Reliance Industries Limited-Final Report
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Transcript of 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
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
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
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,
MW mix = Average Molecular Weight = 7.058 kg/kmol
Pc of Mixture = 16.857 Bar a
Tc of Mixture = 71.902 ⁰ K
Cp of Mixture = 34.60360882 (Refer Figure 10)
At Suction Condition
Pr = P suction/Pc = 16.52 / 16.857 = 0.98
Tr = T suction/Tc = 305 / 71.902 = 4.24
From Nelson - Obert Generalized Compressibility Charts
For Pr = 0.98 & Tr = 4.24
Suction Compressibility = 1.02 (refer Figure 5)
At Discharge Conditions
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,
K-value of Mixture = Cp mix/C v mix = Cp mix/ (Cp mix - R)
= 36.897/ (36.897-8.314)
= 1.286
Average Compressibility = Z avg. = (Z1+Z2)/2 = (1.02+1.03) = 1.025
PR = Pressure Ratio = Pdischarge / Psuction
= 34.65/16.51
= 2.098
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 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,
C = Clearance in % for first stage = 19.97 % (from design data)
PR = Pressure Ratio = 34.65/16.51 = 2.098
k=Ratio of specific heats i.e. k value = 1.286
L = Leakage losses normally 0.03 to 0.05 for Lubricated compressor & 0.07 to 0.1 for non- lubricated compressor.
C-3072/73 is NON-LUBRICATED COMPRESSOR . Therefore L= 0.07
Thus ,
Volumetric Efficiency is
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
Piston displacement = Vd (m3/hr) = (∏/4) x (D2 + (D2 - d2)) x S x N x 60
Where,
20
d= diameter of the rod =2.5 inches= 0.06355 m
D= Cylindrical Bore Diameter = 8.5 inches = 0.2159 m
S= Cylindrical Sroke = 12.0078 inches = 0.305 m
N= Speed of Compressor = 390
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
Capacity of the compressor:
Inlet Capacity of a Compressor Q (m3/hr) = Vd X Ƞ
= 499.18 x 0.7444
= 371.58 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.
21
Z = Compressibility factor at first stage suction condition = 1.02
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 .
R = Universal Gas Constant = 0.08206 m3-atm/k-mol ⁰K
MW = Molecular Weight of gas at first stage = 7.058 kg/k-mol.
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
M (kg/hr) = Q (m3/hr) / V sp (m3/Kg)
= 371.58 / 0.2216
= 1676.8 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
= 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
Adiabatic Efficiency:
It indicates the actual process compressor deviation from the adiabatic compression.It is given by,
Ƞad = Tad – Ts
Td ₋actual - Ts
= (378.06 – 305)/(394.28 – 305)
= 73.06/89.28
= 0.8183
= 81.83 %
Adiabatic Head :
It is defined as the work done by the compressor to raise the pressure of the gas. It is given by ,
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)
Name ofMol Frac. Antioine Constants ln Pi Pi
Ki = Pi/Pt Xi= Ni/Ki
Compone
nt Kpa Yi A B C
H Hydrogen 0.86912.78
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
C5P I-Pentane 0.006913.71
9 2396.4 -37.55 5.4313 228.44 0.06590.10466
1
C5P n-Pentane 0.005713.97
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
Name ofMol Frac. Antioine Constants ln Pi Pi
Ki = Pi/Pt
Xi= Ni/Ki
Compone
nt Kpa Yi A B C
H Hydrogen 0.86912.78
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
Name ofMol Frac. Antioine Constants ln Pi Pi
Ki = Pi/Pt
Xi= Ni/Ki
Compone
nt Kpa Yi A B C
H Hydrogen 0.86912.78
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
C5P I-Pentane 0.006913.71
9 2396.4 -37.55 4.7588 116.61 0.03400.20272
5
C5P n-Pentane 0.005713.97
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
C5P I-Pentane 183.60.006
9 1.267 0.034 0.40618 0.8607 0.4780
C5P n-Pentane 183.60.005
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 = Volumetric flow rate in ft3/min.
QL = 17724.76 / (48.7101 x 60)
45
QL = 6.0647 ft.
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. = 18.485 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
L/D ratio = Total Height / Diameter
=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
Corrosion Allowance = C = 3.2 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
Corrosion Allowance = C = 3.2 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 = Volumetric flow rate in ft3/min.
QL = 16653.7 / (48.7101 x 60)
QL = 5.698 ft = 1.736 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 = 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
L/D ratio = Total Height / Diameter
= 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 = Volumetric flow rate in ft3/min.
QL = 16653.7 / (48.7101 x 60)
QL = 5.698 ft = 1.7367 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 = 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
L/D ratio = Total Height / Diameter
= 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
Corrosion Allowance = C = 3.2 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
Corrosion Allowance = C = 3.2 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
Corrosion Allowance = C = 3.2 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,
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
75
Average Compressibility = Z avg. = (Z1+Z2)/2 = (1.01+0.98) = 0.995
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
76
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
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
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
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.
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
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
Adiabatic Efficiency:
It indicates the actual process compressor deviation from the adiabatic compression.It is given by,
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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
Adiabatic Power :
It is the power required by the compressor to raise the pressure of gas. In mechanical terms it is a brake power.
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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
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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,
MW mix = Average Molecular Weight = 6.019 kg/kmol
Pc of Mixture = 16.588 Bar a
Tc of Mixture = 65.674 ⁰ K
Cp of Mixture = 32.789 (Refer Figure 10)
At Suction Condition
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Pr = P suction/Pc = 16.51 / 16.588 = 0.9952
Tr = T suction/Tc = 287 / 65.674 = 4.37
From Nelson - Obert Generalized Compressibility Charts
For Pr = 0.9952 & Tr = 4.37
Suction Compressibility = 1.02 (refer Figure 5)
At Discharge Conditions
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)
K-value of Mixture = Cp mix/C v mix = Cp mix/ (Cp mix - R)
= 33.653/ (33.653-8.314)
= 1.328
Average Compressibility = Z avg. = (Z1+Z2)/2 = (1.02+1.05) = 1.035
PR = Pressure Ratio = Pdischarge / Psuction
= 34.65/16.51
= 2.098
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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 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,
C = Clearance in % for first stage = 19.97 % (from design data)
PR = Pressure Ratio = 34.65/16.51 = 2.098
k=Ratio of specific heats i.e. k value = 1.328
L = Leakage losses normally 0.03 to 0.05 for Lubricated compressor & 0.07 to 0.1 for non- lubricated compressor.
C-3072/73 is NON-LUBRICATED COMPRESSOR . Therefore L= 0.07
Thus ,
Volumetric Efficiency is
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.
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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
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 = 8.5 inches = 0.2159 m
S= Cylindrical Sroke = 12.0078 inches = 0.305 m
N= Speed of Compressor = 390
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
Capacity of the compressor:
Inlet Capacity of a Compressor Q (m3/hr) = Vd X Ƞ
= 499.18 x 0.7507
= 374.734 m3 / hr.
Specific Volume :
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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 = 1.02
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 .
R = Universal Gas Constant = 0.08206 m3-atm/k-mol ⁰K
MW = Molecular Weight of gas at first stage = 6.019 kg/k-mol.
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
M (kg/hr) = Q (m3/hr) / V sp (m3/Kg)
= 374.734 / 0.2447
= 1531.40 Kg / Hr
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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
= 287 x (2.098) (1.328 - 1/1.328)
= 287 x (2.098)0.2469
= 287 x 1.2
= 344.4 o K
Adiabatic Efficiency:
It indicates the actual process compressor deviation from the adiabatic compression.It is given by,
Ƞad = Tad – Ts
Td ₋actual - Ts
= (344.4 – 287)/(349.3 – 287)
= 57.4/62.3
= 0.9213
= 92.13 %
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
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= 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.)
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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.
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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.
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