20 Oct 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 6

20 Oct 2011Prof. R. Shanthini2 Plate columns are used mainly in distillation You probably know how to calculate the number of plates required for the desired separation.

20 Oct 2011Prof. R. Shanthini3 Packed columns are used in absorption, stripping and adsorption There are no plates here. We need to calculate the height of packing in a packed column where the separation process takes place. How to do that?

20 Oct 2011Prof. R. Shanthini4 Equations for Packed Columns Inlet solvent L s,in, L in, X in, x in Treated gas G s,out, G out, Y out, y out Spent solvent L s,out, L out, X out, x out Inlet gas G s,in, G in, Y in, y in Counter-current packed column used for absorption GsGYyGsGYy LsLXxLsLXx

20 Oct 2011Prof. R. Shanthini5 Notations G s - inert gas molar flow rate (constant) L s - solvent molar flow rate (constant) G - total gas molar flow rate (varies as it looses the solute) L - total liquid molar flow rate (varies as it absorbs the solute) Y - mole ratio of solute A in gas = moles of A / moles of inert gas y - mole fraction of solute A in gas = moles of A / (moles of A + moles of inert gas) X - mole ratio of solute A in liquid = moles of A / moles of solvent x - mole fraction of solute A in liquid = moles of A / (moles of A + moles of solvent) Solute in the gas phase = G s Y = G y Solute in the liquid phase = L s X = L x

20 Oct 2011Prof. R. Shanthini6 Relating Y to y and X to x: Therefore y = Y / (Y+1) y = Similarly, it can be shown x = X / (X+1) moles of solute A moles of solute A + moles of inert gas = (moles of inert gas / moles of solute A ) = Y Y + 1 = (1/Y)

20 Oct 2011Prof. R. Shanthini7 For dilute mixers Y - mole ratio of solute A in gas = moles o fA / moles of inert gas y - mole fraction of solute A in gas = moles of A / (moles of A + moles of inert gas) X - mole ratio of solute A in liquid = moles of A / moles of solvent x - mole fraction of solute A in liquid = moles of A / (moles of A + moles of solvent) For dilute mixtures, y ≈ Y x ≈ X

20 Oct 2011Prof. R. Shanthini8 For dilute mixers G s - inert gas molar flow rate (constant) L s - solvent molar flow rate (constant) G - total gas molar flow rate (varies as it looses the solute) L - total liquid molar flow rate (varies as it absorbs the solute) For dilute mixtures, G s ≈ G L s ≈ L

20 Oct 2011Prof. R. Shanthini9 Equations for Packed Columns for dilute solutions 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 The operating equation for the packed column is obtained by writing the mass balance for solute over the control volume: Control volume L in x in + G y = L x + G out y out If dilute solution is assumed, then L in = L = L out and G in = G = G out. Therefore, the above becomes L x in + G y = L x + G y out y = (L / G) x + y out - (L / G) x in The operating line therefore becomes (74) (75) (76)

20 Oct 2011Prof. R. Shanthini10 Mass transfer of solute from gas phase to liquid phase is assumed to be explained by the Two-film Theory: Equations for Packed Columns for dilute solutions

20 Oct 2011Prof. R. Shanthini11 Rate of mass transfer from vapour phase to liquid phase is therefore given by the following: k y a (y - y i ) = k x a (x i - x) k y and k x are film mass transfer coefficients based on unit area. But the area for mass transfer in a packed column in difficult to determine. Therefore we use a factor ‘a’ which gives the area of mass transfer per unit volume of packed bed. In packed columns, we use ‘k y a’ and ‘k x a’ which are the film mass transfer coefficients based on unit volume of packed bed. Molar fractions y i and x i are the interface properties that are related by the equilibrium ratio K (= y i / x i ). (77) (78) Equations for Packed Columns for dilute solutions

20 Oct 2011Prof. R. Shanthini12 Relating equilibrium mole fractions to operating mole fractions: Equation (77) gives Equilibrium line: K = y i / x i y = (L / G) x + y out - (L / G) x in Operating line: y - y i k y a k x a x - x i = - Equilibrium line (slope = K) x y Operating line (slope = L/G) y out - (L / G) x in (x i,y i ) Slope = - k y a k x a (x,y) (76) (78) Equations for Packed Columns for dilute solutions

20 Oct 2011Prof. R. Shanthini13 x y (x,y) (x i,y i ) (x,y*) (x*,y) y* is the gas-phase mole fraction that would have been in equilibrium with the liquid-phase mole fraction x. y* = K x x* is the liquid-phase mole fraction that would have been in equilibrium with the gas-phase mole fraction y. y = K x* y - y i x* - x i y i – y* x i - x K = = y out - (L / G) x in Equations for Packed Columns for dilute solutions

20 Oct 2011Prof. R. Shanthini14 1 kxakxa 1 kyakya + 1 KxaKxa =(77) y - y i x* - x i 1 kyakya 1 kxakxa + 1 KyaKya = y i – y* x i - x Since the liquid usually has strong affinity for the solute, mass transfer resistance is mostly in the gas. Therefore, determination of the packed height of a column most commonly involves the overall gas- phase mass transfer coefficient based on unit volume of packed base, which is K y a. 1 kxakxa 1/K kyakya += 1 kyakya K kxakxa +=(78) Equations for Packed Columns for dilute solutions

20 Oct 2011Prof. R. Shanthini15 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 Let us now obtain equations required to determine the height of packing required for a specified separation in a packed column operated with dilute solution. G y+dy L x+dx z dz Z Equations for Packed Columns for dilute solutions

20 Oct 2011Prof. R. Shanthini16 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 Mass of solute lost from the gas over the differential height of packing dz = G y - G (y + dy) = - G dy Mass of solute transferred from the gas to the liquid = K y a (y – y*) S dz where S is the inside cross- sectional area of the tower. G y+dy L x+dx z dz Z Therefore, mass balance for solute gives the following: -G dy = K y a (y – y*) S dz(79) Equations for Packed Columns for dilute solutions

20 Oct 2011Prof. R. Shanthini17 Rearranging and integrating (79) gives the following: K y aS G dz dy y – y* 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 y in ∫ 0 Z The packed height is therefore given by: G K y aS Z dy y – y* = ∫ y out y in H OG N OG Equations for Packed Columns for dilute solutions

20 Oct 2011Prof. R. Shanthini18 G K y aS dy y – y* ∫ y out y in Definition of H OG and N OG : H OG ≡ N OG ≡ which is termed as the overall height of a transfer unit (HTU) based on the gas phase which has the dimension of length which is termed as the overall number of transfer units (NTU) based on the has phase which is dimensionless Z = H OG N OG Height of the packed column is written in terms of transfer units as follows: (80) (81) (82) Equations for Packed Columns for dilute solutions

20 Oct 2011Prof. R. Shanthini19 G K y aS dy y – y* ∫ y out y in HTU: H OG = NTU: N OG = The smaller the HTU, the more efficient is the contacting. It represents the overall change in solute mole fraction divided by the average mole-fraction driving force. The larger the NTU, the greater is the extent of contacting required. Equations for Packed Columns for dilute solutions

20 Oct 2011Prof. R. Shanthini20 dy y – y* ∫ y out y in Integration of N OG =(81) Equilibrium relationship gives:y* = K x Operating line gives: x = (G/L) y - (G/L) y out + x in Using these the denominator in (81) can be written as follows: y - y* = y - K x = y - K(G/L) y + K(G/L) y out - K x in = (1 - KG/L) y + (KG/L) y out - K x in (82) Equations for Packed Columns for dilute solutions

20 Oct 2011Prof. R. Shanthini21 dy ∫ y out y in N OG = (1 - KG/L) y + (KG/L) y out - K x in Combining (81) and (82), we get = ln[(1 - KG/L) y + (KG/L) y out - K x in ] (1 - KG/L) y out y in = (1 - KG/L) y in + (KG/L) y out - K x in (1 - KG/L) 1 y out - K x in ln Equations for Packed Columns for dilute solutions

20 Oct 2011Prof. R. Shanthini22 N OG = y in - K x in + KG/L (y out - y in ) (1 - KG/L) 1 y out - K x in ln y in - K x in + KG/L (y out - K x in + K x in - y in ) (1 - KG/L) 1 y out - K x in ln= (1 - KG/L) (y in - K x in ) (1 - KG/L) 1 y out - K x in ln= + KG/L(83) Equations for Packed Columns for dilute solutions

20 Oct 2011Prof. R. Shanthini23 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

20 Oct 2011Prof. R. Shanthini24 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