Thursday 10 March 2011

DEMETHANIZER DESIGN



Feed Compositions


Mass flow rate

Molar flow rate

Mole fractions

H2

CH4
C2H2

C2H4

C2H6
C3H4
C3H6
C3H8

101.04
1894.5
101.04
3953.19
429.42
151.56
1528.28
50.52

50.52
118.40
3.88
141.18
14.3
3.78
36.38
1.148

0.136
0.32
0.01
0.381
0.038
0.01
0.098
0.003

Top Compositions


 

Mass flow rate

Molar flow rate

Mole fractions

H2

CH4

C2H4


101.04

1875.07
37.84

50.50

117.19
1.35

0.29

0.69
0.0079

 

 


Bottom Compositions



Mass flow rate

Molar flow rate

Mole fractions

CH4
C2H2

C2H4

C2H6
C3H4
C3H6
C3H8
9.43
101.04
3915.35
429.42
151.56
1528.23
50.52
1.21
3.60
139.38
14.314
3.789
36.38
1.148
0.006
0.017
0.698
0.071
0.0189
0.181
0.005




Let top temperature is 172k



Mole fractions

K


X/k

H2

CH4

C2H4

0.29

0.69
0.0079
0.00
0.75
0.1
0.00
0.92
0.079
1.00

So top temperature is 172 k


Let bottom temperature is T=280k


Mole fractions

k

kx

CH4
C2H2

C2H4

C2H6
C3H4
C3H6
C3H8
0.006
0.017
0.698
0.071
0.0189
0.181
0.005
4.01
1.53
1.21
0.84
0.34
0.30
0.26
0.02
0.026
0.851
0.063
0.006
0.05
0.0013
1.00

So bottom temperature is 280k
To find feed condition
Let  T=218k


Mole fractions

k

kx

H2

CH4
C2H2

C2H4

C2H6
C3H4
C3H6
C3H8
0.136
0.32
0.01
0.381
0.038
0.01
0.098
0.003

2.40
0.612
0.55
0.356
0.136
0.12
0.104

0.768
0.006
0.209
0.013
0.001
0.011
0.0003
1.00

This temperature is very close to feed temperature which is 215k
So feed is at its boiling point

Pinch temperatures

Upper pinch =Top temperature+1/3(bottom temperature -top temperature)
Upper pinch=208k
Bottom pinch=Top temperature+2/3(bottom temperature-top temperature)
Bottom pinch=244k

Rm using gilliand method


Rm=1/(µab-1)[xda/xna -µabxdb/xnb]

µab=7.1
xda=0.69
xdb=0.0079
xna=rf/(1+ rf)(1+åµxfh)

µ
µxfh

H2

CH4
C2H2

C2H4

C2H6
C3H4
C3H6
C3H8

7.1
1.29
1.00
0.71
0.45
0.429
0.105




0.026
0.0045
0.042
0.0003
0.0728

Rf=0.32/0.381
Rf=0.839
Xna=0.468
Xnb=0.558

Rm=1/7.1-[0.69/0.468-7.1(.0079/0.558)]
Rm=0.163

No of stages at total reflux using fenske’s equation

N+1=log[(xa/xb)d(xb/xa)]/logµab

N+1=5

N=4

Theoratical no of stages using Gilliand relation

 

R

R-Rm/R+1
N-Nm/n+2

N

0.2
0.4
0.6
0.8
1.0
2.0
5.0

0.0308
0.169
0.273
0.358
0.418
0.612
0.806
0.7
0.48
0.46
0.385
0.30
0.19
0.1
18
9.46
8.0
7.75
6.57
5.40
4.60

As we observe that there is slight change in no. of plates after R=0.4
So 9 theoratical plates is optimum answer
Reflux ratio=0.4
No. of theoratical plates(N)=9

R=Ln/D
Ln=805.58Kg/hr
Vn=Ln+D
Vn=2819.53 Kg/hr
Lm=Ln+F
Lm=9015.13 Kg/hr
Vm=Lm-W
Vm=2829.58 Kg/hr

Finding Densities

For top

For liquid phase
For CH4 and C2H4
Density=PM/Tc(0.0653/zc0.773 –0.09T)]
For H2
Density=PM/RT

 

Mole fractions


Density


Mole fractions*Density

H2

CH4

C2H4


0.29

0.69
0.0079

0.005
0.34
0.624

0.0014
0.2346
0.0049
0.240


Avg Density=0.240g/cm3
Avg Density=240kg/ m3
For gaseous phase
P=35atm
T=172k
M=15.46
R=82.05atm. cm3/g-mol.k
Vap.density=PM/RT
Vap.density=0.0323g/ cm3
                   =32.3kg/ m3

For Bottom
Density of liquid phase

Density=PM/Tc(0.0653/zc0.773 –0.09T)]


Mole fractions

Density

Mole fractions*density

CH4
C2H2

C2H4

C2H6
C3H4
C3H6
C3H8
0.006
0.017
0.698
0.071
0.0189
0.181
0.005
0.006
0.017
0.698
0.071
0.0189
0.181
0.005
0.000867
0.0084
0.332
0.028
0.011
0.097
0.0025
0.479g/ cm3

Avg.density=0.479g/ cm3
Avg.density=479kg/ cm3

For vapor phase

Vap.density=0.047g/ cm3

Vap.density=47kg/ m3


Column area and diameter

At Top

Flooding vapor velocity uf=K1[(rl-rg)/ rg]1/2
FLV=Lw/Vw[rg/rl]
FLV=0.114
Plate spacing=0.45m
K1=0.02
Vapor velocity=0.045m/s
Assume 80% flooding
Vapor velocity=0.80*0.045=0.036
Volumetric vapor flow rate=2819.53/38.3*3600
Volumetric vapor flow rate=0.0204
Net area required=0.566m2

At bottom

FLV=1.002
Plate spacing=0.45m
K1=0.0125
Flooding vapor velocity uf=0.037
Assume 80% flooding
Vapor velocity=0.80*0.037=0.0296
Volumetric vapor flow rate=2829.53/47*3600
Volumetric vapor flow rate=0.0167
Net area required=0.0167/0.0296
Net area required=0.60m2

As there is no appreciable difference in column areas of top and bottom

So
Net column area=0.60m2
Take downcomer area 15%of net area
Column cross sectional area=0.69m2

Column Diameter=[A*4/P]

Column Diameter=0.827m


Liquid flow patterns

Maximum volumetric liquid rate=9015.13/479*3600

Maximum volumetric liquid rate=5.22*10-3
A single pass plate can be used

 Plate design data

Column Diameter Dc=0.878m
Column AreaAc=0.69m2

Down comer Area=0.09m2

Net Area An=0.60m2

Active Area Aa=Ac-2Ad

Active Area Aa=0.51m2

Hole Area Ah=0.051m2(take hole area 10% of active area)
(Ad/Ac)*100=13.04
From graph
Lw/Dc=0.76
Weir length lw=0.76*0.878
lw=0.667
Weir height=50mm
Hole diameter=5mm
Plate thickness=5mm

Check weeping

For Top

Maximum liquid rate=805.58/3600=0.318
Minimum liquid rate,at 70% turn down=0.7*0.318=0.223Kg/sec
Minimum how=750[Lw/rl*lw]
Minimum how=1.044mm
how+hw=50+1.044
how+hw=51mm
From graph
K2=30
Uh min=[K2-0.90(25.4-dh)]/ (rv)1/2
Uh min=1.880m/sec
Actual min. vapor velocity=0.783*0.7/0.051=10.74m/sec
Actual min vapor velocity is quite large than min velocity for flooding

For bottom

Maximum liquid rate=9015.13/3600=3.57
Minimum liquid rate,at 70% turn down=0.7*3.57=2.50Kg/sec
Minimum how=750[Lw/rl*lw]
Minimum how=30mm
how+hw=50+30
how+hw=80mm
From graph
K2=30.8
Uh min=[K2-0.90(25.4-dh)]/ (rv)1/2
Uh min=4.33m/sec
Actual min. vapor velocity=0.793*0.7/0.051=10.77m/sec
Actual min vapor velocity is quite large than min velocity for flooding

Plate pressure drop

For Top

Max vapor velocity through holes=0.783/0.051=15.68m/s
Plate thickness/Hole dia=1
Ah/Aa=0.1
From graph
Co=0.84
hd=51(uh/Co)2rv/rl
hd=140.32mm
hr=12.5*103/rl
hr=52mm
ht=hd+(hw+how)+hr
ht=243mm

 For bottom
Max vapor velocity through holes=0.793/0.051=15.78m/s
Plate thickness/Hole dia=1
Ah/Aa=0.1
From graph

Co=0.84

hd=51(uh/Co)2rv/rl
hd=110.32mm
hr=12.5*103/rl
hr=26mm
ht=hd+(hw+how)+hr
ht=216mm of liquid

Downcomer liquid back up

For Top

Hap=hw-10=40mm
Area under apron Aap= Haplw
Area under apron Aap=0.026m2
As it is less than Ad=0.09m2

Use Aap in equation

hdc=166[Lwd/rlAap]2
hdc=6mm
Backup in down comer
hb= hdc+(hw+how)+ht
hb=0.30m
0.30<1/2(plate spacing+weir height)
0.30<0.32
so tray spacing is acceptable
Residance time
tr=Adhbcrl/lwd
tr=25s
For bottom
Hap=hw-10=40mm
Area under apron Aap= Haplw
Area under apron Aap=0.026m2
As it is less than Ad=0.09m2

Use Aap in equation

hdc=166[Lwd/rlAap]2
hdc=7mm
Backup in down comer
hb= hdc+(hw+how)+ht
hb=79+216+7=302mm
hb=0.31m
0.31<1/2(plate spacing+weir height)
0.301 <0.32
so tray spacing is acceptable

Residance time

tr=Adhbcrl/lwd
tr=5 sec
Check Entrainment
Uv=0.783/0.60=1.305
%age flooding=1.305/0.045
%age flooding=30
Flv=0.114
Y=0.02
So thats satisfactory

 

No. of holes per tray

Area of one hole=1.964*10-5m2
No. of holes=0.051/1.964*10-5
No. of holes=2596

 

Overall Efficiancy Calculations

 


Mole fractions

m

Mole fractions*m

H2

CH4
C2H2

C2H4

C2H6
C3H4
C3H6
C3H8
0.136
0.32
0.01
0.381
0.038
0.01
0.098
0.003
0.0075
0.008
0.0082
0.0085
0.0095
0.012
0.15
0.22
0.0012
0.00256
0.000082
0.00323
0.000361
0.000012
0.0147
0.000066
0.022

Eo=0.492[m(aLK/Hk)av]-0.245
Eo=0.492[0.02(7.1)]-0.245
Log Eo=-0.09
Eo=75
Actual no of plates=9/0.75
Actual no of plates=12

Location of feed

Log NA/NB=0.206 log {(B/D) (x HK /x LK) [(x LK) B/(x HK)] 2}
Log NA/NB=0.206 log {200.27/169.06 (0.381/0.32) (0.006/0.0079)2}
NA/NB=0.995
NA+NB=11
NA=6, NB=5
Feed enters the column 6 theoretical stages above the bottom stage

6 comments:

  1. As far as I know Gilliland's correlation is valid for the relative volatility values of 1.1 to 4.05. Although the average volatility of the methane is 10 while ethylene is heavy key, you applied Gilliland's correlation. Is there a reference stating the applicability of Gilliland's correlation in this case?

    ReplyDelete
  2. How did you arrive at 7.1 for the relative volatility ?

    ReplyDelete
  3. how did you find the k values at such low temps?

    ReplyDelete
  4. if i have to design a demethanizer by an external reflux and the pressure of entering feed streams are so different i.e of one is 30 bar and other is of 60 bar...so I have to use valve to enter the feed streams at same pressure or not?

    ReplyDelete
  5. if there are two side reboilers with two feed streams and one external reflux then which method should be use

    ReplyDelete
  6. No references or workings? Very hard to understand

    ReplyDelete