Ammonia

INTRODUCTION Ammonia is a colorless alkaline gas, it is lighter than air and has a distinct

pungent odour.

Ammonia was first prepared in the 16th century by distilling animal substances

such as hoofs and horns in the form of a solution. It was first isolated by Pristly

in 1774, and collected over mercury. In 1785, it was shown by Berthollet to be a

compound containing Nitrogen and Hydrogen.

Traces of ammonia and ammonium compounds can be found in the atmosphere

and in natural water and also in the soil in the form of ammonium compounds.

PROPERTIES AND USES OF AMMONIA

Physical Properties

Molecular weight 17

Boiling point 0C -33 to 35 0C

Freezing point 0C -77.7

Critical temperature 0C 133

Critical pressure bar 112.8

Specific heat kJ/(kg K)

0 0C

100 0C

200 0C

2.0972

2.2262

2.1056

Standard heat of

formation, kJ/kmol

-46 222

Solubility in water (wt%)

0 0C

20 0C

40 0C

60 0C

42.8

33.1

23.4

14.1

Specific gravity of

anhydrous ammonia

-40 0C

0 0C

40 0C

0.69

0.639

0.580

Chemical Properties

1) Ammonia is a weak base and when dissolved in water ionizes to form

Ammonium ions NH4+and hydroxyl ions OH-

NH3+H2O NH4OH

2) Ammonia forms ammonium salts and water with acids

NH3 + H2SO4 (NH4)2SO4

NH3 + HCl NH4Cl

3) Ammonia solution precipitates most metal hydroxides from solutions of their

salts. In some cases like copper the hydroxide is soluble in excess owing to

formation of complex ion.

CuSO4 + 2NH4OH Cu(OH)2 +(NH4)2SO4

Cu(OH)2 Cu2+ +2OH-

Cu2+ +4NH3 [Cu(NH3)4}2+

4)Ammonia does not burn readily in air, nor is it a supporter of combustion, but

warm ammonia burns in Oxygen to form nitrogen and water.

4NH3 + 3O2 2N2 +6H2O

In the presence of platinum guaze and other cataltsts, oxides of nitrogen and

ammonium nitrate are formed.

5) Ammonia is a relatively stable compound. When strongly heated or sparked it

decomposes into its constituent elements.

2NH3 N2 +3H2

6) Ammonia behaves as mild reducing agent. It reduces many heated metallic

oxides like CuO PbO etc

3CuO + 2NH3 3Cu + N2 +3H2O

7)Ammonia reacts with halogens to give ammonium halides.

8NH3 + 3Cl2 N2 + 6NH4Cl

with excess halogens it forms dangerously explosive nitrogen trihalides.

8) Ammonia reacts with certain metals when heated. Sodium forms Sodamide

and hydrogen.

2Na + 2NH3 2NaNH2 + H2

USES OF AMMONIA

Ammonia is the single most widely used compound in the fertilizer industry. It

is the starting material for the production of a various number of nitrogenous

fertilizers like ammonium phosphates, ammonium sulfate, ammonium nitrate

etc.

It is used directly or indirectly as the source for the production of

hexamethylenedeamine for the manufacture of nylon 6,6. In the manufacture of

rayon, ammonia is used in the preparation of ammoiacal copper hydroxide

soulutio for dissolving the copper linters. Oxidation of propylene with ammonia

gives acrylonitrile, used for the manufacture of acrylic fibres, resins, and

elastomers.

Hexamethylenetetramine, produced from ammonia and formaldehyde, is used

in the manufacture of phenolic thermosetting resins. Toluene 2,4 disocyanate

(TDI), employed in the production of polyurethane foam, indirectly consumes

ammonia because nitric acid is a raw material in the TDI manufacturing process.

Urea produced from ammonia is used in the manufacture of urea-formaldehyde

synthetic resins. Melamine is produced by polymerization of dicyandiamine and

high pressure high temperature pyrolysis of urea, in the presence of ammonia

Lesser known uses of Ammonia are

1. As a refrigerant in both compression and absorption systems

2. In the pulp and paper industry for the pulping of wood

3. As a corrosion inhibitor in petroleum refineries.

4. In rubber production for the stabilization of natural and synthetic latex to

prevent coagulaion.

5. In the food and beverage industry as a source of nitrogen required for the

growth of yeast and micro-organism

6. As a curing agent in tanning industries.

In the manufacture of pharmaceuticals such as sulfanilamide, suflathiazole etc.

MANUFACTURING PROCESSES

The synthesis gas for manufacture of Ammonia is produced by steam reforming

or partial oxidation of Hydrocarbon feed. In most industries steam reforming is

the norm. A flow sheet describing the various operartions in a typical single

train ammonia plant is shown. The various processes used commercially in

industries for production of Ammonia are

1. Braun Purifier process

2. Foster Wheeler AM2 process

3. ICI process

Braun Purifier process

In the secondary reformer 1.5 times the stoichiometric quantity of air is used.

This increases the heat load and reduces the radiant duty of the primary

reformer to less than two thirds its usual duty. Excess nitrogen is removed by a

cryogenic purification unit after methanation occurs. Large air compressors used

are driven by gas turbines. Owing to the high purity of synthesis gases, lower

recylce gas flow, lower refrigeration duty and lower purge duty will suffice.

Foster Wheeler AM2 precess

In the secondary reformer excess air is used as in the previous case. Instead of

treating the whole feed in primary reformer, some bypass is fed to the secondary

reformer. Partial reaction in primary reformer will allow for a lower steam to

carbon ratio. Excess air used is much higher than in the case of Braun purifier

process. A cryogenic unit removes nitrogen form the exit gases leaving the

methanator. CO2 is removed using physical solvents. Absorption system

provides refrigeration for ammonia recovery.

ICI process

Hydrocarbon feed is subjected to steam reforming in two stages to form oxides

of Carbon, methane and hydrogen. In the secondary reformer air is mixed with

the gases to get a N2 : H2 ratio of 1:3. Carbon monoxide is removed by shift

conversion. Carbon dioxide is removed by absorption into MEA or Potassium

Carbonate solution. Traces of CO and CO2 are removed by conversion into

methane. Synthesis gas is used to produce ammonia.

Advantages of this process are

1. Intensive heat recovery

2. Generation of steam which can be imported.

3. Less dependancy on electricity

4. Capital cost is least.

Process Description

The hydrocarbon used is Naphtha. Naphtha used contains sulfur which is a

catalytic poison in the ammonia synthesis process. It must be removed before the

feed can be used for producing hydrogen.

Desulfurizer

The sulfur in the naphtha feed stock is converted to hydrogen sulfide in the

presence of Cobalt Molybdenu catalyst at a temperature of about 673 K

H2 + S H2S

The sulfur content is reduced to less than 5ppm

Reforming

The reformation process is carried out in two stages.

Primary Reformer

In the first stage desulfurized naphtha is mixed with steam in a tubular

reforming furnace. The reformation reactions occur at a temperature of about 673

K. Steam to carbon ration must be maintained between 3.5 and 4.5 to ensure that

Carbon deposition does not occur.

CnHm + nH2O nCO + (n+m/2) H2

CO + H2O CO2 + H2

CO + 3H2 CH4 + H2O

The overall reaction is highly exothermic and the outlet temperature is about

1093 K. The primary reforming process is characterized by a low pressure.

Secondary Reformer

The gases from the primary reformer are mixed with air and steam at the outlet

temperature of the primary reformer. The remaining hydrocarbon like methane

are further subjected to reformation and the overall yeild of hydrogen is

increased. Air is mixed to form a mixture of H2 and N2 for the synthesis process.

CH4 + H2O CO +3 H2

Other side reactions occuring are

H2 + O2 H2O

CO +H2O CO2 +H2

CO + O2 CO2

Initially the carbon monoxide burns in air to produce carbon dioxide and the

temperature increases to about 1200 0C. Then adiabatic reaction of methane with

steam occurs and the outlet temperature decreases to about 1000 0C at the outlet.

Carbon monoxide Conversion

Concentration of CO is reduced to about 0.1 0.3 % by volume in two stage

converter.

CO + H2O CO2 + H2

HTSC

CO concentration is reduced to about 3% by volume by means of water gas shift

reaction at a temperature of about 593 to 693 K. The reaction being exothermic

the exit gases are at a higher temperature. They are cooled before being sent to

the secondary reformer.

LTSC

The reaction occurs at a temperature of about 523 K. Concentration of CO in the

exit gas is reduced to about 0.3 % by volume.

CO2 Absorption

Monoethanolamine solution is used as the absorbing medium. The absorption

operates at atmospheric pressure and a temperature of about 100 0C. The tower is

usually a packed tower. Counter current absorption is practised. The spent MEA

solution is regenerated in a regenerating column by steam stripping.

Methanation

Even trace quantities of Carbon monoxide and Carbon dioxide will act as catalyst

poisons in the synthesis loop. Hence they must be removed by conversion into

methane. Methane though not an inert gas is nevertheless inert in the ammonia

synthesis process. The reaction in the methanator occurs at a temperature of

about 573 to 673 K. The exit gases containg steam are condensed and cooled

befroe being stored for the synthesis loop.

CO + 3H2O CH4 + H2O

CO2 + 4H2 CH4 + H2O

Ammonia Synthesis

Synthesis gas is comprtessed to about 50 MPa and heated t about 673 K before

passing through the catalyst beds.

The reaction being a reversible one high pressure favours the forward phase. As

the concentration of ammonia in the exit gas is low, recycling of gas is necessary.

The converter used is a vertical type of converter.

Catalyst activity inreases with increase in temperature, hence ideal temperature profile is one in which rate of ammonia production is a maximum at all parts of the bed. In the upper part of the converter the concentration of ammonia being low the temperature is high ensuring high reaciton rates. In the lower parts temperature must be low due to increasing influence of equilibriem concentration.

MAJOR EQUIPMENT DESIGN PACKED BED ABSORBER

Equilibrium data for CO2 and Monoethanolamine solution

X

mol CO2/mol

MEA sol

Y

mol CO2 /mol

inerts

0.0038 0.026

0.0047 0.034

0.0057 0.044

0.0067 0.056

0.0076 0.067

0.0086 0.078

0.0095 0.087

0.0105 0.099

0.0115 0.109

0.0124 0.120

0.0133 0.133

0.0145 0.153

0.0160 0.180

0.0170 0.202

Gas flow rate at bottom Gb = 10559.1 kmol/hr

= 41.68 kg/s

Gas flow rate at the top Gt = 8726.9 kmol/hr

= 20.85 kg/s

Yt =0.005

Yb =0.216

Gin = 8683.5 kmol/hr

PROPERTIES

Gas density g = 0.487 kg/m3

Liquid density l = 934.4 kg/m3

Gas viscosity g= 0.0175 cP

Liquid viscosity l = 0.299 cP

Gas diffusivity Dg = 1.65X10-5 m2/s

Liquid diffusivity Dl = 1.96X10-5 m2/s

Gas heat capacity Cpg = 2.094 kJ/kg K

Liquid heat capacity Cpl = 4.145 kJ/kg K

From Graph

(Lin/Gin)min = 11.99

(Lin/Gin) = 1.111.99 = 13.189

Lin = 13.189x 8683.5

= 99652.5 kmol/hr

Top Section

Lt = Ls (1+Xt)

= 99652.5 kmol/hr

Bottom Section

Lb = 99652.5(1+Xb)

= 101117.4 kmol/hr

14.5 wt% MEA souliton has molecular wt.= 20.08 kg/kmol

Lt = 198026.6 kg/hr = 555.00 kg/s

Lb = 92452.82 kg/hr = 577.85 kg/s

Calculation of coloumn diameter Choosing 90 mm pall rings (metal)

Void fraction = 0.97

Packing factor Fp = 53

Surface area a = 92 m2/m

At the bottom

At the top

Hence choosing the larger value of 0.61

From Graph

Where

Gf = gas superficial velocity

Fp = packing factor = 53 m-1

= correction factor for density = 1.029

l = viscosity of liquid in cp = 0.299

g = density of gas = 0.487 kg/m3

l = density of liquid = 934.4 kg/m3

g = acceleration due to gravity

On substituting we obtain

Gf =2.085 kg/m2 s

Operating G = 0.85Gf

= 1.772kg/m2s

Ac = Gb/G

Ac = 20.85/1.772

A = 9.948 m2

Di = 3.20 m

Design dia =5m

038.02.02

=

gFG

lg

lpf

61.0=

l

g

GL

56.0=

l

g

GL

Pressure drop

P = pressure drop expressed in terms of water

C2, C3 are constants for specific type of packing

Utg, Utl = superficial velocities of gas and liquid expressed in ft/s

g = gas density in lb/ft3

Utg = Gavg/gAc = 4.226 m/s = 14.108 ft/s

Utl = Lavg/lAc = 0.0596 m/s =0.197 ft/s

Hence P = 27.43 in of H2O/ft of packing

= 2285.4 mm of H2O /m of packin

Degree of wetting from Nuris & Jackson criterion (Weeping Check)

Where

Lmin = minimum flow rate of liquid = 555.00 kg/s

l = density of liquid = 934.4 kg/m3

a = surface area of packing per unit volume = 92 m2/m3

wetting rate is greater than 0.85 ft3/ft hr hence no weeping will occur

22

310 tggUC UCP tl =

clW

aALL

min

=

Evaluation of tower height

For ring type of packings

= parameter for a given packing = 75

D = column diameter if diameter is less than 0.6m else 0.6m

Scg = Schmidt no for gas phase =2.1778

L = liquid rate = 55.58 kg/m2s

f1 = (l/w) = 0.8243

f2 = (l/w) =1.0885

f3 = (w/l) = 1.0611

substituting we get

Hg = 0.25Z0.33

Liquid phase transfer unit

= correlation parameter for a given packing = 0.11

C = correlation parameter for high gas flow rates = 0.5

Scg = Schmidt no for liquid phase = 163.6

Z = tower height

On substitution

Hl = 0.181Z0.15

m = slope of the equilibrium curve =11.25

Lm = 100449.7 kmol/hr

Gm = 9643.0 kmol/hr

lm

m

gOG

OGOG

HLG

mHH

NHZ

+=

=

( ) 5.03215.033.024.1017.0

ffLfScZD

H gg

=

15.05.0

05.328.3

=

ZScCH ll

Overall number of transfer units in the gas phase

Y Y* 1/(Y-Y*)

0.005 0 200

0.015 0.005 100

0.025 0.010 66.67

0.035 0.015 50

0.045 0.020 40

0.055 0.025 33.33

0.065 0.032 30.30

0.075 0.039 27.78

0.085 0.047 26.32

0.095 0.057 26.32

0.105 0.065 25

0.115 0.075 25

0.125 0.083 23.81

0.135 0.095 25

0.145 0.100 22.22

0.155 0.107 20.83

0.165 0.116 20.41

0.175 0.128 21.17

0.185 0.138 21.17

0.195 0.151 22.73

0.205 0.164 24.49

0.210 0.171 25.64

0.215 0.180 28.57

+

+

= t

bY

YOG Y

YYY

dYNb

t11ln

21

*

Computing by using Simpsons 1/3 rule for numerical integration

Hence

computing the value of Z by trial and error we have

The total pressure drop is P = 11.427 m of H2O

Z = 5.0 m

543.7259.0284.7215.0

005.0*

=+=

YY dY

15.033.0 452.1862.1 ZZHNZ OGOG +==

MINOR EQUIPMENT

Shell and tube heat exhanger

The gases coming out of the NH3 converter is at a very high temperature of

778K These gases are used to heat the inlet gases to the converter. The gases then

are cooled to a temperature of 313 K before it enters the condenser where

Ammonia is condensed. Because the heat load is very high five HE are used in

parallel.

Amount of heat to be removed Q = 4989.1 kW

Mass flow rate of gas mh = 23.8 kg/s

Inlet temperature T1 = 405 K

Outlet temperature T2 = 313 K

It is assumed that cooling water inlet temperature is 293 K and exit temperature

is 308 K.

Amount of cooling water mc = Q/Cp T = 79.4 kg/s

Inlet temperature t1 = 293 K

Outlet temperature t2 = 308 K

= 50.38 K

For these values the LMTD correction factor FT = 1

(LMTD)c = 50.38 K

( ) ( )

=

21

21

1221

lntt

TTttTT

LMTD

12.0

46.6

11

12

12

21

=

=

=

=

tTttS

tt

TTR

Properties

Gas stream at 359 K Water at 300.5 K

(kg/m3) 20.03 996.4

(Pa s) 4.876 x 10-5 9 x 10-4

k (W/m K) 0.125 0.359

Cp (kJ/kg K) 1.991 4.187

Routing of Fluids

The gases are taken on shell side. Cooling water is taken on the tube side.

AREA OF HEAT TRANSFER

Assuming Ud = 350 W/m2K

Area of heat transfer is given by

For the exchanger 1 inch 16 BWG tubes are chosen

do = 25.4 mm

di = 22.1 mm

area of tube per unit length = 0.2168 ft2/ft

Assumption is made that the tubes are of length 16 ft

Hence the area per tube a = 4.1888 ft2 = 0.3333 m2

No of tubes

From the tube count table we have for 1 inch o.d tubes on 1.25 inch square pitch

Nt =738 Np = 4 Ds = 1067 mm

Hence corrected area for heat transfer A =Nt x a = 245.97 m2

Thus corrected coefficient of heat transfer Uac = 341 W/m2 K

( )283.242 m

LMTDUQA

Cd

==

730==a

ANt

Velocity Tube side

Tube cross sectional area for flow of fluid in one tube pass

where

di is the tube inside diameter =22.1 mm

Nt is the number of tubes =738

Np is the number of tube passes =4

Tube side velocity

where

mc is the mass flow rate of cold water

t is the density of tube side fluid

The velocity of a liquid should lie between 1 and 2 m/s which is satisfied.

Shell Side

Shell side cross sectional area for fluid flow

Where

Po = pitch = 31.75 mm

do = outside diameter = 25.4 mm

Ds = shell inside diameter = 1067 mm

ls = baffle spacing assumed to be 0.5 Ds = 534mm

22 0708.04

mNNda

p

tiT ==

sm

a

mv

TT

CT 125.1==

( ) 20 1423.0 mP

DldPa

o

sso

S =

=

Shell side velocity is given by

Where

ms = shell side mass flow rate

s = shell side fluid density

The velocity of flow for gases must lie between 10 and 40 m/s which is satisfied

Film Transfer Coefficient

Tube side

Reynolds number

Where

is the viscosity of fluid

is the density of fluid

dI is the inside diameter of tube

Prandtl number

Where

Cp is the specific heat capacity

k is the thermal conductivity

Nusselts equation is given as

sm

a

mv

ss

s

s 325.10==

8.27524Re ==

ivd

75.10Pr ==kC p

smWh

kdh

Nu

t

it

2

4.08.0

1.2646

167

PrRe023.0

=

=

=

Where

ht is the tube side heat transfer coefficient

Shell side

Reynolds number

Prandtl number

By Deitus Bolter equation Nusselts number

Where

hs is the shell side heat transfer coefficient

Overall heat transfer coefficient

Where

hd represents the dirt coefficient. It is assumed to be 2839 W/m2 K for both shell

and tube side

On substitution we obtain

U = 366.4 W/m2 s

The design value is larger than the assumed value. Hence design is safe.

53759Re ==

os dv

94.0Pr ==kC p

smWh

kdh

Nu

s

os

2

3.08.0

64.681

6.137

PrRe023.0

=

=

=

( )k

dddhhd

dhhU

ioi

sdso

i

tdt

ln11111+

+++=

Pressure Drop Tube side

Re = 27524.8

where

f is the friction factor

l is length of pipe

vt is the tube side velocity

g is the acceleration due to gravity

di is the tube inside diameter

Where

Np is the no of tube passes =4

The pressure drop for liquids must lie between 14 and 70 kPa. Hence the design

is safe from view of pressure drop on the tube side.

Shell Side

Pressure drop in the cross flow section

fk is the ideal tube bank friction factor =0.104

W is the mass flow rate = 23.8 kg/s

22

25.0

38.33842

4

00613.0Re079.0

mNggdfLv

P

f

i

tL =

=

==

22

25.15762

5.2 mNvP tc =

=

( ) kPaPPNP cLpt 842.19=+=

14.0

2

2

=l

w

m

ckbk S

NWfbP

Nc is no of tube rows crossed in one cross flow section

lc is the baffle cut =0.25 Ds

Pp is pitch parallel to flow = 31.75 mm

Ds is the shell inside diameter =1067 mm

is the fluid density = 20.3 kg/m3

Sm is cross flow near centerline for one cross flow section = 0.1423 m2

Pbk = 0.4555 kPa

Pressure drop for an ideal window section

b =2 x 10-3

Sw is for flow through window =0.1598 m2

Pwk = 0.3657 kPa

P = 7.845 kPa

The pressure drop for gases should be between 2 and 20 kPa. Hence the design is

safe on shell side also

( )[ ] 8.1621 ==p

scs

c PDlDN

( )wm

cwwk SS

NWbP6.022 +

=

( )[ ]

81

121

==

+++=

s

b

c

cwbkwkbbkb

lLN

NN

PPNPNP

MECHANICAL DESIGN OF ABSORBER Material for shell is Carbon Steel

THICKNESS OF SHELL

Thickness of shell = ts

cpfJ

pDts +

+=

2

Where,

Di = Inner Diameter of vessel = 3.2m

Working Pressure = 1 atm = 1.054kgf/cm2

p = Design Pressure = 1.1 x 1.054 = 1.137 kgf/cm2

f =Permissible Stress = 950 kgf/cm2

J= Joint Efficiency = 0.85

c= Corrosion allowance =2mm

ts = 6mm

thickness of shell is 6mm

HEAD DESIGN: FLANGED & SHALLOW

Material stainless steel

Permissible stress = f= 130 N/mm2

Design pressure = p = 1.064 x 10 5 N/m2

Stress identification factor W is given by

W = () [3 + ( Rc/R1)1/2]

Crown Radius = Rc= 3.2m

Knuckle radius = R1 = 0.192m

Stress identification factor W is 1.77

Thickness of head = th = fJWpRc

2

th= 6mm

Axial Stress Due to Pressure

Axial stress due to pressure =fap

( )ctpDfs

iap

=

4

= 355.4 kgf/cm2

Stress due to Dead Load

a) Compressive Stress due to weight of shell up to a distance X

Do = Di + 2 ts =3.212m

Density of Shell material = s = 7700 kg /m3

( )( )22

22

4

4

io

sio

ds

DD

XDDf

=

=.77X kgf/cm2

b) Compressive stress due to weight of insulation at height X

Insulator used is asbestos

Thickness of insulation = tins =100mm

Diameter of insulation = Dins

Density of insulation =2300 kg / m3

Mean diameter of vessel = Dm

For large diameter column

Dins = Dm

( ))ctDXtDf

sm

insinsinsins

=

= 1.925 kgf/cm2

c) Compressive stress due to liquid in column up to height X

Density of liquid =l = 934.4 kg/m3

( )ctDD

fsm

li

liq

=

24

= 23.29 kgf/cm2

d) Compressive stress due to attachment

i. Packing weight

ii. Head weight

iii. Ladder

Density of packing (pall ring) = 270 kg /m3

Packing Weight = ( )ctDXD

sm

pi

24

Head weight (approximately) = 2000kg

Weight of Ladder = 140 kg/m

Total compressive stress due to attachments is given by

fattach = ( Packing Weight + Head weight + Ladder ) / [ Di ( ts c ) ]

= (3.1972 + 0.2225 X) kg/cm2

Stress due to Wind

Stress due to wind is given by

( )ctDXPf

sm

wws

=

24.1

Where,

Pressure due to wind = Pw = 125 kgf/m2

=0.2778 X2 kgf/cm2

To determine the value of X

fmax = 950 kgf/cm2

Upwind side

Jftmax = fwx + fap +fdx

807.5 = 365.4+3.1792+39.6345 X +0.2778 X2

Solving for X we have

X = 10.331m

Downwind side

Jftmax = -fwx - fap -fdx

807.5 = -365.4-3.1792-39.645 X +0.2778 X2

Solving for X for compressive stresses

X=167 m

Hence the column will support a height of 5m as the calculated height is much

higher than 5 m in both cases

DESIGN OF GASKET AND BOLT

Width of gasket =N = 10 mm

Gasket material is asbestos

Gasket factor = m =2

Minimum design seating stress = Ya= 11.2 N / mm2

Basic gasket seating width bo = N/2

bo = 10 mm / 2 = 5mm

Effective gasket seating width

b = 2.5 ( bo )1/2

Inner diameter = Di = 4.85 m

Outer diameter = Do = 4.85m + 2 x 8 x 10 3

Flange inner diameter =Dfi = 4.866 m

Flange outer diameter = Dfo = 4.926 m

Mean diameter = G = (Dfi + Dfo) / 2

= 4.896 m

Under atmospheric conditions, the bolt load due to gasket reaction is given by

Wm1 = b G Ya

=962.98 x10 3 N

Design pressure = p

= 1.064 x 10 5 N/m2

After the internal pressure is applied, the gasket which is compressed

earlier, is released to some extent and the bolt load is given by

Wm2 =(2b) G x m x p + (/4)G2 p

=2.039 x 10 6 N

Bolt used is hot rolled carbon steel

fa is permissible tensile stress in bolts under atmospheric condition

fb is permissible tensile stress in bolts under operating condition

fa=58.7 x 10 6 N/m2

fb = 54.5 x 10 6 N /m2

Am is the area of bolt

Am1= Wm1 / fa

Am2 = Wm2 / fb

Am1= 0.016 m2

Am2= 0.037 m2

Number of bolts = mean diameter /bo x 2.5

=392 bolts

To determine the size of bolts, the larger of above two areas should be

considered

Diameter of bolts =[(Am2 /Number of bolts) x (4/)]1/2

=1.096 cm

FLANGE THICKNESS

Thickness of flange = tf

tf= [G(p/K f) ] + c Where,

K=1/[ 0.3 + ( 1.5 Wm hG)/H x G]

Hydrostatic end force = H = ( /4) G2 p

=2.003 x10 6 N

hG is radial distance from gasket load reaction to bolt circle,

hG = ( B G )/ 2

= 0.0155

B = outside diameter of gasket + 2 x diameter of bolt + 12mm

= 4.927 m

Wm = 2.039 x 10 6 N

K= 3.28

Hence the thickness of flange = 90.47 mm

NOZZLE THICKNESS

Material Carbon steel

Considering diameter of nozzle = Dn =150 mm

Thickness of nozzle =tn

Material is Stainless steel ( 0.5 cr 18 Ni 11 Mo 3)

Permissible stress =130 x 10 6 N/m2

J=0.85

pfJpD

tn

=

2

tn = 1.24mm

No corrosion allowance , since the material is stainless steel.

We can use thickness of 3mm

SUPPORT FOR ABSORBER

Material used is structural steel ( IS 800)

Skirt support is used.

Inner Diameter of the vessel = Di =3.2m

Outer Diameter of the vessel = Do =3.212m

Height of the vessel = H = 5.0m

Density of carbon steel= s = 7700 kg /m3

Density of liquid = l = 934.4 kg /m3

Total weight = Weight of vessel+ Weight of Attachments (liquid+ packing + head

+ ladder)

=59430 kg

total stress due to dead loads

ttDwfm

ds00.59

=

=

kgf/cm2

Stress Due to wind load

The force due to wind load acting on vessel is

plw = k p1 h1 Do

.

k=0.7 for cylindrical surface

p1 is wind pressure for the part of the vessel upto a height of 20 m.

p1 = 128.5 kgf/m2

h1=8m

ttDHpf

m

wb054.04.1 21

==

kgf/cm2

Stress due to Seismic Load

Load F= CW

W is total Weight of vessel

C is Seismic Coefficient =0.08

ttDCWHf

o

sb2246.0

38

==

kgf/cm2

Maximum Stress at bottom of Skirt

ftmax = ( fwb or fsb ) fdb

= t

775.58 kgf/cm2

Permissible tensile Stress for structural steel = 950 kgf/cm2

Equating the two we have

mmt 74.0

therefore the thickness of skirt is taken as 6mm

Maximum Compressive Stress

fcmax = ( fwb or fsb ) + fdb

= t

2446.59 kgf/cm2

permissible tensile stress for structural steel is 950 kgf/cm2

Equating the two we have

mmt 8.0

Skirt Bearing Plate

ZM

Awf sc +=

Z VXP RI DOO WKH GHDG ORDGV

A = (/4) ( Dsko2 Dski 2) is the area of contact between skirt plate and the

concrete.

Ms = moment due to wind load

Z = section modulus

fc =5.689 kgf/cm2

Permissible stress f in bending is 950 kgf/cm2

2

2

2max 36

B

c

B t

lfbtMf ==

tB= 63mm

Anchor Bolt,

Wmin=45000 kg (assumed)

fc = (Wmin/ A ) (Mw /Z)

= 1.115 kgf/cm2

Fc is positive ,hence anchor bolts are not required if the safety factor is above 1.5

sMRWj min=

= 7.45

The safety factor is above 1.5. Hence anchor bolts are not required.

MECHANICAL DESIGN OF SHELL AND TUBE HEAT

EXCHANGER

Carbon Steel (Corrosion allowance 3 mm)

SHELL SIDE

Number of pass =1

Fluids in shell are Hydrogen, Nitrogen Ammonia and inerts like Argon and

Methane

Design pressure =50.9 kgf?cm2

Shell diameter = 1067mm

Considering steel dished head (torispherical)

Ri= 1067 mm

ri =0.06 of Ri

Inside depth of head can be calculated as

2

1

222

+

+

= ii

ii

iii rD

RD

RRh

= 136 mm

Effective Length = L = 4.88 m + 2 x (0.136)

= 5.154 m

NOZZLE THICKNESS

Material used is carbon steel

Considering diameter of nozzle to be 150mm

Permissible stress = f = 950 kgf/cm2

Corrosion allowance = 3mm

cpfJ

pDt ih +

=

2

=5mm

HEAD THICKNESS

Thickness of head = cfJWpR

t ih += 2

+=

i

ir

RW 341

=1.77

Hence th = 60 mm thickness

TRANSVERSE BAFFLE

Spacing between baffles as = 534 mm

FLANGE JOINT ( BETWEEN SHELL AND TUBE SHEET )

This will be made to satisfy the requirements of the flange joint between tube

sheet and channel. Gasket width (N) = 22 mm

Flange Thickness

Inner diameter of flange =Dfi = 1143 mm

Outer diameter of flange = 1167mm

Gasket Size G=1155 mm

Material is Steel Jacketed asbestos

Width of gasket =N = 24 mm

Gasket factor = m = 3.75

Minimum design seating stress = ya= 5.34 kgf/cm2

Basic gasket seating width bo = N/2

bo = 24 mm / 2 = 12 mm


b = 2.5 ( bo )0.5 = 9 mm

under atmospheric conditions, the bolt load due to gasket reaction is given by

am bGYW =1 = 100500 kg

Design pressure = p = 50.9 kgf/cm2



pGbGmpWm 22 42 += = 667719 kg

Thickness of flange = tf

ckfpGt f +=

Where,

HGhwk gm5.13.0

1

+

=

Hydrostatic end force = pGH 24

=

hG is radial distance from gasket load reaction to bolt circle,

hG = (B G )/2

=37.5 mm

B = bolt circle diameter

wm is the greater of the two loads in this case Wm2

Hence the thickness of flange = 138.8 mm

TUBE SIDE

Stainless steel ( IS- grade 10)

Tube thickness

Where,

Working pressure = 1atm = 1.054 kgf/cm2

Design pressure = p =1.1 times 1.054 = 1.136 kgf/cm2

Permissible Stress =f = 950 kgf/cm2

pfJpD

tt+

=

2

J=1 for seamless tubes

Thickness of tube =0.2 mm

The thickness of tubes is taken as 2 mm with no corrosion allowance as the

material is made of stainless steel.

Tube Sheet

The tube sheet is held between shell flange and the channel. The joint on the shell

flange side is of male and female facing and on the channel side of ring facing,

since the pressure on the channel

Thickness of Tube Sheet

Where F = 1

K=0.25

Thickness of tube sheet = 18 mm

Channel and Channel Cover

Carbon Steel

Thickness of channel

Permissible stress = f = 95 kgf/cm2

K=0.3

Thickness of channel

Substituting the values the thickness of the channel and channel cover is

obtained as 19mm

Flange Joint

Gasket material is steel jacketed asbestos

Gasket width (N) = 22 mm

Gasket factor = m = 5.5

Minimum design seating stress = Ya= 12.66 kgf/cm2

Basic gasket seating width bo = N / 8

fkpFGtts =

fkpGth =

bo = 2.75 mm


b = bo

Mean diameter = G = 1000mm

Design pressure = p = 1.136 kgf/cm2

Under atmospheric conditions, the bolt load due to gasket reaction is given by

= 10002 kgf



= 109374 kgf

Bolt used is hot rolled carbon steel

fa is permissible tensile stress in bolts under atmospheric condition

fb is permissible tensile stress in bolts under operating condition

For Cr Ni Steel,

fa= 1406 kgf/cm2

fb = 66 kgf/cm2

=7.114 cm2

= 16.572 cm2

Am is larger area of Am1 and Am2 =16.572 cm2

Number of bolts = mean diameter / bo x 2 = 40 bolts

am GbyW =1

42

2

2pGGmpbWm

+=

b

m

m fWA 11 =

a

m

m fWA 22 =

To determine the size of bolts , the larger of above two areas should be

considered

= 36 mm

Using M 48 bolts

Pitch diameter = 44.68 mm

Minor diameter = 41.795 mm

Actual bolt area is given by

= 19.66 cm2

Minimum pitch diameter = G +N+2db = 1118mm

Hence Bolt diameter is chosen as B= 1150 mm

Flange Thickness

=2.91

Hence tf = 25 mm

THICKNESS OF NOZZLE

Considering inlet and outlet diameter to be 200mm

Thickness of nozzle

= 1 mm

Thickness of nozzles is taken as 6mm

4m

bAd =

( )21795.44681.48

+=

bA

ckfpGt f +=

HGhWk Gm5.13.0

1

+

=

pfJpD

tn+

=

2

SUPPORT FOR SHELL AND TUBE HEAT EXCHANGER Length of the heat exchanger = 4.88m

InnerdiameterofShell=1067 mm

Outer diameter of Shell = 1137 mm

Inner diameter of tube = 22.1 mm

Outer diameter of tube = 25.4 mm

Number of tubes =738

Density of Steel =s =7850 kg /m3

Density of Liquid =l = 995.3 kg/m3

Volume of Shell body = V = ( / 4) ( Do2 Di2 ) x L

= 0.591m3

Weight of Shell body = Ws = V x s

= 4479 kg

Volume of Tubes = Vt = ( / 4 ) ( do2 di2 ) x L x N

= 0.44 m3

Total Weight of Tubes = Wt = Vt x s

= 3360 kg

Volume of Head = Vh = 0.087 Di3

= 0.106m3

Weight of Head = Wh = Vh x s

= 801 kg

Weight of Liquid = Wl = ( / 4) (di2) x L x N x l

= 10845 kg

Total Weight = W = Ws+ Wt + Wh + Wl

= 19485 kgf

Q =W/2 =9742.5 kgf

Distance of saddle center line from shell end =A= 0.6 x Ri

=320 mm

Longitudinal Bending Moments

Radius = R =1.067m

Height of head =H = 0.136 m

The bending moment at the supports is

M1 = QA [ 1 {(A/L)+ (R2 - H2) / 2 AL}/{1 + 4H/3L}]

= 1842.6 kgf m

The bending moment at the center of the span is given by

M2 = ( Q L / 4)[{1+ 2 ( R2 - H2 ) / L2 }/{1+ 4H / 3L} - ( 4A / L) ]

= 2131kgf m

Stress in Shell at the Saddle For A > 0.5 Ri , the shell is not sufficiently stiffened by the end

For =120,

k1= 0.107

k2 = 0.192

Thickness of shell t = 35 mm

f1 = M1/( k1R2 t)

= 13.97 kgf/cm2

f2 = M1/( k2 R2 t)

=7.79 kgf/cm2

Stress in the shell at Mid- Span The stress at the mid span is f3, which is either tensile or compressive

depending on the position of the fiber. The resultant tensile stresses ( including

the axial stress due to internal pressure ) should not exceed the permissible

stress, and the resultant compressive stress should not exceed the permissible

compressive stress .

f3 = M2 /(R2 t )

= 18.62 kgf/cm2

Axial Stress in Shell due to internal pressure fp = (p Di )/ (4 t)

= 34.99 kgf/cm2

The combined stresses ( fp + f1 ) , ( fp + f2 ) , and ( fp + f3 ) are will within the

permissible values

ENERGY BALANCE

Desulphurizer

Here Sulphur in Naphtha is made to react with Hydrogen in presence of

catalysts to give Hydrogen Sulphide. This reaction takes place at a temperature

of 623 K

H2 + S

Hydrogen and Naphtha are assumed to be stored at 303 K

Specific Heat capacity of naphtha is assumed to be 3.5235 kcal/kmol

Specific Heat capacity of Hydrogen =(6.62+0.00081T) kcal/kmol K

Hence heat required to raise their temperature from 303 K to 693 K

Assuming that the fuel used is Natural gas (calorific value = 39383.82 kJ/m3)

Thus amount of fuel needed to supply this quantity of heat =152.75 m3/hr

Heat liberated within the reactor due to reaction = 1414.575 kcal/kmol

Hr Hence total heat liberated = 11.1 x 1414.575 = 15560.325 kcal/hr

This heat released is utilized in increasing the temperature of the effluents

Solving the equation by a trial and error process we obtain the outlet

temperature value as 623.8 K. Hence it is assumed that the outlet temperature is

623 K itself.

Primary Reformer

Within the reformer steam reacts with Naphtha and produces Carbon monoxide

and Hydrogen. Side reactions also take place producing Carbon dioxide and

hrkcalxdTCnHi

pii610015915.6==

325.15560623

== T piiisul dTCnnH

( ) 22 2 HnmnCOOnHHC mn +++

Methane. The exit gases from the reformer are at a temperature of 1093 K .

Hr is 53.376 x 103 kcal/kmol

H1 is 10.29823 x 107 kcal/hr

Hr is 7.834 x 103 kcal/kmol

H2 is 0.44489 x 107 kcal/hr

Hr is 49.271 x 103

H3 is 1.66672 x 107 kcal/hr

Heat required to raise the temperature of the reaction products from 673 K to

1093 K is

Thus total heat to be supplied is sum of all the heat requirements and the

enthalpy of reactions =22.11061 x 107 kcal/hr

Fuel required for supplying this amount of heat is 5614.13 m3/hr of natural gas.

Secondary Reformer

In the secondary reformer methane produced is converted into CO and H2.

Here air is mixed with the effluent stream in such a quantity that the exit stream

contains 1:3 ratio of N2: H2. It is assumed that the inlet to the reformer enters at

1093 K and reaction takes place at this temperature.

Heat required to raise the temperature of air from ambient conditions at 303 K to

1093 K is

This heat is supplied by natural gas. Hence fuel required to provide this amount

of heat is 549.3 m3/hr

H is 1.96936 x 107 kcal/hr

222 HCOOHCO ++

OHCHHCO 2423 ++

hrkcalxdTCnnH

i

T

piiiref7

623

1094433.13==

hrkcalxdTCnnH

i

T

piiiair7

623

1016325.2==

224 3HCOOHCH ++

H is 2.88989 x 107 kcal/hr

H is 1.28508 x 107 kcal/hr

Total heat liberated during reaction is 2.26464 x 107 kcal/hr. This heat is utilized

in heating the effluents from 1093 K to the exit temperature.

Thus by a trial and error process the temperature is found to be 1262 K.

Heat Recovery between Secondary Reformer and HTSC

The stream entering the HTSC is at 693 K (value obtained from literature). Hence

the stream is cooled from 1262 K to 693 K. The heat recovered from this stream is

utilized in producing steam in quenchers.

Heat recovered is

Amount of steam at 373 K at 1 atm produced from this heat is

High Temperature Shift Converter

In the HTSC Carbon monoxide percentage is reduced to around 3% by volume of

the exiting stream by converting it to Carbon dioxide by the water shift reaction.

Hr is 7.963 x 103 kcal/kmol at 693 K

H is 0.77180 x 107 kcal/hr .

This heat is utilized in increasing the heat content of the exit stream.

Solving for T by a trial and error process we get T = 700 K

OHOH 222 21 +

2221 COOCO +

hrkcalxdTCnHi

T

pii6

1093

1026464.2==

hrkmol

TCH

mspw

s 87.7401=+

=

222 HCOOHCO ++

hrkcalxdTCnHi

T

pii7

693

1077180.0==

hrkcalxdTCnHi

T

pii7

1093

1026464.2==

Heat recovery from HTSC to LTSC

The effluents from the HTSC are at a temperature of 700 K. The inlet stream into

the LTSC is at a temperature of 523 K. The heat recovered from this stream is

utilized in producing steam at 373 K and 1 atm in quencher.

Heat recovered

Amount of steam produced will be

Low Temperature Shift Converter

Here the Carbon monoxide content is reduced to 0.3% by volume of the total

effluent gases. The reaction being exothermic heat is liberated which is used to

increase the temperature of the outlet stream.


H is 0.27252x 107 kcal/hr

The temperature of the outlet stream can be found out by making a heat balance

By a trial and error process the temperature is found to be 543 K.

Condenser

The exit stream from the LTSC contains steam which would be an additional

unwanted load on the absorber. Hence steam is removed by condensation and

the dry gases leave the condenser at 493 K and are cooled to 373 K which is the

temperature at which the gases enter the absorber. It is assumed that all steam

produced is removed by condensation

Heat removed by condensation

hrkcalxdTCnHi

pii7

700

523

1013555.2== hr

kmolTCH

mspw

s 36.2181=+

=

222 HCOOHCO ++

hrkcalxdTCnHi

T

pii7

523

1027252..0==

hrkcalxnQ ssC 71016939.4==

Heat Exchanger

The gases are cooled from 493 K to 373 K. The heat transferred during this

process is

Methanator

Here the trace quantities of CO and CO2 are converted to Methane by reaction

with H2 in the presence of catalysts. The methanator operates at a temperature of

623K. The entering gases have to be heated from 373 K to 623 K.

Heat supplied

This heat may be supplied by condensing steam and utilizing the heat of

condensation.

Amount of steam at 373 K and 1 atm required to heat the gases is

Reactions taking place in the methanator are


H is 1.36386 x 106 kcal/hr

Hr is 39.433x 103 kcal/kmol at 623 K

H is 0.31561 x 106 kcal/hr

Total heat liberated within the methanator is used to increase the temperature of

the exit gases.

hrkcalxdTCnHi

pii7

493

373

1040919.1==

hrkcalxdTCnHi

pii7

623

373

1017869.3==

hrkmolHm

s

s 6.2970=

=

222 HCOOHCO ++

OHCHHCO 2422 3 ++

The temperature of the exit stream found out by a trail and error process is 632 K.

The gases from the Methanator are cooled, water removed by condensation and

the dry gases cooled to 313 K and stored.

Ammonia Synthesis Loop

The synthesis gas enters the converter at a temperature of 673 K. From the

compressor section it is assumed that the gases come out at 313 K. The inlet

stream is heated by the exit stream to the converter inlet temperature. The

converter exit gas is then cooled to 313 K at which temperature at the high

pressure of 50 MPa ammonia condenses.

Hr is 7.301x 103 kcal/kmol at 673 K

H is 2.73698x 107 kcal/hr

This heat produced is utilized in heating the exit gases to a final exit temperature,

found by doing a heat balance.

The exit temperature from the condenser found out by a trial and error process is

T = 778 K

Ammonia condenses at a temperature of 313 K at the given pressure and mole

fraction of 0.0952

Heat evolved by cooling gases from 778 K to 313 K

Heat needed to raise the temperature of synthesis gases from 313 K to 673 K is

hrkcalxdTCnHi

T

pii6

623

1067949.1==

OHCHHCO 2422 3 ++

hrkcalxdTCnHi

T

pii7

673

1073698.2==

hrkcalxdTCnHi

pii7

778

313

1059567.11==

hrkcalxdTCnHi

pii7

673

313

1045070.9==

Thus the excess heat, which is to be removed by the Heat Exchanger, is

H = 2.14497 x 107 kcal/hr

As the heat load is very high five HE in parallel are used. The inlet temperature

of the hot gases found by a trial and error process is 405 K. The exit temperature

is ammonia saturation temperature of 313 K.

Cooling water is assumed to enter at 293 K and exit at 308 K. The mass flow rate

of cooling water is calculated and found to be 79.4 kg/s per HE.

Condenser heat load may be calculated as

Where N is the latent heat of vaporization of ammonia at the given pressure = 4471.65 kcal/kmol

The condenser coolants used are refrigerants because water cannot be used to

cool the contents.

hrkcalxnQ NNC 71002284.2==

MATERIAL BALANCE

The naphtha has the following composition by weight

Hydrogen 16.2%

Carbon 83.0%

Sulphur 0.8%

Composition of the same naphtha by mole is

Hydrogen (H2) 53.85%

Carbon (C) 45.98%

Sulphur (S) 0.17%

Desulphurizer

Sulphur present is removed as Hydrogen Sulphide in the desulphurizer

Input

Naptha :

Hydrogen (H2) 2454.7 kmol/hr

Carbon (C) 1929.7 kmol/hr

Sulphur (S) 11.1 kmol/hr

Hydrogen gas stream(H2) 209.8 kmol/hr

Output

Desulphurized Naptha :



Hydrogen gas (H2) 198.1 kmol/hr

Hydrogen Sulphide (H2S) 11.1 kmol/hr

Primary Reformer

In the primary reformer the desulphurized naphtha is made to react with Steam

to yield Hydrogen, Carbon monoxide, Carbon dioxide, Methane.

To prevent the formation of carbon the steam to carbon ratio is maintained at 3.9

Input

Desulphurized Naptha:

Hydrogen (H2) 2454,7 kmol/hr


Hydrogen gas 198.1 kmol/hr

Steam (H2O) 7525.8 kmol/hr

Output


Carbon monoxide (CO) 964.8 kmol/hr

Carbon dioxide (CO2) 567.6 kmol/hr

Methane (CH4) 397.4 kmol/hr


Secondary Reformer

Input





Air:

Nitrogen (N2) 2148,4 kmol/hr

Oxygen (O2) 596.4 kmol/hr

Inerts (Ar +others) 9.5 kmol/hr

Water vapour (H2O) 428.4 kmol/hr

Steam (H2O) 5440 kmol/hr

Output Hydrogen (H2) 5418.4 kmol/hr







High Temperature Shift Converter

Input








Output








Low Temperature Shift Converter

Input








Output








Condenser

The exit stream from the Low temperature Converter is sent into a condenser to

remove the steam present which would be an additional excess load on the

Absorber

Within the condenser all the water is removed as condensate. Only water is

removed and the exit stream from this goes to the Absorber

Input








Output







Absorber

Within the absorber Carbon dioxide is absorbed using Monoethanolamine. It is

assumed that only Carbon dioxide is soluble in MEA solution. Solubility of other

gases in MEA is assumed to be negligible. It is further obtained from literature

that the CO2 content in the exit stream is 0.5% of the total gas stream.

Input







Output







Methanator

Within the Methanator Oxides of Carbon are made to react with Hydrogen in the

presence of catalysts because oxides of carbon act as poisons for the catalysts in

the Ammonia synthesis loop. CO and CO2 react with Hydrogen to form Methane

and steam. Steam is removed by condensation and separated from the gaseous

mixture going into the synthesis loop. Methane is an inert material in the

synthesis loop.

Input







Output






Ammonia Synthesis Loop

Make up Gas





Recycle Stream


Nitrogen (N2) 6182.2 kmol/hr

Ammonia (NH3) 774.9 kmol/hr

Inerts (Ar + CH4) 4555.0 kmol/hr

Gas into Converter



Ammonia (NH3) 774,9 kmol/hr


Purge





Storage





PLANT LAYOUT

SITE LAYOUT The location of the plant can have a turning effect on the overall viability of a process plant, and the scope for future expansion. Many factors must be considered when selecting a suitable plant site. The most important factors are as follows :

Location, with respect to the marketing area Raw material supply Transport facilities Availability of labor Availability of suitable land Environmental impact and effluent disposal Local community consideration Climate Political and strategic consideration

In addition to the main plant , we also have to consider the associated services which have to be amalgamated within a particular plant site. Canteens, parks, general utilies, emergency medical services and places for storage must also be taken into consideration while deciding on a particular site. PLANT LAYOUT The economic construction and operation of a process unit will depend on how well the plant equipment specified on the process flow sheet and laid out. The principal factors to be considered are:

1. Economic consideration: construction and operation cost. 2. The process requirement 3. Convenience of operation 4. convenience of maintenance 5. Safety 6. Future expansion

COSTS: The cost of construction can be minimized by adopting a layout that gives

shortest run of connecting pipes between equipment, and adopting the least

amount of structural steel work. However, this will not necessarily be the best

arrangement for operation and maintenance.

PROCESS REQUIREMENT: All the required equipments have to be placed properly within process. Even the installation of the auxiliaries should be done in such a way that it will occupy the least space. OPERATION Equipment that needs to have frequent operation should be located convenient to the control room. Valves, sample points, and instruments should be located at convenient position and height. Sufficient working space and headroom must be provided to allow easy access to equipment. MAINTENANCE Heat exchangers need to be sited so that the tube bundles can be easily withdrawn for cleaning and tube replacement. Vessels that require frequent replacement of catalyst or packing should be located on the outside of buildings. Equipment that requires dismantling for maintenance, such as compressors and large pumps, should be placed under cover. SAFETY Blast walls may be needed to isolate potentially hazardous equipment, and confine the effects of an explosion. At least two escape routes for operator must beprovided from each level in the process building.

ENVIRONMENTAL ASPECTS

In the manufacturing of ammonia, there arises a need to vent gases and to

remove condensates. These liquid and gaseous effluents and the chemicals used

in the process may tend to pollute the environment.

The main source of gaseous ammonia emission are from the inert gas purge and

from the ammonia storage section of the ammonia plant. In the case of non

functioning or breakdown of the equipment large quantity of ammonia emission

increases ammonia concentration in the atmosphere.

The liquid effluents contain dissolved ammonia and dissolved Carbon dioxide.

The toxic effect of ammonia depends upon the concentration of free ammonia. Its

presence in water leads to suffocation of aquatic life forms.

Also there is the possibility of emission of particulate matter like Carbon oxides

of Sulfur etc. These may removed by electrostatic precipitation or vacuum

filtration.

Pollution Control

This may be achieved by

1 . Segregation of effluent streams

2 . Control of particulate matter by

a) Mechanical separators

b) Wet Scrubbers

c) Fabric filters and electrostatic precipitators.

Ammonia may be removed by any of the following methods

Ammonia removal technique Concentration Steam stripping 3.4 wt %

Air stripping >1500 ppm Simple lagooning after PH adjustment 1500 ppm

Biological treatment About 700 ppm

1

COST ESTIMATION

Fixed capital investment for cost index of 300 = Rs 3.6108

Cost index for 2002 = 402

Therefore present fixed capital investment = 3.7253107(402/300) =Rs 482 400 000

Estimation of total investment cost: 1) Direct cost:

a) Purchased equipment cost:(15 40% of FCI ) Assume 30% of FCI =Rs 144 720 000 Installation cost:(35 45% of PEC) Assume 35% =Rs 50 652 000 c) Instrument and control installed:(6 30% of PEC) Assume 25% of PEC =Rs 36 180 000 d) Piping installation cost:(10 80% of PEC) Assume 60% =Rs.86 832 000 e) Electrical installation cost:(10 40% of PEC) Assume 35% of PEC =Rs 50 652 000 f) Building process and auxilliary:(10-70% of PEC) Assume 60% =Rs 86 832 000 g) Service facilities:(30-80% 0f PEC) Assume 50% =Rs 72 360 000 h) Yard improvement:(10-15% of PEC) Assume 10%

2

=Rs 14 472 000 i) Land:(4-8% of PEC) Assume 6% =Rs 8 683 200 Therefore direct cost =538 358 400 Indirect cost: Expenses which are not directly involved with material and labour of actual installation or complete

facility

a) Engineering and supervision:(5-30% of DC) Assume 25% =Rs 134 589 600 b)Construction expenses:(10% of DC) =Rs 53 835 840 c)Contractors fee:(2-7% 0f DC) Assume 6% =Rs 32 301 504 d)Contingency:(8-20% of DC) Assume 12% =Rs 64 603 008 Therefore total indirect cost =Rs 285 329 952 Fixed capital investment: Fixed capital investment(FCI) = DC+IC = Rs 823 688 352 Working capital investment: 10 20% of FCI Assume 16% =Rs 107 079 486

2) Total capital investment: = FCI + WC =Rs 930 767 838

3

Estimation of total product cost(TPC): Fixed charges:

a) Depreciation:(10% of FCI for machinery) =Rs 82 368 835

b) Local taxes:(3-4% of TPC) Assume 3% =Rs 42 625 872

c) Insurances:(0.4-1% of FCI) Assume 0.7% =Rs 5 765 818 d)Rent:(8-12% of FCI) Assume 10% =Rs 82 368 835 Therefore total fixed charges = 213 129 360 But, Fixed charges = (10-20% of TPC) Assume 15% Therefore Total product cost = 1 420 862 400 Direct production:

a) Raw material:(10-50% 0f TPC) Assume 40% =Rs 568 344 960 b)Operating labour(OL):(10-20% of TPC) Assume 15% =Rs 213 129 360 c)Direct supervisory and electric labour:(10-25% of OL) Assume 20% =Rs 42 625 872

b) Utilities:(10-20% of TPC) Assume 15% =Rs 213 129 360

4

c) Maintainence:(2-10% of FCI) Assume 8% = Rs 65 895 068

d) Operating supplies (OS):(10-20% of maintainence) Assume 15% =Rs 9 884 260

e) Laboratory charges:(10-20% of OL) Assume 12% =Rs 7 907 408

f) Patent and royalties:(2-6% of TPC) Assume 4%

=Rs 56 834 496 Plant overhead cost: 50-70% of (OL+OS+M) Assume 65% =Rs 187 790 647 General expenses:

a) Administration cost:(40-60% of OL) Assume 50% =Rs 106 564 680

b) Distribution and selling price:(2-30% of TPC) Assume 20% =Rs 284 172 480

c) Research and development cost:(3% of TPC) =Rs 42 625 872 Therefore general expenses(GE) =Rs 433 363 032 Therefore manufaacturing cost(MC)= Product cost+fixed chages+Plant overhead expenses =Rs1 821 782 407

Total production cost: Total production cost =MC + GE =Rs 2 255 145 439

5

Gross earnings and rate of return: The plant is working for say 320 days a year Selling price =Rs. 35 /kg Total income =1500320100035 = Rs1.68 x 1010

Gross income =Total income total product cost =Rs 1.5379 x 1010 Tax =50% Net profit =Rs7 689 568 800 Rate of return =net profit/total capital investment =82.615 %

Ammonia_IntroductionAmmonia_Properties&usesAmmonia_Methods-2520of-2520ProductionAmmonia_Design-2520of-2520EquipmentsAmmonia_Energy-2520BalanceAmmonia_Material-2520BalanceAmmonia_Plant-2520Location&LayoutAmmonia_Cost-2520Estimation&Economics

Documents

Ammonia