03 Nov 2011Prof. R. Shanthini1 Course content of Mass transfer section LTA Diffusion Theory of interface mass transfer Mass transfer coefficients, overall coefficients and transfer units Application of absorption, extraction and adsorption Concept of continuous contacting equipment Simultaneous heat and mass transfer in gas- liquid contacting, and solids drying CP302 Separation Process Principles Mass Transfer - Set 7

03 Nov 2011Prof. R. Shanthini2 Example 6.10 of Ref 2 Experimental data have been obtained for air containing 1.6% by volume of SO 2 being scrubbed with pure water in a packed column of 1.5 m 2 in cross-sectional area and 3.5 m in packed height. Entering gas and liquid flow rates are and 2.2 kmol/s, respectively. If the outlet mole fraction of SO 2 in the gas is and column temperature is near ambient with K SO2 = 40, calculate the following: a) The N OG for absorption of SO 2 b) The H OG in meters c) The volumetric, overall mass-transfer coefficient, K y a for SO 2 in kmol/m 3.s

03 Nov 2011Prof. R. Shanthini3 Solution to Example 6.10 of Ref 2 Data: y in = x in = 0 S = 1.5 m 2 Z = 3.5 m G = kmol/s L = 2.2 kmol/s y out = K SO2 = 40 Inlet solvent L in, x in Treated gas G out, y out Spent solvent L out, x out Inlet gas G in, y in GyGy LxLx G y+dy L x+dx z dz Z

03 Nov 2011Prof. R. Shanthini4 Solution to Example 6.10 of Ref 2 Data: y in = x in = 0 S = 1.5 m 2 Z = 3.5 m G = kmol/s L = 2.2 kmol/s y out = K SO2 = 40 Operating line: y = (L / G) x + y out - (L / G) x in = (2.2/0.062) x Equilibrium line: y = K x = 40 x

03 Nov 2011Prof. R. Shanthini5 Solution to Example 6.10 of Ref 2 Top of the column (x in = 0, y out = 0.004) Bottom of the column (x out = ?, y in = 0.016)

03 Nov 2011Prof. R. Shanthini6 Solution to Example 6.10 of Ref 2 a) N OG could be calculated using (83) (1 – 1.127) ( ) (1 – 1.127) ln N OG = where KG / L = K SO2 G / L = 40*0.062 / 2.2 = = 3.78 N OG = (1 - KG/L) (y in - K x in ) (1 - KG/L) 1 y out - K x in ln + KG/L

03 Nov 2011Prof. R. Shanthini7 Solution to Example 6.10 of Ref 2 b) H OG could be calculated using (80) G K y aS H OG ≡ kmol/s (K y a)(1.5 m 2 ) = Cannot continue this way since the value of K y a is not known. H OG = Z / N OG Use (82) instead to calculate H OG since Z and N OG are known. = 3.5 m / 3.78 = m

03 Nov 2011Prof. R. Shanthini8 Solution to Example 6.10 of Ref 2 c) K y a can be calculated using (80) G K y aS H OG ≡ kmol/s (K y a)(1.5 m 2 ) = m K y a = kmol/s (0.926 m)(1.5 m 2 ) = kmol/m 3.s

03 Nov 2011Prof. R. Shanthini9 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 30 o C 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 k y a and k x a are 200 lbmol/h.ft 3 and 2643 lbmol/h.ft 3, respectively. Calculate the following: a) K y a b) H OG and N OG c) Y out and x out

03 Nov 2011Prof. R. Shanthini10 Solution to Example 6.11 of Ref 2 (modified) Data: y in = 0.02 x in = 0 G = 2500 lbmol/h L = 3500 lbmol/h S = π (4/2) ft 2 = 12.6 ft 2 Z = 2 x 12 ft = 24 ft K= 0.85 k y a = 200 lbmol/h.ft 3 and k x a = 165 lbmol/h.ft 3 Inlet solvent L in, x in Treated gas G out, y out Spent solvent L out, x out Inlet gas G in, y in GyGy LxLx G y+dy L x+dx z dz Z

03 Nov 2011Prof. R. Shanthini11 a) K y a can be calculated using (78). 1 KyaKya = 1 kyakya K kxakxa += K y a = 98.5 lbmol/h.ft 3 b) H OG can be calculated using (80). G K y aS H OG ≡ 2500 lbmol/h (98.5 lbmol/h.ft 3 )(12.6 ft 2 ) = = 2.02 ft Solution to Example 6.11 of Ref 2 (modified) N OG = Z / H OG Use (82) to calculate N OG since Z and H OG are known. = 24 ft / 2.02 ft = 11.88

03 Nov 2011Prof. R. Shanthini12 c) y out could be calculated using (83) (1 – 0.607) ( ) (1 – 0.607) 1 y out - 0 ln = where KG / L = 0.85 * 2500 / 3500 = N OG = (1 - KG/L) (y in - K x in ) (1 - KG/L) 1 y out - K x in ln + KG/L Solution to Example 6.11 of Ref 2 (modified) (84) y out = = 0.007%

03 Nov 2011Prof. R. Shanthini13 Solution to Example 6.11 of Ref 2 (modified) Enlarge it.

03 Nov 2011Prof. R. Shanthini14 Solution to Example 6.11 of Ref 2 (modified) y out = 0.01 mol%

03 Nov 2011Prof. R. Shanthini15 x out can be determined from the mass balance. x out = (G/L) y in - (G/L) y out + x in = (25/35) (25/35) = = 1.42% Solution to Example 6.11 of Ref 2 (modified)

03 Nov 2011Prof. R. Shanthini16 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 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 K y a = 82 mol/m 3.s and the cross-sectional area of the tower as 1.5 m 2.

03 Nov 2011Prof. R. Shanthini17 Solution to determining the minimum liquid flow rate: Data: y in = 0.03 x in = 0 G = 3540 mol/min K ammonia = Inlet solvent L in, x in Treated gas G out, y out Spent solvent L out, x out Inlet gas G in, y in GyGy LxLx G y+dy L x+dx z dz Z y out can be determined from the data provided as shown in the next page. We need to calculate (L/G) min which can be done graphically as shown in the following pages. Dilute mixtures are assumed.

03 Nov 2011Prof. R. Shanthini18 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 y out = 0.1 x 0.03 x 3540 / (0.1 x 0.03 x x 3540) = 0.1 x 0.03 / (0.1 x ) = ≈ 0.003

03 Nov 2011Prof. R. Shanthini19 Operating line with y out = and x in = 0: y = (L / G) x + y out - (L / G) x in = (L / G) x Equilibrium line: y = K x = x Solution to determining the minimum liquid flow rate: Since L/G is not known, we cannot plot the operating line. Equilibrium line can be plotted.

03 Nov 2011Prof. R. Shanthini20 Top of the column (x in = 0, y out = 0.003) Bottom of the column (x out = ?, y in = 0.03) Solution to determining the minimum liquid flow rate:

03 Nov 2011Prof. R. Shanthini21 Top of the column (x in = 0, y out = 0.003) Bottom of the column (x out = ?, y in = 0.03) Solution to determining the minimum liquid flow rate: Operating line should pass through this point.

03 Nov 2011Prof. R. Shanthini22 Top of the column (x in = 0, y out = 0.003) Bottom of the column (x out = ?, y in = 0.03) Solution to determining the minimum liquid flow rate: Operating line with the slope (L/G) min should connect these two points.

03 Nov 2011Prof. R. Shanthini23 Top of the column (x in = 0, y out = 0.003) Bottom of the column (x out = ?, y in = 0.03) Solution to determining the minimum liquid flow rate:

03 Nov 2011Prof. R. Shanthini24 Top of the column (x in = 0, y out = 0.003) Bottom of the column (x out = ?, y in = 0.03) Solution to determining the minimum liquid flow rate: (L/G) min = ( )/x out x out = 0.03/0.772 = (L/G) min = ( ) / = mol of water/mol of air

03 Nov 2011Prof. R. Shanthini25 Solution to determining the minimum liquid flow rate: (L/G) min = mol of water/mol of air G = 3540 mol/min (given) L min = x 3540 = 2460 mol/min (L/G) operating = 1.5 x (L/G) min = 1.5 x = mol of water/mol of air L operating = x 3540 = 3688 mol/min

03 Nov 2011Prof. R. Shanthini26 Top of the column (x in = 0, y out = 0.003) Bottom of the column (y in = 0.03) Solution to determining the minimum liquid flow rate: x out = x out = ??

03 Nov 2011Prof. R. Shanthini27 Top of the column (x in = 0, y out = 0.003) Bottom of the column (y in = 0.03) Solution to determining the minimum liquid flow rate: x out = ?? (L/G) operating = ( )/x out X out = ( )/1.042 =

03 Nov 2011Prof. R. Shanthini28 NTU (N OG ) could be calculated using (83) (1 – 0.741) ( ) (1 – 0.741) ln N OG = where KG / L = / = = 4.65 N OG = (1 - KG/L) (y in - K x in ) (1 - KG/L) 1 y out - K x in ln + KG/L Solution to determining the minimum liquid flow rate:

03 Nov 2011Prof. R. Shanthini29 HTU (H OG ) could be calculated using (80) G K y aS H OG = 3540 mol/min (82 mol/m 3.s )(1.5 m 2 ) = Solution to determining the minimum liquid flow rate: 3540/60 mol/s 82 x 1.5 mol/m.s = = 0.48 m Z = N OG x H OG = 4.65 x 0.48 m = 2.23 m Height of the tower (Z) can be calculated as follows:

03 Nov 2011Prof. R. Shanthini30 Inlet solvent L in, x in Treated gas G out, y out Spent solvent L out, x out Inlet gas G in, y in GyGy LxLx G y+dy L x+dx z dz Z The packed height is given by: G K y aS Z dy y – y* = ∫ y out y in H OG N OG Summary: Equations for Packed Columns for dilute solutions L K x aS Z dx x* – x = ∫ x in x out H OL N OL

03 Nov 2011Prof. R. Shanthini31 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.

03 Nov 2011Prof. R. Shanthini32 Gas absorption, Stripping and Extraction Gas absorption: N OG and H OG are used Stripping: N OL and H OL are used Extraction: N OL and H OL are used Humidification: N G and H G are used.