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(GAS) ABSORPTION AND (GAS) STRIPPING
Overview
Introduction Absorption and stripping equilibria Operating lines for absorption Stripping analysis Analytical solution: Kremser equation
Introduction
Gas absorption is a mass transfer operation in which a gas mixture is contacted with a liquid to preferentially absorb one or more of the components of the gas stream. In this case, the liquid solvent (absorbing liquid) must be added as a separating agent.
In some cases, a solute is removed from a liquid by contacting it with a gas. This operation is the reverse of gas absorption and is called desorption or gas stripping. Here, the gas stream (stripping agent) must be added as a separating agent.
Absorption can be either physical or chemical. In physical absorption, the gas is removed because it has greater solubility in the solvent than other gases. An example is the removal of butane and pentane from a refinery gas mixture (C4 – C5) with a heavy oil.
In chemical absorption, the gas to be removed reacts with the solvent and remains in solution. An example is the removal of CO2 or H2S by reaction with NaOH or with monoethanolamine (MEA).
The reaction can be either irreversible (as with NaOH) or reversible (with MEA). For irreversible reactions, the resulting liquid must be disposed of, whereas in reversible reactions, the solvent can be regenerated. Thus, reversible reactions are often preferred.
Chemical absorption usually has a much more favorable equilibrium relationship than physical absorption (solubility of most gases is usually very low) and is, therefore, preferred.
Both absorption and stripping can be operated as equilibrium stage operations with contact of liquid and vapor.
In both absorption and stripping a separate phase is added as the separating agent.
Absorption and stripping equilibria
For absorption and stripping in three component systems, we often assume that
1. Carrier gas is insoluble.2. Solvent is nonvolatile.3. The system is isothermal and isobaric.
The Gibbs phase rule is
F = C – P + 2= 3(A, B, and C) – 2(vapor and liquid) + 2= 3
If we set T and p constant, there is one remaining degree of freedom. The equilibrium data are usually represented either by plotting solute composition in vapor versus solute composition in liquid or by giving a Henry’s law constant.
Henry’s law is
PB = HBxB
where HB is Henry’s law constat, in atm/mole frac, H = H(p,T,composition);xB is the mole fraction B in the liquid; andpB is the partial pressure of B in the vapor.
Henry’s law is valid only at low concentrations of B. Since partial pressure is defined as
Henry’s law becomes
This will plot as a straight line if HB is constant. If the component is pure, yB = 1 and pB = ptot.
The Henry’s law constants depend upon temperature and usually follow an Arrhenius relationship. Thus,
A plot of log H versus will often give a straight line.
Operating lines for absorption
The McCabe-Thiele diagram is most useful when the operating line is straight. This requires that
The energy balances be automatically satisfied Liquid flow rate/vapor flow rate = constant
In order for energy balances to be automatically satisfied, we must assume that
1. The heat of absorption is negligible2. Operation is isothermal
These two assumptions will guarantee satisfaction of the enthalpy balances. When the gas and liquid streams are both fairly dilute, the assumptions will probably be satisfied.
We also desire a straight operating line. This will be automatically true if we define
and if we assume that:
3. Solvent is nonvolatile4. Carrier gas is insoluble
Assumptions 3 and 4 are often very closely satisfied. The results of these last two assumptions are that the mass balance for solvent becomes
LN = Li = L0 = L = constant
while the mass balance for the carrier gas is
GN+1 = Gi = G1 = G = constant
Note that we cannot use overall flow rates of gas and liquid in concentrated mixtures because a significant amount of solute may be absorbed which would change gas and liquid flow rates and give a curved operating line. For very dilute solutions (< 1% solute), overall flow rates can be used, and mass or mole fractions can be used for operating equations and equilibria. Since we want to use L = moles nonvolatile solvent (S)/hr and G = moles insoluble carrier gas (C)/hr, we must define our compositions in such a way that we can write a mass balance for solute B.
After some manipulation we find that the correct way to define our compositions is as mole ratios. Define
and
The mole ratios of Y and X are related to our usual mole fractions by
and
Note that both Y and X can be greater than 1.0. With mole ratio units, we have
and
Thus we can easily write the steady-state mass balance, input = output, in these units. The mass balance around the top column using the mass balance envelope is
or
Moles B in/hr = moles B out/hr
Solving for Yi+1 we obtain
This is a straight line with slope and intercept . It is our operating line for absorption.
Thus if we plot ratios Y vs X we have a McCabe-Thiele type of graph.
The steps in this procedure are:
1. Plot Y vs X equilibrium data (convert from fractions to ratios).2. Values of X0, YN+1, Y1 and are known. Point (X0, Y1) is on operating line, since it represents
passing streams.3. Slope is . Plot operating line.4. Starting at stage 1, step off stages: equilibrium, operating, equilibrium, etc.
Note that the operating line is above the equilibrium line, because solute is being transferred from the gas to the liquid.
Equilibrium data must be converted to ratio units, Y vs X. These values can be greater than 1.0, since and . The Y = X line has no significance in absorption. As usual the stages are
counted at the equilibrium curve. If the system is not isothermal, the operating line will not be affected, but the equilibrium line will be. Then the McCabe-Thiele method must be modified to include changing equilibrium curves.
For very dilute systems we can use mole fractions, since total flows are approximately constant.
Example 1: Graphical absorption analysis
A gas stream is 90 mole % N2 and 10 mole % CO2. We wish to absorb the CO2 into water. The inlet water is pure and is at 5C. Because of cooling coils, operation can be assumed to be isothermal. Operation is at 10 atm. If the liquid flow rate is 1.5 times the minimum liquid flow rate, how many equilibrium stages are required to absorb 92% of the CO2? Choose a basis of 1 mole/hr of entering gas.
Solution
A. We need to find the minimum liquid flow rate, the value of the outlet gas concentration, and the number of equilibrium stages required.
B. First we need equilibrium data. Since concentrations are fairly high, the problem should be solved in mole ratios. Thus we need to convert all compositions including equilibrium data to mole ratios.
C. Derive the equilibrium equation from Henry’s law. Convert compositions from mole fractions to mole ratios. Calculate Y1 by a percent recovery analysis. Plot mole ratio equilibrium data on a YX diagram, and determine and hence Lmin. Calculate actual , plot operating line, and step off stages.
The problem appears to be straightforward.
D. Equilibrium:
Change the equilibrium data to mole ratios with a table as shown below.
x y = 87.6x
0 0 0 00.0001 0.0001 0.00876 0.008840.0004 0.0004 0.0350 0.03630.0006 0.0006 0.0526 0.05550.0008 0.0008 0.0701 0.07540.0010 0.0010 0.0876 0.09600.0012 0.0012 0.10512 0.1175
Note that x = X in this concentration range, but y Y. The inlet gas mole ratio is
G = (1 mole total gas/hr)(1-yN+1) =
Percent recovery analysis: 8% of CO2 exits.
(0.1 mole in)(0.08 recovered)= 0.008 mole CO2 out
Thus,
Operating line:
Goes through point (Y1,X0) = (0.008888,0).
is found as the slope of the operating line from point (Y1,X0) to the intersection with the equilibrium curve at YN+1.
Lmin = (89.905)(0.9) = 80.914
Lactual = 1.5Lmin = 121.37and
Plot operating line from (Y1,X0) with this slope. Step off stages on the diagram. Need 4.1 equilibrium stages.
The fraction was calculated as
E. The overall mass balances are satisfied by the outlet concentrations. The significant figures carried in this example are excessive compared with the equilibrium data. Thus they shoud be rounded off when reported. The concentrations used were quite high for Henry’s law. Thus, it would be wise to check the equilibrium.
F. Note that the gas is considerably greater than the liquid concentration. This situation is common for physical absorption (solubility is low). Chemical absorption is used to obtain more favorable equilibrium. The liquid flow rate required for physical absorption is excessive. Thus, in practice, this type of operation uses chemical absorption.
If we had assumed that total gas and liquid flow rates were constant (dilute solutions), the result would be in error. An estimate of this error can be obtained by estimating . The minimum operating line goes from (y1, x0) = (0.00881, 0) to (yN+1, xequil,N+1). YN+1 = 0.1 and xequil,N+1 = yN+1/87.6 = 0.1/87.6 = 0.0011415.
Then
This is in error by more than 10%.
Stripping analysis
Since stripping is very similar to absorption we expect a similar result. The mass balance for the column is the same as for absorption and the operating line is still
For stripping we know X0, XN, YN+1, and . Since (XN, YN+1) is a point on the operating line, we can plot the operating line and step off stages.
Note that the operating line is below the equilibrium curve because solute is transferred from liquid to gas. A maximum ratio can be defined; this corresponds to the minimum amount of stripping gas. Start from the known point (YN+1, XN) and draw a line to the intersection of X = X0 and the equilibrium curve. Alternatively, there may be a tangent pinch point. For a stripper, Y1 > YN+1, while the reverse is true in absorption. Stripping often has large temperature changes, so the method used here may have to be modified.
Murphree efficiencies can be used on these diagrams if they are defined as
For dilute systems the more common definition of Murphree vapor efficiency in mole fractions would be used. Efficiencies for absorption and stripping are often quite low.
Usually the best way to determine efficiencies is to measure them on commercial-scale equipment.
Analytical solution: Kremser equation
When the solution is quite dilute (say less than 1% solute in both gas and liquid), the total liquid and gas flow rates will not change significantly since little solute is transferred. Then the entire analysis can be
done with mole or mass fractions and total flow rates. The operating equation is derived by writing a mass balance around stage I and solving for yi+1. The result is
To use this equation in a McCabe-Thiele diagram, we assume:
1. (total flows) is constant2. Isothermal system3. Isobaric system4. Negligible heat of absorption
These are reasonable assumptions for dilute absorbers and strippers.
If one additional assumption is valid, the stage-by-stage problem can be solved analytically. This additional assumption is:
5. Equilibrium line is straight.
This assumption is reasonable for very dilute solutions and agrees with Henry’s law if m = HA/ptot and b = 0.
An analytical solution for absorption is easily derived for a special case where the operating and equilibrium lines are parallel. Now the distance between operating and equilibrium lines, y, is constant. To go from outlet to inlet concentrations with N stages, we have
since each stage causes the same change in vapor composition. y can be obtained by substracting the equilibrium equation from the operating equation.
For the special case of parallel operating and equilibrium lines, and
Combining this equation and , we get
for
This equation is a special case of the Kremser equation. When this equation is applicable, absorption and stripping problems can be solved quite simply and accurately without the need for a stage-by-stage calculation.
For the more general case, yi varies from stage to stage. The y values can be determined from
This equation is easier to use if we replace xi with the equilibrium equation,
Then
Substracting the first equation from the second,
and solving for (y)I+1
This equation relates the change in vapor composition from stage to stage to (L/mV), which is known as the absorption factor. If either the operating or equilibrium is curved, this simple relationship no longer holds and a simple analytical solution does not exist.
The difference between inlet and outlet gas concentrations must be the sum of the yi values. Thus,
y1 + y2 + + yN = yN+1 – y1
Applying
The summation in this equation can be calculated. The general formula is
for A < 1
Then
If L/mV > 1m then divide both sides of this equation by and do the summation in terms of mV/L. The vapor composition is the value that would be in equilibrium with the inlet liquid, x0. Thus,
Removal of y1 from gives
This equation is one form of the Kremser equation. A large variety of alternative forms can be developed by algebraic manipulation. For instance, if we add 1.0 to both sides of the previous equation and rearrange, we have
which can be solved for N. After manipulation, this result is
where L/mV 1. These last two equations are also known as forms of the Kremser equation.
A variety of forms of the Kremser equation can be developed. Several alternative forms in terms of the gas-phase composition are
where
and
Alternative forms in terms of the liquid phase composition are
where
and
A form including a constant Murphree vapor efficiency is
Which form of the Kremser equation to use depends upon the problem statement. When the assumptions required for the derivation are valid, the Kremser equation has several advantages over the stage-by-stage calculation procedure. If the number of stages is large, the Kremser equation is much more convenient to use, and it is easy to program on a computer or calculator. When the number of stages is specified, the McCabe-Thiele stage-by-stage procedure is trial-and-error, but the use of the Kremser equation is not. Because calculations can be done faster, the effects of varying y1, x0, L/V, m etc. are easy to determine. The major disadvantage of the Kremser equation is that it is accurate only for dilute solutions where L/V is constant, equilibrium is linear, and the system is isothermal.
Example 2: Kremser equation
A plate tower providing six equilibrium stages is employed for stripping ammonia from a waste water stream by means of countercurrent air at atmospheric pressure and 80F. Calculate the concentration of ammonia in the exit water if the inlet liquid concentration is 0.1 mole % ammonia in water, the inlet air is free of ammonia, and 30 standard cubic feet (scf) of air are fed to the tower per pound of waste water.
Solution
A. The column is sketched in the figure.
We wish to find the exit water concentration, x6.
B. Since the concentrations are quite low we can use the Kremser equation. Equilibrium data are available in several sources: we find at 80F.C. We have to convert flow to molar units. Since we want a concentration of liquid, we will use
D. We can calculate ratio V/L,
= 1.43 moles air/mole water
Note that the individual flow rates are not needed.
The Kremser equation is
where xN = x6 is unknown, x0 = 0.001, m = 1.414, b = 0,
= y7/m = 0, V/L = 1.43, N = 6
Rearranging,
1
6
x0 = 0.001 y1
p = 1 atm80F
x6y7 = 0.30 ft3 (std.) air/lb water
xN = 7.45 10-6 mole fraction
Most of the ammonia is stripped out by the air.
E. We can check with a different form of the Kremser equation or by solving the results graphically; both give the same result. We should also check that the major assumptions of the Kremser equation (constant flow rates, linear equilibrium, and isothermal) are satisfied. In this dilute system they are.
F. This problem is trial-and-error when it is solved graphically. Also, the Kremser equation is very easy to set up on a computer or calculator. Thus, when it is applicable, the Kremser equation is very convenient.