x = 0in xout yout y = 0.08in G = 5000 kg/hr
Chapter 3
Example 3.2-5.---Sulfur dioxide produced by the combustion of sulfur in air is absorbed in water. Pure SO2 is
then recovered from the solution by steam stripping. Make a preliminary design for the absorption column. The feed will be 5000 kg/hr of gas containing 8 mole percent SO2. The
gas will be cooled to 20oC. A 95 percent recovery of the SO2 is required2.
Solution ---
Operation is at atmospheric pressure as the solubility of SO2 in water is high. The feed water
temperature will be taken as 20oC.
Table E-1. Equilibrium data for SO2 at 1 atm and 20oC.
x 0 .000564 .000842 .001403 .001965 .00279 .00420
y 0 .0112 .01855 .0342 .0513 .0775 .121
At 95 percent recovery of SO2
yout = (0.05)(0.08) = 0.004
Slope of equilibrium line: y* = mx 0.0775 = m(0.00279) ⇒ m = 27.8
To decide the most economic water flow rate, the stripper should be considered together with the absorption design. For this example, the absorption design will be considered alone.
The number of gas transfer unit may be estimated from
NOG = L mG / 1 1 − + − − − mG L y y y y L mG out out out in / ) / 1 ( ln * * Where G = ) 29 )( 3600 )( 454 . 0 ( 5000 = 0.1055 lbmol/s yin = 0.08, yout = 0.004, y*out = mxin = (27.8)(0) = 0 Evaluate as NOG a function of mG/L mG/L 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.99 NOG 3.5 3.8 4.2 4.7 5.4 6.3 7.8 10.6 17.4 L,lbmol/s 14.66 9.78 7.33 5.87 4.89 4.19 3.67 3.26 2.96
The optimum will be between mG/L = 0.6 to 0.8. Below 0.6 there is only a small decrease in the number of stages required with increasing liquid rate and above 0.8, the number of stages increases rapidly with decreasing liquid rate (Figure E-1)
Figure E-1. Optimum liquid flow rate for Sulfur Dioxide absorption.
Check the liquid outlet composition at mG/L = 0.6 and at mG/L = 0.8. Assuming dilute solution, the material balance is
L(xout – xin) = G(yin – yout) ⇒ xout = G(yin – yout)/L + xin
xout = (0.08 – 0.004) L mG 8 . 27 At mG/L = 0.6, xout = 1.64×10-3 At mG/L = 0.8, xout = 2.19×10-3
Use mG/L = 0.8 as the higher concentration will favor the stripper design and operation without significantly increasing the number of stages needed in the absorber. Therefore
NOG = 8 0.2 0.3 0.4 0 .5 0.6 0.7 0 .8 0.9 1 2 4 6 8 10 12 14 16 18 m G/L L (l b m o l/ s ), N O G L NOG
Estimate column diameter: 2 methods.
First design method: chooses the pre Figure 13.41 to evaluate G'(lb/s Area = lbmol G
Check the percent of flooding (65 to 90%).
Figure E-2 Generalized flooding and pressure drop correlation.
Second design method:
determine G'flood and operate at some percentage of
G' = (.65 to .90) G'flood
Once G' is known the column cross
10
Wankat, P. C., Equilibrium Staged Separations : 2 methods.
chooses the pressure drop per unit length of packing then use '(lb/s⋅ft2
). Note: G in the ordinate of Figure E-2 is actually
2 ' . . ft s lb G lbmol lb vapor W M s lbmol ⋅
Check the percent of flooding (65 to 90%).
Generalized flooding and pressure drop correlation. Second design method: Use the flooding curve of Figure E-2 or the Eq. (3.2
and operate at some percentage of G'flood. flood
' is known the column cross-sectional area A can be determined.
Equilibrium Staged Separations, Elsevier, 1988, pg. 420
ssure drop per unit length of packing then use is actually G'.
Generalized flooding and pressure drop correlation.
log10 c L G g F G
ρ
ρ
ψµ
2 . 0 2 ' = − where Flv = ' ' G L L Gρ
ρ
Gm are mass flow rates in this expression.
The first design method will be applied. The physical properties of the gas can be taken as those for air as the concentration of SO
G = 3600 )( 454 . 0 ( 5000 At mG/L = 0.8,
Packing: Choose 1.5" Ceramic Intalox saddle (picture below) with packing parameters 32, α = 0.13, β = 0.15. Intalox saddle is one type of dumped packings.
F is the parameter in the ordinate of Figure E parameters to determine the pressure drop
∆p = α(10βL')
Air (gas) density at 20oC:
Liquid density:
ρ
L = 62.3 lb/ftThe column will be designed for a pressure drop of 0.5 in of water/ft of packing. Table E-2 shows the recommended design values.
1.6678 − 1.085 log10(Flv) − 0.29655[log10(Flv)]2 2 / 1 = m m G L 1/2 L G
ρ
ρ
= the abscissa of Figure E-are mass flow rates in this expression.
The first design method will be applied. The physical properties of the gas can be taken as those for air as the concentration of SO2 is low.
) 29 )( 3600 5000 = 0.1055 lbmol/s = 0.8, L = mG/0.8 = (27.8)(0.1055)/0.8 = 3.666 lbmol/s
Packing: Choose 1.5" Ceramic Intalox saddle (picture below) with packing parameters = 0.15. Intalox saddle is one type of dumped packings.
r in the ordinate of Figure E-2 c L G g F G
ρ
ρ
ψµ
0.2 2 ' . αparameters to determine the pressure drop ∆p in inches of water per foot of packing given by
G G
ρ
2 ' C:ρ
G = ) 293 )( 359 ( ) 273 )( 29 ( = 0.0753 lb/ft3 = 62.3 lb/ft3, liquid viscosity = 1 cp.The column will be designed for a pressure drop of 0.5 in of water/ft of packing. Table 2 shows the recommended design values.
2
(3.2-11)
-2. Note: Lm and
The first design method will be applied. The physical properties of the gas can be taken
0.8 = (27.8)(0.1055)/0.8 = 3.666 lbmol/s
Packing: Choose 1.5" Ceramic Intalox saddle (picture below) with packing parameters F =
α and β are the in inches of water per foot of packing given by
(3.2-22)
Table E-2. Recommended design value for ∆p (inches of water per foot of packing) Application ∆p (inches of water per foot of packing)
Absorber and stripper 0.2 – 0.6
Distillation (atmospheric & moderate pressure)
0.5 – 1.0
Vacuum columns 0.1 – 0.4
If very low bottom pressures are required, structured packings or special low pressure drop dumped packings should be considered (Hyperfil, Multifil, or Dixon rings).
The column area may be estimated from the pressure drop ∆p using either Eq. (3.2-22) or Figure E-2. The procedure for using Figure E-2 will be discussed
ψ
= liquid of Density water of Density = 1 Flv = ' ' G L 1/2 L Gρ
ρ
= m m G L 1/2 L Gρ
ρ
= ) 29 )( 1055 . 0 ( ) 18 )( 67 . 3 ( 1/2 3 . 62 0753 . 0 = 0.75From Figure E-2 at Flv = 0.75 and ∆p = 0.5 in water/ft of packing
c L G g F G
ρ
ρ
ψµ
0.2 2 ' = 0.018At flooding using the flooding line or Eq. 3.2-21:
c L G f g F G ρ ρ ψµ0.2 2 ' = 0.0294 2 ' ' f G G = 0294 . 0 018 . 0 = 0.612 ⇒ f G G ' '
= 0.78 (O.K., between 65 and 90 %)
G' = 2 / 1 2 . 0 018 . 0
µ
ρ
ρ
F gc L G = 2 / 1 ) 1 )( 52 ( ) 2 . 32 )( 3 . 62 )( 0753 . 0 )( 018 . 0 ( = 0.229 lb/s⋅ft2Note:
µ
in the above expression is the liquid viscosity in centipoises. Gas mass flow rate Gm = (0.1055)(29) = 3.06 lb/s = AcG'Ac = 229 . 0 06 . 3 = 13.38 ft2 ⇒ Dc = 2 / 1 ) 38 . 13 )( 4 ( π = 4.13 ft
Check packing size: Recommend size ranges are
Column diameter Use packing size
< 1 ft 1 to 3 ft > 3 ft 1 in. 1 to 1.5 in. 2 to 3 in.
In general, the largest size of packing that is suitable for the size of column should be used, up to 2 in. Small sizes are appreciably more expensive than the larger sizes. Above 2 in., the lower cost does not normally compensate for the lower mass transfer efficiency. Use of too large a size in a small column can cause poor liquid distribution. Since 1.5 in. ceramic Intalox saddle is used in this example, a larger size could be considered.
The height of packing may be determined from the following formula hp = NOGHOG
The height of overall gas transfer unit, HOG, may be evaluated from the height of gas transfer
unit, HG, and the height of liquid transfer unit, HL
HOG = HG +
L mG
HL
The correlation for HG is
HG = 1.25 0.8 2 2 / 1 3 / 1 1 ] ) / ( ) / ( ' ) 3600 [( ) ( ) 10 / ( ) ' ( b V L V L v p b col L Sc h D − − σ σ µ µ ψ
Where b1 = 1.11 for saddles, b2 = 0.50 for saddles D'col = lesser of column diameter in ft or 2
hp = height of packed bed in ft
Scv = Schmidt number for vapor =
µ
v/ρ
vDvL' = mass flux of liquid, lb/s⋅ft2
σ
= surface tension of liquid (L) or water (W)0
Figure E-3. Packing parameter
Ceramic Berl saddles are used to make conservative estimate of packing parameter for Intalox saddles since mass transfer efficiency of Intalox saddles is higher than that of the equivalent size Berl saddles.
W L µµ = 1 ; W L ρρ 102 103
Figure E-4. Packing parameter
The correlation for HL is
HL = φCfL(hp/10)
where φ = packing parameter given in Figure CfL = vapor load
ScL = Schmidt number for liquid =
Estimate hp by assuming a value for so
2.2 ft, then
11 Wankat, P. C., Equilibrium Staged Separations 12
Wankat, P. C., Equilibrium Staged Separations
20 40 60 80 90 . Packing parameter
ψ
(ft) as a function of percent floodsaddles are used to make conservative estimate of packing parameter for Intalox saddles since mass transfer efficiency of Intalox saddles is higher than that of the
= 1 ; W L σσ = 1 104 105
. Packing parameter φ (ft) as a function of L' Figure E-5. Vapor load coefficient
is
/10)0.15(ScL)1/2
= packing parameter given in Figure E-412 = vapor load coefficient given in Figure E-512
= Schmidt number for liquid =
µ
L/ρ
LDLby assuming a value for so HOG that HG and HL can be evaluated. Let
Equilibrium Staged Separations, Elsevier, 1988, pg. 652 Equilibrium Staged Separations, Elsevier, 1988, pg. 654
90 as a function of percent flood11
saddles are used to make conservative estimate of packing parameter for Intalox saddles since mass transfer efficiency of Intalox saddles is higher than that of the
. Vapor load coefficient
hp = NOG HOG = (8)(2.2) ≈ 18 ft
At 78% flooding and ∆p = 1.5 in, Figure E-3 gives a packing value
ψ
= 65 ft. From the PROP program (T.K. Nguyen Website)Diffusivity of SO2 in water at 20oC: DL = 7.5×10-6 cm2/s = 8.07×10-9 ft2/s
Diffusivity of SO2 in air at 20oC and 1 atm: Dv = 0.122 cm2/s = 1.31×10-4 ft2/s
Viscosity of gas (air) at 20oC:
µ
v = 1.82×10-5 kg/m⋅s = 1.22×10-5 lb/ft⋅sScv =
µ
v/ρ
vDv = (1.22×10-5)/[(7.53×10-2)(1.31×10-4)] = 1.237ScL =
µ
L/ρ
LDL = (6.72×10-4)/[(62.3)(8.07×10-9)] = 1337L' = (3.666)(18)/(13.38) = 4.93 lb/s⋅ft2 = 1.78×104 lb/hr⋅ft2 Flooding ratio = 0.78 ⇒ CfL ≈ 0.70 (Figure 19-8)
HG = 1.25 0.8 2 2 / 1 3 / 1 1 ] ) / ( ) / ( ' ) 3600 [( ) ( ) 10 / ( ) ' ( b V L V L v p b col L Sc h D − − σ σ µ µ ψ = 1/2 2 / 1 3 / 1 11 . 1 )] 93 . 4 )( 3600 [( ) 237 . 1 ( ) 10 / 18 ( ) 2 )( 65 ( = 1.42 ft L' = 1.78×104 lb/hr⋅ft2 ⇒ φ ≈ 0.1 (Figure E-4) HL = φCfL(hp/10)0.15(ScL)1/2 = (0.1)(0.7)(1.8)0.15(1337)1/2 = 2.80 ft HOG = HG + L mG HL = 1.42 + (0.8)(2.80) = 3.66 ft hp = NOG HOG = (8)(3.66) = 29.3 ft
Repeat the calculation with hp = 30 ft
HG = 1/2 2 / 1 3 / 1 11 . 1 )] 93 . 4 )( 3600 [( ) 237 . 1 ( ) 10 / 30 ( ) 2 )( 65 ( = 3 / 1 8 . 1 3 (1.42) = 1.69 ft HL = (0.1)(0.7)(3.0)0.15(1337)1/2 = 15 . 0 8 . 1 3 (2.80) = 3.02 ft HOG = HG + L mG HL = 1.69 + (0.8)(3.02) = 4.1 ft hp = NOG HOG = (8)(4.1 = 32.8 ft
