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Welcome to lecture 4 of module 7. In this module 7, we are discussing on heat exchanger.
Last three lectures we have discussed on various types of heat exchangers, their different
parts and their functions of the various parts of the different heat exchangers, we have
discussed in detail.
Now, today's lecture we will discuss on effectiveness NTU method of heat exchanger
analysis. So, this is in short we will write as epsilon NTU, this epsilon means stands
for effectiveness. Now, NTU means number of transfer unit. Now, this effectiveness NTU method of analysis
that is epsilon NTU Analysis, this method it is based on effectiveness of the heat exchanger
in transferring a given amount of heat. Therefore, what happens is, it is basically the effectiveness
of the heat exchanger to transfer a particular amount of heat. So therefore what happens,
what if it can this effectiveness method can help us to it, can help us to compare the
performances of various types of heat exchangers for similar heat exchange. So, that means
it helps us to understand the basic differences between two heat exchangers to compare them,
that how they are performing. So, in the form of the amount of heat exchange. That is very
important and this will help to identify or select the best suited type of heat exchanger.
So, these finally, helps us to select the best suited heat exchanger for the required
application. So, depending upon the type of application we have, we can now based on this
effectiveness method, based on we can find out the effectiveness of the heat exchanger
and then we can compare their performances and then we can select which one is to be,
which one or which type of heat exchanger is to be used for the set applications. So,
this is very important in case of design purpose so therefore, the heat effectiveness method
is also one important aspect in case of heat exchanger studies.
Now, how to define this effectiveness. So, this effective, heat exchanger effectiveness
is defined as, this equal to actual heat transfer by maximum possible heat transfer. Now, what
is this actual heat transfer? Actual heat transfer means the amount of sensible heat
being released or given or may be in terms of latent heat being given by the hot fluid
and which is being taken up by the cold fluid. That is nothing but, the actual heat transfer
what is being transferred through the heat exchanger.
So, this can be so if you draw that a parallel flow case and then the counter flow case,
single-pass case. Considering a single-pass case and this is one is parallel flow and
another is counter flow case, say if I write. So, this is say T h 1, this is T c 1, this
is T h 2 and this is T c 2, this is the station one and this is the station two and here it
can be area or length whatever you say heat exchanger. This is station one, this is station
two, this is area, so you can see it is like this. So, this it is say T h 1, this is T
h 2 sorry T h 2, this is T c 2 and T c 1. What is happening is that, in this case this
is the parallel flow, this is counter flow. So, actual heat transfer or if you say the
for parallel flow, actual heat transfer this will be m dot c mass flow rate of the cold fluid into C p c specific heat of the,
specific heat capacity of cold fluid m c p C p c into the change in temperature of the
cold fluid is T c 2 minus T c 1. So, this is the actual heat transfer that means the
heat being taken up by the cold fluid. Similarly, the same amount of heat is being
lost by the hot fluid and that will be equal to m dot h that the mass flow rate of hot
fluid C p h, that is the heat capacity, specific heat capacity of the hot fluid into the temperature
difference of T h 1 minus T h 2. So, this becomes the actual heat transfer in case of
parallel flow. And what is the maximum possible heat transfer?
Maximum possible heat transfer would be say, it depends upon that total amount of heat
transfer is depends upon the force into the mass heat capacity, so of the material, so
mass flow rate into this. Now, up this cold and hot fluid, if one of the fluid is having
flow rate m dot into C p. We have two things in this heat transfer rate. One we shall say
1 m dot into C p this part, another part is delta T, that is the temperature gradient.
So, and we have 2 fluids. 1 is hot fluid, another is cold fluid. Now, if we say for
hot fluid or say cold fluid. For cold fluid, if m dot C p is lower than, that means if
m dot C p for cold fluid is lower than m dot C p of hot fluid, then what will happen, delta
T for cold fluid will be greater than delta T for hot fluid. So, this will be understandable
because this is to maintain this. Therefore, the maximum possible heat transfer would be
say, suppose if the cold fluid is among these two, which 1 is having minimum m dot C p,
if that is called as the minimum fluid will tell that as a minimum fluid.
Under that situation, minimum fluid means, it has got minimum m dot into C p with respect
to the other one. If that is the minimum fluid, then it will undergo maximum temperature difference.
So, the maximum possible heat transfer would be any of the fluids, whether hot or cold.
They are heat flow rate, if into that the maximum temperature difference. The maximum
temperature difference here it will be T h 1 minus T c 1. So, that means in this particular
case, that the entry point, the temperature difference is maximum. So, T h 1 minus T c
1 will be the maximum temperature difference into the, so into that.
So, if I say that T h 1 minus T c 1, that will be the delta T max, maximum temperature
difference. And then, delta max into m dot C p of the minimum fluid, that is understandable.
So, minimum fluid only can go maximum temperature difference, other fluid cannot go that much
temperature difference. Because, this is to the, other than it will be law, avoid the
law of thermodynamics. So, total heat exchange from 1 to the other will be remaining the
same. So therefore, so if m dot C p is the minimum fluid, then into the delta T maximum,
that will actually the maximum possible heat transfer.
Now similarly for counter flow, actual heat transfer is, if we see previously that is
what I have written it is actually. Now, for cold fluid m dot c C p c into T c 1 minus
T c 2. Because, the cold fluid is entering in the other side, therefore that changes
will be there. So, it will be m dot c C p c into T c 1 minus T c 2, here the T c 1 is
because T c 1 is higher than T c 2 and that will be equal to m dot h C p h into T h 1
minus T h 2 the same thing. So, this is what is happening for counter flow and the maximum
possible heat transfer that we have seen. That if we say that, Q dot max is equal to,
then we can say that means we can say that Q dot max and this is equal to Q dot actual
for parallel sorry for counter flow or anti parallel flow. So, Q dot max is becoming now, m dot C p into
for the minimum fluid. Minimum fluid means, the fluid which is having m dot into C p this
product minimum into will have the maximum temperature difference, that will be T h inlet
minus T c inlet. So, that definitely the cold fluid inlet temperature and the hot fluid
inlet temperature difference is the maximum temperature difference.
So therefore, now we can have based on these combinations, if we have parallel flow, if
we have counter flow and depending on the situations hot fluid and cold fluid minimum
cases. We can have several combinations to have the expressions for effectiveness.
So, if we write say for parallel flow or co-current flow. We have say effectiveness epsilon for
parallel, there are two cases that hot fluid is minimum fluid. That means, that is m dot
h into C p h is less than m dot c into C p c. Then what will happen, we have epsilon
hot fluid I am sorry write parallel flow for hot fluid, it will be like this. It will be
m dot h is the cold fluid C p h into T h 1 minus T h 2 divided by the maximum possible
heat transfer is again m dot h C p h into T h 1 minus T c 1. So, this gives me this
cancel, so T h 1 minus T h 2 by T c T h 1 minus T c 1.
Similarly, you can have epsilon p cold fluid, when that cold fluid is minimum fluid. So,
I will write here parallel flow case and cold fluid is minimum fluid. So, we have four cases.
That is here m dot c C p c less than m dot h C p h. Mind it, that both the cases we are
considering on the parallel flow or co-current flow situation. So, this is becoming m dot
c C p c into and here, in case of parallel flow it is T c 2 minus T c 1. T c 2 is higher
than T c 1 parallel flow, and here it is m dot c C p c into T h 1 minus T c 1, and this
is equal to T c 2 minus T c 1 by T h 1 minus T c 1.
Similarly, if we have for anti parallel flow slash counter flow or counter current flow.
Under this situation, I will write again the hot fluid as minimum fluid. So, hot fluid
is minimum fluid, that is m dot h C p h is less than m dot c C p c. In that says, effectiveness
for anti parallel flow and for hot fluid and this is becoming like this, m dot h C p h
into again T h 1 minus T h 2 by m dot h C p h into T h 1 minus T c 2. yes in this case,
T c 2 is becoming the, because the hot fluid, cold is coming from the other side. Therefore,
it is becoming T h 1 minus T h 2 by T h 1 minus T c 2.
And then cold fluid as minimum fluid, that is m dot c C p c is less than m dot h C p
h. So, under this situation we will write that anti parallel flow and for cold fluid,
that will be equal to m dot c C p c into now here, it is T c 2 is the lower temperature.
So, T c 1 minus T c 2 divided by m dot c C p c into T h 1 minus T c 2. So, this is becoming
T c 1 minus T c 2 by T h 1 minus T c 2. So, this way we can find out different in
all the cases the finally, it is nothing but, if we just see in all the four cases whatever
we have discussed. It is nothing but, the temperature difference of the cold of the
minimum fluid divided by the maximum temperature difference in the heat exchanger.
So finally, we can say that effectiveness E equals to in any of the cases, we can say
the temperature difference of the minimum fluid divided by maximum possible temperature difference in the heat exchanger. So, this is an indirectly
we can, so this is the another definitions of heat exchanger, that you can say.
Now, what we will do is, we will try to find out would develop the expression for a single-pass
situation for effectiveness factor in case parallel flow and counter flow.
So, expression for effectiveness factor condition is, we are first considering this parallel
flow and we will consider the single-pass situation. Because multi-pass is more complicated,
so we are not going for that, we will just see that how single-pass situation is basically
derived. Now, so to do that first we will again draw
the parallel flow situation, whatever we have drawn already. So, if this is the case, now
here it is T h 1 and this is say T h 2, so this is the parallel flow situation and here
it is the area of heat exchanger, this is the station 1 and this is station 2.
Now what is happening, this is T c 1 and it is T c 2 is the outlet temperature of the
cold fluid and take any differences already we have discussed partly of the similar kind
of situations, we have discussed. While we have derived sometimes back the expressions
for LMTD, log mean temperature difference, that time we have made use of similar kind
of diagram and similar kind of energy balances we have seen. So, this is the area which is
nothing but, the differential area d a in the heat exchanger.
So, what happens is, if I say the differential heat exchange d Q dot and that will be equal
to minus m dot h hot fluid into C p h to d T h. Because, this heat being released by
the heat exchanger and we can say the d T h is the final minus initial, so this is a
negative quantity. Therefore, we have taken a minus sign over here and that is equal to
m dot c into C p c into d T c. d T c is the T c 2 is higher than T c 1. Therefore, d c
d T c is a positive quantity therefore, we have not taken any negative power here.
So therefore, from here we can write that d T h minus d T c is equal to, we can say
that a d Q dot is common into 1 by m dot h C p h plus there will be a minus sign over
here plus 1 by m dot c C p c. And again we know that, d Q dot also can be written as
UA, not A this is happening for this d Q is the differential amount of heat transfer,
that is happening for the area d A. So, it will be U overall heat transfer coefficient
into d A is the differential area into the temperature difference. So, if I say that
in this region, that temperature of the hot fluid is say T h and in this the average temperature
of the cold fluid is the T c. So, UA into it will be d A into T h minus T c. So, this
is the temperature difference. So, we have for the small differential area we will assume
that, it is uniformed here, it is uniform T c over here.
So, therefore we can still write, then the d of T h minus T c will be equal to U into
1 by m dot h C p h plus 1 by m dot c C p c into d A into we have this, T h minus T c.
So therefore, we can write that d of T h minus T c by T h minus T c. Now, we will do the
integration from station 1 to station 2 and that is equal to U of, U into 1 by m dot h
into C p h plus 1 by m dot c into C p c and then, we will integrate these for 0 to area
is A. So, area at this point for station 1 is equal to 0 and there is at a so 0 to A,
that is the situation. So, the A total area of the heat exchanger.
So, under this situation you can say, that l n T h 2 minus T c 2 by T h 1 minus T c 1
equals to minus U into A. We will assume that certain aspects here, that in the heat exchanger
the specific heat capacities, they are not changing with time and overall heat transfer
coefficients are not changing with time. So, these are the assumptions, inherence assumptions
that we have taken, while doing this derivation. And then, this is equal to 1 by m dot h C
p h plus 1 by m dot c C p c. Now, we can write that T sorry T h 2 minus
T c 2 by T h 1 minus T c 1 equals to exponential minus UA by m dot c C p c into 1 plus m dot
c C p c by m dot h C p h correct. Now, we know that for if we take assume, cold
fluid as the minimum fluid. Then, under that situation c m dot c C p c is less than m dot
h C p h. Now, what we will write now is that, m dot c C p c as say capital C c and m dot
and in this case, this is the c minimum fluid and m dot h C p h is equal to C h, capital
C h and this is the heat capacity. That is, it is not specific heat capacity, it is the
heat capacity of the fluid and this is specific heat capacity per unit amount. It is not per
unit amount and this is c, we can say that c max. So, if 1 is minimum, then other 1 is
c maximum. So therefore, we can another thing we can
write over here, is that m dot c C p c by m dot h C p h is equal to c mean equal to
C c by C h equal to c mean by c max. In the present case, we have assumed that cold fluid
as a minimum and this we will write C r is the ratio. So, this is the heat capacity ratio.
Now, also we can from the energy balance, we have seen that, we can write that m dot
h C p h into T h 2 minus T h 1 will be equal to m dot c C p c T c 1 minus T c 2. So, actually
it should be T h 1 minus this, and this will be T c 2 minus T c 1, and then negative sign.
So, we can write like this way also. So, Just one negative sign multiplication if you see
this. yeah T h 1 minus T h 2 T c 1 minus T h 1 minus T h 2 T c 2 minus T c 1. So, T h
2 minus T h 1 and T c 1 minus T c 2, just a reverse of that, by just putting it, if
we multiply with a negative sign then you will finally, get that.
And then, from there we can write that T h 2 is equal to T h 1 plus m dot c C p c by
m dot h C p h into T c 1 minus T c 2. And then, we can write that T h 2 minus T c 2
which is the need, because for we will see that, this is equal to T h 1 plus m dot c
C p c by m dot h C p h into T c 1 minus T c 2, then minus T c 2 again. So, and we know
that T h 2 minus T c 2 by T h 1 minus T c 1 equals to, if we just break it like this,
that T h 1 will put it here plus m dot c C p c by m dot h C p h into T c 1 minus T c
2 minus T c 2 divided by T h 1 minus T c 1.
Then, what we will get is left hand side, it is a T h 2 minus T c 2 by T h 1 minus T
c 1 is equal to, we will just rewrite revise the things rearrange basically this plus T
c 1 minus T c 2 plus m dot c C p c by m dot h C p h into T c 1 minus T c 2 and divided
by, we have T h 1 minus T c 1. And this is equal to, we can write now 1 plus T c 1 minus
T c 2 by T h 1 minus T c 1 into, we have 1 plus m dot c C p c by m dot h C p h.
Now, if we say that cold fluid as the minimum fluid. Then, effectiveness already we have
seen that for parallel flow and cold fluid there is the minimum fluid, we have T c 1
minus T c 2 by T h 1 minus T c 1, that already we have seen yes T h 1 T c 2 minus T c 1 by
T h 1 minus T c 1. So, we will have a negative sign over here, because of that reason, it
is T c 2 minus this will be minus of that. Therefore, we can write say this is equal,
so we can write that T h I am sorry this will be T h 2 minus T c 2 by T h 1 minus T c 1
equals to 1 minus epsilon into 1 plus C r. C r is the ratio, heat capacity ratios, and
if this is effectiveness. And this is equal to, and the right-hand side of their expression
had been exponential minus UA by C c into 1 plus C r, that is part over this case.
Now, we if we rewrite it again, then we will get epsilon equals to 1 minus exponential
minus UA by C c heat capacity of the cold fluid into 1 plus C r heat capacity ratio
of the minimum to the maximum fluid divided by 1 plus C r. Now, this UA by c c is called
number of transfer unit. So, it gives us that UA, gives the amount of heat transferred per
unit temperature difference. So, when unit temperature difference is there,
then how much amount of the heat is transferred that is given by UA. So, it is related to
that heat exchanger size and C c is the flow rate. So, UA by, so this UA by C c says, how
much amount of heat is being transferred by the heat exchanger per unit flow rate of the
heat per unit capacity of the cold fluid, minimal fluid and per unit temperature difference.
Therefore, it is an indicative of the heat size of the heat exchanger. So, it is indicative,
this is also NTU indicative of heat exchanger size.
So, then what is can write epsilon is equal to 1 minus exponential minus NTU into 1 plus
C r divided by 1 plus C r. So, this becomes the expression for parallel flow situation
and cold fluid as the minimal fluid. Exactly in the similar way, if hot fluid has the minimal
fluid, then our expression will becomes this will be C r. C r will be c minimum by c optimum,
but will be UA by C h in place of C c, that will be C h. So, one can try upon and try
and find out that. Now, if C r tends to 0. Then what happens,
epsilon is equal to, epsilon if C r tends to 0. Then, epsilon equal to 1 minus e to
the power minus NTU. So, it is then exponential e to the minus N T is the exponential then
1 minus e to the power NTU. That is called, this is the situation happens, particularly
in boiling and condensation. This is we can get during boiling and condensation heat exchange
heat transfer heat exchanger. Because in that case, that change the process fluid remains almost the
constant. In this case, the process fluid remains almost constant, it does not change
much. Therefore, in that this kind of situation can happen.
And another case, if C r equals to 1 then what will happen, then effectiveness can be
written as, if C r equals to 1, this is one. So, it has nothing to do much on this. So,
it will be 1 minus exponential 2 NTU by 2, it is like that 1 minus exponential C r equals
to 1. So, 2 NTU by this. So, C r equals to 1 means, is the case when a both of fluid
having heat capacity set. Now, similar derivations we can see for, but
the things will be little bit different. That we can see for counter current heat exchanger.
So, for counter current heat exchanger for counter flow case. Now, again let us draw
this, so depending upon the situation I am sorry so if I say that this is T h 1, this
is T h 2 and this is T c 2 and this is T c 1. This is station 1, station 2 and A is the
area. So, under this situation again if you write, exactly similar way, the single-pass
case. That d Q differential change in heat transfer will be equal to m dot minus h as
previously we have done C p h into d T h and that will be equal to here again minus, the
reason is m dot c C p c into d T c. The reason is, we can see that final minus initial d
T c is negative. So, to make it a positive quantity, and then this is equal to again
U into d A into again, for a differential area as we have done for this case. It is
a d A differential area and here it is T h, in this case it is T c, this is T h. So, then
we can get T h minus T c and like this way. Now, we can write that d T h minus T c by
T h minus T c as we have done previously in the similar way, if you try to do over here.
we will get it as minus U into 1 by m dot h C p h minus, it will be minus here m dot
c C p c into d A and this integration will be 0 to A and this is station 1 to station
2. So, then we will get an expression l n T h 2 minus T c 2 by T h 1 minus T c 1, and
that will be equal to minus U into A, then we will have 1 by m dot h C p h minus 1 by
m dot c C p c this. So, then we can write that T h 2 minus T c 2 by T h 1 minus T c
1 equals to exponential and if we just take this 1 this side and there is the negative
sides, if you multiply by the negative sign over here. Then, we will get UA by m dot c
C p c and then, we will write 1 minus m dot c C p c by m dot h C p h, why we are doing
like this. Because, again we will try to see the case for m dot c C p c the cold fluid
as the minimum fluid.
Then, so for cold fluid being the minimum fluid. For cold fluid being the minimum fluid,
what we will have is, that effectiveness factor will be equal to T h 2 minus again effectiveness
factor is this, already we have seen that is equal to for cold fluid as the minimum
fluid. Effectiveness factor is T c 2 minus T c 1 by T h 1 minus T c 1, this is the temperature
difference of the cold fluid, and this is the temperature difference for the hot fluid,
maximum temperature difference. Now, here it will be because it is a counter
current flow, so this will not be T c 2 minus T c 1, it will be T c, it is T c 2 minus T
c 1. Because, we can see that T c 2 is greater than, T c 2 is smaller than T c 1.
So, it should be I am sorry it should be T c 1 minus T c 2 by T h 1 minus T h 2 this,
but our expression had come from the integration, if we see that it is T h 2 minus T c 2 by
T h 1 minus T c 1, this we have to expand. So, T h 2 minus T c 2 by T h 1 minus T c 1,
we will write it as T h 1. Because we can see here, again from energy balance as we
have done previously from similar thing, that m dot h C p h into h 2 minus h 1 will be equal
to m dot c C p c into T c 1 minus T c 2 here, T c 1 is greater than T c 2 therefore.
So, we will have h 2 T h 2 T so T h 2 equals to T h 1 plus m dot c C p c by m dot h C p
h into T c 1 minus T c 2. So, we can write it as this is T h 1 minus C r into T c 1 minus T c 2. This is m dot
h, so this is T h 2. Because, we are writing now this will be here, there is a minor corrections
that we should make, this will be m dot into T h 1 minus T h 2. So, if I do this, then
this will be T h 1. So, if I write now, this will be negative this will be negative minus
of this. So, yeah minus C r into this. Now, if we put it over here. Then we can get,
this can be broken and written like this. This will be equal to T h 1 this 1 minus C
r into T c 1 minus T c 2 minus T c 2 divided by T h 1 minus T c 1. So, this can be written
as T h 1 minus T c 1, just adding subtracting with T c 1, and then we will do that minus
C r into T c 1 minus T c 2, then we will have plus we have subtracted one, so we are adding
1 T c 1 minus T c 2 divided by simple calculations T c and rearrangements this.
Then we can write it as, so T h 2 minus T c 2 by T h 1 minus T c 1 we can write as 1
minus T c 1 minus T c 2 by T h 1 minus T c 1 into C r minus 1. And then, just we have
to do some arrangements if we write. So, this is if I write it as 1 minus say K into 1 minus
C r. For example, for the time being if we write it as K, then we can write as 1 minus,
so this is becoming 1 plus K, so 1 plus K into I will write it as, so if I write this
1 minus say then it will become plus, so 1 plus K into 1 minus C r, that is equal to
if I see the from the derivations here, at this 1. We will put over there, exponential
NTU into 1 minus C r. Now, again we know that epsilon is equal to
T c 1 minus T c 2 by T h 1 minus T c 2 and simply we can say 1 by epsilon is equal to
T h 1 minus T c 2 by T c 1 minus T c 2 and that can be rearranged to give as, T h 1 minus
T c 1 plus T c 1 minus T c 2 by T c 1 minus T c 2 and that is equal to 1 by K plus 1.
So, from there we can write that K equals to epsilon by 1 minus epsilon.
So, just putting the value of K over here, we can write that epsilon is equal to minus
1 plus exponential NTU into 1 minus C r divided by 1 minus C r plus exponential NTU into 1
minus C r minus 1. So, this way we can finally, write this one one cancels also. We can write
that 1 minus exponential minus NTU minus 1 NTU into 1 minus C r by 1 minus C r exponential
minus NTU into 1 minus C r. So, this is becoming the expression for anti parallel flow or counter
current flow. Now, if C r tends to 0, then again here it
will be exponential is equal to 1 minus, it will be 1 minus NTU. And if C r equals to
1, then this becomes NTU by 1 plus NTU. So, there are various like this. So, this is for
co-counter current flow and single-pass case. So, like that there are various expressions
are available of or effectiveness factor for different other process and different various
geometries, whether it is a counter current per cost flow or parallel flow or that number
of passage, it can be 1 2 1 4 like that. It is for various situations or various expressions
for effectiveness factor is available in the literature, and then in the different places,
different books, things are available. So, that can be seen and that can be tried upon
even to find out the expressions to derive the expressions, one can always try. This
is just to have an idea that, how to get an expression for effectiveness factor.
Now, we will take up a small problem, the problem says oil of specific heat 2000 joule
per kg per Kelvin entering at 60 degree centigrade and 0.3 kg per second is to be pulled to 30
degree centigrade by cold water in a parallel flow heat exchanger. Entry and leaving or
exit, it is better to write exit, and exit temperatures of water are 15 degree centigrade
and 25 degree centigrade respectively. Water specific heat is equal to 4200 joule per kg per Kelvin and
heat transfer, overall heat transfer coefficient is equal to U equals to 300 meter square sorry
watt per meter square per Kelvin. So, calculate the NTU, effectiveness factor
and area of heat exchange. Now, solution so from the solution, we know that if we try
to see C c is equal to m dot c C p c is equal to 0.43 sorry m dot c C p c for we have to
find out calculate the value of m dot c. That can be found out from energy balance, m dot
c C p c into delta T c is equal to m dot h C p h into delta T h. So, m dot c is equal
to m dot h is equal to 0.3 into 2000 into delta c is equal to 30 equals to 60 to 30,
so 30 divided by m dot c you have to calculate C p c is equal to 4200 and delta c is equal
to 10. So, this is equal to 0.43 kg per second. So, once you know m dot c, then C p c is equal
to we can calculate m dot, so this is equal to 0.43 kg into this 1806 joule per Kelvin
and C h heat capacity of hot fluid is equal to m dot h C p h and that is becoming 600
joule per kg Kelvin. So, c h is less than C c. So, C h is c minimum, so it is the minimum
fluid. And therefore, C r is equal to C h by C c. Here, though we have done the calculations,
it evaluations based on minimum fluid, cold fluid here and hot fluid is becoming minimum,
so we are changing this. Now, C r is equal to c h by c c and it is becoming 600 by 1806
and that is becoming 0.33.
So, once we know see here the effectiveness hot fluid as the minimum fluid, we will get
at T h 1 minus T h 2 by T h 1 minus T c 1 and this is becoming 30 by 45 and this will
be 0.67. So, we know the effectiveness factor and then, we have to find out that sorry then
we know effectiveness factor, this equal to 1 for parallel flow, exponential minus NTU,
just we have seen this expression 1 plus C r, divided by 1 plus C r, and this is now
0.67 we got into 1 plus here is equal to 1.33 will be equal to 1 minus exponential minus
NTU into 1.33. So, from here we can find out NTU is equal
to 1.67. So, this is a number of transfer unit and that is equal to we know that, UA
by c mean. And then, we can find out UA is equal to 1.67 into is the hot fluid case into
600. So, this becoming 1002 joule per Kelvin and then area of heat exchanger is equal to
1002 divided by the overall heat transfer coefficient is given as 300 meter square.
So, this is becoming 3.34 meter square. So, this is the area of this lecture.
So, thus we have seen that how to use the expression of effectiveness factor and we
will give some exercises, later on to see that how to we will make use of this effectiveness
factor relationship for comparing various heat exchangers. So, this is all about today.
Next lecture, we will try to discuss on design issues of heat exchangers thank you.