Test 1 Book Notes examples-1.pdf

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. R = 82.057 m3 atm/kgmol K PA1 = 0.6 atm T = 298K PA2 = 0.2 atm z2-z1 = 0.2m P = 1 atm 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. D C  C A2  D P  PA2     AB A1 J AZ  AB A1 Flux, J AZ if C = P/RT z 2  z1 RT z 2  z1  Substitute in values:  J AX 

0.687  10 0.6  0.2  5.63 10 4

82.0572980.2

6

kgmol/m2s

Equimolar Counterdiffusion

I

Example 6.2-1, page 415 Ammonia 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  J AZ 

II d

PCO2 = 456 mmHg

PCO2 = 76 mmHg

L

P = 1.0132x105 Pa z2 – z1 = 0.10 m R = 8314.3 m3 Pa/kgmol K







D AB PA1  PA2  0.230  10 4 1.013  10 4  0.507  10 4   4.70  10 7 kgmol A/m2 s 8314.32980.1 RT z 2  z1 

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, H2O 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

PB 2  PB1 1  0  1  0.0231   0.988 atm  PB 2   1 0  ln  ln    P  1  0.0231  B1  D AB P PA1  PA2  NA  RT z 2  z1 PBM

PBM 

NA 

0.250 10 1.0 4

82.0572930.15240.988

0.0231  0  1.595  10 7 kgmol/m2s

Diffusion 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 t F 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. DAB P PA1  PA2  NA  RT z2  z1 PBM Assuming a cross-sectional area of 1 m2, the level drops dt in dz seconds, and the leftover kgmol of A is PAdz/MA:  dz  1 N A 1  A M A dt Rearranging and integrating: tF  A zF D P zdz  AB PA1  PA2  dt  0 M A z0 RTPBM

tF 

 A z F2  z 02 RTPBM

2M A D AB PPA1  PA2 

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 N A  1 1  DAB P  P  p A,1     ln   4  r1 r2  RT  P  p A, 2  where N A is the molar rate of material A leaving R is the ideal gas constant the drop T is the system temperature r1 and r2 are two radial points away from pA,1 is the partial pressure of A at point 1 the sphere center pA,2 is the partial pressure of A at point 2 DAB is the diffusion coefficient P is the total system pressure Starting with this equation, derive the following approximate equation for the molar flux at the particle surface, NA,1 2 DAB cA,1  cA,2  N A,1  D1 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

N A  1 1  D AB P  P  PA2  N D P  P  PA2        A  AB ln   ln  4  r1 r2  RT 4r1 RT  P  PA1   P  PA1  Now multiply the equation by 1/r1, rewrite the radius as 2/D on the right side, substitute for surface flux, NAs, and PBM:

Stagnant B

r2 r1 Diffusing A

PB 2  PB1 PA1  PA2 N  and N AS  A 2 P  ln  P  PA2   4r1 ln  B 2    P P  P B1  A1    2 D AB PA1  PA2 So we rewrite the equation as: N AS   DRT PBM PBM 

Assume low vapor pressure, so PA1, PA2 << P and use a Taylor Series approximation ln(x) = x-1:  P  PA2  P  PA2  P  PA2  1   is approaching 1, so ln    P  PA1  P  PA1  P  PA1  P  PA2  P  PA1  PA1  PA2 PA1  PA2   P  PA1 P  PA1 P Lastly, PA1 = CA1RT and PA2 = CA2RT:

Now,

so PBM ≈ P.

2 D AB P 2 D AB 2 D AB   PA1  PA2    C A1  RT  C A2  RT    C A1  C A2  RTD PBM RTD D where CA1 = concentration at surface of the drop and CA2 = concentration far from the drop ≈ 0 N AS 

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

PBM 

Vapor pressure, PA1 = 0.555 mmHg = 74 Pa PA2 = 0 Pa R = 8314.3 m3 Pa/kgmol K



 



PB 2  PB1 1.01325  10 5  0  1.01325  10 5  74   1.0129  10 5 Pa 5 P   1.01325  10  0  ln  ln  B 2   5  1.01325  10  74   PB1 



NA 







D AB P 6.92  10 6 1.01325  10 5 PA1  PA2    9.68  10 8 kgmol A/m2s RT r1  r0 PBM 8314.33180.002 1.0129  105





6.2-5.) 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: 1.75

T  DAB(52.6ºC) = DAB(0ºC)  2   T1  2 m /s Now solve for N A :

1

 P2    = (5.16 x 10-6 m2/s)(325.75/273.15)1.75 = 7.023 x 10-6  P1 

N A  1 1  DAB P  P  PA2      ln  4  r1 r2  RT P  P A1  







N A DAB P  P  PA2    ln  4r1 RT P  P A1  



NA 7.023 10 6 1.01325 10 5  1.01325 105  133.322   = 4.34736 x 10-11 kgmol/s  ln  5 8314325.75 4 0.1  1.01325 10  0  Now solve for NA = N A /A = (4.34736 x 10-11 kgmol/s)/(4πr12) = 3.46 x 10-8 kgmol/m2s

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:

NA D P P  PA2  AB  A1 2 RTr PBM 4r

and

NA 

 A dV M A dt

dV Volume of the sphere, V  4 r 3 so the derivative of volume is:  4r 2 3 dr Plug this volume derivative into molar flux equation for dV:  A 4r 2 dr NA  M A dt

Now set N A equal in each equation:

 A 4r 2 dr

D P P  PA2 1  AB  A1 2 M A dt RTr PBM 4r Separate variables and integrate both sides: t r  A  RT  PBM  A  r 2  RT  PBM  dt  rdr t  0 M A DAB  P  PA1  PA2  0 2M A  D AB  P  PA1  PA2  (b) Now use the equation to find the time in seconds: 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

PA1 = 3840 Pa PA2 = 0 Pa P = 1.01325 x 105 Pa MA = 92.14 kg/kgmol Solve for PBM:

PBM 

tevap =



PA1  PA2 = 99392.6 Pa   P  P   A 2 ln  P  PA1  

8660.0022 8314299.0599392.6  1388.23 seconds 2M A  DAB  P  PA1  PA2  292.140.086 10 4 3840  0  A  r 2  RT  PBM

=

6.2-10 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 equilateral triangle, the length of each side of the triangle being 0.0610m at the left end and tapering

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. .061m

.0305m

1.22m

Area = 1 b 2 tan (60º) 4 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 2 So now Area = 1  0.025  L  0.061 tan (60º) = 0.000271 L2 – 0.001321 L + 0.001611 4 NA  DAB dPA N A  D AB dPA    2 RT dL A RT dL 0.000271 L - 0.001321 L  0.001611

Now separate and integrate both sides: 1.22



0

NA  DAB dL  2 RT 0.000271 L - 0.001321 L  0.001611



PA 2

PA1

dPA

R = 8314 m3 Pa/kgmol K T = 298.15 K

PA1 = 25.33 kPa PA2 = 5.066 kPa DAB = 0.230 x 10-4 m2/s

Now plug into integrate equation:  DAB PA2  PA1   0.230 10 4 5066  25330  N A 1514.8   N A = 1.24 x 10-10 8314298.15 RT kgmol/s Now solve for NA: 1.24 10 10 NA = N A /A = at any point along the length, L 1  0.025L  0.06102 tan 60  4





 

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: 16.8 6.8 46.05 46.05 X A1   0.0732 X A2   0.0277 16.8 83 . 2 6 . 8   93.2 46.05 18.02 46.05 18.02 X B1  1  X A1  1  0.0732  0.9268 X B 2  1  X A2  1  0.0277  0.9723 Now calculate the molecular weight: 100kg 100kg M1   20.07 kg/kgmol M1   18.75 kg/kgmol 16.8 6.8  83.2  93.2 46.05 18.02 46.05 18.02 Now calculate CAvg:  / M 1   2 / M 2 972.8 / 20.07  988.1 / 18.75 C avg  1   50.6 kgmol/m3 2 2 X  X B1 0.9723  0.9268 X BM  B 2   0.949  X B2   0.9268  ln  ln    0.9723   X B1 

NA 

D AB C avg

z 2  z1 x BM

9 x A1  x A2   0.740  10 50.60.0732  0.0277  8.99  10 7 kgmol/m2s 0.0020.949

6.3-2 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 CA1 = 0.02 ρ2 = 961.7 kg/m3 CA2 = 0.1 M1 = 17.9806 kg/kgmol z1 = 0.004 m M2 = 17.9031 kg/kgmol z2 = 0 m T = 278 K Calculate Cavg:

C avg 

  1  991.7 1  1 961.7  3   2       54.435 kgmol/m 2  M 1 M 2  2  17.9806 17.9031 

Calculate XBM: Xw1 = 1-0.02 = 0.98 Xw2 = 1-0.1 = 0.9

X BM 

X B 2  X B1 0.9  0.98   0.939432 ln  X B 2 X B1  ln 0.9 / 0.98

Now calculate flux: D AB C avg  X A1  X A2  1.24 10 9 54.4350.02  0.1 NA  = = 1.437 x 10-6 kgmol/m2 s z 2  z1 X BM 0  0.0040.939432

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

Prediction 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:





1/ 2



1/ 2

D AB  1.173  10 16 M B 









T 298 1/ 2  1.173  10 16 2.6  18.02  1.277  10 9 m2/s 0.6 0.6  BV A 0.89370.0740

For 50ºC:



D AB  1.173  10 16 M B 

T 323 1/ 2  1.173  10 16 2.6  18.02  2.251  10 9 m2/s 0.6 0.6  BV A 0.54940.0740

Prediction 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: 1 / n  1 / n  1 / 1  1 / 1  1.993  10 5 cm2/s  D AB  8.928  10 10 T  8.928  10 10 298 1 /   1 /   1 / 73.5  1 / 76.3 (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 T 291.7 D AB 18.5  D AB 25   1.993  10 5  1.671  10 5 cm2/s 334 w 1.042 (c) From Table 6.3-3, λ+ (Ca2+/2) = 59.5 and λ- (Cl-) = 76.3, n+ = 2 and n- = 1:  59.5 DCa2   2.662  10 7   2.662  10 7  0.792  10 5 cm2/s n 2

 D AB



76.3  2.031  10 5 cm2/s n 1 n  n 2 1    1.335  10 5 cm2/s 5 5 n / D  n / D 2 / 0.792  10  1 / 2.031  10

DCl   2.662  10 7

 2.662  10 7









Molecular 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: 9.40  10 15 T 9.40  10 15 298 D AB    7.70  10 11 m2/s 1/ 3 1/ 3 3  M A  0.8937  10 67500 This 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.

NA 





D AB C A1  C A2  0.727  10 9 0.2  0   3.63  10 9 kgmol/m2s z 2  z1 0.04

Molecular Diffusion in Solids Diffusion of H2 Through 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. PA1 = 0.10 atm Solubility = 0.051 m3/m3 solid atm PA2 = 0 -10 2 DAB = 1.03x10 m /s z2 – z1 = 0.5 mm The equilibrium concentration at the inside surface of the rubber is: 0.051 0.01  2.28  10 5 kgmol H /m3 solid S C A1  PA1  2 22.414 22.414 Because PA2 on the other side of the rubber is zero, CA2 = 0: D C  C A2  1.03  10 10 2.28  10 5  0 N A  AB A1   4.69  10 12 kgmol H2/m2 s z 2  z1 0.0005







Diffusion 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) P P  PA2  4.17  10 12 0.21  0.01 N A  M A1   2.480  10 10 kgmol/m2s 22.414z 2  z1  22.4140.00015

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. N2

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:

2.0 atm

NA 

DAB C A1  C A2  z2  z1 

CA can be related to the permeability: C A 

S  PA where S is the solubility 22.414

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:   PA1  PA2  1 NA   L L 22.414 1  2   PM 1 PM 2

      1   2.0  0   22.414  0.001 0.008    12 1.52 10 12  0.0152 10 

    1.256 10 12 kmole/m2s   

0 atm

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. τ = 4.0 -9 2 DAB = 1.87x10 m /s from Table 6.3-1 (page 431) ε = 0.30 CA1 = 0.1 gmol/liter = 0.1 kgmol/m3 z2 – z1 = 0.002 m CA2 = 0

NA 

 D AB C A1  C A2  0.30 1.87  10 9 0.1  0   7.01  10 9 kgmol/m2s  z 2  z1 4.0 0.002

6.5-6 Diffusion of CO2 in 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. P1 = 2.026 x 105 Pa τ = 1.87 PA1 = 2.026 x 103 Pa ε = 0.30 PA2 = 0 Pa T = 276K







D  P  PA2  .142  10 4 0.30 2.026  103  0 N A  AB A1  8.3142761.871.25  0 RT z 2  z1  N A  1.609  106 mol/m2s



1.25m

P2 = 0 Pa

Unsteady-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 DAB = 4.72x10-10 m2/s C1 = 0 for pure water t = 10 hours * 3600 seconds = 36000 seconds C = concentration at distance x from the center line and at time t C /K C 0 / 1.0  c c Y 1   C1 / K  C0 0 / 1.0  0.1 0.1 (a) X1 = 10.16mm/2 = 5.08mm = 0.00508 m X = 0 (center) 10 D t X  AB 2  4.72  10 36000  0.658 X1 0.005082 Relative position, n = X/X1 = 0 Relative Resistance, m = 0





On Figure 5.3-5 “Unsteady State Conduction in a Large Flat Plate” when X = 0.685, m = 0, n = 0 c  c = 0.0275 kgmol/m3 Y  0.275  0.1 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) c  c = 0.0172 kgmol/m3 Y  0.172  0.1 (b) If the thickness is halved, X becomes: 10 D t X  AB 2  4.72  10 36000  2.632 X1 0.002542 Relative position, n = 0, m = 0 c Y  0.0020   c = 0.0002 kgmol/m3 0.1





7.1-5 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 c k c 0 1.0  c c Y 1   c1 k  c0 0 1.0  0.1 0.1 (a) r1 = 0.01524 m r0 = 0 m (center line)





DAB t 4.72 10 10 360000   0.731601 0.01524 r12 Relative position at midpoint of the cylinder, n = 0/0.1524 = 0 Relative resistance, m ≈ 0 because kc is assumed to be very large X

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 D t 4.72 10 10 360000 X  AB2   0.468 0.01905 r1 From 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 Concentration = 0.0009 kgmol/m3

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. C1 = 0.10 kgmol/m3 kc = 2x10-7 m/s K = 2.0

t = 3000 seconds DAB = 4x10-9 m2/s C0 = 0.01 kgmol.m3

Kk c D AB

D AB t 

2.02  10 7  4  10

9

4 10 3000  1.095 9

For x = 0.01:

x

0.01

 0.457 2 4  10 -9 3000 From the chart 5.3-3 “Unsteady State Heat Conducted in a Semi-Infinite Solid with Surface Convection”: C  C0 C  0.01 C = 2.04x10-2 kgmol/m3 1Y    0.26  C1 / K  C0 0.1 / 2  0.01 For x = 0 x 0  0 2 D AB t 2 4  10 -9 3000 From the chart 5.3-3 “Unsteady State Heat Conducted in a Semi-Infinite Solid with Surface Convection”: C  C0 C  0.01 C = 3.48x10-2 kgmol/m3 1Y    0.62  C1 / K  C0 0.1 / 2  0.01 This is the same value as Ci. To calculate CLi: C Li  KCi  2.03.48  10 2   6.96  10 2 kgmol/m3 2 D AB t











7.1-6 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:

Y

c1 k  c 0.5 1.0  0.1   0.2 c1 k  c0 0.5 1.0  0.3

Solve for X for the graph: X1 = 25.4 mm = 0.254 m X0 (center) = 0 m

X

DAB t 3.72 10 6 t   0.005766t x12 0.254 2

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 Time, t = 130.073 hours

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 PA2 = 0 atm PA1 = 0.2 atm k'y = 6.78x10-5 kgmol/s m2 mol frac First calculate the mole fraction of A:

YA1  PA / P  0.2 / 2.0  0.1 YA2  0 To calculate ky, we must relate it to k'y:

k y y BM  k y'

y BM 

ky 

k y' y BM

y B 2  y B1 1  0.9   0.95 ln  y B 2 y B1  ln 1.0 / 0.9

6.78  10 5   7.138  10 5 kgmol/m2 s mole fraction 0.95

Now we can calculate kG:

7.138  10 5  3.569  10 5 kgmol/m2 s atm P 2.0 Now we can calculate the vaporization rate, NA: N A  k y  y A1  y A2   7.138  10 5 0.10  0  7.138  10 6 kgmol/m2s or 5 N A  k G PA1  PA2   3.569  10 0.2  0  7.138  10 6 kgmol/m2s k G y BM P  k y y BM



kG 

ky



7.2-1 Flux and Conversion of Mass-Transfer Coefficient

A value of kG was 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. kG' P  kG PBM Solve for PBM: P  PA2 0.2  0.05 PBM  A1   0.872853 atm  P  PA2  1  0.05  ln   ln  1  0.2   P  PA1  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

N A  kG' PA1  PA2   0.92470.2  0.05  .1414





N A  kG' PA1  PA2   1.262 10 8 20265  5066.25  1.92 10 4

NA = 0.1414 lbmol/ft2h NA = 1.92 x 10-4 kgmol/m2s

Mass 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: X BM 

X B 2  X B1 1  0  1  0.65   0.619  X B2   1  0  ln  ln     1  0.65  X B1 

To find the flux, NA, use:

k x'  X A1  X A2   0.20 0.65  0  0.210 lbmol/ft2 hr X BM 0.619 To find the ratios, set up the following: k x kc 1 1  '    1.616 ' k x k c X BM 0.619 NA 

So k x  1.616k x'  1.6160.2  0.323 lbmol/ft2 hr; and

k x0 k c0 1  X A1 1  0.65  '    0.565 X BM 0.619 k x' kc

So k x0  0.565k x'  0.5650.2  0.113 lbmol/ft2 hr

7.2-3 Absorption of H2S 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

kc' P  kG' P RT



k G' 

9.567  10 4  3.846  10 5 kgmol/m2s atm 3 82.057  10 303.15









N A  kG PA1  PA2    3.846  10 5 0.01218  0.05  1.455  10 6

NA = 1.455 x 10-6 kgmol/m2s

1

2

Mass 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: P 74  2.799  10 5 kgmol/m3 C Ai  Ai  RT 8314318 Calculate the Schmidt number:  1.932  10 5 N Sc    2.506 D AB 1.114 6.92  10 6 Calculate the Reynolds number: D 0.020.801.114 N Re    922.6  1.932  10 5 Hence, the flow is laminar. We will use Figure 7.3-2 for laminar flow streamlines: D 0.02  N Re N Sc  922.62.506  33.02 L 4 1.10 4 Using Figure 7.3-2 with rodlike flow: C A  C A0  0.55 C Ai  C A0 If CA0 = 0, C A  C A0 CA  0  0.55   CA = 1.539 x 10-5 kgmol/m3 5 C Ai  C A0 2.799  10  0





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. 1.829 m Pure Water T = 26.1ºC v = 3.05m/s

0.00635 m 1.22 m

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:  8.71  10 4 N Sc    702.408 D AB 996 1.245  10 9 Calculate the Reynolds number: D 0.006353.05996 N Re    22147  8.71  10 4 Calculate the Sherwood number: 0.83 0.33 N Sh  0.23N Re N Sc  0.000049487 Now calculate k’c: k ' 1.245  10 9 k' D 0.000049487  c N Sh  c  0.00635 D AB (b) CA1 = 0 CAi = 0.02948 (solubility) CA2 = ?









So now we plug this into the log mean driving force equation: C A2  C A1  vC A2  C A1   A  K C  C  C A1  ln  Ai   C Ai  C A2  CA1 = 0, so it cancels out of the equation:



k c'  1.59  10 4 m/s

vC A2   A  K C

C A2   C Ai  ln    C Ai  C A2 

Now plug in:

3.05C A2     0.003175 2 1.59  10 4 

C A2   0.02948  ln    0.02948  C A2 

CA2 = 0.001151 kgmol/m3 (c) Rate of benzoic acid dissolved = NA: vC A2  C A1  3.050.001151  0 NA    1.11  10 7 2 A   0.003175

Mass 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:

N Sc  Calculate the Reynolds number:

N Re  Now calculate JD:





D AB D





8.71  10 4  702 996 1.245  10 9





0.240.061996  17000 8.71  10 4

0.5 J D  0.99 N Re, L  0.9917000

0.5

Now solve for

 0.00758

k’c:

JD 

k c'



N Sc 2 / 3



k c'  5.85  10 6 m/s

In 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: k' 5.85  10 6 0.02948  0  1.726  10 7 kgmol/m2s N A  c C A1  C A2   X BM 1.0

Mass 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. v = 0.305 m/s D = 0.0254m DAB = 6.92x10-6 m/s PA1 = 0.555 mmHg = 74 Pa ρ = 1.113 kg/m3 T = 45ºC μ = 1.93 x 10-5 Pa s P = 1 atm Calculate the Schmidt number:

N Sc  Calculate the Reynolds number:

N Re  Now find the Sherwood number:

 D AB D



1.93  10 5   2.505 1.113 6.92  10 6







0.02540.3051.113  446 1.93  10 5

N Sh  2  0.552N Re 

0.53

Now solve for k’c: N sh  k c'

Now solve for k’G:

D D AB



k c'  N sh

N Sc 1/ 3  21.0

D AB 6.92  10 6  21.0  5.72  10 3 m/s D 0.0254

k c' 5.72  10 3 k    2.163  10 9 kgmol/s m2 Pa RT 8314318 ’ Since the gas is dilute, k G ≈ kG. Using that we can solve for the flux: N A  k G PA1  PA2   2.163  10 9 73  0  1.599  10 7 kgmol/m2 s ' G

7.3-1 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. Water at 26.1ºC, velocity, v = 0.152 m/sCA1 Benzoic Acid 0.137 m

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: C B1 XB1 = = [(996 kg/m3)/(18kg/kgmol)]/[(996 kg/m3)/(18kg/kgmol) + 0.02948 kgmol/m3] C B1  C A1 XB1 = 0.999468 XB2 = 1.0 X A1  X A2 X  X B1 1.0  0.999468 X BM   B2   0.999734 1  X A2   X B2   1.0  ln  ln  ln     0.999468  1  X X A1    B1  Calculate Schmidt Number:

N SC  Calculate Reynolds Number:

N Re 

 D AB

L





Calculate

K’c:

8.71  10 4  702 996  1.245  10 9

0.1370.152996  23812.5

Calculate JD:

J D  0.99 N Re



0.5

8.71  10 9

 0.9923812.5

0.5

2 / 3 K C'  J DN SC  0.064160.152702

2 / 3

Calculate Flux:

NA 



 0.06416  1.2346  10 5 m/s



K C' C A1  C A2  1.2346  10 5 0.2948  0   3.641  10 7 kgmol/m2s X BM 0.999734

7.3-4 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

Air at 325.6K, 202.6 kPa, v = 1.524 m/s

0.152 m

P = 202600 Pa R = 8314.34 m3 Pa/kgmol K T = 325.6 K Mw = 28.8 kg/kgmol Correct DAB:

T  D AB 325.6 K ,202.6kPa   o  T  Calculate ρ and μ:

1.75

 P   o P 

n

1

T   325.6      o   0.0171cP  T   273 

 325.6     273 

N Re 

L





0.1521.5242.155  25495.5 1.958  10 5

0.2

JD 

K

' C



2/3 N SC

 3.51182  10 6 m2/s

 0.1958 cP = 1.958 x 10-5 kg/m s

0.2 J D  0.036 N Re  0.036  25495.5

N SC 

1

M wP 28.8202600  2.155 kg/m3  8314.34325.6 RT

(a) Calculate NRe:

Because NRe > 15000:

 202.6     101.3 

0.768

o



1.75

 D AB



 0.004732

5

1.958  10  2.58722 2.155 3.51182  10 6





' C

K 2.587222 / 3  K’C = 0.003827 m/s 1.524 ' K 0.003827 KG  C   1.41366  10 9 kgmol/m2 Pa s RT 8314.34325.6 

0.004732 

(b) Dp = 0.0127 m Calculate Reynolds and Schmidt Numbers: D p 0.0127 1.5242.155 N Re    2130.21  1.958  10 5  1.958  10 5 N SC    2.58722 D AB 2.155 3.51182  10 6





Because 0.6 < NSC < 2.7: 0.53 1/ 3 0.53 1 / 3 N SH  2  0.552 N Re N SC  2  0.5522130.21 2.58722  46.0164

N SH 

K C'  D p D AB KG 



46.0164 

K C' 0.0127  3.51182  10 6



K’C = 0.012725 m/s

K C' 0.012735   4.70  10 9 kgmol/m2 Pa s RT 8314.34325.6

Mass 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 AS = 0.01198 m2 μ26.1ºC = 8.71 x 10-4 Pa s ε = 0.436 -4 μ25ºC = 8.94 x 10 Pa s D = 0.0667m V = 5.514x10-7 m3/s S = 2.948x10-2 kgmol/m3 Dp = 6.375mm DAB(25ºC) = 1.21x10-9 m2/s from Table 6.3-1 Correct DAB: 3  T     299.1  0.8940  10    1.254  10 9 m2/s D AB 26.1  D AB 25 new  old   1.21  10 9   3   298  0.8718  10   Told   new  Calculate the area of the column: A   D 2   0.0667  3.494  10 3 m2 4 4 Now use the area and volumetric flow to find the velocity: v  V / A  5.514  10 7 / 3.494  10 3  1.578  10 4 m/s Calculate the Schmidt Number:  8.71  10 4 N Sc    702.6 D AB 996.7  1.245  10 9 Calculate Reynolds number: D p 0.006375 1.578  10 4 996.7  N Re    1.150  8.71  10 4 Calculate JD: 1.09 N Re 2 / 3  1.09 1.1502 / 3  2.277 JD   .436 ’ Then, assuming k c = kc for dilute solutions, k c' k c' 2/3 702.62 / 3 J D  N Sc  2.277  k c'  4.447  10 6 m/s   4 v 1.578  10 Now set the log mean driving force equation can be set equal to the material balance equation on the bulk stream: C  C A1   C Ai  C A2  Ak c Ai  V C A2  C A1   C Ai  C A1  ln    C Ai  C A2  where CAi = 0.02948 (the solubility) CA1 = 0 A = 0.01198 m2 (the surface area of the bed) V = 5.514x10-7 m3/s CA2 = 2.842 x 10-3 kgmol/m3



















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: N A A  A  K C

C A2  C A1 

 C  C A1  ln  Ai   C Ai  C A2  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: N A  A  vC A2  C A1  So now we plug this into the log mean driving force equation: C A2  C A1  vC A2  C A1   A  K C  C  C A1  ln  Ai   C Ai  C A2  CA1 = 0, so it cancels out of the equation: C A2  vC A2   A  K C  C Ai  ln    C Ai  C A2  We can now plug in given numbers: 1.80  10 4 0.0701 1.80  10 4  0.129K C 0.00184   ln  4   0.00184  1.80  10 







KC = 0.0559 ft/hr

Mass 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 μc,water = 6.947 x 10-4 Pa s = 6.947 x 10-4 kg/ m s -9 2 DAB = 3.25x10 m /s ρc,water = 994 kg/m3 Solubility = 2.26x10-7 kgmol O2/m3 liquid ρp,air = 1.13 kg/m3 Calculate NSc:

N Sc  ’

c  c D AB



6.947  10 4  215 994 3.25  10 9





Now calculate k L:

2 D AB  2 / 3  c g    k   0.31N Sc  2  Dp c  

1/ 3

' L









4 9.806   2.29  10 4 m/s 2 3.25  10 9  2 / 3  994  1.13 6.947  10    k   0 . 31  215 4 2   1  10 994   ’ Assuming k L = kL for dilute solutions, N A  k L C A1  C A2   2.29  10 4 2.26  10 7  0  5.18  10 8 kgmol O2/m2 s ' L







1/ 3

Molecular 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: j A  j B  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:   A A   A   B B   B  0  A A   B B    A   B   0



A   A  B  B and    A   B  

    A A   B B   A  A  B  B    0    

 A A   B B   A A   B B   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. 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

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 (a)

NA 

CD AB









1  X A1 4.062  10 2 0.15  10 4 1  1.0  ln  ln   2.112  10 4 kgmol A/m2s  1  X A2 0.002  1 0 

(b) Substitute X A2 

NA in for XA2: k1' C NA 

NA

0.002

4





4.062  10 0.15  10  ln   2

CD AB

ln

1  X A1 N 1 A ' k1C

1  1.0

 N 1  A 5.63  10 3 4.062  10 2 





  2 4   1.004  10 kgmol A/m s  



7.5-7 Unsteady State Diffusion and Reaction Solute A is diffusing at unsteady state into a semi-infinite medium of pure B and undergoes a first-order reaction with B. Solute A is dilute. Calculate the concentration CA at points z = 0, 4, and 10 mm from the surface for t = 1 x 105 seconds. Physical property data are DAB = 1 x 10-9 m2/s, k’ = 1 x 10-4 s-1, CA0 = 1.0 kg mol/m3. Also calculate the kg mol absorbed/m2. Find CA and Q. For un-steady state diffusion and homogenous reaction in a semi-infinite medium,, the general solution is:    z   z  CA 1 k1'  k1'   2 exp  z  k i t   1 2 exp  z  ki t    erf    erf  C D AB    D AB   2 tD AB   2 tD AB  For z1=0m: CA 1  2 exp 0  erf  1  10 4 1  10 5   1 2 exp 0  erf 1  10 4 1  10 5  1.0 C A  1 2 exp 0 2  1 2 exp 0 0  1.0 kgmol/m3





For z2=0.004m

CA  1.0

1

  1 10 1 10





    erf  0.004  1  10 1  10      2 1 10 1 10       erf  0.004  1  10 1  10     2 1 10 1 10    110    0  0.2822642  0.28226 kgmol/m3 exp 0.004 110   

 1  10  4 exp  0 . 004  2 1  10 9 

4

5

9

5

4  4 5 exp 0 . 004  2 9 5 9  4  1104  1 1 C A  1 2 exp  0.004  2   2 2 9 9 1  10   For z2=0.01m    CA 1 1  10  4  0.01  2 exp  0.01  erf   1  10  4 1  10 5   9  1.0 1  10   2 1  10 5 1  10 9      1  10 4  0.01 4 5 1 exp 0.01    erf  1  10 1  10   2 1  10 9   2 1  10 5 1  10 9     1  10 4  1  10 4  1 1 C A  2 exp  0.01  1.9998  2 exp 0.01  0  1 2 0.0423291.9998 9  9  1  10 1  10     3 C A  0.0.0423 kgmol/m 1

 

 

     

     













 













 

(b) The total amount, Q, of A absorbed up to time t is represented by: D AB  ' 1 k ' t  k 't  ' Q  c A0 e   k t  2  erf k t   k '   Plugging in and solving for Q, we get:



Q 1



110  1 10 110    erf 1 10 110   1 10 110 e   110   9

4

4 '

Q  0.0332 kgmol/m

2

5

1 2

4

4

5

5



 110 4 1105

  

7.5-11 Effect of Slow Reaction Rate on Diffusion

Gas A diffuses from point 1 to a catalyst surface at point 2, where it reacts as follows: 2A  B. Gas B diffuses back a distance δ to point 1. (c) Derive an equation for NA for a slow first-order reaction where k1’ is the reaction velocity constant. (d) Calculate NA and xA2 for part (c) where k1’ – 0.53 x 10-2 m/s. (c) Begin with the convective diffusion equation: dC A C A N A  N B  where: x A  C A , N A   D AB  dz C C Plugging in:

N B   12 N A

 A2 dx A dx  x A  12 N A    N A  dz  CD AB  dz 1  xA 2 0 x A1 x

N A  CD AB

1  x A2 2  ln     1  x A1 2  N For a slow reaction, x A2  ' A . Plugging into the flux equation and replacing C with P/RT: k1 C NA 

2CD AB

NA 

2 PD AB 1  N A RT 2k1' P  ln   RT 1  x A1 2  





(d) Plugging in the values, we get: 2 1.01325  10 5 0.2  10 4 1  N A 8.314298 2 0.53  10 2 1.01325  10 5  NA  ln   8.3142980.0013 1  0.97 2   2 4 N A  1.76  10 kgmol/m s













Solving for the mole fraction: N 1.76  10 4 x A2  ' A   0.81479 k1C 1.01325  10 5 0.0053 8.314298





7.5-10 Diffusion and Chemical Reaction of Molten Iron in Process Metallurgy In a steel-making process using molten pig iron containing carbon, a spray of molten iron particles containing 4.0 wt % carbon falls through a pure oxygen atmosphere. The carbon diffuses through the molten iron to the surface of the drop, where it is assume that it reacts instantly at the surface because of the high temperature, as follows, according to a first order reaction: Calculate the maximum drop size allowable so that the final drop after a 2.0 second fall contains an average of 0.1 wt % carbon. Assume that the mass transfer rate of gases at the surface is very great, so there is no outside resistance. Assume no internal circulation of the liquid. Hence, the decarburization rate is controlled by the rate of diffusion of carbon to the surface of the droplet. The diffusivity of the carbon in iron is 7.5 x 10-9 m2/s (S7). Hint: use figure 5.3-13

For unsteady-state diffusion through a spherical geometry where we are looking at average temperature and time, Figure 5.3-13 (page 377) can be used. So, to find the radius, r, we need to find Y, and X, where: C  C ave 0  0.001 Y 1   0.025 C1  C0 0  004 Because A reacts instantly, we know that C1=0. From the graph for a sphere at Y, X=0.32 Solving for r: D t 7.5  10 9 2  X  0.32  AB2  r  0.000217 m r1 r12





Unsteady State Diffusion and Reaction in a Semi-Infinite Medium Reaction and Unsteady State Diffusion Example 7.5-3, page 498 Pure CO2 gas at 101.32 kPa pressure is absorbed into a dilute alkaline buffer solution containing a catalyst. The dilute, absorbed solute CO2 undergoes a first-order reaction, with k’= 35 s-1 and DAB = 1.5x10-9 m2/s. The solubility of CO2 is 2.961x10-7 kgmol/m3 Pa. The surface is exposed to the gas for 0.010 seconds. Calculate the kgmol CO2 absorbed/m2 surface. P = 101.32 kPa k’= 35 s-1 DAB = 1.5x10-9 m2/s The total amount, Q, of A absorbed in time, t is:

Solubility of CO2 is 2.961x10-7 kgmol/m3 Pa Time, t = 0.010 seconds



Q  C A0 DAB / k ' k ' t  12 erf k ' t  k ' t /  e k t '



k’t = (35s-1)(0.010s) = 0.350 CA0 = Solubility*Pressure = (2.961x10-7 kgmol/m3 Pa)( 101.32 kPa) = 3.00x10-2 kgmol CO2/m3





Q  3.00  10 2  1.5  10 9 / 35  .350  12 erf 0.350  0.350 /  e 0.350  1.458  10 7 kgmol CO2/m2

Multicomponent Diffusion of Gases Diffusion of A Through Nondiffusing B and C Example 7.5-4, page 498 At 298K and 1 atm total pressure, methane (A) is diffusing at steady state through nondiffusing argon (B) and helium (C). At z1 = 0, the partial pressures in atm are PA1 = 0.4, PB1 = 0.4, PC1 = 0.2, and at z2 = 0.005m, PA2 = 0.1, PB2 = 0.6, PC2 = 0.3. The binary diffusivities from Table 6.2-1 are DAB = 2.02x10-5 m2/s, DAC = 6.75x10-5 m2/s, and DBC = 7.29x10-5 m2/s. Calculate NA. PA1 = 0.4, PC2 = 0.3 DAB = 2.02x10-5 m2/s PB1 = 0.4, z2 = 0.005m DAC = 6.75x10-5 m2/s PC1 = 0.2 z1 = 0 DBC = 7.29x10-5 m2/s PA2 = 0.1 T = 298K R = 82.057 m3 atm/kgmol K PB2 = 0.6 P = 1 atm Find DAM:

XB 0.4   0.667 1  X A  1  0.4 XB 0.6 At point 2: X B'    0.667 1  X A  1  0.1 XC 0.2 X C'    0.333 1  X A  1  0.4 1 1 D AM  '   2.635  10 5 m2/s ' 0.667 X XB  0.333  C 2.02  10 5 6.75  10 5 D AB D AC Calculate PiM: Pi1  P  PA1  1.0  0.4  0.6 atm Pi 2  P  PA2  1.0  0.1  0.9 atm At point 1: X B' 



PiM 







Pi 2  Pi1 0.9  0.4   0.740 atm Pi 2  ln 0.9  ln   0.4  Pi1 





Now solve for NA: D AM P 2.635  10 5 1.0 PA1  PA2   0.9  0.6  8.74  10 5 kgmol A/m2s NA  82.0572980.0050.740 RT z 2  z1 PiM





7.5-8 Multicomponent Diffusion At a total pressure of 202.6 kPa and 358 K, ammonia gas (A) is diffusing at steady state through an inert, nondiffusing mixture of nitrogen (B) and hydrogen (C). The mole fractions at z1 = 0 are xA1 = 0.8, xB1 = 0.15, and xC1 = 0.05; and at z2 = 4.0 mm, xA1 = 0.2, xB1 = 0.6, and xC1 = 0.2. The diffusivities at 358 K and 101.3 kPa are DAB = 3.28 x 10-5 m2/s and DAC = 1.093 x 10-4 m2/s. Calculate the flux of ammonia. For multi-component diffusion, the flux can be calculated as: D AM C C  C  NA  z 2  z1 C AM A1 A2 Calculate DAM in m2/s: 1 1 D AM    1.9878  10 5     xC1 0.15 0.05 x B1       5 5  D AB 1  x A1  D AC 1  x A1    1.64  10 1  08 5.456  10 1  0.8  Calculate CAM: P 202.6  10 3 C   68.0647 mol/m3 RT 8314358 C A1  Cx A1  68.0647 0.8  54.45 mol/m3 C A2  Cx A2  68.0647 0.2 13.61 mol/m3

C AM 

C  C A1   C  C A2   C  C A1   ln   C  C A2 



68.0647  54.45  68.0647  13.61  29.4605 mol/m3  68.0647  54.45  ln    68.0647  13.61 

Substituting back in and solving for flux: 5 D AM C C A1  C A2   1.9878  10 68.0647  54.45molm 3  13.61molm 3  4.68  10 4 NA  z 2  z1 C AM 0.004  029.4605



N A  4.68  10 4 kgmole/m2s





 



Knudsen Diffusion of Gases Knudsen Diffusion of Hydrogen Example 7.6-1, page 500 A H2(A) – C2H6(B) gas mixture is diffusing in a pore of a nickel catalyst used for hydrogenation at 

1.01325x105 Pa and 373K. The pore radius is 60 A (angstrom). Calculate the Knudsen diffusivity, DKA, of H2. P = 1.01325x105 Pa T = 373K r = 60 angstroms = 6.0x10-9 MA(H2) = 2.016

DKA

 T    97.0r  MA 

1

2



 97.0 60  10

9



 373     2.016 

1

2

 7.92  10 6 m2/s

Flux Ratios for Diffusion of Gases In Capillaries Transition-Region Diffusion of He and N2 A gas mixture at a total pressure of 0.10 atm abs and 298K is composed of N2 (A) and He (B). The mixture is diffusing through an open capillary 0.010m long having a diameter of 5x10-6 m. The mole fraction of N2 at one end is XA1 = 0.8 and at the other is XA2 = 0.2. The molecular diffusivity DAB is 6.98x10-5 m2/s at 1 atm, which is an average value based on several investigations. (a) Calculate the flux NA at steady state. (b) Use the approximate equations for this case. L = 0.010m r = 2.5x10-6 m P = 0.10 atm T = 298 K XA1 = 0.8

XA2 = 0.2 DAB = 6.98x10-5 m2/s MA = 28.02 kg/kgmol MB = 4.003 kg/kgmol R = 8314.3 m3 Pa/kgmol K

(a) 1

1

 T  2  273  2 2 4   97.0 2.5  10 6  DKA  97.0r    7.91  10 m /s M 28 . 02    A In an open system with no chemical reaction, the ratio of NA/NB is constant: NB MA 28.02    2.645 NA MB 4.003 Now we can solve for the flux factor, α: N   1  B  1  2.645  1.645 NA Now we can solve for the transition region NA: D P 1  x A2  D AB / DKA  N A  AB ln  RTL  1  x A1  D AB / DKA 





6.98  10 1.013  10 

 

 

 

1   1.6450.2  6.98  10 5 / 7.91  10 4  NA  ln   6.40  10 5 kgmol/m2s 5 4   1.6458314.32980.010 1   1.6450.8  6.98  10 / 7.91 10  5

4

(b) If we estimate that equimolar counterdiffusion is taking place at steady state, α = 1-1 = 0. We can estimate the diffusivity, D’NA: 1 1 ' DNA    3.708  10 4 m2/s 1 1  1  1 D AB DKA 6.98  10 5 7.91  10 4 Now we can solve for the approximate flux: D' P 3.708  10 4 1.013  10 4 0.8  0.2  9.10  10 5 kgmol/m2s N A  NA  X A1  X A2   83142980.010 RTL















7.6-4 Transition-Region Diffusion in Capillary A mixture of nitrogen gas (A) and helium (B) at 298 K is diffusing through a capillary 0.10 m long in an open system with a diameter of 10 μm. The mole fractions are constant at XA1 = 1.0 and XA2 = 0.0. DAB = 6.98 x 10-5 m2/s at 1 atm. MA = 28.02 kg/kg mol, MB = 4.003. (a) Calculate the Knudsen diffusivity DKA and DKB at the total pressures of 0.001, 0.1 and 10 atm. (b) Calculate the flux NA at steady state at the pressures. (c) Plot NA versus P on log-log paper. What are the limiting lines at lower pressures and very high pressures? Calculate and plot these lines. (a) The Knudsen diffusivity for A and B will be the same at all pressures (it is independent of pressure):  T   DK  97.0r  M  A

1

2





1

6





1

6

DKA  97.0 5  10

DKB  97.0 5  10

2  298  2    0.001582 m /s  28.02 

2  298  2    0.004185 m /s  4.003 

(b) For the transition region, flux is calculated using: D P 1  x A2  DAB DKA  N A  AB ln  RTL  1  x A1  DAB DKA  For P = 0.001: 1   1.6457 0  6.98  10 2 0.001582 6.98  10 2 0.001 2 7 NA  ln    6.21  10 kgmol/m s 2  1.64570.0821572980.1  1   1.64571  6.98  10 0.001582  For P = 0.1: 1   1.6457 0  6.98  10 2 0.001582 6.98  10 2 0.1 2 5 NA  ln    1.321  10 kgmol/m s 2  1.64570.0821572980.1  1   1.64571  6.98  10 0.001582  For P = 10: 1   1.6457 0  6.98  10 2 0.001582 6.98  10 2 10 2 5 NA  ln    1.68  10 kgmol/m s 2  1.64570.0821572980.1  1   1.64571  6.98  10 0.001582 





 

 





 

 





 

 

(c) Lower limit is representative of Knudsen diffusion and the upper limit is representative of molecular diffusion.

9.3 Vapor Pressure of Water and Humidity 9.3-1 Humidity from Vapor-Pressure Data The air in a room is at 26.7ºC and a pressure of 101.325 kPa and contains water vapor with a partial pressure pA = 2.76 kPa. Calculate the following: (a) Humidity, H (b) Saturation humidity, HS, and percentage humidity, HP. (c) Percentage relative humidity, HR. From the steam tables at 26.7ºC, the vapor pressure of water is pAs = 3.50 kPa. Also pA = 2.76 kPa and P = 101.3 kPa. For part (a): 18.02 p A 18.022.76 H   0.01742 kg H2O/ky dry air 28.97 P  p A 28.97101.3  2.76 For part (b) 18.02 p As 18.023.50 H   0.02226 28.97 P  p As 28.97101.3  2.76 H 0.01742 H p  100  100  78.3% Hs 0.02226 For part (c) p 2.76 H R  100 A  100  78.9% p As 3.50

9.3-2 Use of Humidity Chart Air entering a dryer has a temperature (dry bulb) of 60ºC and a dew point of 26.7ºC. Using the humidity chart, determine the actual humidity H, percentage humidity HP, humid heat cs, and humid volume vH. The dew point of 26.7ºC is the temperature when the given mixture is at 100% saturation. Starting at 26.7ºC, and drawing a vertical line until it intersects the line for 100% humidity, a humidity of H = 0.0225 kg H2O/kg dry air is read off the plot. This is the actual humidity of the air at 60ºC. Stated in another way, if air at 60ºC and having a humidity of H = 0.0225 kg H2O/kg dry air is cooled, its dew point will be 26.7ºC. Locating this point where H = 0.0225 and T = 60ºC on the chart, the percentage humidity HP is found to be 14% by linear interpolation vertically between 10 and 20% lines. The humid heat for H = 0.0225 is from Eq. (9.3-6): cs  1.005  1.880.0225  1.047 kJ/kg dry air K The humid volume at 60ºC from Eq. (9.3-7) is v H  2.83  10 3  4.56  10 3 0.0225 60  273  0.977 m3/kg dry air





9.3-3 Adiabatic Saturation of Air An air stream at 87.8ºC having a humidity H = 0.030 kg H2O/kg dry air is contacted in an adiabatic saturator with water. It is cooled and humidified to 90% saturation. (a) What are the final values of H and T? (b) For 100% saturation, what would be the values of H and T? For part (a), the point H = 0.030 and T = 87.8ºC is located on the humidity chart. The adiabatic saturation curve through this point is followed upward to the left until it intersects with the 90% line at 42.5ºC and H = 0.0500 kg H2O/kg dry air. For part (b), the same line is followed to 100% saturation where T = 40.5ºC and H = 0.0505 kg H2O/kg dry air.

9.3-4 Wet Bulb Temperature and Humidity A water vapo-air mixture having a dry bulb temperature of T = 60ºC is passed over a wet bulb, as shown in Figure 9.3-4, and the wet bulb temperature obtained is TW = 29.5ºC. What is the humidity of the mixture? The wet bulb temperature of 29.5ºC can be assumed to be the same as the adiabatic saturation temperature TS, as discussed. Following the adiabatic saturation curve of 29.5ºC until it reaches the dry bulb temperature of 60ºC, the humidity is H = 0.0135 kg H2O/kg dry air.

9.6 Calculation Methods for Constant-Rate Drying Period 9.6-1 Time of Drying from Drying Curve A solid whose drying curve is represented by Figure 9.5-1a is to be dried from a free moisture content X1 = 0.38 kg H2O/kg dry solid to X2 = 0.25. Estimate the time required. From Figure 9.5-1a for X1 = 0.38, t1 is read off as 1.28 hours. For X2 = 0.25, t2 = 3.08 hours. Hence the time required is t = t2 – t1 = 3.08 – 1.28 = 1.80 hours.

9.6-2 Drying Time from Rate-of-Drying Curve A solid whose drying curve is represented by Figure 9.5-1b is to be dried from a free moisture content X1 = 0.38 kg H2O/kg dry solid to X2 = 0.25 in the constant rate drying period. Estimate the time required. For Figure 9.5-1b, a value of 21.5 for LS/A was used to prepare the graph. From the figure, RC = 1.51 kg H2O/m2hours. Substituting into Eq. 9.6-2: L 21.5 0.38  0.25  1.85 hours t  S X 1  X 2   ARC 1.51

9.6-3 Prediction of Constant Rate Drying An insoluble wet granular material is to be dried in a pan 0.457 by 0.457 m and 25.4 mm deep. The material is 25.4 mm deep in the pan, and the sides and bottom can be considered insulated. Heat transfer is by convection from an air stream flowing parallel to the surface at a velocity of 6.1 m/s. The air is at 65.6ºC and has a humidity of 0.010 kg H2O/kg dry air. Estimate the rate of drying for the constant rate period. For a humidity of H = 0.010 and a dry bulb temperature of 65.6ºC, using the humidity chart, the wet bulb temperature TW = 28.9ºC and HW = 0.026 by following the adiabatic saturation line to the saturated humidity. Using Eq 9.3-7 to calculate the humid volume, vH  2.83  10 3  4.56  10 3 H T  2.83  10 3  4.56  10 3 .01 65.6  273  0.974 m3/kg dry air



 



The density for 1.0 kg dry air + 0.010 kg H2O is 1.0  0.010   1.037 kg/m3 0.974 The mass velocity G is G  v  6.136001.037  22770 kg/h Using Eq. 9.6-9 0.8 h  0.0204G 0.8  0.020422770  62.45 W/m2K At TW = 28.9ºC, λW = 2433 kJ/kg from the steam tables. Substituting into Eq. 9.6-8: h T  TW 3600  62.45 65.6  28.93600  3.39 kg/m2h RC  W 2433  1000 The total evaporation rate for a surface area of 0.457 by 0.457 m2 is RC  A  3.390.457  0.457  0.708 kg H2O / h

9.7 Calculation Methods for Falling-Rate Drying Period Example 9.7-1 Numerical Integration in Falling-Rate Period A batch of wet solid whose drying rate is represented by Figure 9.5-1b is to be dried from a free moisture content of X1 = 0.38 kg H2O/kg dry solid to X2 = 0.04. The weight of the dry solid is LS = 399 kg solid and A = 18.58 m2 of top drying surface. Calculate the time for drying. Note that the LS/A = 399/18.58 = 21.5 kg/m2. From Figure 9.5-1b, the critical free moisture content is XC = 0.195 kg H2O/kg dry solid. Hence the drying occurs in the constant-rate and falling-rate periods. For the constant-rate period, X1 = 0.38 and X2 = XC = 0.195. From Figure 9.5-1b, RC = 1.51 kg H2O/m2h. Substituting Eq. 9.6-2: L 399 0.38  0.195  2.63 hours t  S X 1  X 2   18.581.51 ARC For the falling-rate period, reading values of R for various values of X from Figure 9.5-1b, the following table was prepared. To determine this area by numerical integration using a spreadsheet, the calculations below are provided. The area of the first rectangle is the average height (0.663+0.826)/2 = 0.745, times the width ΔX = 0.045, giving 0.0335. Other values are similarly calculated and all the values are summed to give a total of 0.1889.

X 0.195 0.150 0.100 0.065 0.050 0.040

R 1/R ΔX (1/R)avg (ΔX)( 1/R)avg 1.51 0.663 0.045 0.745 0.0335 1.21 0.826 0.050 0.969 0.0485 0.9 1.11 0.035 1.260 0.0441 0.71 1.41 0.015 2.055 0.0308 0.37 2.70 0.010 3.203 0.0320 0.27 3.70 -Total = 0.1889

Substituting into Eq. 9.6-1 L X1 dX 399 0.1889  4.06 hours t S   X A 2 R 18.58 The total time is 2.63 + 4.06 = 6.69 hours.

Approximation of Straight Line for Falling-Rate Period A batch of wet solid whose drying rate is represented by Figure 9.5-1b is to be dried from a free moisture content of X1 = 0.38 kg H2O/kg dry solid to X2 = 0.04. The weight of the dry solid is LS = 399 kg solid and A = 18.58 m2 of top drying surface. Calculate the time for drying. Note that the LS/A = 399/18.58 = 21.5 kg/m2. As an approximation, assume a straight line for the rate R versus X through the origin from point XC to X = 0 for the falling-rate period. RC = 1.51 kg H2O/m2h and XC = 0.195. Drying in the falling rate region is from XC to X2 = 0.040. Substituting into Eq. 9.7-8 L X X 3990.040 0.195 t  S C ln C  ln  4.39 hours ARC X 2 18.581.51 0.040 This is comparable to the value of 4.06 hours obtained in Example 9.7-1 by numerical integration

10.2 Equilibrium Relations between Phases 10.2-1 Dissolved Oxygen Concentration in Water What will be the concentration of oxygen dissolved in water at 298K when the solution is in equilibrium with air at 1 atm total pressure? The Henry’s law constant is 4.38x104 atm/molfrac. Given: p A  0.21atm p A  Hx A   0.21  3.38  10 4 x A   x A  4.80  10 6 molfrac or 0.000835 parts O2 to 100 parts water

10.3 Single and Multiple Equilibrium Contact Stages 10.3-1 Equilibrium Stage Contact for CO2-Air-Water A gas mixture at 1.0 atm pressure abs containing air and CO2 is contacted in a single-stage mixer continuously with pure water at 293K. The two exit gas and liquid streams reach equilibrium. The inlet gas flow rate is 100kgmol/hr, with a mole fraction of CO2 of yA2=0.20. The liquid flow rate entering is 300kgmol water/hr. Calculate the amounts and compositions of the two outlet phases. Assume that water does not vaporize to the gas phase. Given: L0  L'  300kgmol / h

V '  V (1  y A )  100(1  0.20)  80kgmol / hr From Appendix A.3 at 293K -Henry’s Law Constant: H  0.142  10 4 atm / molfrac H '  H / P  0.142  10 4 / 1.0  0.142  10 4 mol frac gas / mol frac liquid y A1  H ' x A1   0.142  10 4 x A1

Total Material Balance:  x   y   x   y L'  2   V '  1   L'  1   V '  2  1  x2   1  y1   1  x1   1  y2

   4  x A1  1.41  10 , y A1  0.20 4  0.142  10 x A1   x A1   0   0.2     80 300   80   300 4  1  0 . 142  10 x 1  0   1  0.2   1  x A1  A1    Total Flow Rates: L' 300 V' 80 L1    300kgmol / hr V1    100kgmol / hr 4 1  y A1 1  0.20 1  x A1 1  1.41  10   

10.3-2 Absorption of Acetone in a Countercurrent Stage Tower It is desired to absorb 90% of the acetone in a gas containing 1.0 mol% acetone in air in a countercurrent stage tower. The total inlet gas flow to the tower is 30.0kgmol/h, and the total inlet pure water flow to be used to absorb the acetone is 90kgmolH2O/h. The process is to operate isothermally at 300K and a total pressure of 101.3kPa. The equilibrium relation for the acetone (A) in the gas-liquid is yA=2.53xA. Determine the number of theoretical stages required for this separation. Given: y AN 1  0.01

x A0  0

VN 1  30.0kgmol / h

L0  90.0kgmol / hr

Acetone Material Balance: Amount of entering acetone= y AN 1VN 1  0.01(30.0)  0.30kgmol / hr Entering Air = (1  y AN 1 )VN 1  (1  0.01)(30.0)  29.7kgmol air / h Acetone Leaving in V1= 0.10(30)  0.030kgmol / hr Acetone Leaving in LN= 0.90(0.30)  0.27kgmol / hr V1=29.7+0.03=29.73 kg mol air + acetone/hr y A1 

0.030  0.00101 29.73

LN=0.90+0.27=90.27 kg mol water + acetone/h x AN 

0.27  0.00300 90.27

*Since the liquid flow does not vary significantly, the operating line is assumed to be straight. Plot the operating line and equilibrium line and step off stages. 5.2 theoretical stages required.

10.3-3 Number of Stages by Analytical Equations It is desired to absorb 90% of the acetone in a gas containing 1.0 mol% acetone in air in a countercurrent stage tower. The total inlet gas flow to the tower is 30.0kgmol/h, and the total inlet pure water flow to be used to absorb the acetone is 90kgmolH2O/h. The process is to operate isothermally at 300K and a total pressure of 101.3kPa. The equilibrium relation for the acetone (A) in the gas-liquid is yA=2.53xA. Determine the number of theoretical stages required for this separation. Solve using the Kremser analytical equations for countercurrent stage process. Given: y AN 1  0.01

x A0  0

VN 1  30.0kgmol / h

L0  90.0kgmol / hr

Acetone Material Balance: Amount of entering acetone= y AN 1VN 1  0.01(30.0)  0.30kgmol / hr Entering Air = (1  y AN 1 )VN 1  (1  0.01)(30.0)  29.7kgmol air / h Acetone Leaving in V1= 0.10(30)  0.030kgmol / hr Acetone Leaving in LN= 0.90(0.30)  0.27kgmol / hr V1=29.7+0.03=29.73 kg mol air + acetone/hr y A1 

0.030  0.00101 29.73

LN=0.90+0.27=90.27 kg mol water + acetone/h x AN 

0.27  0.00300 90.27

The equilibrium relation is: y=2.53xA so m=2.53

L L 90.0   0   1.20 mV mV1 2.53  29.73   A  A1 A2  1.20(1.19)  1.195 LN 90.27  AN    1.19  mVN 1 2.53  30.0 y  mx0  1  1  0.01  2.53(0)  1  1  ln  N 1 1     ln  1    A  A  y1  mx0   0.00101  2.53(0)  1.195  1.195  N   5.04 stages ln A ln 1.195 A1 

10.4 Mass Transfer between Phases

10.4-1 Interface Compositions in Interphase Mass Transfer The solute A is being absorbed from a gas mixture of A and B in a wetted-wall tower with the liquid flowing as a film downward along the wall. At a certain point in the tower the bulk gas concentration yAG=0.380 mol fraction and the bulk liquid concentration is xAL=0.100. The tower is operating at 298K and 1.013x105Pa and the equilibrium data are as follows (x,y data given). The solute A diffuses through stagnant B in the gas phase and then through a nondiffusing liquid. Using correlations for dilute solutions in wetted-wall towers, the film mass-transfer coefficients for A in the gas phase is predicted as ky=1.465x10-3kgmolA/sm2molfrac and for the liquid phase as kx=1.967x10-3kgmolA /sm2molfrac. Calculate the interface concentrations yAi and xAi and the flux NA. Since system is dilute: (1  y A ) iM  1

(1  x A ) iM  1

Plot the equilibrium data and point P at (0.1,0.38) Estimate the slope: slope  

k x' /(1  x A ) iM k y' /(1  y A ) iM



1.967  10 3 / 1.0  1.342 1.465  10 3 / 1.0

Plotting this, we get yAi=0.183 and xAi=0.247. Use these new values to re-estimate slope:

(1  y Ai )  (1  y AG ) (1  0.183)  (1  0.380)   0.715 ln[(1  y Ai ) /(1  y AG )] ln[(1  0.183) /(1  0.380)] (1  x AL )  (1  x Ai ) (1  0.100)  (1  0.247)    0.825 ln[(1  x AL ) /(1  x Ai )] ln[(1  0.100) /(1  0.247)]

(1  y A ) iM  (1  x A ) iM slope  

k x' /(1  x A ) iM k y' /(1  y A ) iM



1.967  10 3 / 0.825  1.163 1.465  10 3 / 0.715

Plotting this, we get yAi=0.197 and xAi=0.257. Use these new values to re-estimate slope:

(1  y Ai )  (1  y AG ) (1  0.197)  (1  0.380)   0.709 ln[(1  y Ai ) /(1  y AG )] ln[(1  0.197) /(1  0.380)] (1  x AL )  (1  x Ai ) (1  0.100)  (1  0.257)    0.820 ln[(1  x AL ) /(1  x Ai )] ln[(1  0.100) /(1  0.257)]

(1  y A ) iM  (1  x A ) iM

k x' /(1  x A ) iM

1.967  10 3 / 0.820  1.160 k y' /(1  y A ) iM 1.465  10 3 / 0.709 This is essentially the same slope so use the interfacial values obtained before. slope  



Calculate flux through each phase: k y' 1.465  10 3 NA  ( y AG  y Ai )  (0.380  0.197)  3.78  10  4 kgmol / sm 2 (1  y A ) iM 0.820

k x' 1.967  10 3 NA  ( x Ai  x AL )  (0.257  0.100)  3.78  10  4 kgmol / sm 2 (1  x A ) iM 0.820

10.4-2 Overall Mass-Transfer Coefficients from Film Coefficients Using the same data as in Example 10.4-1, calculate the overall mass-transfer coefficient Ky’, the flux, and the percent resistance in the gas and liquid films. Do this for the case of A diffusing through stagnant B. From the graph: yA*=0.052,

xAL=0.10,

yAG=0.380,

y Ai  y *A 0.197  0.052 m'    0.923 x Ai  x AL 0.257  0.100 From the previous example: k y' 1.465  10 3 and  1  y A iM 0.709 (1  y *A )  (1  y AG )

yAi=0.197,

xAi=0.257

k x' 1.967  10 3  1  x A iM 0.820

(1  0.052)  (1  0.380)  0.733 ln[(1  y ) /(1  y AG )] ln[(1  0.052) /(1  0.380)] 1 1 1  '  ' ' K y /(1  y A ) *M k y /(1  y A ) iM k x /(1  x A ) iM

(1  y A ) *M 

* A



1 1 0.923     K y'  8.90  10  4 3 3 K / 0.773 1.465  10 / 0.709 1.967  10 / 0.820 ' y

The percent resistance: Gas film: (484/868.8)(100)=55.7% Calculate the Flux: N A 

K y' (1  y A ) *M

Liquid Film: 44.3%

( y AG  y *A ) 

8.90  10 4 (0.380  0.052)  3.78  10  4 kgmol / sm 2 0.773

10.6 Absorption in Plate and Packed Towers

10.6-1 Pressure Drop and Tower Diameter for Ammonia Absorption Ammonia is being absorbed in a tower using pure water at 25C and 1.0 atm abs pressure. The feed rate is 1140 lbm/h (653.2 kg/h) and contains 3.0mol% ammonia in air. The process design specifies a liquid-to-gas mass flow rate ratio GL/GG of 2/1 and the use of 1-in metal Pall rings.Calculate the pressure drop in the packing and gas mass velocity at flooding. Using 50% of the flooding velocity, calculate the pressure drop, gas and liquid flows, and tower diameter. Size the tower for the largest flows (at the bottom of the tower). Assume all ammonia is absorbed. MW gas  28.97(0.97)  17.0(0.03)  28.61 and x ammonia  0.03(17) /( 28.61)  0.01783

 492  28.64  0.07309lbm / ft 3  460  77 359  

G  

From Appendix A.2-4 for water:   0.8937cp From Appendix A.2-3 for water:   0.99708gm / cm 3

 L  0.99708(62.43)  62.25lbm / ft 3 v   /   0.8937 / 0.99708  0.8963 centistokes From Table 10.6-1 for 1-in Pall Rings: Fp=56ft-1. Pflood  0.115Fp0.7  0.115(56) 0.7  1.925H 2 O / ft packing height For Figure 10.6-5 for Random Packing: (GL / GG )(  G /  G ) 0.5  (2.0)(0.07309 / 62.25) 0.5  0.06853 From Figure 10.6-5 for 0.06853 and 1.925, the ordinate=1.7. Capacity parameter equation from ordinate: (To find volumetric flow) 0.5

1.7  vG [  G /(  L   G )] F v 0.5

0.5 p

0.05

  0.07309  vG  (56) 0.5 (0.8963) 0.05   (62.25  0.07309) 

vG  6.663 ft / s GG  vG  G  6.663(0.07309)  0.4870lbm /( sft 2 )

50% Flooding: GG  0.5GG  0.2435lbm and for the liquid: GL  2.0(0.2435)  0.4870lbm / sft 2 New capacity parameter: 0.5(1.7)  0.85 From Figure 10.6-5 for Random Packing at 0.85 and 0.06853: 0.18 inches of water/ft

feedrate  1140  1   lb / s  lb m /( sft 2 )   1.6427 ft 2 gas flow rate  3600  0.2435  2 2 AC  1.6427 ft  ( / 4) D   D  1.446 ft

Solving for Diameter: AC 

Ammonia in outlet water: 0.01783(1440)  25.68lb Total liquid flow rate: 2(1440)  2880lbm / hr Pure liquid flow rate: 2880  25.68  2858.3 lbm / s

10.6-2 Absorption of SO2 in a Tray Tower A tray tower is to be designed to absorb SO2 from an air stream by using pure water at 293K. The entering gas contains 20 mol% SO2 and that leaving 2mol% at a total pressure of 101.3kPa. the inert air flow rate is 150kg air/hm2, and the entering water flow rate is 6000kgwater/hm2. Assuming an overall tray efficiency of 25%, how many theoretical trays and actual trays are needed? Assume that the tower operates at 293K. Calculate the molar flow rates: 150 6000 V'  5.18kgmol inert air / hm 2 L'   333kgmol inert water / hm 2 29 18.0 From Figure 10.6-7 (drawing of tower trays): y N 1  0.2 , y1  0.02 , x 0  0 Overall Material Balance:  x   y   x   y L'  2   V '  1   L'  1   V '  2  1  x2   1  y1   1  x1   1  y2

  

 xN  0.02   0   0.2    5.18 333   5.18   333  1  0   1  0.2   1  0.02  1  xN  x N  0.00355 For a portion of the tower:  y n 1   x  0.02   0    333 n   5.18 333   5.18  1  0   1  0.02   1  y n 1   1  xn  Choose several y points arbitrarily and plot the operating line. Plot the equilibrium data from Appendix A.3. Step off trays from bottom to top: 2.4 theoretical trays Number of trays with 25% efficiency =0.25(2.4)= 9.6 trays

10.6-3 Minimum Liquid Flow Rate and Analytical Determination of Number of Trays A tray tower is absorbing ethyl alcohol from an inert gas stream using pure water at 303K and 101.3 kPa. The inlet gas stream flow rate is 100.0 kgmol/h and it contains 2.2 mol% alcohol. It is desired to recover 90% of the alcohol. The equilibrium relationship is y=mx=0.68x for this dilute stream. Using 1.5 times the minimum liquid flow rate, determine the number of trays needed. Do this graphically and also using the analytical equations. Given: y1  0.022

V1  100.00 kgmol / h

x2  0

m  0.68

V '  V1 (1  y1 )  100.0(1  0.022)  97.8kgmol inert / hr Moles alcohol/h in V1 : 100-97.8=2.20. Removing 90% moles/hr in outlet gas V2: 0.1(2.20)=0.220

V2  V '0.220  97.8  0.22  98.02 y 2  0.22 / 98.02  0.002244 Plot the equilibrium line and points y1, y2, and x2. Draw the operating line from y2, and x2 to point P for Lmin. At point P on the equilibrium line, xmax is: x1 max  y1 / m  0.022 / 0.68  0.03235 Substitute into the operating line equation to find Lmin:  x1, max   x   y   y   V '  2 Lmin  2   V '  1   L'min   1  x2   1  y1   1  y2  1  x1. max 

  

 0   0.22   0.03235   0.002244  ' L'min   L'min  59.24kgmol / hr   97.8   Lmin    97.8  1  0 1  0 . 22 1  0 . 03235 1  0 . 002244         Solve for L and x: L'  1.5L'min  (1.5)(59.24)  88.86

 x1   0   0.22   0.002244    97.8 88.86  x1  0.0218   97.8   88.86  1  0   1  0.22   1  0.002244   1  x1  Plot the operating line with several points. The solution is dilute, so the line is straight. Step off trays: 4 trays. Flow Rates: V1  100.0

L2  L'  8886

V2  V ' /(1  y 2 )  97.8 /(1  0.002244)  98.02

L1  L' / mV1  88.86 /(1  0.0218)  90.84

Calculating A for analytical solution: A1  L1 / mV1  90.84 /( 0.68)(100)  1.336

A2  L2 / mV2  88.86 /( 0.068)(98.02)  1.333

A  A1 A2  1.355 Calculate N:  y  mx2    0.022  0   1 1 (1  1 / A)  1 / A  N ln  1 ln  (1  1 / 1.335)  1 / 1.335  4.04 ln A  y 2  mx2    ln 1.335  0.002244  0 

10.6-4 Absorption of Acetone in a Packed Tower

Acetone is being absorbed by water in a packed tower having a cross-sectional area of 0.186 m2 at 293K and 101.32 kPa (1 atm). The inlet air contains 2.6 mol% acetone and outlet 0.5%. The gas flow is 13.65kgmol inert air/h. The pure water inlet flow is 45.36 kgmol water/h. Film coefficients for the given flows in the tower are: k y' a  3.78  10 2 kgmol/sm3molfrac and k x' a  6.16  10 2 kgmol/sm3molfrac. Equilibrium data are given in Appendix A.3. a) Calculate the tower height using ky’a. b) Repeat using kx’a c) Calculate Ky’a and the tower height. Given: y1  0.026

L'  45.36kgmol / h

y 2  0.005

V '  13.65kgmol / h

x2  0

From Appendix A.3 for acetone-water and xA=0.0333 mol frac: p A  30.760  0.0395atm   y  0.0395 Equilibrium line: y A  mx A   0.0395  m(0.0333)   y  1.186 x Plot:

Overall Material Balance:

 x L'  2  1  x2

 x    y   y   V '  1   L'  1,   V '  2   1  y1   1  y2  1  x1 

  

 x1   0   0.026   0.0055    13.65 45.36  x1  0.00648   13.65   45.36  1  0   1  0.026   1  0.0055   1  x1  Plot the operating line. Approximate the slope at y1 and x1 using trial and error: k x' a /(1  x) iM k x' a /(1  x1 ) i 6.16  10 2 /(1  0.00648) i slope   '  '   1.60 k y a /(1  y) iM k y a /(1  y1 ) 3.78  10  2 /(1  0.026) y1*=0.0077 Using new points yi1=0.0154 and xi1=0.0130 continue trial and error: (1  y Ai )  (1  y AG ) (1  0.0154)  (1  0.026) (1  y A ) iM    0.979 ln[(1  y Ai ) /(1  y AG )] ln[(1  0.0154) /(1  0.026)] (1  x AL )  (1  x Ai ) (1  0.00648)  (1  0.0130) (1  x A ) iM    0.993 ln[(1  x AL ) /(1  x Ai )] ln[(1  0.00648) /(1  0.0130)] Recalculate the slope: slope  

k x' a /(1  x) iM k y' a /(1  y ) iM



6.16  10 2 / 0.993i 3.78  10  2 / 0.929

 1.61

Repeat this process for y2 and x2: k ' a /(1  x) iM k ' a /(1  x1 ) i 6.16  10 2 /(1  0) i slope   x'   'x   1.62 k y a /(1  y) iM k y a /(1  y1 ) 3.78  10  2 /(1  0.005) The slope changes very little. Plotting this line we get: y2*=0.0, yi2=0.0020, and xi2=0.0018 Using the interfacial concentrations: ( y  y i1 )  ( y 2  y i 2 ) (0.026  0.0154)  (0.005  0.002) ( y  yi ) M    0.00602 ln[( y1  y i1 ) /( y 2  y i 2 )] ln[(0.026  0.0154) /( 0.005  0.002)] Calculate the flow rates: V' 13.65 / 3600  V1    3.893  10 3 kgmol / s  1  y1 1  0.026 V1  V2  V Av  V' 13.65 / 3600 2 V2    3.811  10 3 kgmol / s   1  y2 1  0.005

3.893 103  3.811103  3.852 103 kgmol / s 2 45.36 L'  L1  L2  L Av   1.260  10  2 kgmol / s 3600

VAv 

a) V 3.852  10 3 ( y1  y 2 )  k y' az ( y  y i ) M   (0.0260  0.005)  3.78  10  2 z (0.00602)   z  1.911m S 0.186

b) ( xi  x) M 

( xi1  x1 )  ( xi 2  x 2 ) (0.0130  0.00648)  (0.0018  0)   0.00368 ln[( xi1  x1 ) /( xi 2  x 2 )] ln[(0.0130  0.00648) /( 0.0018  0)]

L 1.260  10 2 ( x1  x 2 )  k x' az ( xi  x) M   (0.00648  0)  6.16  10  2 z (0.00368)   z  1.936m S 0.186

(1  y1* )  (1  y1 ) (1  0.0077)  (1  0.026) c) (1  y) *M    0.983 * ln[(1  y1 ) /(1  y1 )] ln[(1  0.0077) /(1  0.026)] 1 1 m' 1 1 1.186  '  '   '   ' 2 K y a /(1  y ) *M k y a /(1  y ) iM k x a /(1  x) iM K y a / 0.983 3.78  10 / 0.979 1.16  10  2 / 0.993 K y' a  2.183  10  2 kgmol / sm 3 molfrac

(y  y )M *

( y1  y1* )  ( y 2  y 2* ) (0.026  0.0077)  (0.005  0)    0.01025 * * ln[( y1  y1 ) /( y 2  y 2 )] ln[(0.026  0.0077) /( 0.005  0)]

V 3.852  10 3 ( y1  y 2 )  K y' az ( y  y * ) M   (0.0260  0.005)  2.183  10  2 z (0.01025)   z  1.944m S 0.186

10.7 Absorbtion of Concentrated Mixtures in Packed Towers 10.7-1 Design of an Absorption Tower with a Concentrated Gas Mixture A tower packed with 25.4-mm ceramic rings is to be designed to absorb SO2 from air by using pure water at 293K and 1.013X105Pa abs pressure. The entering gas contains 20mol% SO2 and that leaving 2 mol%. The inert air flow is 6.53x10-4kgmol air/s and the inert water flow is 4.20x10-2kgmol water/s. The tower crosssectional area is 0.0929 m2. For dilute SO2, the film mass-transfer coefficients at 293K are, for 25.44 rings are: k y' a  0.0594G y0.7 Gx0.25 and k x' a  0.152Gx0.82 where ky’a is kgmol/sm3molfrac, kx’a is kgmol/sm3molfrac, and Gx and Gy are kg total liquid or gas, respectively, per sec per m2 tower cross section. Calculate the tower height. Given: V '  6.53  10 4 kgmolair / s L'  4.20  10 2 kgmol / s

y1  0.20 y 2  0.02

Overall Material Balance:  x   y   x   y L'  2   V '  1   L'  1   V '  2  1  x2   1  y1   1  x1   1  y2

x2  0

  

x1   0   0.02   4  0.2  2    6.53  10  4  4.2  10  2   x1  0.00355   6.53  10    4.20  10   1  0   1  0.2   1  0.02   1  x1  Operating Line: y   x   4  0.2   2  0.00355  4   4.2  10  2    6.53  10    4.20  10    6.53  10  1  y 1  x   1  0.2   1  0.00355    Set y=0.04, x=0.000332. Choose other values of y and plug in for x. Plot equilibrium data from Appendix A.3 and the operating data. Repeat the following calculations for different y values. For y=0.2: V' 6.53 10 4 ' V1    8.16 10  4 1  y1 1  0.2

Gy 

mass flow of air  SO2 ' AC

   0.2  6.53  10  4 29kgair / s   64.1kgSO2 / s   1  0.2    Gy   0.3164kg / sm 2 2 0.0929m Similarly for liquid flow:    0.00355  4.20  10  2 18   64.1 2 L' 4.2  10 '  1  0.00355    Gx   8.241 L1    0.04215 2 1  x1 1  0.00355 0.0929m

k x' a  0.152G x0.82  0.152(8.241) 0.82  0.857 , k y' a  0.0594G y0.7 G x0.25  0.0594(0.3164) 0.7 (8.241) 025  0.04496 Interface Compositions: Estimate the slope of PM by trial and error: Final yi=0.1685 and xi =000565

(1  y A ) iM 

slope  

(1  y Ai )  (1  y AG )  0.816 ln[(1  y Ai ) /(1  y AG )]

(1  x A ) iM 

(1  x AL )  (1  x Ai )  0.995 ln[(1  x AL ) /(1  x Ai )]

k x' a /(1  x) iM

 15.6 k y' a /(1  y ) iM Calculate (1-y), (1-y)iM, and (y-yi) V 8.16  10 4 Integrate: f ( y )    6.33 0.04496(0.0929) k y' aS (0.8)(0.0315) (1  y )( y  y i ) 0.816 (1  y ) iM After repeating for each y value from y1 to y2, numerically integrate to get z=1.588m.

10.8 Estimation of Mass-Transfer Coefficients for Packed Towers 10.8-1 Prediction of Film Coefficients for CO2 Absorption Predict HG, HL, and HOL for absorption of CO2 from air by water in a dilute solution in a packed tower with 1 ½ in metal Pall rings at 303K (30°C) and 101.32kPa pressure. The flow rates are Gx=4.069 kg/sm2 (3000 lbm/h ft2). From Appendix A.3-18 for CO2 at 1atm: p A  1.86  103 x A  y A  p A / 1.0  1.86  103 x A From Appendix A.3-3 for air at 303K:   1.866  10 5 kg / ms and   1.166kg / m 3 From Table 6.2-1 for CO2 at 276.2K: DAB  0.142  10 4 m 2 / s 1.75

 303  4 2 Correct the diffusivity to 303K: D AB  0.142  10    0.167  10 m / s  276.2   1.866  10 5   0.958 Calculate the Schmidt No: N Sc  D (1.166)(0.1670  10 4 ) From Table 10.6-1 for 1 ½ in metal Pall rings to 1 ½ in. Raschig rings: f p  1.34 4

 0.226  N Sc  0.5  G x  0.5  G y     HG  H y    f  0.660   6.782   0.678  p    

0.35

 0.226  N Sc  0.5  G x  0.5  G y     HG  H y    f  0.660   6.782   0.678     p  3 From Appendix A.2-4 for water at 303K:   0.8007  10 kg / ms and   995.68kg / m 3 At 298K:   0.8937  10 3 kg / ms 0.35

From Table 6.3-1 for CO2 in water at 25C: DAB  2  10 9 m 2 / s

 0.8937  10 3  303  9 2  Adjust to 303K: D AB  2.00  10    2.27  10 m / s 3   0.8007  10  276.2   0.8007  10 5   354.3 Calculate the Schmidt No: N Sc  D (995.68)(2.27  10 4 ) 9

0.3  0.357  N Sc  0.5  Gx /     HL  Hx   f  372   6.782 / 0.8937  10 3   p  0.3

 4.069 / 0.8007  10 3     0.2306m 3  6 . 782 / 0 . 8937  10   For dilute air: V  G y / MW  0.5424 / 28.97  0.01872kgmol / sm2  0.357  354.3  HL  Hx      1.34  372 

0.5

For water: L  Gx / MW  4.069 / 18.0  0.2261kgmol / sm2 H OL  H Ox  H L  ( L / mV ) H G





H OL  H Ox  0.2306  (0.2261 / 1.86  10 3 (0.01872))  0.2426 H OL  H Ox  0.2306  0.001575  0.2322m