LITERATURE STUDY AND COMPARISON OF MEMBRANES FOR CO2
SEPARATION:
TABLE 1.1:
Gas composition CO2 20
(Vol%)
H2 80
N2 -
50
28
20
50
-
40
-
72
40
Permeability of CO2
(Barrers)
1296
PEO-PBT
150
PEO-PBT+PEG-DBE
750
100⁰C
6000
150⁰C
1800
Selectivity
-
51.5
40
1800
280
CO2/H2
Gas temperature
87
107
10.3
20
12.4
300
100
110, 150
Gas Flow Rate
feed
9.9
-----
3.6
sweep
feed
1.8
220
147
1.8
36.74
sweep
feed
15.7
90 %RH
-----
41 mol%
sweep
75 % RH
Gas Pressure
(psi)
Humidity
CO2/N2
90 mol%
Mechanism
Facilitated transport
Solution Diffusion
Facilitated
transport
Membrane Area
(cm2)
5.7
35
45.60
Membrane Thickness
(µm)
32
50-100
20-80
REFERENCES:
Reference 1: Membrane- PVA-POS/FS
Crosslinked polyvinylalcohol–polysiloxane/fumed silica mixed matrix
membranes containing amines for CO2/H2 separation Rong Xinga,1, W.S. Winston Hoa,b,∗
Reference 2: Membrane- PEO-PBT
CO2-Philic Polymer Membrane with Extremely High Separation Performance
Wilfredo Yave,*,†,§ Anja Car,†,§ Sergio S. Funari,‡ Suzana P. Nunes,†, and Klaus-Viktor Peinemann†,
Reference 3: Membrane- AIBA-K/Polyallylamine/PVA
CO2-selective polymeric membranes containing amines in crosslinked
poly(vinyl alcohol) Jian Zou, W.S. Winston Ho
LITERATURE STUDY
Reference 4:Integrated assessment of IGCC power generation technology with
carbon capture and storage (CCS)
Calin-Cristian Cormos*
Babes-Bolyai University, Faculty of Chemistry and Chemical Engineering, 11 Arany Janos Street, RO-400028 Cluj-Napoca, Romania
Introduction:
In this paper an integrated gasification combined cycle (IGCC) with CCS (Carbon capture and storage) is
discussed. An IGCC unit is a coal gasification unit coupled with a gas turbine for power generation. Coal
gasification is an efficient process for power generation as it involves gasification of coal in presence of
pure oxygen (oxy-fuel combustion) and steam; this process leads to the formation of Synthesis gas
(Syngas). Syngas is composed mainly of CO and H2 which are comparatively more efficient energy
sources than coal; also gasification enables use of lower grade coal. Traditional IGCC plants involve
removal of only H2S using AGR (Acid Gas Removal) unit. H2S is removed before combustion in Gas
turbines (See fig1).
In this paper CO2 is removed in addition to H2S in the AGR. This reduces CO2 emissions from flue gases,
The captured CO2 is then pressurized and sent for storage (The preferred method of storage in this
paper is Enhanced Oil Recovery (EOR)). Hence in this paper pre-combustion CO2 capture is carried out.
Traditional IGCC Units:
Traditional IGCC units as stated above have AGR only for H2S removal (See fig: 1 [ref 4]).
Here CO2 removal is negligible and hence there is no CCS, so there are high levels of CO2 emissions.
IGCC with CCS:
IGCC plants with CCS suggested in this paper have AGR for removal of both CO2 and H2S using a solventSelexol ®. This solvent absorbs both CO2 and H2S and from the Syngas and then releases both during
regeneration (pressure flash for CO2 and Thermal regeneration for Sulphur) (see fig:2 [ref 4]). gasifiers
considered in the paper are of entrained-flow type operating at high temperature which gives a high
conversion of solid fuel (w99%). From different entrained-flow gasification technologies, Shell and
Siemens technologies were chosen. The main factors for consideration were the dry fed design (for both
gasifiers) which gives high plant energy efficiency compared with slurry fed coupled with high CGE (cold
gas efficiency) and clean syngas. In addition, syngas water quench configuration of Siemens gasifier
ensures optimal conditions for shift conversion (covering water requirement of WGS (gas shift
conversion)) [ref 4].
Another important difference from traditional IGCC plants is the presence of water gas shift stage which
helps to increase CO2 levels in the syngas at the precombustion stage before the AGR; this is done to
enhance CO2 removal. Also in the IGCC-CCS system we have a hydrogen fueled turbine instead of a
syngas fueled turbine.
In terms of carbon capture rate the CCS design is complying with at least 90% carbon capture rate.
Specific CO2 emissions of evaluated plant concepts with CCS are in the range of 76-87 kg/MWh.
Conventional IGCC power generation technology without carbon capture has specific CO2 emissions in
the range of 740-800 kg/MWh.
AN OPPORTUNITY FOR APPLICATION OF MEMBRANE SEPARATION
IN IGCC FOR CCS
Though the method of solvent absorption results in 90% CO2 removal [ref 4], there is some energy
penalty i.e. additional energy is required in this process for solvent regeneration (see table: 2.1) and CO2
pressurization (up to 200 bar). It can be observed that Selexol ® solvent has the least energy penalty
among the three solvents [ref 4].
Table 2.1 [ REF:4]:
CO2 separation using polymer membranes is highly energy efficient as there is no solvent involved
which requires regeneration.Hence we should now try to design a permeation set up which replaces
the AGR. Using the following data obtained from the paper we get the input to the set up and the
required output from the set up.
THE INPUT TO THE PERMEATION SET UP –
Syngas Exiting after water gas shift reaction [ref 4]:
Pressure: 525.38 psi
Temperature: 366.36 ⁰C
Mass flow rate: 586785.76 Kg/hr
Molar flow: 29874.53 Kmol/hr
Gas composition (See Table 2.2)
Table 2.2 [ref 4]:
Gas Component
Vol %
H2
40.14
CO
1.43
CO2
N2
H2S
29.14
2.38
0.1
Argon
H2O
0.43
26.38
( RH= 3.577%)*
*(vap.pressure- Perry handbook pg: 2.60)
THE REQUIRED OUTPUT FROM THE PERMEATION SET UPGas Exiting the Acid gas removal unit (Selexol ® solvent) [ref 4]:
Pressure: 411.6 psi
Temperature: 30 ⁰C
Mass flow rate: 77310 kg/hr
Molar flow rate: 13562.04 Kmol/hr
Gas Composition (see table 2.3)
Table 2.3 [ref 4]:
Gas component
Vol %
H2
88.04
CO
3.03
CO2
2.95
N2
5.08
Argon
0.88
H2O
0.02
H2S
0.00
THE AVAILABLE PERMEATION DATA [ref 3]:
Optimum operating Temperature = 110 ⁰C
Feed pressure = 56.74 psi
Humidity: Feed= 59 mol%
Sweep= 90 mol%
Gas composition: 20% CO2, 40% N2, 40% H2
Comparing the syngas input data (after water gas shift reaction) with the optimum operating condition
data (above). We observe the following:




High temperature of syngas (336.36 ⁰C [ref 4] while optimum is 110 ⁰C [ref 3])
Presence of H2S (0.1%)
Water content ( 26.38 mol% [ ref 4] while optimum is 59 mol% [ref 3])
High pressure ( 525.38 psi [ref 4] while 56.73 psi [ref 3] is optimum)
These differences necessitate the inclusion of additional processes in order to achieve optimum
conditions for high permeation (6000 barrers [ref 3]).
THE PERMEATION SETUP:
The permeation setup should have
1.
2.
3.
4.
5.
6.
A scrubber (for H2S removal)
waste heat boiler (for cooling the gas)
Humidifier (to increase the water content)
line heaters( to maintain water content)
Multiple membrane cells(to distribute the pressure to optimum limits)
A steam boiler to provide sweep gas [ref 3 mentions that steam can be used as sweep instead of
argon].
7. Condenser to condense the sweep steam
8. Compressor for pressurizing the CO2 Captured
THE PROCESS FLOW CHART
Fig 3:
THE FOLLOWING EQUIPMWNTS WILL BE DESIGNED IN THIS PROJECT:
1. Heat Exchanger
2. H2S Scrubber
3. Gas Permeation Unit
HEAT EXCHANGER DESIGN
Shell side fluid: DEA solution
Tube side fluid: Acid gas
HEAT BALANCE :
Fig ( ):
Gas in
Gas out
366.36 ⁰C
100 ⁰ C
100 ⁰C
32⁰C
Aqueous DEA (2Kmol/m3)
ʃ m Cp dt INLET - ʃ m Cp dt OUTLET = Heat Transferred
Q = 292.7 ×106 KJ/hr
Heat Inlet :
=298ʃ639.3 29874.53 [ 0.001 (7.20 +0.0036 T) + 0.4014 ( 6.62+ 0.00081 T) + 0.0143 (
6.60 + 0.0012 ) + 0.2914 (10.34 + 0.0027 T – 19550 /T2) +0.0238(6.50+0.0010 T) +
0.0043 (4.97) + 0.2638 (7.759+0.019×10-3T + 3.15 × 10-6 T2) ] dt
[Ref : 1.Chemical Engineering Hand book , Perry . 2.Stochiometry, Bhatt & Vora]
Heatinlet = 371.52 × 106 KJ
Heat Outlet :
=298ʃ373 29874.53 [ 0.001 (7.20 +0.0036 T) + 0.4014 ( 6.62+ 0.00081 T) + 0.0143 (
6.60 + 0.0012 ) + 0.2914 (10.34 + 0.0027 T – 19550 /T2) +0.0238(6.50+0.0010 T) +
0.0043 (4.97) + 0.2638 (7.759+0.019×10-3T + 3.15 × 10-6 T2) ] dt
HeatOutlet = 78.82 × 106 KJ
371.52 × 106(Heat Inlet ) – 78.82 × 106 KJ ( Heat Outlet) = 292.7 × 106KJ(Heat
Transferred)
mhot=162.99 kg/sec (ref 1)
mcold = Q/(cp∆T) = 292.7×106/(6030.12) = 283.82 kg/sec
LMTD =
(Thi – Tco) - ( Tho – Tci )
Ln[(Thi – Tco)/ ( Tho – Tci )]
= (366.36-100)-(100-32)/ln[(366.36-100)/(100-32)]
=145.28oC = 418.28oK
The driving force for Heat exchanger is
ΔTm= FT* LMTD
R = (Thi-Tho) / (TCo-TCi) = 3.92
S = ( TCo-TCi) /( Thi-Tho) = 0.203
FT = 0.8(from graph)
ΔTm= 116.224oC = 389.224oK
TO FIND NUMBER OF TUBES
Assumed unit:
Let us assume overall heat transfer co-efficient ,
Uo= 125 btu/ft2 F hr = 709.79 w/m2 °K
We know that
A = Q / UoΔTm
= 292.7×106 /(709.79×389.224)
=1059.48m2
Assume the following tube parameters:
Outer Dia =0.01905m, Inner Dia =0.01483m, L=9.144m
PT= 25mm triangular pitch,
Area per tube = πDoL
=3.14×0.01905×9.144 = 0.547 m2
Number of tubes NT = Area / surface area of the tube
= A / π DoL = 1059.48/0.547
= 1936.89 = 1937 tubes
Bundle diameter, db = do[Nt/k1]1/n = 0.01905[1936.89/0.249]1/2.207 (values for k1 and n from
ref.2)
Db = 1.104 m
Clearance = 72 mm (from ref.2)
SHELL DIAMETER = Db + clearance = 1.176 m
TUBE SIDE COEFFICIENT:
Process gas mean temp = 366.36+100/2 = 233.18oC
𝜋
𝜋
Cross sectional area = 4 × 𝐷i2 = 4 ×(0.01483)2 = 1.73×10-4 m2
Tubes per pass = 1936.89/2 = 968.45 tubes
Total flow area = tubes per pass × c.s.a = 0.1675 m2
Mass velocity = flow rate/flow area = 162.99/0.1675 = 973.07 kg/m2sec
Linear velocity = mass vel/density = 973.07/85.59 = 11.36 m/s
Reynold’s number = DV𝜌/µ
Here, D = 0.01483 m , V = 11.36 m/s , 𝜌 = 85.59 kg/m3 , µ = 0.8 × 10-3 Ns/m2
Substituting above values NRe = 18039.92
Prandtl number = Cpµ/k
Here , Cp = 1.854 KJ/kmolK , k = 0.247 W/mK
Substituting above values , NPr = 7.2
L/D ratio = 9.144/0.01483 = 616.58
From the values of L/d and NRe, jh = 4 × 10-3
hi = (k/di).jh.NRe.NPr0.33{µ/µw}0.14
= (0.2047/0.01483).(4×10-3) ×18039.92×7.20.33 = 1910.69 W/m2K
SHELL SIDE COEFFICIENT:
Baffle spacing,Dbs = shell dia/5 = 1.176/5 = 235.2 mm
Tube pitch = 1.25 Do = 23.81 mm
Cross sectional area = (23.81-19.05) × 1176 × 235.2 × 10-6/ 23.81 = 0.056 m2
Mass velocity = flow rate/area = 283.82/0.056 = 5068.21 kg/sm2
Equivalent diameter De = 1.10/Do{Pt2-0.917Do2}
= 13.58 mm
Shell side mean temp = 32+100/2 = 82oC
Here,
DEA solution density = 961 kg/m3
Viscosity = 2.81 × 10-4 Ns/m2
Cp value = 4.52 KJ/kgK
Reynold’s number NRe = GsDe/µ = 5068.21 × 0.01358/0.00281 = 24580.81
Prandtl number NPr = Cpµ/k = 4.52 × 0.00281/0.5919 = 7.52
Assume 45% cut baffle,
From graph, jh = 3 × 10-3
Shell side coefficient ho = (k/do).jh .NRe. NPr {µ/µw}0.14
= (0.5919/0.01358) ×(3×10-3).24580.81 × 7.520.33
= 6254.86 W/mK
OVERALL COEFFICIENT:
Assume fouling factor values for shell and tube side fluids.
For DEA solution, hod = 6000 W/mK
For acid gas, hid = 5000 W/mK
Overall coefficient Uo is given by
1/Uo = 1/ho + 1/hod + doln(do/di)/2kw + (do/di × 1/ hid) + (do/di × 1/hi)
1/Uo = 0.0013002
Uo = 769.11 W/m2K
PRESSURE DROP CALCULATION:
Shell side pressure drop:
Linear velocity = Gs/𝜌 = 5068.21/961 = 5.27 m/s
NRe = 24580.81
From graph, jf = 2.8 × 10-2
ΔPs = 8 jf (Ds/De) × (L/Dbs) × 𝜌U2/2 {µ/µw}-0.14
= 8×2.8×10-2(1.176/0.01358) ×(9.144/23.52) × 961× 5.272/2
ΔPs = 100.63KN/m2
Tube side pressure drop:
NRe = 18039.92
From graph , jh = 4.1×10-3
ΔPt = 2×{8×jh (L/Di) × (µ/µw)m + 2.5 } 𝜌Ut2/2
= 2 × {8× 4.1× 10-3× (9.144/0.01483) 2.5}× 85.59× 11.372/2
ΔPt = 27.78kN/m2
H2S SCRUBBER DESIGN
Fig 5: (EQUILLIBRIUM MASS BALANCE)
8695.29 kmol CO2
0 kmol H2S,
Rest 21139.29 kmol
8705.44 kmol CO2
29.87 kmol H2S
Rest 21139.22 kmol
Scrubber temperature=100oC
Scrubber pressure=525.38psi
1.296 kmol DEA
36.038 kmol H2O
1.294 kmol DEA
36.027 kmol H 2O
10.077 kmol CO2
29.87 kmol H2S
Fig 6: [REF: 5]
Fig 7: [REF: 5]
Fig 8: [REF: 5]
Fig 9: [REF: 5]
Inlet Gas Composition: [Ref Table ( 1.3)]
Assumptions:
 H2, N2, CO, Ar are insoluble in solvent and water.
 Only H2S and CO2 are being scrubbed.
CO2 - 29.14% (by mol) => 8705.44 kmol
H2S - 0.1% (by mol) => 29.87 kmol
Rest ( H2, N2, Ar, CO, H2O) – 70.76% => 21139.22 kmol
Total Pressure = 3622.36 kpa
Partial Pressure CO2 = 1055.55 kpa
Table: ( 2.1 )
Figure No
Solvent
6
7
8
9
DGA
AMP
MEA
DEA
Temperature Solvent
CO2
( OC )
Concentration Removed
(mol acid
gas/mol
amine)
100
60 wt%
0.51
100
2M
0.98
3
100
2.5 kmol/m
0.30
100
2 kmol/m3
0.27
From the table ( ) we can say that DEA is the best solvent to choose.
From Table ( 2.1 ):
[Moles of CO2] / [Moles of DEA solution] = 0.27
[Moles of H2S] / [Moles of DEA solution] = 0.80
Solute Balance:
For 100% removal, 29.87 kmol of H2S should be removed
[Moles of H2S]/[Moles of DEA solution] = 0.80
[29.87]/[Moles of DEA solution] = 0.80
H2S
Removed
(mol acid
gas/mol
amine)
0.76
1.19
0.8
0.8
Difference
0.25
0.21
0.50
0.53
DEA (2 kmol/m3) required = 37.33 kmol
CO2 removed:
[Moles of CO2]/[Moles DEA solution ] = 0.27
[Moles of CO2]/[37.3375] = 0.27
Moles of CO2 removed = 10.08 kmol
Total moles removed = 10.08 + 29.87
Total moles remaining = 298734.53 – (10.08+29.87) = 298694.53 kmol
Outlet Gas Composition:
CO2 = 8705.44 – 10.08 = 8695.29 kmol
H2S = 29.87 – 29.87 = 0
Rest (CO, N2, H2, Ar, H2O) = 21139.22 kmol
Solvent Balance:
2kmol/m3 = 2kmol/(1000/18 kmol) = 0.036 kmol/1 kmol H2O
Solvent Rate = 37.33 kmol/hr
Mole fraction of DEA = 0.036/1.036 = 0.0347
Mole fraction of H2O = 1- 0.0347 = 0.9652
CO2 absorbed = 10.08 kmol
H2S absorbed = 29.87 kmol
Outlet Solvent Composition:
DEA = 1.294 Kmol
H2O = 36.027 Kmol
CO2 = 10.077 Kmol
H2S = 29.87 Kmol
1.2
1
0.8
Y vs X
0.6
X1,Y1
X2,Y2
0.4
0.2
0
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.45
0.4
0.35
0.3
Y/Y-Y* VS LOG Y
0.25
0.2
LOG Y1
0.15
LOG Y2
0.1
0.05
0
-0.8
-0.6
-0.4
-0.2
0
DESIGN OF THE MEMBRANE
MASS BALANCE:
The membrane selected is a facilitated transport membrane [3]
Fig ( ): (All units are in kmol/hr)
Total inlet= 1.47x 10-4
Vol flow rate= 60
cm3
𝑚𝑖𝑛
Total Retentate = 3.83 x 10-5
N2
H2
1.218 x 10-7
6.76 x 10-6
CO2
1.018 x 10-4
Total feed Pressure= 1.25 atm
Total permeate pressure = 1 atm
Membrane temperature = 100 ⁰C
(Where volumetric flow rate =
Permeability x pressure x area
membrane thickness
)
𝑃𝑉
And molar flow rate found using 𝑛 = 𝑅𝑇 where ideal gas state is assumed.
EXPERIMENTATION
Aim: To study the kinetics of CO2 absorption in the membrane (PVA-FA /AIBA/PAA).
Methodology: The experiment consists of the following steps:
1. Membrane preparation
2. Membrane casting
3. Absorption experiment
Membrane preparation and casting:
Membrane solution was prepared and casted using the method prescribed in the paper [3].
Absorption experiment:
CO2 absorption was tested using humidified CO2 gas. The membrane was suspended in a vessel
where CO2 was circulated, the membrane was weighed at regular intervals and the amount of
CO2 absorbed was calculated.
1. Membrane Preparation:
The required membrane composition was 50wt% PVA, 18.3wt% KOH, 20.7Wwt% AIBA-K, and
11.0wt% PAA.
The membrane was prepared according to the method stated in the paper [3].
40 gm of 10% PVA in water solution was prepared, to which stoichiometric amount (2.25 gm) of
formaldehyde was added to achieve 60% cross linking, and 3.6 gram of 50% KOH was added to
catalyze the cross linking. This was named solution no: 1
Weight fraction in solution (X-linked PVA, FA, KOH, AIBA-K, PAA, H2O)
Hence for above solution wt fraction (0.081, 0.017, 0, 0, 0.892)
10.312gm AIBA taken and dissolved in 0.5610 gm of KOH in 13.46 gm water solution. This gives
us 14.212 gm AIBA-K in 13.62 gm water. This was named solution no: 2
Hence weight fraction basis (0, 0, 0.510, 0, and 0.490)
PAA Solution in methanol was obtained from PAA-HCl [3]. The methanol was finally evaporated
and replaced with water. This was named solution no: 3
The weight fraction basis gives us (0, 0, 0, 0.09, and 0.91)
The required dry membrane composition on weight fraction basis is (0.5, 0.183, 0.207, 0.11,
and 0).The available solutions are used in the following quantities (obtained using component
balance):
24.691 gm of solution no: 1, 2.1176 gm of solution no: 2, 0.7729 gm of solution no: 3
These solutions were mixed in a stirrer for half an hour [3].
2. Membrane casting:
The above solution was taken and cast into a petridish, the solution was then dried at 50 ⁰C
overnight. The next day the solution had dried and a polymer film had formed on the petridish,
this was the heated in an oven at 120 ⁰C to complete the cross linking. The membrane was then
gently removed fig ( ) and small circular pieces of diameter 1.32 x 10-4 m2, and having 200
microns thickness were punched out refer fig ( ).
Fig ( ):
Fig ( ):
3. Absorption Experiment:
CO2 absorption was studied for the membrane using the following apparatus fig ( ).The
apparatus consists of CO2 cylinder followed by a flow meter, humidifier and the absorption
bottle, The CO2 was humidified at 41 ⁰C (Relative humidity of 85 %) and the membrane was
taken and weighed at regular intervals to obtain a mass increase versus time curve fig ( ). Since
the gas was humidified there is also significant H2O absorption hence the same membrane was
placed in a humidified atmosphere (having 85% relative humidity) and allowed to equilibrate
for 4 days, this ensured maximum mass uptake of H2O for the given humidity. The membrane
was then taken and placed in the absorption apparatus with 8 LPM CO2 flow rate and 85%
relative humidity the CO2 absorption curve was then obtained Fig ( ). The system pressure
measure using a pressure gauge and found to be 0.11 psi.
Fig ( ):
HUMIDIFIER
ABSORPTION BOTTLE
MEMBRANE
Fig ( ):
WEIGHT VS TIME
0.051
0.0505
0.05
0.0495
WEIGHT VS TIME
0.049
0.0485
0.048
0.0475
0
20
40
60
80
Approximate Slope= 3.5X 10-5 gm/min
Fig ( ):
WEIGHT VS TIME
0.0605
0.06
0.0595
0.059
0.0585
0.058
0.0575
0.057
0.0565
0.056
0.0555
WEIGHT VS TIME
0
50
100
150
Average slope = 1.48 X 10-4 gm/min
Various points on the curve in fig ( ) were taken and corresponding slopes were calculated. This
was done to obtain a rate versus weight curve fig ( ).
Fig ( ):
RATE VS WEIGHT
0.0002
0.00018
0.00016
0.00014
0.00012
0.0001
0.00008
0.00006
0.00004
0.00002
0
RATE VS WEIGHT
The slope of the line in fig ( ) was found to be 0.627 min-1. This shows that the above process of
CO2 uptake follows first order kinetics.
THE DESIGN OF GAS PERMEATION EQUIPMENT
GAS TEMPERATURE=100⁰C
CO2 MOLAR FLOW RATE= 5284 Kmol/hr
For 92 % CO2 Capture:
CO2 to be removed = 5284 - (0.92 x 5284) = 4862.02 Kmol/hr
Retentate CO2 Flow = 422 Kmol/ Hr
Total pressure= 3623 KPa
Partial Pressure of CO2 = 427.27 KPa
The driving force for permeation ΔP should not be taken as 724.26 KPa. This is because of the
fact that the permeator to be designed will not have a flat membrane; instead there would be a
shell and tube arrangement where the partial pressure of CO2 on feed side would reduce as
the gas progresses along the length of the tube. Hence the driving force will be an average or
mean of the partial pressures at the inlet and at the outlet.
By experimentation we have observed that the CO2 absorption by the membrane follows 1st
order kinetics, Hence it can be taken analogous to heat transfer equation
Heat transfer Q = U A ΔTlm
1st order kinetics Mass uptake = K A ΔP
So the LMPD (Log mean pressure difference) can be taken as pressure difference just
ΔPlm is the Log mean driving force. Refer Fig ( )
Fig ( ):
724.7 KPa Shell
36.23 KPa (92% reduction in partial pressure)
Shell
Tube
10.06 KPa
Hence ΔP lm =
0 KPa
(CO2 Partial pressure is 0 at inlet of sweep)
(724.7−10.06)−(36.23−0)
(724.7−10.06)
𝐿𝑛 (36.23−0)
= 227.50 KPa
Now the CO2 Permeability at 100⁰C is 6000 Barrers [3].
Where 1 Barrer is defined as ‘The molar flux per driving force multiplied by thickness of
membrane at STP’.
𝑐𝑚3 (𝑆𝑇𝑃)𝑐𝑚
1 Barrer = x 10-10 𝑐𝑚2 sec 𝑐𝑚 𝐻𝑔
Converting to S.I units we get:
1 Barrer =
1
𝐾𝑚𝑜𝑙 .𝑚
1.72 𝑥 10−14 𝑚2 sec 𝐾𝑃𝑎
Now we have 6000 Barrers:
(𝐹𝑙𝑢𝑥)𝑋 (𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠)
(𝑑𝑟𝑖𝑣𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒)
1
𝑋 1.72 𝑥 10−14 = 6000
The molar flow rate is
4826.06 Kmol/hr = 1.1905 Kmol/sec
and ΔP lm = 227.50 KPa
Hence after simplification we have:
(𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠)
(𝐴𝑟𝑒𝑎)
= 1.738 x 10-8
For a 90 micron thick membrane
Area = 5178.06 m2
DETERMINATION OF TUBE DIAMETER:
The velocity of the sweep gas should not be over 10 m/sec in the tube this is to avoid
excessive pressure drop.
From ref [3] we have sweep gas flow rate of 30 cm3/min
Using PV = n RT (100⁰C and 1 atm pressure)
We have for 1.08 x 10-4 kmol of CO2 (Permeating the membrane) we need 1.633 x 10-5 kmol
sweep gas (steam).
Hence for 4862.06 Kmol/hr CO2 permeating we need 735.16 Kmol/hr of Steam (Sweep Gas)
Mass flow of Sweep gas (steam) = 13232.88 kg/hr
Again using PV = n RT (100⁰C and 1 atm pressure)
we get Vol.flowrate of gas = 23051 m3/hr = 6.4 m3/sec
let dia of tube = 0.001 m ( 1mm)
And let number of tubes = 15000
Then Gas velocity=
6.4
𝜋
( (0.0012 )(15000)
4
= 543.81 m/sec
This is HIGHER than 10 m/sec.
Assuming 1 cm (0.01 m) dia of tube:
The gas velocity =
6.4
𝜋
( (0.012 )(40000)
4
= 5.43 m/sec
Hence below 10m/sec
Hence
Diameter of tube = 0.01 m ( 1 cm)
Number of tubes = 15000
Therefore
5178.06
(0.01)15000
= 10.98m ~ 11 m
Hence length of tube = 11 m
DETERMINATION OF BUNDLE DIAMETER:
Consider triangular arrangement (this ensures maximum packing efficiency)
15000 tubes arranged are into such triplets
2cm
Hence total number of triplets =5000
Arranging these triplets into units of four:
Hence total number of units= 1250
Arranging the 1250 units into a square:
The obtained square’s side = (1250)(1/2) =35.35 ~ 36 units in a side Fig ( ).
Fig ( ):
36
………………………
36
.
Length of each side = 36 x (0.01+0.01+0.02+0.01+0.02+0.01)= 2.88m
Therefore area= 2.282 =8.29 m2
Equivalent circle π R2 =8.29
Hence Bundle Diameter 2 x R =3.24 m
DETERMINATION OF SHELL DIAMETER:
The velocity of the feed gas should not be over 10 m/sec in the tube this is to avoid excessive
pressure drop.
Bundle diameter (from above)= 3.24 m
Hence the shell diameter should be above 3.24 m
Molar flow rate= 26420 kmol/hr (The total gas flow rate)
Using PV= n RT (at 3623 KPa; 100⁰C)
We get vol.flow rate= 22614 m3/hr = 6.28 m3/sec
Therefore assuming 3.3 m shell dia:
6.28
𝜋
4
𝜋
4
( 3.32 )−( (0.012 )(15000))
= 0.85 m/sec
Which is below 10 m/s. hence is selected as shell diameter.
Hence shell diameter= 3.3 m
DESIGN SPECIFICATIONS OF THE GAS PERMEATION EQUIPMENT:
90 micron thick membrane
Diameter of tube = 0.01 m (1 cm)
Number of tubes = 15000
length of tube = 11 m
Bundle Diameter =3.24 m
shell diameter= 3.3 m
Sweep Gas Required= 13232.88 Kg/hr (at 1 atm and 100⁰C)