Heat Transfer Coefficient Video Lecture | Mass Transfer - Chemical Engineering

if you have followed the other modules
on conduction and convection you will
note that we have considered those modes
of heat transfer separately now we are
going to combine conduction and
convection so that we can then develop
some useful expressions that will help
us in designing food processing
equipment so let's look at this concept
of overall heat transfer we will
consider a pipe and the inside radius of
this pipe is RI the outside radius is R
or let's say that there is a fluid
flowing inside the pipe and the
temperature the bulk temperature of that
fluid is TI for the inside and outside
let's say that again there is perhaps
air around the pipe and the bulk
temperature of air is T infinity the
pipe itself is made of some material
let's say it's thermal conductivity is K
and the convective heat transfer
coefficients for the inside of the pipe
on the inside surface of the pipe is H I
and on the outside surface is H O so if
we draw a temperature profile we will
start from the inside temperature let's
consider that to be our higher
temperature than outside so there is
some bulk temperature and that drops to
right at the surface of the pipe then if
the pipe is highly conductive there is a
very little temperature gradient inside
the pipe and then there is again a
temperature drop to t infinity
so we will use the approach of thermal
resistance values to obtain an
expression for the overall heat transfer
coefficient we can write that Q as the
rate of heat transfer equals TI minus T
infinity divided by RT where RT is a
combination of various thermal
resistances so let's see what what kind
of thermal resistances are there in this
case so RT will be equal to RT for the
inside convection that is inside the
pipe
plus RT for conduction through the pipe
material and then RT for the outside of
the pipe which is again due to
convection so now we will write
individual expressions for these various
thermal resistances so RT for inside
convection will equal 1 or H I ai you
may need to look at the previous module
on convective resistance to find how we
determine this expression then the
thermal resistance for conduction can be
written as RT conduction equals natural
log of RO over RI divided by 2 pi K L
where L is the length of the pipe again
you want to look at the module before
when we talked about conduction to see
how we obtain this expression for the
thermal resistance due to conduction the
third item is thermal resistance due to
outside convection and that will equal 1
over H o AO where again AO is the
outside area now if we substitute all
these different R values into our
earlier equation we have Q equals TI
minus T infinity divided by 1 or H I AI
plus natural log of RO over RI divided
by 2 pi L k plus 1 over H o times a or
to obtain an expression for overall heat
transfer we will write an equation very
similar
to convective heat transfer so overall
heat transfer is from the inside to the
outside which includes both convection
as well as convection so we will write Q
equals UI times AI times TI minus T
infinity where AI is the inside area of
the pipe and this term UI is the overall
heat transfer coefficient which is based
on the inside area of the pipe now UI is
the overall heat transfer coefficient
that we will need to know in designing
various equipment so we can rearrange
our equation as Q equals TI minus T
infinity divided by 1 over u ai ai where
just by comparison of this equation and
the previous one we have 1 over UI a I
equals 1 over H I a I plus natural log
of RO over RI divided by 2 pi L k plus 1
over H o AO so this equation here can be
used to determine UI if we know all the
other terms that are there in this
equation the one thing you will notice
is that we defined our overall heat
transfer coefficient UI as based on
inside area so the selection of this
area to calculate overall heat transfer
is actually quite arbitrary for example
if we were to use the outside area then
instead of writing u I will be writing
uo + uo will be the overall heat
transfer coefficient that is based on
the outside area in that case
the expression will be one over u o AO
equals one over HIA I plus natural log
of RO or RI divided by 2 pi L K plus 1
over H o AO so as you will notice on the
right hand side the items are exactly
the same in both cases but on the left
hand side if you use the U value for
inside area then you I must be
multiplied by AI on the other hand if
you used you Oh which is based on the
outside area then you must be multiplied
by a oh so the overall expression then
will be if you are using UO Q will be
equal to u or a or TI minus T infinity
either of these two equations will give
you the same value of Q this you can see
in a couple of solved examples in the
textbook