(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 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 In physical absorption, 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.
absorption, the gas to be removed reacts with the solvent and remains in solution. An In chemical absorption, 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.
Chemic Chemical al absorp absorpti tion on usuall usually y has a much much more more favora favorable ble equili equilibri brium um relati relationsh onship ip than than physic physical al 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. Carr Carrie ierr gas gas is inso insolu lubl ble. e. 2. Solv Solven entt is nonvol nonvolat atil ile. e. 3. The syst system em is isot isother hermal mal and and isoba isobaric ric.. The Gibbs phase rule is F
= = =
C–P+2 3(A, B, and C) – 2(vapor and liquid) liquid) + 2 3
If we set T and p constant, constant, there is one remaining remaining degree of freedom. freedom. The equilibri equilibrium um 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 xB pB
is Henry’s law constat, in atm/mole frac, H = H(p,T,composition); is the mole fraction B in the liquid; and 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 yB
≡
p B p tot
Henry’s law becomes yB
=
HB p tot
xB
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,
H
A plot of log H versus
1 T
− E = H 0 exp RT
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 heat of abso absorpt rption ion is negl negligi igible ble 2. Oper Operat atio ion n is isot isother herma mall 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 L G
=
moles nonvolatil e solvent hr moles insoluble carrier gas hr
and if we assume that: 3. Solv Solven entt is is non nonvol volat atil ilee 4. Carr Carrie ierr gas gas is is ins insol olubl ublee Assumptions 3 and 4 are often very closely satisfied. The results of these last two assumptions are that the mass balance for solvent becomes L N = Li = L0 = L = constant while the mass balance for the carrier gas is G N+1 = Gi = G1 = G = constant Note Note that that we cannot cannot use overall overall flow rates rates of gas and liquid liquid in concent concentrat rated ed mixtur mixtures es becaus becausee 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. ratios. Define
Y
moles B in gas
=
moles pure carrier gas C moles B in liquid and X = moles pure solvent S
The mole ratios of Y and X are related to our usual mole fractions by Y
=
y 1 −y
and X
=
x 1 −x
Note that both Y and X can be greater than 1.0. With mole ratio units, we have Yi G
= moles B in gas stream moles carrier gas
i moles carrier gas hr
moles =
B in gas stream hr
and Xi L
moles B in liquid stream i moles solvent moles = = moles solvent hr
B in liquid stream i hr
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 Yi 1G + X 0 L +
=
X i L + Y1G
or Moles B in/hr = moles B out/hr Solving for Yi+1 we obtain Yi+1
This This is a stra straig ight ht line line with with slope slope
L G
=
L G
Xi
L + Y1 − X 0 G
and intercept intercept
Y − L X 1 operating line for 0 . It is our operating G
absorption. Thus if we plot ratios Y vs X we have a McCabe-Thiele type of graph. The steps in this procedure are: 1. 2. 3. 4.
Plot Plot Y vs vs X equil equilib ibri rium um data data (conv (conver ertt fro from m fra fract ctio ions ns to rati ratios os). ). L G Values of X0, Y N+1, Y1 and are known. Point (X0, Y1) is on operating line, since it represents passing streams. Slope is L G . Plot operating line. Star Starti ting ng at stag stagee 1, ste step p off off stag stages es:: equi equili libr briu ium, m, ope opera rati ting ng,, equil equilib ibri rium um,, etc. 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 Y = y (1 − y ) and X = x (1 − x ) . 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 ap proximately 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 5 °C. 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 find the minimu minimum m liqui liquid d flow flow rate, the value of the outlet outlet gas concent concentrat ration ion,, and the number of equilibrium stages required. B. First First we need equilibrium equilibrium data. Since concentrat concentrations ions are fairly fairly high, the problem problem should be solved in mole ratios. Thus we need to convert all compositions including equilibrium data to mole ratios. C. Derive Derive the equilibrium equilibrium equation equation from Henry’s Henry’s law. Convert compositi compositions ons from mole fractions fractions to mole ratios. Calculate Y1 by a percent recovery analysis. Plot mole ratio equilibrium data on a YX diagram, and determine ( L G ) min and hence Lmin. Calculate actual L G , plot operating operating line, and step off stages. The problem appears to be straightforward. D. Equi Equili libr briu ium: m: y=
H p tot
x
= 876 x = 87 .6x 10
Change the equilibrium data to mole ratios with a table as shown below. x 0 0.0001 0.0004 0.0006 0.0008 0.0010 0.0012
X
=
x 1 −x
0 0.0001 0.0004 0.0006 0.0008 0.0010 0.0012
y = 87.6x 0 0.00876 0.0350 0.0526 0.0701 0.0876 0.10512
Y
=
y 1 −y
0 0.00884 0.0363 0.0555 0.0754 0.0960 0.1175
Note that x = X in this concentration range, but y ≠ Y. The inlet gas mole ratio is Y N+1
=
y N+1 1 − y N+1
=
0.1 0.9
= 0.1111
moles CO 2 moles N 2
G = (1 mole total gas/hr)(1-y N+1) = 0.9
moles N 2 hr
Percent recovery analysis: 8% of CO2 exits. (0.1 mole in)(0.08 recovered) = 0.008 mole CO2 out Thus, Y1
=
moles CO2 moles N 2
=
0.008 mole CO2 0.9 0.9 mole N 2
= 0.008888
Operating line: Yi+1
=
L G
Xi
L + Y1 − X 0 G
Goes through point (Y1,X0) = (0.008888,0).
( L G ) min is found as the slope of the operating line from point (Y 1,X0) to the intersection with the equilibrium curve at Y N+1.
( L G ) min = 89 .905
∴ Lmin = (89.905)(0.9) = 80.914
moles water hr
Lactual = 1.5Lmin = 121.37 and ( L G ) actual =134 .86 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 Frac
=
− X 4 0.000758 − 0.00071 = = 0.13 X5 − X 4 0.00108 − 0.00071
X out
E. The overall mass balances are satisfied by the outlet concentrations. The significant significant figures carried 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 is considerably greater than the liquid concentration. This situation is common common for phy physi sica call absor absorpt ptio ion n (sol (solub ubil ilit ity y is low) low).. Chem Chemic ical al abso absorp rpti tion on is used used to obta obtain in more more favo favora rabl blee 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 ( L G ) min . The minimum operating line goes from (y1, x0) = (0.00881, 0) to (y N+1, x equil,N+1). Y N+1 = 0.1 and xequil,N+1 = y N+1/87.6 = 0.1/87.6 = 0.0011415. Then
( L G ) min, dilute =
y N+1 − y1 x equil , N +1 − x 0
=
0.1 − 0.00881 0.0011415
−1
886 = 79.886
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 Yi+1
=
For stripping we know X0, X N, Y N+1, and L plot the operating line and step off stages.
L G
G
Xi
L + Y1 − X 0 G
. Since (X N, Y N+1) is a point on the operating line, we can
Note that the operating operating line is below the equilibrium equilibrium curve because solute is transferre transferred d from liquid to L G gas. A maximum ratio can be defined; this corresponds to the minimum amount of stripping gas. Start from the known point (Y N+1, X N) and draw a line to the intersection of X = X0 and the equilibrium curve. Alternative Alternatively, ly, there may be a tangent pinch point. For a stripper, stripper, Y1 > Y N+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 o n these diagrams if they are defined as
− Yi+1 E = MV * Yi − Yi+1 Yi
For dilute systems systems the more common definition of Murphree vapor efficiency in mole fractions fractions would be used. Efficiencies for absorption and stripping are o ften quite low. Usually the best way to determine efficiencies is to measure them o n 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 yi +1
=
L V
xi
L + y1 − x 0 V
To use this equation in a McCabe-Thiele diagram, we assume: 1. L V (total flows) is constant 2. Isot Isothe herm rmal al sys syste tem m 3. Isob Isobar aric ic syst system em 4. Neglig Negligibl iblee heat heat of absorp absorpti tion on These are reasonable assumptions for dilute absorbers and strippers. If one additi additional onal assump assumpti tion on is valid, valid, the stagestage-byby-sta stage ge proble problem m can be solved solved analyt analytica ically lly.. This This additional assumption is: 5. Equili Equilibri brium um line line is straig straight. ht.
= mx i + b
yi
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 equilib equilibriu rium m lines lines are parall parallel. el. Now the distan distance ce betwee between n operati operating ng and equili equilibri brium um lines, lines, ∆ y, is constant. To go from outlet to inlet concentrations with N stages, we have N
∆y = y N +1 − y1
since each stage causes the same change in vapor composition. ∆ y can be obtained by substracting the equilibrium equation from the operating equation.
( ∆y ) i = yi+1 − yi = − m x i + y1 − L
V
L V
x0
− b
For the special case of parallel operating and equilibrium lines, L V = m and
( ∆y ) = ( ∆y ) i = y1 − Combining this equation and N ∆y
= y N +1 − y1 ,
L V
x 0 − b = cons tan t
we get
N =
y N+1 − y1
y − L x − b 1 0 V
for
L mV
=1
This equation is a special case of the Kremser equation. When this equation is a pplicable, 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
( ∆y ) i = yi+1 − yi = − m x i + y1 − L
V
L V
x0
− b
This equation is easier to use if we replace xi with the equilibrium equation, xi
=
yi − b m
Then
( ∆y ) i =
L
L L −1 b − x 0 yi + yi − V mV mV
( ∆y ) i+1 =
L
mV
L L −1 yi+1 + yi − b − x 0 V mV
Substracting the first equation from the second,
( ∆y ) i+1 − ( ∆y ) i =
L
mV
L −1 ( ∆y ) −1 ( yi+1 − yi ) = mV i
and solving for (∆ y)I+1
( ∆y) i+1 =
L mV
( ∆y ) i
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 + ⋅ ⋅ ⋅ + ∆ y N = y N+1 – y1 Applying ( ∆y ) i+1
=
L mV
( ∆y) i
N −1 L L 2 L ∆y1 1 + + + + = y N +1 − y1 m V m V m V
The summation in this equation can be calculated. The general formula is a (1 − A k 1 ) aA = (1 − A ) i =0 +
k
∑
for A< 1
i
Then N
y N +1 − y1
∆y1
L 1− mV = L 1− mV
N If L/mV > 1m then divide both sides of this equation by ( L mV )
1
−
and do the summation in terms of
* mV/L. The vapor composition composition y1 is the value that would be in equilibrium with the inlet liquid, x0. Thus,
y1*
= mx 0 + b
N
Removal of ∆ y1 from
y N +1 − y1
∆y1
L 1− mV = L 1−
gives
mV
N +1
L − mV mV = L 1− L
y N +1 − y1 y1 − y1*
mV
This This equatio equation n is one form form of the Kremser Kremser equati equation. on. A large large variet variety y of altern alternati ative ve forms 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 N +1
y N +1 − y1* y1 − y1*
L 1− mV = L 1− mV
which can be solved for N. After manipulation, this result is
mV y N+1 − y1* mV + ln 1 − * − L y y 1 1 L N = L ln mV 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 y N+1 − y1 y1 − y1*
( L mV ) − ( L mV ) N+1 = N+1 1 − ( L mV ) N
* y N +1 − y N +1
L = * y1 − y1 mV * * ln[( y N+1 − y N +1 ) ( y1 − y1 ) ] N = ln( L mV )
ln[ ( y N+1 − y N+1 ) ( y1 − y1 ) ] *
N =
*
ln[ ( y N+1 − y1 ) ( y N* +1 − y1* ) ]
where * y N +1
= mx n + b
and
*
y1
= mx 0 + b
Alternative forms in terms of the liquid phase composition are
L x 0 − x N* L + ln 1 − * mV x x − N N m V N = ln( m V L) N
=
ln [ ( x N
[( N = ln[ ( x
ln x N * 0
x N* ) ( x 0 ln ( mV L ) −
*
− x N *
− x N
) (x
) (x
−
x *0 ) ] *
0
0
− x0 − x N
)] )]
− x N* 1 − ( mV L ) = N +1 * x 0 − x N 1 − ( mV L )
x N
− x N* x 0 − x*0
x N
where
N
L = mV
* x N
=
y N+1 − b m
and
*
x0
=
y1 − b m
A form including a constant Murphree vapor efficiency is
{[1 − m V L][ ( y + − y ) ( y − y )] + m V L} N = − N 1
* 1
1
* 1
ln[1 + EMV ( m V L − 1) ]
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-bystage 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 80°F. 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 column is sketche sketched d in the the figure. figure.
y1
x0 = 0.001 1
p = 1 atm 80°F 6 x6
y7 = 0.30 ft 3 (std.) air/lb water
We wish to find the exit water concentration, con centration, x6. B. Since Since the concent concentrat ration ionss are quite quite low we can use the Kremse Kremserr equati equation. on. Equilibr Equilibrium ium data are 414 x NH3 at 80°F. available in several sources: we find y NH3 = 1.414 C.
We have have to conv convert ert flow flow to molar molar unit units. s. Since Since we we want want a concent concentrat ration ion of liqu liquid, id, we wil willl use use
− x N* 1 − ( mV L ) = N +1 * x 0 − x N 1 − ( mV L )
x N
D.
We can calculate ratio V/L, V L
=
30 scf scf air air 1 lb mole air air 1 lb water
×
379 379 scf scf air air
×
18 lb water 1 lb mole water
= 1.43 moles air/mole water Note that the individual flow rates are not needed. The Kremser equation is
−x = x0 − x
x N
* N * N
1−
mV
L N +1 mV
1−
L
where x N = x6 is unknown, x0 = 0.001, m = 1.414, b = 0, * x N
= y7/m = 0, V/L = 1.43, N = 6
Rearranging, x N
x N
=
=
1 − mV L N +1
1 − ( mV L )
1 − (1.414 414 ) (1.43) 1 − [ (1.414 414 ) (1.43) ]
7
x0
( 0.001 001)
x N = 7.45 × 10-6 mole fraction Most of the ammonia is stripped out by the air. E. We can check with a different different form of the Kremser Kremser equation equation or by solving solving the results results graphically; 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 problem is trial-and-er trial-and-error ror when it is solved graphical graphically. ly. Also, the Kremser Kremser equation equation is very easy applicable, the Kremser equation is very to set up on a computer or calculator. Thus, when it is applicable, convenient.
