamate -free aqueous urea solution is obtained. Off gas from this
low pressure decomposer is condensed and recycled as an aqueous ammonium carbamate
solution to the synthesis section via the medium pressure recovery section.
Concentrating the urea water mixture obtained from the low pressure decomposer is
preformed in a single or double evaporator depending on the requirement of the finishing
section. Typically, if prilling is chosen as the final shaping procedure, a two stage evaporator is
required, whereas in the case of a fluidized bed granulator a single evaporation step is sufficient to
achieve the required final moisture content of the urea melt. In some versions of the process, heat
exchange is applied between the off gas from the medium pressure decomposer and the aqueous
urea solution to the evaporation section. In this way, the consumption of low pressure steam by the
process is reduced.
The process condensate obtained from the evaporation section is subjected to a
desorption hydrolysis operation to recover the urea and ammonia contained in the process
condensate.
10
CHAPTER 3 PROCESS DESCRIPTION AND FLOW SHEET
11
PROCESS DESCRIPION
The process which is used in formation of urea is Snam Pragetti Process at IFFCO Plant. This is self-stopping process. The basic raw material for the formation of urea is Ammonia & Carbon Dioxide . The formation of urea is taking place in following manner:-
2NH3 + CO2 NH4COONH2 + Heat (1) (ammonium carbamate)
NH4COONH2 NH2CONH2 + H2O - Heat (2)(UREA)
First reaction is takes place at high pressure and temperature that is P=150kg/cm2(g) & T= 1700C. In this reaction carbamate is formed. At high pressure reaction is taking place at in forward direction and at low pressure reaction is taking place in backward direction. It is exothermic reaction. In the 2nd reaction carbamate is dehydrated to form Urea. This is endothermic process. The heat which is generated in reaction first is utilised in reaction two. At a very high temperature reaction two proceed backward direction.
The process root is summarised in the following steps:-
1. COMPRESSION OF CARBON DIOXIDE
In this step carbon dioxide is compressed through compressor. The carbon dioxide enters in the compressor at a 1.4 ata & temp. is around 400C for increasing the pressure up to 155kg/cm2(g). This is achieved by using two centrifugal pumps driven by an extraction cum condensing turbine. Ammonia is comes from the Ammonia Plant or from the Ammonia Storage Tank. The ammonia is passed through the preheated tank to high pressure synthesis loop. The high pressure synthesis loop is combination of booster centrifugal pump and reciprocating pressure pump. The pressure of ammonia comes out from the high pressure synthesis loop is 240kg/cm2. The high pressure liquid ammonia is also provided for motive force for ejector, which recycles carbamate solution to urea reactor. The ammonia is kept in excess for the complete conversion of carbon dioxide. The ration of ammonia to carbon dioxide is 3.33:1.
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2. UREA SYNTHESIS AND HIGH PRESSURE RECOVERY
This section consist of reactor, high pressure stripper, horizontal carbamate condenser (two unit placed in series). The compressed carbon dioxide and excess ammonia is entered in the reactor to form the urea at the temp. 1900C & pressure 150kg/cm2(g). the concentration of urea formed in the reactor is nearly 32%. The effluent of reactor is consisting of ammonia. Carbon dioxide, carbamate,vapour and urea. This effluent is passed to stripper in which CO2 is absorbed according to the henery law. Heat required for stripping is supplied by 26kg/cm2(g) steam obtained from extraction of carbon dioxide compression turbine. The concentration of urea obtained from the stripper is 45%. The off gases obtained from the stripper ammonia, CO2 and vapour is entered into horizontal carbamate condenser where the total mixture ,except for some inert ,is condensed as carbamate and recycled to the reactor by means of ejector.
3. UREA PURIFICATION AND LOW PRESSURE RECOVERIES
Urea purification takes place in two stages at decreasing pressures as follows:
Medium Pressure at 18 ata pressure
Low Pressure at 4.5 ata pressure
Medium Pressure Purification and Recovery at 18ata
The solution, with a low residual CO2 content, leaving the bottom of the stripper at synthesis pressure is let down to18 ata and enters medium pressure decomposer The M.p decomposer and divided in two parts
1.Top separated :where the released flash gases are removed before the solution enters the tube bundle
2. Decomposition section (falling film type): where residual carbamate is decomposed and the heat require for the decomposition is applied by means of 26 ata steam condensate flowing out of the shell side of stripper
The NH3 and CO2 rich gases leaving the top separator are sent to medium pressure condenser where they are partially absorbed in aqueous carbonate solution coming from low pressure recovery section .The absorption heat is removed by tempered cooling water circulation in the tube side of the medium pressure condenser. In the M.P condenser CO2 is almost totally absorbed. The effluents flow to medium pressure absorber. The gaseous phase enters the
13
rectification section of the M.P absorber. The rectification section has bubble trays. The bubble cap trays are fed by pure reflux ammonia at the top trays which eliminates residual CO2 and H2O from gases leaving M.P absorber. The reflux ammonia is pumped to rectification column.NH3 with inert gases leaving the M.P absorber is condensed in ammonia condenser.
The inert gases , saturated with ammonia enter ammonia preheater where an additional amount of ammonia is condensed by heating cold ammonia coming from ammonia storage area and used as make up feed to Urea plant
The inert gases with residual ammonia content are sent to medium pressure ammonia absorber, which is a falling film type and where they meet a condensate flow which absorbs ammonia From bottom of ammonia absorber the water ammonia solution is pumped to medium pressure absorber.The inerts leaving the top are free from ammonia.
Low pressure purification and recovery stage(at 4.5 ata)Low pressure decomposer consists of:
1.Top separator: where the released gases are removed before the solution enters the lower tube bundle
2. Decomposition section (falling film type):where residual carbamate is decomposed and the heat require for the decomposition is applied by means of saturated steam at 4.5 ata
The urea solution from the M.P decomposer bottom enters the L.P decomposer after expansion through a level controller. Consequently most of the residual carbamate is decomposed and in the process urea solution gets concentrated. The remaining carbamate is decomposed in a falling film exchanger, which is a part of L.P. decomposer.
The vapors from the L.P decomposer enter the L.P. condenser where they get cooled and liquefied. Prior to the entry of L.P off gases in L.P condenser the vapor gets mixed with the aqueous solution from waste water section.The vapor thus formed get condensed in L.P condenser goes to carbonate solution tank from where it is send back to MP condenser.The inert gases in the tank contains considerable amount of ammonia and thus are absorbed in cool condensate before being sent to vent stack.
The urea solution at the bottom of the L.P. decomposer is sent to pre vacuum concentrator through a level control valve.
14
Urea concentration section:As it is necessary in order to prill urea ;to concentrate urea solution up to 99.8 % wt, a vaccum concentration section in two stages is provided.
The two concentrator use saturated steam at 4.5 ata the liq. Vapor phase coming out of second vacuum concentrator enters gas- liq. Separator where the vapors are extacted by second vacuum system.
First vacuum system:First evaporator is operated at 130C and 0.3 Kg/cm2 pressure. Over head vapor from the top of the first vacuum separator is directed to the shell side of pre condenser and heat of condensation is removed by cooling water in the tube side. Ammonia vapor and residual CO2 is absorbed in condensate forming dil. Ammonium carbonate sol. and flows down through barometric leg of waste water tank.
Uncondensed gases are sucked by the ejector (motive fluid being 44.5 ata steam) and discharged in the shell side after condenser , which also receives uncondensed gases from second vacuum system.Heat of condensation is removed by cooling water in the tube side.
Second vacuum system:It operates at 1400 C and 0.03 Kg/cm2 pressure. Over head gases from second vacuum separator are sucked by a booster ejector and discharged at slightly higher pressure where heat of condensation is removed by cooling water in the tube side.
Uncondensed gases are drawn by ejector and discharged to shell side of second inter condenser where heat of condensation is again removed by cooling water.
Urea prilling:The molten urea leaving second vacuum separator is pumped to the prilling bucket by means of centrifugal pump.
The molten urea coming out of the prilling bucket in the form of drops fall along the prilling tower and encounters air flow which causes its solidification and subsequent cooling solid prills are sent to the conveyer belt by rotary scraper which carries urea to bagging plant or storage. The heated air containing few ppm of NH3 is released from the top into the atmosphere.
15
PROBLEM STATEMENT:
To design a UREA PLANT of capacity 1000 ton/day using ammonia and carbon di-oxide as raw material
16
CHAPTER 4 MATERIAL BALANCE
17
MATERIAL BALANCE
Plant Capacity =1000 ton/day of urea
Taking 10% over the design factor,
Capacity =1.1*1000 =1100 tons/day
=1100*1000 =45833.33 Kg/hr
24
=763.89 Kmol/hr
1. IInd VACCUM CONCENTRATOR
X3U =0.989 (given) X3
B =0.002 (assume)
: W3 =46343.0107 Kg/hr 4 XNH3=0.2722
X3H2O =0.011 XCO2= 0.128
XU5 =0.94 (given)
Material balance: 5
W3 X3U = W5 XU
5 XU=0.94
46343.1075*0.989 = W5 *0.94 XH2O=0.040
W5 =48758.865 Kg/hr XNH3=0.0135
W4 = W5 – W3 XCO2 =0.00634
=2415.76 Kg/hr XB= 0.0019 3 XU =0.989
Let X4NH3 =0.2722 X4
CO2=0.128 XH2O =0.011
:0.2722*2415.76 = X5NH3 *48758.865 XB =0.002
: X5NH3 = 0.1035
: 0.128 *2415.76 = X5CO2 *48758.865
X5CO2 = 0.00634
X5H2O = 0.040
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2 .Ist VACCUM CONCENTRATOR
XU7 =0.70 , XU
5 =0.94(given)
Material balance
: 0.94 *48758.865 =0.7 * W7
W7 =65476.19 Kg/hr
W6 = W7 – W5
= 16717.325 Kg/hr
W6
X5CO2 = 0.00634 XB=0.001415 XH2O=0.76
X5NH3 =0.0135 XU=0.7 XNH3=0.17
XB5 = 0.0019 XNH3 =0.0534 XCO2=0.0726
Let water evaporated =12683.7973, W7
CO2 from top =1214.65, XCO2=0.023
NH3 from top =2841.945 XH2O=0.22 XU=0.94
X6CO2=0.0126, X6
NH3 =0.17 , X6H2O =0.76 W5 XB=0.0019
: X7NH3 *65476.19 =0.17*16717.325 +0.0135*48758.865 XNH3=0.0135
X7NH3= 0.0534 XCO2=0.00634
X7B*65476.19 =0.0019 *48758.865 XH2O=0.040
X7B =0.001415
X7CO2 *65476.19 =1214.65+0.00634*48758.865
X7CO2 =0.023
X7H2O = 0.22
19
3 L.P DECOMPOSER
XU9 =0.63 , XU
7 =0.70 (given)
Material balance
0.63 *W9 = 0.7 *65476.19
W9 = 72751.322
Let carbamate =72.28 Kmol/hr
As NH2COONH4 2NH3 + CO2
NH3 produced =2 * 72.28
=144.56 Kmol/hr
=2457.58 Kg/hr
:CO2 produced =3180.32 Kg/hr W8
72751.132 =65476.19 +W8 XNH3=0.517
W8 =7275.132 XH2O=0.252
CO2 in vapour =3180.32 -0.023*65476.19 XCO2=0.23
=1674.367 Kg/hr W9
Let H2O evaporated =1838.77 Kg/hr XNH3=0.0997
NH3 evaporated =1304.475 Kg/hr XB=0.00127
W8 =1304.475 +2457.52 +1838.77 +1674.367 XCARB=0.0775
=7275.132 XH2O=0.1928
0.001415 *65476.19 = X9B *72757.322 W7 XNH3=0.0534
X9B =0.00127 XB=0.00415
X9NH3 *72757.322 =0.0534 *65476.19 +1304.475 +2457.52 XCO2=0.023
X9NH3 =0.0997 XH2O=0.22
X9CARB = 72.28*78 XU=0.70
72757.322
20
= 0.0775
: X9H2O = 0.1928
4 .VACCUM SYSTEMS
W10 = W6 + W4 =19133.085 Kg/hr W6 + W4
X10NH3 = 2841.945 +(0.2722*2415.76)
19133.085
=0.183
X10H2O = 12683.7973 +0.599*2415.76
19133.085 XNH3=0.183, XH2O=0.739
= 0.739 W10 XCO2=0.078
: X10CO2 =0.0783
5.DISTILLATION COLUMN
NH3 distilled = 17% W11 XNH3=0.289
= 3252.62 Kg/hr XH2O=0.576
Water distilled = 32.8% XCO2=0.133
=6466.98 Kg/hr W10
CO2 at top =1498.12 XNH3=0.183
W11 =11217.72 XH2O=0.739
W10 – W11 = W12 XCO2=0.078
W12 =19133.085 -11217.72
=7915.365 Kg/hr W12 XNH3=0.031
XNH 3=0.031 XH2O=0.968
XH2O =0.968
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6.REFLUX ACCUMULATOR
Assume CO2 evaporation =52.7% W13
=789.196 Kg/hr
; W13 =789.196 W11 W14
For W15 , NH3 =2663.06 XNH3 =0.289
CO2 =652.88 XH2O=0.576 XCO2=0.076
H2O=5240.22 XCO2=0.133 W15 XH2O=0.615
: W15 =8590.52 XNH3=0.309
W14 = W11 - W15 - W13
= 1838.004 Kg/hr W13
X 14XNH3 = W11*0.29 - W15*0.31 W11 W14 XCO2=0.46
X 14NH3 = 0.32 XNH3 =0.32
W14X14CO2 = W11(0.1325) - W15 (0.076) W15
: X 14CO2 =0.46
7. L. P CONDENSER
Assume:
W8 + W14 =9113.136 XCO2=0.0873
NH3 =4350.16 W16(g) XNH3=0.912
CO2 =2519.85
H2O =2243.13
Assume: W14(g)+ W8(g)
NH3 in vapour = 526.37
CO2 in vapour =50.397
: W16 =576.767 Kg/hr
W17 = W14 + W8 - W16 W17(l) XCO2 =0.29, XNH3=0.145, XH2O =0.565
22
= 8536.369 Kg/hr
W 17XNH3 = W8(0.577) + W14(0.32) -526.37
: X 17NH3 =0.45
W 17XCO2 = W8 X 8CO2 + W14 X 14
CO2 -50.397
: XCO2 =0.29
8. L.P NH3 ABSORBER
NH3 added =526.37
Let H2O added =440 Kg/hr vent(g) 50.397Kg/hr
: W18 =966.37 Kg/hr W16(g) H2O(440 Kg/hr)
W16 +440 =966.37 +vent XNH3=0.912 XNH3=0.544
:vent =50.397 Kg/hr XCO2=0.0873 W18(l) XH2O=0.455
9. CARBAMATE SOLUTION TANK
W21 = W17 + W18 XNH3=0.544
=9502.739 Kg/hr W18(l) XH2O=0.45
W21 XNH3 = W18*0.544 + W17*0.45 W17(L)
X 21NH3 =0.46 XNH3 =0.145 W21(l)
W21 XCO2 = W17 *0.29 XCO2=0.29 XNH3=0.46
X 21CO2 =0.26 XH2O=0.563 XCO2= 0.26
XH2O=0.28
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10.M .P DECOMPOSER
Let 64% of carbamate decomposed
:cabamate decomposed =130 Kmol/hr
=10140 Kg/hr W23(L) W22(G) XNH3=0.307
NH2COONH4 2NH3 +CO2 XH2O=0.299
NH3 formed =2*130 =260 Kmol/hr XCO2=0.393
=4420 Kg/hr
CO2 formed =130 Kmol/hr =5720 Kg/hr
Let 0.5% of urea be converted into biuret W9(L) XNH3=0.0997
2NH2CONH2 NH2CONHCONH2 +NH3 XB=0.00127
Urea converted = 4072 Kmol/hr XCARB=0.0775
=286.46 Kg/hr XH2O=0.1928
Biuret formed =2.385 Kmol/hr W23(L) W22(G)
=245.69 Kg/hr XNH3=0.134
NH3 formed =2.385 *17 =40.545 Kg/hr XCARB=0.18
Total NH3 in vapour =4420+40.545 Kg/hr XH2O=0.156
Water evaporated =4356.4 Kg/hr XU=0.53 W9(L)
W22 =4356.4 +4460.545 +5720
=14536.945 Kg/hr
W23 - W22 = W9
: W23 =72751.322+14536.945
=87288.267 Kg/hr
W23XU = W9*0.63 +urea converted
=45833.33 +286.46
=46119.79
24
: X 23U =0.53
W23Xcarb = W9*0.078 +carbamate decomposed
X 23carb =0.18
W23XNH3 = W9 *0.0997 +4460.545
X 23NH3 = 0.134
11. M.P CONDENSER
CO2 in (W22 + W21 ) XNH3=0.46
=5720 +2470.712 W21(l) XCO2=0.26
=8190.712 W22(g) XH2O=0.28
NH3 =4460.545 +4371.26 XNH3=0.307 W25(l)
=8831.8 XH2O=0.299
W25 = W23 + W22 XCO2=0.393
=9502.739+14536.945
=24039.684 Kg/hr
98% of CO2 converted back to carbamate through condensation
2NH3 +CO2 NH2COONH4
Carbamate formed =0.98 *8190.712/44
=182.45 Kmol/hr W21(l)
=14229.5 Kg/hr W25(l)
W 25Xcarb =14229.5 W22(g) XNH3 =0.27
: X 25carb =0.341 XCO2=0.241
W25XNH3 = W21 X 21NH3 + W22 X 22
NH3 XCARB=0.42
=4460.545 +0.46 *9502.739
25
: X 25NH3 = 0.37
12.M.P ABSORBER
NH3 =2400 Kg/hr W26(g) W28(g)
W26 =2400 Kg/hr XNH3=1
Water in W27 =3900 Kg/hr W25 W27(l)
& NH3 in W27 = 475 Kg/hr XH2O=0.003 XNH3=0.108
: W27 =4375 XNH3=0.27 XH2O=0.891
CO2 absorbed =113.23 Kg/hr XCO2= 0.241 W29(l)
NH3 absorbed =85% XCARB=0.492
Total ammonia in =11769.68
NH3 absorbed =10023.68 Kg/hr
NH3 not absorbed =8894.68 +2400+475 -10013.68 W26(g) W28(g) XNH3=0.98
=1746.0 Kg/hr W25 XH2O=0.011
CO2 not absorbed=9215.74 W27(l)
: W28 =10961.74
W29 = W25 + W26 + W27 + W28 W29(l) XCO2=0.0059 , XNH3= 0.42
=19852.9 Kg/hr XCARB=0.50
W 29X29 CO2 = W 25X25CO2 - W 28X28
CO2
: X 29CO2 =0.0057
W 29X29 carb= W 25X25carb
X29 carb=0.50
W 29X29 NH3= W 26X26 NH3 + W 27X27 NH3 +8894.62 - W 28X28 NH3
=2400+475+8894.62-1746
=10023.68
: X29 NH3 =0.32
26
13.AMMONIA CONDENSER
Let ammonia condensed =25% W30(g) XNH3=0.1244
=0.25*1746 XCO2=0.876
=436.5 W28(g)
: W31 =436. XNH3=0.98 W31(l)
W28 = W30 + W31 XCO2=0.011 XNH3=1
: W30 =10961.74 -436.5
=10525.24
W 30X30 NH3 = W 28X28 NH3 – NH3 condensed
=1746 -436.5
=1309.5
: X30 NH3 =0.1244
W 30X30 CO2 = W 28X28 CO2
=9215.74
: X30 CO2 =0.876
14.AMMONIA RECEIVER
W30 + W31 =1746
Let ammonia leaving top =454
Let CO2 leaving top =45 XNH3=1, W32(l) W34(g) XNH3=0.909
NH3 leaving bottom =34000 Kg/hr XCO2=0.090
: W34 =499 & W33 =34000 W31(l) + W30(g)
W30 + W31 + W32 = W33 +W34
1746+ W32 =34000+499
: W32 =32753 W33(L) XNH3=1
27
15.M.P ABSORBER
Let ammonia absorbed =464.5 Kg/hr
Water added =4000
Vent = W34 +water added – W27
=499 +4000 -4375
=124 Kg/hr
16.STRIPPER
Let 80% conversion of carbamate is taken place at stripper
W23 =87288.267 Kg/hr W35(g) XNH3=0.750
NH323 =11696.63 Kg/hr XCO2=0.24
Carb23 =15711.9 Kg/hr W36(g)
Carbamate decomposed = 930 Kg/hr XNH3=0.222
NH2COONH4 2NH3 +CO2 XN2=0.00193
NH3 formed =70000 Kg/hr XCARB=0.58
CO2 formed =35000 Kg/hr XU=0.199 W23(l) XNH3=0.134
NH3 boiled in vapour =40000 Kg/hr XH2O=0.156
Total NH3 boiled in vapour =40000 +70000 =110000 Kg/hr XCARB=0.18
NH3 in vapour =430 Kg/hr XU=0.53
W35 =110000 +35000 +430
=145430 Kg/hr
W36 = W35 + W23
=145430+87288.267
=232718.27 Kg/hr
W 36XU = W23 (0.53)
: X36U =0.199
28
W 36X36 N2 = W 35X35 N2 =450
: X36 N2 =0.00193
W 36X NH3 = W 23X23 NH3 + W 35X35 NH3 - NH3 formed
=11696.6+110000-70000
=57696.6 Kg/hr
X NH3 =0.222
W 36X CARB = W 23X23CARB
: X36CARB = 0.58
17 CARBAMATE CONDENSER
NH3 + CO2 NH4COONH2 XNH3=0.759 W35(g)
Let CO2 condensed =64000 XCO2=0.24
NH3 condensed =59100 W29 XNH3=0.42
:carbamate formed =113454.54 XCO2=0.0059
W37 = W35 + W29 XCARB=0.50
= 145430+19852.9 XNH3= 0.25 W37
= 165282.9 Kg/hr XCARB=0.73, XCO2=0.00069, XH2O=0.0693
W 37X CARB = W29X29 CARB + W29 (0.32) – NH3 condensed
=11000+6352.9 -59100
=0.25
W 37XCO2 =64000+113.23-64000
=0.00069
29
18.CARBAMATE SEPARATOR
NH3 evaporated =1500 Kg/hr vent(g) XNH3=0.43,XH2O=0.52
H2O evaporated =1800 Kg/hr XCO2=0.048
CO2 in vent = 150 Kg/hr W38 W37
Vent =3450 Kg/hr XNH3=0.17 XNH3=0.25
W 37= W 38 +vent XH2O=0.09 XCARB=0.73
W 38 =165282.9 – 3450 XCARB=0.74 XCO2=0.00069
= 161832.9 XH2O=0.0193
W 38X CARB = W 37X37 CARB +vent X CARB
=41320.725 +1500
X38 NH3 =0.17
X38 H2O =0.09
19.UREA REACTOR
W 40 = W 33 - W 26 W36(g) XU=0.199
= 34000 -2400 XNH3=0.222
=31600 Kg/hr XCARB=0.58
W 36 = W39 + W38 + W 40 W39 XN2=0.00193
= 232718.27 -161832.9 -31600 XCO2=0.9854
=39285.37 Kg/hr XN2=0.01145
Known amount of N2 in W 36 =430
Corresponding O2 =143.3 W40 W38 XCARB=0.74
Air used =573.2 XNH3=0.17, XH2O=0.09
Amount of CO2 coming in reactor with W39 steam=39285.37-573.2
=38712.17 Kg/hr
30
W 39XCO2 =38712.17
: X39CO2 =0.9854
W 39XN2 = W 36X36 N2
: W39XN2=0.01145
2NH3 +CO2 NH2COONH4
For 100% CO2 conversion,
NH3 required =W32
= 32753 Kg/hr
=1926.65 Kmol/hr
UREA FORMED =771.85 Kmol/hr
31
CHAPTER 5 ENERGY BALANCE
32
ENERGY BALANCE
Energy balance across stripper:
Now, here the solution enters at 190˚Cand leaves at 210˚C.Also a part of carbamate decomposes into NH3 and CO2.The decomposition is as follows;
NH4COONH2 2NH3 +CO2: ∆H= -38.Kcal/mol
The vapour and gases product from the top at 190˚C.Now we assume an average temperature of 200˚Cand find the final heat capacity of the solution.
Cp of water =4.278 KJ/Kg˚C
Cp of ammonia =8.851 KJ/Kg˚C
Cp of carbamate =2.682 KJ/Kg˚C
Cp of urea =2.331 KJ/Kg˚C
Cp mean :
[ ( 4.278 *42510.27) +(8.851 *15711.9) +(2.682 *11696.63) +(2.331 *17370.365)]
(42510.27 +15711.9 +11696.63 +17370.365)
Cp mean =4.499 KJ/Kg˚C
Energy consumed in raising solution temperature is
=87288.267 *4.499*(210-190)
=7854198.265KJ/hr
Energy consumed in carbamate decomposition is
=(930/78) *159 *103
=1895769.23 KJ/hr
Total energy consumed =9749967.495 KJ/hr
Saturated steam at 26 bar and266˚C is used for heating, λ=1827.914 KJ/kg
Steam flow rate =5333.93 Kg/hr
33
Energy balance across M.P Decomposer
Now, here the solution enters from the stripper, the solution is at 145 Kg/cm2 and 210˚C, and it is flashed to 17 Kg/cm2before it enters the decomposer.Thus the temperature falls to 147˚C.Here the first solution is heated to155˚C then water is evaporated from the solution.Also carbamate is decomposed producing NH3 and CO2 .The energy is supply by steam condensate at 225˚C at 26 bar, which gets cooled to 210˚C in this process
Cp mean =[ (4.27*14026.45) + (8.85*7253.306) + (2.68*5638.23) + (2.33*45833.33)]
72751.322
Energy consumed in raising solution temperature :
=72751.322*3.38(155-147)
=3.38 KJ/Kg˚C
Energy consumed in carbamate decomposition :
=130*159*103
=20670000 KJ/hr
Total energy consumed =22637894.57 KJ/hr
Hot water at 225˚C is used to provide the heat ,which get cooled to 210˚C
Water flow rate =22637894.57/(4.2*(225-210))
=359331.66 Kg/hr
Energy balance across L.P Decomposer:
Now the solution enters from the M.P Decomposer at 147˚C and 16.5 Kg/cm2 and it is flashed to 13.5 Kg/cm2 before it enters the decomposer.Thus the temperature falls to 100˚C.
Here all the carbamate isdecomposed and liquid stream flows out at a temperature of138˚C.The top gases have NH3, CO2 and water vapour. The energy is supplied by steam condensate at 148˚C at 4.5 bar.
Cp mean=[(2.33*45833.33) +(4.278*14404.76) +(8.85*3496.43)]
65476.19
=3.045 KJ/Kg˚C
Energy consumed in carbamate decomposition :
34
=159*103*72.28
=11492520 KJ/hr
Energy consumed in evaporation of water:
=1838.77*2149
=3951516.73 KJ/hr
Energy consumed in raising solution temperature :
=65476.19*3.045*38
=7576249.9 KJ/hr
Total energy consumed:
=20123979.16 KJ/hr
Energy balance across L.P ammonia absorber:
Ammonia is absorbed in water and the residue O2 is vented to the atmosphere
Here ammonia dissolved is =526.37Kg/hr
Heat released due to ammonia absorption =526.37/17 *34.79*103
=1077200.724 KJ/hr
Cooling water enters at 32˚C and leaves at 40˚C
Cooling water required =1077200.724 / 4.2*(40-32)
=32059.5437Kg/hr
Energy balance across I st vaccum concentrator separator:
In this due to flashing the temperature of the solution drop to 90˚cwhile effluent stream goes out at 130˚C.This heat is required to raise the temoerature and vaporize the water.The system operates at 130˚C and 0.3 Kg/cm2 pressure.
Cp mean=[(4.278*1958.156) +(8.851*658.2) +(2.331*45833.33)]
48449.6891
=2.5 KJ/Kg˚C
Heat to raise the solution temperature =4878.865*2.5(130-90)
35
=4872504.618 KJ/hr
Heat required to vaporize water =12683.7973*2337
=29642034.29 KJ/hr
Total heat required =34514538.9 KJ/hr
Steam at 4.5 bar is used as heating
λ =2119.77KJ/kg
Steam required =34514538.91
2119.7
=16282.75Kg/hr
Energy balance across II nd vaccum concentrator separator:
The unit operates at0.03 Kg/cm2 and 140˚C.The incoming feedis at 130˚C due to flashing.Thusheat is required to first raise the feed temperature to 140˚C and to vaporize water.
Cp mean=[(4.278*509.77) +(2.33*45833.33) ]
46343.10
=2.35 KJ/Kg˚C
Heat to raise the solution temperature =46343.10*2.35*10
=1089062.55 KJ/hr
Heat required to vaporize water =2445.1*1448.97
=3542883.511 KJ/hr
Total heat required =4631946.361 KJ/hr
Steam at 4.5bar is used as heating media
36
λ=2119.7KJ/kg
Steam required = 4631946.361
2119.7
=2185.19Kg/hr
Energy balance across reactor:
2NH3 +CO2 NH2COONH4 38.1Kcal/mol
NH2COONH4 NH2COONH2 +H2O -7.1Kcal/mol
Carbamate formed =1730.47 Kmol/hr
Urea formed =771.85 Kmol/hr
=46313.94Kg/hr
Energy released in carbamate formation:
=1730.47*100*159
=27514460.77 KJ/hr
Energy consumed in urea formation :
= 46310.94*30
=1389328.2 KJ/hr
Net energy released =26125132.57 KJ/hr
37
DESIGN OF AMMONIA PREHEATER
Ammonia inlet flow rate = 32753 kg/hr
Specific heat of ammonia = 1.23 kcal/kg c
Inlet ammonia temp.= 86 F = 30 c
Outlet ammonia temp.= 230 F= 110 c
Heat required by ammonia= (32753 kg/ 3600 sec) * 5.14 KJ/kg c * 80 c
= 3741.12 KJ/sec
= 3741.12 KW
Latent heat of steam = 503.7 cal/mol= 2105.466 J/mol
Therefore, m*2105.466 = 3741.12
m = (3741.12*1000)/2105.466
m = 1776.86 mol/sec =98.7 kg/sec
LMTD = {(304 - 86) – (304 – 230 )}/ ln (218/74) = 133.28
Assume U = 200 W/ m2 c
Area A = Q/(U*LMTD) = (3741.12*1000)/(200*133.28)= 140.35 m2
Choose 20mm O.D.,16mm I.D.,4.88m long tubes, cupro nickel
L= 4.83m
Area of one tube = 3.14*d*l= 3.14*4.83*(20/1000)= 0.303 m2
No.of tubes= 140.5/0.303 = 463
Use 1.25 triangular pitch
Bundle diameter, Db= d0(Nt/k1) where, Nt= no.of tubes
Db= bundle dia in mm
d0= tube outside dia in mm
Therefore, Db= 20*(43/0.249) = 605.897mm
38
Using a split ring floating head type,
From the graph— bundle diameter clearance= 62mm
Therefore, Shell diameter, Ds= 605.897+62 = 667.897mm
COLD FLUID TUBE SIDE:--
Mean ammonia temp.= (230+86)/2 =158 F= 70 c
Tube cross-sectional area =(3.14/4)*162= 201mm2
Tubes per pass =no.of tubes/2 = 463/2 = 231.5 = 231
Tube flow area = (231*201)/1000000= 0.046m2
Ammonia mass velocity= 32753/(60*60*0.046)= 197.78 kg/sec m2
Density of ammonia = 0.618 g/ml= 618 kg/m3
Ammonia linear velocity, ut= 197.78/618 = 0.32 m/sec
From the relation,
Viscosity of ammonia= 0.19 cp=(0.19/100) N/m2
K of ammonia= 0.29 Btu/hr ft2 = 0.29*1.729 = 0.50141 W/m c
Specific heat of ammonia= 1.02 cal/gm c= 4.2636 KJ/kg c
Reynold no.= (di*v*p)/viscosity= (618*0.32*16)/(0.19*10)= 1665.35
Prandtl no.=(cp*viscosity)/Kf=(4.2636*10*0.19)/ 0.50141= 16.156
L/ di = 4.83*1000/16= 302
From the fig., jh= 2.2*10-3
Therefore, hi= 2.2*10-3 * 1665.35* (16.156)0.33*0.50141 / 16*10-3
= 287.575 W/m2 c
39
HOT FLUID SHELL SIDE:--
Choose baffle spacing = DS/5 =667.897/5 = 133.6mm
Tube pitch = 1.25*20 = 25mm
Cross sectional area, AS= (PT-d0) Ds*lb/ Pt={(25-20)*667.897*133.6*10-6}/25
= 0.0178 m2
Mass velocity, GS= 98.7 kg/s*(1/0.0178)m2 = 5544.9 kg/m2 s
Equivalent diameter, de= 1.1/do( Pt2- 0.917d0
2)
= 1.1/20( 252-0.917*202)= 14.4mm
Mean shell side temp.=340 F= 151.11 c
K of steam = 700*10-3 W/m c
Viscosity of steam = 1860*10-6 Ns/m2
Re= (Gs*de)/ viscosity= 5544.9*14.4*10-3/ 1860*10-6= 42928.26
Pr= Cp*viscosity/ K= 4.21*1860*10-6/0.700 = 0.011
Choose 25% baffle cut, from fig.
Jh= 3*10-3
Therefore, (hs*de)/ Kf= Jh*Re* Pr0.33
hs=3*10-3* 42928.26*0.0110.33*0.700/ 14.4*10-3
hs=1413.38 W/m2 c
OVERALL COEFFICIENT;
K of cupro nickel alloys =50 W/m c
Take the fouling coefficient from the table;
Steam condensate= 5000 W/m2c
Ammonia= 10,000 W/m2c
40
1U0
= 1h0
+ 1hod
+
do ln( do
d i)
U0+d0
di× 1hid
+d0
d i× 1hi
1/UO = 1/1413.38+1/5000+20*10-3 ln(20/16)/(2*50)+(20/16)*(1/10000)+1/287.575
UO= 219.58 = 220 W/m2 c
PRESSURE DROP:
TUBE SIDE-
ΔPt=N P [8 jf ( Ldi )(μμw )
−m+2 .5] ρut
2
2
From fig, for Re= 1.67*103
Jf= 7.1*10-3
Therefore, ∆Pt= 2*[8*7.1*103(4.83*103)/16 + 2.5] *618*(0.322)/2
= 1243.29 N/m2= 1.24 KPa
SHELL SIDE-
Linear velocity = Gs/p = 5544.9/618 = 8.97 m/s
From the graph, jf= 4*10-2
41
ΔPs=8 jf (Ds
de)( Ls
IB ) ρu s2
2 ( μμw )
−0 . 14
Therefore, ∆Ps= 8*4*10-2*(667.897/14.4)*(4.83*103/133.6)*(618* 8.972)/2
= 1334.07 N/m2
42
43
BELL’S METHOD
No.of tubes = 463
Shell i.d = 667.897 mm
Bundle diameter= 605.897 mm
Tube o.d.= 20 mm
Tube length = 4830 mm
1.HEAT TRANSFER COEFFICIENT—
As= (25-20)*667.897*133.6*10-6/ 25
= 0.0178 m2
Gs= 98.7/0.0178 = 5544.94 ks/s m2
Re= Gs*do/ viscosity = 5544.94*20*10-3/(1860*10-6)= 59623
From fig., jh= 4.1*10-3
Prandtl no.= 0.011
Neglect viscosity correction factor,
hoc*do/K = jh*Re*Pr0.33
hoc= 4.1*10-3*59623*0.0110.33*0.700/ (20*10-3)
hoc= 1931.636vW/m2 c
2. TUBE ROW CORRECTION FACTOR—
Tube vertical pitch, Pt’= 25* 0.87 =21.8 mm
Baffle cut height, Hc=0.25*667.897= 166.97mm
Height between baffle tips= 667.897-(2*166.97)=333.957 mm
Ncv = 333.957/ Pt’ = 15.32
44
From fig., Fn=1.02
3. WINDOW CORRECTION FACTOR—
Hb =Db/2 –Dsi(0.5 – baffle cut)
=605.897/2 -667.897(0.5 -0.25)
=302.9485-166.974
=135.974 mm
Bundle cut =Hb/Db = 135.97/605.897 =0.22
From the fig.12.41 at cut of 0.22
Ra1 =0.16
Tubes in one window area, Nw=Nt*Ra’
=463*0.16= 74.08=74
Tubes in cross flow area, Nc=Nt-2Nw
=463-2*74
= 315
Rw=2Nw/Nt=2*74/463= 0.32
From fig., Fw= 1.07
4.BYPASS CORRECTION FACTOR,Fb
Ab=(Di-Db)Bs= (667.897-605.897)*267.2*10-6=0.0166 m2
Ab/As=0.0166/0.0178=0.93
Fb= exp[-1.35*0.39]=0.59
Very low sealing strips needed; try one strip for each 5 vertical rows
Ns/Nw=1/5
Fb= exp[-1.35*0.93{1-(2/5)0.33}]=0.72
45
5. LEAKAGE CORRECTION FACTOR,FL
Using clearances as specified in the standards
Tube to baffle 1/32inch =0.8 mm
Baffle to shell 3/16inch =4.8mm
Atb=(0.8/2)*20*π*(463-74)=9771.68 mm2=0.0098m2
From fig., 25% baffle cut,ϴb=2.1 rads
Asb=(Cs*Ds/2)*(2π-ϴb)=(4.8/2)*667.897*(2π-2.1)
= 6700.34mm2=0.006m2
At=Atb+Asb=0.0098+0.006=0.0158 m2
Al/As=0.0158/0.0178=0.89
From fig, βL=0.43
Fl=1-0.43{0.0098+2*0.006}/0.0158=0.41
SHELL SIDE COEFFICIENT:
hs = hoc Fn Fw Fb FL=1931.636*1.02*1.07*0.72*0.41
= 622.34 W/m2 c
Appreciably lower than that predicted by Kern’s method.
PRESSURE DROP--------
1.CROSS FLOW ZONE:
From fig.,at Re=59623,for 1.25∆pitch,
Jf= 5*10-2
Us=Gs/p=5544.94/618=8.97 m/s
∆Pi=8*jf*Ncv*pus2/2=8*5*10-2*15.32*618*(8.972)/2=152356.9 N/m2
Α=4.0 for Re>100,
46
Therefore, Fb’=exp[-4*0.93{1-(2/5)0.33}]=0.38
From fig., βL’=0.7
FL’=1-0.7[(0.0098+2*0.006)/0.0158]=0.034
∆Pc=1523569*0.38*0.034=1968.45N/m2
2. WINDOW ZONE:
From fig. for 25% baffle cut, Ra=0.19
Aw=[(πDs2/4)*Ra – (NW*π*dO
2/4)]
=[(π*667.8972*0.19/4) – (74*π*202/4)]=43297.8 mm2=0.043m2
Uw=(98.7/618)/0.043=3.7m/s
Uz=( Uw*Us)1/2= (3.7*8.97)1/2=5.8m/s
Nwv=Hb/ Pt’=135.97/21.8 = 6.24
∆Pw=FL’(2+0.6Nwv)pUz2/2=2030.055 N/m2
3.END ZONE:
∆Pe=∆Pc[(Nwv+Ncv)/Ncv]*Fb’
= 152356.9[(6.24+15.32)/15.32]*0.38=81477.13N/m2
TOTAL PRESSURE DROP:--
No.of baffles, Nb= (4830/267.2)-1=17
∆Ps=2*81477.13+1968.45(17-1)+17*2030.055
= 228960.39 N/m2
= 228.96 KPa
47
DESIGNING OF UREA REACTOR
The liquid mixture of NH3 and carbamate (100˚C) and gaseous CO2 (140˚C) are fed to reactor where they meet 180˚C temperature and 150 atm pressure and form ammonia carbamate .This carbamate dehydrates and forms Urea.The reactions taking place in the reactor are given as:
2NH3 +CO2 NH2COONH4
NH2COONH4 NH2COONH2 +H2O
KINETIC DESIGN
48
The predominant rate controlling mechanism is the reaction kinetics.The first reaction occurs rapidly and goes to completion, the second reaction occurs slowly and determines the reaction rate. Assuming order of reaction is first order reaction. Then rate of reaction is
-r =kCA
Where ,C A=C AO(1−X A)
1+ɛXA
From material balance;
Mole ratio = NH3 =2.55 , H2O =0.37
CO2 CO2
Assuming overall carbamate conversion to be 42%
Now , ɛ = 2.55 – 3.55 = -0.28
3.55
Performance equation of an ideal plug flow reactor
V=F ao ∫0
X dX−r
V =FAO ∫0
X dX (1+ɛ X )kCa(1−X )
I =∫0
0.42 1−0.28 X1−X
Value of integral is solved by Simpson’s 1/3 rule:
S.No. XA (1-0.28XA)/(1-XA)
1 0 12 0.05 1.04
49
3 0.1 1.084 0.15 1.135 0.20 1.186 0.25 1.247 0.30 1.318 0.35 1.399 0.40 1.4810 0.45 1.59
Using Simpson’s rule:
∫a
b
f ( x )dx = h/3 [y0 +4 (y1 +y3 + y5+ y7 +….. ) +2(y2 +y4 +y6 + y8 + y10…..) +yn]
=0.059/3[1 +4(1.04 + 1.13 + 1.24 + 1.39) +2(1.08 +1.18 +1.31 +1.48) +1.59]
=0.627
FAO =(4333.91/17) +(14564.96/18)+(120727.34/78)
=254.9+809.164+1547.79
=2611.85 Kmol/hr
ʋ0 = 4333.91 +14564.96 +120727.34
890 1000 910
= 4.87 +14.565+132.667
=152.10 m3/hr
CAO = FAO / ʋ0
= 2611.85/152.10
=17.172 Kmol/m3
DETERMINATION OF RATE CONSTANT ‘K’:
Reactor temperature= T2=180ºC= 453 K
R=1.985 cal K-1 mol-1
At temperature T1=140 ˚C =413 K, K=0.0134 min-1 for 1st order reaction(reference)
From the reaction, ∆E=38.1-7.1= 31 Kcal/mol=31000 cal/mol
50
ln K2 =∆E{(1/T1) –(1/T2)}
K1 R
Therefore, K2= 0.0378 min-1 =2.27 hr-1
Volume of reactor V = FAO * I
K* CAO
V= 2611.85*0.627 / (2.27*17.172)
V=42.01 m3
Length and diameter of urea reactor
Assuming L/D =20
V=(π /4) *D2 *20D
Di =1.39 m
L =27.8 m
51
COST ESTIMATION
Acceptable plant design must present a process that is capable of operating
under conditions, which will yield profit. Since net profit equals total value
minus all expenses, it is essential that the chemical engineer be aware of the
many different types of cost involved in the manufacturing processes.
Capital must allocate for the direct, plant expenses, such as those for raw
material, labor and equipment.
Besides direct expenses many others indirect expenses are incurred, and these
must be included if a complete analysis of the total cost is to be obtained. Some
examples of these indirect expenses are administrative salary, product
distribution cost and cost for interplant communication. A capital investment
is
required for every industrial process and determination of necessary investment
is an important part of a plant design process. The total investment for any
process consist fixed capital investment for practical equipment and facilities in
the plant plus working capital, which must be available to pay salaries, keep raw
material and products on hand, and handle other special items requiring the
direct cost outline.
When the cost for any type of commercial process is to be determined, sufficient
accuracy has to be provided for reliable decision. There are many factors
affecting investment and production cost. These are;
1. Source of equipment
2. Price fluctuation
3. Company policies
4. Operating and rate of production
5. Governmental policies
Before an industrial plant can be put into operation, a large sum of money must
52
be supplied to purchase and install the necessary machinery and equipment.
53
Land and service facilities must be obtained, and the plant must be erected completely
with all piping, controls and services.
The capital needed to supply the necessary manufacturing and plant facilities is called the
fixed-capital investment, while that necessary for the operation of plant is termed the
working capital. The sum of the fixed capital investment and the working is known as the
total capital investment. Generally, the working capital amounts 10-20% of the total
capital investment. Following is the breakdown of the fixed capital investment for a
chemical process.
DIRECT COST:
1. purchased equipments
2. purchased equipment installation
3. instrumentation and control
4. piping
5. electrical equipment and material
6. building (including services)
7. yard improvement
8. land
INDIRECT COST:
1. engineering supervision
2. construction expenses
3. contractor’s fee
4. contingency
TYPES OF CAPITAL COST ESTIMATE:
• Order of magnitude estimate (ratio estimate) based on similar cost data;
probable accuracy of this estimate over ± 30%.
54
• Study estimate based on knowledge of major items of equipment,
probable accuracy of this estimate up to ± 30%.
• Preliminary estimate( budget authorization estimate scope method): based
on sufficient data to permit the estimate to the budget, probable accuracy
of this estimate is within ± 20%.
• Detailed estimate based on complete engineering drawing, specifications
and site survey, probable accuracy of this estimate within ± 10%.
COST ESTIMATION OF 1000 TONS/DAY OF UREA PLANT:
Urea plant size = 1000 T/day
Fixed capital investment for cost index of 130 = Rs 2.051×107
Cost index for 2002 = 402
Therefore present fixed capital investment = 2.051×107×(402/130)
=Rs 63,42,32,31
Estimation of total investment cost :
1) Direct cost:
a) Purchased equipment cost:(15 - 40% of FCI ) Assume
40% of FCI
=Rs 25369292
b) Installation cost :(35 - 45% of PEC)
Assume 45%
=Rs 11416181
c) Instrument and control installed :(6 -30% of PEC)
Assume 30% of PEC
=Rs 7610787
d) Piping installation cost :(10 -80% of PEC)
Assume 80%
=Rs.20295433
e) Electrical installation cost:(10 - 40% of PEC)
55
Assume 40% of PEC
=Rs 10147716
f) Building process and auxilliary:(10-70% of PEC)
Assume 70%
=Rs 17758504
g) Service facilities:(30-80% 0f PEC)
Assume 80%
=Rs 20295433
h) Yard improvement :(10-15% of PEC)
Assume 15%
=Rs 3,805,393
i) Land:(4-8% of PEC)
Assume 8%
=Rs 2029543
Therefore direct cost =Rs. 118728282
Indirect cost:
Expenses which are not directly involved with material and labour of actual installation or complete
facility
a) Engineering and supervision :(5-30% of DC)
Assume 30%
=Rs 35618485
b)Construction expenses:(10% of DC)
=Rs 11872828
c) Contractors fee:(2-7% 0f DC)
Assume 7%
=Rs 8310979
d) Contingency :(8-20% of DC)
Assume 18%
=Rs 21371091
Therefore total indirect cost =Rs 77,173,383
Fixed capital investment:
56
Fixed capital investment (FCI) = DC+IC
= Rs 195,901,665
Working capital investment:
10 -20% of FCI
Assume 18%
=Rs 35262299
2) Total capital investment:
= FCI + WC
=Rs 231163965
Estimation of total product cost(TPC):
Fixed charges:
a) Depreciation :(10% of FCI for machinery)
=Rs 19590166
b) Local taxes :(3-4% of FCI)
Assume 4%
=Rs 7836067
c) Insurances :(0.4-1% of FCI)
Assume 0.9%
=Rs 1763115
d) Rent :(8-12% of FCI)
Assume 12%
=Rs 23508199
Therefore total fixed charges =Rs 52697547 But,
Fixed charges = (10-20% of TPC)
Assume 20%
Therefore Total product cost =52697547/0.20
=Rs 263487735
Direct production:
a) Raw material :(10-50% 0f TPC)
Assume 50%
57
=Rs 131743867
b) Operating labour(OL):(10-20% of TPC)
Assume 20%
=Rs 52697547
c) Direct supervisory and electric labour:(10-25% of OL)
Assume 25%
=Rs 13174387
b) Utilities :(10-20% of TPC)
Assume 20%
=Rs 52697547
c) Maintenance :(2-10% of FCI)
Assume 9%
=Rs 17631149
d) Operating supplies (OS):(10-20% of maintenance) Assume
20%
=Rs 3526229
e) Laboratory charges :(10-20% of OL)
Assume 18%
=Rs 9485558
f) Patent and royalties :(2-6% of TPC)
Assume 6%
=Rs 15809264
Plant overhead cost:
50-70% of (OL+OS+M = 73854925)
Assume 70%
=Rs 51698447
General expenses:
a) Administration cost:(40-60% of OL)
Assume 60%
=Rs 31618528
b) Distribution and selling price:(2-30% of TPC) 58
Assume 30%
=Rs 79046320
c) Research and development cost :
(3% of TPC)
Rs= 7904632
Therefore general expenses (GE) =Rs 118569480
Therefore manufacturing cost (MC)= Product cost+fixed charges+Plant
overhead expenses
=Rs 367883729
Total production cost:
Total production cost =MC + GE
=Rs 486453209
Gross earnings and rate of return:
The plant is working for say 340 days a year
Selling price =Rs. 15 /kg for a Urea
Total income =100×340×1000×15
=Rs 510000000
Gross income =Total income - total product cost
=Rs 23546791
Tax =50%
Net profit =Rs 11773395
Rate of return = net profit/total capital investment
=5.1%
59
REFERENCES
IFFCO( Phulpur unit) manual
Peters, Max S. and Timmerhaus, Klaus D., Plant Design & Economics, 4th edition,
McGraw Hill, Inc. (1980)
Wilbrant, Frank C. & Dryden, Charles E., Chemical Engineering Plant Design, 4 th
edition, McGraw Hill, Inc. (1980)
Coulson, J. M. & Richardson, J. F., Chemical Engineering, Volume 6
Perry, J.H., Chemical Engineer’s Handbook, 7th edition, McGraw Hill, Inc. (1985)
Joshi, M.V. and Mahajani, V.V, Process Equipment Design,3rd edition, Macmillan
India ltd.(2007)
McCabe, Warren L., Smith, Julian C., and Harriot, Peter, Unit Operations of
Chemical Engineering, 5th edition, Pergamon Press (1983)
Kern, D.Q, Process Heat Transfer , J. A., 4th Edition , McGraw Hill International
Edition .
Shreeve R . N. & Brink J A , Chemical Process Industries , 5th Edition , McGraw
Hill International Edition .
Sahu, J.N ,Patwardhan,A.V and Meiko, B.C (IIT Kharagpur) Equilibrium and
kinetic studies of generation of urea in areactor
60
APPENDIXES
61
62
63
64
65
66
67
68
69
70
71