Absorber
-
Upload
haiqal-aziz -
Category
Documents
-
view
4 -
download
0
description
Transcript of Absorber
Appendix F - 52
Production of 100,000 MTA Hydrogen
F.9 SIZING AND COSTING FOR ABSORBER
Absorber T-20
(a) Flowrate Data from simulation :Top Column (tray above feed)
= 20331.00 kmol/hr = 6.540E+05 kg/ hr
= 7208.80 kmol/hr = 1.95E+04 kg/ hr
= 838.3
= 2.687Surface tension, s = 0.02530 N/m
Bottom Column (tray below feed)
= 22059.0 kmol/hr = 7.298E+05 kg/ hr
= 7282.2 kmol/hr = 2.18E+04 kg/ hr
= 838.80
= 2.714Surface tension, s = 0.02127 N/m
(b) Column DiameterEstimation of vapor-liquid flow factorFor top column
== 0.1597
For bottom column
== 0.1723
Take tray spacing as = 0.55m
Liquid flowrate, Ln
Vapor flowrate, Vn
Liquid density, rL kg/ m3
Vapor density, rv kg/ m3
Liquid flowrate, Lm
Vapor flowrate, Vm
Liquid density, rL kg/ m3
Vapor density, rv kg/ m3
FLV Ln / Vn [ (rV / rL)1/2 ]
FLV Lm / Vm [ (rV / rL)1/2 ]
Estimation of correction factor, K1
Appendix F - 53
Production of 100,000 MTA Hydrogen
Chemical Engineering Volume 6 Second Edition (R.K. Sinnott, 1996)From Figure 11.27
= 0.0900
= 0.0800Correction for surface tensions,
== 0.0943
== 0.0810
== 1.6635 m/s
== 1.4215 m/ s
Design for 85 % flooding at maximum flowrate
= 1.4140 m/ s
= 1.2083 m/ s
Maximum volumetric flowrates
=
= 2.0136
=
= 2.2322
K1 (top column)
K1 (bottom column)
K1,c (top column) K1 (top column) [ s / 0.02 ] 0.2
K1,c (bottom column) K1 (bottom column) [ s / 0.02 ] 0.2
Estimate the flooding velocity, Uf
Uf (top column) K1,c (top column) [ ( rL - rv ) / rv] 1/2
Uf (bottom column) K1,c (bottom column) [ ( rL - rv ) / rv] 1/2
Therefore, Uf = 0.85 x Uf
Uf (top column)
Uf (bottom column)
Vmax (top column) Vn
3600 × rv
m3/ s
Vmax (bottom column) Vm
3600 × rv
m3/ s
Appendix F - 54
Production of 100,000 MTA Hydrogen
Net area required
=
= 2.01361.4140
= 1.4241
=
= 2.23221.2083
= 1.8473
Take downcomer area as 12% of total column cross-sectional area.Column cross-sectional area
= 0.88
= 1.42410.8800
= 1.6182
= 0.88
= 1.8473 0.88
= 2.0992
Column diameter
== 1.4353 m
== 1.6348 m
Take the larger diameter as a final diameter for further design procedure, so
= 1.6348 m
An (top column) Vmax (top column)
Uf (top column)
m2
An (bottom column) Vmax (bottom column)
Uf (bottom column)
m2
Ac (top column) An (top column)
m2
Ac (bottom column) An (bottom column)
m2
Dc (top column) [ (Ac (top column) × 4) / p ]1/2
Dc (bottom column) [ (Ac (bottom column) × 4) / p ]1/2
Dc
Appendix F - 55
Production of 100,000 MTA Hydrogen
(c) Liquid Flow Pattern
aximum volumetric =
liquid rate,Lmax = 0.2417
Chemical Engineering Volume 6 Second Edition (R.K. Sinnott, 1996)
= 1.6348 mand
= 0.2417
Therefore,The selected recommended liquid flow pattern is single pass tray
(d) Provisional Plate Design
= 1.6348 m
= 2.0992
Downcomer area, =
Ad (at 12%) = 0.2519
=
= 1.8473
=
= 1.5954Take 10% as first trial :
=
= 0.1595
Chemical Engineering Volume 6 Second Edition (R.K. Sinnott, 1996)
Therefore,
= 0.77
= 0.77 × Dc= 1.2588 m
Take the recommended value
= 50 mm (above 1 atm operation)= 0.050 m
= 5 mmPlate thickness = 3 mm (stainless steel)
Lm / (rL × 3600)
m3/ s
From Figure 11.28
For Dc
Lmaxm3/ s
Column diameter, Dc
Column area, Ac m2
12% × Ac
m2
Net area, An Ac - Ad
m2
Active area, Aa Ac - 2Ad
m2
Hole area, Ah 10% × Aa
m2
From Figure 11.31
For (Ad / Ac) x 100% = 12%,
Iw / Dc
Weir length, Iw
Weir height, hw
Hole diameter, dh
Appendix F - 56
Production of 100,000 MTA Hydrogen
(e) Check Weepingaximum liquid rate, = 729820.0 kg / hrLmax = 202.7278 kg /s
Minimum liquid rate =at 70% turn down, = 141.9094 kg/sLmin
== 249.6116 mm liquid
== 196.7872 mm liquid
At minimum rate, = 246.7872 mmhw + how, min
Chemical Engineering Volume 6 Second Edition (R.K. Sinnott, 1996)
= 31.0
Minimum vapor velocity through the holes
=
= 7.6726 m/s
Actual minimum vapor velocity,
=
= 9.7936 m/s
so minimum operating rate will not be well above weep point
Lmax × 70%
Maximum how 750 [ Lmax / ( rLIw) ]2/3
Minimum how 750 [ Lmin / ( rLIw) ]2/3
From Figure 11.30
K2
Uh (min) [ K2 - 0.90( 25.4 - dh ) ]
rv1/2
Ua (min) 0.7 x Maximum volumetric flowrate, Vmax
Ah
Since Ua (min) < Uh (min),
Appendix F - 57
Production of 100,000 MTA Hydrogen
(f) Plate Pressure Drop
Maximum vapor =velocity through = 2.2322 holes, Uh 0.1595
= 13.9909 m/s
Chemical Engineering Volume 6 Second Edition (R.K. Sinnott, 1996)From Figure 11.34
For (i) plate thicknes= 0.6
Therefore, Orifice = 0.7400coefficient, Co
== 58.9862 mm liquid
== 14.902 mm liquid
= 373.500 mm liquid
(g) Downcomer Liquid Back-UpDowncomer pressure loss
= 40 mm= 0.0400 m
=
=
= 0.0101
== 0.010 mm= 0.000010 m
== 112.32 mm= 0.1123 m
Since0.1123 < 1/2 ( plate spacing + weir height )0.1123 < #VALUE!
So, tray spacing is acceptable
Vmax / Ah
(ii) Ah/ Ap @ Ah/ Aa = 0.1
Dry plate drop, hd 51 [ Uh / Co]2 rv / rL
Residual head, hr (12.5 × 103) / rL
Total plate pressure drop, ht = hd + hw + how (max) + hr
Take hap = hw -10 = 50 - 10
Area under apron, Am hap × It
(40 × 10-3) × 1.0383
m2
Head loss in downcomer, hdc 166[ Lmax / (rL x Am) ]2
Back-up downcomer, hb hw + how + ht + hdc
Appendix F - 58
Production of 100,000 MTA Hydrogen
=
= 0.1171 s(since tr > 3s, the retention time is satisfied)
= 0.0020 min
(h) Check EntrainmentActual percentage flooding for design area :
=
= 2.2322
1.8473= 1.2083 m/ s
% flooding == 1.2083 × 100
1.4140= 85.45 %
Chemical Engineering Volume 6 Second Edition (R.K. Sinnott, 1996)From Figure 11.34For % floooding = 85.45 %
= 0.1723Fractional = 0.0350entrainment, y
y well below 0.1, satisfactory (y below 0.1 give small effect only to the column efficiency.)
(i) Perforated AreaFrom Figure 11.32Chemical Engineering Volume 6 Second Edition (R.K. Sinnott, 1996)
= 1.25881.6348
= 0.7700
= 99°
Angle subtended at plate edge by unperforated strip, = 180° - 99° = 81°
Mean length, unperforated edge = (1.6348 - 0.05)π × (81/ 180) strips = 2.2131
Check retention time, tr AdhbrL
Lmax
Uv Maximum volumetric flowrate, Vmax
An
(Uv / Uf) × 100
and FLV
At Iw/ Dc
Therefore, qc
= ( Dc - hw )π × (81 / 180)
Appendix F - 59
Production of 100,000 MTA Hydrogen
Area of unperforated = edge strips = 0.05 × 2.2131
= 0.1107
Area of calming =zones = 2 × 0.05 ( 1.2588 - 2 (0.012) )
= 0.1159
=
= 1.3689
= 0.15951.3689
= 0.1165
hw × mean length
m2
2hw ( weir length, Iw - 2hw )
m2
Total area available for perforation, Ap
Aa - (area of unperforated + area of calming)
m2
Ah/ Ap
Appendix F - 60
Production of 100,000 MTA Hydrogen
Chemical Engineering Volume 6 Second Edition (R.K. Sinnott, 1996)From Figure 11.33
= 2.76 (satisfactory, within 2.5 to 4.0)
(j) Number of Holes
Area of one hole = 1.964E-05
Number of holes =Area of one hole
= 0.15951.964E-05
= 8123.4= 8123 holes
(k) Height of Column
Number of trays, n = 8assume stage efficiency = 0.7
Number of real = (n-1)/0.7stages,N
= 10
Ip / dh
m2
Ah
Appendix F - 61
Production of 100,000 MTA Hydrogen
Plate thickness = 3 mm= 0.003 m= N × plate thickness = 0.024 m
plate, Hp
Single plate spacing, = 0.55 m24"
Number of spacing = N - 1= 9
Tray stack = (N-1) x tray spacing = 5.40 mExtra feed space = 1.5 m
Disengagement space = 2.0 m(top & column)
Skirt height = 1.5 mTotal height of = 10.42 m
column, Ht
= 6.3764Equipment Costing for Absorber T-20From sizing results :
Pressure vessel = 21.7800 atm= 22.0686 bar
Column height = 10.42 m = 34.20 ftInside diameter = 1.63 m = 5.36 ftNumber of trays = 8Type of vessel = Vertically orientedMaterial of = Stainless steelconstructionEconomic data :
= 405.60
= 115
Equipment cost, C = Co
From Table 4.11, page 134, Systematic Methods of Chemical Process Design:With Co = 1000; Lo = 4.0; Do = 3.0; a = 0.81 ; b = 1.05
BC == $10,466.85
MF = 4.23
Total thickness of
Check : Ht/ Dc (satisfactory if Ht/ Dc < 20)
CE Index Year 2003, CEI03
CE Index Year 1968, CEI68
1000 × (34.22/4.0)0.81 × (19.67/3.0)1.05
( LLO
)α
( DDO
)β
Appendix F - 62
Production of 100,000 MTA Hydrogen
MPF =
= 1.93 (Stainless steel)
= 1.0MPF = 1.0 × 1.93
= 1.93
Update factor, UF = 3.53Update bare module = UF (BC) (MPF + MF - 1)cost = 3.57 (40992.46) (1.93 + 4.23 - 1)
= $190,487.22 = RM 725,756.30
Price of sieve trays
Equipment cost,C = Co
From Table 4.11, Systematic Methods of Chemical Process Design:With Co = 180; Lo = 10; Do = 2.0; a = 0.97; b = 1.45
C == $2,037.12
MF = 1
MPF =
= 1.7 (Stainless steel)
= 0 (sieve tray)
= 1 (24 inch.spacing)then,
MPF = 2.7000
Update factor, UF = 3.53Update bare module = Update bare module cost = UF(BC)(MPF+MF-1)cost = 3.57 (13418.05) (2.7 + 1 - 1)
= $19,399.04 = RM 73,910.34
Total equipment = 799,666.65cost, (RM)
Fm × Fp
Fm
Fp
180 × (34.22/10)0.97 × (19.67/2.0)1.45
Fm + Fs + Ft
Fm
Ft
Fs
( LLO
)α
( DDO
)β