Lecture Objectives McCabe-Thiele Graphical Solution to Single Countercurrent Cascades Topic 4a Ch.E. 421

• To apply the principles of the McCabeThiele Method to solve multistage problems in: Gas Absorption Liquid Extraction Stripping

Assumptions of the Mc-Cabe Thiele Method

1. Constant Molar Overflow: There are constant L and V rates in the products leaving the stages (Operating Line based on the compositions attached to the constant rates is linear as in the Kremser Equation). 2. The Equilibrium Line is non-linear but data is available for the Equilibrium Curve to be plotted.

Application to Gas Absorption

A gas absorption plate tower is to treat 9000 L/min of a gas mixture containing 22%NH3, 78% Air by mole at 1.5 atm and 30oC. Pure water will enter the top of the tower at a rate which is 1.5 times the minimum. 95% recovery of the ammonia is desired. Using the McCabe Thiele method, determine: a) Water rate (kg/hr) needed b) Number of ideal plates Obtain the needed equilibrium data from the Handbook.

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Given: a

Pure water Rate = 1.5 x minimum Required: a) Water rate b) No. of ideal plates Solution: L’min V’ b 95% recovery L’ = 1.5 L’min = Yb - Ya Xb * - Xa

To obtain Xb*, use the equilibrium data from the handbook. See the solubility data for Ammonia-air-water: Note that the data gives wt NH3/100 wts H2O,Wr and the equilibrium partial pressure, PA, in Torrs From problem data: PAb = yb x P = 250.8 Torrs From the Handbook, the pertinent equilibrium data for PA will cover 11.5 to 260 Torrs Wr 20 15 10 7.5 5.0 4.0 3.0 2.5 2.0 1.6 1.2 0 PA 260 179 110 79.7 51 40.1 29.6 24.4 19.3 15.3 11.5 0 Converting to X and Y data: PA Wr/17 Y= X= P - PA 100/18

P = 1.5 atm T = 30oC

Yb = 0.22/0.78 = 0.28205 Feed Gas 9000 L/min NH3 - Air yb = 0.22 Ya = (1- r) Yb = 0.0141 Xa = 0

Equilibrium X-Y Data: X 0.21176 0.15882 0.10588 0.07941 0.05294 0.04235 0.03176 0.02647 0.02118 0.01694 0.01271 0 Y 0.29545 0.18626 0.1068 0.07517 0.04683 0.03646 0.02666 0.02187 0.01722 0.0136 0.01019 0

Yb

E.C. This data is plotted and the value of Xb* is obtained using the value of Yb = 0.282 Y

X

Xb* = 0.205

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L’min V’

=

Yb - Ya Xb * - X a

= 1.303

Yb Y4

Graphical Determination of Ideal Plates

V’ = 1.5 x 9/(0.08205*303) x 0.78 x 60 = 25.41275 kmols/hr L’ = 1.5 x L’min = 49.8242 kmols/hr = 896.83636 kg/hr Solving the actual Xb = Xa + V’/L’ (Yb - Ya) = 0.137 The operating line may now be plotted using the boundary points (Xa, Ya) and (Xb, Yb) - straight line Starting with (Xa, Ya), triangular steps are drawn using alternately the equilibrium and operating lines until (Xb, Yb) is reached or passed by. Number of Steps = Number of Ideal Plates

4

Y3

3

Y2 Ya 1 Xa X1

2

X2

X3 Xb

X4

Four plates are needed.

Additional Exercises: 1. What water rate would be needed if six ideal plates are needed to obtain the same recovery? 2. What recovery is attained if six ideal plates are used together with the original water rate? 3. Determine the number of actual plates needed if a Murphree gas based efficiency of 75% is assumed. 4. 6.7, 6.8, 6.9, 6.10/pp. 246-247 Seader and Henley for Submission by August 2, 2011

Application to Liquid Extraction

Example: Pyridine may be removed from water by countercurrent multistage extraction with pure chlorobenzene. 100 kg/hr of a feed with 20% pyridine and 80% water by weight shall be processed with a desired pyridine recovery of 90%. Pure chlorobenzene shall be introduced at a rate which is 1.25 times the minimum. The equilibrium data for the system is as follows: Wt% Pyridine in Raffinate 0 Wt% Pyridine in Extract 0 3.3 6.2 9.1 9.1 13.0 16.7 12.7 20.0 16.7 20.3 23.1 25.9

Assume water and chlorobenzene are completely immiscible.

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A – pyridine B – water 90% Rec.

S - chlorobenzene 1.25 x min

Determination of Minimum Solvent Rate

Vb

Ya*=0.342

0.3

Va a La

La = 100 xa = 0.2 X Y...