24 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 8

24 Nov 2011Prof. R. Shanthini2 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 Control volume We have learned to determine the height of packing in packed columns with dilute solutions in the last lecture class. Today, we will learn to determine the height of packing in packed columns with concentrated solutions.

24 Nov 2011Prof. R. Shanthini3 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

24 Nov 2011Prof. R. Shanthini4 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 was used for packed column with dilute solution assuming G can be taken as a constant for dilute solutions. G y+dy L x+dx z dz Z For concentrated solution we ought to use G s (inert gas molar flow rate) which is constant across the column along with Y (mole ratio of solute A in gas). Equations for Packed Columns

24 Nov 2011Prof. R. Shanthini5 Inlet solvent L in, x in Treated gas G out, y out Spent solvent L out, x out Inlet gas G in, y in GsYGsY LsXLsX Mass of solute lost from the gas over the differential height of packing dz = G s Y - G s (Y + dY) = - G s dY G s Y+dY L s X+dX z dz Z Relate G s to G: G s = G (1 – y) Relate Y to y: From y = Y / (Y+1), we get Y = y / (1 – y) (85) (86) (87) Equations for Packed Columns with concentrated solutions

24 Nov 2011Prof. R. Shanthini6 Inlet solvent L in, x in Treated gas G out, y out Spent solvent L out, x out Inlet gas G in, y in GsYGsY Using (86) and (87), mass of solute lost from the gas over the differential height of packing dz, given by (85), can be written as follows: -G s dY= - G (1 – y) d[y / (1 – y)] G s Y+dY L s X+dX z dz Z Equations for Packed Columns with concentrated solutions = - G (1 – y) (1 – y) 2 dy = - G (1 – y) dy (88) LsXLsX

24 Nov 2011Prof. R. Shanthini7 Inlet solvent L in, x in Treated gas G out, y out Spent solvent L out, x out Inlet gas G in, y in GsYGsY 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 s Y+dY L s X+dX z dz Z Relating (88) to the above at steady state, we get -G = K y a (y – y*) S dz (89) Equations for Packed Columns with concentrated solutions (1 – y) dy LsXLsX Compare (89) with (79) used for packed columns with dilute solutions. What are the differences?

24 Nov 2011Prof. R. Shanthini8 Inlet solvent L in, x in Treated gas G out, y out Spent solvent L out, x out Inlet gas G in, y in GsYGsY LsXLsX G s Y+dY L s X+dX z dz Z Equations for Packed Columns with concentrated solutions Rearranging and integrating (89) gives the following: K y aS G Z = dy (1 – y)(y – y*) ∫ y out y in (90)

24 Nov 2011Prof. R. Shanthini9 Equations for Packed Columns with concentrated solutions K y a(1 – y) LM S G Z = (1 – y) LM dy (1 – y)(y – y*) ∫ y out y in (91) Multiply the numerator and denominator of (90) by (1 – y) LM : where (1 – y) LM is the log mean of (1 – y) and (1 – y*) given as follows: (1 – y) LM y* – y ln[(1 – y )/(1 – y* )] =(92)

24 Nov 2011Prof. R. Shanthini10 Equations for Packed Columns with concentrated solutions K y a(1 – y) LM S G Z = (1 – y) LM dy (1 – y)(y – y*) ∫ y out y in (93) Therefore (91) becomes the following: Even though G and (1 – y) LM are not constant across the column, we can consider the ratio of the two to be a constant and take G / [K y a(1 – y) LM S] out of the integral sign in (91) without incurring errors larger than those inherent in experimental measurements of K y a. (Usually average values of G and (1 – y) LM are used.) H OG N OG

24 Nov 2011Prof. R. Shanthini11 Equations for Packed Columns with concentrated solutions x in (94) can be related to y in (93) using the operating line equation. y* in (93) can be related to the bulk concentration using the equilibrium relationship as follows: y* = K x (94) We will determine the operating line equation next

24 Nov 2011Prof. R. Shanthini12 Equations for Packed Columns with concentrated 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 a 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. (74) We somehow have to relate L to L in and G to G out in (74).

24 Nov 2011Prof. R. Shanthini13 Equations for Packed Columns with concentrated 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 Control volume L in (1 – x in ) = L (1 – x) To relate L to L in, write a mass balance for solvent over the control volume: To relate G to G out, write the overall mass balance over the entire column: L = L in (1 – x in ) / (1 – x)(95) G + L in = G out + L(96)

24 Nov 2011Prof. R. Shanthini14 Equations for Packed Columns with concentrated solutions Use (95) to eliminate L from (96): G + L in = G out + L in (1 – x in ) / (1 – x) G = G out + L in (x – x in ) / (1 – x)(97) Combining (74), (95) and (97), we get the following: L in x in + [G out +L in (x – x in )/ (1 – x)] y = L in (1 – x in )x/(1 – x) + G out y out y = G out + [L in (x – x in ) / (1 – x)] G out y out + [L in (1 – x in ) x / (1 – x)] - L in x in (98) Equation (98) is the operating line equation for packed columns with concentrated solutions. Compare it with equation (76) used for packed columns with dilute solutions. What are the differences?

24 Nov 2011Prof. R. Shanthini15 Equations for Packed Columns with concentrated solutions If the solvent fed to the column is pure then x in = 0. Therefore, (98) becomes y = G out + [ L in x / (1 – x) ] G out y out + [ L in x / (1 – x) ] (99)

24 Nov 2011Prof. R. Shanthini16 Example 1: Draw the operating curve for a system where 95% of the ammonia from an air stream containing 40% ammonia by volume is removed in a packed column. Solvent used in 488 lbmol/h per 100 lbmol/h of entering gas. Solution: L in = 488 lbmol/h; G in = 100 lbmol/h; y in = 0.4; x in = 0 Ammonia removed from the air stream = 0.95 x 40 = 38 lbmol/h In the inlet air stream: ammonia = 40 lbmol/h; air = 60 lbmol/h In the outlet air stream: ammonia = 02 lbmol/h; air = 60 lbmol/h Therefore, G out = (60+2) = 62 lbmol/h, and y out = 2 / 62 =

24 Nov 2011Prof. R. Shanthini17 Example 1: y = 62 + [488 x / (1 – x)] 62 x [488 x / (1 – x)] Using L in = 488 lbmol/h, G out = 62 lbmol/h, y out = and x in = 0 in (98), we get the operating curve as follows: Operating curve y in (bottom of the tower) y out (top of the tower)

24 Nov 2011Prof. R. Shanthini18 Example 2: Draw the equilibrium curve, which is approximately described by K = x , on the same plot as in Example 1. Solution: Equilibrium curve y = K x = x x Operating curve

24 Nov 2011Prof. R. Shanthini19 Example 3: Determine the N OG using the data given in Examples 1 and 2. (1 – y) LM dy (1 – y)(y – y*) ∫ y out y in N OG = where (1 – y) LM y* – y ln[(1 – y )/(1 – y* )] = Given x, calculate y from the operating curve and y* from the equilibrium curve. Using those values, determine the integral above that gives N OG. Solution:

24 Nov 2011Prof. R. Shanthini20 Example 3: y in = 0.4 y out = The shaded area gives N OG as 3.44 (Numerical integration is better suited to get the answer.)

24 Nov 2011Prof. R. Shanthini21 Example 4: Determine the height of packing Z using the data given in Examples 1 and 2. H OG = Solution: K y a(1 – y) LM S G Z = N OG H OG Need more data to work it out. Can be calculated once H OG is known.

24 Nov 2011Prof. R. Shanthini22 Summary with overall gas-phase transfer coefficients for packed column with concentrated solutions where (1 – y) LM is the log mean of (1 – y) and (1 – y*) given as follows: (1 – y) LM y* – y ln[(1 – y )/(1 – y* )] =(92) K y a(1 – y) LM S G Z = (1 – y) LM dy (1 – y)(y – y*) ∫ y out y in (93) H OG N OG

24 Nov 2011Prof. R. Shanthini23 where (1 – x) LM is the log mean of (1 – x) and (1 – x*) given as follows: (1 – x) LM x* – x ln[(1 – x )/(1 – x* )] =(92) K x a(1 – x) LM S L Z = (1 – x) LM dy (1 – x)(x* – x) ∫ x in x out (93) H OL N OL Summary with overall liquid-phase transfer coefficients for packed column with concentrated solutions

24 Nov 2011Prof. R. Shanthini24 Summary: Equations for Packed Columns for dilute solutions Distributed already: 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.

24 Nov 2011Prof. R. Shanthini25 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.