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Differential equations of mass transfer, Study notes of Differential Equations

Rice University Differential Equations

The differential equation for mass transfer is obtained by applying the law of ... Steady state, one dimensional flow (assume in x direction) and constant ...

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Differential equations of mass transfer

Definition:

The differential equations of mass transfer are general equations describing mass transfer

in all directions and at all conditions.

How is the differential equation obtained?

The differential equation for mass transfer is obtained by applying the law of

conservation of mass (mass balance) to a differential control volume representing the

system.

The resulting equation is called the continuity equation and takes two forms:

(1) Total continuity equation [in – out = accumulation] (this equation is obtained if we

applied the law of conservation of mass on the total mass of the system)

(2) Component continuity equation[in – out + generation – consumption = accumulation]

(this equation is obtained if we applied the law of conservation of mass to an

individual component)

(1) Total continuity equation

Consider the control volume, Δx Δy Δz (Fig. 1)

Fig. 1

Partial preview of the text

Download Differential equations of mass transfer and more Study notes Differential Equations in PDF only on Docsity!

Differential equations of mass transfer

Definition:

The differential equations of mass transfer are general equations describing mass transfer in all directions and at all conditions.

How is the differential equation obtained?

The differential equation for mass transfer is obtained by applying the law of conservation of mass (mass balance) to a differential control volume representing the system.

The resulting equation is called the continuity equation and takes two forms:

(1) Total continuity equation [in – out = accumulation] (this equation is obtained if we applied the law of conservation of mass on the total mass of the system) (2) Component continuity equation[in – out + generation – consumption = accumulation] (this equation is obtained if we applied the law of conservation of mass to an individual component) (1) Total continuity equation

Consider the control volume, Δx Δy Δz (Fig. 1)

Fig. 1

Apply the law of conservation of mass on this control volume [in – out = accumulation] direction in out in - out x (^) 𝜌v𝑥 ∆𝑦∆𝑧⃒ (^) 𝑥 𝜌v𝑥 ∆𝑦∆𝑧⃒ (^) 𝑥+Δ𝑥 (𝜌v𝑥 ⃒ 𝑥 − 𝜌v𝑥⃒ 𝑥+Δ𝑥) ∆𝑦∆𝑧

y (^) 𝜌v𝑦 ∆𝑥∆𝑧⃒ (^) 𝑦 𝜌v𝑦 ∆𝑥∆𝑧⃒ (^) 𝑦+Δ𝑦 (𝜌v𝑦 ⃒ 𝑦 − 𝜌v𝑦 ⃒ 𝑦+Δ𝑦) ∆𝑥∆𝑧 z (^) 𝜌v𝑧 ∆𝑥∆𝑦⃒ (^) 𝑧 𝜌v𝑧 ∆𝑥∆𝑦⃒ (^) 𝑧+Δ𝑧 (𝜌v𝑧 ⃒ 𝑧 − 𝜌v𝑧 ⃒ 𝑧+Δ𝑧) ∆𝑥∆𝑦

Accumulation = 𝜕𝑚𝜕𝑡 = 𝜕𝜌∆𝑥∆𝑦∆𝑧𝜕𝑡 = ∆𝑥∆𝑦∆𝑧 𝜕𝝆𝜕𝑡

Write the above terms in the overall equation [in – out = accumulation (rate of change)]

(𝜌v𝑥⃒ 𝑥 − 𝜌v𝑥 ⃒ 𝑥+Δ𝑥) ∆𝑦∆𝑧 + (𝜌v𝑦 ⃒ 𝑦 − 𝜌v𝑦 ⃒ 𝑦+Δ𝑦) ∆𝑥∆𝑧 + (𝜌v𝑧 ⃒ 𝑧 − 𝜌v𝑧⃒ 𝑧+Δ𝑧) ∆𝑥∆𝑦 = ∆𝑥∆𝑦∆𝑧 𝜕𝝆𝜕𝑡

Dividing each term in the above equation by ∆𝑥∆𝑦∆𝑧:

(𝜌v𝑥⃒ 𝑥− 𝜌v𝑥 ⃒𝑥+Δ𝑥) ∆𝑥 +^

(𝜌v𝑦 ⃒𝑦− 𝜌v𝑦 ⃒𝑦+Δ𝑦) ∆𝑦 +^

(𝜌v𝑧 ⃒𝑧− 𝜌v𝑧 ⃒𝑧+Δ𝑧) ∆𝑧 =^

𝜕𝝆 𝜕𝑡

Take the limit as Δx, Δy, and Δz approach zero:

− (^) 𝜕𝑥𝜕 𝜌v𝑥 - (^) 𝜕𝑦𝜕 𝜌v𝑦 - (^) 𝜕𝑧𝜕 𝜌v𝑧 = 𝜕𝜌𝜕𝑡

∴ (^) 𝜕𝑥𝜕 𝜌v𝑥+ (^) 𝜕𝑦𝜕 𝜌v𝑦 + (^) 𝜕𝑧𝜕 𝜌v𝑧 + 𝜕𝜌 𝜕𝑡 = 0

The above equation is the general total continuity equation (the velocity distribution can be obtained from this equation)

It can be written in the following form (this form can be used in all coordination system):

∇. 𝜌v⃗⃗⃗⃗ + 𝜕𝜌𝜕𝑡 = 0

direction in out in - out x (^) 𝑛𝐴,𝑥 ∆𝑦∆𝑧⃒ (^) 𝑥 𝑛𝐴,𝑥 ∆𝑦∆𝑧⃒ (^) 𝑥+Δ𝑥 (𝑛𝐴,𝑥 ⃒ 𝑥 − 𝑛𝐴,𝑥⃒ 𝑥+Δ𝑥) ∆𝑦∆𝑧 y (^) 𝑛𝐴,𝑦 ∆𝑥∆𝑧⃒ (^) 𝑦 𝑛𝐴,𝑦 ∆𝑥∆𝑧⃒ (^) 𝑦+Δ𝑦 (𝑛𝐴,𝑦 ⃒ 𝑦 − 𝑛𝐴,𝑦 ⃒ 𝑦+Δ𝑦) ∆𝑥∆𝑧 z (^) 𝑛𝐴,𝑧 ∆𝑥∆𝑦⃒ (^) 𝑧 𝑛𝐴,𝑧 ∆𝑥∆𝑦⃒ (^) 𝑧+Δ𝑧 (𝑛𝐴,𝑧 ⃒ 𝑧 − 𝑛𝐴,𝑧 ⃒ 𝑧+Δ𝑧) ∆𝑥∆𝑦

Accumulation = 𝜕𝑚𝜕𝑡 = 𝜕𝝆𝑨∆𝑥∆𝑦∆𝑧𝜕𝑡 = ∆𝑥∆𝑦∆𝑧 𝜕𝝆 𝜕𝑡𝑨

If A is produced within the control volume by a chemical reaction at a rate 𝑟𝐴 (mass/(volume)(time)

Rate of production of A (generation) = 𝑟𝐴 ∆𝑥∆𝑦∆𝑧

Put all terms in the equation: in – out + generation – consumption = accumulation

(𝑛𝐴,𝑥 ⃒𝑥 − 𝑛𝐴,𝑥⃒ 𝑥+Δ𝑥) ∆𝑦∆𝑧 + (𝑛𝐴,𝑦⃒ 𝑦 − 𝑛𝐴,𝑦 ⃒𝑦+Δ𝑦) ∆𝑥∆𝑧 + (𝑛𝐴,𝑧 ⃒𝑧 − 𝑛𝐴,𝑧⃒ 𝑧+Δ𝑧) ∆𝑥∆𝑦 + 𝑟𝐴 ∆𝑥∆𝑦∆𝑧 = ∆𝑥∆𝑦∆𝑧 𝜕𝜌 𝜕𝑡𝐴

Dividing each term in the above equation by ∆𝑥∆𝑦∆𝑧:

(𝑛𝐴,𝑥⃒ 𝑥− 𝑛𝐴,𝑥⃒ 𝑥+Δ𝑥) ∆𝑥 +^

(𝑛𝐴,𝑦 ⃒𝑦− 𝑛𝐴,𝑦 ⃒𝑦+Δ𝑦) ∆𝑦 +^

(𝑛𝐴,𝑧⃒ 𝑧− 𝑛𝐴,𝑧⃒ 𝑧+Δ𝑧) ∆𝑧 +^ 𝑟𝐴=^

𝜕𝜌𝐴 𝜕𝑡

Take the limit as Δx, Δy, and Δz approach zero:

− (^) 𝜕𝑥𝜕 𝑛𝐴,𝑥 - (^) 𝜕𝑦𝜕 𝑛𝐴,𝑦 - (^) 𝜕𝑧𝜕 𝑛𝐴,𝑧 + 𝑟𝐴= 𝜕𝜌 𝜕𝑡𝐴

𝜕 𝜕𝑥 𝑛𝐴,𝑥^ +^

𝜕𝑦 𝑛𝐴,𝑦^ +^

𝜕𝑧 𝑛𝐴,𝑧^ +^

𝜕𝑡 −^ 𝑟𝐴^ =^0 (1)

Equation 1 is the component mass continuity equation and it can be written in the form:

∇. 𝑛⃗𝐴 + 𝜕 𝜕𝑡𝜌𝐴 − 𝑟𝐴 = 0 (2)

(The above equation can be written in different coordinate systems since it is written in a vector form)

But from Fick’s law

𝑛𝐴 = −𝜌𝐷𝐴𝐵∇𝜔𝐴 + 𝜌𝐴v

Substitute in equation 2 by this value we can get the equation:

−∇. 𝜌𝐷𝐴𝐵∇𝜔𝐴 + ∇. 𝜌𝐴v + 𝜕 𝜕𝑡𝜌𝐴 − 𝑟𝐴 = 0 (3)

Equation 3 is a general equation used to describe concentration profiles (in mass basis) within a diffusing system.

(ii) Component molar continuity equation

Equations 1, 2 and 3 can be written in the form of molar units to get the component continuity equation in molar basis by replacing:

𝑛𝐴 𝑏𝑦 𝑁𝐴;𝜌 𝑏𝑦 𝑐; 𝜔𝐴 𝑏𝑦 𝑦𝐴 ; 𝜌𝐴 𝑏𝑦 𝑐𝐴 𝑎𝑛𝑑 𝑟𝐴 𝑏𝑦 𝑅𝐴

The different forms of the component molar continuity equation:

𝜕 𝜕𝑥 𝑁𝐴,𝑥^ +^

𝜕𝑦 𝑁𝐴,𝑦^ +^

𝜕𝑧 𝑁𝐴,𝑧^ +^

𝜕𝑡 −^ 𝑅𝐴^ =^0 (^4 )

Equation 4 is the component molar continuity equation and it can be written in the form:

∇. 𝑁⃗⃗𝐴 + 𝜕 𝜕𝑡𝑐𝐴 − 𝑅𝐴 = 0 (5)

But from Fick’s law

𝑁𝐴 = −𝑐𝐷𝐴𝐵∇𝑦𝐴 + 𝑐𝐴V

Substitute in equation 5 by this value we can get the equation:

−∇. 𝑐𝐷𝐴𝐵∇𝑦𝐴 + ∇. 𝑐𝐴V + 𝜕 𝜕𝑡𝑐𝐴 − 𝑅𝐴 = 0 (6)

Equation 6 is a general equation used to describe concentration profiles (in molar basis) within a diffusing system.

Note: you may be given the general form and asked to apply specific conditions to get a special form of the differential equation.

For molar concentration:

𝜕𝑐𝐴 𝜕𝑡 =^ 𝐷𝐴𝐵∇

Equation 7 referred to as Fick’s second law of diffusion

Fick’s second ‘‘law’’ of diffusion written in rectangular coordinates is

Note: what are the conditions at which there is no fluid motion? (bulk motion = 0.0)

The assumption of no fluid motion (bulk motion) restricts its applicability to:

a) Diffusion in solid b) Stationary (stagnant) liquid c) Equimolar counterdiffusion (for binary system of gases or liquids where 𝑁𝐴 is equal in magnitude but acting in the opposite direction to 𝑁𝐵 𝑁𝐴 = −𝑐𝐷𝐴𝐵^ 𝑑𝑦 𝑑𝑧 𝐴+ 𝑦𝐴(𝑁𝐴 + 𝑁𝐵)

If 𝑁𝐴 = −𝑁𝐵 the bulk motion term will be cancelled from the above equation

If there is no fluid motion, no consumption or generation term, constant density and diffusivity and steady state conditions

For mass concentration:

∇^2 𝜌𝐴 = 0

For mass concentration:

∇^2 𝑐𝐴 = 0

Note: see page 438 in the reference book for the differential equation of mass transfer in different coordinate systems.

The general differential equation for mass transfer of component A, or the equation of continuity of A, written in rectangular coordinates is

 Initial and Boundary conditions

To describe a mass transfer process by the differential equations of mass transfer the initial and boundary conditions must be specified.

Initial and boundary conditions are used to determine integration constants associated with the mathematical solution of the differential equations for mass transfer

1. Initial conditions:

It means the concentration of the diffusing species at the start (t = 0) expressed in mass or molar units.

𝑎𝑡 𝑡 = 0 𝑐𝐴 = 𝑐𝐴𝑜 (𝑚𝑜𝑙𝑎𝑟 𝑢𝑛𝑖𝑡𝑠)

𝑎𝑡 𝑡 = 0 𝜌𝐴 = 𝜌𝐴𝑜 (𝑚𝑎𝑠𝑠 𝑢𝑛𝑖𝑡𝑠)

where 𝑐𝐴𝑜 and 𝜌𝐴𝑜 are constant (defined values)

(ii) for solutions where species A is only weakly soluble in the liquid, Henry’s law may be used to relate the mole fraction of A in the liquid to the partial pressure of A in the gas 𝑝𝐴 = 𝐻 ∙ 𝑥𝐴 H: Henry’s constant (iii) at gas solid interface

𝑐𝐴𝑠𝑜𝑙𝑖𝑑 = 𝑆 ∙ 𝑝𝐴 𝑐𝐴𝑠𝑜𝑙𝑖𝑑 : is the molar concentration of A within the solid at the interface in units of kg mol/m^3 and 𝑝𝐴 is the partial pressure of gas phase species A over the solid in units of Pa. 𝑆: solubility constant (partition coefficient) (2) A reacting surface boundary is specified

𝑁𝐴 ⃒ 𝑧=0 = 𝑘𝑐𝑐𝐴^ 𝑛𝑠

where 𝑘𝑐 is a surface reaction rate constant with units of m/s. n is the reaction order

Note: the reaction may be so rapid that 𝑐𝐴𝑠 = 0 if species A is the limiting reagent in the chemical reaction.

(3) The flux of the transferring species is zero at an impermeable boundary

𝑁𝐴 ⃒ 𝑧=0 = −𝐷𝐴𝐵^ 𝜕𝑐 𝜕𝑧⃒𝐴 𝑧=0 = 0

where the impermeable boundary or the centerline of symmetry is located at z = 0

(4) The convective mass flux at the boundary surface is specified

𝑁𝐴⃒ 𝑧=0 = 𝑘 (𝑐𝐴𝑠 − 𝑐𝐴∞ ) Where k is the convection mass transfer coefficient

Solved problems:

Problem 1:

The following sketch illustrates the gas diffusion in the neighborhood of a catalytic surface. Hot gases of heavy hydrocarbons diffuse to the catalytic surface where they are cracked into lighter compounds by the reaction: H → 2L, the light products diffuse back into the gas stream.

a. Reduce the general differential equation for mass transfer to write the specific differential equation that will describe this steady-state transfer process if the catalyst is considered a flat surface. List all of the assumptions you have made in simplifying the general differential equation. b. Determine the Fick’s law relationship in terms of only compound H and insert it into the differential equation you obtained in part (a). c. Repeat the solution for spherical catalyst surface.

Solution

a. The specific differential equation

Assumptions: steady state, unidirectional mass transfer.

𝜕 𝜕𝑥 𝑁𝐴,𝑥^ +^

𝜕𝑦 𝑁𝐴,𝑦^ +

𝜕𝑧 𝑁𝐴,𝑧^ +

𝜕𝑡 − 𝑅𝐴^ = 0

Apply these assumptions on the general equation we get the specific differential equation:

𝜕 𝜕𝑧 𝑁𝐻,𝑧^ + 𝑅𝐻^ = 0

b. Fick’s law relationship in terms of only compound H

𝑁𝐻 = −𝑐𝐷𝐻𝐿^ 𝑑𝑦 𝑑𝑧 𝐻+ 𝑦𝐻(𝑁𝐻 + 𝑁𝐿)

Apply these assumptions on the general differential equation:

∇. 𝑁⃗⃗𝐴 + 𝜕𝑐 𝜕𝑡 𝐴− 𝑅𝐴 = 0

∴ ∇. 𝑁⃗⃗𝐴 = 0

For spherical coordinates and mass transfer in r-direction

1 𝑟^2

b. The simplified differential form of Fick’s equation for water vapor (species A)? Assume water is A and air is B

𝑁𝐴 = −𝑐𝐷𝐴𝐵^ 𝑑𝑦 𝑑𝑟 𝐴+ 𝑦𝐴(𝑁𝐴 + 𝑁𝐵)

𝑁𝐵 = 0

𝑁𝐴 = − (^) (1 − 𝑦𝑐𝐷𝐴𝐵 𝐴)

Problem 3:

A large deep lake, which initially had a uniform oxygen concentration of 1kg/m^3 , has its surface concentration suddenly raised and maintained at 9 kg/m^3 concentration level. Reduce the general differential equation for mass transfer to write the specific differential equation for

a. the transfer of oxygen into the lake without the presence of a chemical reaction; b. the transfer of oxygen into the lake that occurs with the simultaneous disappearance of oxygen by a first-order biological reaction.

Solution:

Assume oxygen = A and water = B

a. the transfer of oxygen into the lake without the presence of a chemical reaction

Basic assumptions:

no chemical reaction occur
For deep lake (stationary liquid) we can assume v = 0
Unidirectional mass transfer (assume in z direction) 𝑑𝑁𝐴 𝑑𝑧 +

But

𝑁𝐴 = −𝑐𝐷𝐴𝐵^ 𝑑𝑦 𝑑𝑧 𝐴+ 𝑐𝐴𝑉

Since the liquid is stationary

∴ 𝑁𝐴 = −𝑐𝐷𝐴𝐵^ 𝑑𝑦 𝑑𝑧𝐴

𝑑𝑁𝐴 𝑑𝑧 = −𝐷𝐴𝐵

𝑑^2 𝑐𝐴

𝑑𝑧^2

𝑑^2 𝑐𝐴

𝑑𝑧^2

b. The transfer of oxygen into the lake that occurs with the simultaneous disappearance of oxygen by a first-order biological reaction.

Basic assumptions:

chemical reaction occur (−𝑅𝐴 = 𝑘𝑟𝑐𝐴)
For deep lake (stationary liquid) we can assume v = 0
Unidirectional mass transfer (assume in z direction)

𝐷𝐴𝐵^ 𝑑

𝑑𝑧^2 +

𝜕𝑡 + 𝑘𝑟𝑐𝐴^ = 0

where 𝑘𝑟is the reaction rate constant

But 𝑉𝑦 = 0 (based on the assumptions)

∴ 𝑁𝐴,𝑦 = −𝐷𝐴𝐵^ 𝑑𝑐 𝑑𝑦𝐴

Substitute by the values of 𝑁𝐴,𝑥 and 𝑁𝐴,𝑦 in equation (i)

−𝐷𝐴𝐵^ 𝜕

𝜕𝑥^2 + ay

𝜕^2 𝑐𝐴

𝜕𝑦^2 = 0

∴ 𝐷𝐴𝐵 [𝜕

𝜕𝑥^2 +

𝜕^2 𝑐𝐴

𝜕𝑦^2 ] = ay

Boundary conditions:

𝑐𝐴 = 𝑐𝐴𝑠 at y = 0
𝑐𝐴 = 0 at y = ∞
𝑐𝐴 = 0 at x = 0

Note:

The specific differential equation of mass transfer for a given system can be obtained by two methods:

Select the general equation according to the system coordinates and omit the unnecessary terms
Make a mass balance over a control volume, divide by the volume and take limits as values of length approaches zero

Supplementary data:

The general differential equation for mass transfer of component A, in rectangular coordinates is

Differential equations of mass transfer, Study notes of Differential Equations

Related documents

Partial preview of the text

Download Differential equations of mass transfer and more Study notes Differential Equations in PDF only on Docsity!

Differential equations of mass transfer

𝜕𝑦 𝑛𝐴,𝑦^ +^

𝜕𝑧 𝑛𝐴,𝑧^ +^

𝜕𝑡 −^ 𝑟𝐴^ =^0 (1)

𝜕𝑦 𝑁𝐴,𝑦^ +^

𝜕𝑧 𝑁𝐴,𝑧^ +^

𝜕𝑡 −^ 𝑅𝐴^ =^0 (^4 )

𝜕𝑦 𝑁𝐴,𝑦^ +

𝜕𝑧 𝑁𝐴,𝑧^ +

𝜕𝑡 − 𝑅𝐴^ = 0

𝑑^2 𝑐𝐴

𝑑𝑧^2

𝑑^2 𝑐𝐴

𝑑𝑧^2

𝑑𝑧^2 +

𝜕𝑡 + 𝑘𝑟𝑐𝐴^ = 0

𝜕^2 𝑐𝐴

𝜕𝑦^2 = 0

∴ 𝐷𝐴𝐵 [𝜕

𝜕𝑥^2 +

𝜕^2 𝑐𝐴