We defined the term called mean temperature difference and we derived expressions for
the mean temperature difference in parallel flow and in counter flow. You would recall
that the logarithm of delta Ti delta Te occurs in the denominator of the expressions, both
the expressions. For this reason the mean temperature difference in a heat exchanger
is also referred to as the logarithmic mean temperature difference and the symbol capital
LMTD - all letters in capital - LMTD is used instead of delta Tm by many authors and many
writers.
Then, we moved on to discussing how to obtain the mean temperature difference in cross flow.
Now, you will recall that the, what emerged after discussing the temperature profiles
on the hot side of the hot side fluid and the cold side was that in cross flow - unlike
the parallel flow or counter flow - the temperature is not a function of one variable, the temperature
is a function of 2 variables x and y. If both fluids are unmixed - temperature Th is a function
of x and y, temperature Tc is function of x and y. If both fluids are mixed, one fluid
is a temperature, one fluid's temperature is a function of x, the other fluid's temperature
is a function of y. And if one is mixed and the other is unmixed, the unmixed is a function
of x and y whereas the mixed one is a function only of one variable. But regardless of whichever
case it is, the fact is the temperature on the two sides are not function of one variable
x as is the case in counter flow or cross flow but are a function of x and y. And therefore,
in order to obtain the mean temperature difference it becomes necessary to perform numerical
integrations to obtain expressions for the mean temperature difference. And you will
also recall that towards the end, I said the mean temperature difference is given, the
values are given in terms of a correction factor F and F is defined as the mean temperature
difference in cross flow divided by the mean temperature difference if the arrangement
had been counter flow; this is where we stopped. So, let us take off from that again.
We were deriving getting results for the mean temperature difference in cross flow; we were deriving the results
for this case and, towards the end, we said F - we are going to express the results in
the form of correction factor F. And F is equal to the delta Tm which we were looking
for - The delta Tm in cross flow, this is the quantity we are looking for, divided by
the delta Tm. If the arrangement had been counter flow, delta Tm if the arrangement
was counter flow and we know how to calculate the denominator and in case the fluid is unmixed
on one side, you have to take the mean temperature of the fluid. The bulk mean temperature of
the fluid leaving the heat exchanger and substitute that in the expression for delta Tm for obtaining
the counter flow delta Tm.
Now the results for given, before I go to the actual presentation of the values of F,
how they are presented, let me just make a statement. For given values of Thi, The, Tci
and Tce that is given values of the inlet and outlet temperatures on the hot side and
cold side, I am making a statement - for given values of Thi, The, Tci and Tce for the given
values, the delta Tm in counter flow is the highest. That is, for given values of this 4 quantities, if I calculate delta
Tm in parallel flow, if I calculate delta Tm in counter flow, in cross flow for any
of the 3 cases, what I will find is the delta Tm in counter flow is the highest of all flow
arrangements, amongst all flow arrangements; this is what we will find. Therefore, it follows
therefore since delta Tm in counter flow is always going to be the highest, it follows
that F - the correction factor F - which we have defined above here must be always in
the range 0 to 1.
It must be a fair number which will range between 0 and 1; 0 less than equal to F less
than or equal to 1 that is follows. And with this definition of F, if we are to calculate
the q for a cross flow arrangement
Therefore, if we are to calculate q for a cross flow arrangement, we will get q for
a cross flow arrangement is equal to U A delta Tm cross flow which is with the definition
of F equal to UA F delta Tm counter flow. So, when we have a cross flow problem, a situation
of cross flow, we will calculate the delta Tm in counter flow; get F which has been obtained
by numerical integration. Then which, I will tell you shortly how we get it from the various
charts or graphs available and substitute into this as our basic performance equation.
Now the quantity F - how do we get it? F based on the numerical integrations, F is plotted;
the results which have been obtained by numerical integration F is plotted as a function of
two parameters R and S, dimensionless parameters R and S. R is defined as, let me put down
the definitions R is equal to T1i minus T1e divided by T2e minus T2i and I will tell you
in a moment what is 1; what are the subscripts 1 and 2? And S is equal to T2e minus T2i divided
by T1i minus T2i. Now, the subscripts 1 and 2 correspond to the 2 fluids, the subscripts
1 and 2 correspond to the 2 fluids correspond to the 2 fluids.
Subscripts 1 and 2 correspond to the two fluids. For cases 1 and 3, what are cases 1 and 3?
Cases 1 and 3 is - case 1 is both fluids unmixed and case 3 is both fluids mixed. For cases
1 and 3 of cross flow, these 2 cases of cross flow it is immaterial which subscript corresponds to the hot side and which to the
cold side. It doesn't matter you can take, either way the results are valid. That means
for this situation, cases 1 and 3, these situation for cases 1 and 3 subscript 1 can be equal
to h and subscript 2 can be equal to c. Or we may have the reverse - subscript 1 can
be equal to c and subscript 2 can be equal to h. It doesn't matter which way you take
it; both are equally acceptable. The results - numerical results - come out right. It is
only for case 2 that you have to be a little careful; for case 2, let me put down what
you do.
For case 2 that is one fluid mixed and the other unmixed, care has to be taken; care
must be taken to see that the mixed fluid has subscript 1. The mixed fluid maybe the hot fluid or the cold fluid, it doesn't matter
which one but take care to see that the mixed fluid is given subscript 1 and the unmixed
fluid automatically becomes subscript 2. That is the only precaution we have to take. So,
these are 2 parameters in terms of which the values of F are found out. Now what are these
parameters? What is the dimensionless parameter R? If you look at it, it is nothing but the
ratio of the change of temperatures of the 2 fluids, that is what is R. And R by definition
therefore will be a positive quantity but unbounded greater than or equal to 0 and the
limit 0.
What is S? The parameter S the second dimensionless quantity - it is the ratio of change in temperature
of one of the fluids, ratio of change of temperature of one of the fluids to the difference of
inlet temperatures of the 2 fluids - Thi minus Tci, that is what it is. So, automatically
by definition you can see the parameter S is going to be a number which is going to
be between 0 and 1. The particular results that we get for R, the result that we get
for R - if you were to plot them. First just let me show the variation; it looks something
like this.
Suppose this is the results that are obtained based on the numerical integration for any
of the cases 1 to 3. If I plot F against S, S can range from 0 to 1, F can also range
from 0 to 1. Then, for a particular value of R a typical variation of S maybe something
like this. This is some particular value of R, some specified value of R. We will get,
this is the kind of variation which we get for F and these has been obtained based on
the numerical integration which I mentioned to you earlier. Now, let us look at actual
results which have been obtained. I am going to show you 3 charts for the 3 cases of cross
flow.
This is the chart which I am showing; you know the correction factor F for the mean
temperature difference in single pass cross flow both fluids unmixed - case 1, what we
have called. These are results based on the numerical integration to obtain the mean temperature
difference in cross and to get the correction factor F. So, on the y axis you have the correction
factor, on the x axis you have the parameter S and the different values for the different
graphs are the values of R ranging here from .2 to 4. The value of R equal to 0 will be
the horizontal line and the vertical line, this will correspond to R equal to 0. So,
this is these are the results for the correction factor F for the case of both fluids unmixed.
Now, let us look at the next results. I am just showing these results which are to be
used for actual calculations. The next figure shows the correct, let me read out, shows
the correction factor F for mean temperature difference in single pass cross flow fluid
with subscript 1 - mixed, fluid with subscript 2 - unmixed. The nature of the curves is of
course the same. But the values are very different for the different cases. Again it is F plotted
against S with R as the parameter varying from 0 or to 4. 0 is this case the vertical
line that is 0 then .2, .4, .6, .8 all the way up to 4. These are typically values encountered
in practice; R greater than 4 is probably not encountered very often so the values are
not plotted there at all.
And then finally, here is the correction factor F - results for correction factor F plotted
for mean temperature difference in single pass cross flow for the case of both fluids
mixed; similar results but of course numerical values are different. So, depending upon which
cross flow case you have, you will go to the appropriate figure and read off the values
of S that is the point I want to mean.
We will do a numerical example so that you will be to use, learn how to use these graphs.
Now, let us just talk briefly about some special cases before we do some numerical examples,
going to do a couple of numerical examples to illustrate all these ideas. First, let
us just look at two special cases; these two special cases are the following.
a m dot h Cph tending to infinity - this is one special case. The value of the product
of the flow rate and the specific heat on the hot side is very high tending to infinity
or automatically it follows that q is finite and m dot h Cph tends to infinity. Then, the
change of temperature on the hot side of the hot side fluid must be tending to 0. b is
the reverse case; the other case that is m dot c Cpc tends to infinity. The product of
the flow rate and the heat capacity on the cold side is tending to infinity and therefore
since it is tending to infinity Tce minus Tci is tending to 0. So, these are 2 special
cases.
Now, if you were to calculate the mean temperature difference for these 2 cases that is either
a or b, if you were to calculate the mean temperature difference for these 2 cases,
what you will find is the value of the mean temperature difference will be the same whether
the arrangement is parallel flow, counter flow or cross flow, any of the cases of cross
flow. You can do that yourself and convince yourself that that is what will happen. The
fluid temperature on one side doesn't change and as a result, the values of mean temperature
difference come out to be the same whatever be the flow arrangement. I would like you
to check it on your own and in your mind you should be able to explain why this is also
happening.
Cases a and b - incidentally one way of looking at them is to say that m dot h Cph is tending
to infinity. The other way of looking at case a is to say it is a case in which a saturated
vapor is condensing. Suppose I have a saturated vapor at some temperature Ts and it is condensing
at the same temperature. Obviously, since it is saturated vapor, the temperature will
not change as it changes from vapor to liquid so automatically the temperature Ts will not
change on the hot side. So, case a also corresponds to a saturated vapor condensing and vice versa.
Case b corresponds to a saturated liquid evaporating. So, remember case a and b can be interpreted
in two ways; one interpretation is the flow rates are very high on one or the other side,
the other interpretation is that case a corresponds to the case of saturated vapor condensing
and case b to a saturated liquid evaporating. So, these are 2 special cases and as I said
it is worth noting that the mean temperature difference is the same in these cases regardless
of the flow arrangement. Now, the same idea, again I have shown in a sketch here so I will
be sort of repeating things here.
This is the figure a corresponding to m dot h Cph tending to infinity; the temperature
on the hot side doesn't change or this could be a vapor condensing. The temperature on
the cold side is increasing. The b is the case of m dot c Cpc tending to infinity so
the temperature on the cold side doesn't change; this could be a saturated liquid evaporating
and the hot side fluid is giving up heat slowly like this. The way I have shown it, it is
parallel flow but remember - even if it is counter flow all that will happen is I will
get a mirror image kind of variation of Tc in the other direction. And therefore you
can see why the main temperature difference is the same whether it is counter flow, cross
flow or parallel flow. So, these are 2 special cases.
Now, let us do a couple of problems to illustrate all these ideas of mean temperature difference
for different flow arrangements. Let us take one example - a numerical example.
We have a parallel flow heat exchanger to be designed, a parallel flow heat exchanger
to be designed to cool 5 kilograms per second of air from 500 centigrade to 350 centigrade by an equal
flow rate of air entering at 90 degrees centigrade. 5 kg per second of air to be cooled from 500
to 350 on the cooling side, the same flow rate of air entering Tci being 90 degree centigrade.
We are given that the heat transfer coefficient on the hot side hh is equal to 60 and on the
cold side hc is equal to 30 - these are the values of h. Calculate the area A of the heat
exchanger, that is the example to be solved. Take Cp; we need the value of Cp; take Cp
of air on both sides to be 1020 joules per Kelvin per kilogram Kelvin, take this value of Cp. So, design
a parallel flow heat exchanger; that means find the area A of a parallel flow heat exchanger
for given conditions. Let us just draw a sketch of what we have, what we are asked to do.
This is, let us say - let me draw the sketch of the heat exchanger.
Let us say this is our heat exchanger, this is the heat transfer surface, the fluid on
the hot side, it is a parallel flow heat exchanger. m dot h equal to 5 kilograms per second, m
dot c is given to be the same - 5 kilograms per second. Thi, you are told the inlet temperature
on the hot side equal to 500 centigrade; Tci equal to 90 degrees centigrade, The given
to be 350 centigrade. The value of hh heat transfer coefficient on the hot side is equal
to 60 in the usual units, Watts per meter squared Kelvin, and the value of hc on the
cold side equal to 30 Watts per meter squared Kelvin. So, find the area, these are the exit
points; find the area A of the heat exchanger, the configuration is parallel flow.
We can sketch the temperature profiles - on the hot side going to be something like this,
on the cold side it is going to be something like this. This is 500, this is 90, this is
350. Now, the flow rates on the hot side and the cold side are equal and the Cps are given
to be equal. Therefore, it follows that the change of temperature on the hot side and
the change temperature on the cold side must be equal. So, the change of temperature on
the hot side is 150 so it follows that the change of temperature on the cold side must
be also 150. So, Tce must be equal to 240, that follows I don't have to even do any real
calculation for that. So, delta Ti is equal to 410 and delta Te is equal to 110 degree
centigrade. These are temperature profiles on the 2 sides; these are the values of delta
Ti and delta Te. It is relatively a very simple problem. We have to remember we have to substitute
into the expression q is equal to U A delta Tm. So, we need to get q, we need to get U,
we need to get delta Tm then we will get the value of A so let us get them one by one.
Let us get the value of, first, the value of q.
q is equal to 1020 that is the specific heat into the flow rate 5 into the change of temperature
on the hot side or cold side - it doesn't matter which side you take - 500 minus 350.
And that comes out to be 765 into 10 to the power of 3 Watts. This is the heat transfer
rate in the heat exchanger. The value of delta Tm in parallel flow is equal to delta Ti minus
delta Te divided by log to the base e delta Ti by delta Te. So, it is 410 minus 110 divided
by logarithm to the base e 410 divided by 110 and that comes out to be 2208.02 degrees
centigrade.
Now, we are given the values of hh and hc; we aren't told anything about fouling so we
will have to make an assumption. First of all about fouling and secondly about the thermal
resistance B by k - we are not told anything about that. But, notice the values of hh and
hc are low; they are low because we have got air flowing as the heat transfer medium. So,
values of hh and hc are bound to be of the order 50, 60, 100 something like that, not
going to be in thousands with air flows. Automatically, it follows that the thermal resistance of
any metal wall you can take, any typical wall, a millimeter thick, 2 millimeters thick, take
a conductivity of a metal of 10, 20. You will immediately see its thermal resistance will
be insignificant compared to 1 upon hh or 1 upon hc.
Similarly, if you look at any of the fouling factors that I have given you, you will remember
those fouling factors are for a situation involving liquids. With gases fouling factors
which are in the region of .30s, 130s, 230s, 4 in the usual units will be there, effect
will be negligible compared to the thermal resistance offered because of the heat transfer
coefficient. So, with gas flows the values of a neglecting fouling is a good assumption.
Neglecting the thermal resistance of the metal wall if that data is not available is also
a good assumption. And we can say in this situation 1 upon U is equal to 1 upon hh plus
1 upon hc. We just leave out B by k saying it is negligible and we will also not consider
fouling because it is bound to be negligible compared to the values of h that we are seeing
for gas flows. So, all that we get is this is equal to 1 upon 60 plus 1 upon 30 and therefore
the value of U comes out to be equal to 20 Watts per meter squared Kelvin. Now substitute
into our basic performance equation to get the value of A.
So, our basic performance equation is q is equal to U A delta Tm; in this case delta
Tm in parallel flow. So, we will get A is equal to q which is 10, q which is equal to
1020 into 5 into 150 divided by A UA. U is 20 sorry U delta Tm so U is 20 and delta Tm
is 2208.02. So, we get the area A of the heat exchanger to be 167.75 square meters, that
is the answer to the problem - 167.75 square meters for the given data that we have got.
Now, let us say instead of a parallel flow heat exchanger, the arrangement instead of
being parallel flow is counter flow. I ask you to calculate the area A of the heat exchanger
if the flow arrangement is counter flow; let us do a further calculation. Calculate the
area of the heat exchanger if the flow arrangement is counter flow. In this case, notice we again have to substitute into the same expression
q is equal to U A delta Tm. q is not going to change here is my basic expression, q is
equal to U A delta Tm. q is not going to change, U is not going to change, what is going to
change is the delta Tm. So, get a new value of delta Tm for counter flow situation and
again substitute into our basic equation. What is delta Tm in counter flow? Now let
us calculate that.
delta Tm in counter flow would be, notice this is a case of counter flow with m dot
h Cph equal to m dot c Cpc, notice that. Therefore, T in this case with m dot h Cph equal to m
dot c Cpc, the temperature profile on the 2 sides are going to be parallel lines like
this. So, we are going to have 500 here, hot side fluid entering temperature, 350 leaving.
Entering Tci 90, leaving Tce 240 and we will get the value of delta Ti and delta Te - both
delta Ti and delta Te. Both in this case will be 260 degree centigrade. So, delta Tm counter
flow will be equal to the same, delta Tm counter flow will also be equal to 260 degree centigrade.
This is the case of the 2 temperature profiles being parallel to each other; this is that
special case which we talked of earlier. As I said, q and U don't change therefore A will
be equal to the value of q which is 765000 Watts divided by the value of U and the value
of delta Tm - 260. So, we get 147.12 square meters; so this is the value for the area
A, the arrangement is counter flow.
I will make a statement here without proving it that if it is a cross flow arrangement,
any kind of cross flow arrangement, then you are going to get since this is the counter
flow arrangement is 147.1 2 and in parallel flow you have got 167.75. A cross flow arrangement,
any cross flow arrangement, would give an area between these 2 extremes; parallel flow
gives the highest, count flow gives the least. You will get a value for cross flow in between
these two; I am just making that as a statement. So, this is a good numerical example to illustrate
how we substitute into our basic performance equation - q is equal to U A delta Tm - and
calculate the area A of a heat exchanger for a given application. Now, let us do one more
problem, another numerical example.
The problem we are going to do now is the following - we say 1000 kilograms per hour
of water at 50 degrees centigrade enters a single pass cross flow heat exchanger and leaves at 40 degree centigrade. A hot
water stream, 1000 kilogram per hour of water, enters at 50, leaves at 40. The heat from
this hot water, the heat is transferred to cooling water entering at 35 centigrade and leaving at 40 degree centigrade. The cooling
water which is cooling this hot water enters at 35 and leaves at 40 degree centigrade.
Calculate the area of the heat exchanger if the fluids on both sides are unmixed.
Take U to be 1000 Watts per meter squared Kelvin with fouling included; that means the
effects of fouling are included in this value of U. Take the value of U to be 1000 Watts
per meter square Kelvin and the value of Cp on both sides, the Cp for water on both sides
to be 4174 Joules per kilogram Kelvin; take the value of Cp to be the following. It is
a straight forward example of calculating area, very similar to the earlier one excepting
that now the flow configuration is cross flow with both fluids unmixed. Now, let us just
draw a sketch.
Here is let us say a schematic diagram. This is the area, the cross flow situation; let
us say this is the heat transfer area A which we have to find out. This is the hot fluid
entering here, m dot h equal to 1000 kg per hour, Thi entering at a temperature of 50
degree centigrade, leaving at a temperature of - hot fluid leaving at a temperature of
40 degrees centigrade, cold fluid Tci entering with the temperature of 35 centigrade and
leaving with a temperature of 40 degrees centigrade. Calculate the area A.
Now again our basic performance equation is q is equal to U A delta Tm. In this case,
we are going to get to the correction factor F so we will write this as U A F which will
come from that graphs, multiplied by delta Tm if the arrangement had been counter flow.
So, if I want to get the value of A, I should get q, I should get U, I should get F; so
let us get each of the quantities. First of all - q; what is q? q is nothing but the flow
rate on the hot side 1000 divided by 3600 so many kilograms per second into 4174, is
the value of Cp, into the change of temperature on the hot side 50 minus 40 and that is equal
to 11594.4 so many Watts; so that is the value of q? What is delta Tm in counter flow? Let
us get the value of delta Tm in counter flow.
delta Tm will be equal to, this is the, let us draw a sketch for. If it is a counter flow
situation starting with 50 going out at 40 on the hot side, cold side entering at 35
and leaving at 40, so we have delta Ti is equal to 10 and delta Te equal to 5. So, we
have 10 minus 5 divided by logarithm 10 divided by 5 which comes out to be 7.21 degrees centigrade.
Now, let us go the charts since both fluids are unmixed; since both fluids are unmixed,
it is immaterial whether subscript 1 corresponds to the hot side
and subscript 2 to the cold side or vice versa. It doesn't matter which you take; so we say
take T1 equal to Th and T2 equal to Tc. Let us take that; it doesn't matter which you
take, you will get the same answers.
Therefore R is equal to Thi minus The divided by Tce minus Tci which is equal to 50 minus
40 upon 40 minus 35 which is equal to 2. And S is equal to Tce minus Tci upon Thi minus
Tci and that will be equal to for40 minus 35 divided by - Thi minus Tci - 50 minus 35
which is .333.
Now, we go to figure which I had shown you earlier for the case of the results obtained
for single pass cross flow; recall I had given you some situations there. And if you recall,
let us just show that figure again. Both fluids unmixed here; we have this is the situation
correction, factor F for both fluids unmixed. Now, in this case R is equal to 2 and S is
equal to .333 so this is the graph of R is equal to 2; here this is the graph of R is
equal to 2. Take S equal to .333, somewhere here, go up here to R is equal to 2 which
means somewhere out here you will go at this point, then go horizontally and read of the
value of F. So, here is the value of R equal to 2; get S equal to about .3, it will come
somewhere here. Go horizontally and you will get F is equal to .91 from the figure, F is
equal to .91. Therefore, now last step, area A is equal to q divided by U delta Tm counter
flow into F so it is equal to 11594.4 divided by the value of U is 1000. Value of delta
Tm in counter flow is 7.21 and the value of F is .91 so we get area A equal to 1.77 square
meters and that is the answer for this particular problem; that is the answer for this particular
problem.
So now, for different flow configurations through the help of these 2 examples, we have
illustrated how to obtain the area A for a given heat exchanger. Now this is one technique,
the q, using the equation q is equal to U A delta Tm - this is one technique for finding
out. The other technique which is also used which is the other method which is also used
for finding out the effect the area A of the heat exchanger or the performance of heat
exchanger is what is called as the effectiveness NTU method.
And the effectiveness NTU method is, there are certain reasons why this particular method
has been developed. So, next time we will look at the effectiveness NTU method; we will
discuss it, derive some relations for it and show how it is also applied for doing calculations
with heat exchanges.
