Absorption Tower

8/11/2019 Absorption Tower

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Rating of an Existing Absorption Tower

Calculating:

No. of Transfer Units.

Flooding velocity and Loading velocity. Diameter of the column.

Height of the packed tower.

Pressure drop across the entire tower.

Packed Tower Specifications:

1 Raschig rings, dumped packing. ( a/3= 158)

Composition of Inlet and Outlet Gas Streams:

Inlet Gas CompositionDry(mol%) Outlet Gas CompositionDry(mol%)

Flow Rate = 662 Nm3/hr Flow RateDry= 503 Nm3/hr, Flow RateH20= 3

Nm3/hr

H2 45.70 H2 60.18

N2 19.67 N2 25.90

Ar 02.22 Ar 02.92

CH4 08.32 CH4 10.96

NH3 24.09 NH3 00.04

Average M.Wt. at inlet gas stream =. . . .. . = 12.7361

Average M.Wt.at outlet gas stream =. . . .. .

= 11.384

Now, since Inlet gas flow rate = 662 Nm3/hr,

Molar flow rate = N/

. =.= 29.553 Kmol/hr.

Similarly, Molar flow rate at outlet = 22.4553 Kmol/hr.

Gas flow rate at inlet, Gin= (Molar flow rate x Avg. M.wt.) = (29.533 x 12.7361) = 376.397 kg/hr.

Gas flow rate at outlet, Gout= (22.4553 x 11.384) = 255.632 kg/hr.

Avg. Gas flow rate=GG

= 316.0145kg/hr.


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Composition of Inlet and Outlet liquid streams:

Mass flow rate of inlet liquid stream (water), Lin= 680 kg/hr.

Amount of NH3absorbed from gas stream = (Mass flow rate of NH3 inin gas Mass flow rate of NH3 outingas )

= [{. . x 17 . . x 17}] kg/hr= 120.666kg/hr.

Given, water flow rate at outlet liquid stream = 843 Nm3/hr = 37.633 Kmol/hr = 677.394 kg/hr.

Total Liquid mass flow rate at outlet, Lout= 120.666 + 677.394 = 798.06kg/hr.

Avg. Liquid flow rate=L

L

= 739.03 kg/hr.Calculating Average Gas Stream Composition:

Assuming half of NH3available for absorption remains in the gas stream,

NH3absorbed = {(24.09 x 662)-(.04 x 503)}/(2 x 100) = 79.63 Nm3/hr. = NH3remaining in the gas

stream.

Also, the Temperature and Pressure inside the packed tower are assumed to be the mean of the inlet and

outlet conditions.

T = 600C ; P = 16.68 atm

Calculating Avg. Density of the Gas stream at these conditions:

Assuming ideal gas behavior of the component gases at these conditions,

From Ideal Gas Law, we know that,

PM = RT

Where, P Pressure (atm)

T Temperature (K)

M Molecular Weight (gm)

R Gas constant = 0.082057 (L atm K-1mol-1)

- Density (gm/L)


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Densities of the various components are:

Composition (Nm3/hr), Fi Density (kg/m3), i

H2 303 01.2200

N2 130 17.0912Ar 15 24.4160

CH4 55 09.7664

NH3 79.63 10.3760

g,avg= {(Fix i)/Fi} = 7.4160 kg/m3.

Calculating Average density of the Liquid stream:

Molar flow rate of H2O = 37.633 Kmol/hr.

Molar flow rate of NH3= (79.63/22.4) = 3.555 Kmol/hr.

Average density of outlet stream at 16.68atm , 600C 1000kg/m3.

(since water= 1000 kg/m3and NH3= 10.376 kg/m

3.(shown in the table above))

l,avg = 1000 kg/m3.

Calculating the Flooding and Loading Velocities:

Now that we have the Average gas and liquid stream flow rates and densities, we can calculate the

flooding and loading velocities using below figure


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(Note : FPS unit system was used in drawing the curves)

Calculating the value on the x - coordinate i.e.,(g/L)

0.5we get,

(g/L)

0.5 =

..(7.4160/1000)

0.5= 0.20139

Curve A of figure gives the Flooding curve for dumped raschig rings.

The line x = 0.20139 intersects the curve A at y = 0.8 (in fps system)

. X = 0.8

Where : G is in lb/hr,

is in lb/ft-s,


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is in lb/ft3,

And g is in ft/s2.

Also, we have :.

X

=

.

x

= 0.8 (for Flooding)

Here,

=

1

at 600C = 0.5 cp ( water) = 0.5 x 10-3pa-s.

g= 7.4160 kg/m3.

L = 1000 kg/m3.

gc= 9.8m/s2.

= 158.

Vg= 0.4845 m/s .

Flooding Velocity = 0.4845 m/s.

Also, from figure , we see that curve C gives the Upper Loading values.

X =0.20139 intersects the Upper Loading curve at y = 0.6

Again from :. X

=

. x

= 0.6 ;

we get Vg= 0.4196 m/s as the loading velocity.

Taking the operating velocity as 40% of the flooding velocity, we get :

(Vg)operating= 0.4 x 0.4845 = 0.1938m/s.

(Vg)loading= 0.4196m/s.

Calculating the Diameter of the Packed Tower:

Volumetric Flow rate of gas in the tower =

. =.

X .=0.011837m3/s


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Cross sectional area = =

.. = 0.0610m

2

= 0.0610

D = 0.2788m = 10.98 in (ID)

Taking the closest suitable standard diameter, i.e., 10

We have, ID of tower = 10.02(for a standard 10 tower)

Internal Area = 0.546 ft2.

Superficial Velocity =.

. . = 0.233m/s.

Percentage Flooding =..

x 100 = 48.16%

And Percentage Loading =..x 100 = 55.6%

Calculating Height of Transfer Unit:

Average Mass Flux of gas stream through the tower = G =

G = .. .= 6229.948 kg/m2-hr = 1273.317 lb/ft2-hr

Average Mass Flux of Liquid stream through the tower = L =


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L =.

. .= 14569.3 kg/m2-hr = 2977.771 lb/ft2-hr

Extrapolating the graph of L vs. G for 1 Raschig rings, we get ,

HOG= 0.7 ft.

Correcting the height of Transfer Unit for inert gases in system:

(HOG)with inerts=

.

Where : G = Average Mass Flu x of the gas stream = 1273.317 lb/ft2-hr.

= .mol/hr-ft3-atm ;where is the average mass transfer coefficient for NH3-air mixture.P = Average Pressure in the tower = 16.68 atm.

Now,


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=.

=... . = 3.8456 mol/hr-ft

3-atm.

Also, Diffusivity (Dv)of a mixture of two components A and B having Mol.Wts. Ma , Mband volumes

Vaand Vbrespectively at temperature T and pressure P is given by :

Dv=

.

Here, T = 600C = 6000R

Ma= 17 (for NH3)

Mb= 29.16 (for air)

P = 16.68 atm

Va= 26.7 for NH3 [from table 9.28, Ludwig Vol.02]

Vb= 29.9 for air

Dvof NH3 through air is given by:

Dv = .

. . . = 0.051157 ft

2/hr.

Also,

Diffusion coefficient of NH3through 3:1 N2-H2mixture :

Ma= 17 for NH3

Mb= 11.2 for inert gas mixture

Va= 26.7 for NH3

Vb= 0.75(14.3) + 0.25(31.2) = 18.5 for 3:1 N2-H2mixture

Dv =. .. . . = 0.075267 ft

2/hr.


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Absorption Tower

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