Chap 1(a) molecular-diffusion_in_gas(2)

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BKF 2432: MASS TRANSFER FKKSA, UMP

Principles of Mass TransferCHAPTER 1CHAPTER 1Molecular Diffusion in Molecular Diffusion in GasesGases

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Topic Outcomes

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It is expected that student will be able to: Apply the diffusivity coefficient of molecular

diffusion in gases. Solve mathematical solution of molecular diffusion

in gases.

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CONTENTS

Mass Transfer

Molecular Diffusion Convective Mass Transfer

Gases Liquid Solid

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Introduction

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Equimolar Counter diffussion in Gases

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Molecular Diffusion in Gases

Equimolar Counter diffussion in Gases

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For a binary gas mixture of A and B, the diffusivity coefficient DAB=DBA

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Problem 6.1-1 (pg 452) Diffusion of Methane Through Helium

A gas of CH4 and He is contained in a tube at 101.32 kPa pressure and 298 K. At one point the partial pressure of methane is pA1 = 60.79 kPa, and at a point 0.02 m distance away, pA2 = 20.26 kPa. If the total pressure is constant throughout the tube, calculate the flux of CH4 (methane) at steady state for equimolar counter diffusion.

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Example 6.2-1 (pg 415) Equimolar Counterdiffusion

o Ammonia gas (A) is diffusing through a uniform tube 0.10 m long containing N2 gas (B) at 1.0132 x 105 Pa pressure and 298 K. At point 1, pA1 = 1.013 x 104 Pa , and at point 2, pA2 = 0.507 x 104 Pa. The diffusivity DAB = 0.230 x 10-4 m2/s.

1. 1. Calculate the flux J*A at steady state

2. 2. Repeat for J*B

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Numerical value

Units

82.057 cm3.atm/kg mol . K

82.057x 10-3 m3.atm/kg mol . K

8314.34 J/kg mol . K

8314.34 m3 .Pa / kg mol .K

8314.34 kg . m2/s2 . kg mol.K

o.7302 ft3.atm/lb mol.0R

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BKF 2432: MASS TRANSFER FKKSA, UMPProblem 6.2-1 (pg 452) Equimolar Counterdiffusion of a Binary Gas Mixture

• Helium and nitrogen gas are contained in a conduit 5 mm in diameter and 0.1 m long at 298 K and a uniform constant pressure of 1.0 atm abs. The partial pressure of He at one end of the tube is 0.060 atm and the other end is 0.020 atm. Calculate the following for steady-state equimolar counterdiffusion:

1. Flux of He in kg mol/s.m2

1. 2. Flux of N2

2. 3. Partial pressure of He at a point 0.05 m from either end.

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Solution: DAB = 0.687 x 10-4 m2/s (Table 6.2-1) z2-z1 = 0.1m pA1 = 0.060 atm pA2 = 0.020 atm R = 82.06 x 10-3 cm3.atm/g mol.K (Table A.1-1) a) Eqn. (6.1-13)

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2A1AABAZ

*

zzRT

ppDJ

010.029810x06.82

02.006.010x687.0J

3

4

AZ*

2

6AZ

*

m.s

kgmol10x124.1J

b) 2

6AZ

*

m.s

kgmol10x124.1J

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Diffusion of Gases A and B Plus Convection (General Case) (pg 416)

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Convection is the concerted, collective movement of ensembles of molecules within fluids (e.g., liquids, gases)

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For equimolar counterdiffussion, NA=-NB , then NA=J*A=-NB=-J*B

Eq 6.17

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‘A’ Diffusing Through Stagnant, Nondiffusing ‘B’

(Special Case)

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Example 6.2-2 (pg 419) Diffusion of Water Through Stagnant, Nondiffusing Air

Water in the bottom of a narrow metal tube is held at a constant temperature of 20oC. The total pressure of air (assumed dry) is1.0 atm and the temperature is 20oC. Water evaporates and diffuses through the air in the tube, and the diffusion path z2 – z1 is 0.1524 m (0.5 ft) long. The diagram is similar to Fig 6.2-2a. Calculate the rate of evaporation at steady state in lb mol/hr.ft2 and kg mol/s.m2. The diffusivity of water vapor at 20oC and 1 atm pressure is 0.250x10-4 m2/s. Assume that the system is isothermal.

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Problem 6.2-3 (pg 453) Diffusion of A Through Stagnant B and Effect of Type of Boundary on Flux

Ammonia gas is diffusing through N2 under steady state conditions with N2 nondiffusing since it is insoluble in one boundary. The total pressure is 1.013 x 105 Pa and the temperature is 298 K. The partial pressure of NH3 at one point is 1.333 x 104 Pa, and at the other point 20 mm away it is 6.666 x 103 Pa. The DAB for the mixture at 1.013 X 105 Pa and 298 K is 2.30 x 10-5 m2/s.

a) calculate the flux of NH3 in kg mol/s.m2

b) do the same as (a) but assume that N2 also diffuses, both boundaries are permeable to both gases and the flux is equimolar counterdiffusion. In which case is the flux greater?

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Diffusion Through Cross Sectional Area (Sphere)

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Example 6.2-4 (pg 421) Evaporation of Naphthalene Sphere

A sphere of naphthalene having a radius of 2.0 mm is suspended in a large volume of still air at 318 K and 1.01325 x 105 Pa. The surface temperature of the naphthalene can be assumed to be at 318 K and its vapor pressure at 318 K is 0.555 mm Hg. The DAB of naphthalene in air at 318 K is 6.92 x 10-6 m2/s. Calculate the rate of evaporation of naphthalene from the surface.

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Problem 6.2-5 (pg 453) Mass Transfer from a Naphthalene Sphere to Air

Mass transfer is occurring from a sphere of naphthalene having radius of 10 mm. The sphere is in large volume of still air at 52.6°C and 1 atm abs pressure. The vapor pressure of naphthalene at 52.6°C is 1.0 mmHg. The diffusitivity of naphthalene in air at 0°C is 5.16 x 10-6 m2/s. Calculate the rate of evaporation of naphthalene from the surface in

kg mol/s.m2.

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Diffusion Coefficient for Gases

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Example 6.2-5 (pg 427) Estimation of Diffusivity of a Gas Mixture

• Normal butanol (A) is diffusing through air (B) at 1 atm abs. Using the Fuller et al. method, estimate the diffusivity DAB for the following temperatures and compare with the experimental data.

• Given MA (butanol) = 74.1 kg (mass)/kg mol,

• MB (air) = 29 kg (mass)/kg mol]

a) For 0oC.

b) For 25.9oC

c) For 0oC and 2.0 atm abs

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Numerical value

Units

82.057 cm3.atm/kg mol . K

82.057 x 10-3 m3.atm/kg mol . K

8314.34 J/kg mol . K

8314.34 m3 .Pa / kg mol .K

8314.34 kg . m2/s2 . kg mol . K

o.7302 ft3.atm/lb mol.0R

Gas Law Constant R (pg 955)

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TEST 1 EP 203

Chap 1(a) molecular-diffusion_in_gas(2)

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Transcript of Chap 1(a) molecular-diffusion_in_gas(2)