Absorber Design (Methanol Vopur Into Water)
Absorber Design (Methanol Vopur Into Water)
Absorber Design (Methanol Vopur Into Water)
3.9969 m 3 MeOH
700 kPa
8.1011 m 3 total gasfeed
3.5379 m MeOH
100 kPa
3
4.2563 m total gas exit
5656.131 kg
1h
=1.5711 kg /s
h
3600 s
MeOH
2.079
4.412
4.506
6.012
6.786
7.639
8.785
8.879
9.319
10.599
11.
Partial Pressure
gas, kPa
0.085
0.164
0.164
0.237
0.271
0.322
0.361
0.360
0.396
0.455
0.5
12.025
13.705
14.532
15.318
16.385
18.091
19.998
20.518
22.744
25.571
Partial Pressure
gas, kPa
0.545
0.661
0.742
0.775
0.895
1.055
1.230
1.235
1.423
1.648
From the data, a linear graph can be approximated and its equation is shown in the graph (Figure 1)
below as well.
20
Partial Prssure of MeOH (kPa)
15
10
5
0
0
0.5
1.5
Mole fractionvapour=
24.5256
=0.035
700
1.5
56
Mole fractionliquid =
=0.00487
1.5 98.5
+
56 18
m=
0.035
=7.1869
0.00487
To decide the most economic water flow-rate, absorption design must be taken
into our consideration. Using Figure 2, the number of stages required at different
water rates will be determined and the optimum rate chosen:
y 1 p1 345.36 kPa
= =
=34.62
y 2 p2
9.43 kPa
Gm
Lm
NOG
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
3.9
4.2
4.6
5.8
7.1
10.1
13.8
It can be seen that the optimum will be between mG m/Lm D 0.5 to 0.7, as would be expected. Below
0.5 there is only a small decrease in the number of stages required with increasing liquid rate; and
above 0.7 the number of stages increases rapidly with decreasing liquid rate. For this case, to ensure
better efficiency for the absorption to occur, 0.7 is chosen.
NOG = 8
Lm=
Since
Gm
=0. 7
Lm
Figure 2: Number of transfer units NOG as a function of y1/y2 with mGm/Lm as parameter
Column diameter
The physical properties of the gas inlet are calculated based on the fraction of each of the gas
components exists in the gas stream.
kg
kg
=1.5711
h
s
By considering all the components present in both the gas and liquid stream, the representative
density can be obtained:
Gas density,
Liquid density,
Liquid viscosity,
L = 999.85 kg/m3
L w Lm 26041.01 kg/ hr
=
=
V w G m 5656.13 kg/ hr
Lw V 26041.01
3.01
=
=0.25
999.85
V w L 5656.13
Design for a pressure drop of 20 mm H2O/m (0.1kPa/m) packing, and from Figure 3,
K4 = 0.65
At flooding K4 = 2.0
Percentage flooding =
0.63
100=56.12
2.0
Figure 3: Generalised pressure drop correlation, adapted from a figure by the Norton Co. with
permission
Based on equation below,
(V w ) =
V w =
K 4 V ( L V )
0.1
L
13.1 F p
L
( )
0.63 3.01(999.853.01)
3 0.1
13.1170
10
( 0.315999.85
)
=1.9475
kg
m2 s
V
Columnarea required= w =
Vw
Diameter=
kg
hr
hr 3600 s
2
=0.8068 m
kg
1.9475 2
m s
5656.13
4
0.8068=1.0135 m
1.1
( 2)
=0.9503 m2
4
D2
Columnarea=
=
4
56.12
0.8068 m
=47.65 flooding (satisfactory )
2
0.9503 m
Estimation of HOG
Cornells method
Gas viscosity,
http://www.methanol.org/Technical-Information/Resources/Technical-Information/Physical-Properties-of-Pure-Methanol.aspx
http://www.gsi-net.com/en/publications/gsi-chemical-database/single/343.html, http://webserver.dmt.upm.es/~isidoro/dat1/Mass%20diffusivity%20data.pdf
viscosity , v
( density , ) (Diffusivity , D)
0.315 x 103
=6.38
3.01 1.64 x 105
4
0.1144 x 10
=6.73
9
999.85 1.7 x 10
26041.01
3600
Liquid mass flowrate per unit area , LW =
=7.6119 kg / m2 s
0.9503
2) HL= 0.305(0.065)(6.73)0.5(0.97)(Z/3.05)0.15
i.
HL=0.0499(Z/3.05)0.15
Gm
Gm
m
m
=0.7
3) HOG = HG +
Lm HL ;
Lm
4) Z = HOG NOG ; NOG = 8
By using goalseek tool in Microsoft Excel, Equation (1) to Equation (4) above can be
solved to obtain the height of the column. HOG is calculated to be around 0.37 m, and the
obtained column height, Z is 2.9722 m.
Height of column , Z=2.9722 m