Interphase Mass Transfer – see Handout At equilibrium a species will distribute (or “partition”) between two phases. Examples: 1. oxygen (species) will partition between air (gas phase) and water (liquid phase). 2. benzene (species) will partition between ground water (liquid phase) and soil (solid phase).
Equilibria are quantified by specific “distribution laws”. An example of a distribution law is given by Henry’s Law, which is used to calculate the liquid phase concentration of a gas (i.e., gas solubility): p O2 = x O2 H O2 If liquid is water, and gas is rather insoluble in liquid (like oxygen): p O2 ≈ c O2 H O2 M H2O /(c H2O M O2 ) see Question 1 (page 2) for an example calculation.
At equilibrium: Interface Gas Phase Liquid Phase c i = c l O2 (8.5 mg/L) O2 Concentration of oxygen p i = p g O2 (160 mm Hg) O2 Distance
Consider now the physical situation: Two phases are suddenly brought into contact in which concentrations of the species are not in equilibrium. Let us contemplate this situation using the specific example of a gas containing oxygen at a concentration of p g and a liquid containing O2 oxygen at a concentration of c l , such that the O2 species oxygen will be transferring from the gas to the liquid.
Not at equilibrium: Interface Gas Phase Liquid Phase Concentration of oxygen p g O2 c l O2 Distance
What happens? Two-resistance theory proposes that rate of diffusion across a microscopic interface is instantaneous, and that therefore equilibrium at the interface “locally” will be achieved immediately (“local equilibrium”). However, in either bulk solution (distant from the interface) the concentration will not have changed. Thus, a concentration gradient will develop in each phase.
Moving towards equilibrium: Interface Gas Phase Liquid Phase Direction of Flux Concentration of oxygen p g O2 c i O2 p i O2 δ L δ G c l O2 Distance
How do we predict the flux? ∝ Flux Concentration Gradient However, we don’t have values for film thicknesses ( δ G , δ L ) ∝ Flux Concentration Difference So, we can write a proportionality in either phase: Flux (Gas Phase) ∝ (p g – p i ) O2 O2 Flux (Liquid Phase) ∝ (c i – c l ) O2 O2 Note also that if system is at steady-state: Flux (Phase I) = Flux (Phase II) = Φ O2
For each phase we will define a proportionality constant called a mass transfer coefficient such that: Φ O2 = k G (p g – p i ) I.3 O2 O2 Φ O2 = k L (c i – c l ) I.4 O2 O2 (p g – p i ) is the “gas phase driving force” O2 O2 for mass transfer (c i – c l ) is the “liquid phase driving force” O2 O2 for mass transfer
Φ O2 = k G (p g – p i ) O2 O2 Φ O2 = k L (c i – c l ) O2 O2 Good News: k L depends only on liquid phase, while k G depends only 1. on gas phase. Knowing properties of that one phase (e.g., viscosity, 2. density, velocity) allows us to calculate k L , k G empirically. (Really) Bad News: We don’t know values for interface concentrations. 1.
We would really like to have a mass transfer coefficient which is related to the total, overall driving force… ∝ We’d like to have something like Φ O2 (p g – c l )! O2 O2 The problem with (p g – c l ) is that these two O2 O2 concentrations are in different phases (one in terms of pressure and the other in terms of liquid concentration). We cannot compare the two concentrations directly.
However, we can relate the gas phase concentration (p g ) to the liquid phase concentration that would O2 theoretically be in equilbrium with that concentration (c* O2 ). So, instead of the overall driving force being (p g O2 – c l ), it is (c* O2 – c l ). We define an overall mass O2 O2 transfer coefficient K L as: Φ O2 = K L (c* O2 – c l ) I.6 O2 Remember, c* O2 is in equilibrium with p g . If the O2 equilibrium relationship is linear, then I.9 p g = m O2 c* O2 O2
Similarly, we can relate the liquid phase concentration (c l ) to the gas phase concentration that would O2 theorectically be in equilbrium with that concentration (p* O2 ). In this case, we define an overall mass transfer coefficient K G as: Φ O2 = K G (p g – p* O2 ) I.5 O2 In this case, p* O2 is in equilibrium with c l . If the O2 equilibrium relationship is linear, then I.8 p* O2 = m O2 c l O2
Φ O2 = K G (p g – p* O2 ) O2 Φ O2 = K L (c* O2 – c l ) O2 K G – overall mass transfer coefficient in terms of gas phase driving force K L – overall mass transfer coefficient in terms of liquid phase driving force Good News: The concentrations at the interface do not appear 1. in the equations. Bad News: K G and K L depend on properties of both phases 1. (strictly, but we often make a simplification!)
Derive relationship between K L and k G , k L : c* O2 – c l = (c* O2 – c i ) + (c i – c l ) O2 O2 O2 O2 Divide each term by Φ O2 c* O2 – c l (c* O2 – c i ) (c i – c l ) O2 O2 O2 O2 = + Φ O2 Φ O2 Φ O2 Note that p g = m O2 c* O2 and p i = m O2 c i O2 O2 O2 c* O2 – c l (p g – p i ) (c i – c l ) O2 O2 O2 O2 O2 = + Φ O2 m Φ O2 Φ O2
c* O2 – c l (p g – p i ) (c i – c l ) O2 O2 O2 O2 O2 = + Φ O2 m Φ O2 Φ O2 From the definitions of mass transfer coefficients… 1 1 1 = + K L mk G k L Similarly… 1 1 m = + K G k G k L
Consider two extreme cases: 1) gas is very soluble in liquid (NH 3 in H 2 O) m is very small 1 1 m = + K G k G k L K G ≈ k G We can use gas phase properties to calculate an overall mass transfer coefficient
2) gas is very insoluble in liquid (O 2 in H 2 O) m is very large 1 1 1 = + K L mk G k L K L ≈ k L We can use liquid phase properties to calculate an overall mass transfer coefficient
Two mass transfer expressions Φ NH3 = K G (p g – p* NH3 ) NH3 Φ O2 = K L (c* O2 – c l ) O2 can be “simplified” to yield: Φ NH3 = k G (p g – p* NH3 ) NH3 Φ O2 = k L (c* O2 – c l ) O2 Note, we can’t write: Φ O2 = k G (p g – p* O2 ) O2
Φ NH3 = k G (p g – p* NH3 ) NH3 Φ O2 = k L (c* O2 – c l ) O2 (Only) Good News: k L depends only on liquid phase, while k G depends only 1. on gas phase. Knowing properties of that one phase (e.g., viscosity, 2. density, velocity) allows us to calculate k L , k G empirically. We can calculate all the concentrations, in particular 3. p* O2 and c* O2 .
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