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

Jamison and in this short oval we are
going to learn about one of the potion
this question already has can get
previously and for two months
alright so in this question what they
have passed that what we're what well is
so well is cool from 80 degree
centigrade to 50 degree sending it in
and well cooler which uses air as the
coolant air temperature rises from 30 to
40 the designer uses the a LM t devalue
or 26 degree okay so what we have to do
we have to find out which type of heat
exchanger exactly they are using alright
so from the title you came to came to
watch this video total most wobbly
because I'm talking about here cross
flow but we don't have any idea or any
formula to find out LM till now as per
our course to find out LM TT we know
only for finding out LM to be only for
parallel flow and counter flow so if we
give this type of question what you have
to do we have to find out LM t be for
parallel flow and counter flow then even
come to know which type of heat
exchanger is this because a lambda T
value is live into in parallel flow and
counter flow heat exchanger that means
Ln this will a parallel flow LM TD in is
24 23 degree centigrade and counter flow
LM today is 27 degree centigrade so the
cross flow in between the value would be
the relative value would be for cross
flow would be in between parallel flow
and counter flow that's I'm talking
about so what you'll have to do
basically in this question we have to
find out LMPD for parallel flow and LM
TD for counter flow so for this as I
have already shared with you couple of
questions regarding and of course
concept for finding out LM TD in case of
heat exchanger for that you have to
follow my Nathan's
so what about these methods let me share
with you so suppose that this is a
parallel flow heat exchanger so this
would be like this and one we are
drawing a diagram for counter flow heat
exchanger okay so the diagonals for
parallel flow this
the parallel flow and this is the
counter flow parallel flow heat
exchanger and this is counter flow heat
exchanger so what is happening here
hot whale is cool from this so 80 to 50
8250 and here 30 to 40 so 30 to 40 and
30 to 40 and temperature they have given
in degrees centigrade so it doesn't
matter for us for that because we have
all temperature that is 2 and degree
centigrade only okay so now what we are
going to do we are going to find out nmt
default panel fluid exchanger and for
counter flow heat exchanger
as well now for that we have to find out
theta 1 and theta 2 for both the type of
heat exchanger so how we can able to
find out theta 1 theta 2 so theta 1 for
here is 50 difference between these two
temperatures here is 10 and here 20 we
are 20 and here how much you're getting
for tea so we got this belt right so now
what we have to do we have to find out
LM telling now how we can able to find
out LM TD that is also matter that is
also matter so how we can able to find
out LM TD let's go for it
so in case of pal I mean LM TD that is
theta my denoted by theta M is equal to
theta 1 minus theta 2 natural log ETA 1
by theta 2 okay just put values theta 1
is 50 minus 10 and then natural log 50
by 10 15 by 10 so we have to use
calculator so keep t minus 10 division
natural log 50 by 10 okay so what we got
basically value yes twenty four point
eight five from this value degree
centigrade of course so from this
weather we can say this is not the Sun
the parallel flow is north off Sun for
the designer now we will go for the
counter flow heat exchanger the formula
for finding out LM did
for the case of counter flow heat
exchanger is also same but for finding
out Lita 1 theta 2 the way is different
that's why theta M for this counter flow
heat exchanger is theta 1 minus theta 2
natural log kita 1 by theta 2 so what is
theta 1 here for T what is Theta 2 here
20 so natural log for T by 20 this is
cancelled - so from here you can say
this is 20 20 degrees and natural log 2
so what we are getting value here
littlez value is yeah value is you can
see yeah of course you can see twenty
eight point eight five degree centigrade
now this is also not the option as I
have shared with you l entity for
parallel flow and counter flow is
suppose that this is this is 24 in doubt
for twenty eight so the cross flow LM TT
in between parallel flow and counter
flow therefore the Hobson is the right
option even double Phi we do not have
such formula for finding Oregon today at
least according to our course you know
so this is all about that finding out LM
TD okay for cross flow how to find out
how to know this is the cross flow cross
flow heat exchanger so this is about the
concept I hope you understand the
concept to find out the sum of this
potion so thank for thanks for watching
thank you very much and if you are
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watching thank you very much bye bye