Overall Heat Transfer Coefficient HMT Tutorial- 26, Heat and Mass Transfer, Mechanical Engineering Video Lecture | Topper Handwritten Notes & Videos for GATE ME

Oh hide my dear so this will calm our
scene and in this video tutorial you
will learn about overall heat transfer
coefficient and if you have been
watching my videos then probably you
have noticed that I am sharing lot more
questions those has been asking gate
previously here with each and every
topic that's why you'll have basic
concept as well as numerical approach to
solve the problem okay so keep watching
and learn something about overall heat
transfer coefficient okay so basically
what is it overall heat transfer
coefficient is denoted by capital u it's
unit is what per meters per Kelvin okay
and what is the formula of this q is
equal to u a theta m and probably if you
have studied heat and mass transfer then
you know what is theta M theta M is L M
tilde logarithmic mean temperature
difference is area you I have already
shared with you and Q is heat transfer
rate which is in what so this is this is
all about heat transfer overall heat
transfer coefficient heat transfer rate
and all terms right so let's go for the
numerical problem but going before going
to numerical problem how we can utilize
this overall heat transfer policy and
and how this overall heat transfer
coefficient is helpful to find out
temperature event okay and even require
heat area heat area so are you
interested to know that problem I am see
or you are interested right so what do
you think should you go for that
numerical problem or we will end this
video here I hope you wanted to watch
that problem and wanted to know that
problem right so let's go for it
to go wait for few second on this one or
two second okay see I is the question
for you so we got this question I hope
right now we can read the whole question
okay that's why I have written like in
that way so what we have even basically
we have to find out he required heat
exchange
in meter square that is it now what we
have given basically we have given a
heat exchanger Soho hot water flowing
this heated from this and this is this
is no work ok book about it
there is nothing but most important
thing is that we haven't given any yes
what type of it external basically we
have given this is a counter flow heat
exchanger so at least in case of a
exchanger after seeing what type of heat
exchanger we have given just draw a
diagram of counter flow heat exchanger
that's why all elements given in this
numerical problem you can put there and
there it will be quite easier for you to
find out the answer for the numerical
problem okay I hope you're getting so
let's store the diagram for counter flow
heat exchanger okay this is a counter
for heat exchanger and it is flowing
like this or clewd and cold fluid this
is TC 1 this is TC 2 this is th one this
is th two well in my previous video so
from one of my viewers told me don't be
in hurry try to make us understand
slowly that's why these days I am
putting lot more time to make videos
more slow and push load - well you can
understand even if you are beginner ok
so what we have given cold water is cold
water already given water is flowing at
this much so what we have given point 1
kg per second cold water so I'm heated
from 2 until he is centigrade to 70
degree centigrade 20 degree centigrade
to 7 to agree centigrade okay now what
water is steam flowing at this same kz
per second and entering at say 90 degree
centigrade maglia centigrade the
specific heat of water is water is both
the for both the gases same kilo Joule
per kg Kelvin here also 4200 Joule per
kg Kelvin so Donnie Kovach
because both the cases further were the
cases we have same right now what we
have the men density of water is this
much so density is equal to mass by
volume that is thousand kg per meter
cube
if overall heat transfer coefficient you
for heat exchanger please we have given
you is called to 2000 word per meter
square Kelvin right and the required
heat exchange area a we need to find out
so we know that Q is equal to you a
theta M now what we required what we
have given you we have given you right
so what we required basically we need to
find out LM today LM today we have to
find out and also Q we need to find out
so there two things we need to find out
and we have given all things what we
require to find out these two things and
after finding out these two things we
can food all values here to find out the
answer of the question has been asked so
first of all let me find out Q so
basically what is Q q is equal to we
know that MCP delta T okay so what is MC
P heat capacity so heat capacity M is
0.1 and here 4200 delta T if will say
this or Delta D means changing temp I
mean the champ temperature difference so
70 minus 20 how much 50 so basically
what we are going to get for 420 420
here into 420 into 15 so the answer will
world for 20 into 54 2251 thousand
21,000 what don't forget to write the
unit because without writing unit you
cannot get the right answer so remember
that unit play most important rule for
any numerical problem if you are at
least if you are preparing for complete
examines and just like gate or SS CJ or
any other examination just like you pay
examination also coming
yeah in front of you and surfer right so
let's go fight so we have given Q we
have find out Q now what we have to do
we have to find out theta a theta M that
is LM TD so Ln PT is equal to how much
theta 1 minus theta 2 natural log beta 1
by theta 2 so it will say what is theta
1 and what is Theta 2 right for this we
need to find out that one value that is
th to we do not know what is th 2 so for
finding out th do you have we have
define a formula that is M CPC T C 2
minus T C 1 is equal to M - H CPH th one
- th tooth now
this value is same that is why will
cancel out and what will have T C 2
minus TC 1 is 50 already we if I find
out from here so 50 th one is 90 minus T
H 2 so th 2 is equal to 90 minus 90
minus 50 40 degree centigrade okay so
this is 40 degree centigrade so theta 1
is difference of these two so 20 degree
centigrade theta 2 is depends of these
two 20 degree centigrade how so not how
it's a 3 so what is it
theta 1 is equal to theta 2 so this is
only possible in case of counter flow
heat exchanger remember that I am
repeating them theta one is called theta
2 only possible in the case of counter
flow heat exchanger if it is the case
then what you need to do you need to say
like this theta 1 is equal to theta 2 is
equal to theta M that means theta 1 is
equal to theta 2 is equal to L MTD so
the value of L MTD here is only 20
degree centigrade so only 20 degree
centigrade now now what we have given
here what we have given a formula and we
have find out QQ is 21,000 if we will
stand
these theta M this side 20 and you is
how much 2,000 is equal to a okay if it
is as it is because we haven't changed
anything so don't need to take care of
it because area will become
automatically in metre square only so
what do we do till now we have find out
value of Q we have find out value of
theta M by using simple formula and most
important thing if we solve this problem
then probably the concept of a depth
sensor you will completely understand
because we have used numeric we have
used one two three four four formulas
here so these four formulas play most
important role for solving any question
related to heat exchanger but here one
thing was amazing that is we are using
only pakka counter flow heat exchanger
we are missing parallel flow heat
exchanger because most of time we also
have seen quotient from parallel flow
heat exchanger okay so what is the
answer of this twenty one thousand
divided twenty into two thousand is
equal to how much zero point five two
five meter square so let me see though
option 0.2 yeah here's of Sun so right
answer for this portion is 0.5 to 5 so I
hope you understand the numerical
problem as well as concept I wanted to
make you understand hopefully you have
enjoyed the video and got something new
from this video thanks for watching see
you soon in my next video bye bye