Convective Mass Transfer Coefficient Video Lecture | Mass Transfer - Chemical Engineering

let's consider convective mass transfer
note that in the previous modules on
diffusion we have noted that molecular
diffusion is quite slow so to enhance
transport of a component we can use
convection then the mass flux of that
component will be higher in other words
it will move at a faster rate so we can
have convective mass transfer in gases
and liquid and also within porous solids
the convective mass transfer coefficient
is given a symbol K M M stands for mass
transfer and K is the symbol for the
mass transfer coefficient and it is
defined as the rate of mass transfer per
unit area per unit concentration
difference so we can write that in a
mathematical expression as km equals m
dot B divided by a times CB 1 minus C B
2 where m dot B is the mass flux of
component B in kilograms per second and
C is the concentration of the component
B and that will be mass per unit volume
which is kilograms per cubic meters and
a is the area or in square meters so the
units of km will be by substituting the
units for the terms on the right hand
side we have kilograms per second
and in the denominator square meters and
then kilogram in the denominator and
cubic metre in the numerator so
kilograms will cancel out square metre
term will cancel out so we are left with
meter per second so the units of km are
meters per second we can get an a better
understanding of the mass transfer
coefficient from its units as we noted
that the units are meter per second we
can also multiply both the numerator and
denominator by square meters so we have
the units expressed as cubic meter
divided by meter square per second and
then we can say that km the mass
transfer coefficient represents the
volume in cubic meter of component B as
it is transported across a boundary of
one square meter per second and so that
gives us a little better understanding
of mass transfer coefficient from its
units so we have km equals m dot B
divided by a times C B 1 minus C B 2 we
can rearrange these terms as m dot B
equals km times a times C B 1 minus CB 2
now you will recall from your perhaps
first course of physical chemistry that
for an ideal gas law we can write PB
equals Rho B times RB times T where PB
is the partial pressure of gas B in
kiloPascals RB is the gas constant for
gas be Rho B is the mass concentration
of be in kilogram per cubic meters and T
is the absolute temperature we can
rearrange the terms as Rho B equals P B
divided by RB times T now note that the
gas constant R B can be written in terms
of the universal gas constant as R u
divided by M B so ru is the universal
gas constant and M B is the molecular
weight of gas B so we can substitute for
R B in our earlier expression the mass
concentration of B Rho B equals P B
times M B divided by
R u times T now note that Rho B is the
mass concentration of B and earlier in
our expression above we call the
concentration as cb1 and cb2 so that's
the same the C the concentration is same
as Rho so we can go ahead and substitute
for that concentration terms in that
expression and we will get m dot B
equals K M times a times the molecular
weight of BM B divided by are you the
universal gas constant times the
absolute temperature and in parentheses
now we have partial pressure P b1 minus
partial pressure P b2 this expression is
useful when we are trying to estimate
the mass flux based on the vapor
pressure gradient instead of the
concentration gradient so note that here
we have the gradient expressed in
partial pressures we will use these
expressions in solving problems
where we are interested in finding out
the mass flux