Snamprogetti Urea Process

Manufacture Of Urea A REPORT SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIRMENT FOR THE DEGREE OF BACHLOR IN CHEMICAL ENGINEERING By

1. Heba Ramadan Mohammed 2. Haneen Mohammed Mahdi 3. Safa Ramadan Mohammed 4. Sarah Deia'a Khaleel 5. Mowafq Khalaf Suliman Supervisor Asst.Prof.Dr.Duraid F.A

University Of Tikrit Department Of Chemical Engineering 2015

CONTENTS Abstract CHAPTER 1: Introduction

PAGE NO.

1.1

Introduction

1

1.2

Physical Properties

2

1.3

Chemical Properties

3

1.4 1.5

Uses of Urea Environmental Effects

4 5

1.5.1 Elimination Method

5

1.6

7

Urea Production Methods

1.6.1 Once- Through Urea Process

7

1.6.2 Partial recycle process

8

1.6.3 Stripping process based plants

9

1.7

Selection of the Process

13

1.8

Process Description

14

Chapter 2: Material Balance

19

2.1 Around Reactor

21

2.2 Around Stripper

23

2.3 Around Medium Pressure Decomposer

24

2.4 Around Low Pressure Decomposer

25

2.5 Around Vacuum Evaporator

27

2.6 Around Prilling Tower

28

Chapter 3: Energy Balance

32

3.1 Around Reactor

33

3.2 Around Stripper

35

3.3 Around Carbamate Condenser

37

3.4 Around Medium Pressure Decomposer

40

3.5 Around Low Pressure decomposer

42

3.6 Around Vacuum Evaporator

42

3.7 Around Prilling Tower

43

Chapter 4: Equipment Design

46

4.1 Reactor Design

48

4.1.1 Introduction

48

4.1.2 Thickness of Shell Design

49

4.1.3 Head Design

50

4.1.4 Piping Design

53

4.1.5 Sieve Plate Design

55

4.2 Carbamate Condenser Design

55

4.2.1 Introduction

55

4.2.2 Design Procedure

56

4.3 Low Pressure Decomposer Design

60

4.3.1 Introduction

60

4.3.2 Design Calculation

60

4.4 Vacuum Evaporator Design

62

4.4.1 Evaporation And Its Mechanism

62

4.4.2 Evaporator Types

63

4.4.3 Evaporator Selection

64

4.4.4 Design Calculation

64

4.5 Prilling Tower Design

71

4.5.1 Introduction

71

4.5.2 Material Of Construction

72

4.5.3 Design Calculation

72

Chapter 5:Process Control

76

5.1 Introduction

77

5.2 Control Objectives

77

5.3 Types Of Control System

78

5.4 Equipment Control

79

5.4.1 Around Reactor

79

5.4.2 Around Carbamate Condenser

81

5.4.3 Around Low Pressure Decomposer

82

5.4.4 Around Vacuum Evaporator

84

5.4.5 Around Prilling Tower

87

Chapter 6: Economy Cost

89

6.1 Introduction

90

6.2 Types Of Cost Involves In Manufacturing

90

6.3 Cost Calculation

90

Chapter 7: Safety Aspect

94

7.1 Introduction

95

7.2 Principle Of Protection And Prevention

96

7.3 Safety Precaution

96

7.4 List Of Safety equipment

97

7.5 Fire Hazards

98

Reference

99

Abstract Urea in many ways the most convenient form of fixed nitrogen. It has The highest nitrogen available in a solid fertilizer(46%). It is easy to produce as prills or granules and easily transported in bulk or bags with no explosive hazard. It leaves no salt residue after use in crops. Its specific gravity gravity 1.335, decomposes on boiling and it is is fairly soluble in water. The principle raw materials required for this purpose are NH3 and CO2. Two reactions are involved in the manufacture of urea ,first , ammonium carbamate is formed under pressure pressure by reaction between CO2 and NH3.



CO2+NH3  NH2COONH4

∆H= -37.4 kcal

The highly endothermic reaction is followed by an endothermic Decomposition of ammonium carbamate.



NH2COONH4 NH2CONH2+H2O

∆H=+6.3 kcal

We selected snamprogetti ammonia stripping process for the Manufacture of urea. The selected capacity of the plant 500,000 ton/year Of urea producing 68055.56 kg/hr of( 98%) purity. Urea material and Energy balance of the plant is done. Urea reactor, vacuum evaporator, Carbamate condenser, low pressure decomposer and prilling tower ar e 3

Designed. The volume of reactor is calculated and found to be 119.97 m The length and diameter of the reactor are 17.65 m and 2.94 m Respectively. The evaporator used is of climbing-film long-tube type.

Chapter 1

Introduction

1.1 Introduction Urea is an organic compound with the chemical formula (CO(NH2)2 , the molecule has two functional group joined by a carbonyl (C=O) functional group. Urea scientific name (Carbonyl di amide); it is composed of 45% (N2), 20% ( C ) 26% (O2) and 9% (H2) it has the highest highest nitrogenous fertilizer fertilizer in common use use Therefore it being the preferred nitrogen fertilizer worldwide. worldwide. It is is used in solid Fertilizer, liquid fertilizer, formaldehyde resins and adhesives. Roulle first Discovered urea in urine in 1773. His discovery followed by the synthesis of urea from ammonia and cyanic acid by Wӧhler in 1828. This is considered to be the first synthesis of an organic compound from inorganic compound. In 1870 Bassarow produced urea by heating ammonium carbamate in a sealed tube in What was the first synthesis of urea by dehydration. Urea is produced worldwide 6

On a large scale and its production volume exceeds 150*10  ton/year in 2010. At the present urea is produced on an industrial scale exclusively by reactions based on NH3 and CO2 as the raw material.

1

1.2 Physical Properties Urea is a white odorless solid. Due to extensive hydrogen bonding with water ( up to Six hydrogen bonds may form – form – two  two from the oxygen atom and one from each hydrogen) Urea is very soluble.

Table 1.1 physical properties of urea

Cas as number

57-13-6

Mole lecula lar weig ight

60.07

Melt ltin ing point (C (C )

132.7

Boilingpoint(C )

135

˚

˚

1.335

3

Densit ity( y(gm/cm ),solid Bulkdensit ity( y(gm/cm3 )

0.74

Viscosity ,v (mpa. Sec)

20

1.9

Heat of fu fusio ion (J/g /gm)

251

Heat of so solutio lutioninwater (J/gm (J/gm)

243

0

o

Specific cificheat ,S (KJ/kg. C )

1.439

Specific fic gravity

1.335

Vapor pressu sure (pa)

<10

Solubility in in water (g/100ml)

2

o

108 (20C ) o

167(40C ) o 251 (60C ) o 400 (80C ) o 733(100C )

1.3 Chemical Properties



Urea decomposes to ammonia, biuret, ammelide and triuret at atmospheric Pressure and it's melting point.



Urea acts as a mono basic and forms salts with acids. Such as with Nitric acid To form urea nitrate CO(NH2)2.HNO3 .



Urea reacts with NOx, both in gas phase at (800— (800—1150)

. And in liquid phase

At lower temperature to form N2,CO2 and H2O. This reaction used industrially To remove NOx from combustion gasses. 

Urea reacts with formaldehyde under acidic conditions to form methylene Urea and it's derivative ( dimethylene, trimethylene, etc) which are used as Slow released fertilizer.



Urea is similar to the amids in the possibility of decomposition by dilution Acids, alkaline or water vapor above temperature of 150



NH2CONH2+H2O

 2NH3+CO2

3

 .

1.4 Uses of Urea 

More than 90% of urea world production is destined for use as a fertilizer



A raw material for the manufacture of plastics, to be specific, Urea – Urea – formaldehyde  formaldehyde resin.



A raw material for the manufacture of various glues ( urea— urea—formaldehyde Or urea melamine— melamine—formaldehyde); the latter is waterproof and is used For marine plywood.



A flame— flame—proofing agent ( commonly used in dry chemical fire extinguishers e xtinguishers As urea potassium bicarbonate).



A reactant in some ready-to-use cold compressors for first-aid use, due to The endothermic reaction it creates when mixed with water.



A cloud seeding agent, along with salts, to expedite the condensation of water In clouds, producing precipitation .



Feed for hydrolyzation into ammonia which in turn is used to reduce emissions From power plants and combustion engines.

4

1.5 Environmental Effect The chemical fertilizer one of the most important factors that has helped to improve The agricultural production, but ammonia, CO2 and urea releases from process Have a negative influence on the environment, because nitrogen fertilizers break Down into nitrates and travel easily through the soil. Because it is water-soluble And can remain in groundwater for a long time, besides their emissions to air And water. Urea, contributes to acid rain, groundwater contamination and ozone Depletion due to release of nitrous oxide by denitrification process. In general There are four types of emission estimation techniques (ETTs) that may be used To estimate emissions from the facility. The four types are: 

Sampling or direct measurement



Mass balance



Fuel analysis or other engineering calculations



Emission factor

1.5.1 Elimination Methods Presently plants are equipped with the following features to keep the effluent and Emissions at extremely low levels: 

N/C ratio meter



Waste water treatment section



Absorbers



Special operational facilities

N/C ratio meter in the synthesis section Instead of using spectrometer in the gas phase of the synthesis section Nitrogen/Carbon (N/C) ratio meters are installed in the liquid phase (reactor Liquid outlet) of the urea synthesis section. This N/C ratio meter allows the Process at all times to be operated at the optimum ratio to achieve higher reactor 5

Efficiency combined with higher energy efficiency. Special procedures are used to Eliminate emissions during start – start  –up. up.

Waste water treatment section The purpose of the process water treatment is to remove ammonia, carbon dioxide And urea from the process condensate. This water is usually discharged from the urea Concentration and evaporation section of plant. Removal of ammonia and urea from Wastewater can be a problem as it is difficult to remove one in the presence of other One method used to overcome this problem is the hydrolysis of urea to ammonium Carbamate, which is decomposed to NH3 and CO2. These gasses can be then stripped From the wastewater. This recovered water can be used for a variety purpose Depending on the required quality such as cooling water, etc.

Absorbers Absorbers are used to eliminate emissions emissions to the atmosphere, can be classified as 1. The vent from the synthesis section of the plant: ammonia is washed out with A large flow of low concentrated and cooled process water and secondly the Remaining ammonia is absorbed in cooled condensate or clean waste water. 2. The vent from the low pressure section of the plant: the ammonia and carbon Dioxide present in the off gases of the recirculation system, the process water Treatment system are washed out in an atmospheric absorber where large Amounts of cooled low concentrated process water are used to absorb all ammonia Present in the said off gases .

6

1.6 Urea Production Method Several processes are used to urea manufacturing. Some of them are used conventional Technologies and others use modern technologies to achieve high efficiency. These Process had several comparable advantage and disadvantage based on capital cost, Maintenance cost, energy cost, efficiency and product quality. Som e of the widely Used urea production processes are.

1.6.1 Once-Through Urea Process It is a conventional process in which the unconverted carbamate is decomposed to NH3 And CO2 by heating the urea synthesis reactant effluent mixture mixture at low pressure. The NH3 and CO2 is separated from the urea solution and utilized to produce ammonium Salts by absorbing NH3.

Advantage 

Simple process

Disadvantage 

Large quantity of ammonia salt formed as a co product



Overall carbon dioxide conversion is low



High production cost



High energy cost



High environment pollution

7

Figure 1.1 Typical once-through urea process

1.6.2 Partial Recycle Process 

Part of the off gas is recycled back to the reactor



The amount of ammonia is reduced to 15% to that of once-through that must be used in other process



High CO2 conversion



High energy cost



High environmental pollution



High production cost

8

Figure 1.2 Typical Typical partial recycle urea process

1.6.3 Stripping Process Based Plants (Internal carbamate recycle) The unreacted carbamate and the excess ammonia are stripped from the urea Synthesis reactor effluent by means of gaseous CO2 or NH3 at the reactor Pressure, instead of letting the reactor effluent down to a much lower Pressure. The NH3 and CO2 gas recovered at reactor pressure, is condensed And returned to the reactor by gravity gr avity flow for recovery.

Snamprogetti Process (Italy) 

Synthesis and high pressure (HP) recovery (154 bar)



Medium pressure (MD) purification purification and recovery (18 bar)



Low pressure (LP) purification and recovery (4.5 bar) 9



Vacuum concentration ( 2 steps: 0.3 and 0.03 )



Process condensate process



Finishing: prilling and granulation

Figure 1.3 Snamprogetti urea process

Stamicarbon CO2 Stripping Process : NH3 and CO2 are converted to urea via ammonium carbamate carbamate at a pressure of 140 bar and a temperature of 180−185 C°, an NH3:CO2 molar ratio of 3:1 is Applied. The greater part of the unconverted carbamate is decomposed in the Stripper, where ammonia and carbon dioxide are stripped off using CO2 as stripping agent. The stripped off NH 3 and CO2 are then partially condensed and recycled to the reactor. The heat evolved from this condensation is used to 10

produce 4.5 bar steam some of which can be used for heating purpose in the downstream sections of the plant. The NH 3 and CO2 in the stripper effluent are vaporized in 4 bar decomposition stage and subsequently condensed to form a carbamate solution. Further concentration of urea solution takes place in the evaporation section, where 99.7% of urea melt is produced.

Figure 1.4 stamicarbon CO2 stripping process

ACES Process ACES ( Advanced Process for Cost and Energy Saving ) process has been developed by Toyo Engineering Corporation. Its synthesis section consists 11

of the reactor, stripper, two parallel carbamate condensers and a scrubber all operated at 175 bar. The reactor is operated at 1900 C° and an NH3:CO2 molar feed ratio of 4:1. Its consist of five main sections 

Synthesis section



Purification section



Concentration and prilling section



Recovery section



Process condensate treatment section

Figure 1.5 ACES urea process

12

1.7 Selection Of The Process Snamprogetti ammonia-stripping ammonia-stripping urea process is selected because it involves a High NH3:CO2 ratio in the reactor, ensuring the high conversion of carbamate To urea. The highly efficient ammonia stripping operation drastically reduces the recycling of carbamate and the size of equipment in the carbamate decomposition. Snamprogetti Snamprogetti differs from other methods in being based on The use of excess ammonia to avoid corrosion as well as promote the Decomposition of unconverted carbamate into urea. The success of any urea manufacture process depens on how economically we can recycle carbamate to the reactor.



NH2COONH4(s)  2NH3(g)+CO2(g)

∆H= +37.4 kcal/gmmol

This reaction involves increase in volume and absorption of heat. Thus this Reaction will be favored by decrease in pressure and increase in temperature Moreover decreasing the partial pressure of either of the products will also favor the forward reaction. The process based on increase/decrease of partial pressure of NH3 or CO2 is called stripping process. According to the above equation we have: 2

K= (PNH3) *(PCO2)

[where, K= equilibrium constant]

The stripping is effected at synthesis pressure itself using CO2 or NH3 as Stripping agent. If CO2 is selected, it is to be supplied to the decomposer /stripper as in stamicarbon CO2 stripping process. While if NH3 , is to be obtained from the system itself because excess NH 3 is present in the reactor as in snam's process. At a practical temperature K is constant so when (PNH3 ) is reduced to keep K constant, carbamate will reduce much faster by 13

decomposition as (PNH3 ) appear in the t he equilibrium equation with a power of st

two. Selection of 1  decomposition should be in such a way that minimum water evaporates because the recovered gases go along with the carbamate to reactor again and if water enters reactor production will be effected adversely due to hydrolysis of urea. So , stage wise decomposition of carbamate is done.

1.8 Process Description The urea production process takes place through the following main Operations: 

Urea synthesis and high pressure recovery



Urea purification and low pressure recovery



Urea concentration



Urea prilling

Urea Synthesis And High Pressure Recovery Urea is synthesized from liquid ammonia and gaseous carbon dioxide under Specific concentration, temperature temperature and pressure conditions according to the following reaction:

 NH2COONH NH2COONH  NH CONH NH3(g)+CO2(g)

4(s)

4(s)

2

2(s)

+ H2O(l)

∆H= − 37.64 kcal/gmmol ∆H=

+ 6.32 kcal/ gmmol gmmol

The carbon dioxide is drawn at about 1.6 atm a tm pressure and about 40



Is compressed in a centrifugal compressor up to 162 atm. A small quantity Of air is added to the CO 2 compressor in order to passivate the stainless Steel surfaces. Thus protecting them from corrosion due to both the 14

reagent and the reaction product. product. The liquid ammonia ammonia is drawn from from ammonia receiver tank where its compressed to 23 atm by means of centrifugal pump. Part of this ammonia is sent to the medium pressure absorber and remaining part enters the high pressure synthesis loop. The NH3 of this synthesis loop is compressed to a pressure of about 240 atm The liquid mixture of ammonia and carbamate enters the reactor where it Reacts with the compressed CO2. The reactor operates at 154 atm, NH3:CO2 molar feed ratio of 3:1 is applied A portion of the ammonium carbamate formed from the reaction Dehydrates. The fraction of carbamate that dehydrates is determined By the ratios of various reactants, operating temperature, the residence Time in the reactor and reaction temperature. st

nd

The 1 reaction occurs rapidly and is completed. The 2 reaction occurs Slowly and determines the reactor volume. Urea reactor is a plug flow type With 10 number of sieve trays to avoid back mixing and escape of gaseous CO2 which must react in the lower part of the reactor. Stage wise Decomposition is carried out to reduce water carry over to the reactor Which could adversely affect conversion. Urea solution containing urea, Carbamate, H2O and unconverted CO2 and NH3 enters the high pressure Stripper Where the pressure is same as that of reactor. The mixture is heated as it Flows down the falling film exchangers. The CO2 content of the solution Is reduced by the stripping action of NH3 as it boils out of the solution The carbamate decomposition heat is supplied by 24 atm steam. The Overhead gases from the stripper and the recovered solution from the

15

MP absorber , all flow to high pressure carbamate condenser through Mixer, where total mixture, except except for a few inerts is condensed and And recycle to reactor by means of carbamate ejector. Condensing the Gases at high temperature and pressure permits the recovery permits the Recovery of condensation in the production of steam at 4.5 atm in the high Pressure carbamate condenser.

Urea Purification And Low Pressure Recovery : Urea purification takes place in two stages at decreasing pressure as follow: st

1 stage at 18 atm pressure , i.e. , MP decomposer nd

2  stage at 4.5 atm pressure , i.e. , LP decomposer st

1 stage purification and recovery stage at 18 atm: It is falling film type MP decomposer. It is divided into 2 parts: top Separator, where the released flash gases. The solution enters the tube Bundle and decomposition section where the residual carbamate is decomposed and heat is supplied supplied by means of 24 atm steam condensate flowin out of the stripper. nd

2 Purification and recovery stage at 4.5 atm: The solution leaving the bottom of MP decomposer is expanded at 4.5 atm And enters the LP decomposer ( falling film type). This is divided into two Parts: top separator where the released flash gases are removed before the solution enters the bundle. Decomposition Decomposition section where the the last residual carbamate are decomposed decomposed and the required heat is supplied by means of steam saturated at 4.5 atm .

16

Urea Concentration Next section is urea concentration and the objective is to reduce the water Content as low as 1%. Vacuum concentrator of two stage is provided for This purpose. The solution leaving the LP decomposer bottom with about st

76% urea is sent to the 1 vacuum concentrator operating at a pressure of 0.23 atm. The mixed phase coming out enters the gas liquid separator, st

where from the vapors are extracted by the 2 vacuum system. The two concentrators are fed by saturated steam at 4.5 atm.

Urea Prilling nd

The molten urea leaving the 2  vacuum separator is sent to the prilling Bucket by means of a centrifugal pump. The urea coming out of the bucket In the form of drops fall along the prilling tower and a countercurrent cold Air flow causes its solidification. The solid prills to the bottom of the prilling tower are sent through the screens to retain lumps only and then to belt conveyor which carries the product to the automatic weighing machine and to the urea storage section. Urea lumps by means of belt conv eyor are recycled to the underground tank, where they are dissolved.

17

Figure 1.6 Snamprogetti Urea process

Figure 1.7 Snamprogetti Urea Process Block Diagram 18

Chapter2

Material Balance

19

Material balance

Selected capacity

500,000 ton/year

No. of working days

300 day

Daily production

500,000/300=1666.667 ton/day =69444.44 kg/hr

Composition Composition of final products: Urea

: 98% (68055.56) kg/hr

Water

: 1%

Biuret

: 1%

(694.4444) kg/hr (694.4444) kg/hr

Assumption: overall conversion to urea is assumed to be 95%

Main reactions:

NH COONH

1) CO2  + 2NH3 (44)

2

4

(17)

2) NH2COONH4

(78)

 NH CONH 2

2

+ H2o

(60) 3) CO2  + 2NH3

 NH CONH 2

(18) 2

+ H2O (overall reaction)

Side reaction: 4)2NH2CONH2

 NH CONHCONH  + NH 2

2

3

(103) 20



Input output= accumulation

(general mass balance eqn.)

Input = output

(accumulation=0)

Urea produced from reaction 4

= 694.4444*(2*60/103)=809.0615 kg/hr

Urea produced from reaction 2

= 809.0615+68055.56= 68864.62 kg/hr

NH3 produced from reaction 3

= 68864.62*(2*17/60)=39023.28 kg/hr

CO2 reacted from reaction 1

= (44/60)*68864.62= (44/60)*68864.62= 50500.72 kg/hr

2.1 Reactor Assuming 95% conversion of urea NH3 input to the reactor = 39023.28/0.95 = 41077.14 kg/hr CO2 input  input to the reactor = 50500.72/0.95 = 53158.65 53158.65 kg/hr kg/hr Urea produced from reaction 3 in reactor = (60/40)*53158.65*0.95=688 (60/40)*53158.65*0.95=68864.62 64.62 kg/hr



Urea converted to NH3 & Biuret Biuret in reaction 4 =68864.62 68055.56=809.0615 kg/hr Biuret produced in reaction 4 in reactor = (103/2*60)*809.0615= (103/2*60)*809.0615= 694.4444 kg/hr Water produced produced in reaction 2 in reactor = (18/60)*68864.62= 20659.39 kg/hr

At reactor's exit (Urea=34%) Flow rate of stream = 68055.56/0.34= 200163.4 kg/hr NH3 reacted in reaction 1 = (2*17/60)*68864.62= 39023.28 kg/hr NH3 produced in reaction 4 = (17/2*103)*809.0615 = 114.617 kg/hr

 = 41077.1439023.28+114.617 = 2168.474 kg/hr

NH3 unreacted= NH3 input NH3 reacted + NH3 produced

21

CO2 reacted in reaction 1 = (44/60)*68864.62= (44/60)*68864.62= 50500.72 kg/hr

 = 53158.65 50500.72= 2657.933 kg/hr Flow rate of carbamate = Total flow rate of stream exit – (flow rate of

CO2 unreacted = CO2 input CO2 reacted

urea+CO2+NH3+Biuret+Water)



=200163.4 (68055.56+2657.933+2168.474+694.4444+20659.39)=10592 7.933 kg/hr

NH3=40177.14 kg/hr CO2=53158.68 Kg/hr Carbamate=105927.6 Kg/hr

 R   e  a  c   t    o  r 

NH3=2168.474 kg/hr CO2=2657.933 kg/hr Urea=68055.56 kg/hr Water=20659.39 kg/hr Biuret=694.4444 kg/hr Carbamate=105927.6 kg/hr

Figure 2.1 flow of materials across reactor Table 2.1 flow of material across reactor Input Material

Flow rate Kg/hr

Output %

Flow rate Kg/hr

%

NH3

41077.4

43.589

2168.474

1.0833

CO2

53158.65

56.410

2657.933

1.3278

 ___

 ___

68055.56

 ___

 ___

20659.39

10.321

 ___

 ___

694.4444

0.3469

100

105927.6

52.920

100

200163.4

100

Urea Water Biuret

Carbamate 105927.6

Total

200163.4

22

34

2.2 Stripper No reaction takes place in stripper. Only carbamate get recycled back to the Reactor. Therefore, the amount of ammonia, NH3,CO2,Water NH3,CO2,Water & biuret in the outlet stream of stripper will be the same as inlet stream.

Carbamate=84742.09 Carbamate=84742.09 kg/hr

NH3=2168.474 NH3=2168.474 kg/hr

 S   t    r   i    p  p  e  r 

CO2=2657.933 CO2=2657.933 kg/hr Urea=68055.56 Urea=68055.56 kg/hr Water=20659.39 Water=20659.39 kg/hr Biuret=694.444 Biuret=694.444 kg/hr Carbamate=105927.6 Carbamate=105927.6 kg/hr

NH3=2168.474 NH3=2168.474 kg/hr CO2=2657.933 CO2=2657.933 kg/hr

Urea=68055.56 Urea=68055.56 kg/hr Water=20659.39 kg/hr Biuret=694.444 Biuret=694.444 kg/hr Carbamate=21185.52 Carbamate=21185.52 kg/hr Figure 2.1 flow of materials across stripper

23

Table 2.2 flow of materials across stripper Output

Input Materials

Flow rate (kg/hr)

%

Feed

Flow rate (kg/hr)

Materials

%

Bottom product

NH3

2168.474

1.0833

NH3

2168.474 1.8787

CO2

2657.933

1.3278

CO2

2657.933 2.3028

Urea

68055.56

34

Urea

68055.56 58.962

water

20659.39 10.321

Water

20659.39

17.899

Biuret

694.4444 0.346

Biuret

694.4444

0.6016

carbamate 105927.6 52.920

Carbamate

21185.52 18.354

Top product carbamate Total

200163.4

100

84742.02

100

200163.4

100

2.3 Medium Pressure Decomposer The amount of ammonia,CO2,Biuret , Water and carbamate will be the same As no Reaction takes place. 50% of ammonia&CO2 and 10% of carbamate are Assumed to escape from the top of separator and the rest goes with the bottom product.

24

NH3=1084.237 kg/hr CO2=1328.966 kg/hr Carbamate=19066.67 kg/hr

 M  e  d   d   e  i    c   u  o  m  m  p  p  r   o  e  s   e  s   s   r   u  r   e

NH3=2168.474 kg/hr CO2=2657.933 kg/hr Urea=68055.56 kg/hr Water=20659.39 kg/hr Biuret=694.444 kg/hr

NH3=1084.237 kg/hr CO2=1328.966 kg/hr Urea=68055.56 kg/hr

Carbamate=21185.552 kg/hr

Water=20659.39 kg/hr Biuret=694.444 kg/hr Carbamate=2118.552 kg/hr Figure 2.3 flow of material across medium pressure decomposer

Table 2.3 flow of materials across medium pressure pressure decomposer Input Materials

Flow rate (kg/hr)

Output Materials

%

Flow rate (kg/hr)

%

NH3

2168.474

1.878

NH3

1084.237

1.154

CO2

2657.933

2.302

CO2

1328.966

1.414

Urea

Urea 68055.56

58.962

68055.56

72.444

Water

20659.39

21.991

Water

20659.39

17.899

Biuret

694.4444

0.7392

Biuret

694.4444

0.6016

Carbamate

2118.552

2.255

Carbamate 21185.52

18.354

Total 1

93941.14

100

NH3

1084.237

5.0476

CO2 Carbamate

1328.966 19066.97

6.1869

Total 2

24180.17

100

Losses

Total

115421.3

115421.3

100

2.4 Low Pressure Decomposer

25

88.765

100

No reaction take place and the remaining NH3, CO2and Carbamate are escaped from the top of .

NH3=1084.237 kg/hr CO2=1328.966 kg/hr L   o  w  p  r   e  s   s   u  r   e  d   e  c   o  m  p  o  s   e  r 

NH3=1084.237 kg/hr CO2=1328.966 kg/hr Urea=68055.56 kg/hr Water=20659.39 kg/hr

Carbamate=2118.552bkg/hr

Urea=68055.56 kg/hr (76.11%) Water=20659.39 kg/hr Biuret=694.444 kg/hr

Biuret=694.444 kg/hr Carbamate=2118.552 kg/hr Figure 2.4 flow of materials across low pressure decomposer

Table 2.4 flow of materials across low pressure decomposer Input Materials

Output Flow rate (kg/hr)

%

Materials

Flow rate (kg/hr)

%

Urea

68055.56

76.11

water

20659.39

23.106

NH3

1084.237

1.154

CO2

1328.966

1.414

Urea

68055.56

72.44

Biuret

694.4444

0.776

Water

20659.39

21.99

Total1

89409.39

100

Biuret

694.4444

0.739

Losses

Carbamate 2118.552

2.255

NH3

1084.237 1328.966

23.925

2118.552

46.749

4531.755

100

93914.14

100

CO2 Carbamate Total 2 Total

93914.14

26

100

29.325

2.5 Vacuum Evaporator Total output from low pressure decomposer = 89409.39 kg/hr Let X

mass fraction of urea in feed(F) =0.7611

Y

mass fraction of feed in product(P) =0.9788

E

water losses in vacuum evaporator

Urea balance Input=output F*X=P*Y

  P=(F*X)/Y

P=(89409.39*0.7611)/0.9788 =69523.38 kg/hr Overall material balance



F=P+E



E=F P



E=89409.39 19886.01=19886.01 kg/hr

Water=19886.01 kg/hr

Urea=68055.56 kg/hr (76.11%) Water=20659.39 kg/hr Biuret=694.444 kg/hr

 V   a  c   u  u  m  e  v   a  p  o  r   a  t    o  r 

Urea=68055.56 kg/hr (97.88%) Water=773.37 kg/hr Figure 2.5 flow of material across vacuum evaporator

27

Biuret=694.4444 kg/hr

Table 2.5 flow of materials across vacuum evaporator Output

Input Materials

Flow rate (kg/hr)

%

Materials

Flow rate (kg/hr)

%

Urea

68055.56 76.11

Urea

68055.56

97.88

Water

20659.39 23.106

Water 773.378

1.112

Biuret

694.4444 0.776

Biuret 694.4444

0.998

Losses

Total

89409.39

Water 19886.01

100

89409.39

100

100

2.6 Prilling Tower Output from vacuum evaporator = 69523.38 kg/hr Let X mass fraction of urea in feed (F) =0.9788 Y mass fraction of urea in product(P) =0.979913 Urea balance Input=output



P=(F*X)/Y

P=(69523.38*0.9788)/0.979913 P=(69523.38*0.9788)/0.9799 13 =69444.44 kg/hr

28

Water=78.938 kg/hr Air

Urea=68055.56 kg/hr (97.88%) Water=773.37 kg/hr

 P   r   i    l    l    i    n  g  t    o  w  e  r 

Air

Biuret=694.4444 kg/hr

Urea=68055.56 kg/hr (98%) Figure 2.6 flow of materials across prilling tower

Water=694.4406 kg/hr Biuret=694.4444 kg/hr

Table 2.6 flow of materials across prilling tower Output

Input Materials

Flow rate (kg/hr)

%

Materials

Flow rate (kg/hr)

% 98

Urea

68055.56

97.88

Urea

68055.56

Water

773.378

1.112

Water

694.44

Biuret

694.4444

0.998

Biuret

694.4444

0.999 1

Losses Water Total

69523.38

100

29

78.938

100

69523.38

100

Table 1.7 a: Process flow diagram of materials 1 NH3 FEED

Item

1A Pumped NH3

2 CO2 feed

Stream no.

2A Compressed CO2

3 Reactor outlet

4 Stripped carbamate

4A Carbamate condenser

component

 M  a  s   s   f    l    o  w  r   a  t    e  o  f    c   o  m  p  o  n  e  n  t    (    k    g  /    h   r   )  

NH3

CO2

UREA

WATER

BIURET

41077.14 (43.589%)

 ___

 ___

 ___

 ___

41077.14 (43.589%)

___

 ___

 ___

 ___

 ___

53158.68 (56.410%)

 ___

 ___

53158.68 (56.410%)

 ___

 ___

 ___

 ___

 ___

CARBAMATE  ___

Temperature

Pressure

 ___

 ___

2168.474 (1.0833%)

2657.933 (1.327%)

68055.56 (34%)

20659.39 (10.321%)

694.4444 (0.3649%)

 ___

 ___

 ___

 ___

 ___

 ___

105927.6 (52.920%)

105927.6 (100%)

 ___

 ___

 ___

 ___

 ___

105927.6 (100%)

40

40

40

40

180

185

185

23

240

1.6

162

154

154

4.5

30

Table 1.7 b: Process flow diagram of materials materials Item

13 5 6 7 8 9 10 11 12 17 15 16 Pumped Evaporated Stripper MPD MPD LPD LPD Pumped Evaporator Air Prilling H2O(v)+ Stream no. outlet outlet1 outlet2 Outlet1 outlet2 LPD outlet water Evaporator feed outlet Air outlet

443951.6 (100%)

component  M  a  s  s  f   l    o  w  r  a  t   e  o  f   c  o  m  p  o  n  e  n  t   (   k   g  /   h  r  ) 

78.938 (100%)

NH3

2168.474 1084.237 1084.237  ___ 1084.237  ___ (1.878%) (5.047%) (1.154%) (23.925%)

 ___

 ___

 ___

 ___

 ___

 ___

CO2

2657.933 1328.9661328.966  ___ 1328.966  ___ (2.302%) (6.186%) (1.414%) (29.325%)

 ___

 ___

 ___

 ___

 ___

 ___

UREA

68055.56  ___ 68055.56 68055.56 68055.56 68055.56  ___ 68055.56  ___ 68055.56  ___  ___ (58.96%) (76.116%) (97.88%) (97.888%) (98%) (72.44%) (76.11%)

WATER

20659.39  ___ 20659.39 68055.56  ___ 20659.39 773.378519886.01773.3785  ___ 694.440  ___ (21.991%) (23.106%) (23.106%) (1.112%) (100%) (1.112%) (0.999%) (17.889%)

BIURET

694.4444  ___ 694.4444694.4444  ___ 694.4444 694.4444  ___ 694.4444  ___ 694.4444  ___ (0.601%) (0.739%) (0.776%) (0.776%) (0.988%) (0.988%) (1%)

Carbamate 21185.52 19067.97 2118.552  ___ 2118.522  ___ (18.354%)(88.765%) (2.55%) (46.749%)

Temperature (C˚) Pressure (atm)

185

154

 ___

___

 ___

 ___

 ___

 ___

140

140

80

80

85

27

27

23.77

20

30

30

18

18

4.5

4.5

4.5

0.03

0.03

0.03

1

1

1

31

Chapter 3

Energy Balance 32



Assumption  : reference temperature =25

Heat in +Generated= Heat Heat out + Consumed ( conservation law of energy)

Table 3.1 specific heat constant Speciefic Heat Constant -2

-5

Component

a

NH3(gas)

8.4017

NH3( liquid)

4.6356

CO2 (gas)

6.393

1.01

− 0.3405

 ___

Urea

38.43

4.98

0.705

−8.61

Water

7.88

0.32

 ___

Carbamate

2.596

Biuret

b *10

c*10

-7

d *10

0.70601 0.10567 −0.01598  ___

 ___

 ___

−4.833

 ___

 ___

 ___

 ___

___

___

183.8

3.1 Reactor



2NH3+CO2 NH2CONH2+H2O

(main reaction)

∆H298°=∑(n ∑(ni ∆Hf )p –(n  –(ni ∆Hf )R ∆Hf (urea)= −333.6 KJ/gm mol , ∆Hf (NH3)=−46.16 KJ/gm mol ∆Hf (CO2)= −393.5 KJ/gm mol , ∆Hf (H2O)= −285.84 KJ/gm mol ∆H298°= (-333.6+(-285.84)− (-333.6+(-285.84)−((2*-46.16)+(-393.6))= −133.62 KJ/gm mol Moles of urea formed during the reaction=68055.56/60=1134.259 kmole/hr =1134259 gmmol/hr ∆H298°=−133.62 KJ/ gmmol*1134259gm gmmol*1134259gm mole/hr = −151559722.2 KJ/hr 33

Inlet stream

  

∆HR°=mi 

where (Tin=

Material

specific heat(KJ/kg)

NH3

−69.534

CO2

−9.01033

 ,T

)

ref =25

flow rate(kg/hr) 40177.14 53158.65

∆HR°=(−69.534*40177.14−9.01033 *53158.65) = −3335235 KJ/hr

Outlet stream

 ∆HP°=mi   Materials



Specific heat (KJ/kg)

UREA

Flow rate(kg/hr)

121.7623

WATER

)

where (Tout=180  , Tref =25

68055.56

292.7034

20659.39

∆HP° =(121.7623 *68055.56 +292.7034*20659.39 ) = 14333675 KJ/hr

∆H = ∆H ∆H298° + ∆HP°+ ∆HR° 151559722.2--3335235 +14333675 =−140561282 kJ/hr =−151559722.2--3335235 Q=∆H =−140561282 KJ/hr o

Assumption : cooling water at 25 C  is used to remove heat from reactor. o

The outlet is steam at an absolute pressure of 4.5 bar (Ts=147.9 C ). Heat gained by cooling water = 140561282 KJ/hr



M(cp∆T+ )= 140561282 KJ/hr

=2120.8 KJ/kg Cp=4.187 KJ/kg

from steam table & by interpolation



, ∆T=Ts−T

Ts=147.9

M(4.187*(147.9−25)+2120.8)= 140561282 M= 53336.2 kg/hr 34



, T=25



180 C˚ Feed Q in=    3335235 KJ/hr M in=200163.4 kg/hr

 R   e  a  c   t    o  r 

Product Q out =     140561282 KJ/hr M out= 200163.4 kg/hr Tout =180 C˚

Tin=40 C˚

Figure 3.1 energy flow across reactor

3.2 Stripper Total heat input=14333675 KJ/hr

Outlet stream 

Liquid

Q =mi Materials

 

specific heat(KJ/kmol )

flow rate(kmole/hr)

mole fraction(x)

NH3

6009.302

2168.474/17=127.5573

0.046413

CO2

4801.186

2657.933/44=60.40756

0.02198

68055.56/60=1134.259

0.41271

UREA

7607.734

WATER

5443.803

20659.39/18=1147.744

0.417617

BIURET

29408

694.444/103=6.742179

0.002453

21185.52/78=271.60

0.098827

CARBAMATE 32398

TOTAL=2748.319

35

CP of mixture=∑Xi CPi =(6009.302* 0.046413+ 4801.186 *0.02198+ 7607.734* 0.41271+5443.803 * 0.417617+ 29408* 0.002453+32398*0.098827) = 9071.611 KJ/kmole Heat carried by outlet outlet stream=2748.319*9071.611=24931683 stream=2748.319*9071.611=24931683 KJ/hr

Vapor Stream: ammonium carbamate



material

specific heat (KJ/kmole)

Carbamate For carbamate

flow rate(kmole/hr)

32398.08

1086.436

=210 KJ/kg

Heat carried by carbamate= m*cp*∆ m*cp*∆T+ m*



=32398.08* 1086.436+ 84722.085*210= 52994311 KJ/hr Here, steam at 24 atm is used(Ts=221.8 C)

  of steam =1855.3 KJ/kg

from steam table

Heat supplied by steam= heat output−heat input input



M  =(52994311+24931683 −14333675)= 52994311 KJ/hr M= 63592318 kg/hr

36

Carbamate Q out=52994311 KJ/hr M out=84742.02 kg/hr

T= 185 C˚

T = 185 C˚

 S   t    r   i    p  p  e  r 

Feed

Q in=14333675 KJ/hr M in= 200163.4 kg/hr T in= 180 C˚

T = 185 C˚ Q out=24931683 KJ/hr M out= 115421.38 kg/hr product

Figure 3.2 energy flow across stripper

5.3 CARBAMATE CONDENSER Energy balance Mv

 =m  Cp *(T −25)+m  v

s

Where

s

s

s

V : vapor of carbamate , S: steam

Putting the values we get 105927.6*210=ms[4.187x (147.9−25)+2120.8] Ms =8440.824 kg/hr

Carbamate vapor Carbamate liquid =105927.6 kg/hr

25 C˚ Water

147.9 C˚ Steam

Figure 3.3 energy flow across carbamate condenser

37

5.4 MEDIUM PRESSURE Decomposer Heat input=24931683 KJ/hr

Outlet stream 

Liquid

Q =mi

 

Material

cp(KJ/kmol)

flow rate(kmol/hr)

mole fraction(x)

NH3

4241.15

63.77865

0.026465

CO2

3348.267

30.20378

0.012533

Urea

5068.511

1134.259

0.470669

Water

3879.049

1147.744

0.476264

Biuret

19655.8

6.742179

0.002798

Carbamate 23286.12

27.16092

0.011271

Total =2409.888 Cp of mixture=∑Xi Cpi =(4241.15*0.026465+ 3348.267 * 0.012533+ 5068.511*0.470669+ 3879.049 * 0.476264+ 19655.8* 0.002798+23286.12* 0.011271) = 4704.691 KJ/kmol Heat output=4704.691*2409.88 output=4704.691*2409.888=11337780 8=11337780 KJ/hr 

For Gasses Escaping From The Top

Material

cp(KJ/kmole)

flow rate(kmole/hr) mole fraction(x)

NH3

4241.15

63.77865

0.188454

CO2

3348.267

30.20378

0.089247

Carbamate 23286.12

244.4483

0.722299

38

Cp of mixture=∑Xi Cpi =(4241.15*0.188454+ 3348.267* 0.089247+23286.12*0.7222 0.089247+23286.12*0.722299) 99) = 17917.63 KJ/kmole Material

 (KJ/kmole)

flow rate(kmole/hr)

mole fraction(x)

NH3

22777

63.77865

0.188454

CO2

20265

30.20378

0.089247

Carbamate

16380

244.4483

0.722299

Total = 338.4308

 of mixture=∑X  i

 i

=(0.188454*22777+0.089247*20265+16380*0.722299) = 17932.26 KJ/kmole



Heat escaping from the top =m(cp*∆ =m(cp*∆T+ ) =338.4308(17917.63 +17932.26)= 12132708 KJ/hr Assumption: cooling water enters at 25

 & leaves at 50

Heat gained by cooling water =heat input−heat output =(24931683−12132708 −11337780)= 1461196 KJ/hr M*cp* ∆T=1461196 M=1461196 /(4.184*25)= 13969.37 Kg/hr

39

Off-gases Q out=12132708 KJ/hr

T=140 C Feed Q in=24931683 KJ/hr M in=115421.3 kg/hr T in= 185 C

˚

 M  e  d   d   e  i    c   u  o  m  m  p  p  r   o  e  s   s   e  r   s   u  r   e

M out =24180.17 kg/hr T=140 C

˚

T=140 C

˚

˚

Q out=11337780 KJ/hr M out=93941.14 kg/hr Products Figure 3.4 energy flow across medium pressure separator

5.5 Low Pressure Decomposer Heat input=11337780 KJ/hr

Outlet stream 

liquid

Q =mi Material

  Cp (KJ/kmole )

flow rate(kmole/hr)

mole fraction(x)

UREA

2226.101

1134.259

0.495581

WATER

1833.484

1147.744

0.501473

Biuret

8195

6.742179 Total = 2288.745

Cp of mixture=∑Xi*Cpi mixture=∑Xi*Cpi

40

0.002946

=(2226.101*0.495581+1833.484 * 0.501473+ 8195*0.002946) =2046.797 KJ/kmole Heat output= 4684597 KJ/hr 

For gasses escaping from the top

Material

Cp(KJ/kmole)

flow rate(kmole/hr)

mole fraction(x)

NH3

1978.934

63.77865

0.526473

CO2

1534.283

30.20378

0.249323

Carbamate

11136.84

27.16092

0.224202

Total =121.1434 Cp of mixture=∑Xi*Cpi mixture=∑Xi*Cpi

=(1978.934* 0.526473+ 1534.283*0.249323+11136.84*0.224202) = 3919.077 KJ/kmole Material

 (KJ/kmole)

flow rate(kmole/hr)

mole fraction(x)

NH3

1851

63.77865

0.526473

CO2

1566

30.20378

0.249323

Carbamate

16380

27.16092

0.224202

Total =121.1434 =∑Xi*Cpi  of mixture =∑Xi*Cpi =(0.526473* 1851 + 0.249323*1566 +0.224202*16380 ) =5073.415 KJ/kmol



Heat escaping from the top =m(cp*∆ =m(cp*∆T+ ) =121.1434(3919.077 +5073.415)= 1085019 1085019 KJ/hr Assumption : cooling water enters at 25 C & leaves at 50C Heat gained by cooling water=heat input−heat output

41

=(11337780− =(11337780−1085019− 1085019−4684597)= 5568163 KJ/hr M*cp M*cp ∆T=5568163 ∆T=5568163 KJ/hr M=5568163 /(4.187*25)= 5394.177 kg/hr

Off-gases Q out=1085019 KJ/hr M out=4531.755 kg/hr

Feed Q in=11337780 KJ/hr Min= 93914.14 kg/hr T=140 C˚

T=80 C˚ L   d   o  e  c   w  o  p  m  r   e  p  s   o  s   s   u  e  r   r   e

T=80 C˚

T=80 C˚ Q out=4684597 KJ/hr M out=89409.39 kg/hr Products Figure 3.5 energy flow across low pressure decomposer

4.6 Vacuum Evaporator st

For product stream coming coming out from 1  evaporator Material

Cp(KJ/kmole)

flow rate(kmole/hr)

mole fraction(x)

Urea

2443.41

1134.259

0.750452

Water

2002.149

370.4335

0.245087

6.742179

0.004461

Biuret

8940

Total =1511.435 Cp of mixture=∑Xi*Cpi mixture=∑Xi*Cpi =(2443.41*0.750452+2002.149 =(2443.41*0.750452+2002.14 9 * 0.245087+ 8940* 0.004461) =2364.243 KJ/kmole

42

M*cp M*cp ∆T= 2364.243*1511.435 =3573399 KJ/hr

Heat balance st

1  Evaporator Heat input(by feed)+Heat input by steam=Heat carried by water vapor + Energy of bottom product Heat input by feed+ S1



s1=

E1HE1 +energy of bottom product

4684597+ 4684597+ S1 * 2123.8= 2123.8= 13991.58*2614.97+ 13991.58*2614.97+3573399 S1= 16687.25 kg/hr nd

2  Evaporator Heat input(by feed)+Heat input by steam=Heat carried by water vapor +Energy of bottom product 3573399+ 3573399+ S1 * 2123.8= 2123.8=5894.424*2545.7+2464.393 *1183.967 S2= 6756.660 kg/hr

E1=13991.58 kg/hr

E2=5894.424 kg/hr

Feed Tin=80 C˚ Q in=4684597 KJ/hr M in=89409.39 kg/hr

T=23.77 C˚

T=63.1 C˚

P=0.03 atm

P=0.23 atm steam

steam T=147.9 C˚

T=147.9 C˚ P=4.5 atm

P=4.5 atm T=27.25 C˚ Q out2=2917759 KJ/hr M out=89409.39 kg/hr P1=75417.8 kg/hr

Product

(90.93%) urea Q out1=3573399 KJ/hr Figure 3.6 energy flow across vacuum evaporator

4.7 Prilling Tower Heat input= energy of bottom product of evaporator=m*cp*∆ evaporator=m*cp*∆T

43

=2464.393 *1183.967=2917759 KJ/hr

Outlet stream

Q =mi

 

Material

Cp(KJ/kmol)

Urea

192.8017

1134.259

0.961578

Water

165.0169

38.58003

0.032707

6.74217

0.005716

Biuret

flow rate(kmol/hr)

745

mole fraction(x)

Total= 1179.581 Cp of mixture=∑Xi*Cpi mixture=∑Xi*Cpi =( 192.8017*0.961578+ 165.0169*0.032707+745*0.005716)= 195.0492 KJ/kmole Heat output=195.0492*1179.58 output=195.0492*1179.581=230076.4 1=230076.4 KJ/hr Heat carried away by air=heat input−heat output = 2917759− 2917759−230076.4 =2687683 KJ/hr (m*cp*∆ (m*cp*∆T)dry air=2687683 KJ/hr Cp air=1.009 KJ/kg



, ∆T=26−20=6

,

M=2687683/(1.009*6)= 443951.6 kg/hr

44

Air Q out=2687683 KJ/hr M out=443951.6 kg/hr T out= 26 C

 P   r   i    l    l    i    n  g  t    o  w  e  r 

Feed Q in=2917759 KJ/hr M in=69523.38 kg/hr T in= 27 C

˚

Air T in= 20 C

˚

˚

T out= 30 C

˚

Q out=230076.4 KJ/hr M out=69444.44 kg/hr Product Figure 3.7 energy flow across prilling tower

45

Chapter 5

Equipment Design

46

4.1 Reactor Design 4.1.1 Introduction The reactor is the heart of a chemical process. It is the only place in the Process where raw materials are converted into products, and reactor Design is a vital step in the overall design of the process. Design of the reactor is no routine matter, and many alternative can be proposed for a process. In searching for the optimum it is not just the cost of the reactor that must be minimized. One design may have lower reactor cost but the materials leaving the unit may be such that their treatment requires a much higher cost than alternative designs. designs. Hence, the economics of the overall process must be considered. Reactor design uses knowledge , information and experience from a variety of areas such as chemical kinetics, thermodynamic, fluid mechanics , heat transfer ,mass transfer and economics. Chemical reaction engineering is the Synthesis of all these factors with the aim of properly designing a chemical to find what a reactor reactor is able to do we need to know the kinetics , the contacting pattern and the performance equation. The selected reactor for this project is plug flow type with ten numbers of sieve trays in a continuous process. The liquid mixture of NH3 and carbamate and gaseous CO2 are fed to the reactor where they meet 180

 temperature and 154 bar bar pressure and form ammonia ammonia carbamate. This carbonates dehydrates and forms urea.

47

4.1.2 Residence Time And Volume



2NH3+CO2 NH2CONH2+H2O

(main reaction)

Let Ca= NH3 , Cb= CO2 , CS= urea , Cd = water Density of liquid NH3 = 618 kg/m Density of CO2 gas at 40

3

 =277.38 kg/m

3

(density=PMwt/RT,P=162 (density=PMwt/RT,P=162 atm)

3

Density of carbamate=1600 kg/m

So, NH3 flowing into the reactor =41077.14/618 =66.46786 m3/hr CO2 flowing into into the reactor =53158.65/277.38 =191.6456 m3/hr 3

Carbamate flowing flowing into the the reactor =105927.6/1600= =105927.6/1600= 66.20475 m /hr Total flow rate into the reactor=66.46786+191.6456+66.20475=322.7663 m3/hr

=Cbo  

Where, = Residence time −rb= rate of the reaction Cbo= initial concentration of the limiting reactant CO2 is the limiting reactant= CB , Xb=0.65

b=(2−1/3)=1/3 b=(2−1/3)=1/3 −rb=K1 CA CB – K  – K2 CS CD CA=CAO− xb*CBO

*xb

CB=CBO(1−xb)/1+ (1−xb)/1+

CS=xb*CBO , CD=xb*CBO −rb=K1* CBO2 (CAO/CBO –xb)(1  –xb)(1 –xb)/(1+1/3 *xb)−K *xb)−K2* CBO2 *xb2 3

3

CAO=7.322205 kmol/m , CBO=3.743114 kmol/m

K1 & k2 from Arrhenius equation k=ko* exp(Ea/RT) 48

For k1

Ea1=139500 j/mol , ko1=2.07 *1010  1/s

For k2

Ea2=98500 j/mol , ko2=9*10   1/s

11

At T=180+273=453 k , R=8.314 j/mol k k2=3.944292121 1/s , k2=3.944292121

K1=1.6978E-06

  –   –      



Now , = 1/CBO *

Using Simpson's rule to solve the above equation we get

=22.30223 min = 0.371704 hr Volume of reactor(V)=*V o

3

Volume of reactor=0.371704 *322.7663= 119.9735 m Assuming L/D= 6



2



3

V= /4 D  *6*D = 3 /2 D   D=2.942308 m ,L=17.65385 m

4.1.3 Thickness Of Shell Data available: Temperature inside the reactor= 180



Pressure inside the reactor= 154 atm Material of construction : low alloy carbon steel

Ts= (P Di/2J f – f –P P ) +c Ts : thickness of the shell ( m) Di : internal diameter (m) P : design pressure

2

(N/m )

F: Allowable stress= 1.05*108 N/m2 J: Joint factor =1 49

C: corrosion allowance (m) 7

2

Internal pressure= 154 atm = 1.56 *10  N/m

7

7

2

Design pressure =(10% extra) = 1.1*1.56 *10 = 1.716 *10  N/m 7

8

7

Ts=( 1.716 *10  *2.942308 /2*1.05*10 *1−1.716 *10 )+0.003 = 0.264827 m = 265 mm

4.1.4 Head Design

For ellipsoidal heads Th = P *Di/(2*J*f −0.2*P) 7

8

7

=(1.716 *10 *2.942308 /2*1.05*10 *1−0.2*1.716 *10 )=0.2474264 m =248 mm

4.1.5 Piping design for reactor Input pipes: Data available: Input temperature for all material =40



2

Design stress(f)=135 N/mm

For liquid ammonia (NH3) 2

Pressure

240 atm =24.312 N/mm 3

Density

618 kg/m

Mass rate

41077.14 kg/hr

Velocity (u)

2 m/s 50

0.0184633 m3/s

Volumetric flow rate

2

Cross sectional area

0.00923165 m

ID

0.10844381 m=4.26944151 in

Pipe thickness

10.7310329 mm=0.42248161 in

From schedule No.=120 D (nominal)

5 in

ID

4.563 in

OD

5.563 in

Thickness

0.5 in

For CO2 (gas) 2

Pressure

162 atm =16.4106 N/mm 3

Density

277.38 kg/m

Mass rate

53158.68 kg/hr

Velocity(u)

20 m/s 3

Volumetric flow rate

0.05323491 m /s

Cross sectional area

0.00266175 m2

ID

0.05823022 m=2.292842 in

Pipe thickness

3.76826823 mm=0.14835702 in

From schedule No.=80 D(nominal)

2.5 in

ID

2.323 in

OD

2.875 in

51

Thickness

0.276 in

For carbamate (liquid) 2

Pressure

240 atm=24.312 N/mm 3

Density

1600 kg/m

Mass rate

105927.6 kg/hr

Velocity (u)

20 m/s 3

Volumetric flow rate

0.01839021 m /s

Cross sectional area

0.0091951 m

ID

0.10822896 m=4.26098281 in

Pipe thickness

10.7097724 mm=0.42164458 in

2

From schedule No.=120 D(nominal)

5 in

ID

4.563 in

OD

5.563 in

Thickness

0.5 in

Output Pipe Data available: Temperature of the outlet materials =185



2

Design stress(f) =105 N/mm Phase : solution

2

Pressure

154 atm =15.6002 N/mm

Density

1283.97 kg/m3

52

Mass rate

200163.4 kg/hr

Velocity(u)

2 m/s 3

Volumetric flow rate

0.04330393 m /s

Cross sectional area

0.02165196 m

2

ID

0.16607866 m=6.53852974 in

Pipe thickness

13.3274841 mm=0.5247041 in

From schedule No.=160 D(nominal)

8 in

ID

7.437 in

OD

8.625 in

Thickness

0.594 in

4.1.6 Sieve Plate Tray The reactor consists of 10 number of sieve plate Diameter of reactor(d)=2.9423 m



Total area of reactor(At) = /4 *(2.9423)2=6.795883 m2 Area of downcomer(Ad) = 0.1* At 2

Ad=0.1*6.795883= 0.6795883 m

Active area(Aa)= At−2* Ad=5.4367 m2 Weir length(Lw)=0.75*d Lw=0.75*2.9423=2.147885 Lw=0.75*2.9423=2.1478 85 m Weir height(Hw)= 40 mm Assuming hole diameter (ho)=12 mm 53

Mass of outlet solution =2000163.4 kg/hr = 55.600 kg/s 3

Density of outlet solution=1283.97 kg/m Assuming (mmin)=70% of mmax mmin=0.7*55.600=38.92066 kg/s



maximum weir crest, HWC =750(mmax /Lw* )

2/3

=750(55.600/2.147885*1283.97)2/3 =55.557 mm



2/3

minimum weir crest, H WC =750(mmin /Lw* ) 2/3

= 750(38.92066 /2.147885*1283.97) =43.997 mm liquid height The constant(k2)of weep point correlation =30.8 =30.8 at HWC +Hw =40+43.997 =83.997 mm

 

U min =(k2−0.9(25.4−ho))/

1/2

=0.5229 m/s the minimum minimum vapor velocity

at the weep point Actual minimum vapor velocity at minimum vapor flow rate = actual vapor flow rate/AH =70% of Q max max/AH =0.7*(0.0433/0.815) =0.03717 m/s The minimum operating velocity is above the weep point velocity Perforated area(AP)=AA−ACZ−AES ACZ: calming zone area AES: area occupied by edge strip

=95°

Lw/d= 2.147885/2.9423=0.73 ,



Angle subtended by the chord(edge plate),  = 180−95=85° The unperforated edge strip(edge plate)mean length from the geometry -3

  /180), =(2.9423−50*10

-3

LES=(d−50*10 )* ( -3

/180)=4.288

)* (

2

AES=50*10 * LES , AES=0.2144 m

LCZ= weir length(Lw)+width of un perforated edge strip 54

=2.147885+50*10-3=2.197 m -3

2

ACZ=2*(50*10 * LCZ)=0.219 m

2

AP=AA−ACZ−AES=5.4367−0.219−0.2144=3.019 m Take total hole area AH=0.15 AA=0.815 m2

 

2

AH= /4*dh *Nh =0.815

dh: hole diameter

Number of holes(Nh)=7214

D=2.942 m    m    3    5    6  .    7    1   =    H

Ts=0.2648 m

Th = 0.248 m Figure 4.1 reactor design

4.2 Carbamate Condenser 4.2.1 Introduction A condenser is a type of heat exchanger in which vapors are transferred 55

Into liquid state by removing the latent heat with the help of a coolant such as water. Condensers may may classified into into two main types types 

Those in which the coolant and condensing vapor are brought into Direct contact.



Those in which the coolant and condensate stream are separated by A solid surface, usually a tube wall.

Condenser Types 

Double pipe and multiple pipe



Air-cooled condensers



Compact condensers



Shell and tube

4.2.2 Design Procedure The selected type for this project is fixed shell and tube condenser : 1 Shell and 4 tube passes. Carbamate is the shell side and cooled Water in the tube side.

Water physical properties at 86.45 Specific heat(Cp) =4.2 KJ/kg





Thermal conductivity (k)= 0.746 W/m K Density =966 kg/m3



-4

Viscosity( )= 3.2*10

Pa. sec

Carbamate physical properties at 185 Specific heat(Cp) =2.596 KJ/kg





Thermal conductivity (k)= 0.531 W/m K 3

Density =1600 kg/m



-4

Viscosity( )= 5*10   Pa. sec 56

Water Flow Rate Mass flow rate of carbamate= 105927.6 kg/hr =29.42433 kg/sec Q=m *



,

 of carbamate= 210 KJ/kg

Q= 29.42433 * 210= 6179.11 KJ/sec Q=m cp ∆T + m



(for water) , m= 6179.11/(4.2*(147.9−25)+2120.8)

M= 2.343 kg/s

Mean temperature difference ∆Tm T1=T2=185



, t1=25

 , t =147.9  2

∆Tlm= ((T1−t2)−(T2−t1))/ln((T1−t2)/

(T2−t1))

=((185−147.9)−(185−25))/ln((185−147.9)/ (185−25)) =84.088 Ft =1

, ∆Tm=∆Tlm* Ft



∆Tm=84.088*1= 84.088

Overall heat transfer coefficient assuming Assuming Uo= 500 W/m2



Total area and number of tubes A=Q*1000/Uo* ∆Tm 2

A=(6179.11*1000)/(500*84.088)= A=(6179.11*1000)/(500*84.088 )= 146.967 m Choose do= 20 mm , di= 16 mm L= 4.88 m



 

Area of one tube =  do L = * 20*10-3 * 4.88 =0.306464 m2 Number of tubes(Nt)= total area/area of one tube = 146.967/0.306464 = 480

Shell side diameter For a triangular triangular pitch (1 shell pass and 4 tube passes)

57



Pt =1.25 do , n=2.285 , k= 0.175 1/n

1/2.285

Db= do(Nt/k)

, Db= 20(480/0.175)

 =639.31 mm

From split ring floating head C=61 , Ds= Db+ C , Ds=700.3138 mm

Tube side heat transfer coefficient (hi) Tmean = 25+147.9/2 = 86.45 C



 

2

2

2

At = /4 *di * Nt/Np Nt/Np , At= /4 *(16*0.001)  *480/4 =0.024115 m 2

Gt=Wt/At , Gt= 2.343/0.024115= 97.17648 kg/m  sec



Ut =Gt/

, Ut=97.17648 /966=0.100597 m/s 0.8

0.2

hi=4200(1.35+0.02 tmean)Ut /di

2

=1182.775 w/m



shell side heat transfer coefficient (ho) As =(pt−do)/pt * LB* Ds , pt=1.25 do , LB= Ds/5 2

As=0.019618 m

Gs =Ws/As ,Gs=29.42433/0.019618=1499.89 ,Gs=29.42433/0.019618=1499.89 kg/m2 sec 2

2

2

2

De=1.1/do (pt −0.917 do ) =1.1/20*(25 −0.917*20 )=14.201 mm



Re=Gs*de/   = 42600.05

 

-4

Pr= cp * /k = 2.596*1000*5*10  /0.531 =2.444 Jh =0.0028 ho*de/k=jh*Re*Pr0.33 ,

ho=5990.226 W/m2



Dirt coefficients

 hod=5000 W/m  2

hid= 5000 W/m

2

Calculated overall heat transfer coefficients(Uo)

58

1/Uo =1/ho+1/hod + do*ln(do/di))/2*kw+(1/hi+1/hid)*do/di Kw=50 W/m K 1/Uo=0.001718 2

Uo =518.9354 W/m Ucal



 Uassume

Tube side pressure drop -3

For

Re=4858.824 , Jf=5.9*10

 u

2 t /2

∆Pt= Np*(8*jf*(L/di)+2.5)* -3

2

∆Pt= 4(8*5.9*10

*(4.88*1000/16)+2.5)* 966*(0.100597) /2

∆Pt=330.3384 Pa

=0.3303384 Kpa

Shell side pressure drop -2

For

Re=42600.05 , jf=4*10

  u

2 s /2)

∆Ps=8*jf* (L/LB)*(Ds/de)*( -2

∆Ps=8*4*10

2

*(4.88/0.140063)*(700.3138/14.201)*(1600*(0.937) /2)

∆Ps=386539 Pa =386.539 kpa

Since the pressure drop of shell is too high it can be reduced by Increasing the baffle pitch, tripling the pitch reducing the shell side Velocity, which reduces the pressure pressure drop by a factor of approximately 2

(1/3)

∆Ps=386.539/9=42.94878 kpa

This will reduce the shell side heat transfer coefficient by a factor of 0.8

(1/3)

0.8

ho=5990.226 *(1/3)

2

=2487.403 W/m

 2

This will give overall coefficient coefficient of 511 W/m 2

value of 500 W/m

. 59

 still above the assumed

L=4.88 m

Ds =0.7003 m

Figure 4.2 carbamate condenser summery

4.3 Low Pressure Decomposer 4.3.1 Introduction The purpose of this section is to further stripping of NH 3, CO2 and the left Carbamate . it is divided into 2 parts: top separator, where the released Flash gases are condensed and recycled to the reactor before the Solution enters the tube bundle, where the last carbamate is Decomposed and required heat is supplied by means of steam saturated At 4.5 atm.

4.3.2 Design Calculation 3

Density of urea=1230 kg/m

3

Density of biuret=1467 kg/m

3

Density of water= 1000 kg/m Density of liquid(

)=(0.7611*1230+0.00776*1467+0.231*1000)

3

=1178.696 kg/m

Density of NH3(gas)= P*Mwt/R*T=(4.5*17)/(0.082*353)=2.64 kg/m3 3

Density of CO2(gas)= P*Mwt/R*T=(4.5*44)/(0.082*353)=6.8 kg/m 60

Density of carbamate= P*Mwt/R*T=4.5*78/(0.082*353)=12.126 kg/m3

 )=2.64+6.8+12.126=21.526 kg/m

3

Density of gases(

Mass flow rate of gases=4531.755 kg/hr

 )= 4531.755/21.526

Qv = mass flow rate of gases/ density of gases( 3

3

=75.854 m /hr=0.0210 m /sec Mass flow rate of liquid=89409.39 liquid=89409.39 kg/hr 3

3

QL=89409.39 /1178.696 =210.524 m /hr=0.0584 m /sec

 )/ )

Ut= 0.07*( −

1/2

1/2

=0.07*(1178.696−21.526/21.526) =0.513m/sec

Without demister Uv=0.15 ut , Uv=0.0769 m/sec



Dv=(4*Qv/ * Uv)=0.983 m 2

A=Qv/uv= 0.0210/0.0769=0.7596 m Setting time=10 min=600 sec

Volume held in the vessel=600* 0.0584=35.087 m Liquid height (HL)=V/A=35.087/ 0.7596=16.643 m Feed height (Hf)=0.5* Dv=0.491 m Vapor or gas height(Hv)= Dv=0.983 m Total height=0.983 + 0.491+16.643=18.8711 m

Wall thickness of vessel T=P*Di/2*J*f −P Where P: design pressure D: diameter of vessel J: welding joint factor

61

3

F: design stress Operating pressure(P)=4.5 atm D=938 mm 2

Design pressure(Pd)=1.2*P=5.472 bar=0.5472 N/mm 2

F=125 N/mm

T=( 0.5472 *983)/(2*125*1−0.5472) =2.157 mm

Dv=0.983 m H=18.87 m

T=0.0215 m Figure 4.3 low pressure decomposer design summery

4.4 Evaporator Design 5.2.1 Evaporation And Its Mechanism The objective of evaporation is to concentrate a solution containing the Desired product or to recover the solvent. Sometimes both may be Accomplished. Evaporator design consists of three principal elements Heat transfer, vapor-liquid separation and efficient utilization of energy In most cases the solvent is water , heat is supplied by condensing steam And the heat is transferred by indirect heat transfer across metallic 62

Surface. For evaporation to be efficient, the selected and used must be Able to accomplish several things. 1 transfer large amounts of heat to the solution with minimum amount of metallic surface area. This requires, determines the type, size, and cost of the evaporation system. 2 achieve the specified separation of liquid and vapor and to do it with the simplest devices available. 3 Make efficient use of the available energy. 4 Meet the condition imposed by the liquid being evaporated or by the solution being concentrated. Factors that must be considered include product quality, salting and scaling, corrosion, foaming , product degradation, holdup, holdup, and the need for special type of construction.

4.4.2 Types Of Evaporators Evaporators are varying in the mechanical specifications and operating parameters , this result in products with different characteristics. The major types of evaporator are. 

Horizontal shell-side



Short tube-vertical



Basket type



Long-tube vertical



Climbing film



Falling film 63



Horizontal tube-side



Plate type

4.4.3 Evaporator Selection The selection of the most suitable evaporator type for a particular application will depend on the following factors. 

The throughput required



The viscosity of the feed and the increase in viscosity during evaporation.



The nature of the product required; solid, slurry or concentrated solution.



The heat sensitivity of the product.



Whether the materials are fouling or non-fouling



Whether the solution is likely foam



Whether direct heating can be used.

The selected type for this project is climbing film long tube vertical Evaporator.

4.4.4 Design Calculation

Vapor space pressure=0.23 atm Vapor space temperature=63.1 BPR=21.9





Energy balance st

For product stream coming out of 1 evaporator Material

Cp(KJ/kmole)

flow rate(kmole/hr)

64

mole fraction(x)

Urea

2443.41

1134.259

0.750452

Water

2002.149

370.4335

0.245087

6.742179

0.004461

Biuret

8940

Cp of mixture=∑Xi*Cpi mixture=∑Xi*Cpi =(2443.41*0.750452+2002.149 =(2443.41*0.750452+2002.14 9 * 0.245087+ 8940* 0.004461) =2364.243 KJ/kmole M*cp M*cp ∆T= 2364.243*1511.435 =3573399 KJ/hr st

1  Evaporator Heat input(by feed)+Heat input by steam=Heat carried by water vapor + Energy of bottom product Heat input by feed+ S1



s1=

E1HE1 +energy of bottom product

4684597+ 4684597+ S1 * 2123.8= 2123.8= 13991.58*2614.97+ 13991.58*2614.97+3573399 S1= 16687.25 kg/hr

Economy = 13991.58/16687.25=0.8384 nd

2  Evaporator Heat input(by feed)+Heat input by steam=Heat carried by water vapor +Energy of bottom product 3573399+ 3573399+S2 * 2123.8= 2123.8=5894.424*2545.7+2464.393 *1183.967 S2= 6756.660 kg/hr

Area of evaporator ∆T1 = (∆T)app – BPR – BPR =(147.165− =(147.165−63.1)− 63.1)−21.9= 62.165 U1=2214.537 W/m2 k st

Area of 1  evaporator A1=S1



s1/U1∆T1

65



A1 = 16704 * 2123.2 / 2214.537* (62.165+273)=47.78 ( 62.165+273)=47.78 m2 nd

Area of 2  evaporator U2= 738 W/m2 k ∆T2 = (∆T)app – BPR – BPR = (147.165 – (147.165 – 23.77)  23.77) –  – 3.48  3.48 = 119.915



A2 = S2 λs2 / U2∆T2 2

A2 =6756.660 * 2123.2 / 738 *( 119.915+273)= 49.472 m

[Ref : values of U1 & U2 from Perry's handbook, 10-35] Number of tubes Assume Length = 6 m Tube OD =25.4 mm Tube ID =21.1836 mm For a triangular pitch(Pt)=1.25 do

 = *25.4*10

Area of one tube =  do L

-3

2

*6=0.47853 m

st

Tubes of 1 evaporator Nt1= A1/A Nt1=47.78/0.47853=100 Tubes of 2nd evaporator Nt2=A2/A Nt2=49.472/0.47853=103

Shell diameter For a triangular pitch , 1 shell−2tube pass K= 0.249 , n=2.207 st

Shell diameter of 1 evaporator 66

Db =do(Nt/k)1/n 1/2.207

Db=25.4(100/0.249)

=384.26

C=55 mm , Ds=Db+ C Ds=439.26 mm nd

Shell diameter of 2 evaporator 1/n

Db =do(Nt/k)

Db=25.4(103/0.249)1/2.207=389.19 mm C=52 mm , Ds=Db+ C Ds=441.19 mm

Wall thickness Material of construction: Mild steel 2

F: 135 N/mm Di=21.16 mm

Ps=4.5 atm=4.413 bar Pd=1.1*ps 2

Pd=0.4854 N/mm C= 3 mm J=1 Tt= P*Di/2f*J−P Where

Tt : thickness of tube 2

P: design design pressure pressure ( N/mm N/mm ) Di : inside diameter (mm) 2

F: allowable stress ( N/mm ) J: joint factor C: corrosion allowance (mm) Tt =(0.4854*21.16)/(2*135*1−0.4854)+3=3.038 1 mm

Drums diameters and height 3

Density of urea=1230 kg/m

67

Density of water=1000 kg/m3 3

Density of biuret=1467 kg/m st

Diameter for 1  drum Density of liquid( kg/m3

)=(0.7611*1230+0.231*1000+0.00776*1000)=1178.53 )= P*Mwt/R*T

Density of water vapor(

3

=(0.23*18)/(0.082*336.1) =0.15 kg/m

Volumetric flow rate of water vapor (Qv)=13991.58/0.15=93277.2 m3/hr 3

Volumetric flow rate of liquid(QL)= 75418.7/1178.53=63.993 m /hr

−/)

1/2

Ut =0.07 *(

1/2

= 0.07*(1178.53−0.15/0.15) =6.204 m/s

Ut=Uv with demister 2

A= Qv/ Uv = 25.901/6.204 = 4.176 m



1/2

Dv =(4* Qv/   *uv)   =2.3 m Dv=Hv Dv: minimum vessel diameter Hv: gas or vapor height Hf =0.5*Dv= 1.15 m Hf : feed height Assume settling time= 10 min= 600 sec 3

Volume held in drum(V)= 0.01777*600=10.665 m HL= V/A=10.665/4.176 = 2.55 m HL: liquid height Total height(H)= 2.55+1.15+2.3 =6 m nd

For 2  drum Density of liquid( 3

)=(0.9023*1230+0.0488*1000+0.0488*1467)=1230.34

kg/m

68

Density of water vapor (

)= P*Mwt/R*T = (0.03*18)/(0.082*296.77)

3

=0.022 kg/m

3

Volumetric flow rate of vapor(Qv)= 6233.488/0.022=283340.4 m /hr 3

Volumetric flow rate of liquid(QL)= 89409.39/1230.34=72.670 m /hr

−/)

1/2

Ut =0.07 *(

=16.55 m/s

Ut=Uv 2

A= Qv/ Uv=4.574 m



1/2

Dv =(4* Qv/   *uv)   =3.028 m Hv=Dv=3.028 m Hf =0.5*Dv =1.514 m 3

Volume held in drum(V)=600*0.02018=12.111 drum(V)=600*0.02018=12.111 m HL=V/A= 12.111/4.574=2.547 m Total height =2.547+1.514+3.028=7.089 m

Drum thickness st

1  drum Drum operating pressure=0.23 atm Design pressure = 0.25 bar Thickness(t) =15 mm (assumed) Atmospheric pressure=1 bar Pa=1/0.23=4.167 Do=Di + 2*t 2*t , Do=2.3+2*0.015=2.33 m L / Do=6/2.33=2.575 m

69

Do/t=155.3 Factor B=10878 Pa =B/(14.22*( Do/t)) =10878/(14.22*155.3)=4.92 bar Pa is greater than design pressure , so the assumed thickness is acceptable. nd

2 drum Operating pressure=0.03 atm Design pressure= 0.032 Thickness(t)=20 mm Pa =1/0.03=33.3 bar Do =Di+2*t , Do=3.028+2*0.02=3.068 m L /Do=7.089/3.068=2.31 Do /t= 3.068/0.02=153.4 Pa =B/(14.22*( Do/t)) =10878/(14.22*153.4)=4.98 bar Pa is greater than design pressure The assumed thickness accepted

70

Ts=0.00303 m

6m

7.089

Product

-E

Figure4.4 vacuum evaporator design summery

4.5 Prilling Tower 4.5.1 Introduction Prilling is the process of spray crystallization. crystallization. A liquid liquid is sprayed to Produce drops falling through a cooling medium and crystallizing into Particles. The urea granulation process consists of following three sections: 

Granulation section



Recycle and product cooling section



Dust removal and recovery section

Aqueous urea solution from urea plant is fed to the granulator to Enlarge recycle particles in the granulator. In the granulator , the Granules are dried and cooled simultaneously. The granulator is

 and at slightly negative pressure. Enlarged Urea particles are cooled to about 90 in the after-cooler inside

Operated at 110 115

The granulator to be transported to the recycle section. The

71

Discharged granules are separated into three sizes, product, small and Large size by the screen. Product size granules are further cooled Below 60

 in the product cooler to be sent to the urea storage or

Bagging facility. Large size granules are crushed by the crusher. The Crushed particles and smaller size particles from the screen are Recycled to the granulator as seed. Urea dust contained in the exhaust Air from the granulator and the product cooler is scrubbed in the dust Scrubber by contacting counter currently with aqueous urea solution 3

Urea dust content in the exit air of the bag filter is 30 mg/m . Or less Urea recovered in the bag filter, approximately 2.5-3.5% of production Rate , is recycled to the urea granulator. The process of crystallization And cooling takes a number of seconds. Congealing towers can have a Maximum free-fall height of 60 m, while the tower diameter may be About 15 m.

4.5.2 Material of construction Prilling tower are usually constructed in concrete(fertilizer), steel, Stainless steel.

4.5.3 Design Calculation Urea physical properties: To=132.6

(To= melting point)

Hf =224457 J/kg

(Hf = heat of fusion)

=1230 kg/m =1335 kg/m

 (= Solid density)

3

( = melt density)

3

Cp (l)=2098 J/kg. k

(melt specific heat) 72

Cp (s)=1748 J/kg. k

(solid specific heat)

K(l)=0.83 w/m. k

(melt thermal conductivity)

K(s)=1.19 w/m. k

( solid thermal conductivity)



-3

Viscosity(  )= 2.16*10  Pa.sec

Air physical properties

 )=1.168 kg/m

3

Density(

-6

Viscosity =18.48*10  Pa.sec



Cp air=1.009 KJ/kg .

Heat carried away by air=heat input−heat output = 2917759− 2917759−230076.4 =2687683 KJ/hr (m*cp*∆ (m*cp*∆T)dry air=2687683 KJ/hr Cp air=1.009 KJ/kg



, ∆T=26−20=6

,

M=2687683/(1.009*6)= 443951.6 kg/hr

Tower diameter Volumetric flow rate=380095.548 m3/hr= 105.5821 m3/sec Assuming an air superficial velocity=1.2 m/sec 2

A=Q/u = 105.5821/1.2=87.985 m

D=(87.985 *4/3.14)^0.5=10.586 m

Absolute velocity -3

Dp=1.5*10  m Assume ur=6.3 m/s

(Dp: particle diameter) (ur:Relative velocity)

/

Re = dp*ur*

73

Re =1.5*10-3*6.3*1.168/18.48*10-6 =597 Cw=0.65

(Cw: resistance coefficient(falling coefficient(falling particle))

/6 *dp *(−)g=cw*/4*dp  *1/2**ut LHS= /6*(1.5*10 ) *(1335−1.168)=2.31112*10 RHS= 0.65* /4*1/2*1.168*6.3 =2.66109*10 3

2

-3 3

-5

-5

LHS=RHS , the assumption ur=6.3 m/sec is correct Ua=ur−superficial velocity velocity from new diameter diameter (ua: absolute velocity) Ua=6.3−1.156=5.144 m/s

Heat transfer coefficient

*cp/k

-6

Re =597 ,Pr=

0.5

,Pr=18.48*10 *1009/0.02606=0.7155

0.33

Nu =2+0.552 Re Pr

  =14.07952 2

Nu=ho*dp/k , ho=244.6082 W/m k

Solidification time



Average air temperature=(20+26/2)=23 temperature=(20+26/2)=23 Ph=Hf + cp(l)*(Tf −To)/cp(s)*(To−Tc)

 T =132.6  T =23 Tf =135

(ph= phase transfer number)

(Tf : melt temperature) (To: melting point)

o

(Tc: average air temperature)

c

Ph=224457+2098*(135−132.6)/1748*(132.6−23)=1.198 Bi=ho*dp/2*k(s)

, Bi=0.154164

Fo =ph(1/6+1/3*Bi)=2.7922

(Bi: Biot number) (Fo: Fourier number)

 =1.19/1748*1335=5.099*10

-7

A=k(s)*/cp(s)*

-3

-7

(A: thermal diffusivity)

Ts1=dp*Fo/4*A =1.5*10 *3.132/4*5.099*10 =3.08 sec (ts1= time for solidification)

/,min=1.15 74

Corrected solidification time Ts2=(

/,min)* T

s1=1.15*3.08=3.542

sec

Tower height for solidification=3.542*5.1 solidification=3.542*5.144=18.220 44=18.220 m

Prill cooling time -3

-4

R=dp/2 =1.5*10 /2=7.5*10 m

(R: sphere radius)

1/ho2 R =1/7.5*10-4*(244.6082)2=7267.859 k/w 2

-4

-4

-4 2

2(R−R/2)/k(s)*R = 2*(7.5*10 −7.5*10 /2)/1.19*(7.5*10 ) =1120.448 2

1/kc R =8388.307+211.93=7267.859+211.93=8388.307 2

Kc=211.93 W/m  k

*dp*cp(s)/6*kc ln(T −T /Ts−T )

Cooling time tc=

o

c

c

-3

Tc =1335*1.5*10 *1748/6*211.93 *ln(132.6−23/60−23)=1.494 sec Prill cooling height=1.494*5.144=7.688 m Total height=7.688+18.22=25.908 m.

D=10.586 m H=25.908 m

Figure 4.5 prilling tower

75

Chapter 5

Process Control 76

5.1 Introduction Control may be defined as a set of organized actions directed towards Achieving or maintaining a specific goal and it is one of the most Important factors in improving process performance , which are Equipment design, operating conditions and process control. Control Action involves dynamic(active) command, regulation and co-ordination Of the systems so as to fulfill the prescribed objective of the system in the Most effective and efficient manner.

Equipment Design

Operation Conditions

Safe and profitable Plant operation

Process Control Figure 5.1 schematic representation of the three critical elements for achieving

5.2 Control Objective The primary objective of the designer when specifying instrumentation And control schems are:

1. Safety: The process control strategies contribute to the overall plant safety by maintaining key key variables near their desired values.

2. Environmental protection: Control can contribute to the proper operation of units which is 77

Deal with toxic components resulting in consistently low effluent Concentration, in addition control systems can direct effluent to Containment vessels when any extreme disturbance occur.

3. Equipment protection: Operating conditions must be maintained within bounds to present damage.

4. Smooth operation and production rate: Key variables in streams leaving the process should be maintained Close to their desired values to present disturbance to downstream unit or maintain desired production rate.

5. Production quality : Product quality which is needed may be expressed as composition, Physical properties, performance properties or combination of all Three. Process control contributes to good plant operation by Maintaining the operating conditions required for excellent prduct Quality.

6. Monitoring and diagnosis: The plant operators require very rapid information so that they can Ensure that the plant conditions remain within acceptable bounds.

7. Profit: The equivalent goal is to provide the product at lowest cost. Before Achieving the profit oriented goal, selected independent variables Are adjusted to satisfy the first five higher priority control objectives.

5.3 Types Of Control Systems There are many types of control systems, the use of any system depend on the process which we want to control it and the variables that effect on the control process, some of these control system are: 

Feed forward control system



Feedback control system 78



Cascade control system



Ratio control system



Inferential control system



Level and inventory control system



Predictive control system

5.4 Equipment Control 5.4.1 Reactor control

products

NH3 Feed

FR Flow rate Set point

PIT

PIC

PI

CO2 Feed

PI

PIT

Flow rate Set point

PIC

Carbamate Feed

Figure 5.1 reactor process control

1. Reactor Pressure Table 5.1 Element of control loop for reactor pressure Process Controller

Reactor PID Controller

Controlled variable

Pressure

Measuring element

Manometer

Regulating element

Valve (pneumatic)

Manipulating element Load variables

79

Stream of CO2, carbamate Reactor temperature,feed temperature temperature and composition

2. Reactor Flow Ratio

Table 5.2 Element of control loop for reactor ratio Process

Reactor

Controller

PID Controller

Controlled variable

Flow ratio

Measuring element

Orifice

Regulating element

Valve

Manipulating element

Stream of CO2, NH3

Load variables

Feed temperature -vacuum pressure

(PID) controller transfer function

 )

G(s)=Kc*(1+1/ *s+ Where

Kc: proportional gain of the controller

: integral time constant (min) : derivative time constant (min) Kc=3

= 2 min =0.5 min Control valve transfer function



Gv(s)= Kv/ v*s+1 Where Kv: the steady state gain of valve

v: time constant of valve 80

5.4.2 Carbamate Condenser Control Objective: control of inlet water to the condenser

Carbamate in

Water out Water in

Condensate TI

TIC

TIT Figure 5.2 carbamate condenser process control

Table 5.3 Element of control loop for carbamate condenser condenser temperature

Process

Carbamate condenser PID Controller

Controller Controlled variable

Temperature of outlet water

Measuring element

Orifice

Regulating element

Valve

Manipulating element Load variables

Steam flow rate Pressure and temperature of the feed

(PID)Controller transfer function

 )

G(s)=Kc*(1+1/ *s+ Where

Kc: proportional gain of the controller

: integral time constant (min) : derivative time constant (min) 81

Kc=2.5

= 2 min =0.5 min Control valve transfer function



Gv(s)= Kv/ v*s+1 Where Kv: the steady state gain of valve

v: time constant of valve 5.4.3 Low Pressure Decomposer Control

Off gases PI PIC

PIT

PIT

PI

PIC Feed

PG

FIT FIC

FI Product

Figure 5.3 Low pressure decomposer process control

82

1. Low pressure decomposer decomposer pressure Table 5.4 Element of control loop loop for low pressure decomposer decomposer pressure

Process

Low pressure decomposer

Controller

PID Controller

Controlled variable

Pressure

Measuring element

Manometer

Regulating element

Valve Stream of feed, off-gases flow rate

Manipulating element

Pressure and temperature of streams

Load variables

3. Low Pressure Decomposer Flow Rate Table 5.5 Element of control loop for low pressure decomposer decomposer flow rate

Process

Low pressure decomposer

Controller

PID Controller

Controlled variable

Flow rate

Measuring element

orifice

Regulating element Manipulating element Load variables

(PID)Controller transfer function

 )

G(s)=Kc*(1+1/ *s+ Where

Kc: proportional gain of the controller

: integral time constant (min) : derivative time constant (min) Kc=1 83

Valve Stream of outlet product Pressure and temperature of streams

= 1 min =0.5 min Control valve transfer function



Gv(s)= Kv/ v*s+1 Where Kv: the steady state gain of valve

v: time constant of valve 5.4.4 Evaporator Process Control

PIC

PIT PI

Vapor

feed

TI

TIT

TIC

Steam

FIT

FIC

FE

Product

1.Vacuum evaporator pressure

84

Table 5.6 Element of control loop loop for vacuum evaporator pressure

Process

Vacuum evaporator PID Controller

Controller Controlled variable

Pressure

Measuring element

Manometer

Regulating element

Valve

Manipulating element

Stream of outlet vapor

Load variables

Pressure and temperature of steam and vacuum pressure

2.Vacuum Evaporator Flow Rate

Table 5.7 Element of control loop loop for vacuum evaporator evaporator flow rate

Process

Vacuum evaporator PID Controller

Controller Controlled variable

Flow rate of urea solution

Measuring element

orifice

Regulating element

Valve

Manipulating element

Stream of the product Pressure and temperature of steam and vacuum composition

Load variables

3. Vacuum Evaporator Temperature

85

Table 5.6 Element of control loop for vacuum evaporator evaporator temperature

Process Controller

Vacuum evaporator PID Controller

Controlled variable

Temperature

Measuring element

Thermocouple

Regulating element Manipulating element Load variables

(PID)Controller transfer function

 )

G(s)=Kc*(1+1/ *s+ Where

Kc: proportional gain of the controller

: integral time constant (min) : derivative time constant (min) Kc=1

= 1 min =0.5 min Control valve transfer function



Gv(s)= Kv/ v*s+1 Where Kv: the steady state gain of valve 86

Valve Steam flow rate Feed temperature -vacuum pressure

v: time constant of valve

5.4.5 Prilling Tower Process Control

Air out

TI

TI T

TIC

FI

PG

feed Air in FI

product Figure 5.5 prilling tower process control

1. Prilling Tower Temperature

87

Table 5.7 Element of control loop for prilling tower temperature temperature

Process

Prilling tower PID Controller

Controller

Temperature of outlet air

Controlled variable Measuring element

Thermocouple

Regulating element

Valve

Manipulating element Load variables

flow rate of inlet air Feed temperature -vacuum pressure

(PID)Controller transfer function

 )

G(s)=Kc*(1+1/ *s+ Where

Kc: proportional gain of the controller

: integral time constant (min) : derivative time constant (min) Kc=1.5

= 2 min =1 min Control valve transfer function



Gv(s)= Kv/ v*s+1 Where Kv: the steady state gain of valve

v: time constant of valve 88

Chapter 6 Economic Cost 89

6.1 Introduction Economical evaluation is a major component of chemical plant design That decides whether the design is economically feasible since projects Are built to make a profit. Also chemical engineers are concerned with Cost as well as design.

6.2 Types Of Costs Involved In Manufacturing Process

1. Total capital investment a. Fixed capital investment (manufacturing and nonmanufacturing) b. Working capital 2. Operating costs (total production cost) a. Direct expenses: variable and fixed charges b. Indirect expenses

6.3 Cost Calculation

90

Number of  equipment 1

Cost of equipment ($) 61854.53

1

12490.32

12490.32

Low pressure  decomposer

1

44210.53

44210.53

Vacuum evaporator

2

148830.5

297661

CO2 compressor

1

166672.2

166672.2

Prilling tower

1

545637.3

545637.3

Ammonia pump

1

6302.65

6302.65

Conveying system

4

393.92

Stripper

1

120684.9

120684.9

Pumps

2

6922..384

13844.768

Equipment Autoclave Reactor Medium pressure  decomposer

Total

Total cost 61854.53

1575.68

1226723.348

Figure 6.1 equipment cost Total capital investment 1. fixed capital investment a. direct cost

91

Item

% purchased equipment cost

Total cost

Equipment cost

100%

1226723.348

Installation

25%

306680.835

Instrumentation And control

10%

122672.334

Piping

25%

306680.835

Electrification

20%

Building Service facilities

Land Requisition

245344.668

30%

368017.002

40%

490689.336

4%

5068.933

3071877.28

Total

b. indirect cost

Item Engineering & supervision

% purchased Equipment cost 10%

Total cost 122672.334

Construction

10%

122672.334

Contactor

5%

61336.167

contigency

5%

61336.167 368017.002

Total

Fixed capital investment(FCI) investment(FCI) =DC+IC =3071877.28+368017.002=3439894 92

b) Working capital investment(WCI) investment(WCI) =15% of FCI= 0.15*3439894=515984

Total capital investment =FCI+WCI =3439894+515984=3955878 c) Maintenance& Repair cost =4% of FCI=0.04*3439894=137 FCI=0.04*3439894=137595 595

Fixed charges 1. Deprecation =10% of FCI =0.1*3439894=343989.4

2. Local taxes =1.5% of FCI =0.015*3439894=51598.41

3. Insurance =1% of FCI =0.01*3439894=34398.94 Fixed charges=34398.94+51598 charges=34398.94+51598.41+343989.4=429986.75 .41+343989.4=429986.75

93

Chapter 7 Safety Aspects

94

7.1 Introduction Safety is the state of being ''safe'', the condition of being protected Against physical , social, financial , emotional ,physiological , educational Or other consequences of failure, damage, error, accidents, harm or any Other events which can be non-desirable. This can take the form of being Protected from the event or from exposure to something that causes Health or economic losses. No industry can afford to neglect the Fundamentals of safety in design and operation of its plant and Machinery. It is important that all the people responsible for management and operation of any industry should have a good knowledge of industrial safety.

Safety Safe use of man, material or machine machine by safe system method of work is to achieve zero accidents which result in higher productivity.

Accidents An accidents is un planned or un expected events which interfere or interrupts the planned process of work work and results in personal injury

Accident factors 1 a personal accident accident injury occurs as a result result of accident 2 an accident due to un safe act and/or unsafe condition 3 unsafe act/unsafe condition condition exists due to faults faults of persons 4 faults of persons due to negligence Thus, if we can remove fault of a person we can prevent 98% of accidents. 95

7.2 Principal Of Protection And Prevention: Industrial accidents are caused by negligence of employer, the worker or both employers ''efforts to reduce the accidents are generally motivated By four considerations": considerations": 1 to lessen human suffering 2 to prevent damage to plant and machinery 3 to reduce the amount of time lost as a result 4 to hold the expenses of workman's compensation compensation to minimum The basic reasons for preventing industrial accidents are human and economic . the most important of these should be to avoid human suffering. Accidents are economic losses and this is a challenging Reason for accident prevention.

7.3 Safety Precaution 1. when taking sample of anhydrous ammonia and when operating or working on ammonia ammonia valves, equipment containing ammonia such as ammonia feed pumps, operators laboratory and maintenance personal must wear safety overalls. 2. goggles and rubber gloves. If any part of the skin has been exposed to ammonia , wash wash immediately immediately and thoroughly with water 3. work on the ammonia equipment should be done from the upwind side of the equipment to avoid or minimize contact with escaping ammonia. 4. the location of fire hydrants, safety showers , eyewash fountains ammonia canisters gas mask, emergency air breathing apparatus should be well known to all person's 5. instruments containing mercury must not be used if ammonia is likely to come in contact with mercury

96

6. heavy leakage leakage of ammonia ammonia can be dealt by spraying large quantity of water with spray nozzles.

7.4 List of safety equipment A. respiratory protective equipment 1 self —contained breathing apparatus sets of 30 min and 10 min 2 continuous airline mask 3 trolley mounted self contained breathing apparatus set 2.5 hours 4 canister gas mask 5 dust mask/cloth mask (air purifying respirator)

B. non—respiratory protective equipments 1 helmets 2 ear muff and ear plugs 3 goggles 4 face shield 5 hand gloves 6 aprons 7 safety shoes 8 suits 9 safety harness

C. warning instrument 1 oxygen, carbon dioxide , chlorine, ammonia indicator with replaceable sensors. 2 explosive meters for measuring explosive range 3 fire fly instrument for confined space entry

D. Gas Leak Instruments 97

1.safety showers 2. manual water sprinklers 3. communication systems

7.5 Fire Hazards The general types of fire are encountered in the process plants. One involves common combustible material such as wood, rags, paper ,etc (class A fires),the next flammable liquids and gasses such as lubrications oil and solvents, ammonia vapors etc. (class B fires) and the third involve Electrical equipment (class C fires) In general three things required to make a fire 1. Something which will burn eg. A combustible material 2.Oxygen— 2.Oxygen—air 3. A source of ignition or existence of a temperature at or above which a material will start burning spontaneously.

98

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rd

Shreve RN . chemical process industries, 3 edition. New York: McGraw hill book company, 1967



Othmer Kirk, encyclopedia of chemical technology , vol. 21. New York : John Wiley & Sons, 2004



th

Perry RH. Chemical Engineering Handbook, 6

edition. New York :

McGraw Hill Book Co, 1984 

WWW.basf.Com



WWW.wikipedia.COM



Kern DQ. Process Heat Transfer. New Delhi : McGraw Hill Companies , 2004



Urea manufacture processes ''Ullmann's Encyclopedia of Industrial Chemistry'', 5th Edition, Volume A27.



th

R.K.Sinnot. Chemical Chemical Engineering Design, Design, vol. 6, 4  edition. Coulson & Richardson's, Elsevier,2005.



Williad . Baasel. Preliminary chemical engineering plant design

99

 

100