CHE3165 Separation Processes S1 2015
GAS ABSORPTION IN PACKED TOWERS
Leaning Outcomes By the end of the lecture you should be able to: 1. Define rate based method for packed columns 2. Define Height Equivalent to a Theoretical Plate (HETP) and explain how it and the number of equilibrium stages differ with the height of a transfer unit (HTU) and number of transfer units (NTU) 3. Explain the differences between loading point and flooding point in a packed column 4. Estimate the packed height
Contents 1
Introduction
2
Mole balance
3
Absorption of of ve very di dilute mi mixtures
4
Transfer unit
5
Absorption of co concentrated mixtures
6
Absorption of semi-dilute mixtures
7
Controlling resistance
1. Introduction
Absorption:: Absorption Solute A absorbed from the gas phase into the liquid phase OR a process involves molecular/mass transfer of solute A through a stagnant, non-diffusing gas B into a stagnant liquid C Gas-Liquid System: Solute transfer from Gas
Liquid
Adsorption: Components of a liquid or gas stream adsorbed on the surface or in the pores of a solid adsorbent. Gas/Liquid – Solid System: Solute transfer from Gas/Liquid Gas/Liquid – Solid
Stripping:: Stripping Reverse of absorption Liquid – Liquid – Gas System: Solute transfer from Liquid
Gas
Tray and packed bed columns
yo, V’, Vo xi, L’, Li
L = Liquid stream total flow rate V = Gas stream total flow rate L’ = Inert (carrier) liquid flow rate V’ = Inert gas flow rate yA = Mass/mole fraction of A in gas stream xA = Mass/mole fraction of A in liquid stream
yi, V’, Vi
xo, L’, Lo
Example 1 (Geankoplis: Example 10.3-2) It is desired to absorb 90% of the acetone in a gas containing 1.0 mol% acetone in air in a countercurrent stage tower. The total inlet gas flow to the tower is 30.0 kg mol/h and the total inlet pure water flow to be used to absorb the acetone is 90kg mol H 2O/h. Identify all the unknowns: x i, L’, Li, yi, V’, Vi, yo, Vo, xo, Lo Solution: Vi = 30 kgmol/h yi = 0.01 V’ = 30 (1-0.01) = 29.7 kgmol/h L’ = 90 kgmol/h xi = 0 Li = 90 kgmol/h Ac in V0 = 30 (0.01) (1-0.9) = 0.03 kgmol/h Ac in L0 = 30 (0.01) (0.9) = 0.27 kgmol/h V0 = 29.7 + 0.03 kgmol/h = 29.73 kgmol/h L0 = 90 + 0.27 kgmol/h = 90.27 kgmol/h y0 = 0.03/29.73 = 0.00101 x0 = 0.27/90.27 = 0.003
yo, V’, Vo xi, L’, Li
yi, V’, Vi
xo, L’, Lo
Example 2 (Geankoplis: Example 10.3-1) A gas mixture at 1.0 atm pressure abs containing air and CO2 is contacted in a single-stage mixer continuously with pure water at 293 K. The t wo exit gas and liquid streams reach equilibrium. The inlet gas flow rate is 100 kgmol/h, with a mole fraction of CO2 of y A2 = 0.20. The liquid flow rate entering is 300 kgmol water/h. Calculate the amounts and compositions of the two outlet phases. Assume that water does not vaporize to the gas phase. Solution:
L=300 xA2=0
V1? yA1?
V=100 yA2=0.2 V’ =? L1? XA1?
V2, y2 L2, x2 Gas Liquid
z2=Z
(2)
V and L = total molar flow rates
Packing A – total interfacial area (m 2) a – interfacial area per unit volume of packed tower (m2/m3) S – cross sectional area of tower (m) Z – Bed height (m)
z1=0
(1)
Gas Liquid V1, y1 L1, x1
A = a x vol. of packed tower = a x (S x Z)
Function of packing is to generate largest possible interfacial area for the smallest possible gas pressure drop.
Packed Tower Operation Flooding velocity: Upper limit to the rate of gas flow. Liquid can no longer flow Optimum economic: V ≥ 0.5 flooding velocity = f(equipment cost, ∆P, processing variables)
Loading point: Gas flow rate where liquid down flow starts to be hindered by gas
Packing Materials
Packing Materials
Packing Materials
http://finepacstructures.tradeindia.com
Packing Materials
Random
Structured
Raschig rings and saddles
“Through flow”
Relative cost
Low
Moderate
High
Pressure drop
Moderate
Low
Very low
Efficiency
Moderate
High
Very high
Vapor capacity
Fairly high
High
High
Example 3 (Geankoplis: Example 10.6-2) A tray tower is to be designed to absorb SO2 from an air stream by using pure water at 293K. The entering gas contains 20 mol% SO 2 and that leaving 2 mol% at a total pressure of 101.3 kPa. The inert air flow rate is 150 kg air/h.m2 and the entering water flow rate is 6000 kg water/h.m 2. Assuming an overall tray efficiency of 25%, how many theoretical trays and actual trays are needed? Assume that the tower operates at 293 K. y1 Solution
L’ xo
V’ yn+1
xn ??
2. Mole Balance Mole balance for solute over differential volume of tower: - We can choose either gas or liquid phase.
V’Y = V’(Y+dY) (in at z)
+
dń
(out at z+dz) (transfer to liquid)
V’dY = - dń
Write ń in terms of flux and area, and Y in terms of y: '
V dY - dn
y - N.dA V d 1- y '
V
'
V
dy 2
(1- y) dy
(1- y)
- NaS dz Volume
- NaS dz
Separating variables
dz
V
dy
NaS (1- y)
Integrating from (1) to (2) y 2
z2
z1
dz
V
dy
NaS (1- y) Z
y1
This is the most general equation relating total packed height (z) to gasphase variables.
If we balance the liquid phase alone, an analogous equation is obtained: x2
Z
L
dx
NaS (1- x)
x1
To evaluate the integrals, we need V(y), N(y) or L(x), N(x)
For gas phase analysis:
N K y y y * or
Solute flux
N k y y y i or
N K G P A - P A* or N k G P A - P Ai Choose overall Ky as typical case
N
K 'y yBLM
K 'y
y y 1- y y y *
*LM
Across gas film
*
1- y *LM
1- y - 1- y * 1- y ln 1- y *
f(y)
1- y*LM
varies up the tower
y 2
z
V
N K y y y*
dy
NaS (1- y)
N
y1
y2
z
V
K aS
y1
' y
' y2
1- y *LM (1- y)
1- y *LM
dy (y y * )
V 1- y *LM dy z S K ' aS(1- y)2 (y y * ) y1 y
K 'y
V
y y *
V'
1 y
a may vary with V (i.e. with y) K’y is a function of flowrate V V’ solvent-only flow and S tower cross section area are the only true constant To further progress, need to make simplifying assumptions!!!
3. Absorption of very dilute mixtures Assumptions: 1.
Very dilute solutions (x, y less than 0.1 or 10%)
2. Linear equilibrium (i.e. Henry) y2
Start with
z
V
K aS ' y
y1
1- y *LM (1- y)
V varies only slightly from (1) to (2)
Use
V AVE
V1 V2 2
dy (y y * )
V = f(y)
V
V'
1 y
K’ya will be nearly constant because V ~ constant
(1 – y) ~ 1 (1 – y)*LM ~ 1
1- y *LM (1 - y) y2
z
z
1 1
V
' K aS y1 y
V AVE
y2
1- y *LM (1- y) - dy
K 'y aS y1 (y
*
y )
dy (y y * )
Approximate
This integral can be evaluated analytically if Henry’s law holds
Mass balance (operating line): straight line if V & L are constant from (1) to (2) (y y*) f(y) is stright line Say (y - y*) αy β d(y - y*) α dy
dy
d(y - y*)
α
put (y y1, y* y *1 ) & (y y 2 , y* y *2 )
α
(y 2 - y *2 ) (y1 - y *1 ) y 2 - y1
Δ(y y*) Δy
z
V AVE K 'y aS
y2
- d(y - y*)
α(y y*)
y1
V 1 y - y * ln 2 2 z AVE K ' aS α y y * 1 1 y V (y 2 y1 )( 1) AVE z ' K aS (y - y * ) (y - y * ) 2 2 1 1 y
α
(y 2 - y *2 ) (y 1 - y *1 )
y 2 - y *2 ln y y* 1 1
These terms make up a log mean
V (-1)(y y ) 2 1 z AVE K ' aS (y - y*)LM (1)-(2) y
y 2 - y1
Similar expressions hold for the various forms of N
V (-1)(y y ) 2 1 z AVE k 'y aS (y - y i )LM (1)-(2) L AVE (-1)(x 2 x1 ) z ' k aS (x - x) LM (1)-(2) x i L AVE (-1)(x 2 x1 ) z ' K aS (x * - x) LM (1)-(2) x We usually choose the phase which is controlling the rate (i.e. the phase with the greatest mass transfer resistance)
Re-arrange
V S L S V S L S
y1 - y 2 k 'y az(y yi )M x1 - x 2 k 'x az(xi x)M y1 - y 2 K 'y az(y y*)M x1 - x 2 K 'x az(x * -x)M
How to find mole fraction at interface and equilibrium ??? y y1,x1
yi1 y*1
y2,x2
Dilute solution
yi2 y*2
xi1 xi2
x
Example 4 (Geankoplis: Example 10.6-4) Acetone is being absorbed by water in a packed tower having a cross-sectional area of 0.186 m2 at 293K and 101.32 kPa. The inlet air c ontains 2.6 mol% acetone and outlet 0.5%. The gas flow is 13.65 kgmol inert air/h. Pure water inlet flow is 45.36 kg mol water/h. Film coefficients for the given flows in the tower at k’ya = 3.78 x 10-2 kgmol/s.m3.mol frac and k’xa = 6.16 x 10-2 kgmol/s.m3.mol frac. (a) Calculate the tower height using k’ya. (b) Repeat using k’xa. (c) Calculate K’ya and the tower height. Solution V S L S V S L S
y1 - y 2 k 'y az(y y i )M x1 - x 2 k 'x az(xi x)M y1 - y 2 K 'y az(y y*)M x1 - x 2 K 'x az(x * -x)M
4. Transfer Units Z = Hy Ny
z (m)
V AVE
y2
- dy
k 'y aS (y y1
yi )
Z = Hx Nx If OVERALL coefficients are used to describe the flux: Z = Hoy Noy
(m) HEIGHT of a gas film transfer unit
Dimensionless Number of a gas film transfer unit
Heights of transfer units have been correlated with operating variables (i.e. V, L, packing type and size)
Z = Hox Nox Depending on concentration units and coefficients used Z = HG NG = HL NL Z = HOG NOG = HOL NOL HG ↔ Hy; NG ↔ Ny; HOG ↔ Hoy; NOG ↔ Noy
HG HL
V k 'y aS L k 'x aS
HOG HOL
V K 'y aS L K 'x aS
V k y a(1- y)iMS L k x a(1- x)iMS
V K y a(1- y)*M S L K x a(1- x)*M S
NG NL
NOG NOL
y1 - y 2 (y - y i )M x1 - x 2 (x i - x)M
y1 - y 2 (y - y*)M x1 - x 2 (x * -y)M
Example 5 (Geankoplis: Example 10.6-5) Repeat Example 5 using transfer units and height of a transfer unit as follows: (a) Use HG and NG to calculate tower height (b) Use HOG and NOG to calculate tower height Solution
Correlation form: β
γ
V L HG α Sc 0.5 S S η
L Sc 0.5 HL θ SμL Constants α, β, γ, θ, η for different packings and sizes (Geankoplis 10.8b)
V Hy α S
β
Slope β ~ 0.43 on log-log graph
eg. correlation for ½ inch Raschig rings with NH 3/H2O
Absorption Column Size Height – Mass transfer (H y Ny) Diameter – Hydraulics – Two phase flow over packing; gas pressure drop - Design (Flooding and loading)
Difficult to locate precisely (deviation from standard line) Loading – curve nearly vertical
Operate close to loading, ~1/2 flooding
Analytical equation to calculate theoretical number of trays
For transfer of the solute from phase V to phase L (absorption)
y1 - mx 2 (1 1/A) 1/A ln y 2 - mx 2 N ln A
For transfer of the solute from phase L to phase V (stripping)
x 2 - y1/m (1 A) A ln x1 - y1/m N ln (1/A)
A1 L1/m1V1, A 2 L 2 /m2 V2 and A A1 A 2
Analytical equation to calculate packed bed height Operating and/or equilibrium lines are slightly curved Absorption
NOG
y1 - mx2 1/A ln (1 1/A) 1- 1/A y 2 - mx2 1
Stripping
NOL
x 2 - y1/m A ln (1 A) 1- A x1 - y1/m 1
Analytical equation to calculate packed bed height When the operating and equilibrium lines are straight
NOG
N lnA
1- 1/A
The height of a theoretical tray or stage, HETP, in m is related to HOG by
HETP HOG
ln(1/A)
1- A /A
z = HOG x NOG z = N x HETP
See Example 10.6-5
Example 6: Experimental data have been obtained for air containing 1.6% of SO 2 being scrubbed by pure water in a packed bed column of 1.5m 2 cross-sectional area and 3.5 m in packed height. Entering gas and liquid flow rates are 0.062 and 2.2 kmol/s, respectively. If the outlet mole fraction of SO 2 in the gas is 0.004. m = 40. Calculate N OG. Solutions
5. Absorption of concentrated mixtures Dilute mixtures y2
z
Concentrated mixtures 1
V
k aS
1- y iLM
' y1 y
(1- y)
Approximate
z
V AVE
y2
dy (y y i )
V 1- y dy iLM z ' k y aS (1- y) (y y i ) y1 y2
The only constant
- dy
K 'y aS y1 (y y * )
Each term (or bracketed terms) needs to be calculated for a range of values of y between y1 and y2. Then, evaluate the integral numerically (or graphically) in order to find yi between y1 and y2.
We need the whole operating curve from (1) to (2). This is obtained from a series of mass balances between (1) and a range of points up to (2).
Sherwood suggests that (1-y)*LM and (1-y*) can often replace the true interface values if the liquid resistance is small. The * values are much easier to get since ky need not be known.
Example 6 (Geankoplis: Example 10.7-1) A tower packed with 25.4 mm ceramic rings is to be designed to absorb SO 2 from air by using pure water at 293K and 1.013 x 105Pa abs pressure. The entering gas contains 20 mol% SO 2 and that leaving 2 mol%. The inert air flow is 6.53 x 10 -4 kg mol air/s and the inert water flow is 4.20 x 10 -2 kg mol water/s. The tower cross-sectional area is 0.0929 m 2. For dilute SO 2, the film mass-transfer coefficients at 293 K are, for 25.4 mm rings: k’ya = 0.0594Gy0.7Gx0.25 k’xa = 0.152Gx0.82 Gx and Gy are kg total liquid/gas per sec per m 2 tower cross section. Calculate the tower height. Solution
6. Absorption of semi-dilute mixtures Dilute enough to use V AVE or L AVE and constant K’ya or K’xa
V 1- y dy *LM z ' K y aS (1- y) (y y*) y1 y2
V y2 1- y (-dy) AVE *LM z ' K aS (1 - y) (y y*) y y1
Always check this for ~ 1.0. If yes, need only evaluate Which is relatively simple (numerical or graphical)
y2
dy
yy*
y1
7. Controlling resistance i) Liquid film controlling (eg. gas almost insoluble – O2/H2O)
