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CP302 Separation Process Principles

CP302 Separation Process Principles. Mass Transfer - Set 8. Treated gas G out , y out. Inlet solvent L in , x in. We have learned to determine the height of packing in packed columns with dilute solutions in the last lecture class. G y. L x. Control volume.

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CP302 Separation Process Principles

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  1. CP302 Separation Process Principles Mass Transfer - Set 8 Prof. R. Shanthini

  2. Treated gas Gout, yout Inlet solvent Lin, xin We have learned to determine the height of packing in packed columns with dilute solutions in the last lecture class. G y L x Control volume Today, we will learn to determine the height of packing in packed columns with concentrated solutions. Inlet gas Gin, yin Spent solvent Lout, xout Prof. R. Shanthini

  3. Notations Gs - inert gas molar flow rate (constant) Ls - 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 = Gs Y = G y Solute in the liquid phase = Ls X = L x Prof. R. Shanthini

  4. Equations for Packed Columns Treated gas Gout, yout Inlet solvent Lin, xin Mass of solute lost from the gas over the differential height of packing dz = G y - G (y + dy) = - G dy L x G y was used for packed column with dilute solution assuming G can be taken as a constant for dilute solutions. dz L x+dx G y+dy Z z For concentrated solution we ought to use Gs (inert gas molar flow rate) which is constant across the column along with Y (mole ratio of solute A in gas). Inlet gas Gin, yin Spent solvent Lout, xout Prof. R. Shanthini

  5. Equations for Packed Columns with concentrated solutions Treated gas Gout, yout Inlet solvent Lin, xin Mass of solute lost from the gas over the differential height of packing dz = Gs Y - Gs (Y + dY) = - Gs dY (85) Gs Y Ls X Relate Gs to G: dz Gs = G (1 – y) (86) Ls X+dX Z Gs Y+dY z Relate Y to y: From y = Y / (Y+1), we get Y = y / (1 – y) (87) Inlet gas Gin, yin Spent solvent Lout, xout Prof. R. Shanthini

  6. Equations for Packed Columns with concentrated solutions Treated gas Gout, yout Inlet solvent Lin, xin 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: Gs Y Ls X • Gs dY = - G (1 – y) d[y / (1 – y)] dz Ls X+dX dy Z = - G (1 – y) Gs Y+dY (1 – y)2 z dy = - G (88) (1 – y) Inlet gas Gin, yin Spent solvent Lout, xout Prof. R. Shanthini

  7. Equations for Packed Columns with concentrated solutions Treated gas Gout, yout Inlet solvent Lin, xin Mass of solute transferred from the gas to the liquid = Kya (y – y*) S dz where S is the inside cross-sectional area of the tower. Gs Y Ls X dz Relating (88) to the above at steady state, we get Ls X+dX Z Gs Y+dY z dy -G = Kya (y – y*) S dz (89) (1 – y) Compare (89) with (79) used for packed columns with dilute solutions.What are the differences? Inlet gas Gin, yin Spent solvent Lout, xout Prof. R. Shanthini

  8. Equations for Packed Columns with concentrated solutions Treated gas Gout, yout Inlet solvent Lin, xin Rearranging and integrating (89) gives the following: yin ∫ dy G (90) Z = Gs Y Ls X KyaS (1 – y)(y – y*) yout dz Ls X+dX Z Gs Y+dY z Inlet gas Gin, yin Spent solvent Lout, xout Prof. R. Shanthini

  9. Equations for Packed Columns with concentrated solutions Multiply the numerator and denominator of (90) by (1 – y)LM: yin ∫ (1 – y)LM dy G Z = (91) Kya(1 – y)LMS (1 – y)(y – y*) yout where (1 – y)LM is the log mean of (1 – y) and (1 – y*) given as follows: y* – y (1 – y)LM = (92) ln[(1 – y)/(1 – y*)] Prof. R. Shanthini

  10. Equations for Packed Columns with concentrated solutions 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 / [Kya(1 – y)LMS] out of the integral sign in (91) without incurring errors larger than those inherent in experimental measurements of Kya. (Usually average values of G and (1 – y)LM are used.) Therefore (91) becomes the following: yin ∫ (1 – y)LM dy G Z = (93) Kya(1 – y)LMS (1 – y)(y – y*) yout NOG HOG Prof. R. Shanthini

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

  12. Equations for Packed Columns with concentrated solutions Treated gas Gout, yout Inlet solvent Lin, xin The operating equation for the packed column is obtained by writing a mass balance for solute over the control volume: Lin xin + G y = L x + Gout yout (74) G y L x If dilute solution is assumed, then Lin = L = Lout and Gin = G = Gout. Control volume We somehow have to relate L to Lin and G to Gout in (74). Inlet gas Gin, yin Spent solvent Lout, xout Prof. R. Shanthini

  13. Equations for Packed Columns with concentrated solutions Treated gas Gout, yout Inlet solvent Lin, xin To relate L to Lin, write a mass balance for solvent over the control volume: Lin (1 – xin) = L (1 – x) L = Lin (1 – xin)/ (1 – x) (95) G y L x To relate G to Gout, write the overall mass balance over the entire column: Control volume G + Lin = Gout + L (96) Inlet gas Gin, yin Spent solvent Lout, xout Prof. R. Shanthini

  14. Equations for Packed Columns with concentrated solutions Use (95) to eliminate L from (96): G + Lin = Gout + Lin (1 – xin)/ (1 – x) G = Gout + Lin (x – xin) / (1 – x) (97) Combining (74), (95) and (97), we get the following: Lin xin + [Gout+Lin(x – xin)/ (1 – x)] y = Lin(1 – xin)x/(1 – x) + Gout yout Gout yout + [Lin (1 – xin) x / (1 – x)] - Lin xin y = (98) Gout + [Lin(x – xin) / (1 – x)] 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? Prof. R. Shanthini

  15. Equations for Packed Columns with concentrated solutions If the solvent fed to the column is pure then xin = 0. Therefore, (98) becomes Gout yout + [ Lin x / (1 – x) ] y = (99) Gout + [ Lin x / (1 – x) ] Prof. R. Shanthini

  16. 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: Lin = 488 lbmol/h; Gin = 100 lbmol/h; yin = 0.4; xin = 0 In the inlet air stream: ammonia = 40 lbmol/h; air = 60 lbmol/h Ammonia removed from the air stream = 0.95 x 40 = 38 lbmol/h In the outlet air stream: ammonia = 02 lbmol/h; air = 60 lbmol/h Therefore, Gout = (60+2) = 62 lbmol/h, and yout = 2 / 62 = 0.0323 Prof. R. Shanthini

  17. Example 1: Using Lin = 488 lbmol/h, Gout = 62 lbmol/h, yout = 0.0323 and xin = 0 in (98), we get the operating curve as follows: 62 x 0.0323+ [488 x / (1 – x)] y = 62 + [488x / (1 – x)] yin (bottom of the tower) Operating curve yout (top of the tower) Prof. R. Shanthini

  18. Example 2: Draw the equilibrium curve, which is approximately described by K = 44.223x + 0.4771, on the same plot as in Example 1. Solution: Operating curve Equilibrium curve y = K x = 44.223x2 + 0.4771x Prof. R. Shanthini

  19. Example 3: Determine the NOG using the data given in Examples 1 and 2. Solution: yin ∫ (1 – y)LM dy NOG = (1 – y)(y – y*) yout y* – y where (1 – y)LM = 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 NOG. Prof. R. Shanthini

  20. Example 3: The shaded area gives NOG as 3.44 (Numerical integration is better suited to get the answer.) yin = 0.4 yout = 0.0323 Prof. R. Shanthini

  21. Example 4: Determine the height of packing Z using the data given in Examples 1 and 2. Solution: G Need more data to work it out. HOG = Kya(1 – y)LMS Z = NOG HOG Can be calculated once HOG is known. Prof. R. Shanthini

  22. Summary with overall gas-phase transfer coefficients for packed column with concentrated solutions yin ∫ (1 – y)LM dy G Z = (93) Kya(1 – y)LMS (1 – y)(y – y*) yout NOG HOG where (1 – y)LM is the log mean of (1 – y) and (1 – y*) given as follows: y* – y (1 – y)LM = (92) ln[(1 – y)/(1 – y*)] Prof. R. Shanthini

  23. Summary with overall liquid-phase transfer coefficients for packed column with concentrated solutions xout ∫ (1 – x)LM dy L Z = (93) Kxa(1 – x)LMS (1 – x)(x* – x) xin NOL HOL where (1 – x)LM is the log mean of (1 – x) and (1 – x*) given as follows: x* – x (1 – x)LM = (92) ln[(1 – x)/(1 – x*)] Prof. R. Shanthini

  24. 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. Prof. R. Shanthini

  25. 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

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