INTRODUCTION

In every hydrocarbon industries off-gases are commonly emitted to the environment viz CO2,

H2S, SO2. Each gas has its own disadvantages. Out of which we have selected the thiopaq process for treating H2S which s patented by Shell Global Solution in collaboration with Paque. Currently H2S removal from natural gas with subsequent sulphur recovery is mainly performed by amine & clausis plant . In small industries generally hydrogen sulphde is treated by liquid-redox reaction process & followed by incineration or injection of the acid gas in empty well. All these treatment having some weakness. So shell-paque THIOPAQ process is a biotechnological process for recovery of sulphur from gaseous stream by absorption into a mild alkaline solution Followed by the oxidation of absorbed sulphide to elemental sulphur by naturally occurring Bacteria known as THIOBACILIUS. These process as following advantages :

Minimal chemical consumption. High turndown ratio. Gas treatment as well as sulphur recovery. H2S below 4 PPM can be guaranteed. Essentially 99.9 % of conversion. No replacement of catalyst.

This process has been commercialized in many petrochemical industries along with Pulp & Mining industries.

In this project my plant capacity for crude natural gas in 2100,000 M3/day. Out of which 13 ton/day of sulphur is recovered .we also covered the Block diagram, process description, process flow diagram, Energy balance & material balance across absorber. Piping & instrumentation diagram of Absorber including line sizing & Equipment list.

pg. 1

BLOCK DIAGRAM

pg. 2

BLOCK DIAGRAM:

Sweet Gases NaOH + Air + Water

NaOH Solution

H2S

Block Diagram of Thiopaq Process

Elemental sulphur recovered

pg. 3

Absorption Tower Bio Reactor De-Watering

Screw Convertor

PROCESS DESCRIPTION

pg. 4

Process Description

A simplified process flow scheme of the Paques unit is included on the next page.

Sour natural gas is routed via a feed gas knockout vessel and a glycol heater to the absorber.

Even though the Shell-Paques process is an exothermic process, the latent heat content of the

feed gas was so low that heating duty needed to be supplied for this project rather than cooling

duty. Application of the heater on the feed gas flow has one major advantage; it allowed the feed

gas temperature to be controlled in relation to the solvent temperature, i.e. hydrocarbon

condensation control was achieved. The sour natural gas is washed in the absorber counter-

currently. This absorber has three packed beds with 2-inch Pall rings. A total draw-off tray

combined with a liquid redistribution tray in-between the packed beds ensures proper liquid

redistribution. Treated natural gas leaves the top of the absorber and is routed to the sales gas

grid via a knockout drum.

The loaded THIOPAQ solvent is routed from the absorber to the horizontal flash

vessel. The flash gas, which contains some H2S, is washed in the small column on top of the

flash vessel. The treated flash gas is routed to flare.

Flashed, loaded THIOPAQ solvent is then sent to the bioreactor. A blower supplies air

to a distributor in the bottom section of this tank. Part of the oxygen is consumed in reactions

with sulfide and sulfur. The distribution of air also ensures that the content of the bioreactor is

continuously mixed.

Regenerated THIOPAQ solvent is recycled to the main absorber, the flash absorber, the solids

separator and to the top of the bioreactor. The last feature is mainly as a pro-active measure to

prevent foaming. The bacteria catalyze the regeneration reactions 5 and 6, the reaction of sulfide

with oxygen to give either elemental sulfur or sulfate. The reaction that gives elemental sulfur is

very much preferred because:

· It produces a hydroxyl molecule, which can capture a molecule of H2S in the absorber.

· It produces elemental sulfur, which can be separated fairly easily from the solvent.

pg. 5

We have to accept a small percentage of over-oxidation however since we need to convert all

sulfides in the bioreactor.

The sulfate production necessitates a continuous bleed from the unit. The bleed

stream is taken from a separate calm section of the bioreactor, to minimize the sulfur content.

The clarified bleed stream is subsequently aerated. Some bacteria will still be present, they will

ensure that all sulfur and sulfide is completely oxidized to sulfate for which the biological

oxygen demand is zero. Local legislation required sterilization of the bleed stream; therefore a

UV source has been installed after the aeration tank. Bleed water is collected and transported by

truck to a nearby Water treating facility. The vent air leaving the bioreactor can contain a little

H2S (typically less than 4 ppmv) and is therefore routed to a bio-polisher. This bio-polisher is a

tank with a layer of compost; the vent air contains sufficient water vapor to keep this compost

filter humid. The vent air is discharged directly into atmosphere.

The regenerated solvent leaving the bioreactor typically contains 10 kg sulfur per cubic meter, or

roughly one weight percent. The solvent is concentrated into sulfur slurry with around 10 % wt.

sulfur content in the solids separator, the clarified water phase is returned to the bioreactor. This

sulfur slurry is further concentrated into a sulfur paste, with around 65 % wt sulfur content, in a

decanter-centrifuge. The sulfur paste produced at the Paques unit is routed to Landfill.

pg. 6

pg. 7

MATERIAL BALANCE ACROSS ABSORBER

pg. 8

By ideal gas behavior we can calculate the kmoles of natural gas

PV=NRT

N=PV/RT

=12*200000/.08206*283

=103345.91 kmoles/day

Basis: 103345.91 kmoles/day of natural gas

Methane in feed=0.84429*103345.91

=87253.91 kmoles/day

Co2 in feed=0.02506*103345.91

=2589.84 kmoles/day

H2s in feed=0.00202*103345.91

=208.75 kmoles/day

Other gas=5.16+63+4249+4586.49+2109.29+512.59+890.84+341.04+290.40+218.05+120.91

=13386.77 kmoles

Reactions:-

H2S+NaOHNaHS+H2O----- 1

CO2+NaOHNaHCO3---------2

NaOH+NaHCO3Na2CO3+H2O-----3

H2S+Na2CO3NaHS+NaHCO3-----4

According to stoichiometry of reaction---1

1kmoles of H2S =1kmole of NaOH

208.75 kmoles of H2S=?

pg. 9

NaOH required=2589.84 kmoles/day.

Total NaOH required=208.75+2589.84 kmoles/day

Assuming 50%excess NaOH is supplied

= 2798.59*1.5

= 4197.88 kmoles/day

Assuming 50% concentration NaOH is used

=4197.88/0.5

=8395.76 kmoles/Day

Amount of water

=0.5*8395.76

=4197.88 kmoles/Day

70% conversion of H2S with NaOH

=.7*208.75

=146.12 kmoles/Day

According to Stoichiometry of reaction 1

NaOH reacted=146.12 kmoles/Day

NaHS produced=146.12 kmoles/Day

H2O produced = 146.12 kmoles/Day

72% conversion of CO2 with NaOH

=.72*2589.84

=1864.68 kmoles/Day

According to stoichiometry of reaction 2

1kmoles of CO2= 1kmoles of NaHCO3

1864.68 kmoles of co2=?

NaHCO3 produced=1864.84 kmoles/Day

pg. 10

NaOH Reacted = 1864.48 kmoles/day

NaOH Unreacted=839.76-(146.12+1864.68)

= 6384.96 kmoles/Day

Comparing the values NaOH Ureacted and NaHCO3 produced it is found that limiting component is NaHCO3

60% conversion of NaOH with NaHCO3

NaHCO3 reacted =10.6*1864.68

=1118.80 kmoles/Day

Na2CO3 Produced =1118.80 kmoles/Day

According of stoichiometry of reaction 3

1kmoles of NaHCO3= 1 kmoles of NaOH

1118.86 kmoles of naHCO3=?

NaOH required =1118.86kmoles/Day

H2O produced =1118.86 kmoles/Day

Total NaOH Unreacted =5266.1 kmoles/Day

60 % conversion of H2S and Na2CO3

According of stoichiometry of reaction 4

Na2CO3 reacted =.6*62.62

=37.57 kmoles/Day

H2S unreacted =25.028 kmoles/Day

NaHCO3 produced=37.17 kmoles/Da

NaHS produced=37.17kmoles/Day

Co2 leaving absorber =0.28*2589.84

=725.15 kmoles/Day

pg. 11

INPUT DATA

SR. NO. COMPONENT Kmoles/day Kg/hr1 H2 5.16 10.322 He 63 1263 N2 4249 594864 Co2 2589.84 113952.965 H2S 208.75 7097.56 C1 87253.91 1396062.567 C2 4586.49 137594.78 C3 2109.29 92808.769 i-C4 512.59 29730.2210 n-C4 890.84 51668.7211 i-C5 341.04 24554.8812 n-C5 290.40 20908.813 C6 218.05 18752.314 C7 120.91 1209115 NaOH 8395.76 335830.416 H2O 4197.88 75561.84

116032.91 2376236.6

OUTPUT OF GASES

SR. NO. COMPONENT Kmoles/day Kg/hr1 H2 5.16 10.322 He 63 1263 N2 4249 594864 Co2 725.17 31905.1485 H2S 25.048 85.66 C1 87253.91 1396062.567 C2 4586.49 137594.78 C3 2109.29 92808.769 i-C4 512.54 29730.2210 n-C4 890.84 51668.7211 i-C5 341.04 24554.8812 n-C5 290.40 20908.813 C6 218.05 18752.314 C7 120.91 12091

101390.8 1876551.02

pg. 12

OUTPUT OF LIQUIDS

SR. NO. COMPONENT Kmoles/day Kg/day1 NaOH 5266.1 21064.42 NaHS 183.69 10286.643 Na2CO3 1081.29 114616.744 NaHCO3 783.31 65798.045 H2O 5462.86 98331.48

12067.31 499676.9

pg. 13

ENERGY BALANCE ACROSS ABSORBER

pg. 14

INPUT DATA:-

SR. NO. COMPONENT Kmoles/day Kg/hr1 H2 5.16 10.322 He 63 1263 N2 4249 594864 Co2 2589.84 113952.965 H2S 208.75 7097.56 C1 87253.91 1396062.567 C2 4586.49 137594.78 C3 2109.29 92808.769 i-C4 512.59 29730.2210 n-C4 890.84 51668.7211 i-C5 341.04 24554.8812 n-C5 290.40 20908.813 C6 218.05 18752.314 C7 120.91 1209115 NaOH 8395.76 335830.416 H2O 4197.88 75561.84

116032.91 2376236.6

OUTPUT DATA:-

SR. NO. COMPONENT Kmoles/day Kg/hr1 H2 5.16 10.322 He 63 1263 N2 4249 594864 Co2 725.17 31905.1485 H2S 25.048 85.66 C1 87253.91 1396062.567 C2 4586.49 137594.78 C3 2109.29 92808.769 i-C4 512.54 29730.2210 n-C4 890.84 51668.7211 i-C5 341.04 24554.8812 n-C5 290.40 20908.813 C6 218.05 18752.314 C7 120.91 12091

101390.8 1876551.02

pg. 15

TOTAL LEAVING THE ABSORBER:-

SR. NO. COMPONENT Kmoles/day Kg/day1 NaOH 5266.1 21064.42 NaHS 183.69 10286.643 Na2CO3 1081.29 114616.744 NaHCO3 783.31 65798.045 H2O 5462.86 98331.48

12067.31 499676.9

(Heat with gas feed)+(Heat with NaOH feed)+(Heat of reaction) = (Heat in Vent stream) +

(Heat in Bottom stream)

Heat In gas feed :-

Q=m⋋ = 103345.91 × 320.156 = 33086813.16 kcal/hr. Heat in NaOH feed:-

Q=mCp∆T = 12593.64 × 0.26 × 40 = 130973.85 kcal/hr.

Heat of Reaction:-

(-13×146.12)+ (-30×1864.68) + (-10×1118.86) + (-12×37.57)

= -69479.4 kcal/hr.

Heat in Vent stream:-Q=m⋋ = 101390.8 × 326.376 = 33091523.74 kcal/hr.

pg. 16

Heat in Bottom:-

Q=mCp∆T =12067 × 0.47 × (50-40) = 56714.9 Kcal/hr.

(33086813.16) + (130973.85) – (69479.4) = (33091523.74) + (56714.9) 33148307.61 = 33148238.6

pg. 17

PIPING & INSTRUMENTATON DIAGRAM OF ABSORBER

pg. 18

pg. 19

LINE SIZING

pg. 20

FOR GAS FEED LINE :

Q = AV

Putting the values

(2.314) = π/4*di2 * 18

= 0.404 m

= 16 inches

Calculate the Reynolds number

Nre = DVρ\µ

= (0.404 * 18*11.89)\(0.01)

= 8646.40

To calculate the friction factor

F = (0.25)\(log(ε/(3.7D) + (5.74\(Nre^0.9)))2

Where ε = 150 mm for plastic (roughness factor)

Putting the values

F = 0.0328

Calculate the pressure drop

ΔP = FLV2ρ\2D * 10-5

Where L = length of pipe line

Putting the values

ΔP = 0.15 bar/100 m

Pressure drop within th given range hence it is corrected.

FOR NaOH FEED:

Q = AV

(4.19* 10-3) = ((π/4)*D2 )* 2

D = 0.05063 m

pg. 21

D = 2 inches

FOR VENT GAS LINE SIZE

Q = AV

(1.989) = ((π/4)*D2 )* 16

D = 0.411 m

= 16 inches

Calculate the Reynolds number

Nre = DVρ\µ

= (0.411 * 16*11.73)\(0.01)

= 7713.64

To calculate the friction factor

F = (0.25)\(log(ε/(3.7D) + (5.74\(Nre^0.9)))2

Where ε = 150 mm for plastic (roughness factor)

Putting the values

F = 0.251

Calculate the pressure drop

ΔP = FLV2ρ\2D * 10-5

Where L = length of pipe line

Putting the values

ΔP = 0.4 bar/100 m

Pressure drop within the given range hence it is corrected.

pg. 22

FOR BOTTOM LIQUIDS:

Q = AV

(8.22* 10-3) = ((π/4)*D2)* 5

D = 0.0457 m

D = 2 inches.

pg. 23

DESIGN OF ABSORBER

pg. 24

The removal of one or more component from gases by using suitable solvent is known as absorption which is on of the most important process in mass transfer. The purpose of absorbing gases are as followed.

For the separation of component having economic value. For removal of undesired compounds.

TYPES OF ABSORBER

Packed column. Plate column.

COMPARISON BETWEEN PACKED & PLATE COLUMN

Packed columns

small-diameter columns (less than 0.6m)

more choices in materials of construction for packing’s especially in corrosive service (e.g. plastic, ceramic, metal alloys)

lower pressure drop (important in vacuum distillation) Less liquid entrainment · low liquid hold-up, especially suitable for thermally sensitive

material.

Plate columns

variable liquid and/or vapour loads low liquid rates · large number of stages and/or diameter high liquid residence time dirty service (plate columns are easier to clean) Presence of thermal or mechanical stress due to large temperature changes which might

lead to cracked packing’s.

Packing’s

The packing are divided in those types which are dumped at random in to the tower & these must be stacked by hand. Dumped packed consisting of unit of ¼ to 2 inches in major dimension & are used in roost small Columns.

The principal requirements of tower packing are as follows :

It must be chemically inert to the fluids in the tower. It must be strong without excessive weight. It must contain adequate passages for the contacting streams without excessive pressure

drop. It must provide good contact between the contacting phases.

pg. 25

It should be reasonable in cost.

Common packing used are:

Berl Saddle. Intalox Saddle. Rasching rings. Pall rings.

DESIGNING OF ABSORBER

Compositions of components in a gas mixture Entrance

Components Kmoles/day Kg/dayH2 5.16 10.32He 63 126N2 4249 59486CO2 2589.84 113952.96H2S 208.75 7097.5C1 87253.91 1396062.56C2 4586.49 137594.7C3 2109.29 92808.76i-C4 512.59 29730.22n-C4 890.84 51668.72i-C5 341.04 24554.88n-C5 290.40 20908.8C6 218.05 18752.3C7 120.91 12091

Total 1033345.91 1964847.34

Compositions of components in a liquid mixture Entrance

components Kmoles/day Kg/dayNaoh 8395.76 335830.4H2O 4197.88 75561.84Total 12593.64 411392.24

pg. 26

Compositions of components gas mixture at Exit

Components Kmole/day Kg/dayH2 5.16 10.32He 63 126N2 4249 59486CO2 725.117 31905.148H2S 25.048 851.632C1 87253.91 1396062.56C2 4586.49 137594.7C3 2109.29 92808.76i-C4 512.59 29730.22n-C4 890.84 51668.72i-C5 341.04 24554.88n-C5 290.40 20908.8C6 218.05 18752.3C7 120.91 12091Total 101390.8 1876551.02

Compositions of components Liquid mixture at Exit

Components Kmoles/day Kg/dayNaOH 5266.1 210644NaHS 183.69 10286.08Na2CO3 1081.29 114616.74NaHCO3 783.31 65798.04H2O 5462.86 98331.48Total 12777.25 614293.08

Flow rate of liquid entering the absorber: 10.54 kg/s

Flow rate of gas entering the absorber: 22.74 kg/s

Density of gas: 11.89 kg/m3

Density of liquid at 40°C: 2840 Kg/m3

FOR FINDING COLUMN DIAMETER

By using the formula

Flv = (Lw\Vw)*( ρv/ρl)0.5

pg. 27

Flv = 0.03

From fig 1 at 42mm H20/ m 0f packing the ordinate found to be

K4 = 1.75

At flooding K4 = 4.5

% Flooding = (1.75/4.5)*100

= 63 %

Vw * = (K4ρv (ρl-ρv)/13.1Fp (µl\ρl)0.1)0.5

= (1.75 * 11.89 (2840-11.89) \ (13.1 * 82 *(0.0023/2840)0.1)0.5

= 13.31 kg/m2s

Column area required = 22.74\13.31

= 1.708 m2

Diameter of column = ((4\π)*1.708)0.5

= 1.47 m2

= 1.5 m2

Column area required = ((π/4)*(1.5)2 )

= 1.767 m2

Packing to column diameter ratio = (1.5\0.005)

= 30

Percentage of flooding at selected diameter = 63 *(1.55/1.767)

= 56 %

Could consider reduced diameter.

pg. 28

TO FIND THE HEIGHT OF COLUMN

As the concentration of solute is very small therefore the flow rate of gas & liquid will be constant throughout the column & the operating line as well as equilibrium curve for the system is a straight line. According to the solubility data of H2S & CO2 with NaOH is a straight line with slope 0.203 at 12 bar & 313 K.

Y*= 0.203X

M=0.203 & Gm = 1.17 kmoles/s

M = slope

Gm = Gas flow rate

Partial pressure of H2S & CO2 Mixture at inlet = (0.02708*760) = 20.58 mm Hg

Partial pressure of H2S & CO2 Mixture at outlet = ((0.0075*20.59)) = 0.1544 mm Hg

Y1\Y2 = P1\P2 = 20.58\0.1544 = 136.

From fig 2 the values obtained are:

MGm/Lm 0.6 0.7 0.75 0.8NOG 9.6 12 13.8 16.2

Generally the optimum value taken in 0.7.

((0.203*1.17)\Lm)) = 0.7

Lm = 0.34

Taking material balance across absorber

G (Y1-Y2) = L(X1-X2)

Assume X2 = 0

1.17(Y1-0.0027) = 0.34(X1-0)

Y1 = 0.3 X1+ 0.0027

X1 = 3.44(Y1-0.0027)

pg. 29

Assuming the value of Y then plot a graph of X v/s Y & Y* v/s X

Y X Y*0.01 0.032 0.0006460.02 0.0761 0.01540.03 0.1212 0.0240.04 0.165 0.03350.05 0.211 0.0430.06 0.254 0.05150.07 0.298 0.0600.08 0.340 0.00690.09 0.384 0.0780.1 0.428 0.087

Then plotting the graph the value of NoG = 12

For two inches plastic pall ring HG=0.43 & HL = 0.5

HoG = HG + (MGm/Lm) * HL

= 0.43 + (0.7)*0.5

= 0.779

Z = HoG * HoG

12 * 0.779

= 9.35 m

Allowance for liquid distribution = 1.2 m

Allowance for gas distribution = 1.2m

Total height of the column = 9.35 + 1.2 + 1.2

= 11.75 m

= 12 m

pg. 30

CALCULATION OF PRESSURE DROP

This bed should in two section threby requiring one intermediate combing support & Redistrbution plate. It require one bottom support plate.

Pressure drop due to one redistributor plate & one Bottom plate = 2 mm of H2O

Total pressure drop = 9.35*0.5 + 2

= 6.67 Kpa

= 0.0667 bar

Calculation of pressure drop at flooding region

ΔP = 3mm H2O / m Packing

= 3* 9.35 + 2

= 30.05 Kpa

= 0.30 bar

CALCULATION FOR LIQUID HOLDUP

Htw = 0.0004(L/Dp)0.6

L = liquid flow rate in lb/hrft2

Dp = size of packing in Ft.

Htw = 0.0004(7771.5/0.164)0.6

= 0.255 ft3 of water/ft3 of volume

pg. 31

pg. 32

EQUIPMENT LIST

Equipments Quantity

KNOCKOUT DRUMS 2

PUMPS 5

COMPRESSORS 2

BIOREACTORS 1

pg. 33

BLEED WATER TANK 1

SEPERATOR 1

SLURRY HEATER 1

DECANTER 1

HEATER 1

FLASH DRUM 1

STORAGE TANKS 5

pg. 34

CONCLUSION

Removal of hydrogen sulfide from industrial gases is required for reasons ofEnvironment, health, safety and corrosion free operations. For this reason, the economic viability of H2S removal technologies is based on the minimization of capital and operating cost. This process recover 99.9% of sulphur from crude natural gas.as this

pg. 35

process is cheap as compared to any other sulphur recovery processes. The bacteria available are also free no need to purchase it. Also there is no need to replace the bacteria.No need for catalyst.

Sulfur is a major nutrient which ranks fifth or sixth in quantity of macronutrients taken up by plants; this is comparable to the demand for phosphorus. Plant leaves can take up sulfur from the atmosphere as very reduced (COS, CS2 and H2S) up to highly oxidized compounds from the atmosphere as very reduced (COS, CS2 and H2S) up to highly oxidized compounds (SO2). However, most of the sulfur is taken up by plant roots as water-soluble sulfate. Because of the adoption of pollution control measures in industrial countries the decrease in SO2 emissions between 1980 and 1987 varied from 22% in Poland to 64% in France. Several countries switched from coal-fired to gas-fired industries at the end of the 1960’s.The improvement in air quality was beneficial to natural ecosystems, but from the late 1980’s onwards, the decreased sulfur supply resulted in a widespread S deficiency in the soils used for cultivation of several highly S-demanding crops, such as oilseed rape and cereals in Denmark, England, F.R.G. and Scotland.

pg. 36

Fig no : 1

pg. 37

Fig No : 2

pg. 38

Table

pg. 39