Topic 3.0 Unsteady State Diffusion
by Engr. Mudono
3.1 Introduction
3.2 Unsteady State Molecular Diffusion and Fick’s
Second Law
3.3 Transient Diffusion in a Semi-infinite Medium
3.4 Diffusion through a varying cross-sectional areas
3.5 Diffusion through a spherical body
3.1 Introduction to Unsteady State Molecular Diffusion
• Unsteady-state molecular diffusion or transient diffusion
describes processes where the diffusion flux and the
concentrations change with time.
• Since solids are not easily transported through equipment
as fluids, the application of batch and semi batch
processes and consequently unsteady-state diffusional
conditions arise much more frequently than with fluids.
• Even in continuous operation, e.g., a continuous drier, the
history of each solid piece as it passes through equipment
is representative of the unsteady state.
• These generally fall into two categories:
1. a process that is in an unsteady state only during its
initial startup, and
2. a process in which the concentration is continually
changing throughout its duration.
• These cases are therefore of considerable importance.
• The time-dependent differential equations are simple
to derive from the general differential equation of mass
transfer.
• The equation of continuity for component A in terms of mass:
𝜕𝜌𝐴
𝛻 ∙ 𝒏𝐴 +
− 𝑟𝐴 = 0 −− −1
𝜕𝑡
• The equation of continuity for component A in terms of moles:
𝜕𝐶𝐴
𝛻 ∙ 𝑵𝐴 +
− 𝑅𝐴 = 0 −− −2
𝜕𝑡
• Where there is no bulk flow, and in the absence of chemical
reaction, Fick's second law, can be used to solve problems of
unsteady-state diffusion by integration with appropriate
boundary conditions.
Mass transfer chart for solid objects
Dimensionless number in mass transfer chart
3.2 Unsteady State Molecular Diffusion and Fick’s
Second Law
• All mass transfer processes will have an initial period
of time with unsteady – state conditions where the
concentration at certain point varies with time until
steady – state is reached.
Fig 3.1 Unsteady state molecular diffusion mass transfer
𝑁𝑒𝑡 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑚𝑜𝑙𝑎𝑟 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
𝑁𝑒𝑡 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑚𝑜𝑙𝑎𝑟 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
•
−
𝑜𝑢𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
𝑖𝑛𝑡𝑜 𝑡ℎ𝑒 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
=
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑎𝑐𝑐𝑢𝑚𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑝𝑒𝑐𝑖𝑒𝑠 𝐴
𝑖𝑛 𝑡ℎ𝑒 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
• The molar flow rate of species A by diffusion at the plane X=X
is given by Fick's law:
𝜕𝐶𝐴
𝑊𝐴𝑥 = −𝐷𝐴𝐵 𝑆
−− −1
𝜕𝑥 𝑥
• The molar flow rate of species A by diffusion at the plane x =
x+Δx, is:
𝜕𝐶𝐴
𝑊𝐴𝑥 = −𝐷𝐴𝐵 𝑆
−− −2
𝜕𝑥 𝑥+∆𝑥
• The accumulation of species A in the control volume is:
𝜕𝐶𝐴
𝑆
∆𝑥 −− −3
𝜕𝑥
• Combining eqns. 1, 2 & 3:
𝜕𝐶𝐴
𝜕𝐶𝐴
−𝐷𝐴𝐵 𝑆
+ 𝐷𝐴𝐵 𝑆
𝜕𝑥 𝑥
𝜕𝑥
𝑥+∆𝑥
𝜕𝐶𝐴
=𝑆
∆𝑥 −− −4
𝜕𝑥
• Rearranging and simplifying:
𝐷𝐴𝐵
𝜕𝐶𝐴 𝜕𝑥
− 𝜕𝐶𝐴 𝜕𝑥
∆𝑥
𝑥+∆𝑥
𝑥
𝜕𝐶𝐴
=
−− −5
𝜕𝑡
• In the limit, as Δx→0
𝜕𝐶𝐴
𝜕 2 𝐶𝐴
= 𝐷𝐴𝐵
−− −6
2
𝜕𝑡
𝜕𝑥
𝐸𝑞𝑛. 6 𝐹𝑖𝑐𝑘 ′ 𝑠 𝑠𝑒𝑐𝑜𝑛𝑑 𝑙𝑎𝑤 𝑓𝑜𝑟 𝑜𝑛𝑒
− 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛
• For the more general three-dimensional case where
concentration gradients are changing in the x, y and z
directions, these changes must be added to give:
𝜕𝐶𝐴
=
𝜕𝑡
2
𝜕 𝐶𝐴
𝐷𝐴𝐵
2
𝜕𝑥
+
2
𝜕 𝐶𝐴
2
𝜕𝑦
+
2
𝜕 𝐶𝐴
2
𝜕𝑧
−− −7
𝐹𝑖𝑐𝑘 ′ 𝑠 𝑠𝑒𝑐𝑜𝑛𝑑 𝑙𝑎𝑤 𝑓𝑜𝑟 𝑢𝑛𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛
𝑖𝑛 𝑡ℎ𝑟𝑒𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
• Examining eqn.6:
𝜕𝐶𝐴
𝜕 2 𝐶𝐴
= 𝐷𝐴𝐵
−− −6
2
𝜕𝑡
𝜕𝑥
• 𝐶𝐴 : Concentration of component A (kg/m3, kmol/m3)
• t: time (s)
• DAB: mass diffusivity (m2/s)
• x: distance (m)
• We need to employ the following boundary conditions:
𝐹𝑜𝑟 𝑡 = 0, 𝐶𝐴 = 𝐶𝐴0 𝑎𝑡 0 ≤ 𝑥 ≤ ∞
𝐹𝑜𝑟 𝑡 = 0, 𝐶𝐴 = 𝐶𝐴𝑠 , 𝑡ℎ𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑡 𝑥 = 0
𝐶𝐴 = 𝐶𝐴0 𝑎𝑡 𝑥 = ∞
• Including these three boundary conditions Fick's second law
can be solved to yield:
𝐶𝐴𝑥 −𝐶𝐴0
𝐶𝐴𝑠 −𝐶𝐴0
= 1 − 𝑒𝑟𝑓
𝑥
2 𝑫𝑡
−− −8
• The error function(erf) is tabulated and it is just a
mathematical function that can only be represented by an
integral, you can use it just by looking up values in a table
and interpolating.
• You will not need to calculate error functions numerically, but
for your curiosity erf(x) is:
𝑥
2
−𝑦 2
𝑒𝑟𝑓 𝑥 =
𝑒
𝑑𝑦 −− −9
∏ 0
• The error function erf(x) can also be calculated from the
infinite series:
erf 𝑥 = 𝑥 −
𝑥3
3
+
1 𝑥5
2! 5
−
1 𝑥7
3! 7
+ ⋯ −− −10
• However, many problems in unsteady-state diffusion can be
solved without the complication of error function calculation.
• For certain problems, one can employ a simple
relationship between the time and distance at which a
certain concentration will occur.
𝑥2
1 − 𝑒𝑟𝑓
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑜𝑟
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 −− −11
𝑫𝑡
2 𝑫𝑡
𝑥
• Factors that Influence Diffusion Rate:
1. Both the diffusing species and the host material affect D.
2. Temperature
𝑄
− 𝑑 𝑅𝑇
𝑫 = 𝑫𝟎 𝑒
−− −12
𝑄𝑑
𝑙𝑛𝑫 = 𝑙𝑛𝑫𝟎 −
𝑅𝑇
𝑜𝑟
𝑄𝑑
𝑙𝑜𝑔𝑫 = 𝑙𝑜𝑔𝑫𝟎 −
−− −13
2.303 𝑅𝑇
• Therefore, a plot of lnD versus 1/T should yield a straight line with
slope -Qd/R and intercept lnD0.
• For one-dimensional diffusion in the radial direction only for
cylindrical coordinates, Fick's second law becomes:
𝜕𝐶𝐴 𝐷𝐴𝐵 𝜕
𝜕𝐶𝐴
=
𝑟
−− −14 (𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙)
𝜕𝑡
𝑟 𝜕𝑟
𝜕𝑟
• For one-dimensional diffusion in the radial direction only for
spherical coordinates, Fick's second law becomes:
𝜕𝐶𝐴 𝐷𝐴𝐵 𝜕
𝜕𝐶𝐴
2
= 2
𝑟
−− −15(𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙)
𝜕𝑡
𝑟 𝜕𝑟
𝜕𝑟
3.3 Transient Diffusion in a Semi-infinite Medium
• The boundary conditions for this case to solve (eqn.6) are:
• At t = 0 0 < x < ∞ 𝐶𝐴 = 𝐶𝐴0
t>0
x = 0 𝐶𝐴 = 𝐶𝐴𝑖 𝐶𝐴𝑖 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛
t>0
x = ∞ 𝐶𝐴 = 𝐶𝐴0
𝜕𝐶𝐴
𝜕 2 𝐶𝐴
= 𝐷𝐴𝐵
−− −6
2
𝜕𝑡
𝜕𝑥
• To solve the above partial differential equation, either the
method of combination of variables or the Laplace method is
applicable.
• The result, in terms of the fractional accomplished
concentration change (θ), is:
𝐶𝐴 − 𝐶𝐴0
𝑥
𝜃=
= 𝑒𝑟𝑓𝑐
−− −10
𝐶𝐴1 − 𝐶𝐴0
2 𝐷𝐴𝐵 𝑡
• Equation 10 is used to compute the concentration in the semiinfinite medium, as a function of time and distance from the
surface, assuming no bulk flow.
• Thus, it applies most rigorously to diffusion in solids, and also
to stagnant liquid and gases when the medium is dilute in the
diffusing solute.
• The instantaneous rate of mass transfer across the surface of
the medium at X = 0 can be obtained by taking the derivative
of (eqn. 10) with respect to distance and substituting it into
Fick's first law applied at the surface of the medium:
𝜕𝐶𝐴
𝑛𝐴 = −𝐷𝐴𝐵 𝑆
𝜕𝑧
= 𝐷𝐴𝐵 𝑆
𝑥=0
𝐶𝐴𝑆 − 𝐶𝐴0
𝜋𝐷𝐴𝐵 𝑡
𝑥2
𝑒𝑥𝑝 −
4𝐷𝐴𝐵 𝑡
− −11
𝑥=0
• Thus:
𝑛𝐴
𝑥=0
=
𝐷𝐴𝐵
𝑆 𝐶𝐴𝑠 − 𝐶𝐴0 −− −16
𝜋𝑡
• We can determine the total number of moles of solute, NA,
transferred into the semi-infinite medium by integrating eqn. 16
with respect to time:
𝑡
𝑁𝐴 =
𝑛𝐴
0
𝑥=0
𝑑𝑡 =
𝐷𝐴𝐵
𝑆 𝐶𝐴𝑠 − 𝐶𝐴0
𝜋𝑡
𝑡
0
𝑑𝑡
𝑡
= 2𝑆 𝐶𝐴𝑠 − 𝐶𝐴0
𝐷𝐴𝐵 𝑡
−− −17
𝜋
Table 3.1 The Error Function
Mass transfer chart
3.3.1 Semi-infinite Medium correlations
3.4 Diffusion through a varying cross-section area
• The mole rate (𝑵A,kmol/s ) through a system of a varying cross
sectional area is constant, while the mole flux (NA,kmol/m2.s ) is
variable.
• The mass transfer through a cone and sphere can be consider as a
mass transfer through a system of varying cross sectional area.
• On the other hand, the transfer through a cylinder can be consider as a
mass transfer through a system of constant cross section area.
𝑚𝑜𝑙𝑒 𝑟𝑎𝑡𝑒
𝑁𝐴 𝑘𝑚𝑜𝑙
𝑁𝐴 =
=
−− −1
2
𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎
𝑆 𝑚 .𝑠
Fig 3.2 Variable cross sectional area
3.5 Diffusion through a spherical body
𝑑𝐶𝐴 𝐶𝐴
𝑁𝐴 = −𝐷𝐴𝐵
+
𝑁𝐴 + 𝑁𝐵 −− −2
𝑑𝑟 𝐶𝑇
𝑁𝐴
𝑑𝐶𝐴 𝐶𝐴 𝑁𝐴 𝑁𝐵
= −𝐷𝐴𝐵
+
+
−− −3
𝑆
𝑑𝑟 𝐶𝑇 𝑆
𝑆
𝑑𝐶𝐴 𝐶𝐴
𝑁𝐴 = −𝐷𝐴𝐵 𝑆
+
𝑁𝐴 + 𝑁𝐵 −− −4
𝑑𝑟 𝐶𝑇
• But: The surface area of sphere: S=4πr2
𝑑𝐶𝐴 𝐶𝐴
𝑁𝐴 = −4𝜋𝑟 𝐷𝐴𝐵
+
𝑁𝐴 + 𝑁𝐵 −− −5
𝑑𝑟 𝐶𝑇
2
Case 1: Diffusion through a stagnant layer 𝑁𝐵 = 0
𝑑𝐶
𝐶
𝐴
𝐴
2
𝑁𝐴 = −4𝜋𝑟 𝐷𝐴𝐵
+
𝑁𝐴 + 0
𝑑𝑟 𝐶𝑇
𝑁𝐴 1 − 𝐶𝐴
𝑟1
𝑁𝐴
𝑟0
𝑑𝐶
𝐴
2
= −4𝜋𝑟 𝐷𝐴𝐵 𝐶𝑇
𝑑𝑟
𝑑𝑟
𝐶𝑇 − 𝐶𝐴2
= −4𝜋𝐷𝐴𝐵 𝐶𝑇 𝑙𝑛
2
𝑟
𝐶𝑇 − 𝐶𝐴1
4𝜋𝐷𝐴𝐵 𝐶𝑇
𝐶𝑇 − 𝐶𝐴2
𝑁𝐴 =
𝑙𝑛
−− −6
1 1
𝐶𝑇 − 𝐶𝐴1
−
• The most important things is to calculate the mass transfer rate for the
𝟐
sphere surface where the surface area is constant 𝟒𝝅𝒓𝟎 :
4𝜋𝐷𝐴𝐵 𝐶𝑇
𝐶𝑇 − 𝐶𝐴2
𝑁𝐴 𝑆 =
𝑙𝑛
1 1
𝐶𝑇 − 𝐶𝐴1
−
𝑟0 𝑟1
𝑁𝐴 𝟒𝝅𝒓𝟐𝟎
4𝜋𝐷𝐴𝐵 𝐶𝑇
𝐶𝑇 − 𝐶𝐴2
=
𝑙𝑛
1 1
𝐶𝑇 − 𝐶𝐴1
−
𝑟0 𝑟1
𝐷𝐴𝐵 𝐶𝑇
𝐶𝑇 − 𝐶𝐴2
𝑁𝐴 =
𝑙𝑛
−− −7
1
𝐶𝑇 − 𝐶𝐴1
2 1
𝑟0
−
𝑟0 𝑟1
• Mole flux from the sphere surface (eqn.7)
• When the mass transfer from surface to a large distance
compare to the sphere surface (𝐫𝟎): 𝑟1 → ∞ 𝑎𝑛𝑑 𝐶𝐴2 = 0
𝐷𝐴𝐵 𝐶𝑇
𝐶𝑇 − 𝐶𝐴2
𝑁𝐴 =
𝑙𝑛
1
𝐶
−
𝐶
2 1
𝑇
𝐴1
𝑟0
−
𝑟0 ∞
𝐷𝐴𝐵 𝐶𝑇
𝐶𝑇 − 𝐶𝐴2
𝑁𝐴 =
𝑙𝑛
−− −8
𝑟0
𝐶𝑇 − 𝐶𝐴1
• In partial pressure form:
𝑁𝐴 =
𝐷𝐴𝐵 𝑃𝑇
𝑃𝑇 −𝑃𝐴2
𝑙𝑛
𝑟0 𝑅𝑇
𝑃𝑇 −𝑃𝐴1
−− −9
Case II: Equimolecular Counter Diffusion (𝑁𝐴 = −𝑁𝐵 ):
𝑑𝐶𝐴 𝐶𝐴
𝑁𝐴 = −𝐷𝐴𝐵
+
𝑁𝐴 + 𝑁𝐵 −− −10
𝑑𝑟 𝐶𝑇
𝑁𝐴
𝑑𝐶𝐴 𝐶𝐴 𝑁𝐴 𝑁𝐵
= −𝐷𝐴𝐵
+
+
−− −11
𝑆
𝑑𝑟 𝐶𝑇 𝑆
𝑆
𝑑𝐶𝐴
𝑁𝐴 = −4𝜋𝑟 𝐷𝐴𝐵
−− −12
𝑑𝑟
2
𝑟1
𝑁𝐴
𝑟0
𝑑𝑟
=
−4𝜋𝐷
𝐴𝐵
2
𝑟
𝐶𝐴2
𝑑𝐶𝐴 −− −13
𝐶𝐴1
1 1
𝑁𝐴
−
= 4𝜋𝐷𝐴𝐵 𝐶𝐴1 − 𝐶𝐴2 −− −14
𝑟0 𝑟1
4𝜋𝐷𝐴𝐵
𝑁𝐴 =
𝐶𝐴1 − 𝐶𝐴2
1 1
−
𝑟0 𝑟1
• For the mass transfer from surface (S=4π𝑟02 ):
𝐷𝐴𝐵 𝐶𝑇
𝑁𝐴 =
1
2 1
𝑟0
−
𝑟0 𝑟1
• In the case of 𝑟1 is very large ⇒
1
𝑟1
𝐶𝐴1 − 𝐶𝐴2
=0
𝐷𝐴𝐵
𝑁𝐴 =
𝐶𝐴1 − 𝐶𝐴2
𝑟0
• In the form of partial pressure:
𝐷𝐴𝐵
𝑁𝐴 =
𝑃𝐴1 − 𝑃𝐴2 −− −15
𝑟0 𝑅𝑇
Case III: Unequimolecular Counter Diffusion (𝑁𝐴 = −𝑛𝑁𝐵 ):
𝐷𝐴𝐵 𝑃𝑇
1
𝑃𝑇 − (1 − 𝑛)𝑃𝐴2
𝑁𝐴 =
𝑙𝑛
−− −16
𝑅𝑇 𝑟0 (1 − 𝑛)
𝑃𝑇 − (1 − 𝑛)𝑃𝐴1