Single Effect Evaporator: Mass and Enthalpy Balance Video Lecture | Study Heat Transfer - Mechanical Engineering | Best Video for Mechanical Engineering

in this tutorial we will look at a heat
and mass balance for a single effect
evaporator let's look at the schematic
so we have the various streams going
into the evaporator the vapours leaving
on the top we have steam coming on the
side the feed entering at the bottom and
the condensate leaving the system along
with the concentrated product
note that the tubes and the heating
chest of this evaporator is called
calendar iam now let's conduct a mass
balance so for the feed we have m dot F
which is the mass flow rate of the
dilute feed in kilograms per second we
have m dot s the mass flow rate of steam
as kilograms per second we have m dot V
as the mass flow rate of vapors exiting
the evaporator in kilograms per second
we have m dot P the mass flow rate of
concentrated product in kilograms per
second and we also have the solid
fraction in the feed as XF which will be
dimensionless
similarly the solid fraction in the
concentrated product is denoted by X P
which will be dimensionless so we can
write the mass balance by looking at
what is entering the system and what is
exiting the system so we have the feed
entering the system and the mass flow
rate is m dot F
and that equals the mass flow rate of
vapors leaving the system and m dot p
the mass flow rate of concentrated
product leaving the system note that in
the evaporator steam and feed remain
separated because this is an indirect
way of heating when steam condenses it
leaves the evaporator as a condensate so
steam does not mix with the feed
therefore we do not include the mass
flow rate terms for either the steam or
the condensate in the mass balance when
we look at the solids balance we note
that xf is the solid fraction in the
feed and it is dimensionless
so xf times m dot F will represent the
amount of solids in the feed and that
will equal X P which is the solid
fraction in the concentrated product
which is dimensionless times the mass
flow rate of the product and that will
represent the amount of solids present
in the concentrated product stream so
that is our equation 2 now for the
enthalpy balance we write m dot F times
HF where HF is the enthalpy of the
entering feed in kilojoules per kilogram
we also have enthalpy entering into the
system with steam so we have m dot s the
mass flow rate of steam times h vs where
hv s is the enthalpy of saturated vapor
at temperature TS that is the saturation
temperature of steam and the units will
be kilojoules per kilogram
so the heat is entering the system
either with the feed or with steam this
is the only two ways that heat is
entering this system and that equals the
heat leaving with the vapors that is m
dot V times H v1 where H v1 is the
enthalpy of saturated vapors at
temperature t1 t1 is the boiling point
inside the evaporator that is maintained
by certain vacuum conditions the units
for H V 1 are kilojoules per kilogram so
in addition to the heat leaving with the
vapors we also have heat leaving with
with concentrated product stream and
that would be denoted by m dot P times h
p1 where H p1 is the enthalpy of
concentrated product stream in
kilojoules per kilogram we also have
heat leaving the system with the
condensate this is the condensate that
formed on the outside of the tubes as
steam condensed into condensate and the
enthalpy content will be m dot s that is
the mass flow rate of the steam times HC
s which is the enthalpy of the
condensate in kilojoules per kilogram
again note that the heating is done in
an indirect manner so all the mass of
the steam entering this heating chamber
leaves as condensate after it condenses
into liquid water so the mass flow rate
of the steam m dot s is the same as the
mass flow rate of the condensate as we
see on our schematic so let's see how we
can determine each term in our enthalpy
balance
so we have the enthalpy of the entering
stream H F equals CP f which is the
specific heat of the feed times T F
which is the temperature of the feed
minus 0 degree C we base all our
calculations for enthalpy with a
reference temperature of 0 degree C note
that enthalpy is always relative rather
than absolute value so H F then the
units will be kilojoules per kilogram
since we have the specific heat
multiplied by the temperature we are
going to assume that we are using
saturated steam then H V s is obtained
from steam tables depending upon the
saturation pressure or saturation
temperature of the steam each V 1 is
obtained also from the steam tables as
enthalpy of saturated vapors at
temperature T 1 H P 1 is the enthalpy of
the concentrated product stream so we
use the specific heat of the
concentrated product CPP and multiply
that with t 1 where T 1 is the
temperature maintained inside the
evaporator - 0 degree C where 0 degree C
of course is the reference temperature H
CS is obtained from steam tables as
enthalpy of saturated liquid at
temperature TS now we are going to
assume that the condensate leaves the
evaporator at temperature TS if the
condensate leaves at a lower temperature
than TS
then we will need to determine that
additional sensible heat leaving with
the condensate the area for heat
transfer in the calendar can be obtained
with the following equation Q the rate
of heat transfer across the tubes will
equal u times a times TS minus t1 equals
m dot s hv s minus m dot s HC s where u
is the overall heat transfer coefficient
in watts per square meter kelvin a is
the area in square meters and of course
q is the rate of heat transfer in watts
so this equation can be used to
determine the area required to
accomplish a certain amount of heat
transfer for concentrating liquid feed
the performance of the evaporator is
expressed using a term called steam
economy which equals m dot V divided by
m dot s where m dot V is the mass flow
rate of vapors and m dot s is the mass
flow rate of steam entering the
evaporator a typical value of steam
economy for a single effect evaporator
is close to 1