Mass Transfer.

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Presentation transcript:

Mass Transfer

Mass Transfer Fick’s First Law Flux Expression Typical mass transfer model

Mass Transfer Flux due to diffusion NA DZ Flux due to bulk molar flow Gas B Diffusion Plane Liquid A Stagnant gas film Gas insoluble in Liquid Equilibrium at interface

Mass Transfer

Mass Transfer

Diffusion Coefficients Gas Phase - 0.20 cm2 /s Liquid Phase - 1.5 E-5 cm2 /s Solid Phase – 1.0E-8 cm2 /s

Gas Diffusion Kinetic Theory of Gases R 3/2 T 3/2 DAB = MA1/2 dA P

Gas Model (Fuller)

Gas Diffusion Example Determine the diffusion coefficient of n-pentane in air at 30 F and P = 1 atm.

Multi-component Gas Diffusion

Liquid Diffusion Eyring / Stokes Einstein Theory K T DAB = 6pr mB

Dilute Liquid Diffusion (Wilke-Chang)

Liquid Diffusion Example Determine the diffusion coefficient of a dilute concentration of n-pentane in water at 40 F.

Concentrated Liquids (Vignes)

Two Resistance Theory xAi Bulk Gas yAG Gas Film Liquid Film Bulk Liquid xAL yAi Interface

Controlling Resistance Gas Film Liquid Film Interface

Two Resistance Theory (Dilute Solution) y = m x yAG yAi yA* xAL xAi xAL*

Dilute Solution Example Ammonia is being absorbed in a solvent stream. The overall gas phase coefficient (Ky) is 0.205 lbmol NH3 / hr ft2 atm. The bulk gas phase ammonia concentration is 8 mole % and the liquid bulk phase concentration is 0.004 mole / ft3. T = 68 F and P = 1 atm. 85 % of the mass transfer resistance in in the gas phase. The Henry’s law constant is 0.215 atm ft3 / lbmol NH3. Find: kx, ky, Pai and Cai

Two Resistance Theory (Flux Expression - General Case)

Two Resistance Theory (Overall Coefficients - General Cases)

Concentrated Solution Example A wetted wall absorption tower (2.54 cm diameter) is fed water and an ammonia-air mixture. At a particular level in the tower, the ammonia concentration in the bulk gas is 0.80 mole fraction and the mole fraction in the bulk liquid is 0.05. The temperature is 26.7 ºC and the pressure is 1 atm. The rates of flow are such that the local mass transfer coefficient in the liquid is kL= 2.87 X 10-5 kmol /m2 s (kmol/ m3). The local Sherwood number for the gas is 40. The diffusivity of ammonia in air is 2.297 X 10-5 m2 /s Compute the local mass transfer flux for the ammonia absorption.

Mass / heat / momentum analogies Laminar Reynolds Turbulent Prandtl Von Karman

Mass / heat / momentum analogies Chilton-Colburn “j” factor Friction factor (f) vs Re for pipe flow

Empirical Models Wetted Wall Tower

Empirical Models Laminar Turbulent Transition Flat Plate

Empirical Models Single Sphere

Example Water is stored in a buffalo bladder that has a diameter of 0.3 meters. The temperature of the water is 25C. Air at 25 C and a relative humidity of 20 % flows over the outside of the bladder at a velocity of 0.1 meter / sec. Calculate the time needed to cool the water from 25 C to 22 C.

Non-Steady State Mass Transfer

Non-Steady State Mass Transfer Sphere – constant surface concentration

Non-Steady State Mass Transfer Figure 3.11 Concentration profiles for un-steady state diffusion in a sphere.

Non-Steady State Mass Transfer Figure 3.9 Average concentrations for non-steady state diffusion.

Non-Steady State Mass Transfer Sphere – convective mass transfer at surface

Non-Steady State Mass Transfer

Example A deadly black widow spider is 1 foot away from my toe moving at 0.1 ft / sec when I spray it with Raid. I spray enough Raid on the spider to coat its outside shell to a concentration of 1 kgmol / m3. Raid researchers have found that the spider will die when the Raid concentration at the center of the spider reaches 0.56 kgmol/m3. Should I call 911 ??? Data: DRaid-Bug = 1.0 E-6 m2 / sec Diameter of spider = 0.01 m

10 Minute Problem A spherical 3 mm diameter pellet containing insulin is immobilized in the lower intestine. The insulin is quickly transferred from the pellets surface such that the surface concentration is maintained at zero. Determine the time required to reduce the average insulin concentration in the pellet from 2 kgmol / m3 to 0.2 kgmol/m3. Deff = 3.0 X 10 –7 cm2 /sec