L I II PCO2 = 76 mmHg PCO2 = 456 mmHg d
Molecular Diffusion in Gases
Molecular Diffusion of Helium in Nitrogen
Example 6.1-1, page 413
A mixture of He and N2 gas is contained in a pipe at 298K and 1.0atm total pressure which is
constant throughout. At one end of the pipe at point 1, the partial pressure PA1 of He is 0.60atm
and at the other end 0.2m, PA2 is 0.20 atm. Calculate the flux of He at steady state if DAB is
0.687 x 10-4 m2/s. PA1 = 0.6 atm PA2 = 0.2 atm P = 1 atm R = 82.057 m3 atm/kgmol K T = 298K z2-z1 = 0.2m DAB = 0.687 x 10-4 m2/s
DAB for a gas is constant; P is constant meaning C is also constant; flux is constant at steady state. Flux,
1 2 2 1 z z C C D J AB A A AZ if C = P/RT
1 2 2 1 z z RT P P D J AB A A AZ Substitute in values:
6 4 10 63 . 5 2 . 0 298 057 . 82 2 . 0 6 . 0 10 687 . 0 AX J kgmol/m2sEquimolar Counterdiffusion
Example 6.2-1, page 415Ammonia gas (A) is diffusing through a uniform tube 0.10m long containing N2 gas (B) at 1.0132x105 Pa pressure and
298K. At point 1, PA1 = 1.013x104 Pa and at point 2,
PA2 = 0.507x104 Pa. DAB = 0.230x10-4 m2/s.
Calculate the flux J*A and J*B at steady state.
PA1 = 1.013x104 Pa PA2 = 0.507x104 Pa DAB = 0.230x10-4 m2/s T = 298K P = 1.0132x105 Pa z2 – z1 = 0.10 m R = 8314.3 m3 Pa/kgmol K
7 4 4 4 1 2 2 1 10 70 . 4 1 . 0 298 3 . 8314 10 507 . 0 10 013 . 1 10 230 . 0 z z RT P P D J AB A A AZ kgmol A/m 2 sH2O
Diffusion of Water Through Stagnant, Nondiffusing Air
Example 6.2-2 page 419
Water in the bottom of a narrow metal tube is held at a constant temperature of 293K. The total pressure of air (assumed dry) is 1.01325x105 Pa (1.0atm) and the temperature is 293K.
Water evaporates and diffuses through the air in the tube, and the diffusion path z2 – z1 is 0.1524 m long. Calculate the
rate of evaporation at steady state. The diffusivity of water vapor at 293K and 1 atm is 0.250x10-4 m2/s. Assume that the system is isothermal.
P = 1.01325x105 Pa = 1.0 atm T = 293K
z2 – z1 = 0.1524m DAB = 0.250x10-4 m2/s
Vapor pressure of water, PA1 = 0.0231 atm Water pressure in dry air, PA2 = 0 atm R = 82.057 m3 atm/kgmol K
988 . 0 0231 . 0 1 0 1 ln 0231 . 0 1 0 1 ln 1 2 1 2 B B B B BM P P P P P atm
1 2
1 2 A A BM AB A P P P z z RT P D N
7 4 10 595 . 1 0 0231 . 0 988 . 0 1524 . 0 293 057 . 82 0 . 1 10 250 . 0 A N kgmol/m2sDiffusion in a Tube with Change in Path Length
Example 6.2-3, page 419
Water in the bottom of a narrow metal tube is held at a constant temperature of 293K. The total pressure of air (assumed dry) is 1.01325x105 Pa (1.0atm) and the temperature is 293K. At a given time, t, the level is z meters from the top. As water vapor diffuses through the air, the level drops slowly. Derive the equation for the time tF for the level to drop from a starting point of zo m at t = 0 and zF at t = tF seconds.
We can assume a pseudo-steady state condition because the level drops slowly. Now, both NA and z are variables.
1 2
1 2 A A BM AB A P P P z z RT P D N Assuming a cross-sectional area of 1 m2, the level drops dt in dz seconds, and the leftover kgmol of A is PAdz/MA: dt M dz N A A A 1 1
Rearranging and integrating:
F tF A A BM AB z z A A dt P P RTP P D zdz M 0 1 2 0
1 2
2 0 2 2 A AB A A BM F A F P P P D M RTP z z t Stagnant B r2
Diffusing A
r1
Diffusion Through a Varying Cross-Sectional Area – Evaporation Derivation
HW1.1 In the lecture we showed that the molar rate of material A evaporating from a spherical drop immersed in material B could be written as
,2 1 , 2 1 ln 1 1 4 A A AB A p P p P RT P D r r N where A
N is the molar rate of material A leaving the drop
r1 and r2 are two radial points away from
the sphere center
DAB is the diffusion coefficient P is the total
system pressure
R is the ideal gas constant T is the system temperature
pA,1 is the partial pressure of A at point 1
pA,2 is the partial pressure of A at point 2
Starting with this equation, derive the following approximate equation for the molar flux at the particle surface, NA,1
,1 ,2
1 1 , 2 A A AB A c c D D N where D1 is the diameter of the spherical drop
cA,1 is the molar concentration of material A at the surface of the drop
cA,2 is the molar concentration of material A far from the drop
Assume point 1 is at the drop’s surface
Assume point 2 is very far away from the drop, so r2 >> r1
1 2 2 1 ln 1 1 4 A A AB A P P P P RT P D r r N 1 2 1 ln 4 A A AB A P P P P RT P D r N
Now multiply the equation by 1/r1, rewrite the radius as 2/D on the right side, substitute for surface flux, NAs, and PBM:
1 2 2 1 1 2 1 2 ln ln A A A A B B B B BM P P P P P P P P P P P and 2 1 4 r N N A AS So we rewrite the equation as:
BM A A AB AS P P P DRT D N 2 1 2
Assume low vapor pressure, so PA1, PA2 << P and use a Taylor Series approximation ln(x) = x-1:
1 2 A A P P P P is approaching 1, so ln 1 1 2 1 2 A A A A P P P P P P P P Now,
P P P P P P P P P P P P P A A A A A A A A 1 2 1 2 1 1 1 2 so PBM ≈ P. Lastly, PA1 = CA1RT and PA2 = CA2RT:
1 2
1 2
1 2
2 2 2 A A AB A A AB A A BM AB AS C C D D RT C RT C RTD D P P P P RTD D N Evaporation of a Napthalene Sphere
Example 6.2-4, page 421
A sphere of naphthalene having a radius of 2.0mm is suspended in a large volume of still air at 318K and 1.01325x105 Pa (1.0atm). The surface temperature of the naphthalene can be assumed to be at 318K and its vapor pressure at 318K is 0.555 mmHg. The DAB of naphthalene in air at 318K is
6.92x10-6 m2/s. Calculate the rate of evaporation of naphthalene from the surface.
r = 2.0mm T = 318K
P = 1.01325x105 Pa DAB = 6.92x10-6 m2/s
Vapor pressure, PA1 = 0.555 mmHg = 74 Pa PA2 = 0 Pa R = 8314.3 m3 Pa/kgmol K
5 5 5 5 5 1 2 1 2 10 0129 . 1 74 10 1.01325 0 10 1.01325 ln 74 10 1.01325 0 10 1.01325 ln B B B B BM P P P P P Pa
5
8 5 6 2 1 0 1 10 68 . 9 10 0129 . 1 002 . 0 318 3 . 8314 10 01325 . 1 10 92 . 6 A A BM AB A P P P r r RT P D N kgmol A/m2s6.2-5.)
Mass Transfer from a Napthalene Sphere to Air
Mass transfer is occurring from a sphere of naphthalene having a radius of 10mm. The sphere is in a large volume of still air at 52.6ºC and 1atm absolute pressure. The vapor pressure of naphthalene at 52.6ºC is 1.0 mmHg. The diffusivity of naphthalene in air at 0ºC is 5.16 x 10-6 m2/s. Calculate the rate of evaporation of naphthalene from the surface in kg mol/s m2. [Note: the diffusivity can be corrected for temperature by using the temperature correction factor from the Fuller et al. equation (6.2-45)] Assumptions:
the system is at steady-state, so the radius of the sphere is not changing point 1 is at the surface of the sphere and point 2 is very far away , so r2 >> r1
R = 8314 m3 Pa/kgmol K T = 325.75 K
r = 0.01 m
PA1=(1.0mmHg)(1.01325 x 105 Pa/760 mmHg) = 133.322 Pa
PA2 = 0 Pa because the air is still First calculate DAB at the new temperature:
DAB(52.6ºC) = DAB(0ºC) 1 1 2 75 . 1 1 2 P P T T = (5.16 x 10-6 m2/s)(325.75/273.15)1.75 = 7.023 x 10-6 m2/s
Now solve for NA:
1 2 2 1 ln 1 1 4 A A AB A P P P P RT P D r r N 1 2 1 ln 4 A A AB A P P P P RT P D r N
0 10 01325 . 1 322 . 133 10 01325 . 1 ln 75 . 325 8314 10 01325 . 1 10 023 . 7 1 . 0 4 5 5 5 6 A N = 4.34736 x 10-11 kgmol/sNow solve for NA = NA/A = (4.34736 x 10 -11
6.2-9
Time to Completely Evaporate a Sphere
A drop of liquid toluene is kept at a uniform temperature of 25.9ºC and is suspended in air by a fine wire. The initial radius r1 = 2.00mm. The vapor pressure of toluene at 25.9ºC is PA1 = 3.84
kPa and the density of liquid toluene is 866 kg/m3.
(a) Derive Eq. (6.2-34) to predict the time tF for the drop to evaporate completely in a large
volume of still air. Show all steps.
(b) Calculate the time in seconds for complete evaporation.
(a) Equations for flux:
BM A A AB A P P P RTr P D r N 1 2 2 4 and M dt dV N A A A
Volume of the sphere, 3 3
4 r
V so the derivative of volume is: 2 4 r dr dV
Plug this volume derivative into molar flux equation for dV:
dt M dr r N A A A 2 4
Now set N equal in each equation: A
BM A A AB A A P P P RTr P D r dt M dr r 1 2 2 2 4 1 4
Separate variables and integrate both sides:
r A A AB A BM A t rdr P P P D M P RT dt 0 2 1 0
2 1 2 2 A AB A A BM A P P P D M P RT r t (b) Now use the equation to find the time in seconds: PA1 = 3840 Pa PA2 = 0 Pa P = 1.01325 x 105 Pa MA = 92.14 kg/kgmol DAB = 0.86 x 10-4 m2/s R = 8314 m3 Pa/kgmol K T = 299.05 K ρ = 866 kg/m3 r1 = 0.002 m Solve for PBM:
1 2 2 1 ln A A A A BM P P P P P P P = 99392.6 Pa tevap =
2 1 2 2 A AB A A BM A P P P D M P RT r =
0 3840 10 086 . 0 14 . 92 2 6 . 99392 05 . 299 8314 002 . 0 866 4 2 1388.23 seconds6.2-10
Diffusion in a Nonuniform Cross-Sectional Area – Changing Pipe Diameter
The gas ammonia (A) is diffusing at steady state through N2 (B) by equimolar counterdiffusion in a conduit
1.22m long at 25ºC and a total pressure of 101.32 kPa abs. The partial pressure of ammonia at the left end is 25.33 kPa and at the other end is 5.066 kPa. The cross section of the conduit is in the shape of an
1.22m
.0305m .061m
uniformly to 0.0305m at the right end. Calculate the molar flux of ammonia. The diffusivity is DAB = 0.230 x
10-4 m2/s.
Area = tan 4 1 2
b (60º)
To find an equation for how b changes with the length, create a linear fit based on two points: At the first end: (0, 0.061) and at the other end: (1.22, 0.0305)
Slope = -0.0305/1.22 = -0.025 b = -0.025L + 0.061
So now Area =
0.025 0.061
tan 4 1 2 L (60º) = 0.000271 L2 – 0.001321 L + 0.001611 dL dP RT D A NA AB A dL dP RT D NA AB A 0.001611 L 0.001321 -L 0.000271 2 Now separate and integrate both sides:
2 1 22 . 1 0 0.000271L2 -0.001321L 0.001611 PA PA A AB A dP RT D dL N PA1 = 25.33 kPa PA2 = 5.066 kPa DAB = 0.230 x 10-4 m2/s R = 8314 m3 Pa/kgmol K T = 298.15 KNow plug into integrate equation:
15 . 298 8314 25330 5066 10 230 . 0 8 . 1514 4 1 2 RT P P D N AB A A A NA = 1.24 x 10 -10 kgmol/sNow solve for NA:
NA = N /A = A
60 tan 0610 . 0 025 . 0 4 1 10 24 . 1 2 10 Molecular Diffusion in Liquids
Diffusion of Ethanol (A) Through Water (B)
Example 6.3-1, page 429
An ethanol(A)-water(B) solution in the form of a stagnant film 2.0mm thick at 293K is in contact at one surface with an organic solvent in which ethanol is soluble and water is insoluble. Hence, NB = 0. At
point 1, the concentration of ethanol is 16.8 wt % and the solution density is ρ1 = 972.8 kg/m3. At
point 2, the concentration of ethanol is 6.8 wt % and ρ2 = 988.1 kg/m3. The diffusivity of ethanol is
0.740x10-9 m2/s. Calculate the steady-state flux NA.
DAB = 0.740x10-9 m2/s T = 293K CA1 = 16.8 CA2 = 6.8 ρ1 = 972.8 kg/m3 ρ2 = 988.1 kg/m3 Meth =46.05 Mwater = 18.02 Calculate the mole fractions, taking a basis of 100 kg:
0732 . 0 02 . 18 2 . 83 05 . 46 8 . 16 05 . 46 8 . 16 1 A X 0.0277 02 . 18 2 . 93 05 . 46 8 . 6 05 . 46 8 . 6 2 A X 9268 . 0 0732 . 0 1 1 1 1 A B X X XB2 1XA2 10.02770.9723
Now calculate the molecular weight: 07 . 20 02 . 18 2 . 83 05 . 46 8 . 16 100 1 kg M kg/kgmol 18.75 02 . 18 2 . 93 05 . 46 8 . 6 100 1 kg M kg/kgmol
Now calculate CAvg:
6 . 50 2 75 . 18 / 1 . 988 07 . 20 / 8 . 972 2 / / 1 2 2 1 M M Cavg kgmol/m3 949 . 0 9723 . 0 9268 . 0 ln 9268 . 0 9723 . 0 ln 1 2 1 2 B B B B BM X X X X X
7 9 2 1 1 2 10 99 . 8 949 . 0 002 . 0 0277 . 0 0732 . 0 6 . 50 10 740 . 0 A A BM avg AB A x x x z z C D N kgmol/m2s6.3-2 Diffusion of Ammonia in an Aqueous Solution
An ammonia (A) – water (B) solution at 278K and 4.0mm thick is in contact at one surface with an organic liquid at this interface. The concentration of ammonia in the organic phase is held constant and is such that the equilibrium concentration of ammonia in the water at this surface is 2.0 wt % ammonia (density of aqueous solution 991.7 kg/m3) and the concentration of ammonia in water at the other end of the film 4.0 mm away is 10 wt% (density = 961.7 kg/m3). Water and the organic are insoluble in each other. The diffusion coefficient of ammonia in water is 1.24 x 10-9 m2/s.
(a) At steady state, calculate the flux NA in kg mol/s m2
(b) Calculate the flux NB. Explain.
(a) ρ1 = 991.7 kg/m3 ρ2 = 961.7 kg/m3 M1 = 17.9806 kg/kgmol M2 = 17.9031 kg/kgmol T = 278 K CA1 = 0.02 CA2 = 0.1 z1 = 0.004 m z2 = 0 m Calculate Cavg: 9031 . 17 7 . 961 9806 . 17 7 . 991 2 1 2 1 2 2 1 1 M M Cavg 54.435 kgmol/m3 Calculate XBM: Xw1 = 1-0.02 = 0.98 Xw2 = 1-0.1 = 0.9
98 . 0 / 9 . 0 ln 98 . 0 9 . 0 ln 2 1 1 2 B B B B BM X X X X X 0.939432Now calculate flux:
BM A A avg AB A X z z X X C D N 1 2 2 1 =
939432 . 0 004 . 0 0 1 . 0 02 . 0 435 . 54 10 24 . 1 9 = 1.437 x 10-6 kgmol/m2 sPrediction of Diffusivities in Liquids
Prediction of Liquid Diffusivity
Example 6.3-2, page 432
Predict the diffusion coefficient of acetone in water at 25ºC and 50ºC using the Wilke-Chang equation. The experimental value 1.28x10-9 m2/s at 298K.
From Appendix A.2, the viscosity of water at 25ºC is μB = 0.8937 x 10-3 Pa s and at 50ºC is 0.5494 x 10-3 Pa s. From Table 6.3-2 for CH3COCH3 with 3 carbons + 6 hydrogens + 1 oxygen.
VA = 3(0.0148) + 6(0.0037) + 1(0.0074) = 0.0740 m3 kgmol
For the water association parameter φ = 2.6 and MB = 18.02 kg mass/kgmol. For 25ºC:
0.6 9 2 / 1 16 6 . 0 2 / 1 16 10 277 . 1 0740 . 0 8937 . 0 298 02 . 18 6 . 2 10 173 . 1 10 173 . 1 A B B AB V T M D m2/s For 50ºC:
0.6 9 2 / 1 16 6 . 0 2 / 1 16 10 251 . 2 0740 . 0 5494 . 0 323 02 . 18 6 . 2 10 173 . 1 10 173 . 1 A B B AB V T M D m2/sPrediction of Diffusivities of Electrolytes in Liquids
Diffusivities of Electrolytes
Example 6.3-3, page 434
Predict the diffusion coefficients of dilute electrolytes for the following cases: (a) For KCl at 25ºC, predict DºAB and compare with the value in Table 6.3-1
(b) Predict the value of KCl at 18.5ºC. The experimental value is 1.7x10-5 cm2/s
(c) For CaCl2 predict DAB at 25ºC. Compare with the experimental value of 1.32x10-5 cm2/s; also,
predict Di of the ion Ca2+ and of Cl- and use Eq. 6.3-12
(a) From Table 6.3-3, λ+ (K+) = 73.5 and λ- (Cl-) = 76.3:
10
5 10 10 993 . 1 3 . 76 / 1 5 . 73 / 1 1 / 1 1 / 1 298 10 928 . 8 / 1 / 1 / 1 / 1 10 928 . 8 n n T DAB cm2/s (b) To change the temperature, we use a simple correction factor:T = 18.5ºC = 291.7K
From Table A.2-4, μw = 1.042 cP
5 5 10 671 . 1 042 . 1 7 . 291 10 993 . 1 334 25 5 . 18 w AB AB T D D cm 2 /s (c) From Table 6.3-3, λ+ (Ca2+/2) = 59.5 and λ- (Cl-) = 76.3, n+ = 2 and n- = 1:5 7 7 10 792 . 0 2 5 . 59 10 662 . 2 10 662 . 2 2 n DCa cm2/s 5 7 7 10 031 . 2 1 3 . 76 10 662 . 2 10 662 . 2 n D Cl cm2/s
5 5 5 1.335 10 10 031 . 2 / 1 10 792 . 0 / 2 1 2 / / D n D n n n DAB cm2/sMolecular Diffusion in Biological Solutions and Gels
Prediction of Diffusivity of Albumin
Example 6.4-1, page 438
Predict the diffusivity of bovine serum albumin at 298K in water as a dilute solution using the modified Polson equation (6.4-1).
T = 298 K
MA = 67500 kg/kgmol from Table 6.4-1 (pg 437) μwater = 0.8937x10-3 Pa s
We can use the equation for the prediction of diffusivities for biological solutes:
3
1/3 11 15 3 / 1 15 10 70 . 7 67500 10 8937 . 0 298 10 40 . 9 10 40 . 9 A AB M T D m 2 /sThis value is different from the experimental value because the shape of the molecule differs greatly from a sphere.
Diffusion of Urea in Agar
Example 6.4-2, page 439
A tube or bridge of a gel solution of 1.05 wt% agar in water at 278K is 0.04 m long and connects two agitated solutions of urea in water. The urea concentration in the first solution is 0.2 gmol urea per liter solution and is 0 in the other. Calculate the flux of urea in kgmol/m2s at steady-state.
T = 278 K
DAB = 0.727x10-9 m2/s from Table 6.4-2 (page 440) CA1 = 0.2 kgmol/m3
CA2 = 0 kgmol/m3
Because XA1 is less than 0.01, the solute is very dilute and XBM ≈1.0.
9 9 1 2 2 1 10 63 . 3 04 . 0 0 2 . 0 10 727 . 0 z z C C D N AB A A A kgmol/m 2 sMolecular Diffusion in Solids
Diffusion of H
2Through Neoprene Membrane
Example 6.5-1, page 442
The gas hydrogen at 17ºC and 0.010 atm partial pressure is diffusing through a membrane of vulcanized neoprene rubber 0.5 mm thick. The pressure of H2 on the other side of the neoprene is zero. Calculate the
steady-state flux, assuming that the only resistance to diffusion is in the membrane. The solubility S of H2
gas in neoprene at 17ºC is 0.051 m3 (STP of 0ºC and 1atm)/m3 solid atm and the diffusivity DAB is 1.03x10-10
m2/s at 17ºC.
Solubility = 0.051 m3/m3 solid atm DAB = 1.03x10-10 m2/s
PA1 = 0.10 atm PA2 = 0
z2 – z1 = 0.5 mm The equilibrium concentration at the inside surface of the rubber is:
5 1 1 0.01 2.28 10 414 . 22 051 . 0 414 . 22 A A P S C kgmol H2/m3 solidBecause PA2 on the other side of the rubber is zero, CA2 = 0:
12 5 10 1 2 2 1 10 69 . 4 0005 . 0 0 10 28 . 2 10 03 . 1 z z C C D N AB A A A kgmol H2/m 2 sDiffusion Through a Packaging Film Using Permeability
Example 6.5-2, page 443
A polyethylene film 0.00015m thick is being considered for use in packaging a pharmaceutical product at 30ºC. If the partial pressure of O2 outside the package is 0.21 atm and inside it is 0.01atm, calculate the
diffusion flux of O2 at steady-state. Use permeability data from Table 6.5-1. Assume that the resistances to
diffusion outside the film and inside the film are negligible compared to the resistance of the film.
From Table 6.5-1, PM = 4.17x10-12 m3 solute/(s m2 atm/m)
10 12 1 2 2 1 10 480 . 2 00015 . 0 414 . 22 01 . 0 21 . 0 10 17 . 4 414 . 22 z z P P P N M A A A kgmol/m 2 s2.0 atm N2
0 atm
6.5-5 Diffusion Through a Membrane in Series
Nitrogen gas at 2.0 atm and 30ºC is diffusing through a membrane of nylon 1.0 mm thick and polyethylene 8.0 mm thick in series. The partial pressure at the other side of the two films is 0 atm. Assuming no other resistances, calculate the flux, NA at steady state.
P1 = 2.0 atm P2 = 0 atm
PN2/Ny = 0.152 x 10-12 m2/s atm PN2/Poly = 1.52 x 10-12 m2/s atm Diffusion of gas through a solid: CA can be related to the permeability:
414 . 22 A A P S
C where S is the solubility The permeability of a gas in a solid, PM = DABS
When there are several solids with thicknesses L1, L2,…, we can write the flux as:
12 12 12 2 2 1 1 2 1 10 256 . 1 10 52 . 1 008 . 0 10 0152 . 0 001 . 0 1 414 . 22 0 0 . 2 1 414 . 22 M M A A A P L P L P P N kmole/m2s
2 1
2 1 z z C C D N AB A A A Diffusion in Porous Solids That Depends on Structure
Diffusion of KCl in Porous Silica
Example 6.5-3, page 445
A sintered solid of silica 2.0mm thick is porous, with a void fraction ε of 0.30 and tortuosity τ of 4.0. The pores are filled with water at 298K. At one face the concentration of KCl is held at 0.1 gmol/liter, and fresh water flows rapidly past the other face. Neglecting any other resistance but that in the porous solid,
calculate the diffusion of KCl at steady-state.
DAB = 1.87x10-9 m2/s from Table 6.3-1 (page 431) CA1 = 0.1 gmol/liter = 0.1 kgmol/m3
CA2 = 0 τ = 4.0 ε = 0.30 z2 – z1 = 0.002 m
9 9 1 2 2 1 10 01 . 7 002 . 0 0 1 . 0 10 87 . 1 0 . 4 30 . 0 z z C C D N AB A A A kgmol/m2sP1 = 2.026 x 105 Pa
P2 = 0 Pa
1.25m
6.5-6 Diffusion of CO
2in a Packed Bed of Sand
It is desired to calculate the rate of diffusion of CO2 gas in air at steady state through a loosely packed bed
of sand at 276K and a total pressure of 1.013 x 105 Pa. The bed depth is 1.25m and the void fraction ε is 0.30. The partial pressure of CO2 is 2.026 x 103 Pa at the top of the bed and 0 Pa at the bottom. Use a τ of
1.87. τ = 1.87 PA1 = 2.026 x 103 Pa ε = 0.30 PA2 = 0 Pa T = 276K
8.314
276
1
.
87
1.25 0
0 10 026 . 2 30 . 0 10 142 . 4 3 1 2 2 1 z z RT P P D N AB A A A 6 10 609 . 1 A N mol/m2sUnsteady-State Diffusion in Various Geometries
Unsteady-State Diffusion in a Slab or Agar Gel
Example 7.1-1, page 463
A solid slab of 5.15 wt% agar gel at 278K is 10.16 mm thick and contains a uniform concentration of urea of 0.1 kgmol/m3. Diffusion is only in the x-direction through two parallel flat surfaces 10.16 mm apart. The slab is suddenly immersed in pure turbulent water, so the surface resistance can be assumed to be
negligible; that is, the convective coefficient kc is very large. The diffusivity of urea in agar from Table 6.4-2
is 4.72x10-10 m2/s.
(a) Calculate the concentration at the midpoint of the slab (5.08mm from the surface) and 2.54mm from the surface after 10 hours.
(b) If the thickness of the slab is halved, what would the midpoint concentration be in 10 hours?
C0 = 0.10 kgmol/m3 C1 = 0 for pure water
C = concentration at distance x from the center line and at time t
DAB = 4.72x10-10 m2/s
t = 10 hours * 3600 seconds = 36000 seconds
1 . 0 1 . 0 0 . 1 / 0 0 . 1 / 0 / / 0 1 1 c c C K C C K C Y (a) X1 = 10.16mm/2 = 5.08mm = 0.00508 m X = 0 (center)
0.658 00508 . 0 36000 10 72 . 4 2 10 2 1 X t D X AB Relative position, n = X/X1 = 0 Relative Resistance, m = 0On Figure 5.3-5 “Unsteady State Conduction in a Large Flat Plate” when X = 0.685, m = 0, n = 0 1 . 0 275 . 0 c Y c = 0.0275 kgmol/m3
On Figure 5.3-5, when X = 0.658, m = 0, n = 0.00254/0.00508 = 0.5: (n is not 0 now because not at center) 1 . 0 172 . 0 c Y c = 0.0172 kgmol/m3 (b)
If the thickness is halved, X becomes:
2.632 00254 . 0 36000 10 72 . 4 2 10 2 1 X t D X AB Relative position, n = 0, m = 0 1 . 0 0020 . 0 c Y c = 0.0002 kgmol/m37.1-5 Unsteady State Diffusion in a Cylinder of Agar Gel – Radially and Axially
A wet cylinder of agar gel at 278K containing a uniform concentration of urea of 0.1 kgmol/m3 has a diameter of 30.48mm and is 38.1mm long with flat parallel ends. The diffusivity is 4.72 x 10-10 m2/s. Calculate the concentration at the midpoint of the cylinder after 100h for the following cases if the cylinder is suddenly immersed in turbulent pure water.
(a) For radial diffusion only
(b) Diffusion occurs radially and axially
Assume surface resistance is negligible; assume k = 1.0 because properties are similar T = 278 K
C0 = 0.1 kgmol/m3 C1 = 0 for pure water D = 4.72 x 10-10 m/s Time, t = 100h = 360000 seconds 1 . 0 1 . 0 0 . 1 0 0 . 1 0 0 1 1 c c c k c c k c Y (a) r1 = 0.01524 m r0 = 0 m (center line)
731601 . 0 01524 . 0 360000 10 72 . 4 10 2 1 r t D X ABRelative position at midpoint of the cylinder, n = 0/0.1524 = 0 Relative resistance, m ≈ 0 because kc is assumed to be very large
From Figure 5.3-7, “Unsteady-State Heat Conduction in a Long Cylinder”: Using X = 0.731601, m = 0, n = 0 Y = 0.024 = c/0.1 Concentration = 0.0024 kgmol/m3 (b) X1 = L/2 = 38.1 mm / 2 = 19.05 mm = .01905 m
468 . 0 01905 . 0 360000 10 72 . 4 10 2 1 r t D X ABFrom Figure 5.3-7, Unsteady-State Heat Conduction in a Long Cylinder: Using X = 0.468, m = 0, n = 0 Y = 0.375
For both axial and radial diffusion, Y = YxYy = (0.375)(0.24) = 0.009
Unsteady-State Diffusion in a Semi-Infinite Slab
Example 7.1-2, page 465
A very thick slab has a uniform concentration of solute A of C0 = 1.0x10-2 kgmol A/m3. Suddenly, the front
face of the slab is exposed to a flowing fluid having a concentration C1 = 0.10 kgmol A/m3 and a convective
coefficient kc = 2x10-7 m/s. The equilibrium distribution coefficient K = cLi/ci = 2.0. Assuming that the slab
is a semi-infinite solid, calculate the concentration in the solid at the surface (x = 0) and x = 0.01m from the surface after t = 30000 seconds. The diffusivity in the solid is DAB = 4x10-9 m2/s.
t = 3000 seconds DAB = 4x10-9 m2/s C0 = 0.01 kgmol.m3 C1 = 0.10 kgmol/m3 kc = 2x10-7 m/s K = 2.0
095 . 1 3000 10 4 10 4 10 2 0 . 2 9 9 7 t D D Kk AB AB c For x = 0.01:
4 10
3000
0.457 2 01 . 0 2 D t -9 x ABFrom the chart 5.3-3 “Unsteady State Heat Conducted in a Semi-Infinite Solid with Surface Convection”: 26 . 0 01 . 0 2 / 1 . 0 01 . 0 / 1 0 1 0 C C K C C C Y C = 2.04x10-2 kgmol/m3 For x = 0
4 10
3000
0 2 0 2 D t -9 x ABFrom the chart 5.3-3 “Unsteady State Heat Conducted in a Semi-Infinite Solid with Surface Convection”: 62 . 0 01 . 0 2 / 1 . 0 01 . 0 / 1 0 1 0 C C K C C C Y C = 3.48x10-2 kgmol/m3
This is the same value as Ci. To calculate CLi:
2
2 10 96 . 6 10 48 . 3 0 . 2 i Li KC C kgmol/m37.1-6 Drying of Wood – Unsteady State Diffusion in a Flat Plate
A flat slab of Douglas fir wood 50.8mm thick containing 30 wt% moisture is being dried from both sides (neglecting ends and edges). The equilibrium moisture content at the surface of the wood due to the drying air blown over it is held at 5 wt% moisture. The drying can be assumed to be represented by a diffusivity of 3.72 x 10-6 m2/h. Calculate the time for the center to reach 10% moisture.
Assume there is no surface resistance and kc = ∞ At t = 0, c0 = 0.3, c1 = 0.05 At t = ?, c = 0.1 DAB = 3.72 x 10-6 m2/h Solve for Y: 0.2 3 . 0 0 . 1 5 . 0 1 . 0 0 . 1 5 . 0 0 1 1 c k c c k c Y
Solve for X for the graph: X1 = 25.4 mm = 0.254 m X0 (center) = 0 m t t x t D X AB 005766 . 0 254 . 0 10 72 . 3 2 6 2 1
Relative position, n = 0 at the center
Relative resistance, m = 0 because kC is large
From Chart 5.3-5, “Unsteady State Heat Conduction in a Flat Plate”: X = 0.75 = 0.005766t
Convective Mass Transfer Coefficients
Vaporizing A and Convective Mass Transfer
Example 7.2-1, page 469
A large volume of pure gas B at 2 atm pressure is flowing over a surface from which pure A is vaporizing. The liquid A completely wets the surface, which is a blotting paper. Hence, the partial pressure of A at the surface is the vapor pressure of A at 298K, which is 0.2 atm. The k’y has been estimated to be 6.78x10-5
kgmol/s m2 mol frac. Calculate NA, the vaporization rate, and also the value of ky and kG.
P = 2.0 atm PA1 = 0.2 atm
PA2 = 0 atm
k'y = 6.78x10-5 kgmol/s m2 mol frac First calculate the mole fraction of A:
1 . 0 0 . 2 / 2 . 0 / 1P P YA A 0 2 A Y
To calculate ky, we must relate it to k'y:
' y BM yy k k
ln
1.0/0.9
0.95 9 . 0 1 ln 2 1 1 2 B B B B BM y y y y y 5 5 ' 10 138 . 7 95 . 0 10 78 . 6 BM y y y kk kgmol/m2 s mole fraction
Now we can calculate kG: BM y BM Gy P k y k 5 5 10 569 . 3 0 . 2 10 138 . 7 P k kG y kgmol/m2 s atm
Now we can calculate the vaporization rate, NA:
5
6 2 1 7.138 10 0.10 0 7.138 10 y A A A k y y N kgmol/m2s or
5
6 2 1 3.569 10 0.2 0 7.138 10 G A A A k P P N kgmol/m2s7.2-1 Flux and Conversion of Mass-Transfer Coefficient
A value of kGwas experimentally determined to be 1.08 lbmol/h ft2 atm for A diffusing through stagnant B.
For the same flow and concentrations it is desired to predict kG’ and the flux of A for equimolar
counterdiffusion. The partial pressures are PA1 = 0.20 atm, PA2 = 0.05 atm, P = 1.0 atm abs total. Use
English and SI Units.
BM G GP k P k' Solve for PBM: 872853 . 0 2 . 0 1 05 . 0 1 ln 05 . 0 2 . 0 ln 1 2 2 1 A A A A BM P P P P P P P atm
Plug into equation:
k’G (1.0 atm) = (1.08 lbmol/h ft2 atm)(0.872853 atm)
k’G(1.01x105Pa) = (1.08lbmol/hft2atm)(.872853atm)(1hr/3600sec)(1ft2/.0929 m2)(.453kgmol/lbmol)
k’G = 0.9427 lbmol/h ft2 atm
k’G = 1.262 x 10-8 kgmol/s m2 Pa
Now solve for flux: PA1 = 0.2 atm = 20265 Pa PA2 = 0.05 atm = 5066.25 Pa
1 2
0.9247
0.2 0.05
.1414 ' G A A A k P P N
8
4 2 1 ' 10 92 . 1 25 . 5066 20265 10 262 . 1 G A A A k P P N NA = 0.1414 lbmol/ft2h NA = 1.92 x 10-4 kgmol/m2sMass Transfer Under High Flux Conditions
High Flux Correction Factors
Example 7.2-2, page 472
Toluene A is evaporating from a wetted porous slab by having inert pure air at 1 atm flowing parallel to the flat surface. At a certain point, the mass transfer coefficient, k’x for very low fluxes has been estimated as
0.20 lbmol/hr ft2. The gas composition at the interface at this point is XA1 = 0.65. Calculate the flux NA and
the ratios kc/ k’c or kx/ k’x and k0c/ k’c or k0x/ k’x to correct for high flux.
First find XBM:
619 . 0 65 . 0 1 0 1 ln 65 . 0 1 0 1 ln 1 2 1 2 B B B B BM X X X X XTo find the flux, NA, use:
0.65 0
0.210 619 . 0 20 . 0 2 1 ' A A BM x A X X X k N lbmol/ft2 hrTo find the ratios, set up the following:
616 . 1 619 . 0 1 1 ' ' BM c c x x X k k k k So kx 1.616kx' 1.616
0.2 0.323 lbmol/ft2 hr; and 565 . 0 619 . 0 65 . 0 1 1 1 ' 0 ' 0 BM A c c x x X X k k k k So kx0 0.565kx' 0.565
0.2 0.113 lbmol/ft 2 hr1 2
7.2-3 Absorption of H
2S by Water
In a wetted-wall tower an air H2S mixture is flowing by a film of water that is flowing as a thin film down a
vertical plate. The H2S is being absorbed from the air to the water at a total pressure of 1.50 atm abs and
30ºC. A value for kc’ of 9.567 x 10-4 m/s has been predicted for the gas-phase mass transfer coefficient. At a
given point, the mole fraction of H2S in the liquid at the liquid-gas interface is 2.0(10-5) and PA of H2S in the
gas is 0.05 atm. The Henry’s law equilibrium relation is PA (atm) = 609xA (mole fraction in the liquid).
Calculate the rate of absorption of H2S. Hint: Call point 1 the interface and point 2 the gas phase. Then
calculate PA1 from Henry’s Law and the given xA. The value of PA2 is 0.05 atm.
P = 1.50 atm
PA1 = 609 (CA1) = 609(2 x 10-5) = 0.01218 atm PA2 = 0.05 atm T = 30 ºC = 303.15 K k’c = 9.567 x 10-4 m/s XA1 = 2 x 10-5 Henry’s Law: PA = 609XA R = 82.057 x 10-3 m3 atm/kgmol K P k RT P k G c ' '
5 3 4 ' 10 846 . 3 15 . 303 10 057 . 82 10 567 . 9 G k kgmol/m2s atm
5
6 2 1 3.846 10 0.01218 0.05 1.455 10 G A A A k P P N NA = 1.455 x 10-6 kgmol/m2sMass Transfer for Flow Inside Pipes
Mass Transfer Inside a Tube
Example 7.3-1, page 479
A tube is coated on the inside with naphthalene and has an inside diameter of 20 mm and a length of 1.10m. Air at 318K and an average pressure of 101.3 kPa flows through this pipe at a velocity of 0.80 m/s.
Assuming that the absolute pressure remains essentially constant, calculate the concentration of naphthalene in the exit air.
v = 0.80 m/s D = 0.02 m T = 318 K z2 – z1 = 1.10 m DAB = 6.92x10-6 m2/s PAi = 74.0 Pa μair = 1.932x10-5 Pa s from Appendix A ρ = 1.114 kg/m3 R = 8314 m3 Pa/kgmol K
Calculate the concentration, CAi:
2.799 10 5 318 8314 74 RT P CAi Ai kgmol/m 3 Calculate the Schmidt number:
1.114
6.92 10
2.506 10 932 . 1 6 5 AB Sc D N Calculate the Reynolds number:
6 . 922 10 932 . 1 114 . 1 80 . 0 02 . 0 5 Re D NHence, the flow is laminar. We will use Figure 7.3-2 for laminar flow streamlines:
33.02 4 10 . 1 02 . 0 506 . 2 6 . 922 4 Re L D N N Sc Using Figure 7.3-2 with rodlike flow:55 . 0 0 0 A Ai A A C C C C If CA0 = 0, 0 10 799 . 2 0 55 . 0 5 0 0 A A Ai A A C C C C C CA = 1.539 x 10-5 kgmol/m3
Pure Water T = 26.1ºC v = 3.05m/s 0.00635 m 1.22 m 1.829 m
7.3-7 Mass Transfer from a Pipe and Log Mean Driving Force
Use the same physical conditions as in problem 7.3-2, but the velocity in the pipe is now 3.05 m/s. Do as follows:
(a) Predict the mass transfer coefficient kc (Is this turbulent flow?).
(b) Calculate the average benzoic acid concentration at the outlet. [ Note: in this case, Eqs. (7.3-42) and (7.3-43) must be used with the log mean driving force, where A is the surface area of the pipe.]
(c) Calculate the total kgmol of benzoic acid dissolved per second.
ρ = 996 kg/m3 μ = 8.71 x 10-4 Pa s T = 26.1ºC DAB = 1.245x10-9 m2/s Solubility = 0.2948 kgmol/m3 D = 0.00635 m As = 2rπL V = Aυ Calculate the Schmidt number:
996
1.245 10
702.408 10 71 . 8 9 4 AB Sc D N Calculate the Reynolds number:
22147 10 71 . 8 996 05 . 3 00635 . 0 4 Re D NCalculate the Sherwood number:
000049487 . 0 23 . 0 Re0.83 0.33 Sc Sh N N N Now calculate k’c: AB c Sh D D k N '
00635 . 0 10 245 . 1 000049487 . 0 9 ' kc ' 4 10 59 . 1 c k m/s (b) CA1 = 0 CAi = 0.02948 (solubility) CA2 = ?So now we plug this into the log mean driving force equation:
2 1 1 2 1 2 ln A Ai A Ai A A C A A C C C C C C K A C C v
2 2 2 ln A Ai Ai A C A C C C C K A C vNow plug in:
2 2 4 2 2 02948 . 0 02948 . 0 ln 10 59 . 1 003175 . 0 05 . 3 A A A C C C CA2 = 0.001151 kgmol/m3 (c)Rate of benzoic acid dissolved = NA:
2 7 1 2 10 11 . 1 003175 . 0 0 001151 . 0 05 . 3 A C C v N A A AMass Transfer for Flow Outside Solid Surfaces
Mass Transfer From a Flat Plate
Example 7.3-2, page 481
A large volume of pure water at 26.1ºC is flowing parallel to a flat plate of solid benzoic acid, where L = 0.24 m in the direction of flow. The water velocity is 0.061 m/s. The solubility of benzoic acid in water is 0.02948 kgmol/m3. The diffusivity of benzoic acid is 1.245x10-9 m2/s. Calculate the mass-transfer coefficient kL and the flux NA.
Solubility = 0.02948 kgmol/m3 T = 26.1ºC
L = 0.24 m
DAB = 1.245x10-9 m2/s v = 0.061 m/s
Because the solution is dilute, we can use the properties of water for the solution: ρ = 996 kg/m3
μ = 8.71 x 10-4 Pa s
Calculate the Schmidt Number:
702 10 245 . 1 996 10 71 . 8 9 4 AB Sc D N Calculate the Reynolds number:
17000 10 71 . 8 996 061 . 0 24 . 0 4 Re D N Now calculate JD:
17000
0.00758 99 . 0 99 . 0 Re,0.5 0.5 L D N JNow solve for k’c:
2/3 ' Sc c D N k J ' 6 10 85 . 5 c k m/sIn this case, A is diffusing through stagnant B. We use the solubility for CA1 and CA2 = 0. Also, since the solution is dilute, xBM ≈ 1:
7 6 2 1 ' 10 726 . 1 0 02948 . 0 0 . 1 10 85 . 5 A A BM c A C C X k N kgmol/m2sMass Transfer From a Sphere
Example 7.3-3, page 482
Calculate the value of the mass-transfer coefficient and the flux for mass transfer from a sphere of
naphthalene to air at 45ºC and 1 atm abs flowing at a velocity of 0.305 m/s. The diameter of the sphere is 0.0254m. The diffusivity of naphthalene in air at 45ºC is 6.92x10-6 m2/s and the vapor pressure of solid naphthalene is 0.555 mmHg. D = 0.0254m PA1 = 0.555 mmHg = 74 Pa T = 45ºC P = 1 atm v = 0.305 m/s DAB = 6.92x10-6 m/s ρ = 1.113 kg/m3 μ = 1.93 x 10-5 Pa s
Calculate the Schmidt number:
1.113
6.92 10
2.505 10 93 . 1 6 5 AB Sc D N Calculate the Reynolds number:
446 10 93 . 1 113 . 1 305 . 0 0254 . 0 5 Re D NNow find the Sherwood number:
21.0 552 . 0 2 Re 0.53 1/3 Sc Sh N N NNow solve for k’c:
AB c sh D D k N '
3 6 ' 10 72 . 5 0254 . 0 10 92 . 6 0 . 21 D D N k AB sh c m/sNow solve for k’G:
9 3 ' ' 10 163 . 2 318 8314 10 72 . 5 RT k k c G kgmol/s m 2 Pa Since the gas is dilute, k’G ≈ kG. Using that we can solve for the flux:
9
7 2 1 2.163 10 73 0 1.599 10 G A A A k P P N kgmol/m2 s7.3-1 Mass Transfer from a Flat Plate to a Liquid
Using the data and physical properties from Example 7.3-2, calculate the flux for a water velocity of 0.152 m/s and a plate length of L = 0.137 m. Do not assume that xBM = 1.0 but actually calculate its value.
DAB = 1.245 x 10-9 m2/s Solubility, S = 0.02948 kgmol/m3 ρ = 996 kg/m3 Mw = 18 kg/kgmol CA1 = 0.02948 kgmol/m3 μ = 8.71 x 10-9 Pa s Solve for XBM: XB1 = 1 1 1 A B B C C C = [(996 kg/m 3
)/(18kg/kgmol)]/[(996 kg/m3)/(18kg/kgmol) + 0.02948 kgmol/m3] XB1 = 0.999468 XB2 = 1.0 999734 . 0 999468 . 0 0 . 1 ln 999468 . 0 0 . 1 ln 1 1 ln 1 2 1 2 1 2 2 1 B B B B A A A A BM X X X X X X X X X
Calculate Schmidt Number:
702 10 245 . 1 996 10 71 . 8 9 4 AB SC D N
Calculate Reynolds Number:
5 . 23812 10 71 . 8 996 152 . 0 137 . 0 9 Re L N Calculate JD:
23812.5
0.06416 99 . 0 99 . 0 Re 0.5 0.5 N JD Calculate K’c:
2/3 5 3 / 2 ' 10 2346 . 1 702 152 . 0 06416 . 0 D SC C J N K m/s Calculate Flux:
7 5 2 1 ' 10 641 . 3 999734 . 0 0 2948 . 0 10 2346 . 1 BM A A C A X C C K N kgmol/m2s7.3-4 Mass Transfer to Definite Shapes – Flat Plate and a Sphere
Estimate the value of the mass-transfer coefficient in a stream of air at 325.6 K flowing in a duct past the following shapes made of solid naphthalene. The velocity of the air is 1.524 m/s at 325.6K and 202.6 kPa. The DAB of naphthalene in air is 5.16 x 10-6 m2/s at 273K and 101.3 kPa.
(a) For air flowing parallel to a flat plate 0.152 m in length (b) For air flowing past a single sphere 12.7 mm in diameter
CA1
0.137 m Benzoic Acid
P = 202600 Pa R = 8314.34 m3 Pa/kgmol K T = 325.6 K Mw = 28.8 kg/kgmol Correct DAB:
6 1 75 . 1 1 75 . 1 10 51182 . 3 3 . 101 6 . 202 273 6 . 325 6 . 202 , 6 . 325 o o AB P P T T kPa K D m2/s Calculate ρ and μ: 1958 . 0 273 6 . 325 0171 . 0 768 . 0 cP T T n o o cP = 1.958 x 10-5 kg/m s
2.155 6 . 325 34 . 8314 202600 8 . 28 RT P Mw kg/m3 (a) Calculate NRe:
5 . 25495 10 958 . 1 155 . 2 524 . 1 152 . 0 5 Re L N Because NRe > 15000:
25495.5
0.004732 036 . 0 036 . 0 Re0.2 0.2 N JD
2.155
3.51182 10
2.58722 10 958 . 1 6 5 AB SC D N 3 / 2 ' SC C D N K J
2/3 ' 58722 . 2 524 . 1 004732 . 0 KC K’C = 0.003827 m/s
9 ' 10 41366 . 1 6 . 325 34 . 8314 003827 . 0 RT K K C G kgmol/m 2 Pa s (b) Dp = 0.0127 mCalculate Reynolds and Schmidt Numbers:
21 . 2130 10 958 . 1 155 . 2 524 . 1 0127 . 0 5 Re p D N
2.155
3.51182 10
2.58722 10 958 . 1 6 5 AB SC D N Because 0.6 < NSC < 2.7:
2130.21
2.58722
46.0164 552 . 0 2 552 . 0 2 Re0.53 1/3 0.53 1/3 SC SH N N N Air at 325.6K, 202.6 kPa, v = 1.524 m/s 0.152 mAB p C SH D D K N '
6 ' 10 51182 . 3 0127 . 0 0164 . 46 KC K’C = 0.012725 m/s
9 ' 10 70 . 4 6 . 325 34 . 8314 012735 . 0 RT K K C G kgmol/m 2 Pa sMass Transfer of a Liquid in a Packed Bed
Example 7.3-4, page 485
Pure water at 26.1ºC flows at the rate of 5.514x10-7 m3/s through a packed bed of benzoic acid spheres having diameters of 6.375mm. The total surface area of the spheres in the bed is 0.01198 m2 and the void fraction is 0.436. The tower diameter is 0.0667m. The solubility of benzoic acid in water is 2.948x10-2 kgmol/m3.
(a) Predict the mass transfer coefficient kc.
(b) Using the experimental value of kc, predict the outlet concentration of benzoic acid in the water.
Because the solution is dilute, we can use the properties of water for the solution: ρ = 996 kg/m3 μ26.1ºC = 8.71 x 10-4 Pa s μ25ºC = 8.94 x 10-4 Pa s V = 5.514x10-7 m3/s Dp = 6.375mm AS = 0.01198 m2 ε = 0.436 D = 0.0667m S = 2.948x10-2 kgmol/m3 DAB(25ºC) = 1.21x10-9 m2/s from Table 6.3-1
Correct DAB:
9 3 3 9 10 254 . 1 10 8718 . 0 10 8940 . 0 298 1 . 299 10 21 . 1 25 1 . 26 new old old new AB AB T T D D m2/s Calculate the area of the column:
3 2 10 494 . 3 0667 . 0 4 4 D A m2Now use the area and volumetric flow to find the velocity:
7
3
4 10 578 . 1 10 494 . 3 / 10 514 . 5 / V A v m/sCalculate the Schmidt Number:
996.7
1.245 10
702.6 10 71 . 8 9 4 AB Sc D N Calculate Reynolds number:
150 . 1 10 71 . 8 7 . 996 10 578 . 1 006375 . 0 4 4 Re p D N Calculate JD:
1.150
2.277 436 . 09 . 1 09 . 1 2/3 2/3 Re N JD Then, assuming k’c = kc for dilute solutions,
2/3 ' Sc c D N v k J 4
2/3 ' 6 . 702 10 578 . 1 277 . 2 kc ' 6 10 447 . 4 c k m/sNow set the log mean driving force equation can be set equal to the material balance equation on the bulk stream:
1 2 2 1 2 1 ln A A A Ai A Ai A Ai A Ai c V C C C C C C C C C C Ak where CAi = 0.02948 (the solubility) CA1 = 0
A = 0.01198 m2 (the surface area of the bed) V = 5.514x10-7 m3/s
7.3-5 Mass Transfer to Packed Bed and Driving Force
Pure water at 26.1ºC is flowing at a rate of 0.0701ft3/h through a packed bed of 0.251-in. benzoic acid spheres having a total surface area of 0.129 ft2. The solubility of benzoic acid in water is 0.00184 lbmol benzoic acid/ft3 solution. The outlet concentration is cA2 is 1.80 x 10-4
lbmol/ft3. Calculate the mass transfer coefficient kc. Assume dilute solution.
μ = 0.8718 x 10-3 Pa s ρ = 996.7 kg/m3 υ = 0.0701 ft3 /hr A = 0.129 CAi = 0.00184 lbmol/ft3 CA1 ≈ 0 CA2 = 1.80 x 10-4 lbmol/ft3
Log mean driving force equation:
2 1 1 2 ln A Ai A Ai A A C A C C C C C C K A A N
where CA1 is the inlet bulk flow concentration, CA2 is the outlet bulk flow concentration, and CAi is the concentration at the surface of the solid (which in this case equals the solubility)
Material Balance on the bulk stream: NAAv
CA2 CA1
So now we plug this into the log mean driving force equation:
2 1 1 2 1 2 ln A Ai A Ai A A C A A C C C C C C K A C C vCA1 = 0, so it cancels out of the equation:
2 2 2 ln A Ai Ai A C A C C C C K A C vWe can now plug in given numbers:
4 4 4 10 80 . 1 00184 . 0 00184 . 0 ln 10 80 . 1 129 . 0 10 80 . 1 0701 . 0 KC KC = 0.0559 ft/hrMass Transfer to Suspensions of Small Particles
Mass Transfer from Air Bubbles in Fermentation
Example 7.4-1, page 488
Calculate the maximum rate of absorption of O2 in a fermenter from air bubbles at 1 atm abs pressure
having diameters of 100 μm at 37ºC into water having a zero concentration of dissolved oxygen. The solubility of O2 from air in water at 37ºC is 2.26x10-7 kgmol O2/m3 liquid. The diffusivity of O2 in water at
37ºC is 3.25x10-9 m2/s. Agitation is used to produce the air bubbles.
Dp = 1 x 10-4 m DAB = 3.25x10-9 m2/s
Solubility = 2.26x10-7 kgmol O2/m3 liquid
μc,water = 6.947 x 10-4 Pa s = 6.947 x 10-4 kg/ m s ρc,water = 994 kg/m3 ρp,air = 1.13 kg/m3 Calculate NSc:
994
3.25 10
215 10 947 . 6 9 4 AB c c Sc D N Now calculate k’L: 3 / 1 2 3 / 2 ' 31 . 0 2 c c Sc p AB L g N D D k
4 3 / 1 2 4 3 / 2 4 9 ' 10 29 . 2 994 806 . 9 10 947 . 6 13 . 1 994 215 31 . 0 10 1 10 25 . 3 2 L k m/sAssuming k’L = kL for dilute solutions,
4
7
8 2 1 2.29 10 2.26 10 0 5.18 10 L A A A k C C N kgmol O2/m2 sMolecular Diffusion Plus Convection and Chemical Reaction
Proof of Mass Flux Equation
Example 7.5-1, page 491
Table 7.5-1 gives the following relation: jA jB 0
Prove this relationship using the definition of the fluxes in terms of velocities.
From Table 7.5-1 (page 490), substituting A
A
for jA and B
B
for jB, and rearranging: 0 A A A B B B AABB
A B
0 B B A A and A B
0 B B A A B B A A AABB
AABB
0 Thus the identity is proved.Diffusion and Chemical Reaction at a Boundary
Example 7.5-2, page 495
Pure gas A diffuses from point 1 at a partial pressure 101.32 kPa to point a distance 2.00mm away. At point 2, it undergoes a chemical reaction at the catalyst surface and A 2B. Component B diffuses back at steady state. The total pressure is P = 101.32 kPa. The temperature is 300K and DAB = 0.15 x 10-4 m2/s.
(a) For instantaneous rate of reaction, calculate xA2 and NA.
(b) For a slow reaction where k’1 = 5.63x10-3 m/s, calculate xA2 and NA.
PA2 = XA2 = 0 because no A can exist next to the catalyst surface XA1 = PA1/P = 1.0 δ = 0.002 m T = 300 K C = P/RT = 101320/(8314*300) = 4.062 x 10-2 kgmol/m3 NB = -2NA DAB = 0.15 x 10-4 m2/s k’1 = 5.63x10-3 m/s (a)
