13 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 5

13 Oct 2011Prof. R. Shanthini2 Liquid phase Liquid film Gas phase p Ab C Ai C Ab p Ai Gas film Mass transport, N A N A = k p (p Ab – p Ai ) = k c (C Ai – C Ab ) p Ai = H A C Ai (53) Summary: Two Film Theory applied at steady-state (52)(59)(51)(62) 1 kpkp HAHA kckc + 1 KGKG p A * = H A C Ab p Ab = H A C A * KLKL = HAHA (60) (57) (58 and 61) = K G (p Ab - p A * ) = K L (C A * - C Ab )

13 Oct 2011Prof. R. Shanthini3 Liquid phase Liquid film Gas phase y Ab x Ai x Ab y Ai Gas film Mass transport, N A N A = k y (y Ab – y Ai ) = k x (x Ai – x Ab ) y Ai = K A x Ai Summary equations with mole fractions y A * = K A x Ab y Ab = K A x A * = K y (y Ab - y A * ) = K x (x A * - x Ab ) 1 kyky KAKA kxkx + 1 KyKy KxKx = KAKA (63) (65) (66) (64) (67)

13 Oct 2011Prof. R. Shanthini4 Notations used: x Ab : liquid-phase mole fraction of A in the bulk liquid y Ab : gas-phase mole fraction of A in the bulk gas x Ai : liquid-phase mole fraction of A at the interface y Ai : gas-phase mole fraction of A at the interface x A * : liquid-phase mole fraction of A which would have been in equilibrium with y Ab y A * : gas-phase mole fraction of A which would have been in equilibrium with x Ab k x : liquid-phase mass-transfer coefficient k y : gas-phase mass-transfer coefficient K x : overall liquid-phase mass-transfer coefficient K y : overall gas-phase mass-transfer coefficient K A : vapour-liquid equilibrium ratio (or equilibrium distribution coefficient)

13 Oct 2011Prof. R. Shanthini5 x Ai y Ai y Ab x Ab yA*yA* xA*xA* xAxA yAyA y Ai = K A x Ai y A * = K A x Ab y Ab = K A x A * Gas-liquid equilibrium ratio (K A ) curve

13 Oct 2011Prof. R. Shanthini6 x Ai y Ai xAxA yAyA y Ai = K A x Ai Gas-liquid equilibrium ratio (K A ) curve How to determine K A ?

13 Oct 2011Prof. R. Shanthini7 x Ai y Ai y Ab x Ab yA*yA* xA*xA* xAxA yAyA slope m y = y Ab - y Ai x A * - x Ai slope m x = y Ai - y A * x Ai - x Ab Gas-liquid equilibrium ratio (K A ) curve - nonlinear

13 Oct 2011Prof. R. Shanthini8 x Ai y Ai y Ab x Ab yA*yA* xA*xA* xAxA yAyA Gas-liquid equilibrium ratio curve (is not linear) slope m x = y Ai - y A * x Ai - x Ab slope m y = y Ab - y Ai x A * - x Ai 1 kxkx 1 mykymyky + 1 KxKx =(67) 1 kyky mxmx kxkx + 1 KyKy =(68) when driving forces for mass transfer are large Derivation is available on page 109 of Reference 2

13 Oct 2011Prof. R. Shanthini9 Gas & Liquid-side Resistances in Interfacial Mass Transfer 1 KLKL 1 H k p =+ 1 kckc 1 KGKG 1 kpkp =+ H kckc f G = fraction of gas-side resistance = 1/K G 1/k p = + H/k c kckc kckc = + H k p f L = fraction of liquid-side resistance = 1/K L 1/k c 1/Hk p 1/k c = + 1/k c + k c /H kpkp = kpkp

13 Oct 2011Prof. R. Shanthini10 If f G > f L, use the overall gas-side mass transfer coefficient and the overall gas-side driving force. If f L > f G use the overall liquid-side mass transfer coefficient and the overall liquid-side driving force. Gas & Liquid-side Resistances in Interfacial Mass Transfer

13 Oct 2011Prof. R. Shanthini11 1 kpkp HAHA kckc + 1 KGKG KLKL = HAHA (58 and 61) The above is also written with the following notations: 1 K OG 1 KGKG =+ H KLKL = H K OL

13 Oct 2011Prof. R. Shanthini12 Tutorial discussed. Drive the relationship between the following under ideal conditions: (i) k p and k y (ii) k c and k x (iii) K G and K y (iv) K L and K x (v) H A and K A

13 Oct 2011Prof. R. Shanthini13 (i) Drive the relationship between k p and k y (y Ab – y Ai ) Therefore, k p = k y (p Ab – p Ai ) Start from N A = k p (p Ab – p Ai ) = k y (y Ab – y Ai ) Since, the partial pressure (p A ) can be related to the mole fraction in vapour phase (y A ) and the total pressure in the vapour phase (P) by p A = y A P for an ideal gas, the above expression can be rewritten as follows: k p = k y (y Ab – y Ai ) (y Ab P – y Ai P) = k y / P k y = k p P Therefore (69)

13 Oct 2011Prof. R. Shanthini14 (ii) Drive the relationship between k c and k x (x Ai – x Ab ) Therefore, k c = k x (C Ai – C Ab ) Start from N A = k c (C Ai – C Ab ) = k x (x Ai – x Ab ) Since, the concentration (C A ) can be related to the mole fraction in liquid phase phase (x A ) and the total concentration in the liquid phase (C T ) by C A = x A C T, the above expression can be rewritten as follows: k c = k x (x Ai – x Ab ) (x Ai C T – x Ab C T ) = k x / C T k x = k c C T Therefore (70)

13 Oct 2011Prof. R. Shanthini15 (iii) Drive the relationship between K G and K y (y Ab – y A * ) Therefore, K G = K y (p Ab – p A * ) Start from N A = K G (p Ab – p A * ) = K y (y Ab – y A * ) using p A = y A P for an ideal gas, the above expression can be rewritten as follows: K G = K y (y Ab – y A * ) (y Ab P – y A * P) = K y / P K y = K G P Therefore (71)

13 Oct 2011Prof. R. Shanthini16 (iv) Drive the relationship between K L and K x (x A * – x Ab ) Therefore, K L = K x (C A * – C Ab ) Start from N A = K L (C A * – C Ab ) = K x (x A * – x Ab ) Since C A = x A C T, the above expression can be rewritten as follows: K L = K x (x A * C T – x Ab C T ) = K x / C T K x = K L C T Therefore (72) (x A * – x Ab )

13 Oct 2011Prof. R. Shanthini17 (v) Drive the relationship between H A and K A Start from the equilibrium relationship p Ai = H A C Ai Since p Ai = y Ai P and C Ai = x Ai C T, the above expression can be written as follows: Therefore y Ai P = H A x Ai C T We knowy Ai = K A x Ai Combining the above, we get the following: K A x Ai P = H A x Ai C T K A = H A C T / P (73) True ONLY for dilute system

13 Oct 2011Prof. R. Shanthini18 Example on calculating Henry’s constant: Use the NH 3 -H 2 O data at 293 K given in the table below to calculate the Henry’s law constant (H A = p A / x A ) at low concentrations of NH 3, where p A is the equilibrium partial pressure of ammonia over aqueous solution having x A mole fraction of ammonia. Wt NH 3 per 100 wts. H 2 O p A (mm Hg)

13 Oct 2011Prof. R. Shanthini19 Solution to Example on calculating Henry’s constant: The mass concentration data given in the table must be converted to mole fraction of ammonia in the liquid. It is done as follows: x A = (moles A) / (moles A + moles water) = (m A / M A ) / [(m A / M A ) + m H2O / M H2O )] = (20 / 17) / [(20 / 17) / 18)] for the first data point = Wt NH 3 per 100 wts. H 2 O x A (mm Hg) p A (mm Hg) H A (mm Hg)

13 Oct 2011Prof. R. Shanthini20 Solution to Example on calculating Henry’s constant:

13 Oct 2011Prof. R. Shanthini21 H A = 591 mm Hg per mol fraction of NH 3 in water Solution to Example on calculating Henry’s constant:

13 Oct 2011Prof. R. Shanthini22 H A = 591 mm Hg mole fraction of NH 3 in water Solution to Example on calculating Henry’s constant: H A = 591 mm Hg per mol fraction of NH 3 in water Henry’s law constant determined above is H A = p A / x A. Use the relationship p A = y A P T, where p A is the partial pressure of NH 3 in air, y A is the mol fraction NH 3 in air and P T is the total pressure. Henry’s law constant therefore becomes H A = y A P T / x A, from which we get K A = y A / x A = H A /P T At 1 atm total pressure, K A = 591 / 760 = mol fraction NH 3 in air / mol fraction NH 3 in water

13 Oct 2011Prof. R. Shanthini23 Example 1

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13 Oct 2011Prof. R. Shanthini25 Example 2

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13 Oct 2011Prof. R. Shanthini27 Example 3

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13 Oct 2011Prof. R. Shanthini32 Example 4

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