test 1 book notes examples-1.doc

Molecular Diffusion in Gases

Molecular Diffusion in GasesMolecular Diffusion of Helium in NitrogenExample 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,

if C = P/RT

Substitute in values:

kgmol/m2s

Equimolar Counterdiffusion

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

P = 1.0132x105 Pa

z2 z1 = 0.10 m

R = 8314.3 m3 Pa/kgmol K

kgmol A/m2 s

Diffusion of Water Through Stagnant, Nondiffusing AirExample 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


atm

kgmol/m2sDiffusion in a Tube with Change in Path LengthExample 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.

Assuming a cross-sectional area of 1 m2, the level drops dt in dz seconds, and the leftover kgmol of A is PAdz/MA:

Rearranging and integrating:

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

where

is the molar rate of material A leaving the drop

r1 and r2 are two radial points away from the sphere centerDAB 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

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 drops surface

Assume point 2 is very far away from the drop, so r2 >> r1

(

Now multiply the equation by 1/r1, rewrite the radius as 2/D on the right side, substitute for surface flux, NAs, and PBM:

and

So we rewrite the equation as:

Assume low vapor pressure, so PA1, PA2 > r1R = 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(52.6C) = DAB(0C)= (5.16 x 10-6 m2/s)(325.75/273.15)1.75 = 7.023 x 10-6 m2/s

Now solve for :

(

= 4.34736 x 10-11 kgmol/s

Now solve for NA = /A = (4.34736 x 10-11 kgmol/s)/(4r12) = 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 25.9C and is suspended in air by a fine wire. The initial radius r1 = 2.00mm. The vapor pressure of toluene at 25.9C 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:

and

Volume of the sphere, so the derivative of volume is:

Plug this volume derivative into molar flux equation for dV:

Now set equal in each equation:

Separate variables and integrate both sides:

(

(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

= 866 kg/m3r1 = 0.002 m

Solve for PBM:

= 99392.6 Pa

tevap ==1388.23 seconds6.2-10 Diffusion in a Nonuniform Cross-Sectional Area Changing Pipe DiameterThe gas ammonia (A) is diffusing at steady state through N2 (B) by equimolar counterdiffusion in a conduit 1.22m long at 25C 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.

Area = (60)

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 = (60) = 0.000271 L2 0.001321 L + 0.001611

(

Now separate and integrate both sides:

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 K

Now plug into integrate equation:

(

= 1.24 x 10-10 kgmol/s

Now solve for NA:

NA = /A = at any point along the length, L

Molecular Diffusion in LiquidsDiffusion 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 1 = 972.8 kg/m3. At point 2, the concentration of ethanol is 6.8 wt % and 2 = 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

1 = 972.8 kg/m3

2 = 988.1 kg/m3Meth =46.05

Mwater = 18.02

Calculate the mole fractions, taking a basis of 100 kg:

Now calculate the molecular weight:

kg/kgmol

kg/kgmol

Now calculate CAvg:

kgmol/m3

kgmol/m2s

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)

1 = 991.7 kg/m32 = 961.7 kg/m3M1 = 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:

54.435 kgmol/m3Calculate XBM:

Xw1 = 1-0.02 = 0.98

Xw2 = 1-0.1 = 0.9

0.939432

Now calculate flux:

== 1.437 x 10-6 kgmol/m2 s

Prediction of Diffusivities in LiquidsPrediction of Liquid Diffusivity


Predict the diffusion coefficient of acetone in water at 25C and 50C using the Wilke-Chang equation. The experimental value 1.28x10-9 m2/s at 298K.

From Appendix A.2, the viscosity of water at 25C is B = 0.8937 x 10-3 Pa s and at 50C 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 = 2.6 and MB = 18.02 kg mass/kgmol. For 25C:

m2/s

For 50C:

m2/sPrediction of Diffusivities of Electrolytes in LiquidsDiffusivities of ElectrolytesExample 6.3-3, page 434

Predict the diffusion coefficients of dilute electrolytes for the following cases:(a) For KCl at 25C, predict DAB and compare with the value in Table 6.3-1

(b) Predict the value of KCl at 18.5C. The experimental value is 1.7x10-5 cm2/s

(c) For CaCl2 predict DAB at 25C. 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, + (K+) = 73.5 and - (Cl-) = 76.3:

cm2/s

(b) To change the temperature, we use a simple correction factor:T = 18.5C = 291.7K

From Table A.2-4, w = 1.042 cP

cm2/s

(c) From Table 6.3-3, + (Ca2+/2) = 59.5 and - (Cl-) = 76.3, n+ = 2 and n- = 1:

cm2/s

cm2/s

cm2/s

Molecular Diffusion in Biological Solutions and GelsPrediction of Diffusivity of AlbuminExample 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)

water = 0.8937x10-3 Pa s

We can use the equation for the prediction of diffusivities for biological solutes:

m2/s

This value is different from the experimental value because the shape of the molecule differs greatly from a sphere.

Diffusion of Urea in AgarExample 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/m3CA2 = 0 kgmol/m3Because XA1 is less than 0.01, the solute is very dilute and XBM 1.0.

kgmol/m2s

Molecular Diffusion in SolidsDiffusion of H2 Through Neoprene MembraneExample 6.5-1, page 442

The gas hydrogen at 17C 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 17C is 0.051 m3 (STP of 0C and 1atm)/m3 solid atm and the diffusivity DAB is 1.03x10-10 m2/s at 17C.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:

kgmol H2/m3 solid

Because PA2 on the other side of the rubber is zero, CA2 = 0:

kgmol H2/m2 sDiffusion Through a Packaging Film Using PermeabilityExample 6.5-2, page 443

A polyethylene film 0.00015m thick is being considered for use in packaging a pharmaceutical product at 30C. 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)

kgmol/m2s

6.5-5 Diffusion Through a Membrane in SeriesNitrogen gas at 2.0 atm and 30C 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:

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:

kmole/m2s

Diffusion in Porous Solids That Depends on StructureDiffusion of KCl in Porous SilicaExample 6.5-3, page 445

A sintered solid of silica 2.0mm thick is porous, with a void fraction of 0.30 and tortuosity 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/m3CA2 = 0

= 4.0

= 0.30

z2 z1 = 0.002 m

kgmol/m2s

6.5-6 Diffusion of CO2 in a Packed Bed of SandIt 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 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 of 1.87. = 1.87

PA1 = 2.026 x 103 Pa = 0.30

PA2 = 0 PaT = 276K

mol/m2sUnsteady-State Diffusion in Various GeometriesUnsteady-State Diffusion in a Slab or Agar GelExample 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/m3C1 = 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

(a)X1 = 10.16mm/2 = 5.08mm = 0.00508 m

X = 0 (center)

Relative position, n = X/X1 = 0Relative 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 = 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)

(c = 0.0172 kgmol/m3(b)

If the thickness is halved, X becomes:

Relative position, n = 0, m = 0

(c = 0.0002 kgmol/m37.1-5 Unsteady State Diffusion in a Cylinder of Agar Gel Radially and AxiallyA 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/m3C1 = 0 for pure water

D = 4.72 x 10-10 m/s

Time, t = 100h = 360000 seconds

(a)

r1 = 0.01524 m

r0 = 0 m (center line)

Relative 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

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/m3Unsteady-State Diffusion in a Semi-Infinite SlabExample 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.m3C1 = 0.10 kgmol/m3kc = 2x10-7 m/sK = 2.0

For x = 0.01:

From the chart 5.3-3 Unsteady State Heat Conducted in a Semi-Infinite Solid with Surface Convection:

(C = 2.04x10-2 kgmol/m3For x = 0

From the chart 5.3-3 Unsteady State Heat Conducted in a Semi-Infinite Solid with Surface Convection:

(C = 3.48x10-2 kgmol/m3This is the same value as Ci. To calculate CLi:

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/hSolve for Y:

Solve for X for the graph:

X1 = 25.4 mm = 0.254 m

X0 (center) = 0 m

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 hoursConvective Mass Transfer CoefficientsVaporizing A and Convective Mass TransferExample 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 ky 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:

To calculate ky, we must relate it to k'y:

kgmol/m2 s mole fraction

Now we can calculate kG:

(

kgmol/m2 s atm

Now we can calculate the vaporization rate, NA:

kgmol/m2s

or

kgmol/m2s

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.

Solve for PBM:

atm

Plug into equation:

kG (1.0 atm) = (1.08 lbmol/h ft2 atm)(0.872853 atm)

kG(1.01x105Pa) = (1.08lbmol/hft2atm)(.872853atm)(1hr/3600sec)(1ft2/.0929 m2)(.453kgmol/lbmol)kG = 0.9427 lbmol/h ft2 atm

kG = 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

NA = 0.1414 lbmol/ft2h

NA = 1.92 x 10-4 kgmol/m2s

Mass Transfer Under High Flux ConditionsHigh Flux Correction FactorsExample 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, kx 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/ kc or kx/ kx and k0c/ kc or k0x/ kx to correct for high flux.

First find XBM:

To find the flux, NA, use:

lbmol/ft2 hr

To find the ratios, set up the following:

So lbmol/ft2 hr; and

So 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 30C. 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 Henrys 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 Henrys 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 C = 303.15 K

kc = 9.567 x 10-4 m/s

XA1 = 2 x 10-5Henrys Law: PA = 609XAR = 82.057 x 10-3 m3 atm/kgmol K

(

kgmol/m2s atm

NA = 1.455 x 10-6 kgmol/m2sMass Transfer for Flow Inside Pipes

Mass Transfer Inside a TubeExample 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

air = 1.932x10-5 Pa s from Appendix A

= 1.114 kg/m3R = 8314 m3 Pa/kgmol K

Calculate the concentration, CAi:

kgmol/m3Calculate the Schmidt number:

Calculate the Reynolds number:

Hence, the flow is laminar. We will use Figure 7.3-2 for laminar flow streamlines:

Using Figure 7.3-2 with rodlike flow:

If CA0 = 0,

(CA = 1.539 x 10-5 kgmol/m37.3-7 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.

= 996 kg/m3 = 8.71 x 10-4 Pa s

T = 26.1C

DAB = 1.245x10-9 m2/s

Solubility = 0.2948 kgmol/m3D = 0.00635 m

As = 2rLV = ACalculate the Schmidt number:


Calculate the Sherwood number:

Now calculate kc:

(

(

m/s

(b) CA1 = 0

CAi = 0.02948 (solubility)

CA2 = ?

So now we plug this into the log mean driving force equation:

CA1 = 0, so it cancels out of the equation:

Now plug in:

CA2 = 0.001151 kgmol/m3(c)

Rate of benzoic acid dissolved = NA:

Mass Transfer for Flow Outside Solid SurfacesMass Transfer From a Flat PlateExample 7.3-2, page 481

A large volume of pure water at 26.1C 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/m3T = 26.1C

L = 0.24 mDAB = 1.245x10-9 m2/sv = 0.061 m/s

Because the solution is dilute, we can use the properties of water for the solution:

= 996 kg/m3 = 8.71 x 10-4 Pa s

Calculate the Schmidt Number:


Now calculate JD:

Now solve for kc:

(

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:

kgmol/m2sMass Transfer From a SphereExample 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 45C 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 45C 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 = 45C

P = 1 atm

v = 0.305 m/s

DAB = 6.92x10-6 m/s

= 1.113 kg/m3 = 1.93 x 10-5 Pa s

Calculate the Schmidt number:


Now find the Sherwood number:

Now solve for kc:

(

m/s

Now solve for kG:

kgmol/s m2 Pa

Since the gas is dilute, kG kG. Using that we can solve for the flux:

kgmol/m2 s

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.

DAB = 1.245 x 10-9 m2/s

Solubility, S = 0.02948 kgmol/m3 = 996 kg/m3Mw = 18 kg/kgmol

CA1 = 0.02948 kgmol/m3 = 8.71 x 10-9 Pa sSolve for XBM:XB1 = = [(996 kg/m3)/(18kg/kgmol)]/[(996 kg/m3)/(18kg/kgmol) + 0.02948 kgmol/m3]

XB1 = 0.999468XB2 = 1.0

Calculate Schmidt Number:

Calculate Reynolds Number:

Calculate JD:

Calculate Kc:

m/s

Calculate Flux:

kgmol/m2s7.3-4 Mass Transfer to Definite Shapes Flat Plate and a SphereEstimate 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

P = 202600 Pa


T = 325.6 K

Mw = 28.8 kg/kgmolCorrect DAB:

m2/s

Calculate and :

cP = 1.958 x 10-5 kg/m s

kg/m3(a)

Calculate NRe:

Because NRe > 15000:

(

(KC = 0.003827 m/s

kgmol/m2 Pa s(b)Dp = 0.0127 m

Calculate Reynolds and Schmidt Numbers:

Because 0.6 < NSC < 2.7:

(

(KC = 0.012725 m/s

kgmol/m2 Pa sMass Transfer of a Liquid in a Packed BedExample 7.3-4, page 485

Pure water at 26.1C 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:

= 996 kg/m326.1C = 8.71 x 10-4 Pa s

25C = 8.94 x 10-4 Pa s

V = 5.514x10-7 m3/sDp = 6.375mm

AS = 0.01198 m2

= 0.436

D = 0.0667m

S = 2.948x10-2 kgmol/m3

DAB(25C) = 1.21x10-9 m2/s from Table 6.3-1Correct DAB:

m2/s

Calculate the area of the column:

m2Now use the area and volumetric flow to find the velocity:

m/s

Calculate the Schmidt Number:

Calculate Reynolds number:

Calculate JD:

Then, assuming kc = kc for dilute solutions,

(

(

m/s

Now set the log mean driving force equation can be set equal to the material balance equation on the bulk stream:

where CAi = 0.02948 (the solubility)

CA1 = 0A = 0.01198 m2 (the surface area of the bed)

V = 5.514x10-7 m3/s

CA2 = 2.842 x 10-3 kgmol/m37.3-5 Mass Transfer to Packed Bed and Driving Force

Pure water at 26.1C 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.

= 0.8718 x 10-3 Pa s = 996.7 kg/m3 = 0.0701 ft3/hr

A = 0.129

CAi = 0.00184 lbmol/ft3CA1 0 CA2 = 1.80 x 10-4 lbmol/ft3Log mean driving force equation:

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:

So now we plug this into the log mean driving force equation:

CA1 = 0, so it cancels out of the equation:

We can now plug in given numbers:

(KC = 0.0559 ft/hr

Mass Transfer to Suspensions of Small ParticlesMass Transfer from Air Bubbles in FermentationExample 7.4-1, page 488Calculate the maximum rate of absorption of O2 in a fermenter from air bubbles at 1 atm abs pressure having diameters of 100 m at 37C into water having a zero concentration of dissolved oxygen. The solubility of O2 from air in water at 37C is 2.26x10-7 kgmol O2/m3 liquid. The diffusivity of O2 in water at 37C 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

c,water = 6.947 x 10-4 Pa s = 6.947 x 10-4 kg/ m s

c,water = 994 kg/m3p,air = 1.13 kg/m3Calculate NSc:

Now calculate kL:

m/s

Assuming kL = kL for dilute solutions,

kgmol O2/m2 s

Molecular Diffusion Plus Convection and Chemical ReactionProof of Mass Flux EquationExample 7.5-1, page 491

Table 7.5-1 gives the following relation:

Prove this relationship using the definition of the fluxes in terms of velocities.From Table 7.5-1 (page 490), substituting for jA and for jB, and rearranging:

(

and

(

Thus the identity is proved.

Diffusion and Chemical Reaction at a BoundaryExample 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 k1 = 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

= 0.002 m

T = 300 K

C = P/RT = 101320/(8314*300) = 4.062 x 10-2 kgmol/m3NB = -2NADAB = 0.15 x 10-4 m2/s

k1 = 5.63x10-3 m/s

(a)

kgmol A/m2s

(b)

Substitutein for XA2:

kgmol A/m2s

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:

For z1=0m:

kgmol/m3For z2=0.004m

kgmol/m3For z2=0.01m

kgmol/m3(b)

The total amount, Q, of A absorbed up to time t is represented by:

Plugging in and solving for Q, we get:

kgmol/m27.5-11 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 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:

where:

Plugging in:

For a slow reaction, . Plugging into the flux equation and replacing C with P/RT:

(d)

Plugging in the values, we get:

kgmol/m2 sSolving for the mole fraction:

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:

Because A reacts instantly, we know that C1=0. From the graph for a sphere at Y, X=0.32

Solving for r:

(

mUnsteady State Diffusion and Reaction in a Semi-Infinite MediumReaction and Unsteady State DiffusionExample 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-1DAB = 1.5x10-9 m2/s

Solubility of CO2 is 2.961x10-7 kgmol/m3 Pa

Time, t = 0.010 seconds

The total amount, Q, of A absorbed in time, t is:

kt = (35s-1)(0.010s) = 0.350

CA0 = Solubility*Pressure = (2.961x10-7 kgmol/m3 Pa)( 101.32 kPa) = 3.00x10-2 kgmol CO2/m3

kgmol CO2/m2Multicomponent Diffusion of Gases

Diffusion of A Through Nondiffusing B and C


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,

PB1 = 0.4,

PC1 = 0.2PA2 = 0.1

PB2 = 0.6

PC2 = 0.3

z2 = 0.005m

z1 = 0

T = 298K

P = 1 atm

DAB = 2.02x10-5 m2/s

DAC = 6.75x10-5 m2/s

DBC = 7.29x10-5 m2/s


Find DAM:At point 1:

At point 2:

m2/s

Calculate PiM:

atm

atm

atm

Now solve for NA:

kgmol A/m2s

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:

Calculate DAM in m2/s:

Calculate CAM:

mol/m3

mol/m3

mol/m3

EMBED Equation.3 mol/m3Substituting back in and solving for flux:

kgmole/m2sKnudsen Diffusion of Gases

Knudsen Diffusion of HydrogenExample 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(angstrom). Calculate the Knudsen diffusivity, DKA, of H2.

P = 1.01325x105 Pa

T = 373K

r = 60 angstroms = 6.0x10-9MA(H2) = 2.016

m2/sFlux Ratios for Diffusion of Gases In Capillaries

Transition-Region Diffusion of He and N2A 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


(a)

m2/s

In an open system with no chemical reaction, the ratio of NA/NB is constant:

Now we can solve for the flux factor, :

Now we can solve for the transition region NA:

kgmol/m2s(b) If we estimate that equimolar counterdiffusion is taking place at steady state, = 1-1 = 0. We can estimate the diffusivity, DNA:

m2/sNow we can solve for the approximate flux:

kgmol/m2s7.6-4 Transition-Region Diffusion in CapillaryA 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 m. 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):

m2/s

m2/s(b)For the transition region, flux is calculated using:

For P = 0.001:

kgmol/m2sFor P = 0.1:

kgmol/m2s

For P = 10:

kgmol/m2s(c)Lower limit is representative of Knudsen diffusion and the upper limit is representative of molecular diffusion. 9.3 Vapor Pressure of Water and Humidity9.3-1 Humidity from Vapor-Pressure Data

The air in a room is at 26.7C 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 26.7C, the vapor pressure of water is pAs = 3.50 kPa. Also pA = 2.76 kPa and P = 101.3 kPa. For part (a):

kg H2O/ky dry air

For part (b)

For part (c)

9.3-2 Use of Humidity Chart

Air entering a dryer has a temperature (dry bulb) of 60C and a dew point of 26.7C. Using the humidity chart, determine the actual humidity H, percentage humidity HP, humid heat cs, and humid volume vH.

The dew point of 26.7C is the temperature when the given mixture is at 100% saturation. Starting at 26.7C, 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 60C. Stated in another way, if air at 60C and having a humidity of H = 0.0225 kg H2O/kg dry air is cooled, its dew point will be 26.7C.

Locating this point where H = 0.0225 and T = 60C 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):

kJ/kg dry air K

The humid volume at 60C from Eq. (9.3-7) is

m3/kg dry air

9.3-3 Adiabatic Saturation of Air

An air stream at 87.8C 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 = 87.8C 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 42.5C and H = 0.0500 kg H2O/kg dry air.

For part (b), the same line is followed to 100% saturation where T = 40.5C 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 = 60C is passed over a wet bulb, as shown in Figure 9.3-4, and the wet bulb temperature obtained is TW = 29.5C. What is the humidity of the mixture?

The wet bulb temperature of 29.5C can be assumed to be the same as the adiabatic saturation temperature TS, as discussed. Following the adiabatic saturation curve of 29.5C until it reaches the dry bulb temperature of 60C, the humidity is H = 0.0135 kg H2O/kg dry air.

9.6 Calculation Methods for Constant-Rate Drying Period9.6-1 Time of Drying from Drying CurveA 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 CurveA 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:

hours

9.6-3 Prediction of Constant Rate DryingAn 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 65.6C 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 65.6C, using the humidity chart, the wet bulb temperature TW = 28.9C and HW = 0.026 by following the adiabatic saturation line to the saturated humidity. Using Eq 9.3-7 to calculate the humid volume,

m3/kg dry air

The density for 1.0 kg dry air + 0.010 kg H2O is

kg/m3The mass velocity G is

kg/h

Using Eq. 9.6-9

W/m2K

At TW = 28.9C, W = 2433 kJ/kg from the steam tables. Substituting into Eq. 9.6-8:

kg/m2h

The total evaporation rate for a surface area of 0.457 by 0.457 m2 is

kg H2O / h

9.7 Calculation Methods for Falling-Rate Drying PeriodExample 9.7-1 Numerical Integration in Falling-Rate PeriodA 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:

hours

XR1/RX(1/R)avg(X)( 1/R)avg

0.1951.510.6630.0450.7450.0335

0.1501.210.8260.0500.9690.0485

0.1000.91.110.0351.2600.0441

0.0650.711.410.0152.0550.0308

0.0500.372.700.0103.2030.0320

0.0400.273.70--Total = 0.1889

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 X = 0.045, giving 0.0335. Other values are similarly calculated and all the values are summed to give a total of 0.1889.

Substituting into Eq. 9.6-1

hours

The total time is 2.63 + 4.06 = 6.69 hours.

Approximation of Straight Line for Falling-Rate PeriodA 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

hours

This is comparable to the value of 4.06 hours obtained in Example 9.7-1 by numerical integration

10.2 Equilibrium Relations between Phases10.2-1 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 Henrys law constant is 4.38x104 atm/molfrac.

Given:

or 0.000835 parts O2 to 100 parts water

10.3 Single and Multiple Equilibrium Contact Stages10.3-1 Equilibrium Stage Contact for CO2-Air-WaterA 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:

From Appendix A.3 at 293K -Henrys Law Constant:

Total Material Balance:

Total Flow Rates:

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:

Acetone Material Balance:

Amount of entering acetone=

Entering Air =

Acetone Leaving in V1=

Acetone Leaving in LN=

V1=29.7+0.03=29.73 kg mol air + acetone/hr

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

*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:

Acetone Material Balance:

Amount of entering acetone=

Entering Air =

Acetone Leaving in V1=

Acetone Leaving in LN=

V1=29.7+0.03=29.73 kg mol air + acetone/hr

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

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

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:

Plot the equilibrium data and point P at (0.1,0.38)

Estimate the slope:

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

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

This is essentially the same slope so use the interfacial values obtained before.

Calculate flux through each phase:

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, yAi=0.197, xAi=0.257

From the previous example:

and

The percent resistance:

Gas film: (484/868.8)(100)=55.7%

Liquid Film: 44.3%

Calculate the Flux:

10.6 Absorption in Plate and Packed Towers 10.6-1 Pressure Drop and Tower Diameter for Ammonia AbsorptionAmmonia 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.

and

From Appendix A.2-4 for water:

From Appendix A.2-3 for water:

From Table 10.6-1 for 1-in Pall Rings: Fp=56ft-1.

For Figure 10.6-5 for Random Packing:

From Figure 10.6-5 for 0.06853 and 1.925, the ordinate=1.7.

Capacity parameter equation from ordinate: (To find volumetric flow)

50% Flooding:

and for the liquid:

New capacity parameter:

From Figure 10.6-5 for Random Packing at 0.85 and 0.06853: 0.18 inches of water/ftSolving for Diameter:

Ammonia in outlet water:

Total liquid flow rate:

Pure liquid flow rate:

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:

From Figure 10.6-7 (drawing of tower trays): , ,

Overall Material Balance:

For a portion of the tower:

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 TraysA 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:

Moles alcohol/h in V1 : 100-97.8=2.20.

Removing 90% moles/hr in outlet gas V2: 0.1(2.20)=0.220

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:

Substitute into the operating line equation to find Lmin:

Solve for L and x:

Plot the operating line with several points. The solution is dilute, so the line is straight.

Step off trays: 4 trays.

Flow Rates:

Calculating A for analytical solution:

Calculate N:

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: kgmol/sm3molfrac and kgmol/sm3molfrac. Equilibrium data are given in Appendix A.3.a) Calculate the tower height using kya.

b) Repeat using kxa

c) Calculate Kya and the tower height.

Given:

From Appendix A.3 for acetone-water and xA=0.0333 mol frac:

Equilibrium line:

Plot:Overall Material Balance:

Plot the operating line.

Approximate the slope at y1 and x1 using trial and error:

y1*=0.0077Using new points yi1=0.0154 and xi1=0.0130 continue trial and error:

Recalculate the slope:

Repeat this process for y2 and x2:

The slope changes very little. Plotting this line we get: y2*=0.0, yi2=0.0020, and xi2=0.0018

Using the interfacial concentrations:

Calculate the flow rates:

a)

b)

c)

10.7 Absorbtion of Concentrated Mixtures in Packed Towers10.7-1 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 cross-sectional area is 0.0929 m2. For dilute SO2, the film mass-transfer coefficients at 293K are, for 25.44 rings are: and where kya is kgmol/sm3molfrac, kxa 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:

Overall Material Balance:

Operating Line:

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:

Similarly for liquid flow:

,

Interface Compositions: Estimate the slope of PM by trial and error: Final yi=0.1685 and xi =000565

Calculate (1-y), (1-y)iM, and (y-yi)

Integrate:

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 in metal Pall rings at 303K (30C) 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:

From Appendix A.3-3 for air at 303K: and

From Table 6.2-1 for CO2 at 276.2K:

Correct the diffusivity to 303K:

Calculate the Schmidt No:

From Table 10.6-1 for 1 in metal Pall rings to 1 in. Raschig rings:

From Appendix A.2-4 for water at 303K: and

At 298K:

From Table 6.3-1 for CO2 in water at 25C:

Adjust to 303K:

Calculate the Schmidt No:

For dilute air:

For water:

r2

Diffusing

A

r1

Stagnant B

.061m

.0305m

1.22m

2.0 atm

N2

0 atm

EMBED Equation.3

P1 = 2.026 x 105 Pa

P2 = 0 Pa

1.25m

1

2

Water at 26.1C, velocity, v = 0.152 m/s

CA1

0.137 m

Benzoic Acid

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

0.152 m

d

PCO2 =

456 mmHg

PCO2 =

76 mmHg

II

I

L

0.00635 m

Pure Water

T = 26.1C

v = 3.05m/s

H2O

1.22 m

1.829 m

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