CP302 Separation Process Principles

1. CP302 Separation Process Principles Mass Transfer - Set 7 Prof. R. Shanthini

2. Example 6.10 of Ref 2 Experimental data have been obtained for air containing 1.6% by volume of SO2 being scrubbed with pure water in a packed column of 1.5 m2 in cross-sectional area and 3.5 m in packed height. Entering gas and liquid flow rates are 0.062 and 2.2 kmol/s, respectively. If the outlet mole fraction of SO2 in the gas is 0.004 and column temperature is near ambient with KSO2 = 40, calculate the following: a) The NOG for absorption of SO2 b) The HOG in meters c) The volumetric, overall mass-transfer coefficient, Kya for SO2 in kmol/m3.s Prof. R. Shanthini

3. Solution to Example 6.10 of Ref 2 Treated gas Gout, yout Inlet solvent Lin, xin Data: yin = 0.016 xin = 0 S = 1.5 m2 Z = 3.5 m G = 0.062 kmol/s L = 2.2 kmol/s yout = 0.004 KSO2 = 40 L x G y dz L x+dx G y+dy Z z Inlet gas Gin, yin Spent solvent Lout, xout Prof. R. Shanthini

4. Solution to Example 6.10 of Ref 2 Data: yin = 0.016 xin = 0 S = 1.5 m2 Z = 3.5 m G = 0.062 kmol/s L = 2.2 kmol/s yout = 0.004 KSO2 = 40 Operating line: y = (L / G) x + yout - (L / G) xin = (2.2/0.062) x + 0.004 Equilibrium line: y = K x = 40 x Prof. R. Shanthini

5. Solution to Example 6.10 of Ref 2 Bottom of the column (xout = ?, yin = 0.016) Top of the column (xin = 0, yout = 0.004) Prof. R. Shanthini

6. Solution to Example 6.10 of Ref 2 a) NOG could be calculated using (83) 1 (1 - KG/L) (yin - K xin) ln NOG = + KG/L (1 - KG/L) yout - K xin where KG / L = KSO2G / L = 40*0.062 / 2.2 = 1.127 1 (1 – 1.127) (0.016 - 0) ln NOG = + 1.127 (1 – 1.127) 0.004 - 0 = 3.78 Prof. R. Shanthini

7. Solution to Example 6.10 of Ref 2 b) HOG could be calculated using (80) 0.062 kmol/s G = HOG≡ (Kya)(1.5 m2) KyaS Cannot continue this way since the value of Kya is not known. Use (82) instead to calculate HOG since Z and NOG are known. HOG = Z / NOG = 3.5 m / 3.78 =0.926 m Prof. R. Shanthini

8. Solution to Example 6.10 of Ref 2 c) Kya can be calculated using (80) G HOG≡ KyaS 0.062 kmol/s 0.926 m = (Kya)(1.5 m2) 0.062 kmol/s = 0.044 kmol/m3.s Kya = (0.926 m)(1.5 m2) Prof. R. Shanthini

9. Example 6.11 of Ref 2 (modified) A gaseous reactor effluent consisting of 2 mol% ethylene oxide in an inert gas is scrubbed with water at 30oC and 20 atm. The total gas feed rate is 2500 lbmol/h, and the water rate entering the scrubber is 3500 lbmol/h. The column, with a diameter of 4 ft, is packed in two 12-ft-high sections with 1.5 in metal Pall rings. A liquid redistributer is located between the two packed sections. Under the operating conditions for the scrubber, the K-value for ethylene oxide is 0.85 and estimated values of kya and kxa are 200 lbmol/h.ft3 and 2643 lbmol/h.ft3 , respectively. Calculate the following: a) Kya b) HOG and NOG c) Yout and xout Prof. R. Shanthini

10. Solution to Example 6.11 of Ref 2 (modified) Treated gas Gout, yout Inlet solvent Lin, xin Data: yin = 0.02 xin = 0 G = 2500 lbmol/h L = 3500 lbmol/h S = π (4/2) ft2 = 12.6 ft2 Z = 2 x 12 ft = 24 ft K = 0.85 kya = 200 lbmol/h.ft3 and kxa = 165 lbmol/h.ft3 L x G y dz L x+dx G y+dy Z z Inlet gas Gin, yin Spent solvent Lout, xout Prof. R. Shanthini

11. Solution to Example 6.11 of Ref 2 (modified) a) Kya can be calculated using (78). 1 1 K 1 0.85 = + = + Kya kya kxa 200 165 Kya = 98.5 lbmol/h.ft3 b) HOG can be calculated using (80). 2500 lbmol/h G = HOG≡ = 2.02 ft (98.5 lbmol/h.ft3)(12.6 ft2) KyaS Use (82) to calculate NOG since Z and HOG are known. NOG = Z / HOG = 24 ft / 2.02 ft =11.88 Prof. R. Shanthini

12. Solution to Example 6.11 of Ref 2 (modified) c) yout could be calculated using (83) 1 (1 - KG/L) (yin - K xin) ln NOG = + KG/L yout - K xin (1 - KG/L) where KG / L = 0.85 * 2500 / 3500 = 0.607 1 (1 – 0.607) (0.02 - 0) ln 11.88 = + 0.607 (84) yout - 0 (1 – 0.607) yout = 0.00007 = 0.007% Prof. R. Shanthini

13. Solution to Example 6.11 of Ref 2 (modified) Enlarge it. Prof. R. Shanthini

14. Solution to Example 6.11 of Ref 2 (modified) yout = 0.01 mol% Prof. R. Shanthini

15. Solution to Example 6.11 of Ref 2 (modified) xout can be determined from the mass balance. xout = (G/L) yin - (G/L) yout + xin = (25/35) 0.02 - (25/35) 0.0001 + 0 = 0.0142 = 1.42% Prof. R. Shanthini

16. Determining the minimum liquid flow rate: Calculate (L/G)min for the removal of 90% of the ammonia from a 3540 mol/min feed gas containing 3% ammonia and 97% air. The inlet liquid is pure water and the temperature and pressure are 293 K and 1 atm, respectively. The equilibrium ratio of mole fraction of ammonia in air to mole fraction of ammonia in water at the column condition can be taken as 0.772. Since a small liquid flow rate results in a high tower (which is costly), and a large liquid flow rate requires a large diameter tower (which is also costly), the optimum liquid flow rate of 1.2 to 1.5 times the minimum flow rate is used in practice. Assuming the liquid flow rate to be 1.5 times the minimum, determine NTU, HTU and the height of tower required. Take Kya = 82 mol/m3.s and the cross-sectional area of the tower as 1.5 m2. Prof. R. Shanthini

17. Solution to determining the minimum liquid flow rate: Treated gas Gout, yout Inlet solvent Lin, xin Data: yin = 0.03 xin = 0 G = 3540 mol/min Kammonia = 0.772 L x G y dz yout can be determined from the data provided as shown in the next page. L x+dx G y+dy Z z We need to calculate (L/G)min which can be done graphically as shown in the following pages. Inlet gas Gin, yin Spent solvent Lout, xout Dilute mixtures are assumed. Prof. R. Shanthini

18. Solution to determining the minimum liquid flow rate: Ammonia in the inlet stream = 0.03 x 3540 mol/min Air in the inlet stream = 0.97 x 3540 mol/min Ammonia removed = 0.9 x 0.03 x 3540 mol/min Ammonia in the outlet stream = 0.1 x 0.03 x 3540 mol/min Air in the outlet stream = 0.97 x 3540 mol/min yout = 0.1 x 0.03 x 3540 / (0.1 x 0.03 x 3540 + 0.97 x 3540) = 0.1 x 0.03 / (0.1 x 0.03 + 0.97) = 0.003083 ≈ 0.003 Prof. R. Shanthini

19. Solution to determining the minimum liquid flow rate: Operating line with yout = 0.003 and xin = 0: y = (L / G) x + yout - (L / G) xin = (L / G) x + 0.003 Since L/G is not known, we cannot plot the operating line. Equilibrium line: y = K x = 0.772 x Equilibrium line can be plotted. Prof. R. Shanthini

20. Solution to determining the minimum liquid flow rate: Bottom of the column (xout = ?, yin = 0.03) Top of the column (xin = 0, yout = 0.003) Prof. R. Shanthini

21. Solution to determining the minimum liquid flow rate: Bottom of the column (xout = ?, yin = 0.03) Operating line should pass through this point. Top of the column (xin = 0, yout = 0.003) Prof. R. Shanthini

22. Solution to determining the minimum liquid flow rate: Bottom of the column (xout = ?, yin = 0.03) Operating line with the slope (L/G)min should connect these two points. Top of the column (xin = 0, yout = 0.003) Prof. R. Shanthini

23. Solution to determining the minimum liquid flow rate: Bottom of the column (xout = ?, yin = 0.03) Top of the column (xin = 0, yout = 0.003) Prof. R. Shanthini

24. Solution to determining the minimum liquid flow rate: Bottom of the column (xout = ?, yin = 0.03) (L/G)min = (0.03-0.003)/xout xout = 0.03/0.772 = 0.0389 (L/G)min = (0.03-0.003) / 0.0389 = 0.695 mol of water/mol of air Top of the column (xin = 0, yout = 0.003) Prof. R. Shanthini

25. Solution to determining the minimum liquid flow rate: (L/G)min = 0.695 mol of water/mol of air G = 3540 mol/min (given) Lmin = 0.695 x 3540 = 2460 mol/min (L/G)operating = 1.5 x (L/G)min = 1.5 x 0.695 = 1.042 mol of water/mol of air Loperating = 1.042 x 3540 = 3688 mol/min Prof. R. Shanthini

26. Solution to determining the minimum liquid flow rate: Bottom of the column (yin = 0.03) xout = ?? xout = 0.0389 Top of the column (xin = 0, yout = 0.003) Prof. R. Shanthini

27. Solution to determining the minimum liquid flow rate: Bottom of the column (yin = 0.03) xout = ?? (L/G)operating = (0.03-0.003)/xout Xout = (0.03-0.003)/1.042 = 0.0259 Top of the column (xin = 0, yout = 0.003) Prof. R. Shanthini

28. Solution to determining the minimum liquid flow rate: NTU (NOG) could be calculated using (83) 1 (1 - KG/L) (yin - K xin) ln NOG = + KG/L (1 - KG/L) yout - K xin where KG / L = 0.772 / 1.042 = 0.741 1 (1 – 0.741) (0.03 - 0) ln NOG = + 0.741 (1 – 0.741) 0.003 - 0 = 4.65 Prof. R. Shanthini

29. Solution to determining the minimum liquid flow rate: HTU (HOG) could be calculated using (80) 3540 mol/min G = HOG= (82 mol/m3.s )(1.5 m2) KyaS 3540/60 mol/s = = 0.48 m 82 x 1.5 mol/m.s Height of the tower (Z) can be calculated as follows: Z = NOG x HOG = 4.65 x 0.48 m = 2.23 m Prof. R. Shanthini

30. Summary: Equations for Packed Columns for dilute solutions Treated gas Gout, yout Inlet solvent Lin, xin The packed height is given by: yin ∫ dy G = Z y – y* KyaS L x G y yout HOG dz NOG L x+dx G y+dy HOL NOL Z z xout ∫ dx L = Z x* – x KxaS Inlet gas Gin, yin Spent solvent Lout, xout xin Prof. R. Shanthini

31. Summary: Equations for Packed Columns for dilute solutions Distributed: Photocopy of Table 16.4 Alternative mass transfer coefficient groupings for gas absorption from Henley EJ and Seader JD, 1981, Equilibrium-Stage Separation Operations in Chemical Engineering, John Wiley & Sons. Prof. R. Shanthini

32. Gas absorption, Stripping and Extraction Gas absorption: NOG and HOG are used Stripping: NOL and HOL are used Extraction: NOL and HOL are used Humidification: NG and HG are used. Prof. R. Shanthini