Reliance Industries Limited-Final Report

INTRODUCTION

This Project is about the study of the Booster – Compressor loop in the Platformer Section of the Paraxylene plant which is installed to optimize the purity of Hydrogen coming from the Product Separators. This is done to use Hydrogen in the Down – Stream process sections like Isomar , NHT , Tatoray and CCR .The purity of hydrogen can be increased upto 90 %. Also the project involves the Design of Two Re- Contact Drums where the Re-cycle feed (Liquid Level) has considerably come down than earlier because of modifications in the process.

This project will give the company an overall view of this loop.The plant is running well about its installed capacity (140 -160 %).The company wants to retain the purity level of nearly 90 % for Hydrogen even after operating above installed capacity. Hence the current project will give the company an idea of the current hydrogen quality (purity) which is obtained & give an idea about changes (if any) to be made to the existing process set up. Also the Re-designing of Re-Contact drums is carried out as the currently they are overdesigned as they are designed according to the earlier process requirements.

1

COMPANY INFORMATION

Reliance Industries Limited, founded by Dhirubhai H. Ambani , is India's largest private sector enterprise, with businesses in the energy and materials value chain. Group's annual revenues are in excess of US $ 44 billion. The flagship company RIL is a Fortune Global 500 company and is the largest private sector company in India.

Backward vertical integration has been the cornerstone of the evolution and growth of Reliance. Starting with textiles in the late seventies, Reliance pursued a strategy of backward vertical integration - in polyester, fibre intermediates, plastics, petrochemicals, petroleum refining and oil and gas exploration and production - to be fully integrated along the materials and energy value chain.

The Group's activities span exploration and production of oil and gas, petroleum refining and marketing, petrochemicals (polyester, fibre intermediates, plastics and chemicals), textiles, retail and special economic zones.

Reliance enjoys global leadership in its businesses, being the largest polyester yarn and fibre producer in the world and among the top five to ten producers in the world in major petrochemical products.

Major Group Companies are Reliance Industries Limited (including main subsidiary Reliance Retail Limited) and Reliance Industrial Infrastructure Limited.

2

RIL, Patalganga

Fibre Intermediates Production in KT (For Whole of Reliance)

Product FY 2007-08 FY 2008-09PX 1878 1879PTA 2035 1978Total 3913 3857

3

Naphtha

Paraxylene

Purified Terephthalic Acid (PTA)

Polyester Staple Fibre(PSF)

Poleyester Filament Yarn

(PFY)

RIL TURNOVER FOR THE LAST 5 YEARS.

FY 2004-05 FY 2005-06 FY 2006-07 FY 2007-08 FY 2008-090

20,000

40,000

60,000

80,000

100,000

120,000

140,000

160,000

4

NET PROFIT AFTER TAX (Rs. Million)

FY 2004-05 FY 2005-06 FY 2006-07 FY 2007-08 FY 2008-090

20,000

40,000

60,000

80,000

100,000

120,000

140,000

160,000

Series 2

NET WORTH IN (Rs. Crore)

FY 2004-05 FY 2005-06 FY 2006-07 FY 2007-08 FY 2008-090

20,000

40,000

60,000

80,000

100,000

120,000

140,000

Series 1

5

Process Description:

The entire process consists of a two-stage Double acting reciprocating Compressor, Two separators with De-mister pads which are referred to as Re-Contact Drums and Two Heat- Exchangers.

Two-stage Double acting Reciprocating compressor:

A reciprocating compressor or piston compressor is a positive-displacement

compressor that uses pistons driven by a crankshaft to deliver gases at high pressure.

The intake gas enters the suction manifold, then flows into the compression cylinder

where it gets compressed by a piston driven in a reciprocating motion via a crankshaft,

and is then discharged. We can categorize reciprocating compressors into many types

and for many applications.

The reciprocating air compressor is said to be double acting when the compression is accomplished using only both the sides of the piston.

Two-stage compressors have a low-pressure stage and a high-pressure stage. The air from the first stage is compressed again by the second stage, which gives more pressure with less stress on the unit. It's easier on the compressor to take air from 2 atmospheres to 4 than it is to go from 1 atmosphere to 4.

6

http://en.wikipedia.org/wiki/Crankshaft

http://en.wikipedia.org/wiki/Piston

http://en.wikipedia.org/wiki/Gas_compressor

http://en.wikipedia.org/wiki/Gas_compressor

http://en.wikipedia.org/wiki/File:ReciprocatingCompressor.jpg

Re-Contact Drum.

A Re-contact Drum here is a vapor –liquid separator which is a vertical vessel used in

several industrial applications to separate a vapor-liquid mixture. Gravity causes the

liquid to settle to the bottom of the vessel, where it is withdrawn.The vapor travels

upward at a design velocity through a De-Mister pad which minimizes the

entrainment of any liquid droplets in the vapor as it exits the top of the vessel

A vapor-liquid separator may also be referred to as a knock-out drum, knock-out pot or

compressor suction drum.

Demister Pad:

A demister, is a device often fitted to vapor liquid separator vessels to enhance the

removal of liquid droplets entrained in a vapor stream. Demisters may be a mesh type

or vane pack or other structure intended to aggregate the mist into droplets that are

heavy enough to separate from the vapor stream.

7

http://en.wikipedia.org/wiki/Entrainment_(engineering)

http://en.wikipedia.org/wiki/Entrainment_(engineering)

Demisters can reduce the residence time required to separate a given liquid droplet size

thereby reducing the volume and associated cost of separator equipment.

Demisters are often used where vapor quality is important in regard to entrained liquids

particularly where separator equipment costs are high (eg. high pressure systems) or

where space or weight savings are advantageous.

What is mist ?

Mist is the fine droplets of liquids of various sizes. They may vary from 0.5 micron to few 100 of microns in size. For example, good spray system generally generate 20 - 1000 micron size droplets, while columns & tray do have 8 - 100 micron size drops.

The separation of entrained liquid droplets from a vapor / gas stream is called mist elimination & the equipment used for the purpose is called De-mister.

How Does it Work?

The following Diagram Demostrated the working mechanism of Demister pad.

Phase 3:Rising vapour free of entrainment

Phase 2: Vapour passes through the demister where the droplets are retainedon the wire surface. They join up and become larger and start to free fall.

8

http://en.wikipedia.org/wiki/Vapor_quality

Phase 1: A vapour breaking free from aliquid has very small drops of liquid"trapped" with it.

Process Description.

Separator & Booster Compressor:

1. Outlet gas from Cold Product Separator (CPS) goes to suction of 1st booster compressor C-3072/73 A/B .

2. Discharge gas from 1st stage is about 17 bar a.3. Due to the compression ,the relatively heavier compounds in the gas stream condense

to form liquid droplets but the temperature of gases at the discharge increases upto 132⁰ C and hence it’s cooled in a Heat Exchanger E-3013 and sent to 1st Re-Contact Drum F-3047 which is provided with a mesh blanket to prevent Liquid entrainment in the gas stream .

4. The gas stream from F-3047 goes to 2nd stage of Booster compressor where it is compressed to about 34 bar a pressure.

5. Again due to increase in pressure of the gas the relatively heavier compounds in the gas stream condense to form liquid droplets but its temperature increases to about 120⁰ C and hence its cooled in a Heat – Exchanger E-3024 and sent to a 2nd Re-contact drum F-3044 which is also provided with a mesh blanket to prevent liquid entrainment in the gas stream flowing out from the top of the Re-contact drum.

6. The separation into Gas and Liquid takes place inside the Re-contact drum due density differences i.e. the liquids being more denser are discharged from the bottom of the Re-contact Drums.

7. The liquid from 1st Re-contact drum viz. F – 3047 is sent to De-Heptanizer & that from 2nd

Re-contact drum F-3044 is sent as a recycle-feed to the 1st stage discharge of compressor C-3072/73.

8. Also the liquid from Cold Product Separator (CPS) is sent as a re-cycle feed to the 2nd stage discharge of compressor C-3072/73

9. The purpose of sending these liquids as re-cycle feed are as follows:a. It helps in pre-cooling the 1st and 2nd stage discharge to about 60 ⁰ C thus helps in

decreasing heat Exchanger loads.b. It helps in increasing the discharge of liquid from both the re-contact drums thereby

improving the purity of the gas stream. (i.e more Hydrogen).

9

Performance Evaluation of compressor C-3072/73

C-3072/73 Compressor is a Two stage Double acting reciprocating compressor.

C-3072 Reciprocating Compressor Calculations

Pc : Critical Pressure.

Tc : Critical Temperature.

Cp : Specific Heat.

MW : Molecular Weight.

P suction : Suction Pressure.

T suction: Suction Temperature.

Tr : Reference Temperature.

Pr : Reference Pressure.

Ƞ : Volumetric Efficiency.

Tad : Adiabatic Temperature.

1. First Stage Calculations

The following are the 1st stage conditions for our calculations

Suction Temp = 47.67 ⁰C = 320.67 0 KSuction Pressure = 6.6 Bar aDischarge Temperature = 132.32 = 405.32 ⁰ KDischarge Pressure = 16.7 Bar a

Thus,

10

MW mix = Average Molecular Weight = 8.25 kg/kmol

Pc of Mixture = 17.18 Bar a

Tc of Mixture = 78.898 ⁰ K

Cp of Mixture = 36.8665 (refer fig. 8)

At Suction Condition

Pr = P suction/Pc = 6.6 / 17.18 = 0.3841

Tr = T suction/Tc = 320.67 / 78.898 = 4.064

From Nelson - Obert Generalized Compressibility Charts

For Pr = 0.3841 & Tr = 4.064

Suction Compressibility = 0.99 (refer fig. 5)

At Discharge Conditions

Pr = P discharge/Pc = 16.7/17.18 = 0.972

Tr = T discharge/ Tc = 405.32/78.898 =5.137

For Pr = 0.972 & T r = 5.137

Discharge Compressibility is = 1.02 (Refer Fig. 5)

Therefore,

K-value of Mixture = Cp mix/C v mix = Cp mix/ (Cp mix - R)

= 39.4562 / (39.4562-8.314)

= 1.25

Average Compressibility = Z avg. = (Z1+Z2)/2 = (1.01+0.98) = 0.995

11

PR = Pressure Ratio = Pdischarge / Psuction

= 16.7/6.6

= 2.53

Volumetric Efficiency:

It is Defined as the actual gas delivered to the piston displacement of the compressor.

Higher the discharge pressure , higher is the quantity that is left back in space between piston and the cyliner end walls.

Thus volumetric efficiency of the compressor can be estimated as per the following relation provided by the vendor at RIL,

Ƞ = 0.97 -C x [PR ^ (1/K) - 1] - L

= 0.97 - 0.1064[( 2.53)^(1/1.25) - 1] - 0.07

= 0.97 – 0.1064(2.1013 -1) – 0.07

= 0.97 – 0.1171 – 0.07

= 0.7828

Where,

C = Clearance in % for first stage = 10.64 % (from design data)

PR = Pressure Ratio = 16.7/6.6 = 2.53

k=Ratio of specific heats i.e. k value = 1.25

L = Leakage losses normally 0.03 to 0.05 for Lubricated compressor and 0.07 to 0.1 for non- lubricated compressor.

C-3072/73 is NON-LUBRICATED COMPRESSOR . Therefore L= 0.07

12

Thus ,

Volumetric Efficiency is

Ƞ = 0.97 -C x [PR ^ (1/K) - 1] - Lc = 78.28 %

Piston Displacement

Piston displacement is defined as the net volume actually displaced by the piston at rated machine speed, as the piston travels its length from top dead center to bottom dead center.

For Double acting Reciprocating Compressor C-3072/73 ;

Bore (D) In m

Stroke (S) In m

Diameter of Rod (d) In m

1st Stage 0.40589 0.305 0.063552nd Stage 0.2159 0.305 0.06355

Piston displacement = Vd (m3/hr) = (∏/4) x (D2 + (D2 - d2)) x S x N x 60

Where,

d= diameter of the rod =2.5 inches= 0.06355 m

D= Cylindrical Bore Diameter = 15.98 inches = 0.40589 m

S= Cylindrical Sroke = 12 inches = 0.305 m

N= Speed of Compressor = 390

Therefore,

Piston Displacement = (3.14/4) x ( 0.40592 + (0.40592 – 0.063552)) x 0.305 x 390 x 60

= (0.785) x ( 0.1647 + 0.1607) x 0.305 x 390 x 60

13

= 0.785 x 0.3254 x 0.305 x 390 x 60

= 1823.468 m3 / hr

Capacity of the compressor:

Inlet Capacity of a Compressor= Q (m3/hr) = Vd X Ƞ

= 1823.068 x 0.7828

= 1427.409 m3 / hr.

Specific Volume :

It is the inverse of Gas Density. i.e. it is the volume of the gas per Kilogram at specified conditions.

It is given by the following formula:

Vsp = Z x Ts x R

Ps MW

Where,

V sp = Specific Volume of Gas in m3/Hr.

Z = Compressibility factor at first stage suction condition = 0.995

Ts = Suction Temp. at first stage = 320.67 ⁰ K

Ps = Suction pressure at first stage = 6.6 Bar a = 6.6/1.01325 = 6.51 atm a .

R = Universal Gas Constant = 0.08206 m3-atm/k-mol ⁰K

MW = Molecular Weight of gas at first stage = 8.25 kg/k-mol.

Therefore,

14

V sp = (0.995 x 320.67 / 6.51) x (0.08206/8.25)

= 48.765 x 0.009947

= 0.487 m3 / kg.

Capacity in Kg/Hr

M = Q * V sp

M (kg/hr) = Q (m3/hr) / V sp (m3/Kg)

= 1427.409 / 0.487

= 2931.02 Kg / Hr

Adiabatic Discharge Temperature

When the Compression takes place by adiabatic process, the discharge temperature can be predicted by the following relation:

Tad = Ts x (PR) ᴷ⁻1⁄ᴷ

= 320.67 x (2.53) (1.25 - 1/1.25)

= 320.67 x 1.203

= 386.08 o K

15

Adiabatic Efficiency:

It indicates the actual process compressor deviation from the adiabatic compression.It is given by,

Ƞad = Tad – Ts

Td ₋actual - Ts

= (386.08 – 320.67)/(405.32 – 320.67)

= 65.61/84.65

= 0.77507

= 77.50 %

Adiabatic Head :

It is defined as the work done by the compressor to raise the pressure of the gas. It is given by ,

H ad (kN-m/Kg) = Zavg x R x Ts x k x [(PR)ᴷ⁻ⁱ⁄ ᴷ - 1]

MWmix k-1

= 0.995 x (8.314/8.25) x 320.67 x (1.25/1.25-1) x [ (2.53)(1.25-1/1.25) - 1 ]

= 0.995 x 1.0077 x 320.67 x 5 x [ 1.203 – 1 ]

= 1607.61 x 0.203

= 326.34 kN-m/kg

16

Adiabatic Power :

It is the power required by the compressor to raise the pressure of gas. In mechanical terms it is a brake power.

Pad (kW) = m x Had

Ƞad x 3600

Where,

m = Total Inlet Feed to the Compressor in Kg/Hr.

Had = Adiabatic Head = 326.34 kN-m/kg

Ƞad = Adiabatic Efficiency = 77.5 % = 0.775

= 3143 x 326.46

0.775 x 3600

= 367.76 KW

Performance Evaluation for C-3073 (2 nd Stage Compression)

C-3072/73 Compressor is a Two stage Double acting reciprocating compressor.1st stage Discharge Pressure is = 1670 KPa whereas 1st Stage Knockout Drum pressure = 1651 KPa.

17

1. Suction Conditions:

The following are the 1st stage suction conditions for our calculations

Suction Temp = 32 ⁰C = 305 ⁰KSuction Pressure = 16.51 Bar aDischarge Temperature = 121.28 = 394.28 ⁰ KDischarge Pressure = 34.65 Bar a

Thus,




Cp of Mixture = 34.60360882 (Refer Figure 10)


Pr = P suction/Pc = 16.52 / 16.857 = 0.98

Tr = T suction/Tc = 305 / 71.902 = 4.24


For Pr = 0.98 & Tr = 4.24

Suction Compressibility = 1.02 (refer Figure 5)


Pr = P discharge/Pc = 34.65/16.857 = 2.055

18

Tr = T discharge/ Tc = 394.28/71.902 =5.483

For Pr = 2.055 & T r = 5.483

Discharge Compressibility is = 1.03 (refer fig. 5)

Therefore,


= 36.897/ (36.897-8.314)

= 1.286



= 34.65/16.51

= 2.098



Higher the discharge pressure , higher is the quantity that is left back in space between piston and the cylinder

End walls.

Thus volumetric efficiency of the compressor can be theoritically estimated as

Ƞ = 0.97 -C x [PR ^ (1/K) - 1] - L

= 0.97 - 0.1997[( 2.098)^(1/1.286) - 1] - 0.07

= 0.97 – 0.1997 (1.779 - 1) – 0.07

= 0.97 – 0.1555 – 0.07

19

= 0.7444

Where,




L = Leakage losses normally 0.03 to 0.05 for Lubricated compressor & 0.07 to 0.1 for non- lubricated compressor.


Thus ,


Ƞ = 0.97 -C x [PR ^ (1/K) - 1] - Lc = 74.44 %

Piston Displacement

Piston displacement is defined as the net volume actually displaced by the piston at rated machine speed

as the piston travels its length from top dead center to bottom dead center.

For Double acting Reciprocating Compressor ,

Bore (D) Stroke (S) Diameter of Rod (d)

1st Stage 0.40589 0.305 0.063552nd Stage 0.2159 0.305 0.06355


Where,

20



S= Cylindrical Sroke = 12.0078 inches = 0.305 m


Therefore,

Piston Displacement = (3.14/4) x (0.2159 2 + (0.2159 2 – 0.063552)) x 0.305 x 390 x 60

= (0.785) x ( 0.0466 + 0.0425) x 0.305 x 390 x 60

= 0.785 x 0.0891 x 0.305 x 390 x 60

Vd = 499.18 m3 / hr


Inlet Capacity of a Compressor Q (m3/hr) = Vd X Ƞ

= 499.18 x 0.7444

= 371.58 m3 / hr.

Specific Volume :



Vsp = Z x Ts x R

Ps MW

Where,


21


Ts = Suction Temp. at first stage = 305 ⁰ K

Ps = Suction pressure at first stage = 16.51/1.01325 = Bar a = 16.308 atm a .



Therefore,

V sp = (1.02 x 305 / 16.308) x (0.08206/7.058)

= 19.076 x 0.01162

= 0.2216 m3 / kg.

Capacity in Kg/Hr

M = Q * V sp


= 371.58 / 0.2216

= 1676.8 Kg / Hr


When the Compression takes place by adiabatic process, the discharge temperature can be predicted by the

following relation:


= 305 x (2.098) (1.286 - 1/1.286)

= 320.67 x (2.098)0.2223

= 320.67 x 1.179

= 378.06 o K

22



Ƞad = Tad – Ts

Td ₋actual - Ts

= (378.06 – 305)/(394.28 – 305)

= 73.06/89.28

= 0.8183

= 81.83 %

Adiabatic Head :


Had (kN-m/Kg) = Zavg x R x Ts x k x [(PR)ᴷ⁻ⁱ⁄ ᴷ - 1]

Mwmix k-1

= 1.025 x (8.314/ 7.0586) x 305 x (1.286/1.286-1) x [ (2.098) (1.286-1/1.286) - 1 ]

= 1.025 x 1.1778 x 305 x 4.4965 x [ 1.1791 – 1 ]

= 1.025 x 1.1778 x 305 x 4.4965 x 0.1791

= 296.5278 kN-m/kg

Adiabatic Power :

It is the power required by the compressor to raise the pressure of gas.In mechanical terms it is a brake power.

23

Pad (kW) = m x Had

Ƞad x 3600

Where,

m = Total Inlet Feed to the Compressor in Kg/Hr. = 1295.3

Had = Adiabatic Head =296.52 kN-m/kg

Ƞad = Adiabatic Efficiency = 0.8183

= 1295.3 x 296.52

0.8183 x 3600

= 130.37 KW

Total Power Required by the two Compressors = 367.76 KW + 130.37 KW = 498.13 KW

Estimation of 1 st Stage outlet from 1 st Re-contact Drum.(Flash Calculations for 1 st Stage)

The quantitative analysis of the feed flashed in the 1st re-contact drum can be obtained by performing flash calculations.This analysis would help us in determining the precise quantity of Vapor and Liquid obtained after Flashing.

24

Here the feed to the 1st Re-contact drum is the vapor obtained from the flashing at the first re-contact drum .This vapor is compressed further to 3465 KPa . As a result of increase in Pressure its temperature increases to 132.32 0 C.thus to cool it, its mixed with the recycle feed from the 2nd Re-contact drum bottoms. It is further cooled upto 34 0 Cin a Heat Exchanger.

Feed = 3143 Kg/Hr of feed. Average Molecular Weight of the feed = 8.25 Kg/ KmolTherefore,Molal Flow rate of feed = Kg. Per Hour of Feed / Avg. molecular weight of feed = 3143 / 8.25 = 380.96 Kmol / Hr.

Gas Properties at 1 st stage Suction Conditions.

Cp = Cp Vap A + Cp Vap B(47.67+273)1 + Cp Vap C(47.67+273)2 + Cp Vap D(47.67+273)3

(The values Cp Vap A,B,C,D are given in fig.8)

Component

sMW Mol %

Mol Frac. Pc Tc

Cp @ 46.67 ⁰C

Wt MW Wt. Pc Wt. Tc

Wtd. Cp

KJ/kmlK Kg/kml

H Hydrogen 2 85.86 0.8586 13 33.2 28.930 1.71711.16

228.50

624.83

9

C1P Methane 16 3.91 0.0391 46.4190.

4 36.505 0.626 1.814 7.445 1.427

C2P Ethane 30 2.80 0.0280 48.8305.

4 54.899 0.840 1.366 8.551 1.537

C3P Propane 44 2.15 0.0215 42.5369.

8 77.528 0.946 0.914 7.951 1.667

C4P I-Butane 58 1.04 0.0104 36.5408.

2 102.378 0.603 0.380 4.245 1.065

C4P n-butane 58 0.96 0.0096 38425.

2 102.764 0.557 0.365 4.082 0.987

C5P I-Pentane 72 0.84 0.0084 33.9460.

4 124.854 0.605 0.285 3.867 1.049

C5P n-Pentane 72 0.74 0.0074 33.7469.

7 125.938 0.533 0.249 3.476 0.932

C6P n-Hexane 86 2.12 0.0212 30.3508.

3 149.998 1.823 0.64210.77

6 3.180

Sum 100.0 1.00 8.250 17.17 78.89 36.68

25

0 7 8 3

Dew Point Calculations at 1 st stage Discharge Pressure of 1670 KPa .

T ⁰C47.2 Pressure(P

t)1670

T ⁰K 320.2 (KPa)

ln Pi = A – B/(T+C)

Comp.Mol Frac. Antioine Constants ln Pi Pi

Ki = Pi/Pt Xi= Ni/Ki

Kpa Yi A B C

HHydroge

n0.858

612.78

4 232.32 8.0812.076

3175661.0

3105.186

20.00816

3

C1PMethan

e0.039

113.58

4 968.13 -3.7210.524

9 37232.75 22.29510.00175

4

C2P Ethane 0.02813.88

01582.1

8-

13.762 8.7166 6103.19 3.65460.00766

2

C3P Propane0.021

513.71

01872.8

2 -25.1 7.3633 1577.05 0.94430.02276

7

C4P I-Butane0.010

413.81

42150.2

3 -27.62 6.4645 641.94 0.38440.02705

5

C4Pn-

butane0.009

613.98

42299.4

4 -27.86 6.1180 453.94 0.27180.03531

7

C5PI-

Pentane0.008

413.71

9 2396.4 -37.55 5.2407 188.80 0.11310.07430

2

C5Pn-

Pentane0.007

413.97

8 2554.6 -36.25 4.9811 145.64 0.08720.08485

2

26

C6Pn-

Hexane0.021

214.05

72825.4

2 -42.71 3.8747 48.17 0.02880.73497

8

sum 1 0.99685

1

Thus Dew Point at above mentioned Conditions = 47.2 ⁰C = 320.2 ⁰ K

Dew Point Calculations at 1 st stage Re-contact drum at a Pressure of 1651 KPa .

T ⁰C 46.8 Pressure(Pt

) 1651T ⁰K 319.8 (KPa)

Name ofMol Frac. Antioine Constants ln Pi Pi

Ki=Pi/Pt

Xi=Ni/Ki

Componen

t Kpa Yi A B C

H Hydrogen0.8586

0 12.78 232.32 8.08 12.08175509.4

4 106.30 0.0081

C1P Methane0.0391

0 13.58 968.13 -3.72 10.52 37088.90 22.46 0.0017

27

C2P Ethane0.0280

0 13.881582.1

8 -13.76 8.71 6062.15 3.67 0.0076

C3P Propane0.0215

0 13.711872.8

2 -25.10 7.35 1563.52 0.95 0.0227

C4P I-Butane0.0104

0 13.812150.2

3 -27.62 6.45 635.51 0.38 0.0270

C4P n-butane0.0096

0 13.982299.4

4 -27.86 6.11 449.07 0.27 0.0353

C5P I-Pentane0.0084

0 13.722396.4

0 -37.55 5.23 186.54 0.11 0.0743

C5P n-Pentane0.0074

0 13.982554.6

0 -36.25 4.97 143.80 0.09 0.0850

C6P n-Hexane0.0212

0 14.062825.4

2 -42.71 3.86 47.47 0.03 0.7374

Sum 1.00 1.00

Thus Dew Point at above conditions = 46.8 ⁰C = 319.8 ⁰K

Thus inorder for the condensation to take place ,The Temperature at Re-Contact Drum should be below the above two dew point temperature

i.e. Re-contact Drum Temperature should be < Dew Point Temperature.

Calculation of Ki at Knockout Drum Conditions :

As Re-contact Drum Temperature should be < Dew Point Temperature,

28

We maintain Re-Contact Drum Temperature at 32 ⁰ C.

T ⁰C 32 Pressure(Pt) 1651T ⁰K 305 (KPa)

Name ofMol Frac. Antioine Constants ln Pi Pi Ki=Pi/Pt Xi=Ni/Ki

Componen

t Kpa Yi A B C

H Hydrogen 0.858612.78

4 232.32 8.0812.042

0169728.1

4102.803

20.00835

2

C1P Methane 0.039113.58

4 968.13 -3.7210.370

6 31907.95 19.32640.00202

3

C2P Ethane 0.02813.88

01582.1

8-

13.762 8.4471 4661.53 2.82350.00991

7

C3P Propane 0.021513.71

01872.8

2 -25.1 7.0187 1117.30 0.6767 0.03177

C4P I-Butane 0.010413.81

42150.2

3 -27.62 6.0618 429.13 0.25990.04001

2

C4P n-butane 0.009613.98

42299.4

4 -27.86 5.6866 294.88 0.17860.05374

9

C5P I-Pentane 0.008413.71

9 2396.4 -37.55 4.7588 116.61 0.07060.11893

1

C5P n-Pentane 0.007413.97

8 2554.6 -36.25 4.4723 87.56 0.05300.13953

4

C6P n-Hexane 0.021214.05

72825.4

2 -42.71 3.2847 26.70 0.01621.31088

8

Sum 1 1.71517

6

29

Flash Calculation for computing the Vapor and liquid discharge from 1 st Re-contact drum.

Component F Mol Frac. Fi = F.Yi Ki=Pi/Pt Assumed LiVi = Fi-

LiVap Mol

Mol Frac.

kmol/hr =Fi/(1+(V/L)Ki % Yi

H Hydrogen 380.96 0.8586 327.092 102.803 0.053 327.040 86.903 0.869C1P Methane 380.96 0.0391 14.896 19.326 0.013 14.883 3.955 0.040C2P Ethane 380.96 0.028 10.667 2.824 0.062 10.605 2.818 0.028C3P Propane 380.96 0.0215 8.191 0.677 0.196 7.995 2.124 0.021C4P I-Butane 380.96 0.0104 3.962 0.260 0.237 3.725 0.990 0.010C4P n-butane 380.96 0.0096 3.657 0.179 0.310 3.347 0.889 0.009C5P I-Pentane 380.96 0.0084 3.200 0.071 0.608 2.592 0.689 0.007C5P n-Pentane 380.96 0.0074 2.819 0.053 0.671 2.148 0.571 0.006C6P n-Hexane 380.96 0.0212 8.076 0.016 4.083 3.993 1.061 0.011

Sum 1.00382.56

0126.20

8 6.234376.32

6100.00

0 1.000

L= 6.2336695 V=F-L 376.32636

Calculated V/L = 60.37

V/L = 60.37

Estimation of 2 nd Stage outlet from 2 nd Re-contact Drum . (Flash Calculations For 2 nd Re-Contact Drum.)

The quantitative analysis of the feed flashed in the 2nd re-contact drum can be obtained by performing flash calculations.This analysis would help us in determining the precise quantity of Vapor and Liquid obtained after Flashing.

30

Here the feed to the 2nd Re-contact drum is the vapor obtained from the flashing at the first re-contact drum .This vapor is compressed further to 3465 KPa. As a result of increase in Pressure its temperature increases to 121.28 0C.thus to cool it, its mixed with the recycle feed from the 2nd Re-contact drum bottoms. It is further cooled upto 34 0 Cin a Heat Exchanger where cooling water is used as the cooling medium.

The cooled components are then sent to the 2nd Re-contact drum which is at 32 0 C and 3426 KPa. Due to this Temperature and Pressure Differential, Flashing takes place and the feed gets separated into Liquid and Vapor.

Feed = 1295.9 Kg/Hr of feed.Average Molecular Weight of the feed = 7.058 Kg/ KmolTherefore,Molal Flow rate of feed = Kg. Per Hour of Feed / Avg. molecular weight of feed = 1295.9 / 7.058 = 183.6 Kmol / Hr.

Gas Properties at 2 nd stage Suction Conditions.

Cp = Cp Vap A + Cp Vap B(32+273)1 + Cp Vap C(32+273)2 + Cp Vap D(32+273)3

(The values Cp Vap A,B,C,D are given in fig.10)

CompMW

Mol % Pc Tc

Cp @ 32 ⁰C

Wt MW Wt. Pc Wt. Tc

Wtd. Cp

Mol Frac

KJ/kmlKKg/kml

H Hydrogen 2 86.9 13 33.2 28.901 1.738 11.29728.85

125.11

5 0.869

C1P Methane 16 3.95 46.4190.

4 35.942 0.632 1.8328 7.521 1.4200.039

5

C2P Ethane 30 2.82 48.8305.

4 53.523 0.8461.3761

6 8.612 1.5090.028

2

C3P Propane 44 2.12 42.5369.

8 75.356 0.9328 0.901 7.840 1.5980.021

2

C4P Iso – Butane 58 0.99 36.5408.

2 99.593 0.57420.3613

5 4.041 0.9860.009

9

C4P n – butane 58 0.89 38425.

2 100.146 0.5162 0.3382 3.784 0.8910.008

9

C5P iso – Pentane 72 0.69 33.9460.

4 121.309 0.49680.2339

1 3.177 0.8370.006

9C5P n – Pentane 72 0.57 33.7 469. 122.511 0.4104 0.1920 2.677 0.698 0.005

31

7 9 7

C6P n-Hexane 86 1.06 30.3508.

3 145.925 0.91160.3211

8 5.388 1.5470.010

6

Sum 438 100 7.058 1

Dew Point Calculations at 2 nd stage Discharge Pressure of 3465 KPa .

T ⁰C 53.7Pressure(P

t) 3465T ⁰K 326.7 (KPa)


Ki = Pi/Pt Xi= Ni/Ki

Compone

nt Kpa Yi A B C


4 232.32 8.0812.090

1178091.3

251.397

20.01690

8

C1P Methane 0.039513.58

4 968.13 -3.7210.586

5 39596.9711.427

70.00345

7

C2P Ethane 0.028213.88

01582.1

8-

13.762 8.8238 6794.10 1.96080.01438

2

C3P Propane 0.021213.71

01872.8

2 -25.1 7.5001 1808.20 0.52180.04062

5

C4P I-Butane 0.009913.81

42150.2

3 -27.62 6.6242 753.12 0.21730.04554

9

C4P n-butane 0.008913.98

42299.4

4 -27.86 6.2890 538.64 0.15550.05725

3


9 2396.4 -37.55 5.4313 228.44 0.06590.10466

1


8 2554.6 -36.25 5.1825 178.12 0.0514 0.11088C6P n-Hexane 0.0106 14.05 2825.4 -42.71 4.1078 60.81 0.0176 0.60397

32

7 2 6

Sum 1 0.99769

Thus Dew Point at the above Conditions = 53.7 ⁰C = 326.7 ⁰K.

Dew Point Calculations at 2 nd stage Re-contact Drum at a pressure of 3426 KPa .

T ⁰C 53.3Pressure(P

t) 3426T ⁰K 326.3 Kpa


Ki = Pi/Pt

Xi= Ni/Ki

Compone

nt Kpa Yi A B C


4 232.32 8.0812.089

2177943.5

451.939

20.01673

1

C1P Methane0.039

513.58

4 968.13 -3.7210.582

8 39450.0711.514

9 0.00343

C2P Ethane0.028

213.88

01582.1

8-

13.76 8.8173 6750.28 1.97030.01431

2C3P Propane 0.021 13.71 1872.8 -25.1 7.4918 1793.35 0.5235 0.0405

33

2 0 2

C4P I-Butane0.009

913.81

42150.2

3-

27.62 6.6146 745.90 0.21770.04547

2

C4P n-butane0.008

913.98

42299.4

4-

27.86 6.2787 533.11 0.15560.05719

5

C5P I-Pentane0.006

913.71

9 2396.4-

37.55 5.4198 225.83 0.06590.10467

8

C5P n-Pentane0.005

713.97

8 2554.6-

36.25 5.1704 175.98 0.0514 0.11097

C6P n-Hexane0.010

614.05

72825.4

2-

42.71 4.0938 59.96 0.01750.60561

7

Sum 1 0.99890

7

Thus Dew point at the above conditions = 53.3 ⁰C = 326.3 ⁰K

Thus inorder for the condensation to take place ,The Temperature at Re-Contact Drum should be below the above two dew point temperature

i.e. Re-contact Drum Temperature should be < Dew Point Temperature.

Calculation of Ki at Knockout Drum Conditions :

As Re-contact Drum Temperature should be < Dew Point Temperature,We maintain Re-Conatct Drum Temperature at 32 ⁰ C.

T ⁰C 32Pressure(P

t) 3426T ⁰K 305 Kpa

34


Ki = Pi/Pt

Xi= Ni/Ki

Compone

nt Kpa Yi A B C


4 232.32 8.0812.042

0169728.1

449.541

20.01754

1

C1P Methane 0.039513.58

4 968.13 -3.7210.370

6 31907.95 9.31350.00424

1

C2P Ethane 0.028213.88

01582.1

8-

13.762 8.4471 4661.53 1.36060.02072

6

C3P Propane 0.021213.71

01872.8

2 -25.1 7.0187 1117.30 0.32610.06500

6

C4P I-Butane 0.009913.81

42150.2

3 -27.62 6.0618 429.13 0.12530.07903

7

C4P n-butane 0.008913.98

42299.4

4 -27.86 5.6866 294.88 0.08610.10340

3


9 2396.4 -37.55 4.7588 116.61 0.03400.20272

5


8 2554.6 -36.25 4.4723 87.56 0.02560.22302

9C6 + n-Hexane 0.0106

14.057

2825.42 -42.71 3.2847 26.70 0.0078

1.360116

Sum 1

2.075824

35

Flash Calculation for computing the Vapor and liquid discharge from 2 nd Re-contact drum.

ln Pi = A – B/(T+C)

Name ofMol flow

FMol Frac. Fi = F.Yi

Ki=Pi/Pt Assumed Li Vi = Fi-Li

Vap.Mol %

Component kmol/hr =Fi/(1+(V/

L)Ki Yi

H Hydrogen 183.6 0.869 159.54849.541

2 0.05166159.496

7 88.5847

C1P Methane 183.60.039

5 7.252 9.3135 0.01247 7.2397 4.0210

C2P Ethane 183.60.028

2 5.178 1.3606 0.06035 5.1172 2.8421

C3P Propane 183.60.021

2 3.892 0.3261 0.18254 3.7098 2.0604

C4P I-Butane 183.60.009

9 1.818 0.1253 0.20635 1.6113 0.8949

C4P n-butane 183.60.008

9 1.634 0.0861 0.25669 1.3773 0.7650


9 1.267 0.034 0.40618 0.8607 0.4780


7 1.047 0.0256 0.40322 0.6433 0.3573

C6 + n-Hexane 183.60.010

6 1.946 0.0078 1.30958 0.6366 0.3536 Sum 1 226.15 2.88905 180.05 100.000 ΣL = 2.889 ΣV = F-L 180.05

Calculated

V/L = 62.32

Estimated V/L

= 62.32

36

Thus under the above operating conditions we get a Hydrogen purity of 88.58 % .

37

Determination of Liquid and Gas flow Rates for 1 st Re-Contact Drum.

Liquid Vapo

r

LiKmol/

hrMol frac. MW

Avg MW Kg/Hr

ViKmol/

hrMol Frac. MW

Avg MW Kg/Hr

0.05270.0084

5 20.0169

1485.8

2 327.04 0.869 21.7380

62656.

9

0.01280.0020

5 160.0328

5 14.882

80.039

5 160.6327

6

0.06220.0099

8 300.2993

4 10.604

70.028

2 300.8453

9

0.19570.0313

9 441.3813

3 7.99490.021

2 440.9347

6

0.23740.0380

8 582.2088

3 3.72460.009

9 580.5740

4

0.31040.0497

9 582.8880

4 3.34680.008

9 580.5158

1

0.60810.0975

5 727.0236

3 2.59190.006

9 720.4958

9

0.67130.1076

9 72 7.7536 2.14780.005

7 720.4109

2

4.0831 0.655 8656.330

4 3.99320.010

6 860.9125

5

sum 6.2337 1 438

77.9349

485.82

376.326 1 438

7.06019

Determination of Liquid and Gas flow Rates for 2 nd Re-Contact Drum.

LIQUID VAPOR

Li

Kmol/hrMol frac. MW

Avg MW Kg/Hr

Vi

Kmol/hrMol Frac. MW

Avg MW Kg/Hr

0.05 0.02 2.00 0.04 206.99 159.50 0.89 2.00 1.77 1088.85 0.01 0.00 16.00 0.07 7.24 0.04 16.00 0.64 0.06 0.02 30.00 0.62 5.12 0.03 30.00 0.85 0.18 0.06 44.00 2.77 3.71 0.02 44.00 0.91 0.20 0.07 58.00 4.13 1.61 0.01 58.00 0.52 0.25 0.09 58.00 5.14 1.38 0.01 58.00 0.44

38

0.40 0.14 72.00 10.11 0.86 0.00 72.00 0.35 0.40 0.14 72.00 10.05 0.65 0.00 72.00 0.26 1.31 0.45 86.00 39.06 0.64 0.00 86.00 0.31

Sum 2.88 1.00 72.00 206.99 180.05 1.00 6.05

Thus we can perform a Material Balance for the loop by using the above method as shown in Figure 2 which is a Generalized Material Balance for the above conditions.

Figure 3 shows the overall material balance wherein we consider the incoming liquid feed of 7347 Kg/Hr.

For the above operating conditions let’s consider the design criteria of the the two Re-Contact Drums.

Instrumentation of Spillback Loop (Figure 4):

Provide 5 Signal multipliers to perform the following functions :

a. Input signal 50-100 % from PRC3016Output signal 0-100 % to PRC3016 valve.

b. Input signal 0-50 % from PRC3016Output signal 0-100 % to low signal selector (LSS), item f.

c. Input signal 50 to 100 % from PRC3022Output signal 100-0 % to LSS item f.

d. Input signal 0-50 % from PRC3022Output signal 0-100 % to LSS item g.

e. Input signal 0-100% from PRC3023Output signal 100-0% to LSS item g.

Provide 2 low signal selectors to perform the following functions.

f. Inputs from multipliers b and c above.Output to PRC3022 control valve.

g. Inputs from multipliers d and e aboveOutput to PRC3023 control valve.

39

Design of 1 st Re-Contact Drum

Vertical vapor-liquid separators are used primarily to disengage a liquid from a vapor when the volume of liquid is small compared to the vapor volume. To reduce liquid entrainment, the maximum allowable vapor velocity in a vertical vessel is a function of liquid and vapor density and the parameter Kv .

The Parameter Kv is itself a function of vapor and liquid density and vapor and liquid flow rates. This polynomial function was developed by Watkins and is valid for a range of Separation factors of 0.006 to 5 .

We will design a vertical Re-Contact Drum for our process as the Flow rate of Liquid is Greater than the flow Rate of Vapor.

Weight of Liquid = 8039.81 Kg = WL = 17724.76 lb/Hr

Weight of Vapor = 2656.93 Kg = WV = 5857.54 lb/Hr

Density of Liquid = 780.261 Kg/m3 = ρL = 48.71 lb/ft3

Density of Vapor = 4.839 Kg/m3 = ρV = 0.302 lb/ft3

Sfac = WL/WV x (ρV/ρL) 0.5

Therefore,

Sfac = 17724.76 / 5857.54 x (0.302/48.71) 0.5

Sfac = 0.2382643 = X

Now,

40

Kv is given by the Stefan’s Relation as follows’

KV = exp (B+DX+EX2+FX3+GX4)

Where,

B = -1.87747

D = -0.81458

E = -0.18707

F = -0.01452

G = -0.00101485

X = Sfac = 0.2382643

Therefore,

KV = exp [-1.87747 -0.81458(0.2382643) -0.18707(0.2382643)2 -0.01452(0.2382643)3 - 0.00101485(0.2382643)4]

= exp (-2.08237)

KV = 0.1246339

Vmax = KV x [(ρL- ρV)/ ρV] 0.5

Vmax = 0.1246339 x [(48.71-0.302)/0.302]0.5

Vmax = 0.1246339 x 12.6606

Vmax = 1.5779445 ft /sec

= 0.4809 m/sec

Calculations For Vapor Space (To calculate Diameter of De-mister.)

The required vapor space is determined as follows :

QV = WV / (ρV x 3600)

41

Where,

QV = Volumetric Vapor Flow in ft3/sec.

QV = 5857.54 / (0.302 x 3600)

QV = 5.38773 ft3/sec

= 0.1525635 m3/sec

AV = QV / Vmax

Where,

AV = Cross – Sectional Area of Vapor Space.

AV = 5.38773 / 1.5779445

AV = 3.4143977 Ft 2

= 0.317 m2

The vessel Diameter is then given By ,

D =[ (4 x AV) / ∏ ] 0.5

D = [(4 x 3.4143977) / ∏ ]0.5

D = 2.085 ft .

The next Larger Commercially available size is chosen & the area is recomputed

Consider the above diameter = 2.0997 ft = 640 mm = 0.64 m

Therefore,

A = 3.4627485 ft 2

= 0.321 m2

The vapor height is given by ,

Vapor Ht = MDH + Nozzle + Mist + MDH above Demister

Where,

42

MDH = Minimum Disengaging Height above feed nozzle and must be 36 inches + Half the Feed Nozzle but Minimum of 48 Inches.

Nozzle = This is the space between the feed nozzle and highest Liquid level and must be 12 inches + Half the Nozzle Diameter or Minimum of 18 inches.

Mist = Height of Demister or Thickness = 150 mm = 6 inches (given)

MDH above Demister for K >= 0.1246 is 100 mm = 4 inches.

The computed values for MDH , Nozzle , Mist , MDH above Demister is as follows:

MDH = 40 inches

Nozzle = 16 inches.

Mist = 6 inches.

MDH above Demister = 4 inches.

But,

MDH must be = 48 inches minimum =

Nozzle must be = 18 inches minimum.

Mist must be = 6 inches.

MDH above Demister must be = 4 inches. (Reference Ludwig Volume – 2, Page 617)

Therefore,

Vapor Ht = 48+18+6+4 = 76 inches

Vapor Ht = 6.3333 ft.

Calculations For Liquid Space.

43

The liquid height is set either by user input or by computing the level required for retention time. Consider a retention time of 10 mins.

Retention time is required to ensure that the vessel can withstand a liquid inventory for a minimum period of time to prevent flooding in case of a process failure.

Calculations for Liquid Height

QL = WL / (ρL x 60)

Where,

QL = Volumetric flow rate in ft3/min.

QL = 17724.76 / (48.7101 x 60)

QL = 6.0647 ft. = 1.848 m

But we must Maintain a provision for 10 mins of liquid inventory i.e. T = 10 mins.

QL eff = QL x T

QL eff = 6.0647 x 10

QL eff = 60.647 ft. = 18.48 m

Liquid Level = LL , LL = QL eff / A

LL = 60.647 / 3.462

LL = 17.514 ft. = 5.338 m

Total Height = LL + Vapor Height

= 17.514 + 6.33

Total Height = 23.84415 Ft. = 7.267 m

44

L/D ratio = Total Height / Diameter

= 7.267 / 0.64

L/D ratio = 11.355

But L/D Has to be Less than 5 for stability considerations

Thus we change the Diameter of the vessel to next best commercially available vessel such that L/D must be below 5

Choose a diameter such that L/D is nearly equal to 5.

Next Commercially available size = 920 mm = 0.92 m.

Therefore new Area , Anew is given by

Anew = (∏/4) x D2

Anew = 0.7854 x 3.0183 2

Anew = 7.155 Ft 2

Anew = 0.664 m2

The vapor height is the same as in the earlier case.

Therefore,

Vapor Ht . = 6.3333 ft = 1.929 m

New liquid level

QL = WL / (ρL x 60)

Where,


QL = 17724.76 / (48.7101 x 60)

45

QL = 6.0647 ft.


QL eff = QL x T

QL eff = 6.0647 x 10

QL eff = 60.647 ft. = 18.485 m

New Liquid Level = LL new , LL new = QL eff / Anew

= 60.647 / 7.155

LL new = 8.476 ft = 2.583 m

Total Height = LL new+ Vapor Height

= 8.476 + 6.33

Total Height =14.806 Ft = 4.512 m


=4.512 / 0.92

L/D ratio = 4.9051 which is acceptable.

Thus,

Diameter of vessel = 3.0183 Ft. = 920 mm

Height of the vessel = 14.806 ft. = 4513 mm

46

Determination of Thickness of Vessel and Selection of Most Economical Head for the 1 st Re- Contact Drum

Di = Internal Diameter of vessel

Do = Outer Diameter of vessel.

P = Design Pressure

f = Permissible stress = 95 N/mm2 for Carbon Steel.

FOS = Factor of Safety = 1.5 for Carbon Steel.

C = Corrosion Allowance = 3.2 mm

ts = thickness of shell

tH = Thickness of Hemispherical head.

tE = Thickness of Elliptical head.

tT = Thickness of Torispherical head.

Rc = Crown Radius of Torispherical head.

Rk = Knuckle Radius of Torispherical head = 6% of Di

W = SIF = Stress Intensification Factor.

J = Joint Efficiency.

Operating Pressure = 1.651 N/mm2

Design pressure = Operating Pressure + 10 % = 1.8161 1.651 N/mm2

Calculations:

47

P = 1.8161 N/mm2

J = 0.85 for Shell &

J = 1 for Head.

f = Permissible Stress = Yield Strength/FOS

Yield Strength for Carbon Steel = 95N/mm2

FOS = 1.5 for Carbon Steel.

Therefore,

f = 95/1.5

= 63.33 N/mm2

Thickness for Shell ,

ts = P x Di / 2fJ – P

= 1.8161 x 920 / (2 x 63.33 x 0.85 – 1.8161)

= 1670.812 / (107.661 – 1.8161)

tS = 15.78 mm

Corrosion Allowance = C = 3.2 mm

Therefore ,

tS = 15.78 + 3.2 = 18.98 mm

= 20 mm

Determination of Thickness of Head.

For Hemispherical Head

tH = PD / 4f = 1.8161 x 920 / 4 x 63.33

= 6.59 mm


48

Therefore,

tH = 6.59 + 3.2 = 9.79

= 10 mm

For Elliptical Head & k = 2

tE = PDV / 2fJ

V = (1/6) x (2 + k2) = (1/6) x (2+ 22)

= (1/6) x 6

= 1

Therefore,

tE = 1.8161 x 920 x 1 / (2 x 63.33 x 1)

= 13.19 mm

Corrosion Allowance = 3.2 mm

Therefore,

tE = 13.19 + 3.2 = 16.39

= 18mm

For Torrispherical Head,

Rc = Inner diameter of vessel = 920 mm

Rk = 6 % of Inner Diameter of Vessel = 0.06 x 920 = 55.2 mm

W = SIF = (1/4) x [ 3+ (920/55.2)1/2 ]

= 1.7706

Therefore,

tT = PRcW / 2fJ

= 1.8161 x 920 x 1.7706 / 2 x 63.33 x 1

49

= 23.35 mm


Therefore,

tT = 23.35 + 3.2 = 26.55

= 28 mm

From the above Calculations we see that the Thickness for Hemispherical heads is the least

(viz. 10mm )

Therefore lower the thickness lower the cost .

Hence we will choose Hemispherical head & thickness of thiis head = 10 mm

50

Design of Flanged joint for 1 st Re-Contact Drum.

P = Design Pressure

Di = Inner Diameter of Vessel.

Do = Outer Diameter of vessel.

fa = Allowable stress at atmospheric conditions for Bolt material. = 57 N/mm2 for Carbon Steel.

fp = Allowable stress at operating conditions for bolt material = 53 N/mm2 for Carbon Steel.

m = Gasket Factor.

Y = Gasket Seating Stress.

Gi = Inner Diameter of Gasket.

Go = Outer Diameter of Gasket.

G = Diameter of Gasket Load Reaction.

N = Gasket Width

bo = Basic Gasket Seating width.

b = Effective Gasket Seating width.

Wm1 = Bolt load under atmospheric condition.

Wm2 = Bolt load under operating condition.

Am1 = Bolt Area under atmospheric condition.

Am2 = Bolt Area under operating Condition.

db = Bolt Diameter.

n = No. of bolts.

51

BCD = Bolt circle Diameter.

Dfo = Outer Diameter of flange.

kf = k = Flange Constant.

f = Permissible stress of flange material.

C = Corrosion Allowance.

t = shell thickness.

Wm = Max. bolt load i.e. the greater of the two values from Wm1 and Wm2 .

hG = Radial distance from gasket load reaction to bolt circle.

Design of Gasket

t = Thickness of shell = 20 mm

Do = Di + 2t = 920 + 2(20)

= 920 + 40

= 960 mm

Gi = Do + 10 mm

= 960 + 10

= 970 mm

To calculate Outer Diameter of gasket,

Go/Gi = [Y – Pm / Y – P(m+1) ] ½

Go / 970 = [ 11 – 1.8161(2) / 11 – 1.8162 (2+1) ] ½

Go = 1117.44

i.e

Go = 1120 mm

52

Gasket Width = N =(Go – Gi) / 2

N = 1120 – 970 / 2 = 150/2

= 75 mm

Basic gasket seating width = bo = N/2

= 75/2

= 37.5 mm

Now,

bo = 37.5 mm > 6.3 mm

Therefore,

b = 2.5 x (bo)1/2

= 2.5 x (37.5)1/2

= 2.5 x 6.123

b = 15.30

i.e. ,

b = 16 mm

Diameter of Gasket Load Reaction,

G = Go – 2b

= 1120 – 2(16)

G = 1088 mm

Design of Bolt

Bolt load under atmospheric conditions

Wm1 = ∏Gb x Y

= 3.1416 x 1088 x 16 x 11

53

= 601577.29 N

Wm2 = ∏G(2b)mP + (∏/4) G2P

= [3.1416 x 1088 x (2x16) x 2 x 1.8161 ] + [ 0.7853 x 10882 x 1.8161]

= 2085728.63 N

Therefore,

Wm = Wm2 = 2085728.63 N (Greater of Wm1 & Wm2)

Bolt Area under Atmospheric Condition

Am1 = Wm1/fa = 601577.29 / 57

= 10553.98 mm2

Am2 = Wm2 / fp = 2085728.63 / 53

= 39353.37 mm2

Selecting larger value,

Am = Am2 = 39353.37 mm2

To calculate bolt Diameter ,

Am = (∏/4) db2 x n

39353.37 = (∏/4) db2 x n

But,

n = G/25

n = 1088/25

= 43.52

But n must be a multiple of 4.Hence we round it off to the nearest multiple of 4 viz. 44.

i.e.

n = 44

54

Therefore,

39353.37 = (∏/4) x db2 x 44

db2 = 1138.77

db = 33.74 mm

i.e.

db = 34mm

It is represented as,

M34 , 44 Bolts.

The minimum BCD is given by,

B = Go + 2db + 12

= 1120 + 2(34) + 12

= 1200 mm

Design of Flange

Outer Diameter of flange

Dfo = B + 2db + 12

= 1200 + 2(34) + 12

= 1280 mm

Now,

k = k = 1/[0.3 + (1.5 x WmhG/H x G )

hG = B-G/2 = (1200 – 1088) /2 = 32/2 = 56 mm

H = Hydrostatic end force = (∏/4) x G2P = 0.7853 x 10882 x 1.8161 = 1688446.991 N

k = 1/ [0.3 + (1.5 x 2085728.63 x 56) / (1688446.991 x 1088) ]

k = 1/0.395

55

= 2.52

Thickness of flange

tf = G (P/Kf) ½

= 1088 (1.8161/2.52) ½

= 1088 x 0.849

= 923.63 mm

= 925 mm

Thus the 1st Re-Contact Drum (F-3047) can be designed as shown in fig. 5

56

Design of 2 nd Re-Contact Drum

Note: Nomenclatures are the same as the ones mentioned for the design of 1st Re-contact drum.

Weight of Liquid = 7553.99 Kg = WL = 16653.7 lb/Hr

Weight of Vapor = 1088.85 Kg = WV = 2400.504 lb/Hr

Density of Liquid = 780.261 Kg/m3 = ρL = 48.71 lb/ft3

Density of Vapor = 4.839 Kg/m3 = ρV = 0.302 lb/ft3

Sfac = WL/WV x (ρV/ρL) 0.5

Therefore,

Sfac = 16653.7/2400.504 x (0.302/48.71) 0.5

Sfac = 0.5462

Now,

KV is given by the Stefan’s relation as follows :

KV = exp (B+DX+EX2+FX3+GX4)

Where ,

B = -1.87747

D = -0.81458

E = -0.18707

57

F = -0.01452

G = -0.00101485

X = Sfac = 0.5462

Therefore,

KV = exp [-1.87747 -0.81458(0.5462) -0.18707(0.5462)2 -0.01452(0.5462)3 - 0.00101485(0.5462)4]

= exp (-2.38072)

KV = 0.09248

Vmax = KV x [(ρL- ρV)/ ρV] 0.5

Vmax = 0.09248 x [(48.71-0.302)/0.302]0.5

Vmax = 0.09248 x 12.66

Vmax = 1.1708 ft /sec

Vmax = 0.356 m/sec

Calculations For Vapor Space (To calculate Diameter of De-mister.)

The required vapor space is determined as follows :

QV = WV / (ρV x 3600)

Where,

Qv = Volumetric Vapor Flow in ft3/sec.

QV = 2400.504 / (0.302 x 3600)

QV = 2.207 ft3/sec.

AV = QV / Vmax

58

Where,

AV = Cross – Sectional Area of Vapor Space.

= 2.207 / 1.1708

AV = 1.885 Ft 2

AV = 0.175 m2

The vessel Diameter is then given By ,

D =[ (4 x AV) / ∏ ] 0.5

D = [(4 x 1.885) / ∏ ]0.5

D = 1.549 ft = 0.3568 m

The next Larger Commercially available size is chosen & the area is recomputed

Consider the above diameter = 1.574 ft = 480 mm = 0.48 m

Therefore,

A = 1.947796 ft 2 = 0.1809 m2

The vapor height is given by ,

Vapor Ht = MDH + Nozzle + Mist + MDH above Demister

Where,

MDH = Minimum Disengaging Height above feed nozzle and must be 36 inches + Half the Feed Nozzle but Minimum of 48 Inches.

Nozzle = This is the space between the feed nozzle and highest Liquid level and must be 12 inches + Half the Nozzle Diameter or Minimum of 18 inches.

Mist = Height of Demister or Thickness = 150 mm = 6 inches (given)

MDH above Demister for K >= 0.1246 is 100 mm = 4 inches.

The computed values for MDH , Nozzle , Mist , MDH above Demister is as follows:

59

MDH = 39 inches

Nozzle = 15 inches.

Mist = 6 inches.

MDH above Demister = 3 inches.

But,

MDH must be = 48 inches minimum = 1220 mm

Nozzle must be = 18 inches minimum = 458 mm

Mist = 6 inches = 152 mm

MDH above Demister must be = 3 inches = 76 mm (Reference Ludwig Volume – 2 , Page 617)

Therefore,

Vapor Ht = 48+18+6+3 = 75 inches

Vapor Ht = 6.25 ft = 1.905 m

Calculations For Liquid Space.

The liquid height is set either by user input or by computing the level required for retention time. Consider a retention time of 10 mins.

Retention time is required to ensure that the vessel can withstand a liquid inventory for a minimum period of time to prevent flooding in case of a process failure.

Calculations for Liquid Height

QL = WL / (ρL x 60)

Where,

60


QL = 16653.7 / (48.7101 x 60)

QL = 5.698 ft = 1.736 m


QL eff = QL x T

QL eff = 5.698 x 10

QL eff = 56.98 ft = 17.36 m

Liquid Level = LL , LL = QL eff / A

LL = 56.98 / 1.9477

LL = 29.25 ft. = 8.9154 m

Total Height = LL + Vapor Height

= 29.25 + 6.25

Total Height = 35.5 Ft = 10.82 m


= 10.82 / 0.48

L/D ratio = 22.54

But L/D Has to be Less than 5 for stability considerations

61

Thus we change the Diameter of the vessel to next best commercially available vessel such that L/D must be below 5

Choose a diameter such that L/D is nearly equal to 5.

Next Commercially available size = 900 mm = 0.9 m = 2.952 ft .

Therefore new Area , Anew is given by

Anew = (∏/4) x D2

Anew = 0.7854 x 2.952 2

Anew = 6.847 Ft 2

= 0.6361 m2

The vapor height is the same as in the earlier case.

Therefore,

Vapor Ht . = 6.25 ft = 1.905 m

New liquid level

QL = WL / (ρL x 60)

Where,


QL = 16653.7 / (48.7101 x 60)

QL = 5.698 ft = 1.7367 m


QL eff = QL x T

QL eff = 5.698 x 10

QL eff = 56.98 ft = 17.367 m

62

New Liquid Level = LL new , LL new = QL eff / Anew

= 56.98 / 6.847

LL new = 8.32 ft = 2.535 m

Total Height = LL new+ Vapor Height

= 8.32 + 6.25

Total Height =14.57 Ft = 4.44 m


= 4.44 / 0.899

L/D ratio = 4.93 which is acceptable.

Thus,

Diameter of the vessel = 2.9527 ft = 900 mm

Height of the vessel = 14.57 = 4440.936 mm

63

Determination of Thickness of Vessel and Selection of Most Economical Head for the 2 nd Re- Contact Drum

Note: Nomenclatures are the same as the ones mentioned for the design of 1st Re-contact drum.

Operating Pressure = 3.465 N/mm2

Design pressure = Operating Pressure + 10 % = 3.8115 N/mm2

Calculations:

P = 3.8115 N/mm2

J = 0.85 for Shell &

J = 1 for Head.

f = Permissible Stress = Yield Strength/FOS

Yield Strength for Carbon Steel = 95N/mm2

FOS = 1.5 for Carbon Steel.

Therefore,

f = 95/1.5

= 63.33 N/mm2

Thickness for Shell ,

ts = P x Di / 2fJ – P

= 3.8115 x 900 / (2 x 63.33 x 0.85 – 3.8115)

64

= 3430.35 / (107.661 – 3.8115)

tS = 33.03 mm


Therefore ,

tS = 33.03 + 3.2 = 36.23 mm

= 38 mm

Determination of Thickness of Head.

For Hemispherical Head

tH = PD / 4f = 3.8115 x 900 / 4 x 63.33

= 13.54 mm


Therefore,

tH = 13.54 + 3.2 = 16.74

= 18 mm

For Elliptical Head & k = 2

tE = PDV / 2fJ

V = (1/6) x (2 + k2) = (1/6) x (2+ 22)

= (1/6) x 6

= 1

Therefore,

tE = 3.8115 x 900 x 1 / (2 x 63.33 x 1)

= 27.083 mm

Corrosion Allowance = 3.2 mm

65

Therefore,

tE = 27.083 + 3.2 = 30.28

= 32 mm

For Torrispherical Head,

Rc = Inner diameter of vessel = 900 mm

Rk = 6 % of Inner Diameter of Vessel = 0.06 x 900 = 54 mm

W = SIF = (1/4) x [ 3+ (900/54)1/2 ]

= 1.7706

Therefore,

tT = PRcW / 2fJ

= 3.8115 x 900 x 1.7706 / 2 x 63.33 x 1

= 47.95 mm


Therefore,

tT = 47.95 + 3.2 = 51.15

= 52 mm

From the above Calculations we see that the Thickness for Hemispherical heads is the least

(viz. 18 mm )

Therefore lower the thickness lower the cost .

Hence we will choose Hemispherical head & thickness of this head = 18 mm

66

Design of Flanged joint for 2 nd Re-Contact Drum.

Calculations for Design of Gasket

t = Thickness of shell = 38 mm

Do = Di + 2t = 900 + 2(38)

= 900 + 76

= 976 mm

Gi = Do + 10 mm

= 976 + 10 = 986

= 990 mm

To calculate Outer Diameter of gasket,

Go/Gi = [Y – Pm / Y – P(m+1) ] ½

Go / 990 = [ 12 – 3.8115(2) / 12 – 3.8115 (2+1) ] ½

=( 3.377 / 0.5655 ) ½

= 2.44

Go = 2419.27

i.e

Go = 2420 mm

Gasket Width = N =(Go – Gi) / 2

N = 2420 – 990 / 2 = 1430/2

= 715 mm

67

Basic gasket seating width = bo = N/2

= 715/2

= 357.5 mm

Now,

bo = 357.5 mm > 6.3 mm

Therefore,

b = 2.5 x (bo)1/2

= 2.5 x (357.5)1/2

= 2.5 x 18.9076

b = 47.269

i.e. ,

b = 48 mm

Diameter of Gasket Load Reaction,

G = Go – 2b

= 2420 – 2(48)

G = 2324 mm

Design of Bolt

Bolt load under atmospheric conditions

Wm1 = ∏Gb x Y

= 3.1416 x 2324 x 48 x 12

= 4199992.47 N

Wm2 = ∏G(2b)mP + (∏/4) G2P

= [3.1416 x 2324 x (2x48) x 2 x 3.8115 ] + [ 0.7853 x 23242 x 3.8115]

= 21509032.05 N

68

Therefore Wm = Wm2 = 21509032.05 N (Greater out of Wm1 & Wm2)

Bolt Area under Atmospheric Condition

Am1 = Wm1/fa = 4199992.47 / 57

= 73684.07 mm2

Am2 = Wm2 / fp = 21509032.05 / 53

= 405830.79 mm2

Selecting larger value,

Am = Am2 = 405830.79 mm2

To calculate bolt Diameter ,

Am = (∏/4) db2 x n

405830.79 = (∏/4) db2 x n

But,

n = G/25

n = 2324/25

= 92.96

But n must be a multiple of 4.Hence we round it off to the nearest multiple of 4 viz. 96.

i.e.

n = 96

Therefore,

405830.79 = (∏/4) x db2 x 96

db2 = 5382.49

db = 73.36 mm

i.e.

69

db = 74 mm

It is represented as,

M 74 , 96 Bolts.

The minimum BCD is given by,

B = Go + 2db + 12

= 2420 + 2(74) + 12

= 2580 mm

Design of Flange

Outer Diameter of flange

Dfo = B + 2db + 12

= 2580 + 2(74) + 12

= 2740 mm

Now,

k = 1/[0.3 + (1.5 x WmhG/H x G )

hG = B-G/2 = (2580 – 2324) /2 = 256/2 = 128 mm

H = Hydrostatic end force = (∏/4) x G2P = 0.7853 x 23242 x 3.8115 = 20585820.02 N

k = 1/ [0.3 + (1.5 x 21509032.05 x 128) / (20585820.02 x 2324) ]

k = 1/0.386

= 2.59

Thickness of flange

tf = G (P/Kf) ½

= 2324 (3.8115/2.59) ½

= 2324 x 1.213

= 2819.012

70

= 2820 mm

Thus the 2nd Re-Contact Drum (F-3044) can be designed as shown in fig. 6.

Recommended Alternate Design:

The new Design to the Booster Loop can be Accomplished as Shown in the Figure 8.

Here, we Cool the vapors up to 14 o C by adding one more Heat Exchanger before sending it to Re-Contact Drum 1. Chilled Brine is the cooling medium for this additional Heat Exchanger.

Performing the same Flash Calculations for the inlet as earlier we get the following composition at the outlet of first Re-Contact Drum (F-3047):

Comp. F Mol Frac. Fi = F.Yi Ki=Pi/Pt Assumed Li Vi = Fi-Li Vap MolMol Fr.

kmol/hr =Fi/(1+(V/L)Ki % Yi

H Hydrogen 380.96 0.8586 327.09 98.254 0.109 326.983 88.282 0.8828C1P Methane 380.96 0.0391 14.90 15.757 0.031 14.865 4.013 0.0401C2P Ethane 380.96 0.028 10.67 1.974 0.175 10.492 2.833 0.0283C3P Propane 380.96 0.0215 8.19 0.427 0.585 7.606 2.053 0.0205C4P I-Butane 380.96 0.0104 3.96 0.152 0.705 3.257 0.879 0.0088C4P n-butane 380.96 0.0096 3.66 0.100 0.902 2.755 0.744 0.0074C5P I-Pentane 380.96 0.0084 3.20 0.037 1.506 1.695 0.458 0.0046C5P n-Pentane 380.96 0.0074 2.82 0.027 1.553 1.266 0.342 0.0034C6P n-Hexane 380.96 0.0212 8.08 0.007 6.609 1.468 0.396 0.0040

Sum 1.00 382.560 116.735 12.175 370.385 100.000 1.000

L= 12.1749

V=F-L 370.3851

Calculated V/L = 30.42

V/L = 30.42

71

Liquid Vapo

r

LiMol frac. MW Avg MW Kg/Hr Vi

Mol Frac. MW

Avg MW

Kg/Hr

0.1090.0089

5 2 0.0179 913.53326.98

30.882

8 21.765

62229.3

6

0.0310.0025

5 16 0.0407 14.8650.040

1 160.642

1

0.1750.0143

7 30 0.4312 10.4920.028

3 300.849

8

0.5850.0480

5 44 2.1142 7.6060.020

5 440.903

6

0.7050.0579

1 58 3.3585 3.2570.008

8 580.510

0

0.9020.0740

9 58 4.2970 2.7550.007

4 580.431

4

1.5060.1237

0 72 8.9061 1.6950.004

6 720.329

5

1.5530.1275

6 72 9.1841 1.2660.003

4 720.246

1

6.6090.5428

3 86 46.6837 1.4680.004

0 860.340

9

sum 12.175 1 438 75.0334 913.53

370.387

1.0000 438

6.0190

72

Similarly for 2 nd Recontact Drum (F-3044)

Name ofMol flow

FMol Frac. Fi = F.Yi Ki=Pi/Pt Assumed Li Vi = Fi-Li

Vap.Mol %

Component kmol/hr =Fi/(1+(V/L)Ki Yi H Hydrogen 144.292 0.8828 127.381 49.5412 0.00156 127.3794 88.3414

C1P Methane 144.292 0.0401 5.786 9.3135 0.00038 5.7857 4.0126C2P Ethane 144.292 0.0283 4.083 1.3606 0.00182 4.0816 2.8307C3P Propane 144.292 0.0205 2.958 0.3261 0.00550 2.9525 2.0476C4P I-Butane 144.292 0.0088 1.270 0.1253 0.00613 1.2636 0.8764C4P n-butane 144.292 0.0074 1.068 0.0861 0.00749 1.0603 0.7353C5P I-Pentane 144.292 0.0046 0.664 0.034 0.01166 0.6521 0.4522C5P n-Pentane 144.292 0.0034 0.491 0.0256 0.01138 0.4792 0.3323C6 + n-Hexane 144.292 0.004 0.577 0.0078 0.04173 0.5354 0.3713 Sum 1 226.15 60.82 0.08765 144.19 100.000 ΣL = 0.088 ΣV = F-L 144.1899

Calculated

V/L = 1645.00

Estimated

V/L = 1645

73

LIQUID VAPOR

Li Mol frac. MWAvg Mol

Wt. Kg/Hr Vi Mol Frac. MWAvg Mol

Wt. Kg/Hr

0.00156 0.017798 2 0.03559612 6.34342 127.3794 0.707467 2 1.41493363 861.97220.00038 0.004335 16 0.0693668 5.7857 0.032134 16 0.51414163 0.00182 0.020764 30 0.62293212 4.0816 0.022669 30 0.68007776 0.0055 0.06275 44 2.76098118 2.9525 0.016398 44 0.7215218

0.00613 0.069937 58 4.05636052 1.2636 0.007018 58 0.40704693 0.00749 0.085454 58 4.95630348 1.0603 0.005889 58 0.34155735 0.01166 0.133029 72 9.57809469 0.6521 0.003622 72 0.26076756 0.01138 0.129835 72 9.34808899 0.4792 0.002661 72 0.19162677 0.04173 0.476098 86 40.9444381 0.5354 0.002974 86 0.25573119

0.08765 1 72.372162 6.34342 180.05 1 4.78740461

Performance Evaluation for C-3072 (1 st Stage Compression) for the Modified Design:

1. First Stage Calculations

The following are the 1st stage conditions for our calculations

Suction Temp = 47.67 ⁰C = 320.67 0 KSuction Pressure = 6.6 Bar aDischarge Temperature = 132.32 = 405.32 ⁰ KDischarge Pressure = 16.7 Bar a

74

Thus,




Cp of Mixture = 36.8665 (refer fig. 8)


Pr = P suction/Pc = 6.6 / 17.18 = 0.3841

Tr = T suction/Tc = 320.67 / 78.898 = 4.064


For Pr = 0.3841 & Tr = 4.064

Suction Compressibility = 0.99 (refer fig. 5)


Pr = P discharge/Pc = 16.7/17.18 = 0.972

Tr = T discharge/ Tc = 405.32/78.898 =5.137

For Pr = 0.972 & T r = 5.137

Discharge Compressibility is = 1.02 (Refer Fig. 5)

Therefore,


= 39.4562 / (39.4562-8.314)

= 1.25

75



= 16.7/6.6

= 2.53



Higher the discharge pressure , higher is the quantity that is left back in space between piston and the cyliner end walls.

Thus volumetric efficiency of the compressor can be estimated as per the following relation provided by the vendor at RIL,

Ƞ = 0.97 -C x [PR ^ (1/K) - 1] - L

= 0.97 - 0.1064[( 2.53)^(1/1.25) - 1] - 0.07

= 0.97 – 0.1064(2.1013 -1) – 0.07

= 0.97 – 0.1171 – 0.07

= 0.7828

Where,




76

L = Leakage losses normally 0.03 to 0.05 for Lubricated compressor and 0.07 to 0.1 for non- lubricated compressor.


Thus ,


Ƞ = 0.97 -C x [PR ^ (1/K) - 1] - Lc = 78.28 %

Piston Displacement

Piston displacement is defined as the net volume actually displaced by the piston at rated machine speed, as the piston travels its length from top dead center to bottom dead center.

For Double acting Reciprocating Compressor C-3072/73 ;

Bore (D) In m

Stroke (S) In m

Diameter of Rod (d) In m

1st Stage 0.40589 0.305 0.063552nd Stage 0.2159 0.305 0.06355


Where,



S= Cylindrical Sroke = 12 inches = 0.305 m


77

Therefore,

Piston Displacement = (3.14/4) x ( 0.40592 + (0.40592 – 0.063552)) x 0.305 x 390 x 60

= (0.785) x ( 0.1647 + 0.1607) x 0.305 x 390 x 60

= 0.785 x 0.3254 x 0.305 x 390 x 60

= 1823.468 m3 / hr


Inlet Capacity of a Compressor= Q (m3/hr) = Vd X Ƞ

= 1823.068 x 0.7828

= 1427.409 m3 / hr.

Specific Volume :



Vsp = Z x Ts x R

Ps MW

Where,



Ts = Suction Temp. at first stage = 320.67 ⁰ K

Ps = Suction pressure at first stage = 6.6 Bar a = 6.6/1.01325 = 6.51 atm a .



78

Therefore,

V sp = (0.995 x 320.67 / 6.51) x (0.08206/8.25)

= 48.765 x 0.009947

= 0.487 m3 / kg.

Capacity in Kg/Hr

M = Q * V sp


= 1427.409 / 0.487

= 2931.02 Kg / Hr


When the Compression takes place by adiabatic process, the discharge temperature can be predicted by the following relation:


= 320.67 x (2.53) (1.25 - 1/1.25)

= 320.67 x 1.203

= 386.08 o K



79

Ƞad = Tad – Ts

Td ₋actual - Ts

= (386.08 – 320.67)/(405.32 – 320.67)

= 65.61/84.65

= 0.77507

= 77.50 %

Adiabatic Head :



MWmix k-1

= 0.995 x (8.314/8.25) x 320.67 x (1.25/1.25-1) x [ (2.53)(1.25-1/1.25) - 1 ]

= 0.995 x 1.0077 x 320.67 x 5 x [ 1.203 – 1 ]

= 1607.61 x 0.203

= 326.34 kN-m/kg

Adiabatic Power :

It is the power required by the compressor to raise the pressure of gas. In mechanical terms it is a brake power.

80

Pad (kW) = m x Had

Ƞad x 3600

Where,

m = Total Inlet Feed to the Compressor in Kg/Hr.

Had = Adiabatic Head = 326.34 kN-m/kg

Ƞad = Adiabatic Efficiency = 77.5 % = 0.775

= 3143 x 326.46

0.775 x 3600

= 367.76 KW

81

Performance Evaluation for C-3073 (2 nd Stage Compression)

C-3072/73 Compressor is a Two stage Double acting reciprocating compressor.1st stage Discharge Pressure is = 1670 KPa whereas 1st Stage Knockout Drum pressure = 1651 KPa.

1. Suction Conditions:

The following are the 1st stage suction conditions for our calculations

Suction Temp = 14 ⁰C = 287 ⁰KSuction Pressure = 16.51 Bar aDischarge Temperature = 76.3 ⁰C = 349.3 ⁰ KDischarge Pressure = 34.65 Bar a

Thus,




Cp of Mixture = 32.789 (Refer Figure 10)


82

Pr = P suction/Pc = 16.51 / 16.588 = 0.9952

Tr = T suction/Tc = 287 / 65.674 = 4.37


For Pr = 0.9952 & Tr = 4.37

Suction Compressibility = 1.02 (refer Figure 5)


Pr = P discharge/Pc = 34.65/16.588 = 2.088

Tr = T discharge/ Tc = 349.3/65.674 = 5.31

For Pr = 2.088 & T r = 4.96

Discharge Compressibility is = 1.05 (refer fig. 5)


= 33.653/ (33.653-8.314)

= 1.328



= 34.65/16.51

= 2.098

83



Higher the discharge pressure , higher is the quantity that is left back in space between piston and the cylinder

End walls.

Thus volumetric efficiency of the compressor can be theoritically estimated as

Ƞ = 0.97 -C x [PR ^ (1/K) - 1] - L

= 0.97 - 0.1997[( 2.098)^(1/1.328) - 1] - 0.07

= 0.97 – 0.1997 (1.779 - 1) – 0.07

= 0.97 – 0.1555 – 0.07

= 0.7507

Where,




L = Leakage losses normally 0.03 to 0.05 for Lubricated compressor & 0.07 to 0.1 for non- lubricated compressor.


Thus ,


Ƞ = 0.97 -C x [PR ^ (1/K) - 1] - Lc = 75.07 %

Piston Displacement

Piston displacement is defined as the net volume actually displaced by the piston at rated machine speed

as the piston travels its length from top dead center to bottom dead center.

84

For Double acting Reciprocating Compressor ,

Bore (D) Stroke (S) Diameter of Rod (d)

1st Stage 0.40589 0.305 0.063552nd Stage 0.2159 0.305 0.06355


Where,



S= Cylindrical Sroke = 12.0078 inches = 0.305 m


Therefore,

Piston Displacement = (3.14/4) x (0.2159 2 + (0.2159 2 – 0.063552)) x 0.305 x 390 x 60

= (0.785) x ( 0.0466 + 0.0425) x 0.305 x 390 x 60

= 0.785 x 0.0891 x 0.305 x 390 x 60

Vd = 499.18 m3 / hr


Inlet Capacity of a Compressor Q (m3/hr) = Vd X Ƞ

= 499.18 x 0.7507

= 374.734 m3 / hr.

Specific Volume :

85



Vsp = Z x Ts x R

Ps MW

Where,



Ts = Suction Temp. at first stage = 287 ⁰ K

Ps = Suction pressure at first stage = 16.51/1.01325 = Bar a = 16.308 atm a .



Therefore,

V sp = (1.02 x 287 / 16.308) x (0.08206/6.019)

= 19.076 x 0.01162

= 0.2447 m3 / kg.

Capacity in Kg/Hr

M = Q / V sp


= 374.734 / 0.2447

= 1531.40 Kg / Hr

86


When the Compression takes place by adiabatic process, the discharge temperature can be predicted by the

following relation:


= 287 x (2.098) (1.328 - 1/1.328)

= 287 x (2.098)0.2469

= 287 x 1.2

= 344.4 o K



Ƞad = Tad – Ts

Td ₋actual - Ts

= (344.4 – 287)/(349.3 – 287)

= 57.4/62.3

= 0.9213

= 92.13 %

Adiabatic Head :



Mwmix k-1

87

= 1.035 x (8.314/ 6.019) x 287 x (1.328/1.328-1) x [ (2.098)(1.328-1/1.328) - 1 ]

= 1.035 x 1.3812 x 287 x 4.0487 x [ 1.2 – 1 ]

= 1.035 x 1.3812 x 287 x 4.0487 x 0.2

= 332.21 kN-m/kg

Adiabatic Power :

It is the power required by the compressor to raise the pressure of gas.In mechanical terms it is a brake power.

Pad (kW) = m x Had

Ƞad x 3600

Where,

m = Total Inlet Feed to the Compressor in Kg/Hr. = 868.47 kg/hr

Had = Adiabatic Head =332.21 kN-m/kg

Ƞad = Adiabatic Efficiency = 0.8183

= 868.47 x 332.21

0.9213 x 3600

= 86.98 KW

Total Power Required for the Two Compressors = 367.76 KW + 86.98 KW

= 454.74 KW

Power Required under current operating condition = 498.13 KW (From Page 24.)

88

Therefore the Net Power Saved in case of the new Recommended Design

= Power Required under current operating condition – Power required under the Recommended Design

= 498.13 – 454.74

= 43.39 KW per hour.

Cost of One unit of Power = Rs 5 per KW Hr.

Generally a plant operates for about 8000 hours per year

Thus Annual Savings = 43.39 x 5 x 8000 = Rs 17,35,600 /-

Cost of Power Required for cooling of Chilled Brine which is used in the additional Heat Exchanger which we are suggesting to provide.

For this we first calculate the total Heat which is to be removed in order to cool the feed further to a temperature of 14 o C . This is given by,

Q = [ Vapor flow rate at 32 0 C x Cpv x (Ti – Tf )]+ [ Liquid flow rate at 320 C x CpL x (Ti – Tf )] +

[( Liquid Flow Rate at 14 o C – Liquid Flow Rate at 32 o C) x Һ]

Where,

Q = Heat that is to be removed.

Vapor flow rate at 32 0 C = 2656.935 Kg/Hr.

Cpv = Specific Heat Capacity of Vapor = 1.165 Kcal / Kg 0 C

CpL = Specific Heat Capacity of Liquid = 0.574 Kcal/Kg 0 C

Ti = Intital temperature = 32 0 C.

Tf = Final Temperature = 14 o C

Liquid flow rate at 32 0 C = 485.82 Kg/Hr

Liquid flow rate at 14 o C = 913.53 Kg/Hr.

89

Һ = Sensible Heat = 88.17 Kcal / Kg 0 C

Therefore,

Q = [2656.935 x 1.165 x (32-14)] + [485.82 x 0.574 x (32-14)] + [(913.53 – 485.82) x 88.17]

= 55715.92 + 5019.49 + 37711.19

= 98446.6 Kcal/Hr

Now,

1 TR of Refrigeration = 3000 Kcal/Hr

Therefore,

98446.6 Kcal/Hr of Refrigeration = 98446.6/3000

= 32.81 TR

1 TR of Refrigeration requires 1 HP power = 0.746 KW/Hr

Therefore,

Power requirement for 32.81 TR of Refrigeration = 32.81 HP = 32.81 x 0.746 KW/Hr

= 24.47 KW/Hr.

Suppose That the Plant Operates for 8000 Hours per Year.

Therefore Annual Power Requirement = 24.47 x 8000

= 195760 KW

Cost of Power = Rs 5 per KW Hr.

Thus total operating cost of 32.81 TR of Refrigeration unit = 195760 x 5

= Rs 978800/-

The cost Needed for operating the needed Refrigeration system will be subtracted from the savings of Rs 17,35,600 /- as mentioned on Page 86 as this would be the expense to operate the required TR of Refrigeration System.

Thus Net savings in operating cost from the proposed design

90

= Annual Savings from the compressor section – Annual operating cost of 32.81 TR of Refrigeration

= 17,35,600 – 978800

= Rs 7,56,800/-

Conclusions & Recommendations

Conclusion

1. The entire Booster-Compressor Loop is running efficiently as a hydrogen purity of almost 89 % is being obtained as per our calculations.

2. Also the Reciprocating Compressor C-3072/73 is operating satisfactorily at an efficiency of 77.5 % for the 1st stage and at 81.83 % for the 2nd stage.

3. The two Re-contact Drums viz. F-3047 & F-3044 are overdesigned as the comparison between their dimensions as shown below justifies the same.

Recommendations

1. The recommended design should be taken under consideration by the company as it helps saving the operating costs to the tune of Rs 7,56,800/- as shown through the detailed calculations on page nos. 68 – 87. Also the load of 2nd Heat Exchanger comes down considerably.

2. As per our observations it’s clear that the two Re-Contact Drum are over designed so we recommend new design dimensions for them which can be tabulated as shown below (Based on our calculations as shown on Page Nos. 37-54) :

3. Although the height which is obtained for the 2 Re-contact drums is higher than the present value but it is not considered to be overdesigned as the cost of a vessel is a direct function of its Diameter and not its length i.e. as the Diameter of the vessel increases, its fabrication cost increases irrespective of its height as Larger diameter also

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1st Re-Contact Drum 2nd Re-Contact Drum Current Recommended Current Recommended

Diameter 1500 mm 920 mm 1000 mm 900 mm

Length 4300 mm 4513 mm 3700 mm 4442 mmL/D ratio 2.86 4.905 3.7 4.935

means larger Heads which eventually necessitates use of higher number of bolts and hence higher overall fabrication cost.

4. Keeping in mind the above point, I thus recommend that whenever the company Re-designs the Two Re-contact Drums then they should keep a L/D ratio near to 5and not as low as 2.86 and 3.7 for the 1st and 2nd re-contact drums respectively as in the present case to avoid unnecessary higher cost of fabrication and installation.

BIBLIOGRAPHY

Books

1. The properties of Gas and Liquid,

-Robert C. Reid, John. M. Prausnitz, Bruce E Poling

Constants to calculate the isobaric heat capacity of ideal gas. J/molK. Pg 658 – 732,

Compressibility Chart , Pg 31

2. Applied Process Design for Chemical and Petrochemical Pants , Volume2, Third Edition

-Ernest E. Ludwig.

Flash Vapourisation (Calculation), Pg 15

3. Stoichiometry (SI Unit), (3RD Edition) Pg206,

B.I. Bhatt and S.M. Vora

Antoine Constants, Pc, Tc values

4. Process Equipment Design – M.V. Joshi

(2nd Edition) Pg 123 -179

5. Platforming Section Manual at RIL.

92

Online Reference

1. www.wikipedia.org 2. www.ril.com 3. Intranet at RIL, Patalganga.

93

http://www.ril.com/

http://www.wikipedia.org/

Reliance Industries Limited-Final Report

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