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

so welcome to distort Russ this is
virtual Commerce in and today in this
video taught well I am going to share
with you how to find out LM today in
case of counterflow alright
in case of counter flow heat exchanger
so the same question right now you can
see in front of you this question
already asked in gate previous here
that's why I am sharing with you so how
to solve this problem first of all if
they have ass they have given this is a
counter flow heat exchanger then you
need to draw a counter flow heat
exchanger first at least in case of heat
exchanger numerical problems you need to
draw diagrams if we draw a diagram then
things become quite easy of oil why it
is counter flow because it's flow will
be in reverse direction
alright so now now I will read the
question and put values according to
they have given so this one is th one
this one is th two this one is TC 2 this
one is TC 1 right and we have to find
out what theta M we have to find out
theta M that is LM TD and L M TT 4
counter flow heat exchanger is what
theta M is equal to theta 1 is equal to
theta 2 and what is theta 1 theta 1 is
equal to th 1 minus T C 2 and theta 2 is
equal to t h2 minus T C 1
I mean defense of these two is equal to
theta 2 difference of these two is equal
to theta 1 got it so this is these are
things we require to find out theta M
that is why I am sharing with you do you
know I haven't even read the quotient
till I have shared with you what you
need to know to solve the quotient and
this is not only for counter flow heat
exchanger even fill the condenser case
you will get evaporator case 4 or even
you will get parallel flow heat
exchanger if you know these things you
know the basic the diagram in heat
exchanger you can solve the numerical
problem smoothly I mean you can get the
right answer this is a simple topic and
the most
of the time if you are talking about
create even university examination we
got quotient from this topic that's why
I am sharing with you I hope you are
getting my point right now let's go for
the numerical problem so what helps if
they have said in this quotient that
water CP is this at 80 degree centigrade
enters a counter flow it has dinner with
a mass flow rate this much okay and air
enters at it this much so if water is
entered this mark this and this my
temperature and air is entering with
disma temperature automatically we can
say which one is hot fluid and which one
is cold fluid here so definitely hot
fluid is water and cold fluid is here so
water is 80 degree centigrade and cold
fluid is 30 degree centigrade got it now
we got these two arrows again we got few
or few other information that is CP of
water is we have given one a four point
one eight kilo Joule per kg Kelvin again
we have given another information that
is mass flow rate of the water is 0.5 kg
per say per second C is eco it's M des
okay
now let's go for the air air CP is equal
to 1 kilo joule kilo joule per kg Kelvin
and it's mass flow rate is equal to two
point zero nine kg per second getting my
point so these are data they have given
so we know that C is equal to M multiple
CP let's say this is for water so zero
point five into one point okay now use
calculator averaging state so it's two
point zero nine we got its heat capacity
you can see now from here it's is M dash
and CP right is equal to one multiple
two point zero nine so it's two point
zero nine now if we check the unit let's
check unit right kilo Joule per kg
Kelvin multiple kg per second so this
one this one get cancelled
what is remaining 2.09 kilo Joule per
Kelvin second got it so in this way you
can get even unit so we don't need to go
there about that unit and all that right
you can find out quite easy so what we
got basically now why we require this
that is important I am sharing with you
here effectiveness they have given that
is epsilon epsilon is what Q by Q max
that is actual heat transfer by the heat
exchanger actually transfer head you can
say and the maximum possible heat
transfer is in the denominator are you
getting my point
so from this equation we know that heat
capacity of hot fluid by pH 1 minus th 2
division C minimum that is it capacity
meenie minie mo what is the minimum
minimum will compare heat capacity of
cold fluid will with the heat capacity
or what weight capacity of hot field
then we will consider the minimum one so
that is th one minus is in one so what
is yes basically here we got that 2.09
what is th one here th one is a t t s2
will find out from here what is C
minimum and there is both the same so we
can consider any one th one is ATTC one
is 30 now what is effectiveness we have
given they have given point eight now
multiple knife just at this here this
cancel this will come fifty okay now
what we have to do 80 minus th 2 is
equal to 0.8 multiple fifty point eight
multiple 50 are forty so 40 point 0 now
th to this side and this is on this side
so 80 minus 40 is equal to 40 now th to
equal code is equal to 40 as I have
already shared with you in case of
counter flow heat exchanger theta 1 is
equal to theta theta 2 is equal to theta
M based on that data we can say theta 2
is equal
to theta 2 is equal to 40 minus 30 is
equal to 10 right 10 degrees centigrade
therefore LM TD LM TD for this question
is 10 degree centigrade cause because
theta M is equal to theta 2 is equal to
theta 1 so this condition is applicable
here that's why I am sharing with you so
the answer for this question is 10
degrees centigrade which one is this
option of some B I hope you understand
the total if you are getting help for my
daughterís then don't forget to share
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watching thank you very much for to see
you again in my next video tutorial bye
bye