# Topic 3.0 Unsteady State Diffusion ECE 2102 ```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
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
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 &amp; 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
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 &lt; x &lt; ∞ 𝐶𝐴 = 𝐶𝐴0
t&gt;0
x = 0 𝐶𝐴 = 𝐶𝐴𝑖 𝐶𝐴𝑖 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛
t&gt;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
```