Urea Plant Design

July 22, 2017 | Author: Aamli Agarwal | Category: Urea, Ammonia, Fertilizer, Organic Compounds, Nitrogen
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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


Melting Point


Sp. Gravity

- [email protected]

Heat of Fusion

- 57.08 Cal/gm

Heat of Combustion

- 2531 Cal/gm

Crystal Form

- Tetragonal

Nitrogen Content


Boiling Point

- Decompose on boiling


- White

Bulk Density

- 740 to 750 kg/m3

Angle of Repose

- 23 to 300C


- [email protected]

Triple point

- 1020C

Dielectric Constant

-3.52 to 0.2 3

- 0.42Kcal/cm2

Sp. Heat

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 (NH4)+



and hydroxide



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



+ 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





the reaction of hydrogen and nitrogen under high pressures





catalyst. N2 +3 H2


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 Anhydrous ammonia is usually stored as a liquid in refrigerated tanks at 4

-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 crystal is nucleated and can thus be forced by seeding. This property has been used to separate 5

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 Ureapotassium 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.


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.




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


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


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.




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 (ammonium carbamate) NH2CONH2 + H2O - Heat (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.


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/cm 2(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 CO 2 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 CO 2 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.


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.


PROBLEM STATEMENT: To design a UREA PLANT of capacity 1000 ton/day using ammonia and carbon di-oxide as raw material




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) X3B =0.002 (assume) : W3 =46343.0107 Kg/hr


X3H2O =0.011

XNH3=0.2722 XCO2= 0.128

XU5 =0.94 (given) Material balance:


W3 X3U = W5 XU5


46343.1075*0.989 = W5 *0.94


W5 =48758.865 Kg/hr


W4 = W5 – W3

XCO2 =0.00634

=2415.76 Kg/hr

XB= 0.0019

Let X4NH3 =0.2722


:0.2722*2415.76 = X5NH3 *48758.865


XU =0.989 XH2O =0.011 XB =0.002

: X5NH3 = 0.1035 : 0.128 *2415.76 = X5CO2 *48758.865 X5CO2 = 0.00634 X5H2O = 0.040


2 .Ist VACCUM CONCENTRATOR XU7 =0.70 , XU5 =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


X5NH3 =0.0135



XB5 = 0.0019

XNH3 =0.0534


Let water evaporated =12683.7973,


CO2 from top =1214.65,


NH3 from top =2841.945


X6CO2=0.0126, X6NH3 =0.17 , X6H2O =0.76


XU=0.94 W5


: X7NH3 *65476.19 =0.17*16717.325 +0.0135*48758.865


X7NH3= 0.0534


X7B*65476.19 =0.0019 *48758.865


X7B =0.001415 X7CO2 *65476.19 =1214.65+0.00634*48758.865 X7CO2 =0.023 X7H2O = 0.22


3 L.P DECOMPOSER XU9 =0.63 , XU7 =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


72751.132 =65476.19 +W8


W8 =7275.132


CO2 in vapour =3180.32 -0.023*65476.19


=1674.367 Kg/hr


Let H2O evaporated =1838.77 Kg/hr


NH3 evaporated =1304.475 Kg/hr


W8 =1304.475 +2457.52 +1838.77 +1674.367




0.001415 *65476.19 = X9B *72757.322



X9B =0.00127


X9NH3 *72757.322 =0.0534 *65476.19 +1304.475 +2457.52


X9NH3 =0.0997


X9CARB = 72.28*78


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


Water distilled = 32.8% =6466.98 Kg/hr

XCO2=0.133 W10

CO2 at top =1498.12


W11 =11217.72


W10 – W11 = W12


W12 =19133.085 -11217.72 =7915.365 Kg/hr XNH 3=0.031

W12 XNH3=0.031 XH2O=0.968

XH2O =0.968


6.REFLUX ACCUMULATOR Assume CO2 evaporation =52.7%


=789.196 Kg/hr ; W13 =789.196 For W15 , NH3 =2663.06



XNH3 =0.289

CO2 =652.88






: W15 =8590.52

XH2O=0.615 XNH3=0.309

W14 = W11 - W15 - W13 = 1838.004 Kg/hr X 14XNH3 = W11*0.29 - W15*0.31

W13 W11

W14 XCO2=0.46

X 14NH3 = 0.32

XNH3 =0.32

W14X14CO2 = W11(0.1325) - W15 (0.076)


: X 14CO2 =0.46 7. L. P CONDENSER Assume: W8 + W14 =9113.136


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 14CO2 -50.397 : XCO2 =0.29 8. L.P NH3 ABSORBER NH3 added =526.37 Let H2O added =440 Kg/hr :

W18 =966.37 Kg/hr

vent(g) 50.397Kg/hr W16(g)

H2O(440 Kg/hr)

W16 +440 =966.37 +vent



:vent =50.397 Kg/hr


W18(l) XH2O=0.455



=9502.739 Kg/hr W21 XNH3 = W18*0.544 + W17*0.45

W18(l) XH2O=0.45 W17(L)

X 21NH3 =0.46

XNH3 =0.145

W21 XCO2 = W17 *0.29



X 21CO2 =0.26


XCO2= 0.26




10.M .P DECOMPOSER Let 64% of carbamate decomposed :cabamate decomposed =130 Kmol/hr =10140 Kg/hr NH2COONH4


W22(G) XNH3=0.307

2NH3 +CO2


NH3 formed =2*130 =260 Kmol/hr


=4420 Kg/hr CO2 formed =130 Kmol/hr =5720 Kg/hr Let 0.5% of urea be converted into biuret 2NH2CONH2

W9(L) XNH3=0.0997



Urea converted = 4072 Kmol/hr


=286.46 Kg/hr


Biuret formed =2.385 Kmol/hr =245.69 Kg/hr

W23(L) XNH3=0.134

NH3 formed =2.385 *17 =40.545 Kg/hr


Total NH3 in vapour =4420+40.545 Kg/hr


Water evaporated =4356.4 Kg/hr




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



11. M.P CONDENSER CO2 in (W22 + W21 )


=5720 +2470.712

W21(l) XCO2=0.26



NH3 =4460.545 +4371.26






W25 = W23 + W22


=9502.739+14536.945 =24039.684 Kg/hr 98% of CO2 converted back to carbamate through condensation

2NH3 +CO2


Carbamate formed =0.98 *8190.712/44 =182.45 Kmol/hr


=14229.5 Kg/hr W 25Xcarb =14229.5

W25(l) W22(g)

XNH3 =0.27

: X 25carb =0.341


W25XNH3 = W21 X 21NH3 + W22 X 22NH3


=4460.545 +0.46 *9502.739 25

: X 25NH3 = 0.37 12.M.P ABSORBER NH3 =2400 Kg/hr


W26 =2400 Kg/hr



Water in W27 =3900 Kg/hr & NH3 in W27 = 475 Kg/hr




: W27 =4375


CO2 absorbed =113.23 Kg/hr

XCO2= 0.241

NH3 absorbed =85%


XNH3=0.108 XH2O=0.891 W29(l)

Total ammonia in =11769.68 NH3 absorbed =10023.68 Kg/hr NH3 not absorbed =8894.68 +2400+475 -10013.68 =1746.0 Kg/hr

W26(g) W25

W28(g) XNH3=0.98 XH2O=0.011

CO2 not absorbed=9215.74


: W28 =10961.74 W29 = W25 + W26 + W27 + W28


=19852.9 Kg/hr

XCO2=0.0059 , XNH3= 0.42 XCARB=0.50

W 29X29 CO2 = W 25X25CO2 - W 28X28CO2 : 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 =436.5

XCO2=0.876 W28(g)

: W31 =436.



W28 = W30 + W31




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)

NH3 leaving bottom =34000 Kg/hr : W34 =499 & W33 =34000

W34(g) XNH3=0.909 XCO2=0.090

W31(l) + W30(g)

W30 + W31 + W32 = W33 +W34 1746+ W32 =34000+499 : W32 =32753

W33(L) XNH3=1


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


Carb23 =15711.9 Kg/hr


Carbamate decomposed = 930 Kg/hr




2NH3 +CO2

NH3 formed =70000 Kg/hr


CO2 formed =35000 Kg/hr


W23(l) XNH3=0.134

NH3 boiled in vapour =40000 Kg/hr


Total NH3 boiled in vapour =40000 +70000 =110000 Kg/hr


NH3 in vapour =430 Kg/hr


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


Let CO2 condensed =64000




NH3 condensed =59100

W29 XNH3=0.42

:carbamate formed =113454.54


W37 = W35 + W29


= 145430+19852.9 = 165282.9 Kg/hr

XNH3= 0.25


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


18.CARBAMATE SEPARATOR NH3 evaporated =1500 Kg/hr

vent(g) XNH3=0.43,XH2O=0.52

H2O evaporated =1800 Kg/hr


CO2 in vent = 150 Kg/hr



Vent =3450 Kg/hr



W 37= W 38 +vent



W 38 =165282.9 – 3450



= 161832.9


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


=31600 Kg/hr


W 36 = W39 + W38 + W 40


= 232718.27 -161832.9 -31600


=39285.37 Kg/hr



Known amount of N2 in W 36 =430 Corresponding O2 =143.3


Air used =573.2

W38 XCARB=0.74 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


For 100% CO2 conversion, NH3 required =W32 = 32753 Kg/hr =1926.65 Kmol/hr UREA FORMED =771.85 Kmol/hr




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


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 Ist 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 IIndvaccum 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


λ=2119.7KJ/kg Steam required = 4631946.361 2119.7 =2185.19Kg/hr Energy balance across reactor: 2NH3 +CO2






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


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


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


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- 0.917d02) = 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


d d o ln  o 1 1 1  di    U 0 h0 hod U0

    d0  1  d0  1 d i hid d i hi

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


  L    m  u 2 Pt  N P 8 jf     2.5 t   di  w   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 SIDELinear velocity = Gs/p = 5544.9/618 = 8.97 m/s From the graph, jf= 4*10-2

 D  L  u P s  8 jf  s  s   de  I B  2

2 s

     w 

 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



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 h


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 A l/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*dO2/4)] =[(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


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




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

1+ɛXA From material balance; Mole ratio = NH3

=2.55 , H2O




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 o∫

V =FAO ∫ I =∫

Value of integral is solved by Simpson’s 1/3 rule: S.No. 1 2 3 4

XA 0 0.05 0.1 0.15

(1-0.28XA)/(1-XA) 1 1.04 1.08 1.13 49

5 6 7 8 9 10

0.20 0.25 0.30 0.35 0.40 0.45

1.18 1.24 1.31 1.39 1.48 1.59

Using Simpson’s rule: ∫

= 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






= 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


ln K2

=∆E{(1/T1) –(1/T2)}



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=( /4) *D2 *20D Di =1.39 m L =27.8 m


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 be supplied to purchase and install the necessary machinery and equipment.


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%.


• 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)


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:


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%


=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)


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%



 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 , 5 th 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














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