Multiple Effect Evaporator: Mass and Enthalpy Balance Video Lecture | Heat Transfer - Mechanical Engineering

in this tutorial we will consider mass
and enthalpy balances for a three effect
evaporator as we see here in this
schematic we have identified various
quantities related to the feed stream
the streams for vapors for the
concentrated product for our steam as
well as for condensate
so let's first write the mass balance so
for the feed the mass flow rate is m dot
F entering the bottom of the first
effect so m dot F equals the mass flow
rate of vapours leaving the first effect
m dot v1 note that those vapors are used
as heating medium for the second effect
and therefore they are leaving the
system later as a condensate so in our
mass balance we must account for the
mass of those vapors so we have m dot v1
similarly we have m dot v2 vapors coming
from the second effect n m dot v3
vapours coming from the third effect
plus of course the mass flow rate of the
concentrated product stream the solid
balance can be written as X F times m
dot F where XF is the solid fraction
which will be dimensionless so xf m dot
F tells us the amount of solid present
in the feed stream and that equals X P
the solid fraction in the concentrated
product stream times m dot P now for the
enthalpy balance we have to account for
enthalpy of each of the streams so the
enthalpy
entering with the feed is H F so we have
m dot F times H F plus m dot s times H V
s where H V s represents the enthalpy of
the saturated steam that we obtain from
steam tables that equals m dot v1 that's
the mass flow rate of the vapors coming
out from the first effect times H V 1 H
v1 is the enthalpy of the vapors from
first effect plus m dot F 1 which is the
amount of partially concentrated feed
coming out of Effect 1 times H F 1 where
H F 1 is the enthalpy of that pre
concentrated stream plus m dot s times h
c s where m dot s represents now the
mass flow rate of condensate leaving the
first effect times the enthalpy of that
condensate the second equation is for
the second effect of the evaporator for
which we have now our feed as m dot F 1
times H F 1 which represents now the
enthalpy of the feed stream into second
effect plus m dot V 1 times H V 1 note
that the vapors leaving from first
effect are now the heating medium for
the second effect that equals the mass
flow rate of vapours leaving the second
effect m dot v2 times the enthalpy of
those vapors H v2 plus m dot F 2 that is
the enthalpy with the mass flow rate of
this product stream leaving the second
effect times the enthalpy of that stream
HF 2 plus m dot v1
times h c1 the
enthalpy of that condensate so m dot v1
times HC 1 so our third equation is for
the third effect m dot F 2 which is the
mass flow rate of the feed that came
from the second effect and now entering
the third effect times its enthalpy H f2
plus m dot V 2 times H v2 representing
the enthalpy of the vapours that came
from the second effect and now being
used as heating medium in the third
effect that equals m dot V 3 times H v3
which represents the enthalpy leaving
with the vapors from the third effect
plus m dot P times HB 3 representing the
enthalpy leaving with the final
concentrated product stream plus m dot V
2 times HC 2 representing the enthalpy
leaving with the condensate from the
third effect so these three equations
for the enthalpy balances for first
second and third effect plus the solid
balance and the mass balance give us 5
equations so those 5 equations are
solved simultaneously to obtain the
unknowns in a given problem for heat
transfer areas we can use the following
equations for the three effects q1 the
rate of heat transfer in the first
effect will equal u 1 a 1 times TS minus
t1 equal m dot s h vs minus m dot s HC s
so the left hand side gives us the rate
of heat transfer across the tubes the
heating tubes in the first effect by
using the overall heat transfer
coefficient the area
and the temperature difference that is
between the temperature of the steam and
the boiling point inside the first
effect t1 and on the right hand side we
have the enthalpy associated with the
steam minus the enthalpy that leaves
with the condensate so that's the amount
of heat that gets transferred from steam
into the liquid stream second equation
for the second effect is q2 equals u to
a to t1 minus t2 equals m dot v1 h v1
minus m dot v1 HC 1 again this is
similar to our discussion for the first
equation the only difference here of
course we have used t1 minus t2 because
T 1 is the temperature of the vapors
coming in as heating medium and T 2 is
the boiling point temperature maintained
inside the second effect the third
equation for the third effect is Q 3
equals u 3 a 3 times T 2 minus T 3
equals m dot v 2h v2 minus m dot v2 HC 2
and again here T 2 minus T 3 represent a
temperature difference between the
heating medium which is the vapors
coming from the second effect at
temperature T 2 minus T 3 where T 3 is
the temperature maintained inside the
third effect we can obtain the Steam
economy as m dot v1 plus m dot v2 plus m
dot V 3 where these are the mass flow
rates of the vapor streams exiting from
each of the effects so that total amount
of vapors that are produced in the
triple effect evaporator
divided by
the mass flow rate of steam m dot s
typically the steam economy of a triple
effect evaporator is between two and
three now note that the units for these
various quantities that we have used we
can summarize them here m dot s m dot V
1 m dot V 2 m dot V 3 are in kilograms
per second and so are the M units 4 m
dot F m dot F 1 m dot F 2 m dot P they
are all kilograms per second as they are
mass flow rates XF and XP the solid
fractions are dimensionless and then the
enthalpy terms hv s hv 1 HB 2 h v3 hf h
f1 h f2 h p3 its CS h c1 and its c2 are
all in kilojoules per kilogram