Ammonia Production(Project)

1200 TPD Ammonia production By Vikrant Rana(DCET)

Introduction Ammonia is an important nitrogenous material used as fertilizer. Most of ammonia is made synthetically.It is also obtained as by-product in some cases. Ammonia gas is used directly as fertilizer in heat treatment, paper pulping, nitric acid and nitrate manufacturing, nitric acid esters and nitrocompound manufacture, explosive of various types and as a refrigerant. The most important field is of fertilizer. Importance of ammonia in this field is due to the fact that it is by far the most simple form of actual nitrogen which can be administrated and the most economical source of other nitrogen chemicals. Hence ammonia is the backbone of Nitrogen fertilizer processes. Its production is the first consideration in the initiation of a fertilizer industry. Great strides have been made in the last few year in the economy and efficiency of ammonia processes. The synthesis of ammonia from Nitrogen and hydrogen was first chemical reaction to be carried out under high pressure on commercial scale. This synthesis reaction has probably been studied more extensively and is better understood than any other high presure reactions. The name of Haber, Corl Bosch, Claude, Casale, Fauser, Mont Cenis stand out from those of early engineers and scientists, worked for the development of first commercial synthesis. In India, till 1948, the only factory which was producing synthetic ammonia was small concern, Mysore Chemicals and Fertilizers Company

2

Ltd. at Belegula, Mysore. The Fertilizer Factory at Sindri, India’s first major state-owned enterprise, went into production on October 30, 1961. a gradual expansion followed and soon fertilizer plants were commissioned at Nangal (1961), Trombay (1968), Namrup (1969), Durgapur (1974) and Barauni (1976).

In 7 t h decade of 20 t h century,a revolution in ammonia manufacturing came about as the reciprocating compressors common at that time were replaced by centrifugal units. At that time, the ammonia process was regarded as highly efficient, a mature process, but careful examination of the process step by step and the severe economic pressure led to major improvements and cost drop approaching to 50 percent. Some of the important developments are – better catalysts, better converters, replacement of reciprocating compressors to centrifugal units, lone pressure techniques, energy conservation, digital computer control process and the most recent development on which researches have been made and commercial

backgrounds

are

being

studied

is

to

commissioned

a

Compressor Ammonia Plant. High pressure ammonia synthesis becomes economically attractive with this technique of compressorless manufacture process of ammonia. As without application of compressor, high pressure synthesis gases are obtained which have many advantages over low pressure techniques e.g. at high pressure ammonia condenses at normal cooling water temperature, reduced amount of recirculation thus replacing

3

recirculating compressure by a static ejector. Trend in again in favour of high pressure techniques for ammonia manufacture.

Process And Process Selection Ammonnia manufacturing consists of six phases :1.

Manufacture of Reactant gases.

2.

Purification

3.

Compression

4.

Catalytic reaction

5.

Recovery of ammonia formed

6.

Recirculation

RAW MATERIALRaw material requirement consists of a sources of hydrogen, the nitrogen is obtained from air, the feed stock supplies the hydrogen. The production of ammonia depend upon cost and operability of available raw material for hydrogen. Different raw materials for hydrogen can be – Wood Coal

Refinery gas

Electrolysis of water

Naphtha

Coke oven gas

Fuel oil

Natural gas

Crude oil

Liquid petroleum gas

MANUFACTURING PROCEDURESThe principal manufacturing process that are used for synthetic ammonia production are steam water gas process, the steam hydrocarbon process, the coke oven gas process & electrolysis of water.

4

Much research work is still diverted to find cheaper manufacturing methods.Many new plants use ethanol amine to remove carbon dioxide from gas streams instead of high pressure water. Nitrogen required for synthesis gas is obtained usual from an air liquefaction plant. In water gas and natural gas process, nitrogen is obtained in same manufacturing process as hydrogen. Air is added in secondary reforms of natural gas process and reacts with carbon monoxide to furnish carbon dioxide and nitrogen. Cost is greatly influenced by the pressure, temperature, catalysts and raw material used. Raw material selection depends upon the availability and cost of raw material. If the plant is near some refinery then raw material can be naphtha, natural gas or LPG, but in case it is near a coal mine, one should depend on coal as raw material for hydrogen. There may be some other cases where electricity is readily available to meet the requirements

for

electrolysis

then

hydrogen

can

be

obtained

by

electrolysis of water

Comparsion of processes Many variation of the original Haber process for the synthesis of ammonia are now used in commercial practice, some varying to such an extent that they are identified by a name, often that of group of men developing them. Important among these are modified Haber Bosch, Claude, Casale, Fauser, and Mont Cenis processes. All of them are fundamentally the same in that nitrogen is fixed with hydrogen as ammonia in the presence of a catalyst , but have variation in arrangement and construction of equipment , composition of catalyst, temperature and pressure used. Table C gives a condensed comparison of different processes 40% conversion of the gas upon passes through a single converter and 85% conversion after passes through a series of converter . gas is vented after one pass through the converter.

5

Table C : Synthesis Ammonia Systems Designation

Pressure

Temp.

Catalyst

Recir-

Conver.

Doubly

culation Yes

% 8

Yes

20-22

No

40-85

500

iron Promoted

Yes

15-18

500

iron Promoted

Yes

12-23

C 550 0

Haber-Bosch

200-350

Promoted

Modified

200-300

iron 500-550 -Do-

Haber-Bosch Claude

900-1000

500-650 Promoted

Casale

Fauser

600

200

iron

Modified Haber- Bosch Process The ammonia concentration in the circulation gas leaving the catalyst is 10-11 mole %. In the condenser the ammonia concentration is reduced by condensation to equilibrium at the exit temperature of the condenser. The condensed ammonia is separated from the circulating gas, and the gas is boosted in pressure by a circulating compressor to overcome the pressure drop in the synthesis loop.

6

The Claude Process This process depart from the Haber process more than any of the other ammonia syntheses process, the residual gas is wasted to the atmosphere or utilized for its heat content. The large amount of heat evolved in operating at space velocities of 100,000 with as much as 40% of the hydrogen nitrogen mixture converted to ammonia in one pass, called for a special converter design. In his original process, Claude used -+ from the liquefaction of air. Among advantages claimed for the Claude process are the following: 1.

Greater

compactness,

simplicity,

and

case

of

construction of the converter, since under the high pressure used the gases have smaller volume. 2.

Elimination of the expensive heat exchangers required in processes operated at lower pressure.

3.

Removal of ammonia with water cooling alone, rather than by ammonia refrigerators or scrubbing processes.

Cited against these advantages are the shorter life of converters, high apparatus upkeep in the high-pressure operation, and the efficiency loss in wasting approximately 20% of the makeup gas, which is unconverted. Modifications of the Claude process include recycling of the gas through the synthesis converters.

7

The Casale Process Pressure of 500-900 atm are used in this process which is otherwise distinguished by the method used for controlling catalyst temperature in the specially designed converter. A method involving recirculating gas around a synthesis loop, similar to the Haber process, is used. As in the Claude process, the higher pressures allow liquefaction of the ammonia at temperatures that can be attained by water cooling. The basis for heat control of the catalyst is to leave 2 or 3% ammonia in the gas to the converter, thereby showing down the rate of formation of ammonia and eliminating excessive heating of the catalyst.

The Fauser Process This pressure incorporates some features not previously mentioned. Electrolytic hydrogen from Fauser cells and nitrogen from liquid air unit or from a purification unit utilizing tail gases from absorption towers in the ammonia oxidation plant are used. The mixture of hydrogen and nitrogen is compressed to 200-300 atm and, after passing through an oil separator goes to an oxygen burner. In the oxygen burner any oxygen contained in the gas mixture combines with hydrogen in the presence of a copper catalyst the water formed is condensed out in a cooler and removed in a water separator.

The Mont Cenis Process This process was originally developed to use hydrogen separated from coke oven gas by a liquefaction process, and nitrogen was obtained by the liquefaction of the air. The essential characteristics of the Mont Cenis process are its operating pressure of 100 atm or less. Mixed hydrogen and nitrogen, after being compressed to 100 atm and heated to about 3000C in interchangers, passes through a carbon monoxide purifier. In the purifier, while in contact with a nickel catalyst, carbon monoxide and oxygen contained in small quantities in the gas react with hydrogen to form methane and water.

8

9

Material Balance Ammonia Production = 1200 TPD=2941.18 kmol/hr N2 + 3H2

2 NH3

Let % conversion=20% Which means, 1 kmol N2 forms 2x0.2=0.4kmol NH3 Assuming losses=8% Amoonia production=2941.18/0.92=3196.93kmol/hr

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Hence, N2 required=3196.93/0.4=7992.34kmol/hr Air required=7992.34x100/79=10,117kmol/hr=10117x22.4=226621 Nm3/hr Siimilarly, H2 required=3x7992.34=23977kmol/hr Reaction taking place in the gasifier is CnHm + n/2O2 nCO + 2H2O CnHm + nH2O nCO + (m/2+n)H2O At high temperature, CnHm + ( n + m/4)O2

nCO2 + m/2H2O

Composition is given in weight percent. C

= 84.98%

H

= 12.07%

S

= 1%

N

= 0.4%

O

= 1.5%

Composition of gases coming out of the gasifier is given asCO

= 42%

H2

= 51.23%

H2S

= 0.23%

CO2

= 5.7%

N2

= 0.18%

Ar

= 0.02%

CH4

= 0.5%

Material balance across H2S absorber

11

Basis: 100 kmoles or Raw gas CO H2

= 42 kmole = 51.23 kmole

H2S

= 0.23 kmole

CO2

= 5.7 kmole

N2

= 0.18 kmole

Ar

= 0.09 kmole

Now, maximum concentration of H2S in H2S free raw gas in limited to 0.3 ppm. Amount of H2S to be removed = 0.229997 kmole In H2S free gas H2 = 51.23 kmole

% of H2

= 51.23/99.77x100 = 51.34

CO = 42 kmole

% of CO =

H2S = 0.3 ppm

% of H2S = 0.3 ppm

42/99.31 X 100 = 42.09

CO2 = 5.77 kmole

% of CO2 = 5.77/99.31x 100 = 5.78

CH4 = 0.5 kmole

% of CH4 = 0.5/ 99.31x 100 = 0.501

N2 = 0.18 kmole

% of N2 = 0.18/99.31x 100 = 0.18

Ar = 0.09 kmole

% of Ar = 0.09/99.31x 100 = 0.09

Total no. of moles = 51.23 + 42.09 + 0.00003 + 5.77 + 0.5 +0.18 + 0.09 = 99.77 kmol Material balance across CO shift converter Basis 100 kmole of H2S free gases Now since 97% of CO is converted in converter Therefore kmoles of CO2 in product = 5.78 + 42.09 x 0.97 = 5.78 + 40.8 = 46.6 kmoles

12

similarly, kmoles of H2 in the Product = 51.34 + 40.8 = 92.14 Kmole of CO = 42.09 – 40.4 = 1.26 Kmole of H2S = 0.3ppm Kmole of CH4 = 0.501 kmole Kmole of Ar = 0.09 kmole Kmole of N2 = 0.18 kmole Total no of moles in the product stream = ( 46.4 + 92.14 + 1,26 + 0.501 + 0.09 + 0.18 ) = 140.77 kmole H2 in the product = 92.14/ 140.77 x 100 = 65.45% CO2 in the product = 46.6/140.77 x 100 = 33.1% H2S = 0.3 ppm CH4 = 0.501/140.77 x 100 = 0.36% Ar = 0.09/140.77 x 100 = 0.063% N2 = 0.18/140.77 x 100 = 0.13% Material balance across CO2 Absorber Basis 100 mole of converted gas Taking efficiency of the absorber to be 92% CO2 to be removed = 0.92 x 33.1 = 30.45 kmole CO2 in the product stream = 2.65 kmole Total no of moles in the product stream

13

= ( 2.65 + 65.45 + 0.89 + 0.13 + 0.06 + 0.36 ) = 69.54 kmole CO2 = 2.65/69.54 X 100 = 3.8 H2 = 65.45/69.45 X 100 = 94.2 CO = 0.89/69.45 X 100 = 1.3 CH4 = 0.36/69.45 X 100 = 0.518 N2 = 0.13/69.45 X 100 = 0.19 Ar = 0.064/69.45 x 100 = 0.09 H2S = 0.3 ppm Material balance across Adsorber The main function of absorber is to adsorb H2S, CO & CO2. They are reduced to an amount which is negligible. Moles remained = 100 – 3.8 – 1.3 = 94.9 kmole % of gas after adsorption H2

= 94.2/95 X 100 = 99%

CH4

= 0.54

Ar

= 0.095

N2

= 0.19/95 = 0.2

N2 is added in such a way that H2 : N2 becomes 3 : 1 Suppose x mole of N2 is added 99 =

( 0.18 + x ) x 3

14

x

=

32.80

Total no of moles = 99 + 32.80 + 0.54 + 0.095 = 132.43 kmole % of N2 in the product = 32.80/132.43 x 100 = 24.76 % % of H2 in the product = 99/132.43 x 100 = 74.76% % of Ar % of CH4

= =

0.072% 0.54/132.43 X 100 = 0.4%

Material balance across NH3 Separator Now in NH3 Separator 3 streams are there out of which 2 are going out and one is making in the stream that moves in F13 & stream that gives main product is F14 & F 15. F16 is recycle stream moving to compressor. Suppose the moles of F 13 stream are as follows N2 = a H2 = 3a NH3 = b Ar = c CH4 = d Now solubility of H2 , N2 , Ar, NH4 in liquid NH3 is given as Solubility of H2 = 0.0998 cm3 of H2 at NTP/gm of liq NH3 Solubility of N2 = 0.1195 cm3 of N2 at NTP/gm of liq NH3 Solubility of Ar = 0.154 cm3 of Ar at NTP/gm of liq NH3

15

Solubility of CH4 = 0.304 cm3 of CH4 at NTP/gm of liq NH3 For F 14 stream H2

= 0.0998/22414 x 17 x 3a x 22 x 298/273 = 5.45 x 10-3 a

N2

= 0.1195/22414 x 17 x a x 22 x 298/273 = 2.18 x 10-3 a

Ar

=

0.154/22414 x 17 x c x 22 x 298/273

=

2.8 x 10-3 c

CH4 = =

0.304/22414 x17 x d x 22 x 298/273 5.54 x 10-3 d

Now we want 1200 TPD production Therefore NH3 production in F 14 = 1200 x 1000/(17x24) = 2941.18 kmole/hr Now for F15 stream Equilibrium of NH3 in gas phase with liq. NH3 in separator = 5.9% Assuming that total inert conc. In stream from the separator = 5% H2 + N2 = 100 – 5.9 – 5 = 89.1% H2

=

N2 =

¾ x 89.1 = 66.83% ¼ x 89.1

= 22.28 %

NH3 = 5.9% Ar + CH4 = 5% Since CH4 is 2.5 times the Ar.

16

Ar = 1.43% CH4 = 3.57% Now applying material balance on NH 3 separator F 13 = F 14 + F 15 H2 balance : 3a = 0.6683 F15 + 5.45 x 10-3 a

(1)

= 0.2228 F 15 + 2.18 x 10-3 a

(2)

N2 balance : a

NH3 balance : b = 2450.98 + 0.059 F 15

(3)

CH4 balance : d = 5.54 x 10-3d + 0.0357 F 15

(4)

: c = 2.8 x 10-3c + 0.0143 F 15

(5)

Ar balance From (5)

0.9945d = 0.0357 F15

(6)

d = 0.0359 F15 From eqn. (1) 2.99478 a = 0.6683 F15 a = 0.223 x F15

(7)

From eqn. (4) 0.9972 c = 0.0143 F 15 c = 0.01434 F 15

(8)

Now total moles in product stream F14 = 2941 + ( 2.18 + 5.45 ) x 10-3 x 0.223 x F15 + 2.8 x 10-3 x 0.0143 F 15 + 5.54 x 10-3 x 0.0359 F15 = 2941 + 1.9404 x 10-3 F15

17

η of NH3 separator = 98.5% Therfore 2941

=

0.985

2941 + 1.9404 x 10-3 F15 2941= 2414.22 + 1.9113 x 10-3 F 15 F 15 =

2302kmol/hr 0.0019113

From (7) a = 5133.7 kmol/hr From (8) c

= 330.12 kmol/hr

From (6) d =

826.45 kmol/hr

& from (3) b

= 4299.24 kmol/hr

Now for F-13 stream H2 = 3 x 5133.7 = 15401.1 kmol/hr N2 =

5133.7 kmol/hr

NH3 = 4299.24 kmol/hr Ar

=

CH4 =

330.12kmol/hr 826.45 kmol/hr

Total no of moles = 25990.61kmol/hr

18

H

=

15401.1

=

59.26%

25990.61 N2

=

19.75%

NH3

=

16.52%

Ar

=

1.28%

CH4

=

3.18%

For F14 stream NH3

=

2941 kmol/hr

H2

=

5.45 x 10-3 a

=

5.45 x 10-3 x 5133.7

=

27.98kmol/hr

N2

=

2.18 x 10-3 x a

=

2.18 x 10-3 x 5133.7

=

11.19 kmol/hr

CH4

Ar

=

5.54 x 10-3 x d

=

5.54 x 10-3 x 826.45

=

4.58 kmol/hr

=

2.8 x 10-3c

=

2.8 x 10-3 x 330.12

=

0.924 kmol/hr

Total no of moles in product =

2985.67 kmol/hr

19

NH3

=

2941

=

98.5%

2985.67 H2

=

27.98

=

0.94%

2905.67 N2

=

11.19

=

0.38%

2985.67 CH4

=

4.58

=

0.15%

2985.67 Ar

=

0.924

=

0.03%

2985.67 Material balance across NH3 converter Suppose in F12 stream N2

=

x kmole

H2

=

y kmole

NH3 = z kmole Assuming 20% conversion For N2 balance X [1 – 0.2] = 5133.7 X = N2 =

6417.125 kmol/hr

H2 = 3 x 6417.125 = NH3 formed

NH3 in F12

19251.4 kmol/hr

=

2 x 0.2 x 6147.125

=

2566.85kmol/hr

= 4299.24 - 2144.48

20

= 1732.4 kmol/hr CH4 in F12

= 826.45 kmol/hr

Ar in F12

= 330.12kmol/hr

Total no of moles in F12 = 28557.5 kmol/hr N2

=

6417.125

X 1000

=

22.47%

28557.5

H2

=

19251.4

X 1000

=

67.42%

28557.5 NH3 =

1732.4

X 1000

=

6.04%

28557.5 CH4 =

826.45

X 1000

=

2.89%

X 1000

=

1.17%

28557.5 Ar

=

330.12 28577.5

Applying material balance across compressor F11 + F 16

=

F 12

F11 + F 16

=

28557.5

Applying H2 balance 0.7476 F11 + 0.6683 F 16 = 0.6742 x 28557.5 0.7476 [285557.5 – F16] + 0.6683 F 16 = 0.6742 X 23855.25 Therefore F16 = 26432.79 kmol/hr

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Recycle = 26432.79 kmol/hr F11 = 28557.5-26432.79 =

2124.79 kmol/hr

Suppose the kmole of feed gases coming from absorber is A A x 0.99 = 2124.71 x 0.7476 A = 1588.43 kmol/hr Amt of N2 Added = 2124.71 x 0.2476 = 526.1kmol/hr Across the absorber applying overall material balance. H balance, F10 x 0.942 kmol/hr F10 = 1422.8025 kmol/hr Applying material balance across CO2 removal F8 x 0.6545 = F10 x 0.942 F8 =

1422.8025 x 0.942

= 2047.79 kmol/hr

0.6545

Applying material balance across shift converter F7 x 0.534 = 2286.17 x 0.6545 F7 =

2914.49 kmol/hr

Applying material balance across H2S removal F5 x 0.5123 = 2914.49 x 0.5134 F5 =

2920.75 kmol/hr

H2S removed = 2920.75 x 0.0023 – 2914.49 x 0.3 x 10-4

22

=

6.63 kmol/hr

For H2S gas W = 0.094, P

TC

=49 bar

Tr =

T

257

= 373.K

PC

=89.63 bar

= -220C = 2510 K =

0.67

=

0.55

373.5 Pr

=

49 39.63

B0

= 0.083 -

O.42 2

=

- 0.59

( 0.67)1.6 kg/m2B

= 0,139 -

0.172

= -0.78

( 0.67)4.2 Z

=

1 -

0.59 x 0.55

= 0.52

0.67 PV

=

Z n RT

49 x V

=

0.082 x 0.55

=

1.297 m3

V

x 251 x 5.94

Solubility of H2S in Methanol =

60 m3 of H2S m3 of methanol

Weight of methanol required

at - 220C & 49 atm

23

=

1.534 60

=

kmol of methanol required

2.56 x 10-2 x 1215

=

=

31.1 kg

31.1

= 0.97 kmol/hr

32 3% by weight of carbon is converted into coke, therefore only 97% of carbon is present in raw gas. Coke formed

= 20556.22 x 0.8489 x 0.03 =

523.5 kmol/hr

Applying mass balance across carbon recovery unit; F3*=F5*- mass of coke formed (* denotes mass per unit time) F3*=2920.75x14.6-523.5x12=36361 kg/hr Applying mass balance across gasifier F1*=F3*-F2*=36361-462.42x32=21564 kg/hr Hence, Fuel Oil requirement=21,564 kg/hr

Energy Balance Data for Specific heats For Gases Cp/R= A+ BT +CT2 +DT2

24

A

103 B

106 C

10-5 D

CH4

1.702

9.081

-2.164

CH 3 OH

2.211

12.216

-3.450

NH3 (g)

3.578

3.020

-0.186

CO

4.376

1.257

- 0.031

CO 2

5.457

1.045

- 1.157

H2

4.249

0.422

0.083

H2S

3.931

1.490

- 0.231

N2

3.280

0.593

0.040

O2

3.639

0.506

- 0.227

SO2

5.699

0.801

- 1.015

1.450

0.121

H2S (vap)

3.470

For liquids Cp/R =A +BT +CT2 A

103 B

106 C

NH3 (l)

22.626

- 100.75

192.77

Methanol

41.653

- 210.32

427.20

8.712

1.25

- 0.18

Water

For Solids Cp/R = A +BT + CT2 A

103 B

10-5 C

Carbon (s)

1.771

0.771

- 8.67

S

4.114

- 1.728

- .78

25

Since, the assumed conversion=97% , from graph Inlet Temperature=3500C Reaction is, CO +H2O

CO2 +H2

(∆ H)298 = (HCO2 + HH2 ) – (HCO +HH2O) = (-393509)- (- 110525 –285830) = 2846 kj/kmole Mole, of CO converted =583.6 kmole/hr (∆ H)298 = 2846x 583.6 = 16.6 x 105 kj/hr (∆ H)R = -2.27 x 107 kj/hr For adiabatic condition, ∆ HR + ∆ HF8 + ∆ H298 = 0 7 ∆ Hp =2.17 x 10 kj/hr 460 (T- 298) + 1.17/2 x 10 –3 (T2 – 2982 ) +0.3 x 105 (1/T – 1/298) = 1954.07 Solving this equation by heat & trail we get

26

T = 622.50 K i.e.t = 349.50C In cooler converted gas is cooled from 349.5 to 42.50C, i.e. from Heat lost by converted gas = 1335.5 x 8.314 {4.6 (622.5-315.5) + 1.17x 10-3/2 (622.52 – 315.52 –6332 ) –0.3x105 (315.5 – 622.5)} =22.9 x 107 kj/hr If m be amount of water required for cooling M 4.18 x 25 =22.9 x 106 Therefore M =2.19 x 107 kg/hr

Taking refernce temp. =298 k for balance around the Converter . Feed to converter is at 3500C ( ∆ H)F12 = 12585.8x 8.314 [4.12(298-923) + 0.8319/2 x

27

10-3(298 2 –623 2) + 0.0565 x 10 5 (1/623 – 1/298)] =

- 15.36 x107 kj/hr

NH3 formed =1167.66 kmole/hr (∆ Hrxn) = -46110 KH/kmole (∆ Hrxn)298 =700.6 x 46110 =- 3.23 x 107 kj/hr If product is heated to a tamp . of T1 then ( ∆ H)F13 = 8.134 x 7145.79 [3.845 (T- 298) + 1.1299 x 10-3 (T2 – 2982) + 0.040 x 105 (1/T – 1/298)] For adiabatic condition, (H)F12 + (∆ H)F13 + (∆ Hrxn)298 = 0 ( ∆ H)F13 =12.45 x107 =3.845(T – 298) + 1.1299 x 10-3 /2 + (T2 – 2982 ) + 0.040 x 105 (1/T-1/298) = 12.45 x 107 /7145.79 x 8.314 = 2095.60 Solving this equation by hit and trial wet dot, T= 773 k Temperature of outlet gas = 500.50C Product gases are cooled from 500.50C to 4100C while heating The feed gases from 2580C to 350 0C

28

29

30

31

32

33

34

35

Specification Sheet for Heat Exchanger design Number required

:

5

Function

:

To cool the gases coming out of shift Converter

Type

:

1- 4 shell and tube with floating head.

Shell side: Number of shell passes : 1 Shell side fluid

: Hot Gases

Shell internal diameter

:

1067 mm

Baffle spacing

:

534 mm

Entering temp. of gases :

132 0C

Leaving temp. of gases :

400C

Tube side: Number of passes

:

4

Number of tubes

:

738

Tube length

:

4.88 m

Tube side fluid

:

Cooling Water

36

Tube pitch

:

31.75 mm

Outside diameter of tube :

25.4 mm

Inside diameter of tube

22.1 mm

:

Entering temperature of Water :

200 C

Leaving temperature of Water

350 C

:

Reactor Design CATALYST A triply promoted ( K2O-CaO-Al2O3) iron oxide catalyst will be used. The iron oxide (Fe2O3-FeO) is in the form of nonstoichiometric magnetite. It is made by fusing the magnetite with the promoters. The catalyst is reduced in situ, and the removal of oxygen yields a highly porous structure of iron with promoters present as interphases between iron crystals and as porous clusters along the pore walls. The pores range from 500A to 1000A and intraparticle diffusion is diffusion to occur by the bulk mechanism. Alumina prevents sintering and corresponding loss of surface area and also bonds the K2O, preventing its loss during use. The K2O and CaO neutralize the acid character of Al2O3. Both K2O and CaO decrease the electron work function of iron and increase its ability to chemisorb nitrogen by charge transfer to the nitrogen.

PROPERTIES Particle size: Granules, in size range 6-10 mm Bulk density: 1200kg/m3 Particle density: 305 lb/cu ft (4.9 g/cm3)

37

Activity loss in service, 30-35% in 3yr depending on severity of operating conditions and presence of poisons. Catalyst is slowly deactivated at operating temperatures above 9850F.

CATALYSTS POISIONS In addition these poisons, hydrocarbons such as lubricating oils and olefins can crack and plug pores. Sulphur, phosphorous and arsenic compounds should not exceed 15ppm. Though temporary poisons, they cause crystals growth and attendant area decline. Chlorine compounds form volatile alkali chlorides with promoters.

CHEMISTRY AND KINETICS The overall stoichiometric equation is 1/2N2 + 3/2H2

NH 3Extensive studies of ammonia synthesis on iron catalysts

suggest that the reaction occur through the following steps N2 (g)

2N (ads)

H2 (g)

2H (ads)

N (ads) + H (ads)

NH (ads)

NH (ads) + H (ads)

NH2 (ads)

NH2 (ads) + H (ads) Nh3 (ads)

NH3 (ads) NH3 (g)

A rate equation based on nitrogen adsorption as the slow step is the most commonly used although other forms have been development that also correlates the data Since the effectiveness factor of ammonia catalyst is less that unity in a commercial size pellets, it is desirafinely ground catalyst and employ and effectiveness factor correction for other sizes. Ammonia synthesis is another example of an old reaction with sufficient data existence to make this procedure feasible. The following equation in term of activity has been recommended. Design of shift converter

F12=28557.5kmol/hr

Plug Flow Reactor

F13=25990.61 kmol/hr

38

Step -1 Reaction occurring in PFR N2 + 3H2 Or

A + 3B

k k

2NH3 C

Reactor Kinetics The only reaction that needs to be considered is the formation of ammonia from H2 and N2, N2 + 3H2 ‹—› 2NH3 A common rate equation for ammonia synthesis is the Temkin-Pyzhev equation given

where R is the rate of nitrogen consumption per unit volume of catalyst, f is a catalyst activity factor, and Pi is the partial pressure of component i in the gas. Values for K1 and K2: K1 = ko1exp(-E1/RT) K2=ko2exp(-E2/RT) Ko1 = 1.78954 x 104 kgmol/m3-hr-atm1.5 E1 = 20,800 kcal/kgmol Ko2 = 2.5714 x 1016 kgmol-atm0.5/m3-hr E2 = 47,400 kcal/kgmol One limitation of this rate equation is that the rate will be infinite if the amount of ammonia in the gas is zero, and this may occur in a reactor being fed with fresh make-up gas. To avoid this numerical problem, the rate equation may be multiplied by a factor K3PNH3/(1+K3PNH3) ; this will avoid the approach to infinity at low PNH3 while having little effect at high ammonia pressures. This modification re-casts the rate equation into the

39

Langmuir-Hinshelwood-Haougen-Watson (LHHW) form, which is one of the options for the RPLUG reactor block in Aspen:

Assume that f = 1.0 and K3 = 2 atm-1. Assume that the bulk density of the catalyst is 1200 kg/m3 and that the catalyst costs $12/kg. Now, reactor temperature is 450 degree Celsius Therefore, k1= 0.00873 kmol/m3-hr-atm1.5 K2=106.3 kmol-atm0.5/m3-hr Also,

pi=pio(1-Xi)/(1+EiXi) ; where i=A,B,C EA=(2-4)/2=-0.5 & XB=3XA,XC=2XA

Using pAO=0.2247*200=44.94atm,pBO=0.6742*200=134.84atm,pCO=0.06*200=12atm Finally, we get pA=44.94(1-XA)/(1-0.5XA) pB=134.84(1-3XA)/(1-1.5XA) pC=12(1-2XA)/(1-XA)

1228.6(1-XA)(1-3XA)1.5 (1-0.5XA)(1-1.5XA)1.5

(-rA)

9.78(1-2XA)2(1-1.5XA)1.5 (1-XA)2(1-3XA)1.5

= 24(1-2XA) 1

+ (1-XA)

40

Now plug flow reactor design equation dxA

=

V

(-rA)

FAO

Where FAO -- feed rate of A From V

=

FAO

∫

dxA

( limits from 0 to 0.2)

(-rA)

XA -rA 1/-rA

0 48.7 0.02

0.04 25 0.04

0.08 24.5 0.04

0.12 23.9 0.04

0.16 23.5 0.04

0.20 23.23 0.04

41

1/(-rA)

1/(-rA) vs XA 0.045 0.04 0.035 0.03 0.025 0.02 0.015 0.01 0.005 0

Series1

0

0.05

0.1

0.15

0.2

XA

Area under curve = 30x0.005x0.05=7.6 x 10-3 V V

FAO x (R+1) x 7.6 x 10-3

=

28557.5 x 10.4 x 7.6 x 10-3

=

V

16.362m3

=

Take factor of safety = 10% New volume

= 1.1 x 16.362 = 18 m3

Assume L/D = 4 Volume

=

Π /4 x D2 x L

Π /4 x D2 x (4D)

18

=

D3

= 5.729

D

= 1.212 m

Hence length (L) = 4 x 1.212 = 4.848 m Now catalyst is divided in 2 beds in ratios 1:2.5 (as upper bed : lower bed) Volume of catalyst in 1st bed = 18/3.5

0.25

42

= 5.14 m3 Volume of catalyst in 2nd bed = (18 – 5.14) = 12.86 m3 Bulk density of catalyst

= 1200 kg/ m3

Weight of catalyst in 2nd bed = 12.86 x 1200 = 15.432 tonne Weight of catalyst in 1st bed = 5.14 x 1200 = 6.168 tonne Now giving allowance for space for gas movement upward and downward and insulation be 0.5 m Hence diameter of reactor becomes = 1.212 + 0.5 = 1.712 m Pressure at which reactor works = 200 kg/m2 Let factor of safety

= 20%

Design pressure = 1.2 x 200 = 240 kg/m2 We use carbon steel with internal lining of titanium Allowable stress = 66,000 psi {hesse & ruston} = 4494 kgf/cm2

Now Allowable stress = PD(1/k2 – 1) (from Hesse) Where PD = design pressure 4494 = k

=

240(1/k2 – 1) 1.0337

Do = kDi Do = 1.0337 x 1.212 = 1.252 m Thickness of shell = (1.252 – 1.212)/2=20mm

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Specification sheet for Ammonia Convertor Operating pressure

:

200 kg/m2

Design pressure

:

240 kg/m2

Temperature

:

300 – 520oC

Allowable stress

:

4494 kgf/cm2

Material of construction

:

Cr – Mo steel

Diameter inside of converter

:

1.212 m

Bulk density of catalyst

:

1200 kg/m3

Total volume of catalyst

:

18 m3

Diameter of catalyst bed

:

1.252 m

Thickness of the shell

:

20mm

Volume of 1st catalyst bed

:

5.14 m3

Volume of 2nd catalyst bed

:

12.86m3

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Weight of 1st catalyst bed

:

6.168 tonne

Weight of 1st catalyst bed

:

15.432 tonne

MECHANICAL DESIGN OF SHELL & TUBE HEAT EXCHANGER Carbon steel(corrosion allowance=3mm) Allowable stress=11,000/14.7=749atm [ref.PED by hesse & ruston,pg.60,table3.1] SHELL SIDE No. of pass=1 Fluids in shell are hydrogen ,nitrogen ,ammonia & inerts like argon & methane Design pressure=51 kgf/sq. cm=51atm Shell diameter=1067mm THICKNESS OF SHELL= [pD/2fJ-p]+c Where; P=design pressure D=shell ID

45

F=allowable stress J=joint efficiciency C=corrosion allowance Shell thickness= [51x1067/(2x749x.85-51)]+3 =48mm SHELL HEAD : assuming dished heads(because Pressure