Chapter 11 Mass Transfer by Diffusion

Chapter 11 Mass Transfer by Diffusion Theory As with heat transfer, the rate of the transferred quantity in mass transfer is proportiona to the driving force and inversey proportiona to the resistance. For mass transfer by diffusion, the driving force is the chemica potentia difference; but concentration (mo/m 3 ), moe fraction, or pressure differences are usuay used. Thus, the mass transfer rate is: For gases: n i ¼ driving force resistance ¼ c i R Driving force Resistance, R z Equimoar Counterdiffusion p i D ij A R GT Diffusion of i through stagnant j p z p jm i D ij A P R GT For iquids: Driving force Resistance, R Equimoar Counterdiffusion x i z D ij A 1 Diffusion of i through stagnant j x i z x jm D ij A c m c m S. Yanniotis, Soving Probems in Food Engineering. Ó Springer 2008 141

142 11 Mass Transfer by Diffusion For soids: Mass transfer rate For a singe wa For a composite wa n i ¼ c i R n i ¼ c i P R Fat wa Cyindrica wa Spherica wa Resistance, R Surface area, A z A DA r A LM ¼ A 1 A 2 DA LM n A 1 =A 2 r p A DA G ¼ ffiffiffiffiffiffiffiffiffiffiffi A 1 A 2 G Fat packaging fims: For gases For water vapor Resistance, R 22414 z P M A z P MWV A Surface area, A A A where A = surface area perpendicuar to the direction of transfer, m 2 A LM = ogarithmic mean of surface area A 1 and A 2,m 2 A G = geometric mean of surface area A 1 and A 2,m 2 c i = concentration of i, mo/m 3 c m = mean concentration of i and j, mo/m 3 D = diffusion coefficient, m 2 /s D ij = diffusion coefficient of i in j, m 2 /s n i = mass transfer fux, rate, mo/s P = tota pressure, Pa p i = partia pressure of i, Pa p jm = ogarithmic mean pressure difference of j, Pa P M = gas permeabiity, cm 3 cm/s cm 2 atm P MWV = water vapor permeabiity, g cm/s cm 2 Pa R G = idea gas constant, m 3 Pa/mo K T = temperature, K

Review Questions 143 x i = mo fraction of i x jm = ogarithmic mean mo fraction of j z and r = wa thickness, m Review Questions Which of the foowing statements are true and which are fase? 1. Fick s 1 st aw refers to mass transfer by diffusion at steady state. 2. Moecuar diffusion is a phenomenon anaogous to heat transfer by conduction. 3. The resistance to mass transfer increases ineary with diffusivity. 4. Mass diffusivity has the same units as therma diffusivity and kinematic viscosity. 5. Diffusivity in gasses is about 10000 times higher than in iquids. 6. Diffusivity does not vary with temperature. 7. Diffusivity in soids may vary with concentration. 8. The driving force for mass transfer by moecuar diffusion is the difference in chemica potentia. 9. There is buk fow in equimoar counterdiffusion. 10. In equimoar counterdiffusion N i = N j, where N i and N j are the fuxes of gas i and gas j with respect to a fixed position. 11. There is no buk fow in the case of diffusion of gas i through stagnant nondiffusing gas j. 12. For the same driving force, the fux N i in equimoar counterdiffusion is smaer than N i in diffusion of i through stagnant nondiffusing j due to the buk motion of i. 13. When concentrations are diute, the buk fow may be negigibe. 14. Permeabiity refers to the diffusion of a gas in a soid and is used extensivey in cacuating mass transfer in packaging materias. 15. Permeabiity is equa to the product of the diffusion coefficient and the soubiity of the gas in the soid. 16. The difference in partia pressure inside and outside the packaging materia is being used as the driving force for mass transfer cacuations. 17. Permeabiity decreases as the temperature increases. 18. Layers of different materias may be combined in aminates to give a composite materia with good barrier properties for water vapor, gasses, and ight. 19. Poyethyene is a good water vapor barrier and serves as an adhesive to the next ayer. 20. Auminum foi is a good gas barrier

144 11 Mass Transfer by Diffusion Exampes Exampe 11.1 Water vapor is diffusing through a stagnant fim of air at 308C towards a cod surface at 68C, where it is condensed. Cacuate the water vapor diffusion fux if the pressure is 1 atm, the water vapor pressure 10 mm from the cod surface is 3 kpa, and the water vapor diffusion coefficient in the air is 0.26 cm 2 /s. Draw the process diagram: p A1 = 3 kpa p B1 Z =10 mm P =1 atm p A2 p B2 6 C air moecue water moecue State your assumptions: The system is at steady state. There are no eddies. Seect the appropriate equation to cacuate the water vapor diffusion fux. Water moecues diffuse towards the cod pate; air moecues diffuse in the opposite direction. There is a buk movement towards the cod pate to keep the system at constant pressure. Therefore the equation to use is: N w ¼ n w A ¼ p w1 p w2 RA ¼ p w1 p w2 z p am D aw A P R G TA ¼ D awp R G T zp am ðp w1 p w2 Þ (with subscript w for water vapor and a for air).

Exampes 145 Step 4 Find the vaues of partia pressure to use in the above equation: i) The water vapor partia pressure at the interphase at 68C is ii) The partia pressure of air is p w2 ¼ 0:935 kpaðfrom steam tabesþ: p a1 ¼ 101325 3000 ¼ 98325 Pa and p a2 ¼ 101325 935 ¼ 100390 Pa p am ¼ p a1 p a2 98325 100390 n p ¼ a1 p n 98325 ¼ 99353:9 Pa a2 100390 Comment: The arithmetic mean instead of the og mean coud have been used with very itte error, since p a1 and p a2 vaues differ from one another by a sma percentage. Step 5 Substitute vaues and cacuate the water vapor diffusion fux: D awp N w ¼ ðp RTzp w1 p w2 Þ¼ am 0:26x10 4 m 2 =s ð101325 PaÞ ¼ 8314:34 m 3 3000 935 ð Pa=kmo KÞð303 KÞð0:01 mþð99353:9paþ ð ÞPa ¼ ¼ 2:17x10 6 kmo=s m 2 Exampe 11.2 A food product is seaed in a fexibe aminated package with 0.1 m 2 surface area that is made of a poyethyene fim ayer 0.1 mm thick and a poyamide fim ayer 0.1 mm thick. The package is stored at 218C and 75% reative humidity. Cacuate the transfer rate of oxygen and water

146 11 Mass Transfer by Diffusion vapor through the fim at steady state if the partia pressure of O 2 inside the package is 0.01 atm, that outside the package is 0.21 atm, and the water activity of the product inside the package is 0.3. The permeabiity (P M ) of poyethyene and poyamide to O 2 are 228010 11 and 510 11 cm 3 / (s cm 2 atm /cm) respectivey, and the water vapor transmission rate (WVTR) for these materias measured at 37.88C using 90% RH water vapor source and 0% RH desiccant sink are 610 11 and 3710 11 g/(s cm 2 /cm) respectivey. State your assumptions: The convective resistance to mass transfer on the two sides of the package are negigibe compared to the diffusion resistance of the fim. The WVTR at 218C does not differ appreciaby from that at 37.88C. Cacuate the diffusion rate of oxygen: i) Seect the equation to use: n O2 ¼ p O P 2 R ii) Substitute vaues and cacuate the diffusion rate: X R ¼ R poyethyene þ R poyamid ¼ ¼ 22414 z poyeth: þ 22414 z poyam: ¼ 22414 z poyeth: þ z poyam: ¼ AP M poyeth: AP M poyam: A P M poyeth: P M poyam: ¼ 22414 cm3 =mo ð0:01 cmþ ð1000 cm 2 Þ 2280 10 11 cm 3 cm=s cm 2 atm! ð0:01 cmþ þ 5 10 11 ¼ cm 3 cm=s cm 2 atm ¼ 9:83 10 6 þ 4:48 10 9 ¼ 4:49 10 9 atm s=mo n O2 ¼ p O P 2 0:21 0:01 atm ¼ R 4:49 10 9 atm s=mo ¼ 4:45 10 11 mo=s

Exampes 147 Cacuate the diffusion rate of water vapor. i) Seect the equation to use: n w ¼ p w P R with p w the water vapor difference between the inside and the outside of the package and R ¼ z P MWV A ii) Find the water vapor pressure inside and outside the package. The water vapor pressure p w at 218C is 2487 Pa (from steam tabes). The water vapor partia pressure outside the package for 75% reative humidity is: p wo ¼ 75 2487 ¼ 1865:3Pa 100 The water vapor partia pressure inside the package for water activity 0.3 is: p wi ¼ 0:3 x 2487 ¼ 746:1Pa iii) Cacuate the water vapor permeabiity. Water vapor permeabiity for poyethyene can be cacuated from the Water Vapor Transmission Rate as: P M WV poyethyene ¼ WVTR poyeth: p Simiary for poyamide: ¼ 6 10 11 gcm=s cm 2 0:90 6586 0Pa ¼ 1:01 10 14 gcm=s cm 2 Pa P M WV poyamid ¼ WVTR poyeth: ¼ 37 10 11 gcm=s cm 2 P 0:90 6586 0Pa ¼ 6:24 10 14 gcm=s cm 2 Pa

148 11 Mass Transfer by Diffusion iv) Cacuate the tota resistance to water vapor transfer for the aminate: X R ¼ Rpoyet h y e ne þ R poyamid ¼ 0:01 cm ¼ 1:01 10 14 gcm=s cm 2 Pa ð 1000 cm 2Þ 0:01 cm þ 6:24 10 14 gcm=s cm 2 Pa ð 1000 cm 2Þ ¼ ¼ 9:90 10 8 þ 1:60 10 8 ¼ 1:15 10 9 Pa s=g z poyeth: P M WV poyeth: A þ z poyamid P MWV poyamid A ¼ v) Cacuate the water vapor transfer rate: n w ¼ p w 1865:3 746:1 Pa P ¼ R 1:15 10 9 Pa s =g ¼ 9:73 10 7 g=s Comment: Notice that the poyethyene fim ayer contributes the main resistance to water vapor transfer (86.1%), whie poyamide contributes the main resistance to oxygen transfer (99.8%). Exercises Exercise 11.1 Water evaporates from the fat surface of the eaf of a vegetabe and is diffusing away through a stagnant ayer of air. The tota pressure is 101325 Pa and the temperature is 248C. Cacuate the evaporation rate in g/s under the foowing conditions: water activity (a w ) of the eaf surface is 0.98, partia water vapor pressure 5 mm away from the surface of the eaf is 2100 Pa, surface area of the eaf is 50 cm 2. The diffusion coefficient of water vapor in air is 2.610 5 m 2 /s. State your assumptions: The time interva the cacuations are based on is sma so that the water vapor pressure at the surface of the eaf is constant and the system is at steady state.

Exercises 149 Seect the equation to use: Since water vapor diffuses through a stagnant ayer of air, the equation to be used is: N w ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::: Find the water vapor pressure p w at 248C from the steam tabes:... Cacuate the partia water vapor pressure p w1 at the surface of the eaf as: Step 4 Cacuate the partia pressure of air: p w1 ¼ a w : p w ¼ ::::::::::::::::::::::::::::::::::::: p a1 ¼ P p w1 ¼ ::::::::::::::::::::::::::::::::::::::::::::::: p a2 ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: p am ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 5 Cacuate the mass transfer fux: N w ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::kmo=s m 2 Step 6 Cacuate the evaporation rate: n w ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::g=s Exercise 11.2 An aroma compound is encapsuated in a nonporous spherica partice made of a homogenous biopoymer fim. Cacuate the aroma reease rate if the partice diameter is 1 mm, the fim thickness is 0.1 mm, the concentration of the aroma compound is 0.1 g/cm 3 and 0.01g/cm 3 on the inside and outside surface of the partice respectivey, and the diffusion coefficient of the aroma compound in the fim is 110 12 m 2 /s.

150 11 Mass Transfer by Diffusion Draw the process diagram: C 2 C 1 r 1 r 2 A 1 A 2 State your assumptions: The aroma reease rate is diffusion-controed. The reease rate is constant (steady state). Cacuate the resistance to mass transfer in a spherica wa. i) Cacuate the geometric mean area of A 1 and A 2 : A 1 ¼ pd 2 ¼ ::::::::::::::::::::::::::::::::::::::::::::::::: A 2 ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: A G ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ii) Cacuate the resistance: R ¼ r DA G ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 4 Cacuate the diffusion rate through the wa of the sphere: n ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

Exercises 151 Exercise 11.3 The soubiity and the diffusivity of O 2 in poyviny choride are 3.84x10 4 cm 3 O 2 at STP/cmHg.cm 3 soid and 1.18x10 12 m 2 /s respectivey. Cacuate the permeabiity of poyviny choride to O 2. The permeabiity is the product of soubiity and diffusivity. Therefore: P M ¼ S:D ¼:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: cm3 cm scm 2 Pa Exercise 11.4 A suppier of frozen hamburger patties is thinking of packaging them in 0.1 mm thick poypropyene fim. The permeabiity of poypropyene to O 2 and water vapor is 1x10 8 cm 3 /(s cm 2 atm /cm) and 8x10 15 g/(s cm 2 Pa/cm) respectivey. The partia pressure of O 2 is 0 atm and that of water vapor is 0.12 Pa inside the package, whie outside the package the partia pressure of O 2 is 0.21 atm and that of water vapor is 0.04 Pa. Cacuate how much oxygen and how much water vapor wi pass through the poypropyene fim of 600 cm 2 surface area in one week. Sate your assumptions: The convective resistance to mass transfer on the two sides of the package is negigibe compared to the diffusion resistance of the fim.... Cacuate the transfer rate of O 2. i) Cacuate the resistance: R ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ii) Cacuate the transfer rate of oxygen: n O2 ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::mo=s Cacuate the tota amount of O 2 that wi diffuse into the package in one week: m O2 ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: g

152 11 Mass Transfer by Diffusion Step 4 Cacuate the transfer rate of water vapor. i) Cacuate the resistance: R ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ii) Cacuate the transfer rate of water vapor: n w ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::g=s Step 5 Cacuate the tota amount of water vapor that wi diffuse out of the package in one week: m w ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: g Exercise 11.5 Meat is packaged in a pastic disk covered with PVC fim. Cacuate the required thickness of the fim so that 110 4 mo of oxygen enters the package through the fim in 24 hours. The permeabiity of PVC to oxygen is 45610 11 cm 3 / (s cm 2 atm /cm), the surface area of the fim is 300 cm 2, and the partia pressure of oxygen inside and outside the package is 0 atm and 0.21 atm respectivey. State your assumptions:... Cacuate the mass transfer rate in mo/s: n O2 ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Cacuate the required resistance: R ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Step 4 Cacuate the required thickness of the fim z ¼ P M A 22414 R ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

Exercises 153 Exercise 11.6 Potato chips are packed in a pastic bag in a nitrogen atmosphere. Cacuate the amount of N 2 that wi diffuse out of the pastic bag in 3 months if the thickness of the fim is 0.05 mm, the absoute pressure of N 2 in the bag is 1.05 atm, the surface area of the bag is 1000 cm 2, and the permeabiity of the pastic fim to N 2 is 1.510 10 cm 3 /s cm 2 atm/cm. State your assumptions:... Cacuate the resistance: R ¼ ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Cacuate the transfer rate of N 2 : n N2 ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::mo=s Step 4 Cacuate the tota amount of N 2 that wi diffuse out of the package in three months: m N2 ¼ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: g Verify that the pressure in the bag wi not change appreciaby due to N 2 diffusion. Hint: Use the idea gas aw.