Double Pipe Counter Flow Heat Exchanger (GATE Previous Year Question), Heat and Mass Transfer, GATE Video Lecture | Topper Handwritten Notes & Videos for GATE ME - Mechanical Engineering

- Marcin until now in case of heat
exchanger I have shared with you a lot
of questions lot of theoretical
questions and now this is a question
which has been asking gay people s year
in numerical later type so whenever you
are getting quotient
at least in case of heat exchanger what
you need to do we need to just draw the
diagram of that is what type of heat
exchanger is this so this is a case of
counter flow so draw a diagram of
counter flow heat exchanger alright so
first of all what do you have to do you
have to draw a diagram of counter flow
heat exchanger so this is th one this is
th - this is TC - this is TC one because
of counter flow heat exchanger right now
we will write what they have given in
this heat exchanger and we will find out
what they wanted to further get right
from this quotient so they said that
counter flow you get sender victors will
be doing water Aslam skate so both the
streams are water we got from this
question we took till now so tube side
water add this is heated from 10 degree
centigrade to 38 degree centigrade so it
said that it is hitting from 10 degree
centigrade to 38 degree centigrade right
it's it's flow rate is 19 litre per
second now shell typer heats sunshine in
heat so you sell side water at it's flow
rate is 25 liter per second and then
it's entering temperature and thing
temperature is 46 teary-eyed centigrade
so they are they said to us the density
is 1000 kg per meter cube and CP is 4
point 1 8 6 kilo joule per kg Kelvin all
right so what we have to find out
basically what is LM TD LM TD is equal
to what we have to find out so here is
that LM today we know that the formula
for LM TT is theta 1 minus theta 2 by
natural log theta 1 by theta 2 we know
this now what is Theta
one difference between these two so what
is that theta one is equal to degree
centigrade but can you find theta 2
theta do from here no because we don't
know the value of pH 2 so first of all
what we have to do we have to find out
what is the edge 2 then only we can able
to find out what theta 2 are you adding
my point so for that we are having a
good relationship what's that let me
share with you what is that about the
relationship that is if activeness of
heat exchanger which is denoted by
epsilon if the dimness of a lectionaries
you got to see edge e H 1 minus T edge
to see minimum T H 1 minus TC 1 is equal
to CC TC 2 minus TC 1 I will share with
you why I am writing this C minimum T H
1 minus TC 1 ok so let me share with you
what is see is what is he see what is
seeming even what you see this is this C
means heat capacity that is mass flow
rate multiple CP so if I am talking
about hot fluid I am talking about hot
fluid the mass flow rate or and CP of
the hot fluid so hot view the fluid is
what water so CP for both the cases
would be worth four point one eight six
kilo Joule per kg Kelvin for this case
also and for these case also okay so
this is the thing now look at this
equation from this equation let us take
this one and this one so the denominator
denominator will get cancelled it will
not become 0 it will become 1 okay so
what we get basically see as th 1 minus
th to CC TC 2 minus TC 1 all right so we
don't know what is ch basically we have
to find out so what is C as basically 15
multiple this so sorry 25 multiple this
not started 15 25 multiple 4.18 6 th one
is 46 th 2 is
we have to find out and what is CC
nineteen multiple 4.18 six TC 2 is 30
late and TC 1 is 10 so this one this one
get cancelled right you can easily see
this this one this one get cancelled now
we can say that 46 - eh 2 is equal to 19
multiple this will become 2890 multiple
28th division 25 then we can also we
have to calculate this for these I have
to use calculator 19 multiple 28
dividend 25 is equal to this so this is
becoming what 21 point to it now send
this th to the side and this one to this
side so we can see the 46 - answer is
this so tears - is word from here we can
find out we find out actually for twenty
four point seven two degrees centigrade
right twenty four point seven degrees
centigrade so what is the value of th to
twenty four point seven - that means
theta two is equal to this one - this
one so fourteen point seven - now put
the values of theta one and theta two
get the answer simple so what is theta 1
basically 8 - 4 sorry
8 - fourteen point seven - natural log
beta 1 is 8 and theta 2 is 14 point 7 -
now find out this eight minus fourteen
point seven - eight - fourteen point
seven two TVs are natural log a division
fourteen point seven - yeah so it's
eleven point zero two degree centigrade
is the lmpd for this quotient okay so
this is the question which has been
asking gate previous here in numerical
type okay so calculation
you must give you focus on calculation
that's why you can get most accurate
answer otherwise you know the question
you may have
even you are not going to get marks of
it right so what's the use of that
nothing isn't it
so you need to focus on this also
alright so thanks for watching thank you
very much and hope to see you again in
my next video tour well Baba