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Differential equations of mass transfer, Study notes of 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 โ€“ 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 โ€“ out + generation โ€“ 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, ฮ”x ฮ”y ฮ”z (Fig. 1)
Fig. 1
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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 โ€“ 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 โ€“ out + generation โ€“ 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, ฮ”x ฮ”y ฮ”z (Fig. 1)

Fig. 1

Apply the law of conservation of mass on this control volume [in โ€“ out = accumulation] direction in out in - out x (^) ๐œŒv๐‘ฅ โˆ†๐‘ฆโˆ†๐‘งโƒ’ (^) ๐‘ฅ ๐œŒv๐‘ฅ โˆ†๐‘ฆโˆ†๐‘งโƒ’ (^) ๐‘ฅ+ฮ”๐‘ฅ (๐œŒv๐‘ฅ โƒ’ ๐‘ฅ โˆ’ ๐œŒv๐‘ฅโƒ’ ๐‘ฅ+ฮ”๐‘ฅ) โˆ†๐‘ฆโˆ†๐‘ง

y (^) ๐œŒv๐‘ฆ โˆ†๐‘ฅโˆ†๐‘งโƒ’ (^) ๐‘ฆ ๐œŒv๐‘ฆ โˆ†๐‘ฅโˆ†๐‘งโƒ’ (^) ๐‘ฆ+ฮ”๐‘ฆ (๐œŒv๐‘ฆ โƒ’ ๐‘ฆ โˆ’ ๐œŒv๐‘ฆ โƒ’ ๐‘ฆ+ฮ”๐‘ฆ) โˆ†๐‘ฅโˆ†๐‘ง z (^) ๐œŒv๐‘ง โˆ†๐‘ฅโˆ†๐‘ฆโƒ’ (^) ๐‘ง ๐œŒv๐‘ง โˆ†๐‘ฅโˆ†๐‘ฆโƒ’ (^) ๐‘ง+ฮ”๐‘ง (๐œŒv๐‘ง โƒ’ ๐‘ง โˆ’ ๐œŒv๐‘ง โƒ’ ๐‘ง+ฮ”๐‘ง) โˆ†๐‘ฅโˆ†๐‘ฆ

Accumulation = ๐œ•๐‘š๐œ•๐‘ก = ๐œ•๐œŒโˆ†๐‘ฅโˆ†๐‘ฆโˆ†๐‘ง๐œ•๐‘ก = โˆ†๐‘ฅโˆ†๐‘ฆโˆ†๐‘ง ๐œ•๐†๐œ•๐‘ก

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

(๐œŒv๐‘ฅโƒ’ ๐‘ฅ โˆ’ ๐œŒv๐‘ฅ โƒ’ ๐‘ฅ+ฮ”๐‘ฅ) โˆ†๐‘ฆโˆ†๐‘ง + (๐œŒv๐‘ฆ โƒ’ ๐‘ฆ โˆ’ ๐œŒv๐‘ฆ โƒ’ ๐‘ฆ+ฮ”๐‘ฆ) โˆ†๐‘ฅโˆ†๐‘ง + (๐œŒv๐‘ง โƒ’ ๐‘ง โˆ’ ๐œŒv๐‘งโƒ’ ๐‘ง+ฮ”๐‘ง) โˆ†๐‘ฅโˆ†๐‘ฆ = โˆ†๐‘ฅโˆ†๐‘ฆโˆ†๐‘ง ๐œ•๐†๐œ•๐‘ก

Dividing each term in the above equation by โˆ†๐‘ฅโˆ†๐‘ฆโˆ†๐‘ง:

(๐œŒv๐‘ฅโƒ’ ๐‘ฅโˆ’ ๐œŒv๐‘ฅ โƒ’๐‘ฅ+ฮ”๐‘ฅ) โˆ†๐‘ฅ +^

(๐œŒv๐‘ฆ โƒ’๐‘ฆโˆ’ ๐œŒv๐‘ฆ โƒ’๐‘ฆ+ฮ”๐‘ฆ) โˆ†๐‘ฆ +^

(๐œŒv๐‘ง โƒ’๐‘งโˆ’ ๐œŒv๐‘ง โƒ’๐‘ง+ฮ”๐‘ง) โˆ†๐‘ง =^

๐œ•๐† ๐œ•๐‘ก

Take the limit as ฮ”x, ฮ”y, and ฮ”z approach zero:

โˆ’ (^) ๐œ•๐‘ฅ๐œ• ๐œŒv๐‘ฅ - (^) ๐œ•๐‘ฆ๐œ• ๐œŒv๐‘ฆ - (^) ๐œ•๐‘ง๐œ• ๐œŒv๐‘ง = ๐œ•๐œŒ๐œ•๐‘ก

โˆด (^) ๐œ•๐‘ฅ๐œ• ๐œŒv๐‘ฅ+ (^) ๐œ•๐‘ฆ๐œ• ๐œŒv๐‘ฆ + (^) ๐œ•๐‘ง๐œ• ๐œŒv๐‘ง + ๐œ•๐œŒ ๐œ•๐‘ก = 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):

โˆ‡. ๐œŒvโƒ—โƒ—โƒ—โƒ— + ๐œ•๐œŒ๐œ•๐‘ก = 0

direction in out in - out x (^) ๐‘›๐ด,๐‘ฅ โˆ†๐‘ฆโˆ†๐‘งโƒ’ (^) ๐‘ฅ ๐‘›๐ด,๐‘ฅ โˆ†๐‘ฆโˆ†๐‘งโƒ’ (^) ๐‘ฅ+ฮ”๐‘ฅ (๐‘›๐ด,๐‘ฅ โƒ’ ๐‘ฅ โˆ’ ๐‘›๐ด,๐‘ฅโƒ’ ๐‘ฅ+ฮ”๐‘ฅ) โˆ†๐‘ฆโˆ†๐‘ง y (^) ๐‘›๐ด,๐‘ฆ โˆ†๐‘ฅโˆ†๐‘งโƒ’ (^) ๐‘ฆ ๐‘›๐ด,๐‘ฆ โˆ†๐‘ฅโˆ†๐‘งโƒ’ (^) ๐‘ฆ+ฮ”๐‘ฆ (๐‘›๐ด,๐‘ฆ โƒ’ ๐‘ฆ โˆ’ ๐‘›๐ด,๐‘ฆ โƒ’ ๐‘ฆ+ฮ”๐‘ฆ) โˆ†๐‘ฅโˆ†๐‘ง z (^) ๐‘›๐ด,๐‘ง โˆ†๐‘ฅโˆ†๐‘ฆโƒ’ (^) ๐‘ง ๐‘›๐ด,๐‘ง โˆ†๐‘ฅโˆ†๐‘ฆโƒ’ (^) ๐‘ง+ฮ”๐‘ง (๐‘›๐ด,๐‘ง โƒ’ ๐‘ง โˆ’ ๐‘›๐ด,๐‘ง โƒ’ ๐‘ง+ฮ”๐‘ง) โˆ†๐‘ฅโˆ†๐‘ฆ

Accumulation = ๐œ•๐‘š๐œ•๐‘ก = ๐œ•๐†๐‘จโˆ†๐‘ฅโˆ†๐‘ฆโˆ†๐‘ง๐œ•๐‘ก = โˆ†๐‘ฅโˆ†๐‘ฆโˆ†๐‘ง ๐œ•๐† ๐œ•๐‘ก๐‘จ

If A is produced within the control volume by a chemical reaction at a rate ๐‘Ÿ๐ด (mass/(volume)(time)

Rate of production of A (generation) = ๐‘Ÿ๐ด โˆ†๐‘ฅโˆ†๐‘ฆโˆ†๐‘ง

Put all terms in the equation: in โ€“ out + generation โ€“ consumption = accumulation

(๐‘›๐ด,๐‘ฅ โƒ’๐‘ฅ โˆ’ ๐‘›๐ด,๐‘ฅโƒ’ ๐‘ฅ+ฮ”๐‘ฅ) โˆ†๐‘ฆโˆ†๐‘ง + (๐‘›๐ด,๐‘ฆโƒ’ ๐‘ฆ โˆ’ ๐‘›๐ด,๐‘ฆ โƒ’๐‘ฆ+ฮ”๐‘ฆ) โˆ†๐‘ฅโˆ†๐‘ง + (๐‘›๐ด,๐‘ง โƒ’๐‘ง โˆ’ ๐‘›๐ด,๐‘งโƒ’ ๐‘ง+ฮ”๐‘ง) โˆ†๐‘ฅโˆ†๐‘ฆ + ๐‘Ÿ๐ด โˆ†๐‘ฅโˆ†๐‘ฆโˆ†๐‘ง = โˆ†๐‘ฅโˆ†๐‘ฆโˆ†๐‘ง ๐œ•๐œŒ ๐œ•๐‘ก๐ด

Dividing each term in the above equation by โˆ†๐‘ฅโˆ†๐‘ฆโˆ†๐‘ง:

(๐‘›๐ด,๐‘ฅโƒ’ ๐‘ฅโˆ’ ๐‘›๐ด,๐‘ฅโƒ’ ๐‘ฅ+ฮ”๐‘ฅ) โˆ†๐‘ฅ +^

(๐‘›๐ด,๐‘ฆ โƒ’๐‘ฆโˆ’ ๐‘›๐ด,๐‘ฆ โƒ’๐‘ฆ+ฮ”๐‘ฆ) โˆ†๐‘ฆ +^

(๐‘›๐ด,๐‘งโƒ’ ๐‘งโˆ’ ๐‘›๐ด,๐‘งโƒ’ ๐‘ง+ฮ”๐‘ง) โˆ†๐‘ง +^ ๐‘Ÿ๐ด=^

๐œ•๐œŒ๐ด ๐œ•๐‘ก

Take the limit as ฮ”x, ฮ”y, and ฮ”z approach zero:

โˆ’ (^) ๐œ•๐‘ฅ๐œ• ๐‘›๐ด,๐‘ฅ - (^) ๐œ•๐‘ฆ๐œ• ๐‘›๐ด,๐‘ฆ - (^) ๐œ•๐‘ง๐œ• ๐‘›๐ด,๐‘ง + ๐‘Ÿ๐ด= ๐œ•๐œŒ ๐œ•๐‘ก๐ด

๐œ• ๐œ•๐‘ฅ ๐‘›๐ด,๐‘ฅ^ +^

๐œ•๐‘ฆ ๐‘›๐ด,๐‘ฆ^ +^

๐œ•๐‘ง ๐‘›๐ด,๐‘ง^ +^

๐œ•๐‘ก โˆ’^ ๐‘Ÿ๐ด^ =^0 (1)

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

โˆ‡. ๐‘›โƒ—๐ด + ๐œ• ๐œ•๐‘ก๐œŒ๐ด โˆ’ ๐‘Ÿ๐ด = 0 (2)

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

But from Fickโ€™s law

๐‘›๐ด = โˆ’๐œŒ๐ท๐ด๐ตโˆ‡๐œ”๐ด + ๐œŒ๐ดv

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

โˆ’โˆ‡. ๐œŒ๐ท๐ด๐ตโˆ‡๐œ”๐ด + โˆ‡. ๐œŒ๐ดv + ๐œ• ๐œ•๐‘ก๐œŒ๐ด โˆ’ ๐‘Ÿ๐ด = 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:

๐‘›๐ด ๐‘๐‘ฆ ๐‘๐ด;๐œŒ ๐‘๐‘ฆ ๐‘; ๐œ”๐ด ๐‘๐‘ฆ ๐‘ฆ๐ด ; ๐œŒ๐ด ๐‘๐‘ฆ ๐‘๐ด ๐‘Ž๐‘›๐‘‘ ๐‘Ÿ๐ด ๐‘๐‘ฆ ๐‘…๐ด

The different forms of the component molar continuity equation:

๐œ• ๐œ•๐‘ฅ ๐‘๐ด,๐‘ฅ^ +^

๐œ•๐‘ฆ ๐‘๐ด,๐‘ฆ^ +^

๐œ•๐‘ง ๐‘๐ด,๐‘ง^ +^

๐œ•๐‘ก โˆ’^ ๐‘…๐ด^ =^0 (^4 )

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

โˆ‡. ๐‘โƒ—โƒ—๐ด + ๐œ• ๐œ•๐‘ก๐‘๐ด โˆ’ ๐‘…๐ด = 0 (5)

But from Fickโ€™s law

๐‘๐ด = โˆ’๐‘๐ท๐ด๐ตโˆ‡๐‘ฆ๐ด + ๐‘๐ดV

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

โˆ’โˆ‡. ๐‘๐ท๐ด๐ตโˆ‡๐‘ฆ๐ด + โˆ‡. ๐‘๐ดV + ๐œ• ๐œ•๐‘ก๐‘๐ด โˆ’ ๐‘…๐ด = 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:

๐œ•๐‘๐ด ๐œ•๐‘ก =^ ๐ท๐ด๐ตโˆ‡

Equation 7 referred to as Fickโ€™s second law of diffusion

Fickโ€™s second โ€˜โ€˜lawโ€™โ€™ 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 ๐‘๐ด is equal in magnitude but acting in the opposite direction to ๐‘๐ต ๐‘๐ด = โˆ’๐‘๐ท๐ด๐ต^ ๐‘‘๐‘ฆ ๐‘‘๐‘ง ๐ด+ ๐‘ฆ๐ด(๐‘๐ด + ๐‘๐ต)

If ๐‘๐ด = โˆ’๐‘๐ต the bulk motion term will be cancelled from the above equation

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

For mass concentration:

โˆ‡^2 ๐œŒ๐ด = 0

For mass concentration:

โˆ‡^2 ๐‘๐ด = 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

๏ƒ˜ 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.

๐‘Ž๐‘ก ๐‘ก = 0 ๐‘๐ด = ๐‘๐ด๐‘œ (๐‘š๐‘œ๐‘™๐‘Ž๐‘Ÿ ๐‘ข๐‘›๐‘–๐‘ก๐‘ )

๐‘Ž๐‘ก ๐‘ก = 0 ๐œŒ๐ด = ๐œŒ๐ด๐‘œ (๐‘š๐‘Ž๐‘ ๐‘  ๐‘ข๐‘›๐‘–๐‘ก๐‘ )

where ๐‘๐ด๐‘œ and ๐œŒ๐ด๐‘œ are constant (defined values)

(ii) for solutions where species A is only weakly soluble in the liquid, Henryโ€™s law may be used to relate the mole fraction of A in the liquid to the partial pressure of A in the gas ๐‘๐ด = ๐ป โˆ™ ๐‘ฅ๐ด H: Henryโ€™s constant (iii) at gas solid interface

๐‘๐ด๐‘ ๐‘œ๐‘™๐‘–๐‘‘ = ๐‘† โˆ™ ๐‘๐ด ๐‘๐ด๐‘ ๐‘œ๐‘™๐‘–๐‘‘ : is the molar concentration of A within the solid at the interface in units of kg mol/m^3 and ๐‘๐ด is the partial pressure of gas phase species A over the solid in units of Pa. ๐‘†: solubility constant (partition coefficient) (2) A reacting surface boundary is specified

๐‘๐ด โƒ’ ๐‘ง=0 = ๐‘˜๐‘๐‘๐ด^ ๐‘›๐‘ 

where ๐‘˜๐‘ is a surface reaction rate constant with units of m/s. n is the reaction order

Note: the reaction may be so rapid that ๐‘๐ด๐‘  = 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

๐‘๐ด โƒ’ ๐‘ง=0 = โˆ’๐ท๐ด๐ต^ ๐œ•๐‘ ๐œ•๐‘งโƒ’๐ด ๐‘ง=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

๐‘๐ดโƒ’ ๐‘ง=0 = ๐‘˜ (๐‘๐ด๐‘  โˆ’ ๐‘๐ดโˆž ) 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 โ†’ 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 Fickโ€™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.

๐œ• ๐œ•๐‘ฅ ๐‘๐ด,๐‘ฅ^ +^

๐œ•๐‘ฆ ๐‘๐ด,๐‘ฆ^ +

๐œ•๐‘ง ๐‘๐ด,๐‘ง^ +

๐œ•๐‘ก โˆ’ ๐‘…๐ด^ = 0

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

๐œ• ๐œ•๐‘ง ๐‘๐ป,๐‘ง^ + ๐‘…๐ป^ = 0

b. Fickโ€™s law relationship in terms of only compound H

๐‘๐ป = โˆ’๐‘๐ท๐ป๐ฟ^ ๐‘‘๐‘ฆ ๐‘‘๐‘ง ๐ป+ ๐‘ฆ๐ป(๐‘๐ป + ๐‘๐ฟ)

Apply these assumptions on the general differential equation:

โˆ‡. ๐‘โƒ—โƒ—๐ด + ๐œ•๐‘ ๐œ•๐‘ก ๐ดโˆ’ ๐‘…๐ด = 0

โˆด โˆ‡. ๐‘โƒ—โƒ—๐ด = 0

For spherical coordinates and mass transfer in r-direction

1 ๐‘Ÿ^2

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

๐‘๐ด = โˆ’๐‘๐ท๐ด๐ต^ ๐‘‘๐‘ฆ ๐‘‘๐‘Ÿ ๐ด+ ๐‘ฆ๐ด(๐‘๐ด + ๐‘๐ต)

๐‘๐ต = 0

๐‘๐ด = โˆ’ (^) (1 โˆ’ ๐‘ฆ๐‘๐ท๐ด๐ต ๐ด)

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:

  1. no chemical reaction occur
  2. For deep lake (stationary liquid) we can assume v = 0
  3. Unidirectional mass transfer (assume in z direction) ๐‘‘๐‘๐ด ๐‘‘๐‘ง +

But

๐‘๐ด = โˆ’๐‘๐ท๐ด๐ต^ ๐‘‘๐‘ฆ ๐‘‘๐‘ง ๐ด+ ๐‘๐ด๐‘‰

Since the liquid is stationary

โˆด ๐‘๐ด = โˆ’๐‘๐ท๐ด๐ต^ ๐‘‘๐‘ฆ ๐‘‘๐‘ง๐ด

๐‘‘๐‘๐ด ๐‘‘๐‘ง = โˆ’๐ท๐ด๐ต

๐‘‘^2 ๐‘๐ด

๐‘‘๐‘ง^2

๐‘‘^2 ๐‘๐ด

๐‘‘๐‘ง^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:

  1. chemical reaction occur (โˆ’๐‘…๐ด = ๐‘˜๐‘Ÿ๐‘๐ด)
  2. For deep lake (stationary liquid) we can assume v = 0
  3. Unidirectional mass transfer (assume in z direction)

๐ท๐ด๐ต^ ๐‘‘

๐‘‘๐‘ง^2 +

๐œ•๐‘ก + ๐‘˜๐‘Ÿ๐‘๐ด^ = 0

where ๐‘˜๐‘Ÿis the reaction rate constant

But ๐‘‰๐‘ฆ = 0 (based on the assumptions)

โˆด ๐‘๐ด,๐‘ฆ = โˆ’๐ท๐ด๐ต^ ๐‘‘๐‘ ๐‘‘๐‘ฆ๐ด

Substitute by the values of ๐‘๐ด,๐‘ฅ and ๐‘๐ด,๐‘ฆ in equation (i)

โˆ’๐ท๐ด๐ต^ ๐œ•

๐œ•๐‘ฅ^2 + ay

๐œ•^2 ๐‘๐ด

๐œ•๐‘ฆ^2 = 0

โˆด ๐ท๐ด๐ต [๐œ•

๐œ•๐‘ฅ^2 +

๐œ•^2 ๐‘๐ด

๐œ•๐‘ฆ^2 ] = ay

Boundary conditions:

  1. ๐‘๐ด = ๐‘๐ด๐‘  at y = 0
  2. ๐‘๐ด = 0 at y = โˆž
  3. ๐‘๐ด = 0 at x = 0

Note:

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

  1. Select the general equation according to the system coordinates and omit the unnecessary terms
  2. 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