Absorption 2nd1415

ABSORPTION

ABSORPTION AND STRIPPING

Absorption (or scrubbing) is the removal of a component (the solute or absorbate) from a gas stream via uptake by a nonvolatile liquid (the solvent or absorbent).

Desorption (or stripping) is the removal of a component from a liquid stream via vaporization and uptake by an insoluble gas stream.

Thus, absorption and stripping are opposite unit operations, and are often used together as a cycle.

Both absorption and stripping can be operated as equilibrium stage processes using trayed columns or, more commonly, using packed columns.

Absorber/Stripper Cycle

Absorption Systems Physical

Examples:CO2 and water Acetylene and acetic acidCO and water NH3 and acetoneH2S and water Ethane and carbon disulfideNH3 and water N2 and methyl acetateNO2 and water NO and ethanol

Physical absorption relies on the solubility of a particular gas in a liquid.

This solubility is often quite low; consequently, a relatively large amount of liquid solvent is needed to obtain the required separation.

This liquid solvent containing the solute is typically regenerated by heating or stripping to drive the solute back out.

Because of the low solubility and large solvent amounts required in physical absorption, chemical absorption is also used

Absorption Systems Chemical

Chemical absorption relies on reaction of a particular gas with a reagent in a liquid.

Examples:CO2 / H2S and aqueous ethanolaminesCO2 / H2S and aqueous hydroxidesCO and aqueous Cu ammonium saltSO2 and aqueous dimethyl anilineHCN and aqueous NaOHHCl / HF and aqueos NaOH

This absorption can often be quite high; consequently, a smaller amount of liquid solvent/reagent is needed to obtain the required separation.

However, the reagent may be relatively expensive, and it is often desirable to regenerate when possible.

Absorption and Stripping the Problem

The principal difference in handling adsorption and stripping, compared todistillation, how the equilibria (equilibrium curve) and mass balances (operating lines) are represented.

In absorption, consider removal of the solute from the gas stream and uptake by the solvent liquid stream; thus, the total liquid and gas stream amounts or flow rates can change.

If we use mole fractions of the solute and assume that the gas and liquid stream amounts or flow rates remain constant, significant error can result if the solute concentration in the inlet gas stream is greater than about 1%.

If we can set up our equilibrium curve and operating line to account for this change in the overall gas and liquid flow rates, we can use the McCabe-Thiele method to solve absorption and stripping problems.

Absorption and Stripping Assumptions

We assume that: The carrier gas is insoluble (or it has a very low solubility), e.g, N2 or Arin water. The solvent is nonvolatile (or it has a low vapor pressure), e.g., water

in air at low temperatures. The system is isothermal. e.g., the effects of heat of solution or

reaction are low or there is a cooling or heating system in the column. The system is isobaric. While the total gas and liquid streams can change in absorption, the

flow rate of the carrier gas, which we assume to be insoluble in the solvent, does not change.

define the equilibrium curve and operating line in terms of mole ratios with respect to the carrier gas and solvent, instead of mole fractions

Similarly, the flow rate of the solvent, which we assume to be nonvolatile, does not change.

The concentration of the solute is low, say

MATERIAL BALANCE IN AN ABSORPTION TRAY TOWER

V1, y1 Lo, x0

1

2

Vn+1, yn+1n

Ln, xn

n + 1 N 1

N

VN+1, yN+1 LN, xN

DESIGN OF PLATE ABSORPTION TOWERS

OPERATING-LINE DERIVATION

L(xo) V (yn+1) L(xn) V (y1)

(1 xo ) (1 yn+1) (1 xn ) (1 y1)+ +=

Where: x is the mole fraction A in the liquidy is mole fraction of A in the gasL is the total moles liquidV the total moles gas

yn+1= xn (L/V) + y1 xo(L/V)

DESIGN OF PLATE ABSORPTION TOWERS

Graphical determination of the number of trays

If x an y are very dilute, the denominators 1 x and 1 y will be close to 1.0 and the OL will be approximately straight with a slope L/V

The number of theoretical trays is determined by stepping off the number of trays.

McCabe-Thiele Plot Absorber

EXAMPLE 1

A tray tower is to be designed to absorb SO2 from an air stream by using pure water at 293K. The entering gas contains 25%mol SO2 and that the leaving 3 mol% at a total pressure of 101.3kPa. The inert air flow rate is 100 Kg air/h-m, and the entering water flow rate is 4000 kg water/h-m. Assuming an overall tray efficiency of 30%, how many theoretical trays and actual trays are needed? Assume that the tower operates at 293K.

*Equilibrium data at A.3-19

2

2

00.05

0.1

0.15

0.2

0.25

0.3

0 0.002 0.004 0.006 0.008 0.01

DESIGN OF PACKED TOWERS FOR ABSORPTION

V2, y2 L2, x2

V, y

dz

L, x

z

V1, y1 L1, x1

DESIGN OF PACKED TOWERS FOR ABSORPTION

OPERATING-LINE DERIVATION

L(x) V (y1) L(x1) V (y)

(1 x ) (1 y1) (1 x1 ) (1 y)+ +=

Where: x1 is the mole fraction A in the liquidy1 is mole fraction of A in the gasL is Kgmol inert liquid/s or kgmol inert liquid/s-mV is Kgmol inert liquid/s or kgmol inert liquid/s-m

2

2

LOCATION OF OPERATING LINES

y1

y2

x2 x1

Tower bottom

Operating line

Equilibrium line

Top

Mole fraction, x

Mo

le f

ract

ion

,y

Absorption of A from V to L stream

y2

y1

x2x1

Tower bottom

Operating lineEquilibrium line

Top

Mole fraction, x

Stripping of A from V to L stream

DESIGN OF PACKED TOWERS FOR ABSORPTION

OPERATING LINE DERIVATION

The equation of the line can also be written in terms of partial pressure p1 of A. Where

y1/(1 y1) = p1/(P p1)

If x and y are very dilute, (1 x) and (1 y) can be taken as 1.0 and the equation of OL becomes

Lx + Vy1 Lx1 + Vy

LIMITING AND OPTIMUM L/V RATIOSABSORPTION

y1

y2

x2 x1

Operating line for actualliquid flow

Equilibrium line

Operating line for minimum liquid flow

Mole fraction, x

Mo

le f

ract

ion

,y

x1 max

LIMITING AND OPTIMUM L/V RATIOSSTRIPPING

y2

y1

x2x1

Operating line for minimumgas flows

Operating lineFor actual gas flow

Equilibrium line

Lecture 21

Minimum Absorbent Rate Lmin

ANALYTICAL EQUATIONS FOR THEORETICAL NUMBER OF STEPS OR TRAYS

y1 mx2y2 mx2

(1 1/A) + 1/A

ln AN = ABSORPTION

ln

x2 y1/mx1 y1/m

(1 A) + 1/A

ln (1/A)N = STRIPPING

ln

Where : A = L/mV

ANALYTICAL EQUATIONS FOR THEORETICAL NUMBER OF STEPS OR TRAYS

For absorption, at the bottom concentrated end tray, the slope m1 or tangent at the point x1 on the equilibrium line is used. Then A1 = L1/m1V1

At the dilute top tray, the slope m2 of the equilibrium line at point y2 is used. Then A2 = L2/m2V2

A = (A1A2)1/2

ANALYTICAL EQUATIONS FOR THEORETICAL NUMBER OF STEPS OR TRAYS

For stripping, at the top concentrated stage, the slope m2 or tangent at point y2 on the equilibrium line is used. Then A2 = L2/m2V2

At the bottom or dilute end, the slope m1 at point x1 on the equilibrium line is used. Then A1 = L1/m1V1

A = (A1A2)1/2

Analytical Minimum Absorbent Flowrate

L = V(yN+1 y1)xN x0

absorbent flowrate

KN = yN+1 /(1 + yN+1)xN /(1 + xN)

Minimum absorbent rate

Lmin =V(yN+1 y1)

yN+1 x0

KN

Minimum absorbent flowrate for Dilute solutions

Lmin = V KN (fraction of solute absorbed)

Minimum absorbent flow rate if xo = 0

Example 2

It is desired to absorb 90% of the acetone in a gas containing 1 mol% acetone in air in a tray tower. The total inlet gas flow to the tower is 30.0 kmol/h. The process is operating isothermally at 300K and a total pressure of 101.3 kPa. The gas-liquid equilibrium relation for this dilute stream is y = mx = 2.53x. Using 1.5 times the minimum liquid flowrate, determine the number of trays needed a) graphically b) analytically.