DIFFUSION: DEFINITIONS
 Diffusion is a process of mass
transport that involves the atomic or
molecular motion.
 In the simplest form, the diffusion can
be defined as the random walk of an
ensemble of particles from regions of
high concentration to regions of lower
concentration
Diffusion: Driving Force
In each diffusion process (heat flow, for example, is also a diffusion process ),
the flux (of mater, heat, electricity, etc.) follows the general relation:
Flux = (Conductivity) x (Driving Force)
• In the case of atomic or molecular diffusion,
the “conductivity” is referred to as the
diffusivity or the diffusion constant, and is
represented by the symbol D, which reflects the
mobility of the diffusing species in the given
environment. Accordingly one can assume larger
values in gases, smaller ones in liquids, and
extremely small ones in solids.
• The “driving force” for many types of diffusion
is the existence of a concentration gradient.
The term “gradient” describes the variation of
a given property as a function of distance in the
specified direction.
MODELING DIFFUSION: FLUX
• Diffusion Flux (J) defines the mass transfer
rate:
• Directional Quantity
• Flux can be measured for:
--vacancies
--host (A) atoms
--impurity (B) atoms
STEADY-STATE DIFFUSION
(Fick’s First Law)
SSD takes place at constant rate. It means that throughout the
system dC/dx=const and dC/dt=0
DC/ Dx dC/dx
DC
Dx
 c 
J  D   
 x 
- Fick’s first law
The diffusion flux is proportional
to the existing concentration gradient
FIRST FICK’s LAW

C 
C 
C 
J  D  grad C  D( e x 
ey 
ez )
x
y
z
where D is the diffusion constant or diffusivity [m2/s] and
C is a concentration [kg/m3]
ey
ez
Jy
Jz
D=D0exp(-Qd/RT)
Jx
ex
here Q is the activation energy for the process: [J/mol];
Do is temperature-independent pre-exponential
constant: [m2/s]
NONSTEADY-STATE DIFFUSION
• Concentration profile, C(x), changes with time:
dC/dt0.
Example: diffusion from a finite volume through a membrane into a finite volume.
The pressures in the reservoirs involved change with time as does,
consequently, the pressure gradient across the membrane.
FICK’s SECOND LAW
Consider a volume element (between x and
x+dx of unit cross section) in a diffusion system.
The flux of a given material into a volume
element (Jx) minus the flux out of the element
volume (Jx+dx) equals the rate of material
accumulation in the volume:
x
Using a Taylor series we can expand Jx+dx:
J x dx
J x
1  2J x 2
 Jx 
dx 
dx  
2
x
2 x
Accordingly, as dx0 and using
Fick’s First Law:

C
C
(D ) 
x
x
t
J x  J x dx 
C
 dx
t
C is the average species concentrat ion
in the volume element
And if D does not vary with x we have
The formulation of Fick’s Second Law:
C
 2C
D 2
t
x
EX: NON STEADY STATE DIFFUSION
• Copper diffuses into a semi-infinite bar of aluminum.
Cs
C(x,t)
Co to
C
 C
D 2
t
x
2
t1
t3
t2
Boundary conditions:
For t=0, C=C0 at 0  x  
For t>0, C=Cs at x=0
C=C0 at x = 
position, x
• General solution:
 x 
C(x, t)  Co  

1 erf 
2 Dt 
Cs  Co
Error Function
erf (z) 
2
z
e
0
 y2
dy
z  x / 2 Dt
 The terms erf and erfc stand for error function and complementary error
function respectively - it is the Gaussian error function as tabulated (like
trigonometric and exponential functions) in mathematical tables. (see Table
5.1, Callister 6e).
 Its limiting values are:
THERMALLY ACTIVATED PROCESSES
Examples: rate of creep deformation, electrical conductivity in
semiconductors, the diffusivity of elements in metal alloy
Arrhenius Equation
Rate = K·exp(-Q/RT)
Energy
where K is a pre-exponential constant (independent of temperature),
Q- the activation energy, R the universal gas constant and T the absolute temperature
Atom must overcome an activation
energy q, to move from one stable
position to another
Mechanical analog: the box must
overcome an increase in potential energy DE,
to move from one stable position to another
Table 1
Table 2
First Fick’s Law
𝑗 = −𝐷 ∙ 𝑔𝑟𝑎𝑑𝐶 ≡ −𝐷 ∙
𝑑𝐶
𝑑𝑥
(1)
Assumptions:
 Molecular transfer without convection (natural or forced),
i.e. no motion as a whole (otherwise additional term
should be added);
 Λ – mean free pass << L – characteristics length of the
problem (otherwise Knudsen diffusion);
 T=constant (otherwise Onsager's equations for thermodiffusion);
 The concentration of this admixture should be small and
the gradient of this concentration should be also small; or
we have to consider only binary system; or diffusion
coefficients are equal to all components;
It means that ji depends on the gradient of only Ci !!
(otherwise Onsager's multicomponent diffusion) .
Convection
 Now, if we have motion of the system as a whole,
which is characterized by linear velocity 𝑣 than:
𝑗 = −𝐷 ∙ 𝑔𝑟𝑎𝑑𝐶 + 𝑣 ∙ 𝐶
(2)
 𝑣 – average molar velocity, which is related to the flux
of particles (for ideal gases equals to the average
volume velocity); for binary system (2) stands for wide
range of concentrations with almost constant D and 𝑣should be expressed as average volume velocity of
the mixture.
Stephan’s Flux
 If the heterogeneous transformations involve the change of the volume it leads
to the overall motion of the mixture in the direction normal to the boundary
on which transformation (reaction) takes place. As a result this convection flux
influences the diffusion flux and formula (2) should be used to describe the
process. This importance of accounting the change of the position of the
boundary in overall diffusion flux was for the first time outlined by Josef
Stefan (Stefan, J., 1890.Uber die Verdampfung und die Auflosung als Vorgange der Diffusion. Ann. Phys. — Berlin 277, 727–747).
 The special importance the Stefan Flux has for the process of evaporation
and condensation.
The average velocity for each component (i) can be defined as:
𝑗
𝑢𝑖 = 𝐶𝑖
(3)
𝑖
 The average velocity of the Brownian Motion equals to zero, thus 𝑢𝑖 is an
average velocity of directional movements of the media. The value of this
velocity depends on the coordinate system and only 𝑢𝑖 − 𝑢𝑘 is invariant.
 The overall flux rate is a sum of the directional rates of each component, thus
molar velocity:
𝑉𝑉 =
𝑗𝑖
𝐶𝑖
=
𝑢𝑖 𝐶𝑖
=
𝐶𝑖
𝑥𝑖 𝑢𝑖 (4)
Diffusion Flux
 We can also represent the flux using partial pressures:
𝑗𝑖 = −
𝐷𝑖
𝑔𝑟𝑎𝑑𝑝𝑖
𝑅𝑇
+
𝑣
𝑃
𝑅𝑇 𝑖
(5)
Summation over all components gives:
𝑖 𝑗𝑖 = −
1
𝑅𝑇
𝑖 𝐷𝑖 𝑔𝑟𝑎𝑑𝑝𝑖 +
𝑣
𝑃
𝑅𝑇
(6)
and if all Di = D, or as for binary diffusion D12=D21=D:
𝑖 𝑗𝑖
=−
If P=const than:
𝑖 𝑗𝑖 =
𝐷
𝑔𝑟𝑎𝑑𝑃
𝑅𝑇
𝑣
𝑃
𝑅𝑇
+
and 𝑣 =
𝑣
𝑃
𝑅𝑇
𝑅𝑇
𝑃
𝑖 𝑗𝑖 =
(7)
𝑗𝑖
𝐶𝑖
thus we obtained again average volume velocity.
(8)
Binary system
Near the surface when the transformation (reaction) occurs one may
consider 1D-case when we can neglect the gradients along the directions
parallel to the surface:
𝑗𝑖 =
𝐷𝑖 𝑑𝑃𝑖
−
𝑅𝑇 𝑑𝑥
+
𝑣
𝑝
𝑅𝑇 𝑖
(9)
 In the binary system it is only one coefficient of mutual diffusion
D12=D21=D
𝑢1 =
𝑗1
𝐶1
𝑢2 =
𝑗2
𝐶2
=
𝐷 𝑑𝑃1
−
𝑅𝑇𝐶1 𝑑𝑥
+𝑣
(10)
=
𝐷 𝑑𝑃2
−
𝑅𝑇𝐶2 𝑑𝑥
+𝑣
(11)
and
𝑢1 − 𝑢2 = −
𝐷 1
𝑑𝑃
(𝐶1 2
𝑅𝑇 𝐶1 𝐶2
𝑑𝑥
− 𝐶2
𝑑𝑃1
)
𝑑𝑥
(12)
Maxwell-Stefan formula
 Since P=P1+P2=const:
d𝑃1 d𝑃2
=d𝑥
d𝑥
we get famous Maxwell-Stefan formula:
𝐷 1
𝑑𝑃2
𝑑𝑃1
𝐷 ∙ 𝑃 𝑑𝑃1
𝑢1 − 𝑢2 = −
𝐶1
− 𝐶2
=−
∙
𝑅𝑇 𝐶1 𝐶2
𝑑𝑥
𝑑𝑥
𝑝1 𝑝2 𝑑𝑥
𝑢1 − 𝑢2 =
𝐷∙𝑃
𝑝1 𝑝2
∙ 𝛻𝑝1 = −
𝐶1 +𝐶2
D∙
𝐶1 𝐶2
𝛻𝐶1
* which for multicomponent case has the following general form:
(13)
Josef Stefan’s evaporation– diffusion tube
 A vertical tube is partly filled
with a liquid, which in turn
evaporated and the vapor flowed
out of the tube at the open end.
 The tube portion above the
liquid level contained a (binary)
mixture consisting of the
surrounding gas (air) and the
vapor, generated on the liquid–
gas interface.
 Due to evaporation, the liquid
level fell, and the process was
unsteady, even if all other
parameters were kept constant.
 Under these conditions, Stefan
derived equations for calculating
the concentration distribution
along tube length and the
evaporation rate of the liquid.
The Stefan solution (1)
 Stefan started by applying the momentum and continuity equations to diffusion in the
gaseous area above the liquid level. The gas mixture and its components were treated as ideal
gases, and the total pressure and temperature are constant through out the whole
system. The origin of the coordinate x is the plane of the upper tube end, and the position of
the liquid level is denoted by L. The gas mixture contained two components, and the
Stefan equations describing the diffusion process are:
0
(14)
(15)
The indices1and 2 refer to the mixture components1(vapor) and2 (gas); t is the time,
while A12 and A21 represent the resistance coefficients.
The Stefan solution (2)
 Adding equations (15) and
accounting (16):
 Adding equations (14):
(16)
(17)
0
Now to solve (14) and (15) Stefan used the boundary condition at the
moving liquid–gas interface where evaporation is taking place: As the
components are not consumed while passing the interface, their fluxes
(𝒏𝟏 𝒂𝒏𝒅 𝒏𝟐 ) are the same on both sides of the interface:
(18)
2G
where uI denotes the velocity of the interface; the indices L, G and I stand
for liquid, gas, and interface, respectively.
 Adding equations (18):
and considering that the interface is impermeable to gas(component 2) and the liquid does
not move, so c 2L=0, and u1L =0 , hence
(19)
Boundary condition
(17)
 Eq. (14) for component 1 is written as:
(20)
 And substituting c2u2 term from (20)
 We can rewrite (14) as:
(21)
where, for reason of simplicity we have omitted some indices, and set:
Note that Eq. (20) is obtained from the boundary conditions at the interface, but due to Eq.
(17) it is valid in the whole diffusion space.
The Stefan solution (3)
(21)
(15)
 Differentiating Eq.(21) with respect to x and combining with Eq. (15) for
component 1 gives the relation :
(21a)
With the boundary conditions:
(22)
This is known as the non-stationary Stefan diffusion equation
Additional Boundary Conditions
(21)
(22)
(20)
At x=0, the equation is satisfied automatically; Let us consider boundary condition at x=L ;
it is assumed that the concentration c1I depends only on temperature (saturation
concentration of vapor),it is independent of time t when the temperature is constant. Thus:
(23)
Integrating Eq.(23) with respect to x,
And comparing with (21) we may conclude that: 𝜑 𝑡 = −(𝑐1 𝑢1 )𝐼 at x=L and from (20):
(24)
Finally, combining Eqs. (23) and (24) gives the additional boundary condition relation:
𝑥=𝐿
𝐷12
𝜕𝑐1𝐼
𝜕𝑥
-
𝑐−𝑐1𝐼
𝑐
∙ 𝑐𝐿
𝑑𝐿
𝑑𝑡
=0
(25)
The Stefan solution (4)
So we have:
(21)
𝑥=𝐿
𝐷12
𝜕𝑐1𝐼
𝜕𝑥
-
𝑐−𝑐1𝐼
𝑐
∙ 𝑐𝐿
𝑑𝐿
𝑑𝑡
=0
(25)
0
Stefan gave the solution of Eq. (21) in the form:
(26)
where the constants B and a are to be determined from the boundary conditions at x=L,
and (26) becomes:
and with (25)
And thus
(27)
The Stefan solution (4)
So we have:
(27)
Stefan then proceeded to discuss the case when the molar liquid density was much larger than the gas
density, namely, cL>>c, and b=(cL-c)/c becomes very large. He simplified the integral (27),
From this equation the following expression for the position of the interface was deduced :
(27)
Stefan compared this expression with an expression he derived in an earlier paper (Stefan, 1873) under
the assumption of a constant evaporation rate (fixed interface):
(28)
Eq. (27) rests on the assumption c10=0, that is, p10=0. Assuming in this equation cL>>c1L, cL>>c1I, it
becomes identical with Eq.(28).
Slattery & Mhetar Solution
 Let us consider a vertical tube, partially filled
with a pure liquid A.
For simplicity, let us replace the finite
gas phase with a semi-infinite gas that
occupies all space corresponding to z2 > 0:
 For time t < 0, this liquid is isolated from the
remainder of the tube, which is filled with a
gas mixture of A and B, by a closed diaphragm.
Z2
at t =0 for all z2 > 0:
G(gas)
 The entire apparatus is maintained at a
constant temperature and pressure (neglecting
the very small hydrostatic effect).
X (A)=X (Ao)
X (A) - mole fraction of species A
in the gas phase;
X (Ao) - initial mole fraction of
species A in the gas phase
 At time t = 0, the diaphragm is carefully
opened, and the evaporation of A commences.
 Let us assume that A and B form an ideal-gas
mixture. This allows us to say that the molar
density c in gas phase is a constant
throughout the gas phase.
(1)
0
Z3
The liquid gas phase interface
is a moving plane
 We also assume that B is insoluble in A.
 We wish to determine the concentration
distribution of A in the gas phase as well as the
position of the liquid-gas phase interface as
functions of time.
L(liquid)
Z2 = h(t)
(2)
and at z2 = h for all t > 0:
X(A)= X(A)e q.
(3)
Slattery & Mhetar Solution (2)
Equations (1) and (2) suggest that we seek a solution to this problem of the form
V1= V3 =0
V2= V2 (t, z2)
(4)
X(A)= X (A)(t, Z2) .
V - molar averaged velocity of gas phase; V2 - Z2 component of the molar averaged velocity
of gas phase. Since c can be taken to be a constant and since there is no homogeneous chemical
reaction, the overall differential mass balance requires:
𝝏𝑽𝟐
𝝏𝒁𝟐
Z2
=𝟎
which implies:
G(gas)
V2=V2(t)
The overall mass balance at interface requires at Z2=h:
-c(L)U2= c(V2-U2)
U2
L(liquid)
(6)
(7)
where U2 is the Z2 component of the speed of displacement of the interface; and
c(L) is a molar density in a liquid phase.
The mass balance for species A at interface at Z2=h demands:
V2
0
(5)
Z3
-c(L)U2= J(A)2- c∙X(A)∙U2
(8)
where J(A)2 - Z2 component of the molar flux of species A with respect to the
laboratory (fixed) frame of reference and X(A) - mole fraction of species A in the
gas phase.
Slattery & Mhetar Solution (3)
The overall mass balance at interface requires at Z2=h:
-c(L)U2= c(V2-U2)
(7)
Where U2 is the Z2 component of the sped of displacement of the
interface
The mass balance for species A at interface at Z2=h demands:
Z2
G(gas)
-c(L)U2= J(A)2- c∙X(A)∙U2
(8)
It means that at Z2=h:
𝒄
𝑼𝟐 = − 𝒄(𝑳) −𝒄 ∙ 𝑽𝟐
(9)
and
V2
0
U2
L(liquid)
𝑱
Z3
𝑨 𝟐
=
𝒄(𝒄∙𝑿 𝑨 −𝒄(𝑳) )
𝒄−𝒄(𝑳)
∙ 𝑽𝟐
(10)
Slattery & Mhetar Solution (3)
It means that at Z2=h:
𝒄
𝑼𝟐 = − 𝒄(𝑳) −𝒄 ∙ 𝑽𝟐
𝑱
𝑨 𝟐
=
(9)
𝒄(𝒄∙𝑿 𝑨 −𝒄(𝑳) )
∙
𝒄−𝒄(𝑳)
𝑽𝟐
(10)
Now, from Fick’s law for binary diffusion:
𝑱
𝑨 𝟐
= 𝒄 ∙ 𝑿(𝑨) ∙ 𝑽𝟐 -𝒄 ∙ 𝑫𝟏𝟐
𝝏𝑿(𝑨)
𝝏𝒁𝟐
(11)
which accounting (10) allows to conclude that:
𝑽𝟐 =
at Z2=h
𝝏𝑿(𝑨)
(𝒄−𝒄 𝑳 )
𝑫
𝒄 𝑳 (𝟏−𝑿 𝑨 𝒆𝒒 ) 𝟏𝟐 𝝏𝒁𝟐
(12)
and based on (6) we may conclude that everywhere
𝑽𝟐 =
𝝏𝑿(𝑨)
(𝒄 𝑳 −𝒄)
− 𝒄 𝑳 (𝟏−𝑿
𝑫
𝑨𝑩 𝝏𝒁 ∣𝒛𝟐=𝒉
𝟐
𝑨 𝒆𝒒 )
(13)
Slattery & Mhetar Solution (3)
We concluded that everywhere :
𝑽𝟐 =
𝝏𝑿(𝑨)
(𝒄 𝑳 −𝒄)
− 𝒄 𝑳 (𝟏−𝑿
𝑫
𝑨𝑩 𝝏𝒁 ∣𝒛𝟐=𝒉
𝟐
𝑨 𝒆𝒒 )
(13)
Now equation, which reflects the continuity of the fluxes, or so-called differential
mass balance for A species, equires:
𝜕𝑋(𝐴)
𝜕𝑡
+ 𝑉2
𝜕𝑋(𝐴)
𝜕𝑍2
= 𝐷𝐴𝐵
𝜕2 𝑋(𝐴)
𝜕𝑍 2
Or in view of eq.(13):
𝝏𝑿(𝑨)
𝝏𝒕
+
𝒄 𝑳 −𝒄
𝝏𝑿
− 𝒄 𝑳 𝟏−𝑿
𝑫𝑨𝑩 𝝏𝒁𝑨
𝟐
𝑨 𝒆𝒒
∣𝒛𝟐=𝒉 ∙
𝝏𝑿 𝑨
𝝏𝒁𝟐
− 𝑫𝑨𝑩
𝝏𝟐 𝑿(𝑨)
𝝏𝒁𝟐
Which must be solved with conditions (1) and (3):
at t =0 for all z2 > 0: X (A)=X (Ao)
at z2 = h for all t > 0: X(A)= X(A)e q.
(1)
(3)
=0 (14)
Slattery & Mhetar Solution (4)
𝝏𝑿(𝑨)
𝝏𝒕
+
𝒄 𝑳 −𝒄
𝝏𝑿
− 𝑳
𝑫𝑨𝑩 𝑨
𝒄
𝟏−𝑿 𝑨 𝒆𝒒
𝝏𝒁𝟐
∣𝒛𝟐=𝒉 ∙
𝝏𝑿 𝑨
𝝏𝒁𝟐
− 𝑫𝑨𝑩
𝝏𝟐 𝑿(𝑨)
𝝏𝒁𝟐
=0 (14)
Which must be solved with conditions (1) and (3):
at t =0 for all z2 > 0: X (A)=X (Ao)
at z2 = h for all t > 0: X(A)= X(A)e q.
With substitutions: 𝜂 =
𝑧2
4𝐷𝐴𝐵 𝑡
and 𝜆 =
(1)
(3)
ℎ
(15)
4𝐷𝐴𝐵 𝑡
eq. (14) becomes:
(16)
where
𝝋=
𝝏𝑿(𝑨)
(𝒄 𝑳 −𝒄)
− 𝒄 𝑳 (𝟏−𝑿
∙
∣𝒛𝟐=𝒉=
𝝏𝒁𝟐
𝑨 𝒆𝒒 )
And (1) and (3) in the form:
𝜂 → ∞; 𝑋(𝐴) → 𝑋
With recognition that:
𝐴 𝑜
- 𝟐𝑽𝟐
𝒕
𝑫𝑨𝑩
and at 𝜂=𝜆; 𝑋(𝐴) = 𝑋
l=constant
𝐴 𝑒𝑞
(17)
(18)
(19)
Solution (5)
(16)
𝜂 → ∞; 𝑋(𝐴) → 𝑋
𝐴 𝑜
and at 𝜂=𝜆; 𝑋(𝐴) = 𝑋
𝐴 𝑒𝑞
(18)
The solution of (16)-(18) is:
(20)
From (17) and (19), we see that:
(21)
Solution (6)
𝑐
𝑈2 = − 𝑐 𝐿 −𝑐 ∙ 𝑉2 (9)
(21)
Let us characterized the rate of evaporation by the position of phase interphase z2=h(t)
From (15) and (19) if follows:
𝑑ℎ
𝑑𝑡
= 𝑈2 = 𝜆
𝐷𝐴𝐵
𝑡
(22)
From (9), (17) and (22):
𝜆
𝜑 = 2(𝑐 𝐿 − 𝑐) 𝑐
(23)
Now we have to solve for a given system, (21), (23) we can find 𝝋 and
𝝀 𝐚𝐧𝐝 𝐭𝐡𝐚𝐧 𝐟𝐢𝐭 𝐭𝐡𝐞 𝐞𝐱𝐩𝐞𝐫𝐢𝐦𝐞𝐧𝐭𝐚𝐥 𝐡 = 𝐡 𝐭 𝐝𝐚𝐭𝐚
Notes:
(21)
 From eq. (21), we see that, in the limit X (A)e q → X (A)o, the dimensionless molar averaged
velocity 𝜑 →0, and the effects of diffusion-induced convection can be neglected (V2=0).
 If one simply says that V2=0 and uses Fick's second law, even though the overall jump
mass balance (7) suggests that this is unreasonable, it can be shown that:
(22)
 And only this equation should be solved to follow experimental h=h(t) data.
 Let us consider two liquids to emphasize this effect. Methanol has a relatively low vapor
pressure at room temperature, and we anticipate that the effects of diffusion-induced
convection will be small. Methyl formate has a larger vapor pressure, and the effects of
diffusion-induced convection can be anticipated to be larger.
Example (1)
 For Evaporation of
Methanol
T = 25,4 C
p = 1.006 × 105 Pa
X (methanol)eq.= 0.172
X(methanol)o = 0
D (methanol,air) == 1.558 x 10 -5 m2/s
Solving eqs. (21) and (23) simultaneously, we found
l= -1.74 x 10 -4
𝝋= -0.208.
From Fig. 1 below , it can be seen that the predicted
height of the phase interface follows the
experimental data closely up to 2500 s. As the
concentration front begins to approach the top of
the tube, we would expect the rate of evaporation to
be reduced. Note that neglecting diffusion-induced
convection results in an under prediction of the rate
of evaporation.
 (corrected from 1.325 × 10 -5 m2/s at
0ºC and 1 atm (Washburn, 1929, p. 62)
using a popular empirical correlation
(Reid et al., (1987) The Properties of
Gases and Liquids, 4th Ed. McGrawHill, New York, U.S.A.1987, p. 587);
c(L)= 24.6 kgmol/ 3
(Dean, J. A. (1979) Lange's Handbook of
Chemistry, 12th ed. McGraw-Hill, New York,
U.S.A. pp. 7 271. 10 89),
c=0.0411 kg mol/m 3
(Estimated for air from Dean, 1979, p. 10- 92)).
The lower curve gives the position of the phase interface h (mm)
as a function of t (s) for evaporation of methanol into air at
T = 25.4ºC and p = 1.006 x 105 Pa. The upper curve is the same
case derived by arbitrarily neglecting convection.
Example (2)
 For Evaporation of
Methyl Formate:
T = 25,4 C
p = 1.011 × 105 Pa
X (mformate)eq.= 0.784
X(methanol)o = 0
D (mformate,air) == 1.020 x 10 -5 m2/s
Solving eqs. (21) and (23) simultaneously, we found
l= -1.84 x 10 -3
𝝋= -1.44.
 From Fig. 2, it can be seen that the predicted
height of the phase interface follows the
experimental data closely over the entire range
of observation. Once again, neglecting diffusioninduced convection results in an
underprediction of the rate of evaporation
 (corrected from 0.872 × 10 -5 m2/s at C
and 1 atm (Washburn, 1929, p. 62)
using a popular empirical correlation
(Reid et al., (1987) The Properties of
Gases and Liquids, 4th Ed. McGrawHill, New York, U.S.A.1987, p. 587);
c(L)= 16.1 kgmol/ 3
(Dean, J. A. (1979) Lange's Handbook of
Chemistry, 12th ed. McGraw-Hill, New York,
U.S.A. pp. 7 271. 10 89),
c=0.0411 kg mol/m 3
(Estimated for air from Dean, 1979, p. 10- 92)).
The lower curve gives the position of the phase interface h (mm)
as a function of t (s) for evaporation of methyl formate into air
at T = 25.4ºC and p = 1.011 x 105 Pa. The upper curve is the same
case derived by arbitrarily neglecting convection.