Psuedo steady state diffusion through a stagnant film

Slides:



Advertisements
Similar presentations
ME 525: Combustion Lecture 27: Carbon Particle Combustion
Advertisements

1st & 2nd Law Analysis for Combustion Process
Dept of Chemical and Biomolecular Engineering CN2125E Heat and Mass Transfer Dr. Tong Yen Wah, E , (Mass Transfer, Radiation)
Introduction to Mass Transfer
Introduction to Mass Transfer
Equation of Continuity. differential control volume:
1 Boyle’s Law (T and n constant) Charles’ Law (p and n constant) Combined Gas Law (n constant) Summary of Gas Laws p 1 ×V 1 = p 2 ×V 2.
NOTES: 14.4 – Dalton’s Law & Graham’s Law
Divide yourselves into groups of three (3). Write your names and your complete solution into your answer sheet, and box / encircle your final answer.
Diffusion Mass Transfer
Gases.  Define pressure, give units of pressure, and describe how pressure is measured.  State the standard conditions of temperature and pressure and.
Gas Notes I. Let’s look at some of the Nature of Gases: 1. Expansion – gases do NOT have a definite shape or volume. 2. Fluidity – gas particles glide.
Chapter 5 The Gaseous State. 5 | 2 Gases differ from liquids and solids: They are compressible. Pressure, volume, temperature, and amount are related.
Preview Lesson Starter Objectives Measuring and Comparing the Volumes of Reacting GasesMeasuring and Comparing the Volumes of Reacting Gases Avogadro’s.
GASES and the Kinetic Molecular Theory A.Gas particles DO NOT attract or repel each other B.Gas particles are much smaller than the distances between them.
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L21-1 Review: Heterogeneous Catalyst.
Chapter 10; Gases. Elements that exist as gases at 25 0 C and 1 atmosphere.
Gas Stoichiometry. Equal volumes of gases at the same temperature and pressure contain equal amounts of particles The coefficients in a balanced equation.
L20-1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. Review: Heterogeneous Catalyst.
Review: Steps in a Heterogeneous Catalytic Reaction
Recycle packed column reactor: - allow the reactor to operate at high fluid velocities. - a substrate that cannot be completely processed on a single.
Chapter 5 Gases.
Chapter 5 – Gases. In Chapter 5 we will explore the relationship between several properties of gases: Pressure: Pascals (Pa) Volume: m 3 or liters Amount:
Ideal Gas Law.
1 CHEM-E7130 Process Modeling Exercise. 2 Exercises 1&2, 3&4 and 5&6 are related. Start with one of the packages and then continue to the others. You.
© 2016 Carl Lund, all rights reserved A First Course on Kinetics and Reaction Engineering Class 38.
Chapter 5 Gases. Air Pressure & Shallow Wells Gases Are mostly empty space Occupy containers uniformly and completely The densities of gases are much.
Derivation of Oxygen Diffusion Equations:
Heterogeneous Catalysis: Kinetics in Porous Catalyst Particles
© 2016 Carl Lund, all rights reserved A First Course on Kinetics and Reaction Engineering Class 40.
CHAPTER 15 CHEMICAL REACTIONS Lecture slides by Mehmet Kanoglu Copyright © The McGraw-Hill Education. Permission required for reproduction or display.
Chapter 15 Chemical Equilibrium. Equilibrium - Condition where opposing processes occur at the same time. - Processes may be physical changes or chemical.
Stoichiometry is the study of quantitative (i. e
Mass Transfer transport of one constituent from a region of higher concentration to that of a lower concentration.
MAE 5310: COMBUSTION FUNDAMENTALS
Droplet evaporation Liquid fuel combustion
Solution of Thermodynamics: Theory and applications
Condensational Growth
Topic 5 Gases. Topic 5 Overview 5.1 Describing Gases - Phases of matter - Units of pressure 5.2 Ideal and Non-Ideal (Real) Gases - Ideal gas behaviour.
Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which.
REACTOR THEORY The basic principles are simple but detailed calculations are difficult Use very simplified models to understand principles of calculations.
Diffusion Mass Transfer
A First Course on Kinetics and Reaction Engineering
Gas Laws II.
A First Course on Kinetics and Reaction Engineering
Chapter 10 Gases: Their Properties and Behavior
Steady-state Nonisothermal reactor Design Part I
Section 11.3 – Stoichiometry of Gases
Chapter 5 Gases.
Diffusion of gases A and B Plus Convection
Gas Volumes and the Ideal Gas Law
Volume and Moles (Avogadro’s Law)
Gas Laws Unit 9 Chapter 11.
Convective Mass Transfer Coefficient
Mustafa Nasser, PhD, MSc, BSc Chemical Engineering
Review: Steps in a Heterogeneous Catalytic Reaction
Gas Stoichiometry.
Gas Laws II.
Basic Chemistry Chapter 11 Gases Chapter 11 Lecture
Gas Laws II.
Chapter 10 Gases: Their Properties and Behavior
Kinetic Molecular Theory
Stoichiometry and the combustion of fuels 2
Calculations Based on Chemical Equations
Gas Laws II.
Chapter 11 Preview Lesson Starter Objectives
Chapter 11 Gas Volumes and the Ideal Gas Law Section 3.
GASES and the Kinetic Molecular Theory
ChE 149: TRANSPORT PHENOMENA
Presentation transcript:

Psuedo steady state diffusion through a stagnant film In many mass transfer operations, one of the boundaries may move with time. If the length of the diffusion path changes a small amount over a long period of time, a pseudo steady state diffusion model may be used. When this condition exists, the equation of steady state diffusion through stagnant gas’ can be used to find the flux If the difference in the level of liquid A over the time interval considered is only a small fraction of the total diffusion path, and t0 – t is relatively long period of time

The related amount (kg) of A leaving the liquid is: at any given instant in that period, the molar flux in the gas phase may be evaluated by where z equals z 2 – z1, the length of the diffusion path at time t. Assume the level drops dz (m) in dt (s) The related amount (kg) of A leaving the liquid is:

Above Equation may be integrated from t = 0 to t and from z = z t0 to z = zt as Solving the integral

Mass diffusion through varying cross-sectional area We know that for the constant cross-sectional area both and are constant In the case of varying cross-sectional area both parameters are variable. Then it Reasonable to define another paramater a) Diffusion from Spherical Shell This situation appears often in such cases as the evaporation of a drop of liquid, the evaporation of a ball of naphthalene, and diffusion of nutrient to spherical-like microorganism in a liquid. As shown down, the sphere radius r1 m in finite gas medium component A at partial pressure at the surface is diffusing in to gas medium (B), where at point 2 the partial pressure pA2 = 0 since large volume of air is passing by. Steady state diffusing is applicable.

The final formula for Diffusion from Sphere is given as: Time required for complete evaporation of the droplet may be evaluated from making mass balance. Above equation gives the total time t required for complete evaporation of spherical droplet of initial radius r1.

c) Diffusion through conduit of non-uniform cross-sectional area z1 z2 z r changes with z , and this can be evaluated This integration need to be solved to find the flux

Example: A sphere of naphthalene having a radius of 2mm is suspended in a large volume of shell air at 318 K and 1 atm. The surface pressure of the naphthalene can be assumed to be at 318 K is 0.555 mm Hg. The D AB of naphthalene in air at 318 K is 6.92 * 10 –6 m2/sec. Calculate the rate of evaporation of naphthalene from the surface.

Solution: The final formula for Diffusion from Sphere is given as:

Combustion of a coal particle The problem of combustion of spherical coal particle is similar to evaporation of a drop with the exception that chemical reaction (combustions) occurs at the surface of the particle. During combustion of coal, the reaction C + O 2  CO 2 cccurs. According to this reaction for every mole of oxygen that diffuses to the surface of coal (maximum of carbon), react with 1 mole of carbon, releases 1 mole of carbon dioxide, which must diffuse away from this surface. This is a case of equimolar counter diffusion of CO 2 and O 2. Normally air (a mixture of N 2 and O 2) is used for combustion, and in this case N 2 does not takes part in the reaction, and its flux is zero. . The molar flux of O 2 could be written as (1) Where is the diffusivity of O 2 in the gas mixture

and from stoichiometry, Since and from stoichiometry, equation (1) becomes (2) This equation is exactly the equmolar counter The molar flux for steady-state equimolar counter diffusion: For spherical evaporating drop where the drop radius r1 is much smaller than the Reference r2 (please see the spherical drop flux section in Geankoplis book), then the Flux become This flux of O2 diffusing to the surface drop (3) In term of mole fraction:

For fast reaction of O 2 with coal, the mole fraction of O 2 at the surface of particle iz zero. (i.e.,) .And also at some distance away from the surface of the particle (because air is a mixture of 21 mole % O 2 and 79 mole % N 2)With these conditions, equation (3) becomes

first-order heterogonous catalytic surface x y yA,L yA,s NA,y Assuming depletion of species A at the catalytic surface (x = 0) Our task is to find Flux at catalyst surface NA,y(0) First order reaction: where

rearrange the equation Hence, at the surface, Substitute Equation 2 in 1 Flux at catalyst surface

Limiting Cases: Process is reaction limited: Process is diffusion limited: