MCEN 4131 Lecture 28 Absorption tower sizing example In the previous lectures we derived the pertinent equations required for analyzing and sizing a tower. The following is a point po int by point example of how a tower is sized. Problem statement ( Fundamentals Fundamentals of air pollution engineering , RC Flagen, JH Seinfeld, pg 490, 1988): A packed tower is to be designed for absorption of SO2 from air by contact with fresh water. The entering gas has a mole fraction of SO2 of 0.10 and the exit gas must contain amole fraction of no greater then 0.005. The water flow rate used is to -1 be 1.5 times the minimum, and the inlet airflow rate (on an SO2 free basis) is 500 kg m h-1. The column is to be operated at 1 atm and 303 K. Determine the height of the packed bed section. The following equilibrium data are available for SO2 absorption in water at this temperature: PSO2 (mm Hg) 0.6 1.7 4.7 11.8 19.7 36.0 52.0 79.0
Conc. (g SO2/100 g H2O) 0.02 0.05 0.10 0.20 0.30 0.50 0.70 1.00
There also exist mass transfer coefficient correlations for SO2 absorption in a column packed with 1-inch rings (at this temperature): k x a
=
~0.82 0.6634L ~
~
k y a
=
~ ~ 0.09944L0.25 V 0.7
where L and V are the liquid and gas mass fluxes, mass fluxes, respectively, in kg m-2 h-1 and the -3 -1 -1 units of k xa and k ya are kg-mol m h (mol fraction) .
Step 1: Plot the equilibrium curve (xi and yi) on the x-y graph. To do this, we need to get the equilibrium data into mol fraction format: y x
PSO 2
=
=
Patm
(e.g.)
0.6 mmHg 760mmHg
c / MWSO 2
=
c / MWSO 2
+
c / MWH 2O
=
=
7.89x10 ! 4 (e.g.)
0.02 / 64 0.02 / 64 + 100 / 18
=
5.625x10 !5
thus, P 0.6 1.7 4.7 8.1 11.8 19.7 36 52 79
y 0.0008 0.0022 0.0062 0.0107 0.0155 0.0259 0.0474 0.0684 0.1039
c 0.02 0.05 0.1 0.15 0.2 0.3 0.5 0.7 1
x 5.6E-05 1.4E-04 2.8E-04 4.2E-04 5.6E-04 8.4E-04 1.4E-03 2.0E-03 2.8E-03
See Figure on the next page Step 2: Determine the minimum liquid flowrate, (Lm)min: Using the equation for the operating line of the absorption tower, we input the appropriate values so that the bottom of the tower is at equilibrium conditions: yt 1' yt
=
& y b $$ % 1 ' y b
# & L'm !! ' $ ' $ " % Vm
# & x b* xt ! $ ' ! $1 ' x* 1 ' x t b " min %
# ! ! "
yt = 0.005 (given in problem statement) xt = 0 (given) y b = 0.1 (given) -2 -1 -2 -1 V'm=500 kg m h = 17.2 kgmol m h x b* can be determined by either a) interpolating from table above, or b) taking the value off the graph on the following page. Interpolation: ( (0.1 ! 0.0684) % x b* = & " 2.8 " 10!3 ! 2.0 " 10!3 # + 2.0 " 10!3 = 2.71 " 10!3 ' 0.1039 ! 0.0684 $ -2 -1 Solving, we calculate (L'm )min= 667 kg-mol m h . Thus L'm = 1.5 _ 667 = 1000 kg-mol -2 -1 m h .
(
)
Step 3. Plot the operating line for the column Before we do this, we need to determine the mole fraction of SO2 in the liquid phase at the tower bottom with the given liquid/gas flowrates. We c an use the same equation, just inserting the right values again:
& y b # & L'm #& x b xt # $$ !! ' $ ' !$$ ! ' 1 ' y t % 1 ' y b " $% Vm !"% 1 ' x b 1 ' x t !" 0.005 0 # & 0.1 # & 1000 #&$ x b !! ' $ !'$ !$ ' ' ' 1 ' 0.005 1 0 . 1 17 . 2 1 x 1 0 % " % "% b " yt
=
=
thus, x b = 0.00183 Now returning to the more general form of the operating line equation, we can plot the operating line in the x-y plane:
=
& y b # & L'm # $$ !! ' $ ' ! $ ! 1 ' y b " % Vm " %
=
x # & 0.1 # & 1000 # & 0.00183 ' $ !'$ ! $ ! % 1 ' 0.1 " % 17.2 " % 1 ' 0.00183 1 ' x "
y 1' y y 1' y y
& x b x # $$ !! ' 1 ' x 1 ' x b % "
& %
= (0.111)' (58.1) $ 0.00183 '
1' y
Step 4. Calculate the value of
x #
!
1' x "
" !k x a
k ya
. For this problem, we can assume that ! = 1
throughout the column(we can check it later). k x a
=
~0.82 0.6634L
k y a
=
~ ~ 0.09944L0.25 V 0.7
Since mass is transferred between the gas and liquid phases, ~ L
the column. We first need to calculate material balance on the SO2:
and
~ V
~ ~ L and V
change through
at the bottom and the top. Let's do a
The SO2 enters at the bottom at a mass rate: y b 0.1 Vm' MWSO 2 = 17.2 kgmol m ! 2 h !1 (64 kg kgmol !1 ) 1 ! y b 1 ! 0.1
=
122.6 kg m ! 2 h !1
The SO2 leaves the top at a mass rate: Vm'
yt 1! yt
2
1
MWSO 2 = 17.2 kgmol m ! h !
0.005 1 ! 0.005
(64 kg kgmol !1 )
=
5.5 kg m ! 2 h !1
Since the water enters the top completely free of SO2, then the mass flowrate of SO 2 in -2 -1 the water stream at the tower bottom = 122.6-5.5 = 117.1 kgmol m h . Thus the total mass flowrate at the bottom and top of the column are: ~ Lt ~ L b ~ V b ~ Vt
=
18000 kg m !2 h !1
=
18000 + 117.1 = 18117 kg m !2 h !1
=
500 + 122.6 = 622.6 kg m !2 h !1
=
500 + 5.5 = 505.5 kg m !2 h !1
To determine the mass transfer coefficients, calculate the value at top and bottom and take the average:
(k x a )top
= 0.6634(18000 )
0.82
= 2047
(k a ) = 0.6634(18117) = 2058 (k x a )average = 2052.4 kg-mol m-3 h-1 (mol fraction)-1 0.82
x
bottom
Similarly,
(k y a )top = 0.09944(18000)0.25 (505.5)0.7 = 89.95 (k y a ) bottom = 0.09944(18117)0.25 (622.6)0.7 = 104.2 k y a
average
= 97.1 kg-mol m-3 h-1 (mol fraction)-1
Thus, " !k x a
k ya
=-21.1 which is the slope of a line connecting any point (x,y) on the operating
line with a corresponding point (xi,yi) on the equilibrium line. I've drawn several lines on the plot that are at (about) a slope of -21. From these lines I am able to pull off 8 operating line points and 8 corresponding equilibrium line points: (y,x) (0.1, 0.0018) (0.088, 0.00157) (0.076, 0.00135) (0.066, 0.00114) (0.053, 0.0009) (0.043, 0.0007) (0.033, 0.00047) (0.019, 0.00025)
(yi,xi)_____ (0.067, 0.00192) (0.0565, 0.00165) (0.048, 0.00140) (0.040, 0.00120) (0.031, 0.00095) (0.022, 0.00075) (0.013, 0.00052) (0.007, 0.00028)
0.12 equilibrium line
yt 0.1 operating line
0.08
0.06
0.04
0.02
yb 0 0
0.0005
0.001
0.0015
0.002
0.0025
x*b
0.003
Figure 1: Operating line, equilibrium line and correspondence lines for graphical/numerical integration of absorption column height. Step 5. Prepare a table with the following headings (we’ll use this information to perform a numerical integration to determine ZT) y 0.1 0.088 0.076 0.066 0.053 0.043 0.033 0.019 0.005
1-y 0.9
0.912
0.924
0.934
0.947
0.957
0.967
0.981
0.995
yi 0.067 0.0565 0.048 0.040 0.031 0.022 0.013 0.007 0.0005
!
1-yi
!/(1-y)(y-yi)
0.933
0.916401
30.85525
0.9435
0.927661
32.29117
0.952
0.93793
36.25272
0.96
0.946941
38.99442
0.969
0.957958
45.98051
0.978
0.967462
48.13962
0.987
0.976966
50.5153
0.993
0.986988
83.84198
0.9995
0.997248
222.7244
The problem states to assume ! = 1. We can check if this makes sense by recalling that
( = (1 ' y) LM =
(1 ' y i )' (1 ' y) & 1 ' yi % 1' y
ln$$
# !! "
! ~ 0.9 to 0.99 throughout the column (see table above for more precise values). Thus, assuming that ! = 1 is a bit imprecise, but usually good enough for a first pass estimate of
the column height. If ! is calculated to be significantly less than 1, then it must be taken into account. Step 6: Integrate to find ZT The reason we want the tabular values is because we want to integrate the following expression:
) G my & #dy $ ! y ' k y a $ y (1 " y)(y " y i ) ( % avg
ZT = '
t
b
The book suggests plotting the function !/(1-y)(y-yi) as a function of y and determining the area under the curve from y b to yt. You can also perform a numerical integration over the “y” scale using a differencing approach using the average value of “!/(1-y)(y-yi)” over a range of y’s:
yt
)dy
' 30.86 + 32.29 $ ' 32.29 + 36.25 $ "(0.1 ! 0.088)+ % "(0.088 ! 0.076) 2 2 # & #
* y (1 ! y)(y ! y ) ( % & i b
' 83.84 + 222.7 $ ! 1 "(0.019 ! 0.005) = 5.8 (mol fraction) 2 & # (" G "G % % "G %+ my m,(top) m,(bottom) $$ '' = *$$ '' + $$ ''- *1/2 k a k a k a * # y & avg )# y (top) & # y (bottom) &-, "G % )" 17.3kgmolm(2h(1 % " 19.1 kgmolm(2h(1 %, my ( $$ '' *1/2 0.188 m (molfraction) Th +$ (3 (1 ' $ (3 (1 '. k a 89.95kgmolm h 104.2 kgmolm h # & # & # y & avg * +L+ %
=
us the height of the absorption column is, ZT = 0.188* 5.8 = 1.1 meters
=
