TKP 4105. Exercise on mass transfer rate-basedabsorption and stripping

Problem 1. Absorption using mass transfer rate-based method:Dilute mixture

In a heat exchanger with constant heat capacities and constant UA, we have the followingsimple formula for the total heat transfer from the hot to the cold stream [J/s]

Q = UA∆TM

where ∆TM = ∆T1−∆T2ln(∆T1/∆T2)

is the log-mean temperature difference.

Countercurrent absorption is similar to a heat exchanger, but the driving force is theconcentration difference between y in the gas phase and y∗(x) in equilibrium with the liquidphase. With constant slopes for the equilibrim and operating lines (which is reasonable fordilute mixtures), the total mass transfer of A from the gas to liquid stream is [mol A/s] (asderived by Sigurd in the lecture)

N̄A = Ky(aSz)(y − y∗)M

where (aSV ) [m2] is the total mass transfer area inside the column and (y − y∗)M is the log

mean vapor concentration difference between the phases, (y − y∗)M =(y−y∗)1−(y−y∗)2

ln((y−y∗)1/(y−y∗)2)

(1=top of column, 2=btm of column). Sigurd thinks it should be easy to remember this

formula at the exam and possibly for the rest of your life!

We want to apply this easy-to-remember formula to Example 10.6-2 (p. 663)/Example 10.7-1 (p. 681) in the book (Absorption of SO2 (A) using water). Itis not really a dilute mixture so we expect some error.

Given data (from the book): The mole fraction of SO2 in air drops from 0.2(btm) to 0.02 (top). The air flow (inert) is 6.53 · 10−4 kmol/s. The enteringliquid is 4.20 ·10−2 kmol/s of pure water. The cross-sectional area of the columnis 0.0929 m2.

(i) Find the total mass transfer of A from the vapor to liquid, N̄A =ybtmVbtm − ytopVtop [mol A/s]

(ii) The column operates at 1 atm and 20C. Use the equilibrium data inAppendix A3-19 to find the slope m of the equilibrium line for the concentrationrange of interest.

(iii) ComputeKya using the data kxa = 0.85kmol/s,m3 and kya = 0.037kmol/s,m3(average values from Example 10.7-2).

(iv) Find the loagarithmic mean driving force (y − y∗)M(v) Use the easy-to-remember formula to find the packing height z.(vi) Compare with z = 1.588m found in Example 10.7-1 and comment on

the difference.

Problem 2. Stripping using mass transfer rate-based method:Absorption of CO2 in water-amine solution (part 2)

This continues a previous exercise, but now you should use the rate-basedmethod (mass transfer) rather than equilibrium stages. Note that since wedo not really have stages any more I choose to use a different notation with”top” and ”btm”, so x0 is now xtop etc.. (see Figure).

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Ltop

xtop

Vtop

ytop

Lbtm

xbtm

Vbtm

ybtm

Let us first repeat the basic data for the column:

Vbtm = 22 kmol/s

ybtm = 0.1176 (CO2 mole fraction) (feed)

ytop = 0.0116

xtop = 0.0219 (feed)

xbtm = 0.0532

Assuming constant inert flows for air (V ′) and water-amine (L′) through thecolumn, we have

V ′ = Vbtm(1− ybtm) = 19.41 kmol/s

L′ = 69.80 kmol/s

The operating line is not quite linear but it goes through the points (xtop, ytop)and (xbtm, ybtm). To simplify you may assume that the operating line is astraight line between these two points (which corresponds to using a constantLV =

ytop−ybtm

xtop−xbtm).

Table: Equilibrium data for CO2 and water-MEAmine at 1.2 bar and about60C (inert air present).xCO2(%) yCO2(%)2.227 1.1623.001 2.9173.339 3.7834.101 6.0574.700 8.4485.324 11.60

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Now finally to the new part of the problem: The overall mass transferrate referred to the gas side is with pressure as the driving force (A=CO2):

NA = KG · a(pA − p∗A)

where p∗A is the partial pressure of A that would result of we had gas in equi-librium with the liquid. The following data is given:

KG · a = 12.5 lbmol/(h ∗ ft3 ∗ atm)

(a) Use the data to find Kya [kmol/(s*m3 packing)] as used in the book,

NA = Kya(yA − y∗A)

where y∗A is the mole vapor mole fraction of a (imagined) gas in equilibriumwith the liquid (which has mole fraction xA).

(b) Find the required packing height when S = 253.5 m2 (cross sectionalarea of column) and you can assume constant molar flows in the column.

Note from the McCabe-Thiele diagram from the previous exercise that theequiibrium line is far from linear so we cannot use the simple method fromProblem 1.

(Hint: Make a table of 1/(yA − yA∗) as a function of yA; you can thenintegrate graphically, for example, by counting the number of squares on a mm-paper).

0.01 0.02 0.03 0.04 0.05

0.02

0.04

0.06

0.08

0.1

0.12

Absorber: CO2 − Water/MEA (1.2 bar)

Liquid mole fraction CO2

Va

po

r m

ole

fra

ctio

n C

O2

T

B

N = 13.1

X0

= 0.0219

XN

= 0.0532

Y1

= 0.0116

YN+1

= 0.1176

L/V = 3.39

L’/V’ = 3.6

Figure 1: McCabe Thiele diagram: Operating and equilibrium lines

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Solution

Problem 1

(i) V ′ = 6.53 · 10−4 kmol air/s. From mass balance gas phase side (A=SO2):NA = V ′( ybtm

1−ybtm− ytop

1−ytop) = 0.653 ((0.2/0.8) - (0.02/0.98) = 0.149 mol A/s

(ii) For y in the range from 0.02 to 0.2 we use (see also Figure 10.6-8):m = 30 (approximately).

(iii) Kya = 1 / (1/0.037) + 30/0.85) ) = 0.016 kmol/s,m3(iv)

TOP BTM

y 0.02 0.20

x 0 0.00355 (0.00355 os found from the total mass balance; see book)

y* 0 0.10 (0.10 is interpolated from equlibrium data in Table A3-19)

y-y* 0.02 0.10

Log mean y-y* = (0.02 - 0.10) / ln(0.02/0.10) = 0.050(v) Packing height z = 0.149 mol/s / (16 mol/s,m3 * 0.0929 m2 * 0.05) ) =

2.00 m(vi) The packing height is 25% higher than that found in Example 10.7-1

using a more exact method. This is expected because the mixture is not diluteand the assumption of linear equilibrium curve does not hold; see Figure 10.6-8in the book).

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Solution

Problem 2. Stripping using mass transfer rate-based method:Absorption of CO2 in water-amine solution (part 2)

(a) Since pA = yAp (where A is CO2 in our case) we have that Kya = KGa · pwhere where p=1.2 bar = 1.185 atm in our case. Furthermore, with 1ft =0.3048m, 1 lbmol=0.454 kmol and 1h= 3600s, we get

Kya = 12.5 · 1.185 · 0.454/(3600 · (0.3048)3) = 0.066 kmol/(s,m3)

(b) From a mass balance, the amount of A-component (here CO2) trans-ferred from vapor to liquid over a small volume of packing is [mol A/s]

Ky(yA − y∗A)︸ ︷︷ ︸NA

aSdz︸ ︷︷ ︸dA

= d(V yA)

To simplify, we assume constant V through the column, and rearrange to get

dz =V

KyaS

dy

(yA − y∗A)

Integrated from the top to the bottom we obtain the packing height:

z =

∫ btm

top

dz =V

KyaS︸ ︷︷ ︸HOG

·∫ yA,btm

yA,top

dyA(yA − y∗A)︸ ︷︷ ︸

NOG

Here yA and xA are related through the mass balance (operating line). If weassume the operating line is a straight line between the points (xbtm, ybtm) and(xtop, ytop) then we get

y =L

V(x− xbtm) + ybtm,

L

V=

ytop − ybtmxtop − xbtm

= 3.387

and we getyA = 3.387xA − 0.06257 (oper)

Furthermore, y∗A and xA are related through the equilibrium.We can set up the following table (we could in principle read off from the

McCabe-Thiele diagram but this is not sufficiently accurate in this case):

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xA y∗A(eq) yA(oper)1

yA−y∗A

0.02227 0.01162 0.01285 813.00.03001 0.02917 0.03907 101.10.03339 0.03783 0.05051 78.90.04101 0.06057 0.07632 63.50.04700 0.08448 0.09660 82.50.05324 0.1160 0.11774 576.2

0.023 0.01328 0.01533 4880.025 0.01781 0.02210 2330.051 0.10469 0.11015 1830.052 0.10974 0.11355 262

Figure 2: Determining the number of transfer units NOG by integration

The last four data points were obtained to get better accuracy and y∗A washere obtained by linear interpolation of the equilibrium data. Using graphicalintegration I count 13 squares (but I have to admit it is a bit inaccurate; youwould get better results on a calculator which has integration or with Excel or

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with Matlab):

N0G =

∫ yA,btm=0.11774

yA,top=0.01285

dyA(yA − y∗A)

= 13 (graphically)

The average value of V is 20.82 kmol/s so we get get

H0G =V

KyaS=

20.82

0.066 · 253.5= 1.245m

The packing height is then

z = NOGHOG = 13 · 1.245m = 16.2m

Comment: In the previous exercise we found that we need N = 13.1 theoreticalstages; thus we have in this case that N0G ≈ N . This is expected from (10.6-54)since we in this case have that the absorption factor A = (L/V )/m is very closeto 1 (see McCabe Thiele diagram)

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