Triple-Effect Evaporator - Introduction

WEBVTT
Kind: captions
Language: en

00:00:03.330 --> 00:00:09.950
In this screencast I am going to explain
how a triple-effect evaporator works. And
00:00:09.950 --> 00:00:16.770
its objective is to concentrate a dilute solution,
usually some solute in water, where we use
00:00:16.770 --> 00:00:22.400
significantly less energy than we would use
in a single-effect evaporator. So there is
00:00:22.400 --> 00:00:26.779
a separate screencast on the single-effect
evaporator that it would be worth watching
00:00:26.779 --> 00:00:33.720
first that describes the behavior for one
of these evaporators in the triple-effect.
00:00:33.720 --> 00:00:39.830
The advantage of the triple-effect evaporator
is that we can use the vapor from this first
00:00:39.830 --> 00:00:45.760
evaporator as the source of the heat for evaporating
the liquid in the second evaporator, and then
00:00:45.760 --> 00:00:52.119
we can use the vapor from the second evaporator
as the source of heat for evaporating the
00:00:52.119 --> 00:00:57.430
liquid in the third effect. And I will go
through in detail to show you what we are
00:00:57.430 --> 00:01:01.769
referring to here. So what we are going to
have in the triple effect in the first effect,
00:01:01.769 --> 00:01:08.340
we are going to have a feed coming in. A liquid
at low concentration of solute. We are going
00:01:08.340 --> 00:01:14.330
to vaporize some fraction of this and then
we are going to create a more concentrated
00:01:14.330 --> 00:01:23.740
solution of liquids. We carry out this evaporation
by having steam being fed to these heat transfer
00:01:23.740 --> 00:01:32.420
coils. And then we have liquid steam leaving
so we're assuming that the feed is at saturation
00:01:32.420 --> 00:01:40.000
pressure. We condense and drip out the liquid,
also at saturation temperature and pressure.
00:01:40.000 --> 00:01:46.200
So we have the feed coming at a feed temperature
T1, would be the temperature in the evaporator
00:01:46.200 --> 00:01:53.260
which also means it is the temperature of
the liquid and the steam is coming in at a
00:01:53.260 --> 00:02:00.630
temperature TS where TS must be greater than
the evaporator temperature for evaporation
00:02:00.630 --> 00:02:08.819
to occur. So what we do now is feed this
liquid leaving the first evaporator into
00:02:08.819 --> 00:02:16.990
the second evaporator and we create the evaporation
by using the vapor, from the first evaporator
00:02:16.990 --> 00:02:24.400
and we feed that into the heating coils and
again, we get liquid at saturation conditions
00:02:24.400 --> 00:02:34.120
leaving. And so we are evaporating steam in
the vapor phase V2 and the liquid, more concentrated
00:02:34.120 --> 00:02:41.860
from the second effect. And then we are going
to take this vapor and feed it into the third
00:02:41.860 --> 00:02:49.250
heat transfer coil. Again what is leaving
is condensed steam so liquid at saturation
00:02:49.250 --> 00:02:57.340
conditions. The concentrated liquid from the
second evaporator is then fed into the third
00:02:57.340 --> 00:03:06.739
evaporator where we get a more concentrated
solution, L3. And we have a final vapor stream
00:03:06.739 --> 00:03:14.190
coming off. So we can get a significant improvement in energy efficiency by using the vapor from
00:03:14.190 --> 00:03:20.640
previous effects to vaporize to a liquid in
the next effect. So the temperature in this
00:03:20.640 --> 00:03:29.200
first effect is T1. Temperature here is T2,
and here is T3. And when I say T3, that means
00:03:29.200 --> 00:03:36.020
both the liquid and the vapor are at T3. Then
T1 must be greater than T2 and must be greater
00:03:36.020 --> 00:03:44.680
than T2 and must be greater than T3 because
we are using the temperature T1, the steam
00:03:44.680 --> 00:03:53.140
here at T1, to vaporize the liquid in the
second effect so T1 must be greater than T2
00:03:53.140 --> 00:04:00.660
for heat transfer. So if we ignore the boiling
point rise due to the solute in the liquid
00:04:00.660 --> 00:04:08.690
then we can use steam tables to also determine
the pressures. So if the first effect is at
00:04:08.690 --> 00:04:16.239
atmospheric pressure or 1 bar, then the pressure
must be greater than the pressure in the second
00:04:16.239 --> 00:04:21.769
must be greater than the pressure in the third
which would mean we would be running these
00:04:21.769 --> 00:04:30.339
effects under vacuum in order to create this
temperature difference. And the concentrations
00:04:30.339 --> 00:04:37.740
we would feed a low concentration of solute,
there would be no solute in the vapor phase,
00:04:37.740 --> 00:04:46.430
we will have concentrated that solute which
is now fed to the second phase, because more
00:04:46.430 --> 00:04:52.409
concentrated and then more concentrated, since
that is the objective of triple effect evaporators.
00:04:52.409 --> 00:05:01.889
And if we assume that the feed coming in is
at T1 or very close to T1 then our heat transfer,
00:05:01.889 --> 00:05:07.659
all the energy goes to vaporize the liquid
and we can write the heat transfer in the
00:05:07.659 --> 00:05:13.659
first effect. Heat transfer coefficient, area
for heat transfer, temperature of the steam
00:05:13.659 --> 00:05:21.300
minus temperature T1. But this should be approximately
equal to the heat transfer in the second effect
00:05:21.300 --> 00:05:30.080
because now we are going to condense back all
that vapor that we formed. So q2 which is U2
00:05:30.080 --> 00:05:36.369
area for heat transfer. Temperature in the
second effect, temperature in the first effect,
00:05:36.369 --> 00:05:41.509
which is now the heating for the second. And
then the same argument means this should be
00:05:41.509 --> 00:05:49.879
equal to q3 which then is the temperature
we are heating and the temperature in the
00:05:49.879 --> 00:05:57.900
third effect. And typically the heat transfer
areas are the same, these are identical evaporators
00:05:57.900 --> 00:06:06.909
and the heat transfer coefficient is approximately
the same also. And so TS minus T1 then is
00:06:06.909 --> 00:06:15.869
approximately T1 minus T2, approximately T2
minus T3 and these temperatures will then
00:06:15.869 --> 00:06:22.939
determine which pressures we have to operate
the effect at to get a steady-state operation.
00:06:22.939 --> 00:06:34.770
The total amount of heat transferred is then is
q1+q2+q3 which is going to be equal to the
00:06:34.770 --> 00:06:43.930
UA, T of the steam minus T3, this total temperature
difference. And so to a first approximation,
00:06:43.930 --> 00:06:54.800
also 1 kilogram of steam fed in here, will vaporize
three kilogram of our original liquid. And the triple
00:06:54.800 --> 00:07:02.259
effect evaporator of this type certainly is
best used if the feed is hot coming in. So
00:07:02.259 --> 00:07:08.279
we don't have to use much of our steam energy
to heat the feed. And if this liquid here
00:07:08.279 --> 00:07:16.099
leaving our product, the concentrated solution,
has some temperature sensitivity as we increase
00:07:16.099 --> 00:07:20.610
its concentration and we would like to be
operating that at a lower temperature. The
00:07:20.610 --> 00:07:26.710
actual calculations are really very similar
to those for a single effect evaporator.
Office location
Engineering company LOTUS®
Russia, Ekaterinburg, Lunacharskogo street, 240/12

Phone: +7 343 216 77 75

E-mail: info@lotus1.ru

Sales phone

Russia: +7 343 216 77 75

WhatsApp: +79122710308