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Welcome to lecture 5 of module 7. In this module 7, we have been discussing on heat
exchanger and the last 4 lectures we have discussed various types of heat exchangers,
their configurations, their structure and their various parts and also we have discussed
their effectiveness into method of analysis of the heat exchanger.
And now, today's lecture and the next lecture, that means in a couple of lectures will be
discussing on design calculations for the heat exchangers and for that reason will be
taking up to most important heat exchanger types. These are double pipe heat exchanger
and shell and tube heat exchanger. In today, we are going to discuss on design calculations
for double pipe heat exchanger.
So, in today's lecture we are going to discuss specifically on double pipe heat exchanger
and how to design, how to do the calculations for designing this heat exchanger, but before
we go for these design calculations, we would like to discuss some important issues and
some important parameters, which may be or sometimes is required for the design calculations.
The first thing, we should discuss on the overall heat transfer coefficient. So, important
parameters or relations. Now, in this first we will see that overall heat transfer coefficient
already we have discussed. So, basically what is happening overall heat
transfer coefficient means, it is the heat transfer coefficient based on the total resistances,
that may lie in the heat exchanger. We have discussed partly on that, and if we just for
an example, if we take a case of the double pipe heat exchanger. So, this is the double
pipe, this is the concentric pipe, this is the central line and this is the another pipe
over here. So, possibly this is the thickness of the inner pipe and this is the central
line and this is the outer pipe. So, what are the resistances that we know
that, there are three resistances can exist. One is the film inside here, another is the
resistance for this wall thickness and another is the film outside the inner part. So, these
are the three resistances that may exist. Resistances for heat transfer from one fluid
to the other fluid and there are two fluids. One fluid is passing through the in case of
double pipe heat exchanger. One fluid is passing through the inner tube, another fluid is passing
through this annulus, and in most of the situation it happens to be a counter flow situation.
And in that case the, we know that, there are three if there are three resistances from
the resistances, it is module of we have discussed long back, that driving forces or the current
is driving force by the total resistances. So, current is equal to in this case is the
Q dot and that will be the driving force is the delta T between the two fluid stream and
divided by the summation of resistances, that is the r i total resistances.
And there are 3 resistances possible as we know, this is the inlet, there is the resistances
due to the inside film of the inner tube, outside film of the inner tube and the resistances
due to the thickness of the tube. And that is in that case, we can write like
that way and this equations also can be written as, now A i is the surface area of the inner
tube into U i, inner U i is the heat transfer coefficient based on the inner tube, area
into and this is T i minus T o. T i is the inner tube temperature say T i, and outer
tube temperature is the, or the annular temperature is T o. So, that is the temperature depends
on the driving force and that also can be written as, A o U o into T i minus T o. So,
this is basically is the delta T. And here, that means 1 by A i U i or 1 by A o U o, these
are the total resistance and that is equal to summation of R i, and these total resistances
based on the that U i and U o are the overall heat transfer coefficients. These are based
on internal area, inside area of the inner pipe or based on the outside area of the inner
pipe.
So, based on that if we just do this, then what we will get is that U i already we have
discussed, this will be equal to 1 by h i plus A i l n r o by r i divided by or twice
pi K into L. k is the thermal conductivity of the wall material plus we have A i by A
o into 1 by h o, this is what. So, this is in terms of, or U o can be written as in the
similar fashion, this by A o by A i into 1 by h i plus A o l n r o by r i divided by
twice pi K L plus 1 by h o. So, these two cases this is the resistance of the inner
field, this is the resistance of the tube length and this is the resistance of the outer
film. Here the resistance of the inner film, this is the resistance of the tube thickness
and this is the resistance of the outer film. So, the only thing is that they are based
on the different area these have been calculated. And we know that A i is equal to twice pi
r i into L and A o is equal to twice pi r o into L for the tubes fine.
So, in that case we can finally, write that U i is equal to 1 by h i plus r i by K W into
l n r o by r i plus r i by r o into 1 by h o fine. And then similarly, U o also can be
written as 1 by r o by r i into 1 by h i plus r o by K W sorry r o by K, K W is not this,
K W means K or K W, what is that is thermal conductivity of the wall material for the
tubes by K into l n r o by r i and plus 1 by h o. We have seen in case of conduction,
that A at l n r o by r i by twice pi K L, this becomes the resistance in case of cylindrical
body. Now, if we so this is the becoming the final
expression in terms of radius. Now, here to one thing we have not taken into consideration
which is called scale factor or the dirt factor, whatever it is that happens due to the deposition
of scales on the surface of the tube. So, we have seen in the first phase, that this
is tube and there is another possibility, the addition of resistance due to formation
of the scale here on the surface and on this surface. So, there is a possibility of formation
of the scales and which is nothing but, is equal to dirt also. So that, if we have to
take into consideration of this dirt factors of the scales.
Then we can write that the resistances due to scales, we can write as 1 by A i h s i
and 1 by A o h s o, what are these. a h s i is the inside surface of inner pipe, the
scale for a heat transfer coefficient a h s i. So, it is expressed in terms of equivalent
to an heat transfer coefficient, because it is though the scale is a solid material. So,
the resistance could be retain in terms of a the thickness in terms of the solid material
bodies difficult to measure the thickness of the scale, that has been found as the very
fine amount. So therefore, many times it is expressed in
terms of equivalent to heat transfer coefficient like this h s i is that, is the heat transfer
coefficient in the inside as i indicates that, inside surface of inner pipe for these scales
formed. Similarly, h s o is the heat transfer coefficient in outside surface of inner pipe due to scale
formation. So, these are the coefficients for this. So, these things are to be these
that 1 by h A, h s A i into h s i, 1 by A o into h s o. These are the two additional
resistances which can come along with the 3 resistances. We have seen in that case,
we will be having in total five resistances in a double pipe heat exchanger.
There are similar kind of view that can be in case of shell and tube heat exchanger also
the, in place of the annulus part that will be a shell basically. Otherwise, there will
be the resistances again these 5 resistances will be involved.
So therefore, considering the scale formation. So, expressions can be, so U i dash that is the changed to one, which
should be 1 by, as you have seen 1 by h i plus 1 by h s i is the based on inside heat
area plus A i l n r o by r i, that is already we have seen, and then 2 pi K L plus A i by
A naught outside area, 1 by h s o this per scale plus A i by A naught into 1 by h o fine.
And then U o dash would be 1 by r o by h i into 1 by h i, in terms of, this is in terms
of radius now. This is similarly, if I write this is in terms of radius. So, if I write
in terms of fine, this will be plus r o by r i 1 by h s i, this will be r o by r i in
terms of radius, r o by r i into 1 by h i plus r o by r i into 1 by h s i. Here, I have
written in terms of area. Now, this part is, I am writing in terms of radius this 2 pi
K L you have to multiply that is all, 2 pi L to multiply you will be getting in terms
of the area. So, this plus will be getting r o by K l n r o by r i plus 1 by h s o plus
1 by h o. So, this part again, I am repeating the, this part is written in terms of radius
and this part is written in terms of the area. So, this area part can be written again in
terms of the radius, just by multiplying with 2 pi F in up and down, if I multiply with
2 pi L then we will get that, in terms of this.
Now, some typical values of this following factors. So, this 1 by h s following factors
which is nothing but, 1 by h s, it can be h s i for inside case and h s o for the outside
case. And this following factor is, it is actually dependent upon the type of fluid
that has been used, and it is unit is meter square per Kelvin per watt. So, under that
situation for different types of fluids, if we see that for distilled water, then liquid
gasoline, then fuel oil, crude oil, that 1 by h s is nearly equal to 0.000086, that is
meter square per Kelvin per watt. And per cases like sea water, then boiler
feed water and similarly, say diesel exhaust gas, etcetera. They have 1 by h s is nearly
equal to 0.000172 meter square per Kelvin per watt. So, these are some representative
values for different types of fluids, it can be liquid or gaseous fluids and how much is
becoming the scaling factor. So, this is what just to have an idea.
The next part, what I am going to discuss is this caloric temperature. Caloric temperature
is a temperature, that is calculated when there is a continuous change and there is
actually what happen is in a heat exchanger. There is a continuous change of heat transfer,
sorry continuous change of temperature from one point of one end of heat exchanger to
the another end of the heat exchanger, there is continuous change, but if the change in
temperature is very high and if the density is the viscosities of the fluids are also
is a strong function of this temperature under these two situations.
That means, what I am saying is that, if the change in temperature in a heat exchanger
along the length is high, that means greater than say 30 degree centigrade or so, if the
changes in high or second case is that, if viscosity of fluid is highly sensitive to
temperature change. Then, under these two cases, when these two cases satisfy or any
of the cases satisfied then, caloric temperature is used. And this is proposed by, proposed
and details are available in the book of D Q kern on process heat transfer; Tata McGraw
hill edition in year 1997. Now, this is proposed by kern and details
are available in the book of D Q kern. How to calculate this caloric temperature? Then,
in that case and otherwise, when these two things are not there, otherwise the average
temperature of the fluid are being taken.
So, caloric temperature is used to estimate the properties, rather I should say, thermo
physical properties of fluids. Otherwise, that means when it is not needed. Otherwise
means, when caloric temperature is not needed, as we have seen the cases previously. The
two conditions, then when caloric temperature not needed. Then average bulk temperature
between two ends are, or rather is used for the estimation.
If we see that, whether it is parallel flow or counter flow, neither of the situations.
If we just see the cases like this, just to have a little sense on that, this is T h 2
and here it is T h 1. So, hot temperature is T h 1, and cold temperature is entering
at T c 1 say, and cold temperature is going at T c 2. So, here it is entering at T c 1
and this is T h 2 and here it is T c 2. So, what happens is that, T h c is the caloric
temperature for the hot fluid, and that is given as T h 2 plus, this is the caloric factor
fraction, rather T h 1 minus T h 2, and T c c there is the caloric temperature for the
cold fluid is given as T c 1 plus F c into T c 2 minus T c 1. If you see that in both
the cases, the T h 2 and T c 1 are the cold lower temperatures, and T h 1 T c 2 are the
higher temperature for the hot and cold fluids. So, and F c is called caloric fraction, and
that needs to be calculated. So, then we will get T h c and T c c, these are the caloric
temperatures of the hot fluid and the cold fluid, and the details are available in the
book of kern's . Now, what will be some other points, we are
going to see that will be discussing during the calculations.
Now, some important points while doing the calculations for double pipe heat exchanger.
So, double pipe heat exchanger design calculations. So, some important point that we have to see,
what are these that, in this case what are the, to estimate heat transfer coefficient,
inside heat transfer coefficient based on outside area, this is inside heat transfer
coefficient, inside film heat transfer coefficient, this is a inside film heat transfer coefficient
based on outside area, and then overall heat transfer coefficient. And then to estimate
U D, that has to be done. That is the for, this is for say clean case, and this is for
dirt case, where there is a fouling factor. So, we know that 1 by U D is equal to 1 by
U c plus R d. R d is a dirt factor, already we have seen that kind of factor. And this
is the clean overall heat transfer coefficient, this is the over heat transfer coefficient
and U D is also overall heat transfer coefficient, where which takes care of the dirt. So, that
means the resistance, another additional resistance due to the formation of the scales, and then,
this is the first thing. Second point is a placement of fluids, where
to put which fluid. So, if we have in the double pipe heat exchanger. There are two
flow regions. One is the inner tube region, another is the annulus region, where should
we put which fluid. So, it depends upon that we have to see this, check the pressure drop,
delta P case and if delta P allowed. So, we have to see that, what is the allowed or allowable
pressure drop for two streams. If we see that both the fluid streams are of similar pressure
drop, the allowable ranges of similar nature. Then, so if delta P allowed is similar for
both fluids. Then once should, see that both the fluids are getting similar kind of delta
P. So, that accordingly the flow distribution should be made to see that, similar mass velocities
and similar delta P are desired. So, in that case similar mass velocity and delta P are
desired for tube side and annulus side fluids. So, this is one consideration, that how to place the fluids.
And then, we have to know the various process conditions. That includes, thermo physical
properties, flow rates, temperatures, etcetera. So, this has to be known.
And then, order of calculations: If you say that, in case of double pipe heat exchanger,
order of calculation says that, first we have to calculate the energy balance for the cold
fluid, heat fluid and overall energy balance. Then, we have to see this LMTD estimation
and in double pipe, it is usually counter flow in double pipe. Because, we have already
discussed at some point of time. That counter flow situation usually, gives better heat
exchange, that is due to maintenance of the uniform delta P all through the heat exchanger,
and that is particularly true, when we have a double pipe heat exchanger. And then, the
third point is that this is the another point, that caloric temperature whether to find caloric
temperature or not. Whether we have to find out the caloric temperature or not, that to
be decided based on need to find out based on delta T and viscosity viscosity or delta
T and nature of fluid, that is very important.
Then, so this is in general, and then after that comes the calculation for inner pipe.
And calculation for inner pipe, what we have to calculate the flow area, then we have to
calculate mass velocity. This is equal to say, tube side is equal to pi d t square,
d t flow i, d t i square by 4, and this will be equal to G t, and that is will be equal
to W mass flow rate divided by a t. And then, caloric temperature calculation if needed. Then, comes Reynolds
number calculation for tube side, that will be d t i into G t by mu t 0 by mu, that we
know that. And then, calculations of h i from sieder-tate or other correlations. From correlations,
then once we know that, then we can calculate h i o. So, these are the calculations for
inner pipe.
Then, we have calculations for annulus. In the annulus, what are their flow area again,
that is a area, and that is to be calculated, this is pi by 4 into say, if D i square. That
means, inner diameter of the bigger pipe minus, that tube outer diameter whole square. Then,
comes that equivalent diameter, that is very important. This equivalent diameter is different
for pressure drop calculation, and this is different for heat transfer calculations.
So, this is equal to D e needs to be calculated 4 into flow area by wetted perimeter. And
then estimation of mass velocity G a and that will be equal to W annulus by a annulus. Then,
calculation of caloric temperature if needed, and accordingly estimation of properties,
thermo physical properties.
This is similar for the previous case also here, caloric temperature if needed and accordingly
estimation of thermo physical properties.
And once we know that, then we have to go for Reynolds number estimation of, Reynolds
number annulus region it is, D e into G a by mu. And then, estimation of h o from correlation.
And then, after we do that, then we have overall coefficient computation of overall coefficient.
So, what is there is compute U c and that is equal to h i o h o by h i o plus h o. And
then compute U D and that is equal to that is from like this, one by U D is equal to
1 by U c plus R d, already we have discussed is a dirt factor. Then compute A, the area
of heat transfer from Q equal to U A delta T L m, temperature difference. And then, compute
length of the heat exchanger.
And then comes the calculation of pressure drop. So, that happens to the for shell, sorry
not for shell, tube side fluid. There are 2 type of pressure drops delta P total is
equal to delta P t plus delta P r. This delta P t is for, this is for frictional losses,
and this is for return loss. This return loss is due to bending of pipe and return of liquid
by changing the direction of flow. So, that means if we have a tube and tube bent. That
means, the tube fluid is going back. So, it is a return mode. Therefore, that because
of this bent, the change in it is direction, there will be loss, and that loss is called
return loss. And so these values has to be calculated and for shell side.
Similarly, we have there are two. First we have to calculate the shell side or here it
is annular side. So, annular side again that D e, estimate new D e. That means, here we
have to take that because during the pressure though the frictional loss can happen for
the fluid touching the inner surface of the outer pipe. And therefore, that frictional
losses has to be taken care of from that and therefore, that D e will be a different. So,
effective sorry that effective diameter will be different or equivalent diameter will be
different. And then, delta P total in this case, also
will be delta P t here also we can say that delta P loss. So, here this is again the frictional
loss and if I say r again, this is frictional loss. This is also it can be, it is usually
that entrance and exit losses. The reason is, in case of double pipe heat exchanger,
that material one counts from one side and then goes. So, that again entrance and exit
losses.
Now, we will take up a problem to discuss this problem. Say benzene, the problem statement
is like this, that benzene is cooled from 70 degree centigrade to 50 degree centigrade
in a counter current double pipe heat exchanger. And the flow rate of benzene is 800 kg per
hour. Tubes are carbons made of carbon steel, and inner tubes I D is equal to 21 millimeter,
O D is equal to, I D means inner diameter, O D is equal to outer diameter, 25.4 millimeter.
Outer tube I D is equal to, inner diameter is equal to 40 millimeter and O D is equal to 48 millimeter,
inner tube specification equation is this and outer tube specification is this.
Now, coolant water enters through the inner tube at 30 degree centigrade and leaves the
inner tube at 40 degree centigrade. So, do the necessary calculations to find the length
of the heat exchanger and the corresponding pressure losses.
So, what happens is that, this problem says that, we have a double pipe heat exchanger
in the annular side, there is the benzene is there, and in the through the inner pipe,
we have a water. The benzene is at higher temperature, it is at from 70 degree centigrade
to 50 degree centigrade to be cooled down, and that is being done by a coolant, that
is the water again and it is entering at a temperature of say 30 degree centigrade, and
it is leaving at a temperature of 40 degree centigrade and the flow is counter current
in nature. So, what we have to find out, we have to find out that heat exchanger area
and from that area, we have to find out the length of the heat exchanger. At the same
time, we have to find out that pressure drops, what other there is happening in both sides
of the double pipe heat exchanger.
Now, to start with the solution. Just a rough sketch of the temperature profiles, if you
see this length, and this is station one, and this is the station two double pipe heat
exchanger and this is a counter current flow. So, here it is 50 degree centigrade and then
it is 70 degree centigrade and then here it is 30 degree centigrade and here it is 40
degree centigrade. So, this is delta T 1 its I am sorry this is 30 degree centigrade and
here it is 20 degree centigrade fine. Now, average temperature of hot fluid, that
means of benzene is equal to 60 degree centigrade and average temperature of water is equal
to 35 degree centigrade and if you see that temperature difference maximum is for benzene
and that is equal to 20 degree centigrade. Both the fluids are not highly sensitive fluids
viscosity, are not highly sensitive to what is that called temperature. So, calculation
of caloric temperature is not necessary, that is one point.
Second point is, that LMTD is equal delta T 1 minus delta T 2 by l n delta T 1 by delta
T 2, and that is equal to we know say 30 minus 20 by l n 30 by 20, and that is equal to 24.66
degree centigrade.
In addition to that, if I use just do the energy balance. We will try to see the energy
balance, it gives that Q dot is equal to W hot fluid C p h into T h. So, before we go for the energy
balance, this is equal to T h 1 minus T h 2 and that will be equal to W c into C p c
into T c 2 minus T c 1. Now, we have to find out this, all this physical properties, thermo
physical properties. So, for benzene at 60 degree centigrade which
is the average temperature, we should say that C p is equal to 1880 joule per kg per
degree centigrade, then mu equals to 0.36 C p and that is equal to 33.6 into 10 to the
power minus 4 kg per meter per second. mu is this. rho equals to 860 kg per meter cube
and K is equal to thermal conductivity equal to 0.153 watt per meter per Kelvin.
And for water at 30 degree centigrade. We have mu is equal to 0.8 C p and that is equal
to 8 into 10 to the power minus 4 kg per meter per second. K is equal to 0.623 and C p is
equal to 4180 joule per kg per degree centigrade. And then, from the energy balance we can say
that, Q dot is equal to 800 into 1800 into 70 minus 50 and that is equal to 28800000
joule per hour. And then, W c we can find out from here, W c is becoming 689 kg per
hour and that is equal to basically 28800000 divided by 4180 into 10 so that is we get
the w.
Then, comes inner pipe calculation says that flow area is 3.64 into 10 to the power minus 4 meter square, it is
equal to pi into d t i square by 4. Then flow rate equal to 689 kg per hour, mass velocity
that is G t is equal to W by a t, and that is equal to 689 by a t is known. So, this
is becoming 552.78 kg per meter square per second. Remind it, it was converted into seconds
divided by 3600 fine. And then, Reynolds number for tube side flow
is equal d t i G t by mu T and that happens to be 14510, if you do the calculations properly
we will get this. And prandtl number we will be getting as, mu C p by K and that is equal
to 5.37. And then, what we will do is, in this case we will use the sieder-tate relation,
that says Nussle number based on diameter is equal to 0.027 Reynolds diameter to the
power 0.8 into prandtl to the power 0.33 into mu by mu W to the power 0.14, and this is
mu by mu W is equal to 1.0, there is no correction. And then, we have Nussle number is like it
is based on diameter will be equal to 29 and Nussle number is equal to h i into d t i by
K. From here, we can find out that, h i is equal to 29782978 watt per meter square per
Kelvin or per degree centigrade, whatever it is. And then, h i o will be equal to h
i into I D by O D, and that is becoming 2462 watt per meter square per Kelvin.
And then comes, that annulus calculation says that flow area is equal to pi by 4 into, as
I have said, minus d t o square and that is equal to 7.5 into 10 to the power minus 4
meter square. And then, equivalent diameter is equal to flow area by this. So, it is becoming
0.0376 meter, and then mass velocity is coming as 296 kg per meter square per second. So,
it was initially per hour, now it is converted into second. And then, we will get a Reynolds
number is equal to D G a by mu, and that happens to be 309156 and prandtl is equal to mu C
p by K, and that is equal to 40.424. Again applying sieder-tate equation for turbulent
flow, what we get is h o, we get as 4427 watt per meter square per Kelvin, this is what
we get.
And then after that, once we get h o, then we have to find out overall heat transfer
coefficient calculation. Here we will get, that 1 by U o is equal to r o by r i into
1 by h i plus 1 by h o plus r o by K l n r o by r i, and that is equal to 1 by h i o
plus 1 by h o plus r o by K l n r o by r i. Then, if we put these values over here, we
will be getting 6.643 into 10 to the power minus 4, and from there U o is equal to 1505
watt per meter square per Kelvin. Here, one thing I should say, that wall resistance
if we do the calculations, inside calculations is about 5.1 percent of the film, total film
resistances. So, that is much less, the wall resistance is nothing but, this one and total
these are the total film resistances. This is nothing but, the total film resistances.
Now, I have got U o now. Now, what you have to find out is that R d
will take as for two different fluids. We will take R d as 0.000172 for one side fluid
plus 0.000088 for the 86 for the other side fluid 86. So, this is total becoming actually
0.000258 meter square per Kelvin per watt. So, one side fluid is the water side and one
is water, another is for benzene. So, we will take something like that. And then, what we
will get is U D, we will get from the relationship of this, and U D we can find out 1084 watt
per meter square per Kelvin.
Now once I know U D, then we can find out area based on the outside diameter will be
28800000 by U a delta T. So, 3600 is the area per hour, 1084 into delta T l n is 224.66
meter square, and that is equal to 0.3 meter square. So, length of heat exchanger is basically
0.3 by pi into 0.0254 outside diameter of the tube, and that happens to be 3.76 meter.
Now, to go for the pressure drop calculations. So, pressure drop calculation for tube side fluid will make use of this relationship, that f
is equal to 0.0014 plus 0.125 by R e to the power 0.32. This relationship is given by
drew, koo and macadams for relatively smooth tubes. So, if we use this relationship for
pressure drop calculations. Then delta P is nothing but, it is the Darcy's
friction factor into G tube side whole square L by 2 G, then rho T into d t naught. So,
here this G t we know f d is Darcy's friction factor we can calculate from here, L is the
length of this, and G is the due to gravity rho T we know, and d t also we know, then
if you put all these values 0.0289, that you can put from here factor, friction factor.
So, f d Darcy's friction factor is 4 into f fanning's friction factor. So, if we put
these values here into 552.78 square. So, if I doing the next page.
So, delta P here is 0.0289 into 552.78 square into 3.78 length of the heat exchanger by
2 into 9.8 into 1000 in the density of water, approximately into 0.0254. So, this become
actually 66.7 kg per meter square, and this is nearly equal to 0.0667 meter of water.
So, this is the pressure drop and there is no return pressure drop for the tubes. Assuming
no tube bend, because length of tube is 3.76 meter. So, if there is a tube bend then we
have to calculate this. So, we are not calculating that part.
And then, for shell side or say annular side, for annular side again. So, delta P you can
similarly, f d G a square, this for annular side G a square into L by again 2 G rho a
annulus into D e dash, this is the equivalent diameter. And that equivalent diameter has
to be calculated based on the total these things. So, equivalent diameter is 4 into
flow area 7.5 into 10 to the power minus 4 by weighted perimeter is changing here, pi
into d t o outside diameter plus D i, inside diameter of the outer pipe. So, this is becoming
now 4 into 7.5 into 10 to the power minus 4, this is very important actually, pi into
0.0254 plus 0.04. So, this gives 0.0145 meter. So, this bend becomes f d is equal to after
calculation f d is becoming, f d is from that relationship. So, this is becoming now roughly
40.44 kg per meter square. So, this is nearly equal to 0.047 meter of benzene. So, if I
express in terms of benzene. But, there is another that exit entrance and
exit losses. So, this is as we said that it is V a square by 2 G into rho a. So, nearly
equal to G a square by twice rho a into G, and this is equal to 5.19 kg per meter square.
So, total loss equals to 45 point add of this 2, 40 plus 5, 45.63 kg per meter square which
is equivalent to 0.053 meter of benzene. So, this is the pressure here. So, the pressure
drop we have in the annular side we have calculated, based on these frictional loss. This is the
frictional loss and this is the entry or exit, entrance or exit losses. And in case of tube
side, we have calculated only due to the frictional losses. So, this way we can find out the pressure
drops and see that whether these pressure drops is lying in our limit or not.
So, this is all about today and in the next lecture, we are going to discuss on shell
and tube heat exchanger calculations. Thank you very much