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Welcome to lecture three on module eight. In the module eight, we are discussing on
evaporation. Now, today's lecture this is the last lecture on evaporation chapter. We
are going to discuss the calculation procedures and we will demonstrate with some problems.
So, this is basically, this lecture is on evaporation or evaporator calculations.
Now in the evaporator calculation, we will be discussing two things. We will first start
with we will we will see first single effect calculations single effect calculation, and
then we will go for multi effect or multiple effect calculation. Now, one thing is pretty
clear that when we have to go for evaporation calculations or evaporator calculations. Basically
our target is to find out that or the design of the evaporator our target is to find out
the area of heat exchanges, that is the most important thing, and while doing this that
we have to see that while going for this calculations, we have to see that we are taking the mass
balance and energy balance, these are the two important issues that has to be taken
into considerations. That one is called mass balance, another one is the energy balance,
so these are the two things we have to do and all these things calculations we will
be showing in in terms of steady state process.
Now, if we just start with the single effect calculation, it is like this say let us draw
a simple single single effect evaporator like this, and here, so this is part of this things
the evaporator tubes kind of things. So, this is the feed here, and here it is the steam.
So feed they say m dot is the flow rate, mass flow rate and suffix indicate that that kind
of stream it is, so f indicates the feed and m dot m is the mass dot, means m dot is the
mass flow rate and h f is the enthalpy of of this feed steam. Similarly, m dot s is
the mass flow rate of the steam and h s small h s is the enthalpy of the steam. That is
entering into and then there will be a condensate, and this condensate if I say that m dot s
again will be the same mass. Whatever, the steam is entering that is getting condensed
so our consideration over is is the saturated steam.
So, heating medium is that condensing steam is nothing but the heating medium so only
the latent heat is involved in this case so there is no extra loss, so there is no mass
loss even. So, m dot that is entering and that m dot s is coming here and in our calculation
we will see that no other energy loss loss of energy is neglected. So, these are the
assumptions, and m dot s is this steam again and then, say in the condensate we have say
h l is the latent heat, and then we will have a product here there is the product and we
can vapour here. So, this is the vapour, it is also the product is the main product and
this is the vapour. We will write it to be vapour m dot v and h v is the latent heat
and here m dot p is the product flow rate and h p is the latent heat.
In that case, so from simple mass balance we will get that m dot f is equal to m dot
that is the feed that is entering in so mass balance of solution. What is to be done for
mass balance of steam is not necessary, because we have assumed that whatever steam is entering
that is getting condensed. So that part is not important for us, what is important is
that energy that transport from the steam part to the fluid part is important, so m
dot f is equal to m dot say v plus m dot p that is the product and then energy balance.
Says, that m dot f h f this is the simple 1 plus m dot s h s is equal to m dot p h p
plus m dot v h v plus m dot s h l this is the lead condensate which come with an enthalpy,
now if these two combines if these two combines it will give you the latent heat of condensation.
So, we are writing at m dot f h f plus m dot s into h l h h minus h s this is h s hl minus
will be the lambda, so m dot into lambda s that is the latent heat of condensation of
steam and that will be equal to m dot p h p plus m dot v h v. So, this is the general
mass balance equation that you can get, and once we know and also once we know that how
how much amount of the heat transfer this is nothing but the heat that is getting transported,
and if we assume that there is no heat loss.
Then we can find out that q dot is the rate of heat transfer that is equal to m dot into lambda s m dot
s into lambda s that will be the rate of heat transfer, and if delta T is the driving force
driving force for heat transfer, and that will be nothing but the T s minus T b where
this T s is saturation temperature of steam in shell, and T b is nothing but he boiling
point of the solution of the solution at the pressure of the vapour chamber. Therefore,
this is becoming that driving force, and then we can say that q dot is equal to U D is the
overall heat transfer coefficient into area into delta T, so from this relationship if
U D is known if U D is known. Then a can be calculated.
So, this is the bottom line for this case and this driving force this includes. This
driving force includes the boiling point or rather b p e boiling point elevation. We have
discussed in the last class. So, actually the boiling point could be so if we see that,
what is the boiling point elevation in this case it will be that say T b minus T w that
is nothing but the boiling point elevation. So, this will be the boiling point of the
solution and this is the boiling point of of water. At that pressure pressure of vapour
chamber, so this is nothing but the boiling point ignition that elevation that we have
already discussed. So, this is is the basic conception.
Now, if we take a problem on this quickly we will see problem in this that a single
effect evaporator concentrate. A 6 percent sucrose solution at the rate of 24000 kilogram
per hour 6 percent at the rate of two concentration is 48 percent using steam at or the saturated
steam at 1.8 atmosphere absolute pressure. Now the feed is at 30 degree centigrade and
a vacuum of 640 millimeter mercury is maintained in the evaporator; that means, it is a single
effect evaporator and it is under vacuum the 640 millimeter. So, absolute pressure is much
lower, now the heat transfer coefficient, which is overall heat transfer coefficient,
is 1800 watt per meter square per degree centigrade then. What is the area of the evaporator?
So, this is to be calculated now given is that specific heat of feed equals to 3.9 kilo
joule per kilogram per degree centigrade, and specific heat of product equals to 3.15
kilo joule per kilo gram per degree centigrade. As we can see in that with increase in concentration
of the product your solute the specific heat is decreasing specific heat it is decreasing,
and boiling point elevation is given as 5 degree centigrade. So, these are the certain
data being provided.
So, if we have to go ahead with the solution. So, first what we have to do that absolute
pressure in the evaporator is equal to 760 minus 640 that is 120 millimeter mercury. So, it is under
as a huge vacuum and boiling point of water at 120 millimeters. So, this has to be found
out from the steam table is equal to 55 degree centigrade and boiling point elevation is
5 degree centigrade. So, total other boiling point of the concentrated
solution in the evaporator in the evaporator and that is equal to 55 plus 5 equal to 60
degree centigrade fine, and say 0 degree centigrade is is taken I will take as reference temperature
or day time temperature. Fine then enthalpy of of feed and that is equal to 24000 is the
feed flow rate into 3.9 into 30 minus 0 that is the delta T, and that is equal to 2.81
into 10 to the power 6 that is kilo joule per kilogram.
Similarly, enthalpy of
vapour generated. So to find out the vapour generated, what we have to do we have to do
the mass balance, so basis if I do take the basis equal to one hour and then feed equals
to 24000 kilogram. Then sucrose in feed that is equal to 1 that is how much is given it
is 6 percent, so 1440 kilogram so 0.06 into 24000. So concentration, because final concentration
is equal to 48 percent, so total mass of product is equal to 1440 by 0.48 and that and that
is equal to 3000 kilogram that is per hour, because basis is hour, so water evaporated
that will be equal to 24000 minus 3000 this is equal to 21000 kilogram and that is definitely
basis is per hour. So, it will be in power hour.
Now, we will go for enthalpy of vapour generated. Now here this will be 21000 is the amount
of the vapour generated, and 4.18 that is the c p of the water vapour into 60 the temperature
difference actually this is the amount of water that has taken that is being there minus
0 plus 2370 into 21000; this is the latent heat of vaporization 2370 is that we should
say latent heat of vaporization. And this is that enthalpy of sorry specific heat of
concentrated solution concentrated solution, but this is the specific heat of water because
we are considering the water heat balance for the water so 21000of the water.
So, what was the specific heat with this kind of water this much 4.18 into the 21000 into
this is the sensible heat which was present with the water that is the enthalpy that part
this is the latent heat, so this gives then it becomes 5.50 into 10 to the power 7 latent
heat of and kilo joule per hour and this becomes latent heat of vaporization of water at 60
degree centigrade and in kilo joule per kilogram, and we can see that this operation is happening
at 60 degree centigrade that the vaporization takes place at 60degree centigrade.
So, this is the latent heat of vaporization; therefore, this is the sensible heat for this
or whatever amounts are present with respect to 0 degree centigrade. The sensible heat
present with the water vapour with the water liquid water, and then this is liquid, and
then this is the latent heat, so this two is becoming harder and its giving it to vaporize.
So, enthalpy of vapour generated is this, and then enthalpy of product equals to 3000 into 3.515 is the specific
heat into 60 minus 0 there is the driving force, so this becomes 5.67 into 10 to the
power 5 kilo joule per hour. So from energy balance or enthalpy balance then enthalpy of steam is nearly equal to 5.1 into 10 to the power
7 kilo joule per.
And then once I know this enthalpy, so at 1.8 bar say atmosphere pressure steam temperature
- saturated steam temperature is equal to 117 degree centigrade and heat of condensation
equals to 2210 kilo joule per kilogram fine, then amount of steam supplied would be 5.1
into 10 to the power 7 divided by 2210, and that is equal to 2.31into 10 to the power
kilogram. So, this is the steam requirement, so area of heat transfer is equal to q dot
a you know that is equal to q dot by U D into delta T.
So, U D is given, so it is given 5.1 into 10 to the power 7 that is the heat transfer
by the steam by 1.8 into 3600 that per unit hour this is 1800 is the joule, so this watt
per meter square per second so it is per hour it is made, and it is in kilo joule its maintained,
so it is the unit conversion into 117 minus 60. So, this gives as 1838 meter square, so
this is becoming the area of the heat exchanger that is needed for. So, this is just to give
an idea about the heat transfer calculations evaporator calculations.
Now, let us come to the multiple effect evaporator calculations. So, what will do is I will just
quickly draw such evaporator sections three such has to be drawn. The idea behind the
evaporator the multiple effect evaporator - the evaporator areas are usually tried to
maintained similar, so that it is easy to do the design, so size of the evaporator maintained
is more or less constant. So, this is the feed that is coming m dot f T f and this is
the steam that is coming m dot s T s, and then p s pressure of the saturated steam,
and this is here; P 1 is the pressure, T b 1 is the boiling temperature, and this is
the vapour that is coming that is m dot s 1 which is with h s 1.
And then it is at the pressure of T 1, and this vapour is coming and this is coming as
the steam fed to this, and bottom product is going to there and it is coming as the
feed over here and it has got, so this will be m dot f minus m dot s 1 which is gone as
the vapour this much amount is going there, and it is having say h 1 and T b 1 is the
boiling point true boiling point of the concentrated solutions.
Similarly, here it will be like this it will go like this and here it will be m dot f minus
m dot s 1 minus m dot s 2, so these are actually that this itself takes care of the mass balance,
and from here it is coming like this is the steam that is fed into... So, here it is m
dot s 2; that is quite clear and h s 2 is the latent heat, it is the P 2 is the pressure
here, and here T b 2 is the temperature here here, P 3 is the temperature and T b 3 is
the temperature over here. Again the condensate the product is coming out and this product
would be m f minus m dot s 1minus m dot s 2 minus it should be m dot s 3, and which
is having enthalpy h 3 and say T b 3 is the boiling point.
Here also m enthalpy it should be h 2 and T b 2 and here the vapour is going out, and
it is now left to atmosphere this is at say P 3 and it has got enthalpy its h 3, and all
these so here in all the cases. We will have some condensate this will be the condensate,
this will be the condensate. So, steam is coming and this is the condensate, so this
is the typical diagram of the triple effective operator, so it is a triple effect forward
feed evaporator system. Now, one thing I should say that while mass balance is already taken
care of while giving these values and boiling point.
So, if I say that this is the effect 1, this is effect 2 and this is effect 3. Steam is
put into the fast effect the saturation is external seen so this is effect 1.
And we should say that BPE boiling point elevation in effect 1 equals to Tb 1 minus Tw1 and Tw1
is the temperature of saturation Tw1 is the boiling point of pure water at P1 pressure
similarly, Tw2 is boiling point of water at P 2 and Tw3 is the boiling point of water
at pressure P3, that it means these pressures are maintained in in the subsequent evaporators
so we can say that P3 is less than p2 is less than p1. So, we know that Tw3 is less than
Tw2 is less than Tw1 and that also should be less than Ts which is the steam and this
less than say Ps. P s is the say if Ii say the steam pressure, so this is P s so we can
say that this is the steam temperature that we know external stem temperature, and this
is the boiling point elevation. So, if we say boiling point elevation in effect
1 is this BPE in effect 2 that will be equal to Tb2 minus Tw2 and BPE boiling point elevation
in effect 3 will be equal to Tb3 minus Tw 3 that will be the boiling point elevation,
and these are the amount of super heat also, and we will assume that that when the superheated
vapours enters from one effect to the other effect, it immediately loses the super heat.
And therefore, the driving forces driving force for effect 1 that will be equal to Ts
steam steam temperature minus Tb1 boiling point temperature, and this driving force
for effect 2, and that will be equal to the steam temperature; the steam is coming with
the temperature of say Tw1 minus it should be for effect 2 Tw1 minus Tb 2, and for effect
3 will be equal to Tw2 minus Tb3, and also we know that Tb1 is greater than Tb2 is greater
than Tb3 that already we have discussed fine.
Now, if we go for the Energy balance mass balance already being taken care of if we
do, and this is drawn neglecting any energy loss loss to the surroundings or to the environment
for effect 1, we will have m dot f h f plus m dot s h s; that is equal to m dot f minus
m dot s 1 and that is h 1 plus m dot s 1 h s 1, that is the vapour this is the concentrated
liquid is vapour, and this is the m dot s h l this is the condensate. So, this is feed,
this is steam, this is your vapour, this is the product, this is vapour, and this is condensate,
and if we just write for each and every cases this condensate will be m dot s h l this will
be m dot as this. So, I am not writing these things again fine
now, like this if we go ahead this is becoming this, so from there we can write as this implies
that m dot f in terms of latent heat, if we write h f plus m dot into lambda s m dot s
lambda s. So, latent heat for this first effect, and and that will be equal to m dot f minus
m dot s 1 into h 1 plus m dot s 1 into h s n h s 1 this is the for vapour. So, this is
becoming that and this is the latent heat of condensation, now comes so effect 2 for
effect 2, it is m dot f minus similar way m dot s 1 into h 1 plus m dot s 1 lambda s
1 will be is equal to m dot f minus m dot s 1 minus m dot is 2 into h 2 plus m dot s
two h s 2, exactly the similar way that is to be done.
And the for effect 3 it will be again that m dot f minus m dot s 1 minus m dot s 2 into
say h 1 plus m dot s 2 into lambda s 2. So, it is a another latent, so latentive of preparation
will change, because the temperatures are different for this steams latentive or condensation
this change in different. And that is be equal to m dot f minus m dot s 1 minus m dot s minus
m dot s 3 into h 3 then plus m dot s3 into h s, so this is what is being done. Now say
UD1, UD2 and UD 3 are overall heat exchangers coefficients coefficients for effects one,
two and three. So, under that situation we can write that
Qdot1 is equal to m dot s lambda s is equal to UD 1 A1 T s minus T v1, and that is equal
to UD 1 A1 delta t1 and Qdot2 is equal to m dot s lambda s1 lambda s1 and that is equal
to UD 2 A2, and this is equal to tw1 minus T b2 and this is equal to UD 2 A2 delta T
2 and Qdot3. So, this is the heat heat transfer m dot s 2 lambda s2, so it is a similar not
much complicacy is that this is equal to UD 3 A 3 into T w 2 minus T b 3, and that is
UD 3 A3 into delta T 3. Now, question is in usual situation as I told previously also
that in because from from design point of view.
It is tried to have A1 is equal to A2 is equal to A3 is equal to A its is tried to have similar
areas for this, and then under that situation U D 1 delta T 1 will be equal to U D 2 delta
T 2 will be equal to U D 3 delta T 3, and from there we can write that delta T summation
of delta T would be equal to delta T1 plus delta T 2 plus delta T three. And that is
equal to T s steam temperature saturation, external steam temperature saturation into
minus Tw3 minus summation of all the boiling point elevation this can be proved.
If we out the values of this we can prove it. And summation of boiling point elevation,
we have already seen, so then from their as delta T is proportional to 1 by UD. We can
write we can calculate that delta T values, so delta T by summation of delta T that is
equal to 1 by UD 1 by 1 by UD 1 plus 1 by UD 2 plus one by UD 3 provided in all the
cases area constant for or same area is same for same for all effects, now what is being
done in this case this way we can find out the delta T. What is being done in this cases,
in case of design of the evaporators first that a all the data are calculated, then some
area is assume to be same for all the effects. And then based on that, we have to calculate
the values of the temperatures delta T, and once delta T is known then we will recalculate
the areas.
So, by knowing delta T calculate areas like this delta 1, sorry area will be m m s lambda
s by UD one delta T1, then area 2 will be equal to m s dot lambda s 1 by UD 2 delta
T2 and area 3 also will be calculating m s 2 lambda s 2 by UD3 A delta T there, so by
this way you can calculate the areas again. And see that wither the areas are correct
or not if not then revise delta T values recalculate areas till A1 is nearly equal to A2 is nearly
equal to A3. So, this is the basic motto or basic objective or basic ways of doing the
heat exchangers design.
Now, if I solve the problem not heat exchangers evaporators design, if you solve the problem
too quickly. I say an aqueous solution of a high molecular weight solute is concentrated
from 2 percent to 30 percent by weight in a forward feed double effect, so just in place
of having we are writing it for a double effect we are doing the problem for the double effect
evaporator. This is just to show the principals how to do this the feed rate is 5000 kilogram
per hour feed at 50 degree centigrade enters the first effect saturated steam at an absolute
pressure of 2.0 bar is supplied to the first effect. Naturally the pressure in the second effect is quite a low value 1 bar,
so that almost in vacuum the overall heat transfer coefficient or one. And two effects
are say 1800 and 1400 Watt per meter square per kelvin respectively, now calculate the
areas of the effects and and and specific heat of the liquid is 4.1 kilo joule per kilogram
per kelvin.
Now, if you see the solution how to go for the solution, we know that feed rate equal
to 5000 kilogram per hour in this solid is 5000 into 0.2, and that is equal to 100 kilogram
per hour. So, water in it is equal to 4900 kilogram per hour. Product has got 30 percent
solid so mass flow rate m dot product mass mass flow rate of product will be is equal
to 333.33 kilogram per hour. So, water out is equal to 333.33minus 100 is the solid so
233.33 kilogram per hour that is the water out as product with product, so water evaporated
that is equal to m dot s 1 plus m dot s 2 this coming from the two the two effects total
water. So, this is total water evaporated, and that will be equal to 4900 minus 233.33
kilogram per hour and this becoming 4666.67 kilogram per hour so this side say equation
1 and then we do this that say boiling temperature in first effect.
In first effect pressure is equal to 2 at mos 2 bar temperature is will be equal to
T s is equal to 120 degree centigrade the steam temperature has steam and its lambda
value would be equal to 2200 sorry 2200 kilo joule per kilogram that is the lambda value
and then absolute pressure in second effect is equal to 0.0140 bar, so boiling point will
be equal to of water equal to 285 kelvin and that is 12 degree centigrade and then lambda
s2 will be equal to 2470 kilo joule per kilogram. Now, boiling point elevation is neglected
in this case because high molecular weight solute solute, when now total temperature
drops that means what I do not known is that pressure of the or the temperature of the
first effect that is not know, so temperature of first effect is not known is it is to be
calculated. Boiling temperature so this is what is to
calculated. So, what we we need to know is the delta T is equal to delta T1 plus delta
2 as we have seen and that will be equal to Ts minus T2 and the boiling point elevation
is equal to 0. So, this is happen to be 108 degree centigrade, and then from here delta
T1 by delta T two that is also 1400 by 1800 that is based on UD 1 delta T1, it will be
equal to UD2 delta T 2 A1 is equal to A2 that is from this relationship so we can have delta
T 1 by delta T2 that will be equal to 0.78 or something like that. And then from here
we can find out delta T1 is equal to 47 degree centigrade and deltaT2 is equal to 61 degree
centigrade sorry 61 degree centigrade. So, we know that then T1 is becoming boiling temperature
will be equal to73 degree centigrade, and then this cube as that T2 will be equal to
12 degree centigrade already its matching whatever is there 2degree centigrade is matching.
Now, for effect 1 if we do the calculations then m dot f h f plus m dot s lambda s is
equal to m dot f minus m dot s1one into h1 plus m dot s1 h s1one and here we will take
the reference temperature is equal to T1 and that is equal to T1, and that is equal to
73 kelvin sorr 73 degree centigrade. And from there so if this is there then we can write
that this is equal to 5000 into into latent 4.1 of the feed that is why c p of feed of
feed is equal to 4.1into 50 minus 73. So, it is negative quantity plus say n dot
s into 2200 will be equal to h1one is equal to this part is 0 and this part is n dot s1
into 2330, so from this we can find out that m dot s1 will be equal to 0.944 n dot s minus
202. Similarly for effect 2 will have say m dot f minus m dot s1 into h1 plus m dot
s1 lambda s1 will be equal to m dot a f minus m dot is 1 minus m dot s 2 h 2 plus m dot
s 2 h s 2 here reference temperature will take as t2 is equal to 12 degree centigrade
12 degree centigrade. So, then we can write that so this also equation 2 we can write
as sorry 5000 minus m dot is 1 into 73 minus 12into 4.1 plus 2330 m dot s1 and this part
is equal to 0 will be equal to 2470 m dot lambda s2 m dot s2. We can find out that m
dot s2 is equal to 0.842 m dot s1 plus 506.
Now, this is equation 3 if we solve all the three equations. We will get m dot s is equal
to s1 is equal to 2407.89 kilogram per hour m dot s 2 is equal to sorry this is m dot
s2m dot s1 is equal to or 2258178 kilogram per hour, and m dot s that is equal to 2606.76kilogram
per hour. So from there we can find out area 1 is equal to q1 by v1 delta T1 is equal to
m dot s lambda s by v delta T 1, and that is is equal to 2606.76 into 2200 by 1800 into
47, and that is 67.79 meter square and a v is equal to again Q2 dot by U 2 delta T 2,
and that is equal to m dot s1 lambda s1 by U 2 this UD 1 UD 2 delta T 2 and that is equal
to 2258.78 divided by divided by into 2330 here, a 1400 into 6163 meter square.
So, actually this should be into 3600 by 1000 to maintain the unit consistency and this
will become then divided by 3.6 meter square and that that is coming to be 18.83 meter
square. And similarly, here also it will be divided by 3600 divided by 1000 this1000is
to make thi into in terms of a kilo a kilo joule. So, this divided by 3.6, and then it
is becoming 17.11 meter square, so this way we have got this, they are very close A1 and
A are close A 2 are close, but we have to make to be moral less equal. So for that what
we have to do what we have to redo the calculation for refining or improvement, then delta T1
will take as a 48 degree centigrade and there will take two will take say 60 degree centigrade,
and redo till A1 is nearly equal to A2. So, this is the thing that is being done in case
of evaporation calculation. Now last thing that I am going to discuss is that vapour
recompression.
So, what is the idea is that in case of evaporators what we have seen that multiple effects is
being used to increase the economy, because by this flow that from from the same amount
of energy coming from the external steam, some steam is generated from different affection.
So, we are getting at good amount of economy the similar economy can also be found out
can also done, can also maintain by putting the evaporator a single effective evaporator
also it can be done, but in that case the vapour which is coming out from the top which
is disconnect. In case of single effective evaporator that is to be compressed, that
means that is why it is called recompression vapour from the single effectiveness evaporate
can be recompressed. And then circulated by recompression is needed because the pressure
of the vapour in the evaporator is low, then the pressure of the saturated steam which
is entering in to the evaporator from the external supply.
Therefore, vapour has to be compressed back and is circulated, and the why compressing
then it what is this doing it is increasing the temperature of the evaporator sorry, it
is increasing the temperature of the vapour that is being back as well as it is increasing
the pressure. So, this this is called vapour recompression, and then many a times this
vapour recompression become, and alternative issue while a single effective evaporator
with an arrangement of vapour recompression can act as a substitute to multiple effective
evaporator. Therefore, the vapour recompression can be done by two ways; one is called thermal
recompression it is called thermal recompression, and this is called mechanical recompression.
So, there other two usual two wedge by which is being done thermal recompression and mechanical
recompression. And in case of thermal recompression what is being done is that there is a steam
jet ejector steam jet ejector is there, and in what is being there is that this ejector
is used changer ejector is used to increase the pressure as well as temperature in this
case, and here there is A1 is called motive fluid that is high pressure steam and that low pressure steam this called load.
This two are mixed to or generate steam of intermediate and they required pressure and temperature,
so this is being done in case of a thermal recompression the steam ejector and motive
fluid steam ejector steam ejector is used and motive fluid there is the high pressure
steam. And low pressure steam there is the load, so this is the external supply and this
is the load this is coming from this single effective evaporator. This two are somehow
mixed and the details can be seen from a different places mixed to generate a steam of intermediate
characteristic in the form of pressure and temperature, which is entering back or recirculate
back to the a single effect evaporator, but in case of mechanical recompression what is
being, and it is simply it is the mechanical devise is used for compression and for large
steam flow it is centrifugal compression is used, and for small steam flow it is that
positive displacement compression is used.
Now, centrifugal used as used for ah for large volume to be handled and for a small volume
to be handled for steam the positive displacement compression is used, and the compression ratio
is usually is nearly equal to 2.1 is very common, so what is meant is that vapour recompression
unit associate with a single effective evaporator sometimes work as an alternative to multiple
effective evaporator, in this case that this that vapour which is coming is compressed,
and then reticulate back and this come and again put into the same evaporator.
So, this way it can work and this is sometimes acting as a replacement to multiple effect
evaporator in terms of improving economy, and then they vapour compression can be done
as two ways. As I discussed, one is called thermo and mechanical recompression, and usually
the compression issue which is desirable is 2 is to 1, which is very common when a when
a an evaporator operates at large driving force at large driving force vapour recompression
may not be economical. At the same time when operates at low driving force large area is
needed, this is also not this is also not this is also not desired
So, what is done to be, so it trade off so it trade off between these the two factors
two opposing factors need to be considered as a or consider while deciding on type of
evaporator to be used. So, this is what about this evaporation, we have discussed we have
discussed about the basic concepts of the evaporation, and then what we have discussed
is that different type of evaporators available. Then we also discuss that calculations procedure
for single effective operator as well as multiple effective operator, and with some problem
we have seen that how to calculate this double effective operator or single effective operator
we have seen. In addition to this there is one important
aspects of which are very common and very important aspect which is the vapour compression
many times, it is used for practice is being discussed in brief. So, this is all about
this model in the next model we will we are going to discuss on radiation heat transfer.
Thank you very much.