Kuwait University
College of Engineering & Petroleum
Chemical Engineering Department
Prof. Mohamed A. Fahim
Spring 2008/2009 Semester
Plant Design
ChE 491/51
Equipment Design
Production of Syngas by
Partial oxidation of natural
gas
- Group II :
Mariam H. AL-Shamma'a
204112285
Dalal B. AL-Othman
204111379
Bibi Y. AL-Motawa
204112571
1. Abstract
The aim of this report is to determine the design aspects of the different equipment
used in the production of synthesis gas by the partial oxidation of natural gas. This
includes dimensions, materials of construction as well as an approximate cost of our
major, minor, and auxiliary equipment.
Detailed calculations of the design as well as a theoretical background are given. The
design of each equipment is implemented using appropriate programs including Excel
and POLYMATH.
Design specification sheets that summarize the operating conditions, design variables,
and costs are provided at the beginning of each section.
2
Contents Page
Contents
Page
1. Abstract
2
2. Absorber
6
2.1 Design
6
2.1.1 T-100 Dalal
9
2.1.2 T-101 Mariam
15
2.1.3 T-102 Bibi
27
3. Reactor
35
3.1 Theory
35
3.2 Design
38
3.2.1 CRV-100 Dalal
40
4. Heat Exchangers , Coolers and 42
Heaters
4.1 Theory
42
4.2 Design
60
3
Contents
Page
4.2.1 E-106 - Mariam
65
4.2.2 E-102 - Mariam
80
4.2.3 E-100 – Bibi
94
4.2.4 E-103 – Bibi
106
4.2.5 E-104 – Bibi
118
4.2.6 E-105 – Bibi
129
5. Separators
140
5.1 Theory
140
5.2 Design
148
5.2.1 V-100 – Dalal
151
5.2.2 V-103 - Mariam
153
5.2.3 V-104 – Bibi
157
6. Pumps
162
6.1 Theory
162
4
Contents
Page
6.2 Design
166
6.2.1 P-100 – Mariam
167
7. Compressors
170
7.1 Theory
170
7.2 Design
173
7.2.1 K-100 Dalal
174
5
2. Absorber
2.1 Design
Equipment Name
Absorber
Objective
Removes small amounts of particulate carbon
from the raw syngas product
Equipment Number
T-100
Designer
Dalal Al-Othman
Type
Packed Bed
Location
Partial Oxidation and Heat Recovery Section
Material of Construction
Stainless-Steel
Insulation
Foam Glass
Cost ($)
Operating Condition
Operating Temperature (oC)
149
Operating Pressure (psia)
980
908.53
Gas Density (kg/m3)
29.048
2.283
Height (m)
13.6
Design Considerations
Liquid Density (kg/m3)
Dimensions
Diameter (m)
6
Equipment Name :
Absorber
Objective :
Removes CO2 from the product syngas by
using MDEA
Equipment Number :
T-101
Designer :
Mariam Hussain AL-Shamma’a
Type :
Packed Column
Location :
CO2 Separation Section
Material of Construction :
Carbon Steel
Insulation :
Foam Glass
Cost :
737400 $
Operating Condition
Operating Temperature (oC) :
49
Operating Pressure (psig) :
943.3
Design Considerations
Liquid Density (kg/m3) :
977.13
Gas Density (kg/m3) :
29.891
Gas Flow rate (MMSCFD) :
235.4
Liquid Flow rate (barrel/day) :
2.5612E6
1.593615262
Height (m) :
Dimensions
Diameter (m) :
7
14.47084386
Equipment Name
H2O Absorber
Objective
To remove H2O from MDEA solution
Equipment Number
T-102
Designer
Bibi A-Motawa
Type
Packed absorber
Location
Before cooler E-105 and after V-103
Material of Construction
carbon Steel
Insulation
Quartz wool
Cost ($)
2452500 $
Operating Condition
Operating Temperature (oF)
592
Gas Feed Flow Rate (kg/s)
46
Operating Pressure (psia)
1090.2.7
Liquid Flow Rate (kg/s)
96.6
Diameter (m)
2.73
Inert Type
Pall ring metal
packing
Height (m)
9.949588565
Thickness (m)
0.236437222
8
2.1.1 T-100 – Dalal
The particulate carbon formed in the oxidizer reactor is absorbed in water in the packed
column T-100. The feed is 54.23 kg/s of gas and 25.07 kg/s of water. Almost all the
carbon is to be absorbed in water.
Note: It is assumed that carbon can be dissolved in water.
Estimating NOG
The first step in the design of the column is to determine the number of overall gasphase transfer units, NOG using the Figure 11.40 below.
The number of transfer units NOG is a function of the ratio of between the top and
bottom mole fractions of the solute in the gas phase y1/y2 and mGm/Lm, where m is the
slope of the equilibrium line and Gm/Lm is the slope of the equilibrium line. Since no
water-carbon equilibrium data are available and therefore those slopes are unknown to
us, it is assumed that mGm/Lm = 0.64
9
NOG = 10
Column Diameter
The liquid-vapor flow factor is given by:
𝐹𝐿𝑉 =
𝐿𝑤 𝜌𝑣
√
𝑉𝑤 𝜌𝐿
Where liquid mass flow-rate Lw = 25.07 kg/s
Vapor mass flow-rate Vw = 54.23 kg/s
Liquid density ρL=908.53 kg/m3
Vapor density ρV = 29.05 kg/m3
FLV = 0.08
Designing for a pressure drop of 83 mm H2O/m packing, from Figure 11.44,
10
K4 = 2.5 and at flooding K4 = 3.9
2.5
Percentage flooding = √3.9 × 100 = 80.06%
The gas mass flow-rate per unit cross-sectional area 𝑉𝑤∗ , kg/m2s, is given by:
1⁄2
𝑉𝑤∗
𝐾4 𝜌𝑣 (𝜌𝐿 − 𝜌𝑣 )
=[
]
1.31𝐹𝑝 (𝜇𝐿 ⁄𝜌𝐿 )0.1
where the packing factor is 130, liquid viscosity µL = 0.0001829 N.s/m2, and all other
variables have been defined previously.
𝑉𝑤∗ = 13.24 kg/m2s
The column area required = 54.23/13.24 = 4.096 m2
11
4
Diameter = √𝜋 × 4.096 = 2.28 m
Estimation of HOG by Cornell's Method
𝑯𝐺 =
𝐷𝑐 1.11 𝑍 0.33
)
(
) ⁄ (𝐿∗𝑤 𝑓1 𝑓2 𝑓3 )0.5
0.305
3.05
0.011𝜓ℎ (𝑆𝑐)0.5
𝑣 (
𝑯𝐿 =
0.305𝜙ℎ (𝑆𝑐)0.5
𝐿 𝐾3 (
𝑍 0.15
)
3.05
Where HG is the height of a gas-phase transfer unit, m
HL = height of a liquid-phase transfer unit, m
Gas Schmidt number (𝑆𝑐)𝑣 = (𝜇𝑣 ⁄𝜌𝑣 𝐷𝑣 )
o Vapor viscosity µv = 1.61×10-5 N.s/m2, Dv = 1.45×10-5m2/s
1.161×10−5
o (𝑆𝑐)𝑣 = (29.05×1.45×10−5 ) = 0.0383
Liquid Schmidt number (𝑆𝑐)𝐿 = (𝜇𝐿 ⁄𝜌𝐿 𝐷𝐿 )
o Liquid viscosity µL = 1.83 ×10-4 N.s/m2, DL = 1.7×10-9m2/s
1.83×10−4
o (𝑆𝑐)𝑣 = (
908.53×1.7×10−9
) = 0.0383
Column diameter Dc = 2.28m
Column height, as a first assumption, Z = 12 m
The percentage flooding correction factor K3 is taken from Figure 11.41
K3 = 0.5
12
HG factor ψh is taken from Figure 11.42 for 80% flooding and 1 inch ring packing.
ψh = 45
13
HL factor ϕh is taken from Figure 11.43 for 𝐿∗𝑤 = 6.12 kg/m2s and 1 inch ring packing.
Liquid mass flow-rate per unit area column cross-sectional a 𝐿∗𝑤 = 6.12 kg/m2s
Liquid viscosity correction factor 𝑓1 =(𝜇𝐿 ⁄𝜇𝑤 )0.16
Liquid density correction factor 𝑓2 =(𝜌𝑤 ⁄𝜌𝐿 )1.25
Surface tension correction factor 𝑓3 = (𝜎𝑤 ⁄𝜎𝐿 )0.8
Since the liquid is water,
𝑓1 = 𝑓2 = 𝑓3 = 1
HG = 0.58m , HL = 1.22m
𝐻𝑂𝐺 = 𝐻𝐺 + 𝑚
𝐺𝑚
𝐻
𝐿𝑚 𝐿
HOG = 1.36
Z = HOG NOG
Z = 13.61 m
14
2.1.2 T-101 Mariam
Detailed Calculations :
Removes CO2 from the product syngas by using MDEA .
T-101 is a Packed Column :
* Data from Hysys :
Properties
Liquid
Gas
Mass Flow-rate (kg/h)
24189.1321
141616.8355
Molar Flow-rate (kgmole/h)
424.9
11514.5
Density ρ (Kg/m3)
977.13
29.891
Viscosity µ (Ns/m2)
2.08E-03
1.25E-05
Surface Tension (N/m)
0.06791
0.067851
15
For Operating line ( CO2 - MDEA ) :
CO2 mole fraction in Liquid phase
CO2 mole fraction in Vapor phase
0.00003
0.045954
0.001756
0.000591
For Equilibrium line ( CO2 - MDEA ) :
CO2 W/W% in liquid phase
CO2 mol fraction in vapor
phase
0
0
0.0001
2.76E-06
0.0005
1.38E-05
0.001
2.76E-05
0.0015
4.14E-05
0.002
0.0000552
0.0025
6.9E-05
0.003
8.28E-05
0.0035
9.66E-05
0.004
0.0001104
0.0045
0.0001242
16
Plotting the Equilibrium and operating lines :
Packed bed Diagram for Absorber
vapor fraction (mol)
0.00014
0.00012
0.0001
0.00008
0.00006
0.00004
Equlibrium data
0.00002
Operating Line
0
0
0.001
0.002
0.003
liquid fraction (mol)
m ( slop of the equilibrium curve ) = 0.02767597
m Gm/Lm = 0.02767597 * (11514.5/424.9) = 0.75
y1/y2 = 0.045954/0.000591 = 77.75634518
17
0.004
0.005
Using Figure (11.40) :
Number of Transfer units , NOG = 12
18
* Column Diameter :
Gas Flow-rate =141616.8355 / 3600 = 39.3380099 Kg/s
Liquid Flow-rate = 424.9/3600 = 6.71920336 Kg/s
Select 38mm ( 1.5 in. ) Cascade Mini Ring Packing :
From Table (11.3) :
Fp = 80 m-1
FLV = (Lw* / Vw*) (ρv / ρl)0.5 = (6.71920336 /39.3380099 ) (29.891/977.13)0.5
FLV = 0.02987442
19
Design for pressure drop 125 mm H2O/m packing :
Using Figure (11.44) :
K4 = 3.9
At Flooding : K4 = 5.5
Percent Flooding = 84.20753583 %
Vw* = ( 3.9 * 29.891 (977.13-29.891)/ 13.1*80 (2.08E-03/977.13)0.1 )1/2
= 19.72220837 Kg/m2s
20
Column Area required = Gas Flow-rate/ Vw* = 39.3380099 /19.72220837
= 1.99460472 m2
Diameter = ( 4/π * Column Area required )0.5 = ( 4/π * 1.99460472 )0.5
= 1.593615262 m = 5.23 ft
Column Area = π/4 * Diameter2 = π/4 * (1.593615262)2
= 1.994604717 m2 = 41.93724 ft2
Packing size to column diameter ratio = 1.593615262/(38/1000) = 41.9372437
Percentage flooding at selected diameter = Percentage Flooding * (Column Area
required/Column Area)
= 84.20753583 (1.99460472 /1.994604717) = 84.20753583 %
21
* Estimation of HOG :
Cornell's Method :
m2/s
DL =
1.5E-09
DV =
0.00000139 m2/s
From Mass Transfer book
Gas Schmidt number (Sc)v = μv /ρv * Dv = 1.25E-05/(29.891*0.00000139) = 0.301671613
Liquid Schmidt number (Sc)L = μL /ρL * DL = 2.08E-03/(977.13*1.5E-09) = 1418.506324
L*w = Liquid Flow-rate/Column Area = 6.71920336 /1.994604717
= 3.368689196 Kg/s m2
Using Figure (11.41) :
K3 =
0.43
22
Using Figure (11.42) :
ψh =
83
Using Figure (11.43) :
φh =
0.048
23
HOG can be expcted to be around =
The Z (height) can be taken as =
The Diameter Correction term will be taken as =
1
12
2.3
m
m
For Water :
Surface Tension =
ρ=
μ=
0.067912
989.63
0.0005535
f1 = (μ Liquid / μ water)0.16 = (2.08E-03/0.0005535)0.16 = 1.235836908
f2 = (ρ water / ρ Liquid)1.25 = (989.63/977.13)1.25 = 1.016016196
f3 = (Surface Tension of water / Surface Tension of Liquid)0.8
= (0.067912/0.06791)0.8 = 1.000023561
Assume f1 = f2 = f3 = 1
24
N/m
Kg/m3
Ns/m2
HG = (0.011) (83) (0.301671613)0.5 (1.593615262/0.305)1.11 (12/3.05)0.33 / (6.71920336
* 1)0.5
= 0.987520747 m
HL = (0.305) (0.048) (1418.506324)0.5 (0.43) (12/3.05)0.15 = 0.291177211 m
HOG = HG + m (Gm/Lm) HL = 0.987520747 + 0.75 * 0.291177211
= 1.205903655 m
Z = Zestimated * HOG = 12*1.205903655
= 14.47084386 m = 47.48 ft
25
* Cost and Material of Construction :
- The Material of Construction is Carbon Steel with 38 mm ( 1.5 in ) Cascade Mini Ring
Packing.
Cost of Column = 737400 $
from www.matche.com
Using Table (16.28) :
Cost of Packing = 19.9 $/ft3
26
2.1.3 T-102 Bibi
A major application of absorption technology is the removal of H2O from
MDEA solution. There are two types of column used for absorption packed
column and plate column. In a packed bed the gas liquid contact is continuous,
not stage wise as in plate column. The liquid flows down the column over the
packing surface and the gas or vapor, counter currently, up the column. In some
gas absorption columns co-current flow is used. The performance of a packed
column is very dependent on the maintenance of good liquid and gas
distribution thought the packed bed, and this is an important consideration in
packed column design.
27
Top
Property
Gas 111
Liquid 20
Flow rate (Kg/s)
6.93E+01
1.20E+02
Density р (Kg/m3)
Viscosity µ (N s/m2)
Molecular wieght (g/mol)
66.033
1.55E-02
34.588
274.36
8.19E-02
62.353
1.05
0.43791
Property
Gas steam
Liquid 112
flow rate (Kg/s)
Density р (Kg/m3)
Viscosity µ (N s/m2)
Molecular wieght (g/mol)
Surface tension (N/m)
volumetric flow rate m3/s
22.7
55.683
2.09E-02
18.015
7.32E+01
950.35
0.82369
62.191
18.2/1000
7.74E-02
Sarface tension (N/m)
volumetric flow rate m3/s
Bottom
0.40764
Gas mass flow rate =
4.60E+01
kg/s
Gas density (ρv) =
2.78E+01
kg/m3
Gas viscosity (μv) =
Gas volumetric flow rate (m3/s)
gas mw
3.30E+01
9.02E+00
26.3015
N.s/m2
Liquid mass flow rate =
9.66E+01
kg/s
Liquid density (ρL) =
4.75E+02
kg/m3
Liquid viscosity (μL) =
liq volumetric flow rate(m3/s)
liq mw
1.38E+02
3.11E+01
62.272
N.s/m2
V*w : gas mass flow rate per unit cross-sectional area, kg/m2s
L*w : liquid mass flow rate per unit cross-sectional area, kg/m2s
FLV 
Lw
Vw
V
L
28
=
Choose:
0.508287704
Pressure drop per meter of packing height (∆P) =
21
mmH2O/
m packing
height
0.076
m
66
m2/m3
From Figure (2):
K4 at design ∆P =
0.48
K4 at flooding =
0.9
% flooding 
K 4 @ design P
 100
K 4 @ flooding
73.0296743
3
=
%
Choose type of packing
From Table:
Pall rings metal packing
Size of packing (dp) =
76
mm
=
Packing factor (Fp) =
52
m-1
Actual area of packing per unit volume (a) =
V w 
=
K 4  V  L   V
13.1F p  L  L 

0.1
3.15179795
8
kg/m2.s
Column area required (A) = Gas mass flow rate / V*w
14.5996033
=
4
m2
Diameter 
4

A
=
4.3114721 m
29
T (K)
590
∑ν
9.44
∑ν
20.1
vm
0.1569
Gas diffusivity (Dv) =
6.40E-09
m2/s
Liquid diffusivity (DL )=
1.90E-11
m2/s
T (k)
580
Critical surface tension for the packing material (σc) =
0.056
Liquid surface tension (σL) =
0.0182
N/m
Gravity acceleartion (g) =
9.81
m/s2
Gas constant (R) =
0.08314
bar.m3/kmol.K
L*w = Liquid mass flow rate / A =
6.61833049
5
kg/m2.s
aw : Effective interfacial area of packing per unit volume


aw
 1  exp  1.45 c

a
 L




0.75
 Lw

 a L



0.1
 
 L 2 a 
 w

  L2 g 


30
0.05
 
 L 2 
 w 
  L L a 


0.2




N/m
aw =
49.1346172
m2/m3
kL : Liquid film mass transfer coefficient
  
k L  L 
 L g 
kL =
1
3
 L
 0.0051 w
 aw  L
1.09952E09




2
3
 L

  L DL



1
2
ad 
0.4
p
m/s
kG : Gas film mass transfer coefficient
 V
k G RT
 K 5  w
a Dv
 a v




0.7
 v

  v Dv
P=
T=
150
584
K5 =
5.23
kG =
1.0602E-08



1
3
ad 
2
p
bar
K
kmol/sm2bar
Vw
Gm 
M .wt
Lm 
M.wt of gas
=
26.3015
Lw
M .wt
M.wt
of
solvent =
62.272
Gm : molar gas flow rate per unit cross sectional area =
0.119833
kmol/m2s
Lm : molar liquid flow rate per unit cross sectional area =
0.106281
kmol/m2s
HG 
Gm
kG aw P
31
HG : height of a gas phase transfer unit =
HL 
1533.59453
m
Lm
k L a w Ct
Ct : Total concentration = ρL /M.wt of solvent =
257813.328
3
HL : height of liquid phase transfer unit =
asuum
mGm/L
=
kmol/m3
m
then m 0.011838108
=
0.005
m : slope of equiliprium line =
H OG  H G  m
7.630637
0.011838108
Gm
HL
Lm
HOG : height of overall gas phasetransfer unit =
4974.794
y1: mole fraction of the solute in the gas at bottom =
0.5632
y2 : mole fraction of the solute in the gas at top =
0.8355
NOG : Number of overall gas phase transfer unit =
NOG : Number of overall gas phase transfer unit =
#NUM!
0.002
Z  H OG N OG
Z : height of column =
 mGm
H OG Ln
 Lm
HETP 
 mGm 

  1
 Lm 
9.949588565



32
m
m
HETP : height equivalent to theoritical plate =
NT 
26490.49
Z'
HETP
Z' : height of tray column section = Z/2 = 4.974794283
m
NT : number of theoritical trays =
tray
NA 
m
0.000187795
NT
m
m : Murphree tray efficiency for absorbtion less than
0.6 ≈
0.55
NA : number of actual trays =
0.000341446
Z'' : actual height of packing =
4.974794283
h: height of total column 12.9495885
=
7
m
≈
4
m
For cylindrical shape vessel the thickness is calculated as follows:
t
PR
SE  0.6 P
P : design pressure =
1090.2
R : inside radius of shell
=
84.8714995
From Table (3):
psia
inch
oF
at Tavg =
592
S : maximum allowable stress value, for stainless steel
304 =
12622.83
E : the joint efficiency, for spot 0.85
33
psi
tray
examined=
t : shell thickness =
9.18355204
4
inch
t actual  t  C
C:corrosion allowance, for stainless
steel=
1.25E-01
9.30855204
4
tactual =
inch
=
0.236437222
inch
m
Do : outside diameter
Di : inside diameter
Vo : volume based on outside diameter
Vi : volume based on inside diameter
4.78434661
8
m
931.217470
2
m3
756.235425
9
m3
Vmetal = Vo - Vi =
174.982044
3
m3
metal density =
7854
Do = Di + 2tactual =
Vo
2
o
Vi
2
i
h=
h=
kg/m3
1374308.976
3029829
2126177$
Cost =
34
kg
lb
3. Reactor
3.1 Theory
Reactor design is the heart of a producing almost all industrial chemicals. The chemical
reactor is where high value products are produced through chemical transformation.
Reaction engineers are concerned with the reactor yield, selectivity, safety,
environment, product quality and purity, and plant economic viability of many different
types of reactors, products and operational conditions.
In chemical engineering, chemical reactors are vessels designed to contain chemical
reactions. The design of a chemical reactor deals with multiple aspects of chemical
engineering. Chemical engineers design reactors to maximize net present value for the
given reaction. Designers ensure that the reaction proceeds with the highest efficiency
towards the desired output product, producing the highest yield of product while
requiring the least amount of money to purchase and operate.
The design of an industrial chemical reactor must satisfy the following requirements:
1. The chemical factors: the kinetics of the reaction. The design must provide
sufficient residence time for the desired reaction to proceed to the required
degree of conversion.
2. The mass transfer factors: with heterogeneous reactions the reaction rates may
be controlled by the rates of diffusion of the reacting species; rather than
chemical kinetics.
3. The heat transfer factors: the addition or the removal of the heat of reaction.
4. The safety factors: the confinement of hazardous reactants and products, and
the control of the reaction and the process conditions.
35
Principal types of reactors
The following characteristics are normally used to classify reactor design:
1. Mode of operation: batch or continuous.
2. Phases present: homogeneous or heterogeneous
3. Reactor geometry:
i.
Stirred tank reactor
ii.
Tubular reactor
iii.
Packed bed: fixed and moving
iv.
Fluidized bed
Reactor geometry (type)
Continuous Stirred Tank Reactor (CSTR)
The CSTR is normally run at steady state and is operated so as to be quite well mixed.
The CSTR is normally modeled as having no spatial variations in concentration,
temperature, or reaction rate throughout the vessel.
A CSTR is used when intense agitation is required and can either be used by itself or as
part of a series or battery of CSTRs. It is relatively easy to maintain good temperature
control but there is however the disadvantage that the conversion of the reactant per
volume is smallest of the flow reactors.
Figure R.1 CSTRs in series
36
Some important considerations for CSTR are:
At steady-state, the inlet flow rate must equal the outlet mass flow rate;
otherwise the tank will overflow or go empty (transient state). While the reactor
is in a transient state the model equation must be derived from the differential
mass and energy balances.
2. The reaction proceeds at the reaction rate associated with the final (output)
concentration.
3. Often, it is economically beneficial to operate several CSTRs in series. This allows,
for example, the first CSTR to operate at a higher reagent concentration and
therefore a higher reaction rate. In these cases, the sizes of the reactors may be
varied in order to minimize the total capital investment required to implement
the process.
4. It can be seen that an infinite number of infinitely small CSTRs operating in series
would be equivalent to a PFR.
1.
Plug Flow Reactor (PFR)
A PFR consists of a cylindrical pipe and is normally operated at steady state. The
reactants are continually consumed as they flow down the length of the reactor.
The PFR is relatively easy to maintain and it usually produces highest conversion per
reactor volume of any of the flow reactors. Te disadvantage of the tubular reactor is
that it is difficult to control the temperature within the reactor and hot spots can occur
when the reaction is exothermic.
Some important aspects of the PFR:
All calculations performed with PFRs assume no upstream or downstream
mixing, as implied by the term "plug flow".
2. Reagents may be introduced into the PFR at locations in the reactor other than
the inlet. In this way, a higher efficiency may be obtained, or the size and cost of
the PFR may be reduced.
3. A PFR typically has a higher efficiency than a CSTR of the same volume. That is,
given the same space-time, a reaction will proceed to a higher percentage
completion in a PFR than in a CSTR.
1.
37
Figure R.2 PFR
Packed bed reactors
There are two types of packed bed reactor: those in which the solid is the reactant and
those in which the solid is the catalyst. Many types of the first example can be found in
the extractive metallurgical industries.
In the chemical process industries the designer will normally be concerned with the
second type: catalytic reactors. Industrial packed-bed catalytic reactors range in size
from small tubes, a few centimeters in diameter, to large diameter packed beds. Packed
bed reactors are used for gas and gas-liquid reactions.
Fluidized bed reactors
The essential feature of a fluidized bed reactor is that the solids are held in suspension
by the upward flow of the reacting fluid which promotes high mass and heat transfer
rates and good mixing. Fluidized bed reactors are useful where it is necessary to
transport large quantities of solids as part of the reaction process, such as where
catalyst are transferred to another vessel for regeneration.
38
3.2 Design
Equipment Name
Reactor
Objective
Convert natural gas into synthesis by partial
oxidation
Equipment Number
CRV-100
Designer
Dalal Al-Othman
Type
Plug flow
Location
Partial Oxidation and Heat Recovery
Material of Construction
Stainless Steel
Insulation
Foam Glass
Operating Condition
Operating Temperature (oC)
371
Volume of Reactor (m3)
95
Operating Pressure (psia)
1000
Catalyst Type
-
Feed Flow Rate (lbmole/h)
19579.66
Catalyst Density (Kg/m3)
-
Conversion (%)
96
Catalyst Diameter (m)
-
Number of Beds
-
Reactor Diameter (m)
3.11
Height of Bed/s (m)
-
Reactor Thickness (m)
0.0514
Length of Reactor (m)
12.46
Cost ($)
515900
39
3.2.1 CRV-100
CH4 + 0.5 O2→2 H2 + CO
Desired Conversion of methane = 96 %
Design Equation
𝑑𝑋 −𝑟𝐴
=
𝑑𝑉 𝐹𝐴0
Rate Law
Assuming the reaction is elementary the rate law is:
−𝑟𝐴 = 𝑘𝐶𝐴2 𝐶𝐵
k = 1200s-1
Stoichiometry
This is a gas phase reaction with no pressure drop.
𝐶𝐴 = 𝐶𝐴0
𝐶𝐵 = 𝐶𝐴0
(1 − 𝑋) 𝑇0
(1 + 𝜀𝑋) 𝑇
𝑏
(𝜃𝐵 − 𝑎 𝑋) 𝑇𝑜
(1 + 𝜀𝑋) 𝑇
𝜀 = 𝑦𝐴0 𝛿 = 0.43(1 + 2 − 0.5 − 1) = 0.65
𝜃𝐵=
𝑦𝐵0
= 0.48
𝑦𝐴0
Energy Balance
𝑜 )
𝑑𝑇
−𝑟𝐴 (−∆𝐻𝑅𝑥
=
𝑑𝑉 𝐹𝐴0 (𝐶𝑝𝐴 + ∆𝐶𝑝 )
Calculation of Mole Balance Parameters
Mass flow-rate of entering methane FA0
𝐹𝐴0 =
61398000 𝑔/ℎ
𝑚𝑜𝑙
= 3837375
= 1065.94 𝑚𝑜𝑙/𝑠
16 𝑔/𝑚𝑜𝑙
ℎ
40
𝐶𝐴0 =
𝐹𝐴0
𝑣
=
10.66
0.88
= 12.11 mol/m3
Calculation of Energy Balance Parameters
a. The standard heats of formation are
𝑜
Methane: 𝐻𝑅𝑥
=-74.81 kJ/mol
𝑜
Carbon Monoxide: 𝐻𝑅𝑥
=-1.106 kJ/mol
𝑜
∆𝐻𝑅𝑥 = −1.106 − (−74.81) = 73.704 kJ/mol
b. The heat capacity ∆Cp =51.851 J/mol.K3
With the aid of POLYMATH to solve two ordinary differential equations, the volume
achieved is:
V = 95 m3 = 23370 gallons
Calculation of Reactor Dimensions
Assuming the reactor length L and the reactor diameter D are:
L/D = 4
𝑉=
𝜋 2
𝐷 𝐿 = 𝜋𝐷3
4
D = 3.112m
L = 12.46m
Wall thickness, 𝑡 = 𝑆𝐸
𝑃𝑟𝑖
𝑗 −0.6𝑃
+ 𝐶𝐶
Where the maximum allowable pressure P = 900 psig
The inside radius of the reactor ri = 61.26 inch.
The maximum allowable working stress S = 34809.0943psi
The efficiency of joints Ej = 0.85
Allowance for corrosion Cc = 0.125 inch.
t = 2.02 in = 5.14 cm
41
4. Heat Exchangers, Coolers and Heaters
4.1 Theory
A shell and tube heat exchanger is a class of heat exchanger designs. It is the
most common type of heat exchanger in oil refineries and other large chemical
processes, and is suited for higher-pressure applications. As its name implies,
this type of heat exchanger consists of a shell (a large vessel) with a bundle of
tubes inside it. One fluid runs through the tubes, and the second runs over the
tubes (through the shell) to transfer heat between the two fluids. The set of
tubes is called a tube bundle, and may be composed by several types of tubes:
plain, longitudinally finned, etc.
Figure E.1 Heat exchanger
Two fluids, of different starting temperatures, flow through the heat exchanger. One
flows through the tubes (the tube side) and the other flows outside the tubes but inside
the shell (the shell side). Heat is transferred from one fluid to the other through the
tube walls, either from tube side to shell side or vice versa. The fluids can be either
liquids or gases on either the shell or the tube side. In order to transfer heat efficiently,
42
a large heat transfer area should be used, so there are many tubes. In this way, waste
heat can be put to use. This is a great way to conserve energy.
Heat exchangers with only one phase (liquid or gas) on each side can be called onephase or single-phase heat exchangers. Two-phase heat exchangers can be used to heat
a liquid to boil it into a gas (vapor), sometimes called boilers, or cool a vapor to
condense it into a liquid (called condensers), with the phase change usually occurring on
the shell side. Boilers in steam engine locomotives are typically large, usually
cylindrically-shaped shell-and-tube heat exchangers. In large power plants with steamdriven turbines, shell-and-tube (see Condenser (steam turbine) ) condensers are used to
condense the exhaust steam exiting the turbine into condensate water which can be
recycled back to be turned into steam, possibly into a shell-and-tube type boiler.
Figure E.2 Shell and tube heat exchanger
Shell and tube heat exchanger Design :
There can be many variations on the shell and tube design. Typically, the ends of each
tube are connected to plenums (sometimes called water boxes) through holes in tube
sheets. The tubes may be straight or bent in the shape of a U, called U-tube.
43
Figure E.3 U-tube heat exchanger
In nuclear power plants called pressurized water reactors; large heat exchangers called
steam generators are two-phase, shell-and-tube heat exchangers which typically have
U-tubes. They are used to boil water recycled from a steam turbine condenser into
steam to drive the turbine to produce power. Most shell-and-tube heat exchangers are
either 1, 2, or 4 pass designs on the tube side. This refers to the number of times the
fluid in the tubes passes through the fluid in the shell. In a single pass heat exchanger,
the fluid goes in one end of each tube and out the other.
Figure E.4 Straight-tube Heat Exchanger (one pass tube-side)
Steam turbine condensers in power plants are often 1-pass straight-tube heat
exchangers. Two and four pass designs are common because the fluid can enter and exit
on the same side. This makes construction much simpler.
44
There are often baffles directing flow through the shell
side so the fluid does not take a short cut through the
shell side leaving ineffective low flow volumes.
Counter current heat exchangers are most efficient
because they allow the highest log mean temperature
difference between the hot and cold streams. Many
companies however do not use single pass heat
exchangers because they can break easily in addition
to being more expensive to build. Often multiple heat
exchangers can be used to simulate the counter
current flow of a single large exchanger.
Figure E.5 Straight-tube Heat
Exchanger (two pass tube-side)
Selection of tube material:
To be able to transfer heat well, the tube material should have good thermal
conductivity. Because heat is transferred from a hot to a cold side through the tubes,
there is a temperature difference through the width of the tubes. Because of the
tendency of the tube material to thermally expand differently at various temperatures,
thermal stresses occur during operation. This is in addition to any stress from high
pressures from the fluids themselves. The tube material also should be compatible with
both the shell and tube side fluids for long periods under the operating conditions
(temperatures, pressures, pH, etc.) to minimize deterioration such as corrosion. All of
these requirements call for careful selection of strong, thermally-conductive, corrosionresistant, high quality tube materials, typically metals. Poor choice of tube material
could result in a leak through a tube between the shell and tube sides causing fluid
cross-contamination and possibly loss of pressure.
Applications
:
Shell and tube heat exchangers are frequently selected for such applications as:







Process liquid or gas cooling.
Process or refrigerant vapor or steam condensing.
Process liquid, steam or refrigerant evaporation.
Process heat removal and preheating of feed water.
Thermal energy conservation efforts, heat recovery.
Compressor, turbine and engine cooling, oil and jacket water.
Hydraulic and lube oil cooling.
45
Figure E.6 Removable Bundle
Basic components of shell and tube heat exchanger :
Tubes :
The tubes are the basic component of the shell and tube exchanger. They provide heat
transfer surface between one fluid flowing inside the tubes and the other fluid flowing
across the outside of the tubes. They are made of copper and steel alloys. Other alloys
of
nickel,
titanium,
or
aluminum
may
also
be
used.
46
Shell and shell nozzle :
The shell is simply the container for the shell side fluid and the nozzlesare the inlet and
exit ports. The shell normally has a circular cross section and is commonly made by
rooling a metal plate of appropriate dimensions into a cylinder into a cylinder and
welding the longitudinal joint ("rolled shells"). In large exchangers, the shell is made out
of low carbon steel wherever possible for reasons of economy, through other alloys can
be and is used when corrosion or high temperature strength demands must be met.
Figure E.7 Components of shell and tube heat exchanger
Baffles :
Baffles serve two functions :
o They support the tubes in the proper position during assembly and and
operation and prvent vibration of the tubes caused by flow-induced
eddies.
o They guide the shell side flow back and forth across the tube field
,increasing the velocity and the heat transfer coefficient.
47
The most common baffle shape is the single segment . The segment sheared off must be
less than half of the diameter in order to insure that adjacent baffles overlap at least
one full tube row. For liquid flows on shell side, a baffle cut of 20 to 25 percent of the
diameter is common; for low pressure gas flow, 40 to 50 percent is more common in
order to minimize pressure drop.
For many high velocity gas flow s, the single segment baffle configuration results in an
undesirably high shell side pressure drop. One way to retain the structural advantages
of the segment baffle and reduce the pressure drop is to use the double segment baffle.
It halves the local velocity and therefore reduces the pressure drop by a factor of 4 from
a comparable size single segment unit.
Figure (E-8) : Single Segment Baffle
Figure (E-9): Double Segment Baffle
48
Figure E.10 different types of segments
Pattern of tubes :
There are four types of tubes layouts with respect to the shell side cross flow direction
between baffle tips :
1.
2.
3.
4.
Triangular (30o).
Rotated triangle (60o).
Square (90o).
Rotated square (45o).
49
Figure E.11 Pattern of tubes
50
Theory of Shell and Tube Heat Exchangers
Calculation
Q  mC p T
where ;
Qh = Heat load transfer in the hot side, KW.
m  Mass flow rate in Kg/s.
T  Temperature difference of the inlet and
outlet.
Tlm 
(T1  t 2 )  (T2  t1 )
(T  t )
ln 1 2
(T2  t1 )
where ;
TLM  Log means Temperature.
T1  Inlet shell side fluid temperature (oC).
T2  Outlet shell side fluid temperature (oC).
t1  Inlet tube side temperature (oC).
t 2  Outlet tube temperature (oC).
R
(T1  T2 )
(t 2  t1 )
S
(t 2  t1 )
(T1  t1 )
Tm  Ft Tlm
51
where ;
Tm  True temperature difference.
Ft  Temperature correction factor.
A
Q
UTm
Where ;
A  Provisional area in m2.
Q  Heat load in W.
Tm  True temperature difference.
A  DL
Where ;
A  Area of one tube, m2.
N t  Provisional area/Area of one tube.
1
N
Db  d 0 ( t ) n1
K1
Where ;
Db  Bundle diameter (mm).
d 0  Outside diameter (mm).
N t  Number of tubes.
K1 & n1 are constant.
52
Ds  Db  Clearance
where ;
Ds  Sell diameter.
Db  Bundle diameter (mm).
Ac 

4
(d i ) 2
where ;
Ac  Tube cross-sectional area.
di  Tube inner diameter.
Tubes N t

Pass
4
where ;
N t  Number of tubes.
At  Ac
Tubes
Pass
where ;
At  Total flow area.
Um 
m
At
53
Where ;
U m  Tube mass velocity.
At  Total flow area.
m  Mass flowrate in Kg/s.
Um
Ut 
 ref
where ;
U t  Tube linear velocity.
 ref  Density.
(4200 * (1.35  0.02t ) * U t )
0.8
hi 
di
0.2
where ;
hi  Inside coefficient (W/m2 oC).
U t  Tube linear velocity.
t  Mean temperature (oC).
Re 
U t d i

where ;
Re  Reynolds number.
54
  Fluid viscosity at the bulk fluid temperature, Ns/m2.
Pr 
Cp
kf
where ;
Pr  Prandtl number.
C p  Heat capacity.
k f  Thermal conductivity of stream.
hi 
k f j h Re(Pr) 0.33
di
where ;
hi  Inside coefficient (W/m2 oC).
j h  Tube side heat transfer factor.
k f  Thermal conductivity of stream.
Pr  Prandtl number.
lB 
Ds
5
where ;
l B  Baffle spacing.
55
Ds  Shell diameter.
pt  1.25d 0
where ;
pt  Tube pitch.
d 0  Outside diameter (mm).
As 
( p t  d 0 ) Ds l B
pt
where ;
As  Cross-flow area.
pt  Tube pitch.
d 0  Outside diameter (mm).
Ds  Shell diameter.
Gs 
m
As
where ;
Gs  Mass velocity.
As  Cross-flow area.
m  Mass flowrate in Kg/s.
56
de 
1. 1 2
2
( pt  0.917 d 0 )
d0
where ;
d e  Equivalent diameter (mm).
d 0  Outside diameter (mm).
pt  Tube pitch.
Re 
Gs d e

where ;
Re  Reynolds number.
d e  Equivalent diameter (mm).
Gs  Mass velocity.
  Fluid viscosity at the bulk fluid temperature, Ns/m2.
1
1
1



U 0 h0 hod
d 0 ln(
d0
)
di
2k w

d0 1 d0 1

d i hid d i hi
where ;
U 0  The overall heat transfer coefficient.
hod  Outside coefficient (fouling factor).
hid  Inside coefficient (fouling factor).
57

L
Pt  N p 8 j f 

 di
  


  w 
m
 u 2
 2.5 t
 2
where ;
Pt  Tube- side pressure drop (N/m²) (pa).
N p  Number of tube -side passes.
u t  Tube-side velocity, m/s.
L  Length of one tube.
j f  Friction factor.
 w  Fluid viscosity at the wall.
  Fluid viscosity at the bulk fluid temperature, Ns/m2.
D
Ps  8 j f  s
 de
 L  u s 2
 
 l B  2
 

 w



0.14
where ;
Ps  Shell-side pressure drop (N/m²) (pa).
j f  Friction factor.
L  Length of tube.
58
t
Pri
 Cc
SEj  0.6 P
where ;
t  Shell thickness (in).
P  Maximum allowable internal pressure (psig).
ri  Internal radius of shell before allowance corrosion is added (in).
E j  Efficiency of joints.
S  Working stress (psi).
C c  Allowance for corrosion (in)
59
4.2 Design
Equipment Name :
Heat Exchanger
Objective :
To decrease the reactor’s effluent
temperature from 1350 oC to 149 oC before
entering the water scrubber
Equipment Number :
E-106
Designer :
Mariam Hussain AL-Shamma’a
Type :
Shell and Tube
Location :
Partial Oxidation and Heat Recovery Section
Material of Construction :
Carbon Steel
Insulation :
Foam Glass
Cost :
289200 $
Operating Conditions
Shell Side
Inlet temperature (oC) :
25
Outlet temperature (oC) :
315
Inlet temperature (oC) :
1350
Outlet temperature (oC) :
149
Tube bundle Diameter (m) :
3.20066032
Number of Tubes :
2413.44
Q total (KW) :
78126.1
Shell Diameter (m) :
3.30066
LMTD (oC) :
429.3372
Heat Exchanger Area (m2) :
758.204
Tube Side
2o
U (W/m C) :
314
60
Equipment Name :
Cooler
Objective :
Cooling of T-100’s overhead vapor outlet
(stream 11) from 149oC to 49oC is carried out
in cooler E-102 to adjust its temperature
before it enters the CO2 Separation Section
Equipment Number :
E-102
Designer :
Mariam Hussain AL-Shamma’a
Type :
Shell and Tube heat exchanger
Location :
Partial Oxidation and Heat Recovery Section
Material of Construction :
Stainless-Steel
Insulation :
Foam Glass
Cost :
302400 $
Operating Conditions
Shell Side
Inlet temperature (oC) :
25
Outlet temperature (oC) :
35
Inlet temperature (oC) :
148.35
Outlet temperature (oC) :
49
Tube bundle Diameter (m) :
3.27368728
Number of Tubes :
2536.644623
Q total (KW) :
21368.38622
Shell Diameter (m) :
3.37368728
LMTD (oC) :
57.55506
Heat Exchanger Area (m2) :
1354.747699
Tube Side
2o
U (W/m C) :
320
61
Equipment Name
Heat exchanger.
Objective
Heating natural gas and prepare it to enter
the reactor (CRV-101).
Equipment Number
E-100.
Designer
Bibi al-motawa
Type
Shell and tube H.E
Location
After the k-100
Utility
Steam.
Material of Construction
Carbon steel
Glass wall and quartz
Insulation
Cost
54,000$
Operating Condition
Process
Inlet temperature (oC)
105
Outlet temperature (oC)
371
Steam
Inlet temperature (oC)
577
Outlet temperature (oC)
Q total (Duty) (kW)
15456.9716
LMTD (oC)
2
Number of tube passes
U (W/m2oC)
Tube
(m)
bundle
Number of Tubes
1000
Heat Exchanger Area (ft2)
Diameter 0.62423059
Shell Diameter (m)
62
297
198.91789
202
1640.03
0.68523059
Equipment Name
Heat exchanger.
Objective
Heating CO2 rich amine solution stream
and prepare it to enter the (V-103).
Equipment Number
E-103
Designer
Bibi al-motawa
Type
Shell and tube H.E
Location
After the T-101
Utility
Steam.
Material of Construction
Carbon steel
Glass wall and quartz
Insulation
Cost
58,500$
Operating Condition
Process
Inlet temperature (oC)
49.64
Outlet temperature (oC)
314.6
Steam
Inlet temperature (oC)
577
Outlet temperature (oC)
Q total (Duty) (kW)
20478.478
LMTD (oC)
2
Number of tube passes
U (W/m2oC)
Tube
(m)
bundle
Number of Tubes
600
Heat Exchanger Area (ft2)
Diameter 0.79692699
Shell Diameter (m)
63
297
254.80603
112
1897.11
0.86392699
Equipment Name
Heat exchanger.
Objective
Cooling stream 19 and prepare it to enter
the (V-104).
Equipment Number
E-104.
Designer
Bibi al-motawa
Type
Shell and tube H.E
Location
After the V-103
Utility
Cooling water
Material of Construction
Carbon steel
Glass wall and quartz
Insulation
Cost
88,500$
Operating Condition
Process
Inlet temperature (oC)
314.6
Outlet temperature (oC)
49
Cooling water
Inlet temperature (oC)
Q total (Duty) (kW)
2
Outlet temperature (oC)
12146.5097
2
Tube
(m)
bundle
123.8
LMTD (oC)
Number of tube passes
U (W/m2oC)
57
Number of Tubes
340
Heat Exchanger Area (ft2)
Diameter 1.18287634
Shell Diameter (m)
64
416
4034.26
1.18287634
Equipment Name
Heat exchanger.
Objective
Cooling stream 112 to recycle it to the
( T-101 )
Equipment Number
E-105.
Designer
Bibi al-motawa
Type
Shell and tube H.E
Location
After the T-102
Utility
Cooling water
Material of Construction
Carbon steel
Glass wall and quartz
Insulation
Cost
268,900$
Operating Condition
Process
Inlet temperature (oC)
312.9
Outlet temperature (oC)
49
Cooling water
Inlet temperature (oC)
Q total (Duty) (kW)
2
Outlet temperature (oC)
3444421.15
LMTD (oC)
2
Number of tube passes
U (W/m2oC)
Tube
(m)
bundle
Number of Tubes
500
Heat Exchanger Area (ft2)
Diameter 11.9038791
Shell Diameter (m)
65
57
123.271082
125837
1055310.56
11.9808791
4.2.1 E-106 Mariam
To decrease the reactor’s effluent temperature from 1350 oC to 149 oC before entering
the water scrubber.
Data from Hysys :
Shell Side ( Cooling Water )
Flow rate
2.44E+05
Kg/h
Inlet Temperature , t1
25
oC
Outlet Temperature , t2
Heat Capacity of inlet stream, Cpin
315
4.30236
oC
Heat Capacity of outlet stream, Cpout
3.64846
KJ/kgoC
Average Heat Capacity, Cpavg
3.97541
KJ/kgoC
Mass Density of inlet stream , ρin
1010.4
kg/m3
Mass Density of outlet stream , ρout
55.196
kg/m3
Average Mass Density, ρavg
532.798
kg/m3
Average Viscosity of stream, µavg
0.45565
mNs/m2
Average Thermal conductivity, Kf
Inlet Stream Pressure
0.34269
1500
W/moC
Psig
66
KJ/kg°C
Qc = mc *Cp * (t1-t2)
where ;
Qc = heat load in the cold side (KW)
mc = mass flowrate of cold fluid (Kg/h)
Cp = heat capacity of hot fluid (kJ/kgoC)
t1 = inlet temperature (oC)
t2 = outlet temperatue (oC)
Qc = 2.44E+05 * 3.97541 * (315-25) = 78126.1 KW > 1000 KW
- The type of the heat exchanger is shell and tube heat exchanger.
Tube Side ( Reactor Effluent stream)
Flow rate
5.43E+01
Kg/s
Average Heat Capacity , Cp
20.0815
kJ/kgoC
Average Mass Density , ρ
18.0752
kg/m3
Average Viscosity of stream , µ
0.032
mNs/m2
Average Thermal conductivity , Kf
0.183
W/moC
inlet Temperature , T1
1350
oC
outlet Temperature , T2
Inlet Stream Pressure
149
977.7
oC
∆Tlm = [(T1-t2) - (T2-t1)] / ln[(T1-t2 ) / (T2-t1)]
where ;
∆Tlm = log mean temperature difference
T1 = hot fluid temperature , inlet (oC)
T2 = hot fluid temperature , outlet (oC)
t1
= cold fluid temperature , inlet (oC)
t2 = cold fluid temperature , outlet (oC)
67
Psig
∆Tlm = [(1350-315) - (149-25)]/ ln [(1350-315) - (149-25)] = 429.3372 oC
Use one shell pass and two tube passes :
R = (T1-T2) / (t2-t1) = (1350-149)/(315-25) = 4.1414
S = (t2-t1) / (T1-t1) = (315-25)/(1350-25) = 0.2189
From figure (12.19) :
Ft = 0.8
∆Tm = Ft * ∆Tlm
where ;
∆Tm = true temperature difference
Ft = the temperature correction factor
∆Tlm = log mean temperature difference
∆Tm = 343.4698 oC
68
From Table (12.1) :
Taking Gases as the Hot fluid and Water as the Cold fluid :
Assuming U = 300 W/m2 oC
69
Provisional area ( A ) = Q / U * ∆Tm
where ;
Q = heat load (W)
U = overall heat transfer coefficient (W/m2 oC)
A = 78126.1 / (300*343.4698 ) = 758.204 m2 = 8161.2 ft2
Choosing :
Tube outside diameter (do) =
50
mm =
1.9685
in
Tube inner diameter (di) =
Tube length (L) =
46
2
mm =
m=
1.811
78.74
in
in
Area of one tube = L* do * π = 2 * (50/1000) * π = 0.31416 m2
Number of tubes = Provisional area / Area of one tube
= 758.204/0.31416 = 2413.44 tubes
As the shell-side fluid is relatively clean use 1.25 Triangular pitch :
From Table (12.4) :
K1 = 0.249
n1 = 2.207
70
Bundle Diameter ( Db ) = (do)*( Nt / K1)(1/n1)
where ;
do = outer diameter (mm)
Nt = Number of tubes
K1 and n1 are constants
Db = (50/1000) * (2413.44/0.249)(1/2.207)
= 3.20066032 m
Using a split-ring floating head type :
Using figure (12.10) :
Bundle diametrical clearance = 100 mm
Shell Diameter ( Ds ) = Db + Bundle diametrical clearance = 3.20066032 + (100/1000)
= 3.30066 m
71
Taking Cylindrical Head :
Shell length = (Ds/2) + (Ds/2) + L
= (3.30066 /2) + (3.30066 /2) + 2 = 5.30066 m
Tube-side coefficient :
First Method :
Mean Tube-side temperature ( t ) = (T1+T2)/2 = (1350+149)/2
= 749.5 oC
Tube cross-sectional area = π/4 * di2 = π/4 * (46)2
= 1661.9 mm
Tube per pass = Number of tubes /4 = 2413.44/4 = 603.36 tubes
Total flow area = Tubes per pass * Cross-sectional area
= 603.36 * (1661.9*10-6) = 1.00273 m2
mass velocity = mass flow rate / Total flow area = 5.43E+01/1.00273 = 54 kg/s.m2
Linear velocity ( ut ) = mass velocity / density = 54/18.0752 = 2.99E+00 m/s
hi = 4200 (1.35 + 0.02t) ut0.8 / di0.2
where ;
hi = inside coefficient (W/m2 oC)
t = mean tube-side temperature (oC)
ut = linear velocity (m/s)
di = tube inside diameter (mm)
hi = 4200 (1.35 + 0.02 * 749.5) (2.99E+00)0.8 / (46)0.2 = 76729.1 W/m2 oC
72
Second Method :
Reynolds number ( Re ) = ρ * ut * di / µ
= 18.0752 * 2.99E+00 * (46/1000) / (0.032/1000) = 78997.6
Prandtl number ( Pr ) = Cp * µ / kf = (20.0815/1000) * (0.032/1000) / 0.183 = 3.44899
L / di = 2 / (46/1000) = 43.4783
From figure (12.23) :
The Tube-side heat-transfer factor ( jh ) = 0.0027
Assuming that the viscosity of the fluid is the same as at the wall's :
(hi * di / kf) = jh Re Pr0.33 * (µ/µwall)0.14
hi = 1280 W/m2 oC
Take hi from the method which gives the lower value :
hi = 1280 W/m2 oC
73
Shell-side coefficient :
Choose baffle spacing (Lb) = Ds/5 = 3.30066/5 = 660.132 mm
Tube Pitch (Pt) = 1.25 * do = 1.25 * 50 = 62.5 mm
Cross-flow area ( As ) = [ (Pt - do) * Ds * Lb ] / Pt
= [ (62.5-50) * 3.30066 * 660.132 * 0.000001 ]/62.5
= 0.43577 m2
Mass velocity (Gs) = mass flow rate / cross-flow area
= 2.44E+05/0.43577 = 155.509 Kg/s.m2
Equivalent diameter ( de ) = (1.1/do) * (pt2 - 0.917do2)
= (1.1/50) * (62.5-0.91*502) = 35.5025 mm
Mean Shell-side temperature = (t1 + t2) / 2 = (25+315)/2 = 170 oC
Reynolds number ( Re ) = (Gs * de) / µ
= 155.509 *( 35.5025/1000)/ (0.45565/1000) = 12116.6
Prandtl number ( Pr ) = Cp * µ / kf = 3.97541*0.032/0.34269 = 5.28581
74
Choose 25% baffle cut :
From figure (12.29) :
Heat transfer factor ( jh ) = 0.0035
Without the viscosity correction term , (µ/µw) = 1 :
hs = kf * jh * Re * Pr(1/3) / de = 713.067 W/m2 oC
Overall Coefficient :
Thermal conductivity of Carbon Steel = 45 W/moC
Taking fouling coefficients from Table (12.2) :
75
For the cooling water ( shell-side ) :
hod = 6000 W/m2oC
For the gases coming out of the reactor ( tube-side ) :
hid = 1500 W/m2oC
1/Uo = (1/ho) + (1/hod) + ( do * [ln(do/di)] /2kw) + [ (do/di) * (1/hid) ] + [ (do/di) * (1/hi) ]
Where ;
Uo = the overall coefficient based on the outseide area of the tube
ho = outside fluid film coefficient (W/m2oC)
hi = inside fluid film coefficient (W/m2oC)
hod = outside dirt coefficient , fouling factor (W/m2oC)
hid = inside dirt coefficient , fouling factor (W/m2oC)
kw = Thermal conductivity of the tube wall material (W/moC)
di = tube inside diameter (m)
do = tube outside diameter (m)
1/Uo = 3.19E-03
Uo = 314 (W/m2oC) , which is very close to the value assumed
Pressure Drop :
Tube-side :
Using figure ( 12.24 ) :
76
Friction Factor ( jf ) = 0.003
ΔPt = Np * [ 8 * jf * (L/di) * (µ/µw)-m + 2.5 ] * (ρ * ut²/2)
Where ;
ΔPt = Tube-side pressur drop ( N/m² = Pa )
Np = number of Tube-side passes
ut = tube-side velosity (m/s)
L = length of one tube
Neglecting the viscosity correction term : (µ/µw) = 1
ΔPt =
N/m2
kPa
psi
11483.34849
11.48334849
1.665629358
Shell-side:
Linear velocity ( us ) = Gs /ρ = 155.509/18.0752 = 0.29187 m/s
Using figure (12.30) :
77
jf = 0.028
Neglect viscosity correction :
ΔPs = 8 * jf * (Ds/de) * (L/lb) * ( ρ * us2/2) * (µ/µw)-0.14
where ;
L = Tube length
Lb = Baffle spacing
∆Ps =
28637.5
28.6375
4.1538
N/m2
kPa
psi
Thickness :
t = ( Pri / (SEJ - 0.6P) ) + Cc
where ;
t = shell thickness (m)
P = Maximum allowable internal pressure (kPa)
ri = internal radius of shell before allowance corrosion is added (m)
EJ = efficiency of joints
S = working stress (kPa)
Cc = allowance for corrosion (m)
78
ri =
P=
S=
1.65033016
6842
94500
EJ =
0.85
Cc =
0.003
t=
m
kPa
kPa
m
0.151 m
Cost and Material of Construction :
Heat transfer area =
Cost =
8161.24
289200
ft2
$
- The material of construction for the Shell is Carbon Steel while for the tubes 316
Stainless-Steel is used.
79
4.2.2 E-102 Mariam
Cooling of T-100’s overhead vapor outlet (stream 11) from 149oC to 49oC is carried out
in cooler E-102 to adjust its temperature before it enters the CO2 Separation Section.
Data from Hysys
Shell Side ( Cooling Water )
Flow rate
1.78E+06
Kg/h
Inlet Temperature , t1
25
oC
Outlet Temperature , t2
Heat Capacity of inlet stream, Cpin
35
4.313794061
oC
Heat Capacity of outlet stream, Cpout
4.314571191
KJ/kgoC
Average Heat Capacity, Cpavg
4.314182626
KJ/kgoC
Mass Density of inlet stream , ρin
1007.3
kg/m3
Mass Density of outlet stream , ρout
999.77
kg/m3
Average Mass Density, ρavg
1003.535
kg/m3
Average Viscosity of stream, µavg
0.804465
mNs/m2
Average Thermal conductivity, Kf
Inlet Stream Pressure
0.614376996
0.304
W/moC
psig
80
KJ/kg°C
Qc = mc *Cp * (t1-t2)
where ;
Qc = heat load in the cold side (KW)
mc = mass flowrate of cold fluid (Kg/h)
Cp = heat capacity of hot fluid (kJ/kgoC)
t1 = inlet temperature (oC)
t2 = outlet temperatue (oC)
Qc = 1.78E+06 * 1003.535 * (35-25) = 21368.38622 KW > 1000 KW
- This cooler can be designed as a shell and tube heat exchanger.
Tube Side ( gas stream)
Flow rate
4.25E+01
Kg/s
Average Heat Capacity , Cp
2.564609601
kJ/kgoC
Average Mass Density , ρ
28.853
kg/m3
Average Viscosity of stream , µ
0.015
mNs/m2
Average Thermal conductivity , Kf
0.097
W/moC
inlet Temperature , T1
148.35
oC
outlet Temperature , T2
Inlet Stream Pressure
49
945.3
oC
∆Tlm = [(T1-t2) - (T2-t1)] / ln[(T1-t2 ) / (T2-t1)]
where ;
∆Tlm = log mean temperature difference
T1 = hot fluid temperature , inlet (oC)
T2 = hot fluid temperature , outlet (oC)
t1
= cold fluid temperature , inlet (oC)
t2 = cold fluid temperature , outlet (oC)
81
psig
∆Tlm = [(148.35-35) - (49-25)]/ ln [(148.35-35) - (49-25)] = 57.55506 oC
Use one shell pass and two tube passes :
R = (T1-T2) / (t2-t1) = (148.35-49)/(35-25) = 9.935
S = (t2-t1) / (T1-t1) = (35-25)/(148.35-25) = 0.081070126
From figure (12.19) :
Ft = 0.945
∆Tm = Ft * ∆Tlm
where ;
∆Tm = true temperature difference
Ft = the temperature correction factor
∆Tlm = log mean temperature difference
82
∆Tm = 54.38952985 oC
From Table (12.1) :
Taking Gases as the Hot fluid and Water as the Cold fluid :
Assuming U = 290 W/m2 oC
83
Provisional area ( A ) = Q / U * ∆Tm
where ;
Q = heat load (W)
U = overall heat transfer coefficient (W/m2 oC)
A = 21368.38622 / (290*54.38952985) = 1354.747699 m2 = 14582.38291 ft2
Choosing :
Tube outside diameter (do) =
50
mm =
1.9685
Tube inner diameter (di) =
Tube length (L) =
40.6
3.4
mm =
m=
1.598425197 in
133.858
in
Area of one tube = L* do * π = 3.4 * (50/1000) * π = 0.534070751 m2
Number of tubes = Provisional area / Area of one tube
= 1354.747699 /0.534070751 = 2536.644623 tubes
As the shell-side fluid is relatively clean use 1.25 Triangular pitch :
From Table (12.4) :
84
in
K1 = 0.249
n1 = 2.207
Bundle Diameter ( Db ) = (do)*( Nt / K1)(1/n1)
where ;
do = outer diameter (mm)
Nt = Number of tubes
K1 and n1 are constants
Db = (50/1000) * (2536.644623 /0.249)(1/2.207)
= 3.27368728 m
Using a split-ring floating head type :
Using figure (12.10) :
Bundle diametrical clearance = 100 mm
85
Shell Diameter ( Ds ) = Db + Bundle diametrical clearance = 3.27368728 + (100/1000)
= 3.37368728 m
Taking Cylindrical Head :
Shell length = (Ds/2) + (Ds/2) + L
= (3.37368728 /2) + (3.37368728 /2) + 3.4 = 6.77368728 m
Tube-side coefficient :
First Method :
Mean Tube-side temperature ( t ) = (T1+T2)/2 = (148.35+49)/2
= 98.675 oC
Tube cross-sectional area = π/4 * di2 = π/4 * (40.6)2
= 1294.618917 mm
Tube per pass = Number of tubes /4 = 2536.644623 /4 = 634.1611557 tubes
Total flow area = Tubes per pass * Cross-sectional area
= 634.1611557 * (1294.618917 *10-6) = 0.820997028 m2
mass velocity = mass flow rate / Total flow area = 4.25E+01/0 .820997028 = 52 kg/s.m2
Linear velocity ( ut ) = mass velocity / density = 52/28.853 = 2 m/s
86
hi = 4200 (1.35 + 0.02t) ut0.8 / di0.2
where ;
hi = inside coefficient (W/m2 oC)
t = mean tube-side temperature (oC)
ut = linear velocity (m/s)
di = tube inside diameter (mm)
hi = 4200 (1.35 + 0.02 * 98.675) (2)0.8 / (40.6)0.2 = 10622.45407 W/m2 oC
Second Method :
Reynolds number ( Re ) = ρ * ut * di / µ
= 28.853 * 2 * (40.6/1000) / (0.015/1000) = 141763.8483
Prandtl number ( Pr ) = Cp * µ / kf = (2.564609601/1000) * (0.015/1000) / 0.097
= 0.392544759
L / di = 3.4/ (40.6/1000) = 83.74384236
87
From figure (12.23) :
The Tube-side heat-transfer factor ( jh ) = 0.003
Assuming that the viscosity of the fluid is the same as at the wall's :
(hi * di / kf) = jh Re Pr0.33 * (µ/µwall)0.14
hi = 745 W/m2 oC
Take hi from the method which gives the lower value :
hi = 745 W/m2 oC
Shell-side coefficient :
Choose baffle spacing (Lb) = Ds/5 = 3.37368728 /5 = 674.737456 mm
Tube Pitch (Pt) = 1.25 * do = 1.25 * 50 = 62.5 mm
Cross-flow area ( As ) = [ (Pt - do) * Ds * Lb ] / Pt
= [ (62.5-50) * 3.37368728 * 674.737456 * 0.000001 ]/62.5
88
= 0.455270635 m2
Mass velocity (Gs) = mass flow rate / cross-flow area
= 1.78E+06/0.455270635 = 1087.936533 Kg/s.m2
Equivalent diameter ( de ) = (1.1/do) * (pt2 - 0.917do2)
= (1.1/50) * (62.5-0.91*502) = 35.5025 mm
Mean Shell-side temperature = (t1 + t2) / 2 = (25+35)/2 = 30 oC
Reynolds number ( Re ) = (Gs * de) / µ
= 1087.936533 *( 35.5025/1000)/ (0.804465/1000) = 48012.61304
Prandtl number ( Pr ) = Cp * µ / kf = 4.314182626*0.804465/ 0.614376996
= 5.648989052
Choose 25% baffle cut :
From figure (12.29) :
Heat transfer factor ( jh ) = 0.0035
Without the viscosity correction term , (µ/µw) = 1 :
hs = kf * jh * Re * Pr(1/3) / de = 5179.12399 W/m2 oC
89
Overall Coefficient :
Thermal conductivity of Stainless-Steel = 16 W/moC
Taking fouling coefficients from Table (12.2) :
For the cooling water ( shell-side ) :
hod = 3000 W/m2oC
For the gases coming out of the reactor ( tube-side ) :
hid = 2000 W/m2oC
1/Uo = (1/ho) + (1/hod) + ( do * [ln(do/di)] /2kw) + [ (do/di) * (1/hid) ] + [ (do/di) * (1/hi) ]
Where ;
Uo = the overall coefficient based on the outseide area of the tube
ho = outside fluid film coefficient (W/m2oC)
hi = inside fluid film coefficient (W/m2oC)
hod = outside dirt coefficient , fouling factor (W/m2oC)
1/Uo =
3.12E03
hid = inside dirt coefficient , fouling factor (W/m2oC)
kw = Thermal conductivity of the tube wall material (W/moC)
di = tube inside diameter (m)
do = tube outside diameter (m)
Uo = 320 (W/m2oC) , which is very close to the value assumed
90
Pressure Drop :
Tube-side :
Using figure ( 12.24 ) :
Friction Factor ( jf ) = 0.0019
ΔPt = Np * [ 8 * jf * (L/di) * (µ/µw)-m + 2.5 ] * (ρ * ut²/2)
Where ;
ΔPt = Tube-side pressur drop ( N/m² = Pa )
Np = number of Tube-side passes
ut = tube-side velosity (m/s)
L = length of one tube
Neglecting the viscosity correction term : (µ/µw) = 1
ΔPt =
N/m2
kPa
psi
7008.252511
7.008252511
1.016528511
91
Shell-side :
∆Ps =
63253.21508
63.25321508
9.17471173
1087.936533 / 1003.535 = 1.084104224 m/s
Using figure (12.30) :
jf = 0.028
Neglecting viscosity correction :
ΔPs = 8 * jf * (Ds/de) * (L/lb) * ( ρ * us2/2) * (µ/µw)-0.14
where ;
L = Tube length
Lb = Baffle spacing
92
N/m2
kPa
psi
Linear
velocity ( us )
= Gs /ρ =
Thickness :
t = ( Pri / (SEJ - 0.6P) ) + Cc
where ;
t = shell thickness (m)
P = Maximum allowable internal pressure (kPa)
ri = internal radius of shell before allowance corrosion is added (m)
EJ = efficiency of joints
S = working stress (kPa)
Cc = allowance for corrosion (m)
ri =
1.68684364
m
P=
6619
kPa
S=
274890
kPa
EJ =
0.85
Cc =
0.003
m
t=
0.052 m
Cost and Material of Construction :
Heat transfer area =
Cost =
14582.38
302400
ft2
$
The material of construction is 316 Stainless-Steel for both the shell and tubes.
93
4.2.3 E-100 Bibi
Shell Side
69183
kg/hr
Flow rate=
Inlet Temperature (T1) =
105
o
371
o
2.6458
kJ/kgoC
3.4017
kJ/kgoC
3.02375
kJ/kgoC
39.457
kg/m3
21.624
kg/m3
Average mass density =
30.5405
kg/m3
average viscosity stream
0.01807
mNs/m2
Thermal conductivity of stream (kf) =
7.18E-02
W/moC
Outlet Temperature (T2) =
C
C
Heat Capacity of inlet stream =
Heat Capacity of outlet stream =
Average Heat Capacity =
Mass Density of inlet stream =
Mass Density of outlet stream=
94
Qh = mh *Cp * (T1-T2)
where:
Qh = heat load in the hot side (kW)
mh = mass flow rate of hot fluid (kg/hr)
Cp = heat capacity of hot fluid (kJ/kgoC)
T1 = inlet temperature
(oC)
T2 =outlet temperature
(oC)
Heat load =
15456.9716
kW
Tube Side
Heat capacity =
3.928
kJ/kgoC
Density =
22.05895
kg/m3
Viscosity =
Thermal
conductivity =
inlet
temperature
(t1) =
outlet temperature
(t2) =
2.47E-05
mNs/m2
7.02E-02
W/moC
577
o
297
o
14.0538365
kg/s
C
C
Q = (mt Cp ΔT)hot =(ms Cp ΔT)cold
Q =heat load (kW)
m
=
mass flowrate
(kg/hr)
DT = temperature difference (oC)
Tube flow =
95
DTlm =(T1-t2)-(T2-t1) / ln((T1-t2)/(T2-t1))
where:
DTlm = log mean temperature difference
T1 =inlet shell side fluid temperature (oC)
T2 =outlet shell side fluid temperature (oC)
t1 =inlet tube side temperature (oC)
t2 =outlet tube side temperature (oC)
ΔTlm =
198.917896
o
C
Using one shell pass and two tube passes
R = (T1-T2) / (t2-t1)
R=
0.95
S=
0.59322034
Ft =
0.51
S = (t2-t1) / (T1-t1)
DTm = Ft * DTlm
where:
DTm = true temperature difference
Ft
= the temperature correction factor
DTlm = log mean temperature difference
ΔTm =
101.448127
o
Assuming U =
1000
W/m2 oC
A= Q / U * DTm
96
C
where:
A = provisional area (m2)
Q = heat load (kW)
U = overall heat transfer coefficient (W/m2 oC)
Provisional area =
152.363302
m2
Choosing
Tube
outside
diameter(do) =
30
Tube
inner
diameter(di) =
26
Tube length(L) =
8
mm
mm
m
Take tube material is cupro- nickel
Area of one tube = L* do
*π
Area of one tube =
0.75398224
Number of tubes = provisional area / area of one
tube
Number of tubes =
Using 1.25 triangular pitch
97
202
m2
K1 =
0.249
n1 =
2.207
Db = (do)*( Nt / K1)^ (1/n1)
where;
Db =bundle diameter
(mm)
do = outer diameter (mm)
Nt : number of tubes
K1 & n1 are constant
Bundle
(Db) =
diameter
624
0.62423059
mm
m
Using split ring floating head type
Bundle diametrical
clearance =
61
mm
Ds = Db +
78
61
Shell diameter(Ds) =
685
0.68523059
mm
m
Mean Tube temperature=(t1+t2)/2 =
437
o
Tube cross-sectional area = p/4 *di2=
530.929158
mm2
Tube side coefficient
Method 1
98
C
Tube per pass=(Nt/2) =
101
Total flow area = tubes per pass * cross sectional
area
Total flow area =
0.05364458
mass velocity = mass flow rate / total flow area
Tube mass velocity
=
261.980553
m2
kg/s.m2
linear velocity (ut ) = mass velocity / density
Tube linear velocity (ut) =
11.8763837
m/s
159916.458
W/m2 oC
hi = (4200* (1.35+0.02t)* ut0.8)/(di0.2)
where:
hi =inside coefficient (W/m2 oC)
t
=mean temperature
o
( C)
ut =linear velocity (m/s)
di =tube inside diameter
(mm)
hi =
Method 2
Reynolds number (Re) = ρ* ut *di/ µ
Re =
2.76E+08
99
Prandtl number (Pr) = Cp µ / kf
Pr =
0.00138179
L/di =
307.692308
jh =
1.70E-03
where jh is the heat transfer factor
assume that the viscosity of the fluid is the same
as at the wall
(µ/µwall) = 1
(hi di / kf) = jh Re Pr0.33 * (µ/µwall)0.14
hi =
144142.61
W/m2 oC
Using hi from method 1 as it has low value
hi =
144142.61
W/m2 oC
Shell-side coefficient
Choose baffle spacing (Lb)= (Ds/5) =
137.046118
37.5
100
mm
mm
Tube pitch (pt) =1.25 * do=
Cross flow area (As) =((pt - do)* Ds* Lb)/pt
Cross flow area As =
0.01878164
m2
Mass velocity (Gs) = mass flow rate / cross flow area
Mass velocity (Gs) =
1023.20678
kg/s.m2
Equivalent diameter de =(1.1/do)(pt2-0.917do2)
de =
21.3015
mm
238
o
Mean Shell side temperature =(T1+T2)/2
Mean Shell side temperature =
Reynolds number (Re) = (Gs de)/ m
Re =
1.E+06
Prandtl number (Pr) = Cp µ / kf
Pr =
7.61E-01
Choose 25% baffle
cut
101
C
jh =
6.90E-04
Without the viscosity correction term, (µ/µw) = 1
hs = kf * jh *Re *Pr^(1/3) /
de
2561.88484
W/m2 oC
50
W/moC
Outside coefficient(fouling factor)=hod
5000
W/m2 oC
Inside coefficient(fouling factor) =hid
3220
W/m2 oC
hs =
Overall Heat Transfer Coefficient
Thermal conductivity of cupro-nickel alloy=
Taking
coefficients
fouling
1/Uo =(1/ho) + (1/hod) + ( do(ln(do/di))/2kw) + (do/di) * (1/hid) +
(do/di) * (1/hi)
1/Uo =
0.00099961
Uo =
Close to initial value assumed
1000.39008
102
W/m2 oC
Pressure Drop
Tube side
Re =
2.76E+08
jf =
1.70E-03
where jf is the friction factor
ΔPt = Np [ 8jf (L/di)(µ/µw)^(-m) +2.5 ] ρut²/2
ΔPt = tube side pressure drop
where ,
(N/m²)(pa)
Np = number of tube side passes
ut = tube side velocity (m/s)
L = length of one tube
Neglecting the viscosity correction term,
(µ/µw) = 1
Δpt=
20798.3891
20.7983891
3.01675138
N/m2
kPa
psi
33.5032754
1.E+06
m/s
Shell side
Linear velocity =Gs /ρ=
Re =
jf =
7.50E-01
ΔPs = 8jf (Ds/de)(L/Lb)( ρus^2/2)(µ/µw)^(0.14)
where :
L : tube length
Lb : baffle
spacing
103
DPs =
193117548
193117.548
28011.1901
N/m2
kPa
psi
Shell Thickness Calculations:
t =(Pri/(SEJ-0.6P))+Cc
where:
t = shell thickness (in)
P = internal pressure
(psig)
ri = internal radius of shell
(in)
EJ = efficiency of joints
S = working stress (psi)
Cc = allowance for corrosion (in)
ri =
P
=
S
=
EJ
=
Cc
=
13.48879847
in
101.32
psi
13700
psi
1.25E-01
in
t=
t=
2.43E-01
6.2
in
mm
0.85
104
(for carbon steel)
(for
spot
examined)
Cost Calculations:
Heat transfer area =
1640.03
Cost =
$54,000
105
ft2
4.2.4 E-103 Bibi
Shell Side
69183
kg/hr
49.64
o
314.6
o
2.9953
kJ/kgoC
5.0483
kJ/kgoC
4.0218
kJ/kgoC
975.75
kg/m3
317.56
kg/m3
Average mass density =
646.655
kg/m3
average viscosity stream
45.215
mNs/m2
Thermal conductivity of stream (kf) =
2.50E-01
W/moC
Flow rate=
Inlet Temperature (T1) =
Outlet Temperature (T2) =
C
C
Heat Capacity of inlet stream =
Heat Capacity of outlet stream =
Average Heat Capacity =
Mass Density of inlet stream =
Mass Density of outlet stream=
106
Qh = mh *Cp * (T1-T2)
where:
Qh = heat load in the hot side (kW)
mh = mass flow rate of hot fluid (kg/hr)
Cp = heat capacity of hot fluid (kJ/kgoC)
T1 = inlet temperature
(oC)
T2 =outlet temperature
(oC)
Heat load =
20478.4779
kW
Tube Side
Heat capacity =
3.928
kJ/kgoC
Density =
22.05895
kg/m3
Viscosity =
Thermal conductivity
=
inlet temperature (t1)
=
outlet temperature
(t2) =
2.47E-05
mNs/m2
7.02E-02
W/moC
577
o
297
o
18.6195064
kg/s
C
C
Q = (mt Cp ΔT)hot =(ms Cp ΔT)cold
Q =heat load (kW)
m
=
mass flowrate
(kg/hr)
DT = temperature difference (oC)
Tube flow =
107
DTlm =(T1-t2)-(T2-t1) / ln((T1-t2)/(T2-t1))
where:
DTlm = log mean temperature difference
T1 =inlet shell side fluid temperature (oC)
T2 =outlet shell side fluid temperature (oC)
t1 =inlet tube side temperature (oC)
t2 =outlet tube side temperature (oC)
ΔTlm =
254.806026
o
C
Using one shell pass and two tube passes
R = (T1-T2) / (t2-t1)
R=
0.94628571
S=
0.5309466
Ft =
0.76
S = (t2-t1) / (T1-t1)
DTm = Ft * DTlm
where:
DTm = true temperature difference
Ft
= the temperature correction factor
DTlm = log mean temperature differace
ΔTm =
193.65258
o
Assuming U =
600
W/m2 oC
A= Q / U * DTm
108
C
where:
A = provisional area (m2)
Q = heat load (kW)
U = overall heat transfer coefficient (W/m2 oC)
Provisional area =
176.24757
m2
50
mm
46
10
mm
m
1.57079633
m2
Choosing
Tube outside diameter(do) =
Tube inner diameter(di) =
Tube length(L) =
Take tube material is cupro- nickel
Area of one tube = L* do
*π
Area of one tube =
Number of tubes = provisinal area / area of one
tube
Number of tubes =
Using 1.25 triangular pitch
109
112
K1 =
0.249
n1 =
2.207
Db = (do)*( Nt / K1)^ (1/n1)
where;
Db =bundle diameter (mm)
do = outer diameter (mm)
Nt : number of tubes
K1 & n1 are constant
Bundle diameter (Db)
=
797
0.79692699
mm
m
Using split ring floating head type
Bundle diametrical
clearance =
67
mm
Ds = Db +
78
67
Shell diameter(Ds) =
864
0.86392699
mm
m
437
o
Tube side coefficient
Method 1
Mean Tube temperature=(t1+t2)/2 =
Tube cross-sectional
area = p/4 *di2=
1661.90251
Tube per pass=(Nt/2) 56
110
C
mm2
=
Total flow area = tubes per pass * cross sectional
area
Total flow area =
0.09323496
m2
199.705191
kg/s.m2
9.0532501
m/s
114823.956
W/m2 oC
mass velocity = mass flow rate / total flow area
Tube mass velocity =
linear velocity (ut ) = mass velocity / density
Tube linear velocity (ut) =
hi = (4200* (1.35+0.02t)* ut0.8)/(di0.2)
where:
hi =inside coefficient (W/m2 oC)
t
=mean temperature
o
( C)
ut =linear velocity (m/s)
di =tube inside diameter
(mm)
hi =
Method 2
Reynolds number (Re) = ρ* ut *di/ µ
Re =
3.72E+08
Prandtl number (Pr) = Cp µ / kf
111
Pr =
0.00138179
L/di =
217.391304
jh =
1.70E-03
where jh is the heat transfer factor
assume that the viscisity of the fluid is the same as
at the wall
(µ/µwall) =
1
(hi di / kf) = jh Re Pr0.33 * (µ/µwall)0.14
hi =
109878.49
W/m2 oC
Using hi from method 1 as it has low value
hi =
109878.49
W/m2 oC
Shell-side coefficient
Choose baffle spacing (Lb)= (Ds/5) =
172.785399 mm
Tube pitch (pt) =1.25
* do=
62.5
112
mm
Cross flow area (As) =((pt - do)* Ds* Lb)/pt
Cross flow area As =
0.02985479 m2
Mass velocity (Gs) = mass flow rate / cross flow
area
Mass velocity (Gs) =
643.698964 kg/s.m2
Equivalent diameter de =(1.1/do)(pt2-0.917do2)
de =
35.5025
mm
182.12
o
Mean Shell side temperature =(T1+T2)/2
Mean Shell side temperature =
Renolds number (Re) = (Gs de)/ m
Re =
5.E+02
Prandtl number (Pr) = Cp µ / kf
Pr =
7.26E+02
Choose 25% baffle
cut
jh =
2.60E-02
Without the viscosity correction term, (µ/µw) = 1
hs = kf * jh *Re *Pr^(1/3) /
de
113
C
833.100716 W/m2 oC
hs =
Overall Heat Transfer Coefficient
Thermal conductivity of cupro-nickel alloy=
50
W/moC
Outside coefficient(fouling factor)=hod
5000
W/m2 oC
Inside coefficient(fouling factor) =hid
5000
W/m2 oC
Taking
coefficients
fouling
1/Uo =(1/ho) + (1/hod) + ( do(ln(do/di))/2kw) + (do/di) * (1/hid) +
(do/di) * (1/hi)
1/Uo =
0.00166931
Uo =
Close to initial value assumed
599.050078
114
W/m2 oC
Pressure Drop
Tube side
Re =
3.72E+08
jf =
1.70E-03
where jf is the friction factor
ΔPt = Np [ 8jf (L/di)(µ/µw)^(-m) +2.5 ] ρut²/2
ΔPt = tube side pressur drop
where ,
(N/m²)(pa)
Np = number of tube side passes
ut = tube side velosity (m/s)
L = length of one tube
Neglecting the viscosity correction term,
(µ/µw) = 1
Δpt=
9865.28788
9.86528788
1.43093394
N/m2
kPa
psi
0.99542873
5.E+02
m/s
Shell side
Linear velocity =Gs /ρ=
Re =
jf =
8.30E-02
ΔPs = 8jf (Ds/de)(L/Lb)( ρus^2/2)(µ/µw)^(-0.14)
where :
L : tube length
Lb : baffle
spacing
115
DPs =
299600.223
299.600223
43.4562208
N/m2
kPa
psi
in
psi
psi
(for carbon steel)
Shell Thickness Calculations:
t =(Pri/(SEJ-0.6P))+Cc
where:
t = shell thickness (in)
P = internal pressure
(psig)
ri = internal radius of shell
(in)
EJ = efficiency of joints
S = working stress (psi)
Cc = allowance for corrosion (in)
ri =
P=
S=
EJ
=
Cc
=
17.00644607
1.25E-01
in
t=
t=
2.74E-01
7.0
in
mm
101.32
13700
0.85
(for spot examined)
116
Cost Calculations:
Heat transfer area
=
1897.11
Cost =
$58,500
117
ft2
4.2.5 E-104 Bibi
Shell Side
5.04E+04
kg/hr
314.6
o
49
o
Heat Capacity of inlet stream =
3.0638
kJ/kgoC
Heat Capacity of outlet stream =
3.4668
kJ/kgoC
3.2653
kJ/kgoC
58.102
kg/m3
126.61
kg/m3
92.356
kg/m3
Flow rate=
Inlet Temperature (T1) =
Outlet Temperature (T2) =
Average Heat Capacity =
Mass Density of inlet stream =
Mass Density of outlet stream=
Average mass density =
C
C
1.92E-02 mNs/m2
viscosity of inlet stream=
4.15E-02 W/moC
Thermal conductivity of stream (kf) =
Qh = mh *Cp * (T1-T2)
where:
Qh = heat load in the hot side (kW)
mh = mass flow rate of hot fluid (kg/hr)
Cp = heat capacity of hot fluid (kJ/kgoC)
T1 = inlet temperature (oC)
T2 =outlet temperature (oC)
Heat load =
118
12146.5097 kW
Tube Side
Heat capacity =
4.1975
kJ/kgoC
Density =
992.06
kg/m3
Viscocity =
1.07E-03
mNs/m2
Thermal conductivity =
6.12E-01
W/moC
inlet temperature (t1) =
2
o
outlet temperature (t2) =
57
o
C
C
Q = (mt Cp ΔT)hot =(ms Cp ΔT)cold
Q
=heat load
(kW)
m = mass flow rate
(kg/hr)
DT = temperature difference (oC)
Tube flow =
52.6136105
k
g
/
s
DTlm =(T1-t2)-(T2-t1) / ln((T1-t2)/(T2-t1))
where:
DTlm = log mean temperature difference
T1 =inlet shell side fluid temperature (oC)
T2 =outlet shell side fluid temperature (oC)
t1 =inlet tube side temperature (oC)
t2 =outlet tube side temperature (oC)
ΔTlm =
123.790574
Using one shell pass and two tube passes
R = (T1-T2) / (t2-t1)
119
o
C
R=
4.82909091
S=
0.1759437
Ft =
0.77
S = (t2-t1) / (T1-t1)
DTm = Ft * DTlm
where:
DTm = true temperature difference
Ft
= the temperature correction factor
DTlm = log mean temperature difference
ΔTm =
95.3187417
o
Assuming U =
340
W/m2 oC
374.795426
m2
C
A= Q / U * DTm
where:
A = provisional area (m2)
Q = heat load (kW)
U = overall heat transfer coefficient (W/m2 oC)
Provisional area =
Choosing
Tube
outside
diameter(do) =
41
Tube inner diameter(di)
=
37
Tube length(L) =
7
Take tube materail is cupro- nickel
120
mm
mm
m
Area of one tube = L* do *π
Area of one tube =
0.90163709
m2
Number of tubes = provisinal area / area of one tube
Number of tubes =
416
K1 =
0.249
n1 =
2.207
Bundle diameter (Db) =
1183
1.18287634
Using 1.25 triangular pitch
Db = (do)*( Nt / K1)^ (1/n1)
where;
Db =bundle diameter (mm)
do = outer diameter (mm)
Nt : number of tubes
K1 & n1 are constant
mm
m
Using split ring floating head type
Bundle
diametrical 77
121
mm
clearance =
Ds = Db +
78
77
Shell diameter(Ds) =
1260
1.25987634
mm
m
Tube side coefficient
Method 1
Mean
Tube
temperature=(t1+t2)/2 = 29.5
Tube
cross-sectional
area = p/4 *di2=
1075.21009
Tube per pass=(Nt/2) =
o
C
mm2
208
Total flow area = tubes per pass * cross sectional area
Total flow area =
0.2234734
m2
235.435669
kg/s.m2
mass velocity = mass flow rate / total flow area
Tube mass velocity =
linear velocity (ut ) = mass velocity / density
Tube linear velocity (ut)
=
0.23731999
hi = (4200* (1.35+0.02t)* ut0.8)/(di0.2)
where:
hi =inside coefficient (W/m2 oC)
t
=mean temperature (oC)
ut =linear velocity (m/s)
122
m/s
di
=tube inside diameter
(mm)
hi =
1252.21278
W/m2 oC
Method 2
Reynolds number (Re) = ρ* ut *di/ µ
Re =
8.14E+06
Pr =
0.0073422
L/di =
189.189189
jh =
1.70E-03
Prandtl number (Pr) = Cp µ / kf
where jh is the heat transfer factor
assume that the viscisity of the fluid is the same as at the
wall
(µ/µwall) = 1
(hi di / kf) = jh Re Pr0.33 * (µ/µwall)0.14
hi =
45207.29
W/m2 oC
Using hi from method 1 as it has low value
hi =
1252.21
123
W/m2 oC
Shell-side coefficient
Choose baffle spacing
(Lb)= (Ds/5) =
251.975269
mm
Tube pitch (pt) =1.25
* do=
51.25
mm
Cross flow area (As) =((pt - do)* Ds* Lb)/pt
Cross flow area As =
0.06349154
m2
220.589333
kg/s.m2
29.11205
mm
181.8
o
Mass velocity (Gs) = mass flow rate / cross flow area
Mass velocity (Gs) =
Equivalent diameter de =(1.1/do)(pt2-0.917do2)
de =
Mean Shell side temperature =(T1+T2)/2
Mean
Shell
temperature =
side
Reynolds number (Re) = (Gs de)/ m
Re =
3.E+05
Pr =
1.51E+00
Prandtl number (Pr) = Cp µ / kf
124
C
Choose 25% baffle cut
jh =
1.30E-03
Without the viscosity correction term, (µ/µw) = 1
hs = kf * jh *Re *Pr^(1/3) / de
710.878711
W/m2 oC
50
W/moC
Outside coefficient(fouling factor)=hod
5000
W/m2 oC
Inside coefficient(fouling factor) =hid
2900
W/m2 oC
hs =
Overall Heat Transfer Coefficient
Thermal conductivity of cupro-nickel alloy=
Taking fouling coefficients
1/Uo =(1/ho) + (1/hod) + ( do(ln(do/di))/2kw) + (do/di) * (1/hid) +
(do/di) * (1/hi)
1/Uo =
0.00291582
Uo =
342.956209
Close to initial value
assumed
125
W/m2 oC
Pressure Drop
Tube side
Re =
8.14E+06
jf =
1.70E-03
where jf is the
friction factor
ΔPt = Np [ 8jf (L/di)(µ/µw)^(-m) +2.5 ] ρut²/2
where ,
ΔPt = tube side pressure drop (N/m²)(pa)
Np = number of tube side passes
ut = tube side velocity (m/s)
L = length of one tube
Neglecting the viscosity correction term, (µ/µw) = 1
Δpt=
283.445214
0.28344521
0.04111298
N/m2
kPa
psi
Shell side
Linear velocity =Gs
/ρ=
2.38846781
Re =
3.E+05
jf =
3.00E-02
ΔPs = 8jf (Ds/de)(L/Lb)( ρus^2/2)(µ/µw)^(-0.14)
where :
L : tube length
126
m/s
Lb :
spacing
baffle
DPs =
76011.6921
76.0116921
N/m2
kPa
11.0252951
psi
Shell Thickness Calculations:
t =(Pri/(SEJ-0.6P))+Cc
where:
t = shell thickness (in)
P = internal pressure (psig)
ri = internal radius of shell (in)
EJ = efficiency of joints
S = working stress (psi)
Cc = allowance for corrosion (in)
ri =
P=
S=
24.8007288
EJ =
0.85
Cc =
1.25E-01
151.99
13700
127
in
psi
psi
in
t=
t=
4.51E-01
11.5
in
mm
Heat transfer area =
4034.26
Cost =
$88,500
Cost Calculations:
128
ft2
4.2.6 E-105 Bibi
5. Cooler E-105
Shell Side
1.31E+07
kg/hr
312.9
o
49
o
4.3541
kJ/kgoC
2.847
kJ/kgoC
3.60055
kJ/kgoC
609.78
kg/m3
975.52
kg/m3
Average mass density =
792.65
kg/m3
viscosity of stream
1.70E+01
mNs/m2
Thermal conductivity of stream (kf) =
2.14E-01
W/moC
Flow rate=
Inlet Temperature (T1) =
Outlet Temperature (T2) =
C
C
Heat Capacity of inlet stream =
Heat Capacity of outlet stream =
Average Heat Capacity =
Mass Density of inlet stream =
Mass Density of outlet stream=
129
Qh = mh *Cp * (T1-T2)
where:
Qh = heat load in the hot side (kW)
mh = mass flow rate of hot fluid (kg/hr)
Cp = heat capacity of hot fluid (kJ/kgoC)
T1 = inlet temperature (oC)
T2 =outlet temperature (oC)
Heat load =
3444421.15
kW
Tube Side
Heat capacity =
4.1975
kJ/kgoC
Density =
992.06
kg/m3
Viscocity =
1.07E-03
mNs/m2
Thermal conductivity =
6.12E-01
W/moC
inlet temperature (t1) =
2
o
outlet temperature (t2) =
57
o
Tube flow =
14919.7949
kg/s
C
C
Q = (mt Cp ΔT)hot =(ms Cp ΔT)cold
Q =heat load (kW)
m = mass flow rate (kg/hr)
DT = temperature difference (oC)
DTlm =(T1-t2)-(T2-t1) / ln((T1-t2)/(T2-t1))
where:
DTlm = log mean temperature difference
T1 =inlet shell side fluid temperature (oC)
130
T2 =outlet shell side fluid temperature (oC)
t1 =inlet tube side temperature (oC)
t2 =outlet tube side temperature (oC)
ΔTlm =
123.271082
o
C
Using one shell pass and two tube passes
R = (T1-T2) / (t2-t1)
R=
4.79818182
S=
0.17690576
Ft =
0.57
S = (t2-t1) / (T1-t1)
DTm = Ft * DTlm
where:
DTm = true temperature difference
Ft
= the temperature correction factor
DTlm = log mean temperature difference
ΔTm =
70.2645165
Assuming U =
500
A= Q / U * DTm
where:
A = provisional area (m2)
Q = heat load (kW)
U = overall heat transfer coefficient (W/m2 oC)
131
o
C
W/m2 oC
Provisional area =
98041.5527
m2
Choosing
Tube outside diameter(do)
=
31
mm
Tube inner diameter(di) =
Tube length(L) =
29
8
mm
m
Area of one tube =
0.77911498
m2
Take tube material is cupro- nickel
Area of one tube = L* do *π
Number of tubes = provisional area / area of one tube
Number of tubes =
125837
Using 1.25 triangular pitch
K1 =
0.249
n1 =
2.207
Db = (do)*( Nt / K1)^ (1/n1)
where;
Db =bundle diameter (mm)
do = outer diameter (mm)
132
Nt : number of tubes
K1 & n1 are constant
Bundle diameter (Db) =
11904
11.903
8791
mm
77
mm
11981
11.9808791
mm
m
m
Using split ring floating head type
Bundle diametrical clearance =
Ds = Db + 78
77
Shell diameter(Ds) =
Tube side coefficient
Method 1
Mean
Tube
temperature=(t1+t2)/2 =
29.5
Tube cross-sectional area =
p/4 *di2=
660.519855
Tube per pass=(Nt/2) =
o
C
mm2
62919
Total flow area = tubes per pass * cross sectional area
Total flow area =
41.5589445
m2
359.003221
kg/s.m2
mass velocity = mass flow rate / total flow area
Tube mass velocity =
133
linear velocity (ut ) = mass velocity / density
Tube linear velocity (ut) =
0.36187652
m/s
hi =
1842.55197
W/m2 oC
Re =
9.73E+06
Pr =
0.0073422
L/di =
275.862069
jh =
1.70E-03
hi = (4200* (1.35+0.02t)* ut0.8)/(di0.2)
where:
hi =inside coefficient (W/m2 oC)
t
=mean temperature (oC)
ut =linear velocity (m/s)
di =tube inside diameter (mm)
Method 2
Reynolds number (Re) = ρ* ut *di/ µ
Prandtl number (Pr) = Cp µ / kf
where jh is the heat transfer factor
assume that the viscosity of the fluid is the same as at the wall
(µ/µwall) = 1
134
(hi di / kf) = jh Re Pr0.33 * (µ/µwall)0.14
hi =
68934.17
W/m2 oC
Using hi from method 1 as it has low value
hi =
1842.55
W/m2 oC
Shell-side coefficient
Choose baffle spacing (Lb)=
(Ds/5) =
2396.17583
mm
Tube pitch (pt) =1.25 * do=
38.75
mm
5.74165859
m2
631.35067
kg/s.m2
22.01155
mm
Cross flow area (As) =((pt - do)* Ds* Lb)/pt
Cross flow area As =
Mass velocity (Gs) = mass flow rate / cross flow area
Mass velocity (Gs) =
Equivalent diameter de =(1.1/do)(pt2-0.917do2)
de =
Mean Shell side temperature =(T1+T2)/2
135
Mean Shell side temperature
=
180.95
o
C
Reynolds number (Re) = (Gs de)/ m
Re =
8.E+02
Pr =
2.86E+02
Prandtl number (Pr) = Cp µ / kf
Choose 25% baffle cut
jh =
1.90E-02
Without the viscosity correction term, (µ/µw) = 1
hs = kf * jh *Re *Pr^(1/3) / de
hs =
994.465444
W/m2 oC
50
W/moC
5000
W/m2 oC
Overall Heat Transfer Coefficient
Thermal conductivity of cupro-nickel alloy=
Taking fouling coefficients
Outside coefficient(fouling factor)=hod
136
Inside coefficient(fouling factor) =hid
5000
W/m2 oC
1/Uo =(1/ho) + (1/hod) + ( do(ln(do/di))/2kw) + (do/di) * (1/hid) + (do/di) * (1/hi)
1/Uo =
0.00202019
Uo =
Close to
assumed
495.003516
initial
W/m2 oC
value
Pressure Drop
Tube side
Re =
9.73E+06
jf =
1.70E-03
where jf is the friction
factor
ΔPt = Np [ 8jf (L/di)(µ/µw)^(-m) +2.5 ] ρut²/2
where ,
ΔPt = tube side pressure drop (N/m²)(pa)
Np = number of tube side passes
ut = tube side velocity (m/s)
L = length of one tube
Neglecting the viscosity correction term, (µ/µw) = 1
Δpt=
137
812.191717
0.81219172
0.11780626
N/m2
kPa
psi
Shell side
Linear velocity =Gs /ρ=
Re =
0.79650624
8.E+02
jf =
7.50E-02
m/s
ΔPs = 8jf (Ds/de)(L/Lb)( ρus^2/2)(µ/µw)^(-0.14)
where :
L : tube length
Lb
:
baffle
spacing
ΔPs =
ShellThickness Calculations:
t =(Pri/(SEJ-0.6P))+Cc
where:
t = shell thickness (in)
P = internal pressure (psig)
ri = internal radius of shell (in)
EJ = efficiency of joints
S = working stress (psi)
Cc = allowance for corrosion (in)
138
274151.388
274.151388
39.7649345
N/m2
kPa
psi
ri =
P=
S=
235.8442047
EJ =
0.85
Cc =
1.25E-01
in
t=
t=
3.23E+00
82.0
in
mm
Heat transfer area =
1055310.56
Cost =
$268,900
151.99
13700
in
psi
psi
(for carbon steel)
(for spot examined)
Cost Calculations:
139
ft2
5. Separators
5.1 Theory
Separators are mechanical devices for removing and collecting liquids from natural
gas. A properly designed separator will also provide for the release of entrained
gases from the accumulated hydrocarbon liquids. A wellstream separator must
perform the following:
1. Cause a primary-phase separation of the mostly liquid hydrocarbons from those
that are mostly gas.
2. Refine the primary separation by removing most of the entrained liquid mist
from the gas.
3. Further refine the separation by removing the entrained gas from the
accumulated liquid.
4. Discharge the separated gas and liquid from the vessel and insure that no reentrainment of one into the other takes place.
140
If these functions are to be accomplished, the separator design must:
1. Control and dissipate the energy of the wellstream as it enters separator;
2. Insure that the gas and liquid flow rates are low enough so that gravity
segregation and vapor-liquid equilibrium can occur;
3. Minimize turbulence in the gas section of the separator and reduce velocity;
4. Eliminate re-entrainment of the separated liquid into the gas;
5. Accumulate and control froths and foams in the vessel;
6. Provide outlets for gases and liquids with suitable controls to maintain pre-set
operating pressure and liquid levels;
7. Provide relief for excessive pressure in case the gas or liquid outlets should
become plugged or valves malfunction;
8. Provide equipment (pressure gauges, thermometers, and liquid-level gauge
assemblies) to visually check the separator for proper operation;
9. Provide cleanout opening at points where solids will accumulate when solids are
present in the inlet stream.
141
Separator Selection: Basic Considerations
The goal for ideal separator selection and design is to separate the wellstream into
liquid-free gas and gas-free liquid. Ideally, the gas and liquids reach a state of
equilibrium at the existing conditions of pressure and temperature within the
vessel. As it is generally not economically justifiable to separate to the state of true
equilibrium, industry consensus standards as to liquid retention time for solution
gas break-out and liquid carry-over in the gas have been set. In some cases, the
process equipment and conditions downstream of a separator will dictate the
necessary degree of separation and the actual design.
Wellstream Characteristics
The following characteristics influence vessel selection, in addition to the obvious
quantities of liquids and gas to be separated:
1. Proportions of gas and liquids composing the inlet stream.
2. Differences between the densities of the gas and liquids.
3. Differences between the viscosities of the gas and liquids.
4. Temperature and pressure at which separation is to be made.
5. Particle sizes of liquids in the gas phase or gas in the liquid phase.
6. Identification of impurities or special conditions such as H2S, CO2, pipe scale,
dust, foam, fogs, etc.
7. Instantaneous flow rates (slugs or heading).
142
Vertical Separators
Vertical separators are capable of handling large slugs of liquid and are therefore
most often used on low to intermediate gas-oil ratio wellstreams. They are ideally
suited as inlet separators to processing plants since they can smooth out surging
liquid flows. They are well suited for handling production that contains sand and
other sediment. When excessive sand production is expected, a cone bottom is
placed in the vertical separator to properly handle the sand. Vertical separators
occupy less floor space than comparably sized other types. This is an important
consideration where floor space can be very expensive, as on an offshore platform.
However, because the natural upward flow of gas opposes the falling liquid
droplets, vertical separators may be larger and more expensive than a horizontal
separator for the same gas handling capacity.
Applications
1. Wellstreams having large liquid to gas ratios.
2. Wellstreams having sizable quantities of sand, mud, or other related substances.
3. Areas having horizontal space limitations, but little or no vertical height
limitations.
4. Wellstreams or process flow streams which are characterized by large
instantaneous volumes of liquid.
5. Upstream of other process equipment tolerating essentially no entrained liquid
droplets in the gas.
6. Downstream of equipment causing liquid formation.
143
Horizontal Separators
Horizontal separators are ideally suited to wellstreams having high gas-oil ratios,
constant flow, and small liquid surge characteristics. Horizontal separators are
smaller and less expensive than vertical separators for a given gas capacity. Liquid
particles in the wellstream travel horizontally and downward at the same time as a
result of two forces acting upon them-the horizontal force of the gas stream and the
downward force of gravity. Therefore, higher gas velocities can be permitted in
horizontal separators and still obtain the same degree of separation as in vertical
separators. Also, the horizontal separators have a much greater gas-liquid interface
area than other types, which aids in the release of solution gas and reduction of
foam. A special de-foaming section is used when severe foaming of the inlet stream
is anticipated. The horizontal configuration is best suited for liquid-liquid-gas, or
three phase, separations because of the large interfacial area available between the
two liquid phases. In addition to being easier to hook-up, easier to service, and
easier to skid-mount, horizontal separators can be stacked in piggy-back fashion to
form stage separation assemblies and minimize horizontal space requirements.
144
Applications
1. Areas where there are vertical height limitations.
2. Foamy production where the larger liquid surface area available will allow
greater gas breakout and foam breakdown.
3. Three phase separation applications for efficient liquid-liquid separation.
4. Upstream of process equipment, which will not tolerate entrained liquid droplets
in the gas.
5. Downstream of equipment causing liquid formation.
6. Wellstreams having a high gas to oil ratio and constant flow with little or no
liquid surges.
7. Applications requiring bucket and weir construction for three phase operation.
145
The Separation Process
All separators have at least three and usually four sections comprising the
separation process:

The primary separation section

The secondary separation section

The liquid accumulation section

The mist extractor section
Primary Separator Section

The primary separation section is the portion of the vessel around the inlet
where the energy of the entering wellstream is dissipated. The purpose of this
section and its mechanical components is to make the initial separation of
liquid from gas using deflectors or impingement baffles. The bulk of the liquid
is diverted to the liquid accumulation section. The larger quantities of liquid
and large liquid drops immediately start falling as a result of the gravitational
force. In vertical separators the inlet deflector forces the liquid to change
direction toward the vessel shell where it spreads out in a thin film, allowing
solution gas to break out. In horizontal separators, the liquid is usually directed
against a deflector plate which may or may not be dish shaped. The liquid is
thrown against the vessel shell to divert it from the main gas stream and allow
rapid release of solution gas. In some cases, impingement baffles are used in
horizontal separators to break the liquid stream into smaller streams and
droplets so that solution gas can be more readily released.
Secondary Separator Section

The area of the separator immediately beyond the inlet deflector, between the
liquid accumulation section, and the mist extractor (or outlet head where a
mist extractor is not used) is called the secondary separation section. In this
section the velocity of the gas and liquid is reduced because of the increased
cross-sectional area.

This allows the liquid particles to begin falling toward the liquid accumulation
section as a result of gravitational force on the mass of the liquid particle. In
vertical separators the upward gas velocity tends to counter the gravitational
146
force effect on the liquid particles as it exerts a drag force on the particle. If
the particle is large, the gravitational force will be the greater force and the
particle will settle to the bottom. Very small particles will be carried along with
the gas as entrainment and will leave the separator, if not removed by some
other device such as a mist extractor. In horizontal separators the drag force is
exerted at right angles to the gravitational force and does not hinder the
particles' fall to the liquid accumulation section. The resultant path of the
particle is a diagonal path or trajectory toward the outlet of the separator. The
horizontal separator must be large enough in cross-section and long enough so
the reduction of the gas velocity and the diagonal paths for the bulk of the
liquid particles will carry them into the liquid accumulation section.
Liquid Accumulation Sections

All separators must provide an area where the collected liquid from the
primary separation section, the secondary separation section, and the mist
extractor can be held for a short period of time and then dumped to storage.
The liquid retention time is normally one minute for two phase (i.e., liquid-gas)
separation. This allows time for the solution gas to break out of the

accumulated liquid. In vertical separators a baffle plate is positioned between
the liquid accumulation section and the secondary separation section. This is
to insure little, if any, re-entrainment of liquid into the gas. It also minimizes
wave action and turbulence on the liquid surface which could upset the liquid
level control system.
Horizontal separators normally utilize approximately half of the cross-section
for liquid accumulation. Because of their configuration, horizontal separators
have less instantaneous surge capacity than vertical separators. However, the
large surface area at the gas liquid interface provides excellent release of
solution gas. Wave breakers or stilling baffles are provided to stop wave action
caused by gas eddy currents near the gas-liquid interface to prevent liquid reentrainment into the gas stream.
 Liquid outlet connections in either vertical or horizontal separators are usually
located as far away from the inlet as possible to assure maximum liquid
retention time for release of solution gas. These connections are also
designed with antivortex baffles or siphon type drains to prevent vortex
development.
147
5.2 Design
Equipment Name :
Separator
Objective :
Separates water from the gas product
Equipment Number :
V-101
Designer :
Dalal Al-Othman
Type :
Horizontal Separator
Location :
Partial Oxidation and Heat Recovery Section
Material of Construction :
Stainless Steel
Cost :
$ 400,000
Operating Condition
Operating Temperature (oC) :
49
Operating Pressure (psia) :
965
Design Considerations
Liquid Density (kg/m3) :
991.98
Gas Density (kg/m3) :
Gas Flow rate (kg/h) :
144940
Liquid Flow rate (kg/h) :
30.7
8434.6
Dimensions
Diameter (m) :
2.67
Length (m) :
148
3.2
Equipment Name :
Separator
The heated CO2 rich amine solution is fed into
separator V-103 that separates the light gases
from the CO2 rich amine stream.
Objective :
Equipment Number :
V-103
Designer :
Mariam Hussain AL-Shamma’a
Type :
Vertical Separator
Location :
CO2 Separation Section
Material of Construction :
Carbon Steel
Insulation :
Foam Glass
Cost :
77000 $
Operating Condition
Operating Temperature (oC) :
314.6
Operating Pressure (psig) :
938.3
Design Considerations
Liquid Density (kg/m3) :
535.92
Gas Density (kg/m3) :
Gas Flow rate (MMSCFD) :
928.4
Liquid Flow rate (barrel/day) :
58.102
2.3488E6
Dimensions
Diameter (m) :
1.699760404
149
Height (m) :
12.35909066
Equipment Name
separator
Objective
To separate CO2 from the other gases and purge
it to the air
Equipment Number
V-104
Designer
Bibi Al-Motawa
Type
Two phase horizontal seperator
Location
After heat exchanger (E-104)
Material of Construction
stainless Steel 316
Insulation
Glass wall and quartz
Cost ($)
217,200 $
Operating Condition
Operating Temperature (oC)
49
Operating Pressure (psia)
16
Design Considerations
Liquid Density (kg/m3)
1018.6
Gas Density (kg/ m3)
688
Height (m)
17.5521
Dimensions
Diameter (m)
5.8507
150
5.2.1 V-101 Dalal
The horizontal separator V-102 is to separate 8434.6 kg/h of liquid, density 991.98
kg/m3, from 144940 kg/h of vapor, density 30.7 kg/m3. The vessel operating pressure is
21 bar.
The settling velocity of the liquid droplets is given by:
𝑢𝑡 = 0.07[(𝜌𝐿 − 𝜌𝑣 )/𝜌𝐿 ]1⁄2
Where ut is the settling velocity, m/s
The liquid density ρL = 991.98 kg/m3
The vapor density ρV = 30.7 kg/m3
ut =0.392m/s
A separator without a demister pad:
ua = 0.15 ut
ua = 0.0587 m/s
Vapor volumetric flow-rate =
144940
3600×30.7
= 1.31 m3/h
Taking the liquid height at half the vessel diameter:
hv= 0.5 Dv and Lv/Dv = 1.2
Cross-sectional area for vapor flow =
𝜋𝐷𝑣2
4
× 0.5 = 0.393𝐷𝑣2
1.31
Vapor velocity, 𝑢𝑣 = 0.393𝐷2 =3.33𝐷𝑣−2
𝑣
Vapor residence time required for the droplets to settle to liquid surface
= ℎ𝑣 ⁄𝑢𝑎 = 0.5Dv/0.0587 = 8.51Dv
Actual residence time = vessel length/vapor velocity
=𝐿𝑣 ⁄𝑢𝑣 =1. 12𝐷𝑣3
For satisfactory separation, required residence time = actual
So, 8.51Dv = 1.12Dv3
151
Diameter of the vessel Dv = 2.67 m
Length of the vessel Lv = 3.2 m
Volume of the Vessel = 8.92 m3
152
5.2.2 V-103 Mariam
The heated CO2 rich amine solution is fed into separator V-103 that separates the light
gases from the CO2 rich amine stream.
V-103 is a Vertical Separator :
Data from Hysys
Property
Value
Unit
ρv
ρL
Vv
58.102
535.92
0.06833
Kg/m3
Kg/m3
m3/s
LV
0.02916
m3/s
Settling Velocity :
Ut = 0.07 [ (ρL-ρv)/ρv ]0.5
where ;
Ut : Settling Velocity (m/s)
ρL : Liquid Density (Kg/m3)
ρv : Vapour density (Kg/m3)
153
Ut = 0.07 [ (535.92-58.102)/58.102 ]0.5 = 0.200739922 m/s
Minimum Vessel Diameter :
Dv = ( 4 Vv / π Us )0.5
where ;
Dv : Minimum Vessel Diameter (m)
Vv : Vapor Volumetric Flow-rate (m3/s)
Us = 0.15 Ut
(m/s)
Us = 0.15 * 0.200739922 = 0.030110988 m/s
Dv = ( 4 * 0.06833 / π * 0.030110988 )0.5 = 1.699760404 m = 5.576641752 ft
Allow a minimum of 10 seconds hold-up :
time = 10 sec
Volume held in vessel = Liquid Volumetric Flow-rate * time
= 0.02916 * 10 = 17.49452655 m3
Liquid Depth required ( hv ) = Volume held -up / Vessel cross-sectional Area
Vessel cross-sectional Area = π/4 * Dv2 = π/4 * 1.699760404 = 2.26916093 m2
hv = 17.49452655 / 2.26916093 = 7.709689655 m
Htotal = hv + Dv/2 + Dv + 0.4 + DV/2 + Dv/2
= 7.709689655 + (1.699760404/2) + 1.699760404 + 0.4 + (1.699760404/2) +
(1.699760404/20
= 12.35909066 m = 40.54819778 ft
154
Thickness
t = [ Pri / (SEJ - 0.6P) ] + Cc
where ;
t = shell thichness (in)
P = Maximum allowable internal Pressure (psig)
ri = Internal raduis of shell before allowance corrosion is added (in)
EJ = Efficiency of joints
S = working stress (psi)
Cc = Allowance for corrosion (in)
P=
150
psig
ri =
33.45978
in
S=
13700
psi
EJ =
0.85
Cc =
0.125
in
t = [ (150*33.45978) / (13700*0.85 - 0.6*150) ] + 0.125
= 0.559354611 in = 0.014207636 m
Surface Area :
Surface Area of the vessel = 2 π (Dv/2) Htotal
= 2 π (1.699760404 /2) * 12.35909066 = 65.99698547 m2
Volume of Metal :
V = Surface Area of the vessel * thickness
= 65.99698547 * 0.014207636 = 0.937661116 m3
155
Weight of Vessel :
W = Volume of the metal * Density of the Metal
Density of Carbon Steel = 7900 kg/m3
W = 0.937661116 * 7900 = 7407.522817 Kg = 16330.77295 lb
Cost of Separator:
Cost = 77000 $
Material of Construction is Carbon Steel
156
5.2.3 V-104 Bibi
Sample calculation
liquid
Vv
vapour
2.83E+00
m³/s
Vl
4.13E-01
-
m³/s
ρl
1018.6
-
kg/m³
ρv
-
688
ml
-
1.45E+06
kg/h
mv
1.18E+04
-
kg/h
kg/m³
ut= 0.07*((ρl-ρv)/ρv)^(0.5)
Where:ut
settling velocity in m/s
ρl
liquid density in kg/m³
ρv
vapour density in kg/m³
m/s
ut
0.04852385
assume there is no dimester
ua = 0.15*ut
ua
0.00727858
m/s
Vapor volumetric flow rate= mass flow rate of vapor(kg/h)/(3600*ρv)
157
Vapor volumetric flow rate=
0.5870478
m^3/s
hv = .5 Dv
because I have the operating pressure less than 20 bar then Lv= 3 Dv
Cross-sectional area for vapour flow =
(π*0.5)/4*Dv^2
Cross-sectional area for vapour flow =
0.39269908 Dv^2
Vapour velocity,uv =
Vapour volumetric flow rate/Cross-sectional area for vapour flow
Vapour velocity,uv =
1.49490496 Dv^-2
Vapour residance time = hv/ua
Vapour residance time =
68.6947425 Dv
Actual residence time =
vessel length/vapor velocity
Actual residence time =
2.00681655 Dv^3
For satisfactory seperation required residence time = actual
68.69474*Dv-2.006817Dv^3 = 0
Dv =
5.85070113
m
Liquid volumetric flow rate =
mass flow rate of liquid/(3600*ρl)
158
Liquid volumetric flow rate =
m^3/s
0.00321247
Liquid cross-sectional area =
(π*Dv^2)/4*(0.5)
Liquid cross-sectional area =
Lv = 3Dv
13.4423659
m^2
Length, Lv =
17.5521034
m
Hold-up volume =
liquid cross-sectional area*Lv
Hold-up volume =
235.941796
Hold-up time =
m^3
liquid volume/liquid volumetric flow-rate
Hold-up time =
73445.5966
s
1224.09328
min
there will be no changes cause the factor =1 ; the hold up time exceed 10 min which is
reasonable with my high flow rate
factor =
1
New Dv =
5.85070113 m
new liquid volume
=
new residence
time=
m^3
235.941796
s
73445.5966
min
1224.09328
A=
(π/4)*D^2
Where:A
Vessel cross sectional area in m²
D
min vessal diameter
159
26.8847318 m2
A
calculate the thichness
(p*ri)/((s*Ej)-(0.6*p))+co
t=
Where:p
16
ri
2.92535056
m
C0
115.171052
in
s
13700
Ej
0.85
co
0.125
t
psia
psia
0.28337331
in
0.0071977
m
surface area of the vessal=2*π*(lv/2)*Dv
A
322.616782
m²
volume of metal=t*area
volume
2.32209768
m³
weight of vessal= v* density
Where:-
density of stainless
steels 316=
weight of vessel
8238
19129.4407
kg/m³
kg
42173.1475
Ib
160
217200
cost =
$
161
6. Pumps
6.1 Theory
Pump is a machine or device used for moving an incompressible liquid from lower to
higher pressure and overcoming this difference by adding energy to the system.
Figure P.1 Pump
Pump selection :
The type of the pump to be designed in this process is a Centrifugal pump because the
pump operates at Low P, low viscosity, high flow rates.
Centrifugal Pump:
A centrifugal pump is a rotodynamic pump that uses a rotating impeller to increase the
pressure of a fluid. Centrifugal pumps are commonly used to move liquids through a
piping system. The fluid enters the pump impeller along or near to the rotating axis and
is accelerated by the impeller, flowing readily outward into a diffuser or volute chamber,
from where it exits into the downstream piping system A centrifugal pump works on the
principle of conversion of the kinetic energy of a flowing fluid (velocity pressure) into
static pressure. This action is described by Bernoulli's principle. The rotation of the
162
pump impeller accelerates the fluid as it passes from the impeller eye (centre) and
outward through the impeller vanes to the periphery. As the fluid exits the impeller, a
proportion of the fluid momentum is then converted to (static) pressure. Typically the
volute shape of the pump casing, or the diffuser vanes assist in the energy conversion.
The energy conversion results in an increased pressure on the downstream side of the
pump, causing flow.The two main components of a centrifugal pump are the impeller
and the volute. The impeller produces liquid velocity and the volute forces the liquid to
discharge from the pump converting velocity to pressure. This is accomplished by
offsetting the impeller in the volute and by maintaining a close clearance between the
impeller and the volute at the cut-water.
Figure P.2 Centrifugal Pump
163
The main components of the pump are:

Shaft: used to spin the impeller.

Coupling: attaches the shaft to the motor driver.

Bearings: sport the shaft.

Seal: prevent the liquid inside the pump from leakage out around the shaft.

Impeller wear ring: minimizes internal liquid leakage, from the pump discharge
back to the pump section.

Impeller: accelerates the liquid. The function of it is to increase the velocity of
the liquid.

Volute: Convert the velocity imparted to the liquid by impeller to feet of head.
Figure P.3 The main components of the centrifugal pump
164
Multistage Centrifugal Pumps :
A centrifugal pump containing two or more impellers is called a multistage centrifugal
pump. The impellers may be mounted on the same shaft or on different shafts. A
multistage centrifugal pump has the following two important functions:


To produce a high head.
To discharge a large quantity of liquid.
If a high head is to be developed then the impellers are mounted on same shaft (series)
while for large quantity of discharge of liquid, the impellers are mounted on different
shafts (parallel).
FigureP.4 General components of Centrifugal Pump
165
6.2 Design
Equipment Name :
Pump
To increase the pressure of the MDEA before
being recycled to the first absorber in the CO2
separation section
Objective :
Equipment Number :
P-100
Designer :
Mariam Hussain AL-Shamma’a
Type :
Centrifugal Pump
Location :
CO2 Separation Section
Material of Construction :
Stainless- Steel
Insulation :
Foam Glass
Cost :
20700 $
Operating Conditions
Inlet Temperature (oC) :
Inlet Pressure (psia) :
Efficiency (%) :
Outlet Temperature (oC) :
49
947.7
Outlet Pressure (psia) :
75
Power (hp) :
166
49.02
962
655.2
6.2.1 P-100 Mariam
To increase the pressure of the MDEA from 6534 kPa to 6633 kPa before being recycled
to the first absorber in the CO2 separation section.
Data from Hysys :
Parameter
Inlet Pressure
Outlet pressure
Mass Flow-rate
Mass Density
Brake horse
Power
Value
Unit
136470
lb/ft2
138530
28765000
lb/ft2
lb/hr
61
lb/ft3
655.2
hp
Actual Head of Pump
ha 
p2  p1

Assume Gravity = 32.174 ft/sec2
Spec. weight = Mass Density * Gravity = 61 * 32.174 = 1959 lb/ft3
Actual Head ( ha ) = 1.05134407 ft
Water Horse Power
Pf 
Qha
550
Mass Flow rate = 28765000/3600 = 7990 lb/sec
Volumetric Flow-rate = Mass Flow-rate / Mass Density = 7990/61 = 131 ft3/sec
167
Water Horse Power = Spec. Weight * Volumetric Flow-rate * Actual Head
= 1959 * 131 * 1.05134407 = 270279/(550) lb.ft/sec = 491 hp
Overall Efficiency :

WHP
BHP
Efficiency = 491/655.2 = 75%
Cost of Pump :
Cost = 20,700 $ From www.matche.com
Pump Type :
Volumetric Flow-rate = 13375 m3/h
From HYSYS Total head = 10.31 m
Cost of Pump :
Cost = 20,700 $ From www.matche.com
Pump Type :
Volumetric Flow-rate = 13375 m3/h
From HYSYS Total head = 10.31 m
Using Figure 5.6
168
Therefore the type of this pump is a single-stage centrifugal pump.
Material of construction
Stainless-Steel is selected as the material of construction of this pump to avoid
corrosion.
169
7. Compressors
7.1 Theory
A gas compressor is a mechanical device that increases the pressure of a gas by
reducing its volume.Compressors are similar to pumps: both increase the pressure of a
fluid and both can transport the fluid through a pipe. As gases are compressible, the
compressor also reduces the volume of a gas. Liquids are relatively incompressible, so
the main action of a pump is to transport liquids.
Types of Compressors
170
Reciprocating, centrifugal and axial flow compressors are the principal types used in the
chemical process industries, and the range of application of each type is shown below.
Reciprocating Compressors
A reciprocating compressor is also known as a piston compressor and it is a positivedisplacement compressor that uses pistons driven by a crankshaft to deliver gases at high
pressure.
Small reciprocating compressors from 5 to 30 horsepower (hp) are commonly seen in
automotive applications and are typically for intermittent duty. Larger reciprocating
compressors up to 1000 hp are still commonly found in large industrial applications, but
their numbers are declining as they are replaced by various other types of compressors.
Discharge pressures can range from low pressure to very high pressure (>5000 psi or 35
MPa). In certain applications, such as air compression, multi-stage double-acting
compressors are said to be the most efficient compressors available, and are typically
larger, noisier, and more costly than comparable rotary units.
171
Centrifugal Compressors
Centrifugal compressors which are also known as radial compressors s are rotating disk
or impeller in a shaped housing to force the gas to the rim of the impeller, increasing the
velocity of the gas. A diffuser (divergent duct) section converts the velocity energy to
pressure energy.
They are primarily used for continuous, stationary service in industries such as oil
refineries chemical and petrochemical plants and natural gas processing plants.Their
application can be from 100 hp (75 kW) to thousands of horsepower. With multiple
staging, they can achieve extremely high output pressures greater than 10,000 psi (69
Mpa).
Figure C.1 Centrifugal Compressor
Axial Flow Compressors
Axial flow compressors are rotating compressors in which the working fluid flows
parallel to the axis of rotation.
Axial flow compressors produce a continuous flow of compressed gas, and have the
benefits of high efficiencies and large mass flow capacity, particularly in relation to their
cross-section. They are however, complex and expensive relative to other designs.
Axial compressors are widely used in gas turbines, such as jet engines, high speed ship
engines, and small scale power stations. They are also used in industrial applications
such as large volume air separation plants, blast furnace air, fluid catalytic cracking air,
and propane dehydrogenation.
172
7.2 Design
Equipment Name
Compressor
Material of Construction
Carbon Steel
Power (Hp)
35839.2981
173
7.2.1 K-100 Dalal
Compressor K-100
First, the appropriate type of compressor is selected by the aid of Figure 10.60 shown
below.
flow-rate at inlet conditions = 2832 m3/h, discharge pressure = 69 bar→ centrifugal
compressor
𝑃𝑜𝑤𝑒𝑟 =
𝑘
𝑝2 (𝑘−1)⁄𝑘
𝑝1 𝑣1 [( )
− 1]
𝑘−1
𝑝1
(𝑘−1)⁄
𝑘
3.03 × 10−5 𝑘
𝑝2
ℎ𝑝 =
𝑝1 𝑞𝑓𝑚1 [( )
𝑘−1
𝑝1
− 1]
𝑣1 𝑘
𝑇2 𝑘⁄(𝑘−1)
𝑝2 = 𝑝1 ( ) = 𝑝1 ( )
𝑣2
𝑇1
174
𝑣1 𝑘−1
𝑝2 (𝑘−1)⁄𝑘
𝑇2 = 𝑇1 ( )
= 𝑇1 ( )
𝑣2
𝑝1
p1 = intake pressure = 72000 lbf/ft2
p2= final delivery pressure = 143200 lbf/ft2
v1 = specific volume of gas at intake conditions = 0.6557 ft3/lbm
v2 = specific volume of gas at final delivery conditions = 0.4060 ft3/lbm
T1 = absolute temperature of gas at intake conditions = 560.1oR
T2 = absolute temperature of gas at discharge conditions = 680.7oR
qfm1 = cubic feet of gas per minute at intake conditions = 56465.97 ft3/min
𝑝2
𝑙𝑛 ( ) = 0.688
𝑝1
𝑇2
𝑙𝑛 ( ) = 0.195
𝑇1
𝑘⁄(𝑘 − 1) =
𝑝
𝑙𝑛 (𝑝2 )
1
𝑇
𝑙𝑛 (𝑇2 )
1
= 3.53, 𝑘 = 1.396
Substituting the parameters into the first two equations above gives:
Power = 35839.29811 ft.lbf/lbm
hp = 93522.41883
175