CHAPTER SIX

EVAPORATION

Evaporation is an operation in which a weak solution/liquor is concentrated by

vaporising a portion of the solvent.
The weak liquor/solution to be concentrated is composed of a non-volatile solute and a
volatile solvent.
In this operation, the solvent to be evaporated is generally water and the concentrated
solution/thick liquor is the desired product. The vapour generated usually has no value, it is
condensed and discarded.
It differs from drying in that the residue obtained is a liquid rather than a solid.
It differs from distillation in that the vapour is usually a single component rather than a
mixture of components.
It differs from crystallisation in that the purpose is to concentrate a solution rather than to
form and build crystals.
Evaporation is generally followed by crystallisation and drying.
Evaporation is carried by supplying heat to a solution to vaporise the solvent. The
common heating medium (heat source) is generally a low pressure steam but in some
situations other sources that might be used are : solar energy, fuel, electricity, hot oil and
flue-gas.
In an evaporation operation, heat is utilized to :
(i) increase the temperature of the solution to its boiling point (sensible heat) and
(ii) supply the latent heat of vaporisation of the solvent.
In this operation, rate processes that occur are : heat transfer from a heating medium to a
solution through the solid surface and simultaneous heat and mass transfer from a liquid to a
vapour phase. Irrespective of this, the operation can usually be considered in terms of heat
transfer from a heater to the solution (heat transfer to boiling liquids) for the design of
evaporators.
A wide variation in properties of the solution to be concentrated requires judgement and
experience in the design and operation of evaporators. Some of the properties of evaporating
liquids that have effects on the process of evaporation are :
(i) Concentration : The viscosity and density of the solution increase with increase in
concentration of the solution. The boiling point of the solution increases considerably as the
concentration increases so that it may be much higher than B.P. of water at the same
pressure.

(6.1)
Unit Operations – II 6.2 Evaporation

(ii) Foaming : Some materials have a tendency to foam that causes heavy entrainment
(the carry over of a portion of the liquid by the rising vapour is called an entrainment) and
thus the loss of solution.
(iii) Scale : Some solutions deposit scale on the heat transfer surfaces. This causes
reduction of the heat transfer coefficient and hence the rate of heat transfer. Therefore, it is
necessary to clean the tubes after definite intervals of service.
(iv) Temperature sensitivity : Some materials such as pharmaceuticals and food
products get thermally degraded when heated to moderate temperatures even for short
periods. For concentrating such materials special techniques are to be used which reduce the
temperature of operation and also the time of heating.
(v) Corrosiveness : Whenever contamination and corrosion is a problem, special
materials such as copper, nickel, stainless steels may be used, otherwise mild steel is
normally used for evaporators. Other liquid properties which must be considered in the
design are specific heat, freezing point, toxicity, etc.
The selection of an evaporator for a particular application is based on the analysis of the
factors such as the properties of the solution to be concentrated, operating cost, capacity,
hold-up and residence time. High product viscosity, heat sensitivity, scale formation and
deposition are the major problems that are occurred during the operation of evaporators and
those should be taken into account while designing evaporators for a new installation.
Usually, the desired product of an evaporation operation is the concentrated solution
(thick liquor), but occasionally the evaporated solvent is the primary product as in the
evaporation of sea water to obtain potable water.
Common examples of evaporation are :
Concentration of aqueous solutions of sodium chloride, sodium hydroxide, glycerol,
sugars, milk, fruit juices, etc.
Performance of tubular evaporators :
The performance of a steam heated tubular evaporator is evaluated in terms of
(i) capacity and (ii) economy
Capacity : The capacity of an evaporator is defined as the number of kilogram of water
vaporised / evaporated per hour.
The rate of heat transfer Q, through the heating surface of an evaporator, is defined as the
product of the area of heat transfer surface A, the overall heat transfer coefficient U, and the
overall temperature drop ∆T.
Q = U · A ∆T … (6.1)
where ∆T is the temperature difference between the heating medium and the boiling
solution (saturation temperature of steam minus the boiling point of solution).
Unit Operations – II 6.3 Evaporation

If the feed solution is at the boiling temperature corresponding to the pressure in the
vapour space of an evaporator, then all the heat that is transferred through the heating surface
is available for evaporation (i.e., for converting liquid to vapour) and the capacity is
proportional to the heat transfer rate. If the cold feed solution is fed to the evaporator, heat is
required to increase its temperature to the boiling point and it may be a quite large and thus,
the capacity for a given rate of heat transfer will be reduced accordingly as heat used to
increase the temperature to the boiling point is not available for evaporation. When the feed
solution to the evaporator is at a temperature higher than the boiling point corresponding to
the pressure in the vapour space, a portion of the feed evaporates adiabatically and the
capacity is greater than that corresponding to the heat transfer rate. This process is called
flash evaporation.
Evaporator economy : The economy of an evaporator is defined as the number of
kilogram of water evaporated per kilogram of steam fed to the evaporator. It is also called as
steam economy.
In a single-effect evaporator the amount of water evaporated per kg of steam fed is
always less than one and hence economy is less than one. The fact that the latent heat of
evaporation of water decreases as the pressure increases tends to make the ratio of water
vapour produced, i.e., water evaporated per kg of steam condensed less than unity.
The economy of an evaporator can be increased by reusing the vapour produced.
The methods of increasing the economy are :
(i) use of multiple effect evaporation system
(ii) vapour recompression.
In a multiple effect evaporation system, the vapour produced in the first effect is fed to
the steam chest of the second-effect as a heating medium in which boiling takes place at low
pressure and temperature and so on. Thus in a triple-effect evaporator, 1 kg of steam fed to
the first-effect evaporates approximately 2.5 kg of water.
Another method to increase the economy of an evaporator is to use principle of thermo
compression. Here, the vapour from the evaporator is compressed to increase its temperature
so that it will condense at a temperature higher enough to make possible its use as a heating
medium in the same evaporator.
Boiling point elevation :
In actual practice, the boiling point of a solution is affected by a boiling point elevation
and a liquid head.
As the vapour pressure of most aqueous solutions is less than that of water at any given
temperature, the boiling point of a solution is higher than that of pure water at a given
pressure. The difference between the boiling point of a solution and that of pure water
at any given pressure is known as the boiling point rise/elevation of the solution. The
boiling point elevation is small for dilute solutions and large for concentrated solutions of
inorganic salts.
Unit Operations – II 6.4 Evaporation

The boiling point elevation can be obtained from an empirical rule known as Duhring's
rule. It states that the boiling point of a given solution is a linear function of the boiling point
of pure water at the same pressure. Thus, when the boiling point of a solution is plotted
against the boiling point of water, we get a straight line. Fig. 6.1 shows such a plot for an
aqueous solution of sodium hydroxide of different concentrations.

OH
Na

Boiling point of solution

%
70 60
0
50 4 20
30
%
0

Boiling point of water

Fig. 6.1 : Duhring's plot for NaOH

Material and enthalpy balances for single-effect evaporator :
·
Consider an evaporator as shown in Fig. 6.2. It is fed at a rate of mf kg/h of a weak
solution containing w1 % solute and the thick liquor containing w2 % solute by weight is
· ·
withdrawn at a rate of m' kg/h. Let mv be the kg/h of water evaporated from it.
mv

mf
w1 % Evaporator
solids
(solute)
m'
Q. heat w2 %
solids
(solute)
Fig. 6.2 : Block diagram of evaporator
Let us take the overall material balance and the material balance of the solute.
Overall material balance over evaporator :
kg/h weak solution = kg/h thick liquor + kg/h water evaporated
· = m· + m'
m · … (6.2)
f v
Material balance of solute :
Solute in feed = Solute in thick liquor
w1 × m· ·
w2 m'
f
100 = 100

w ×m · = w m' · … (6.3)
1 f 2
Unit Operations – II 6.5 Evaporation

Knowing three out of the above mentioned five quantities, we can find the values of the
other two with the help of the above two equations.
Let Tf, T and Ts be the temperatures of the feed entering the evaporator, solution in the
evaporators and condensing steam, respectively.
Let 'λs' be the latent heat of condensation of steam at its saturation temperature and
assume that only the latent heat of condensation is used. Then, the rate of heat transfer
through the heating surface from the steam side is given by
Q = m · λ … (6.4)
s s s
·
where ms is the mass flow rate of steam to the evaporator in kg/h.
Assuming negligible heat losses, the enthalpy balance/heat balance over the evaporator
is :
Heat associated with feed + Heat (latent) associated with steam =
Heat associated with vapour leaving + Heat associated with thick liquor
· · H +m · λ = m · · H'
m f f s s H + m' v v … (6.5)
· · H +m
m · λ = (m · – m')
· H + m'· H' … (6.6)
f f s s f v
where Hv, Hf and H' are the enthalpies of the vapour, feed solution and thick liquor,
respectively.
Rearranging Equation (6.6), we get
m· λ = (m · – m'
· ) H + m'
· H' – m
· H … (6.7)
s s f v f f
Heat transfer rate on the steam side = Heat transfer rate on the liquor side.
In case of solutions having negligible heats of dilution, the enthalpy balance or heat
balance can be written in terms of specific heats and temperatures of the solutions.
Heat transferred to the solution in the evaporator by condensing steam (in the absence of
heat losses) is utilised to heat the feed solution from Tf to T and for the evaporation of water
from the solution. Therefore,
Qs = Q
= m· C (T – T ) + (m · – m'
· )λ … (6.8)
f pf f f v

·
With Qs = ms λs, it becomes
· λ = m
m · · C (T – T ) + (m
· – m'
· )λ … (6.9)
s s f p f f f v

where Cp = specific heat of feed solution

f
λv = latent heat of evaporation of water from thick liquor
For a negligible boiling point rise, λv = λ
where λ = latent heat of vaporisation of water at a pressure in the vapour
space and can be read from steam tables
Unit Operations – II 6.6 Evaporation

With this Equation (6.9) becomes

m· λ = m · C (T – T ) + (m
· – m'
· )λ … (6.10)
s s f p f f f

m · · C (T – T ) + m
· λ = m · λ … (6.11)
s s f p ff

The boiling point of a solution (T) corresponding to a pressure in the vapour space can be
obtained by knowing the boiling point elevation and boiling point of pure water at that
pressure, e.g. if T' is the boiling point of water at a certain pressure of the operation and 'P' is
the boiling point elevation, then
T = T' + P
The heat transfer surface/area of an evaporator is calculated with the help of the
following equation :
Q = U · A · ∆T … (6.12)
· λ = U · A · ∆T
Q =m … (6.13)
s s
where U = overall heat transfer coefficient
A = area of heat transfer
and ∆T = temperature difference
∆T = Ts – T
∆T = Condensing steam temperature – Boiling point of solution
When 'Q' is in W, 'U' is in W/(m2·K) and '∆T' is in K, then 'A' will be in m2.
λs = Specific enthalpy of saturated steam
– Specific enthalpy of saturated water (i.e., of condensate)
Evaporator Types :
Evaporators used in the chemical process industries can be classified as :
(i) Natural circulation evaporators.
(ii) Forced circulation evaporators.
Natural circulation evaporators are commonly used for simpler evaporation operations
singly or in multiple effect, e.g., horizontal tube evaporator, vertical tube evaporator etc.,
whereas forced circulation evaporators are commonly used for salting, viscous and scale
forming solutions. The forced circulation units may be provided with an external horizontal
or vertical heating element.
Open pan evaporator / Jacketed pan evaporator :
The simplest method of concentrating a solution makes use of jacketed pans. Such a type
of evaporator is particularly suitable when small quantities are to be handled. In an
open/jacketed pan evaporator, condensing steam is fed to a jacket for evaporating a part of
the solvent.
Unit Operations – II 6.7 Evaporation
Pan
Steam

Jacket
Condensate

Product
Fig. 6.3 : Jacketed pan evaporator
Pan is made of a single sheet of metal (for small sizes) or several sheets joined by
welding / brazing. A jacket is welded to the pan. The jacket is provided with a steam inlet at
the top, while a condensate drain is provided at its bottom. The pan is provided with an outlet
at the bottom for draining its contents. The jacket is usually constructed out of mild steel,
while the pan is constructed out of mild steel, stainless steel, copper or aluminium as per
process requirements.
The solution to be concentrated is taken into the pan and steam is admitted in the jacket.
Evaporation is carried out for a predetermined time to achieve a desired concentration level.
The thick liquor is then drained from the outlet.
Horizontal tube evaporator :
It consists of a vertical cylindrical shell incorporating a horizontal square tube bundle at
the lower portion of the shell. Channels are provided on either ends of the tube bundle for
introduction of steam and withdrawal of condensate. The shell is closed by dished heads at
both the ends. A vapour outlet is provided on the top cover and a thick liquor outlet is
provided at the bottom. Feed point is located at a convenient point. In this evaporator, steam
is inside the tube and the liquor to be concentrated surrounds the tubes. Steam which is
admitted through one of the steam chests/channels and flows through the tubes. Steam gets
condensed by transferring its latent heat and the condensate is removed from the outlet
provided at the bottom of the opposite steam chest. This type of evaporator is shown in
Fig. 6.4.
Vapour

Evaporator body

Tube
Feed
Steam

Steam chest

Condensate

Thick liquor
Fig. 6.4 : Horizontal tube evaporator
Unit Operations – II 6.8 Evaporation

Heat given out by the condensing steam will be gained by the solution in the evaporator
and the solution boils. Vapours formed are removed from the top, while the thick liquor is
removed from the bottom. In this evaporator as evaporation occurs outside the tubes,
it eliminates a scale formation problem inside the tubes.
Main advantages : Very low head room requirements and large vapour-liquid
disengaging area.
It is not suited for salting and scaling liquids as deposits form on the outside of the
tubes.
It is commonly used for small capacity services and for simpler problems of
concentrations, i.e., for processes wherein the final product is a liquor, such as industrial
sirups.
Calendria - type / Standard vertical tube evaporator / Short tube evaporator :
Construction :
It consists of a vertical cylindrical shell incorporating a short vertical tube bundle at the
lower portion with horizontal tube sheets bolted to the shell flanges. Vapour outlet is
provided at the top cover while a thick liquor discharge is provided at the bottom. Usually
the tube bundle is not more than 150 cm high and tube diameter (outside) not more than
75 mm (25 mm to 75 mm). A downtake is provided at the centre of the tube bundle having a
flow area about 40 to 100 per cent of the total cross sectional area of the tubes for circulating
the cooler liquid back to the bottom of the tubes. In this evaporator, the solution to be
evaporated is inside the tubes and steam flows outside the tubes in a steam chest. Baffles are
incorporated in the steam chest to promote uniform distribution of steam. The condensate is
withdrawn at any convenient point near the lower tube sheet, while the non-condensable gas
such as air is vented to the atmosphere from a point near the top tube sheet.
Vapour outlet

Vapour
disengagement
Feed space

Steam in Vent

Tube
Down take

Condensate
Thick liquor

Fig. 6.5 : Calendria type evaporator

Unit Operations – II 6.9 Evaporation

Working : Thin solution/liquor is introduced to the tube side and steam into the steam
chest. The liquor covers the top of the tubes. Heat transfer to the boiling liquid inside the
tubes takes place from condensing steam on the outside of the tubes. Vapours formed will
rise through the tubes, come to a liquid surface from which they are disengaged into a vapour
space and removed from the vapour outlet. Circulation of the cold liquor is promoted by a
central downtake and the concentrated solution / thick liquor is removed from the bottom of
the evaporator.

Advantages :

(i) Relatively inexpensive.

(ii) As scaling occurs inside the tubes, it can be easily removed by mechanical or
chemical means.

(iii) Provides moderately good heat transfer at a reasonable cost.

(iv) Can be put into more rigorous services than horizontal tube evaporators.

(v) High heat transfer coefficients.

(vi) Requires low head room.

Disadvantages :

(i) Floor space required is large.

(ii) Amount of liquid hold up in the evaporator is large.

(iii) Since there is no circulation, these units are not suitable for viscous liquid.

Long tube vertical evaporator :

Construction : A long tube evaporator consists of a long tubular heating element

incorporating tubes, 25 mm to 50 mm in diameter and 4 to 8 metres in length. The tubular
heating element projects into a vapour space/separator for removing an entrained liquid from
the vapour. The upper tube sheet of tubular exchanger is free and a vapour deflector is
incorporated in the vapour space just above it. A return pipe connecting the vapour space to
the bottom of the exchanger is provided for natural circulation of a unvaporised liquid. It is
provided with inlet connections for feed, steam and outlet connections for vapour, thick
liquor, condensate, etc. In this evaporator, the solution/liquor to be concentrated is in the
tubes and condensing steam surrounds the tubes. Fig. 6.6 shows a long-tube vertical
evaporator.
Unit Operations – II 6.10 Evaporation

Vapour outlet

Vapour space

Deflector

Natural circulation
of unevaporated
Steam

liquid
Heating
element

Tube
Thick
Condensate liquor

Feed

Fig. 6.6 : Long-tube vertical evaporator

Working :
In this evaporator, feed enters the bottom of the tubes, gets heated by the condensing
steam, starts to boil part way up the tubes and a mixture of vapour and liquid issues from the
top of tubes and finally impinges at a high velocity on a deflector. The deflector acts both as
a primary separator and a foam breaker. Some part of the separated liquid is removed as a
product and the remaining part is returned to the bottom of the evaporator.
This type of evaporator is widely used for handling of foamy, frothy liquors.
It is used for the production of condensed milk and concentrating a black liquor in the
pulp and paper industry.
Forced Circulation Evaporators :
Whenever we are dealing with concentration problems involving solutions of high
viscosities or of scale forming tendencies we have to make use of forced circulation
evaporators as increasing the velocity of flow of liquor through tubes increases remarkably
the liquid film heat transfer coefficients and the high velocity resulting by use of a centrifugal
pump prevents the formation of excessive deposits on the heat transfer surfaces. In a natural
circulation evaporator, the liquid enters the tube at a velocity of 0.3 to 1 metre per second,
whereas in a forced circulation evaporator, the velocity is of the order of 2 to 6 metres per
second. In the forced circulation evaporator, smaller diameter tubes are used compared to
those in the natural circulation evaporator, generally not larger than 50 mm.
Unit Operations – II 6.11 Evaporation

Forced circulation evaporators with a horizontal external heating element :

Construction :
Fig. 6.7 shows a forced circulation evaporator with a horizontal external heating element.
It consists of a circulating pump, a separating space (separator), an evaporator body with a
vapour outlet at the top, a deflector plate, an outlet for the discharge of thick liquor and an
external heating surface - a horizontal shell and tube heat exchanger having two passes on
the tube side.

Working :
A centrifugal pump forces the solution to be concentrated through the tubes at high
velocity and is heated as it passes through the tubes due to heat transfer from the condensing
steam on the shell side. Boiling does not take place in the tubes as they are under a sufficient
static head, which raises the boiling point of the solution, above that in the separating space.
Vapour outlet

Steam in
Deflector

Evaporator
body
1-2
shell and tube
heat exchanger Condensate
Concentrated liquor
Pump
Pump Feed

Fig. 6.7 (a) : Forced circulation evaporator with horizontal external heating element
Vapour outlet

Steam

Condensate Concentrated
liquor

Pump
Feed
Fig. 6.7 (b) : Forced circulation evaporator with vertical heating element
Unit Operations – II 6.12 Evaporation

The solution becomes superheated and flashes into a mixture of vapour and liquid just
before entering the separator due to reduction in the static head when it flows from the
exchanger to the separator. The two-phase mixture impinges on a deflector plate in the
separating space, and the vapours are removed from the top, and the liquid is returned to the
centrifugal pump. Some part of the liquid/solution leaving the separating space is withdrawn
as a concentrated solution and the makeup feed is continuously introduced at the pump inlet.
In case of a vertical heating element, single pass shell and tube heat exchanger is used.
Advantages of forced circulation evaporators :
1. High heat transfer coefficients are obtained even with viscous materials.
2. Positive circulation and close control of flow.
3. Whenever there is a tendency to form scale or deposit salts, use of forced circulation
units prevents the formation of excessive deposits due to high velocities.
4. Residence time of liquid in the tube is very small (1– 3 s) because of high velocities
in these units so that moderately heat sensitive liquids can be handled.
Main disadvantage : High pumping cost.
The forced circulation evaporators are commonly used for crystalline products, viscous,
salting, scaling and corrosive and foam forming solutions.
Multiple-effect evaporation :
Most of the evaporators use a low pressure steam for heating purpose. Due to addition of
heat to a solution in the evaporator by condensation of steam, the solution in the evaporator
will boil. If the vapours leaving the evaporator are fed to some form of a condenser, then the
heat associated with the vapours will be lost and the system is said to make poor use of
steam.
The vapour coming out of an evaporator can be used as a heating medium for another
evaporator operating at a lower pressure and temperature in order to provide a sufficient
temperature difference for the heat transfer in that evaporator.
When a single evaporator is put into service and the vapours leaving the evaporator are
condensed and discarded, the method is known as single-effect evaporation. The economy of
a single effect evaporator is always less than one. Generally, for evaporation of one kg of
water from a solution, 1 to 1.3 kg of steam is required.
The method of increasing the evaporation per kilogram of steam by using a series of
evaporators between steam supply and condenser is known as multiple-effect evaporation.
It is the one way to increase the economy of evaporator systems.
The method of re-using the latent heat is called multiple-effect evaporation.
A multiple effect evaporation system is commonly used in large scale operations. In such
a system, evaporators are arranged in series so that the vapour from one evaporator is used as
a heat medium for the next one that is operating under at a pressure and temperature lower
than the previous one. Each unit in such a series is called an effect. In case of a tripple-effect
evaporator, if the first effect is operating at atmospheric pressure, then the second and third
Unit Operations – II 6.13 Evaporation

effect operate under vacuum. Steam is fed to the first effect and the vapour from the third-
effect is condensed in a condenser connected to a vacuum pump (which is used to maintain
vacuum in the last two effects).
With a multiple-effect evaporation system, it is possible theoretically to evaporate N kg
of water from 1 kg of live steam fed, where N is the number of effects.
If the vapour from one evaporator is fed to the steam chest of the second evaporator as
steam supply, the system is called as double-effect evaporator system and so on.
The methods used for feeding a multiple evaporation system are :
1. Forward feed, 2. Backward feed, and 3. Mixed feed.
1. Forward feed :
In this arrangement, the thin liquid feed flows in the same direction as the vapour flow.
Fresh feed enters the first effect and steam is also fed to a steam chest of the first effect. The
vapours produced in the first effect are fed to the steam chest of the second effect as a
heating medium and the concentrated liquor from the first effect is fed to the next effect in
series, as shown in Fig. 6.8. The pressure in the second effect is less than in the first effect
and so on. Thus, this arrangement does not require pumps to transfer concentrated solution
from effect to effect. This is used when the feed is hot and the product may get damaged at
high temperatures.
To condenser and vacuum
Vapour Vapour generating system

st nd rd
1 2 3
Effect Effect Effect

Steam P1 P2 P3

Feed

P = Pressure
P1 > P2 > P3 Thick liquor

Condensate Condensate
Fig. 6.8 : Forward feed arrangement for feeding multiple effect evaporator system
2. Backward feed :
In this arrangement the feed solution and the vapour flow in opposite directions to each
other. Fresh feed (thin liquid) is admitted to the last effect and then pumped through the other
effects. The steam is admitted to a steam chest of the first effect and the vapours produced in
the first effect are fed to the steam chest of the second effect (evaporator) and so on. The
pressure in the first evaporator is highest and that in the last effect is lowest. If the solution is
very viscous then we have to adopt a backfeed arrangement as the temperature of the first
effect is highest and thus the corresponding viscosity of the solution will be less. Fig. 6.9
shows this type of arrangement.
Unit Operations – II 6.14 Evaporation

Since the feed flows in the direction of increasing pressure, pumps are used for the
transfer of solution from one effect to another effect. It is used when the feed is cold, since it
saves a large quantity of steam and increases the economy.
To condenser and
Vapour Vapour vacuum generating system

P = Pressure
st nd rd
P1 > P2 > P3
1 2 3
Effect Effect Effect

Steam P1 P2 P3
Liquor Liquor

Condensate
Thick liquor
Feed
Pump

Fig. 6.9 : Backward feed arrangement for feeding multiple-effect evaporation system
3. Mixed feed :
To condenser and
vacuum generating
Vapour Vapour system

st nd st
1 2 3
Effect Effect Effect

Steam
P1 P2 P3

Condensate

Feed Thick
Pump liquor
rd Pump
Liquor to 3 effect
P1 > P2 > P3

Fig. 6.10 : Mixed feed arrangement for feeding multiple-effect evaporation system
In this feed arrangement, steam is admitted to a steam chest of the first effect and the
vapours leaving the first effect are fed to the steam chest of the second effect and so on. Feed
solution is admitted to an intermediate effect and flows to the first effect from where it is fed
to the last effect for final concentration. This arrangement is a combination of the forward
and backward feed adopted for the best overall performance. Fig. 6.10 shows this type of
arrangement.
Unit Operations – II 6.15 Evaporation

Comparison of forward feed and backward feed arrangements :

1. In the forward feed, the flow of the solution to be concentrated is parallel to the
steam/vapour flow.
1. In the backward feed, the flow of the solution to be concentrated is in opposite
direction to that of steam/vapour (counter current).
2. Forward feed arrangement does not need pumps for moving the solution from effect
to effect as vacuum is maintained in the last effect.
2. Backward feed arrangement needs pumps for moving the solution from effect to
effect as transfer of the solution is to be done from the evaporator operating at a low
pressure to that operating at a higher pressure.
3. In the forward feed as heating of the cold feed solution is done in the first effect, less
vapour is produced per kilograms of steam fed, resulting into a lower economy.
3. In the backward feed, the solution is heated in each effect which usually results in a
better economy than that with a forward feed.
4. With a forward feed, the most concentrated solution is in the last effect wherein the
temperature is the lowest and the viscosity is the highest. These conditions lead to
reduction in the capacity of the system as a whole due to low overall coefficient in
the last effect in case of thick liquors which are very viscous.
4. In the case of backward feed, the concentrated solution is in the first effect wherein
the temperature is the highest as steam is admitted to that effect and the viscosity is
the lowest, thus the overall coefficient can be moderately high inspite of high
viscosity.
5. In the case of forward feed, the maintenance charges and power cost are less.
5. In the case of backward feed, the maintenance charges and power cost are more for
the same duty.
6. The forward feed arrangement is less effective thermally.
6. The backward feed arrangement is more effective thermally. (At high feed
temperatures).
7. Forward feed is more economical in steam.
7. At low values of feed temperature, backward feed arrangement gives higher
economy.
8. Forward feed is very common (largely used) as it is simple to operate.
8. Backward feed is not common as it necessitates the use of pump between effects.
The choice of optimum number of effects will be determined by an economic balance
between the savings in steam obtained by using a multiple-effect evaporation system and the
additional investment cost resulting from the added heat transfer area.
Unit Operations – II 6.16 Evaporation

To
tal
co
st
Co
st
of es
ste a rg
ch

Cost
am
d
xe
Fi

Cost o
f wate
r
Labour

Number of effects
Fig. 6.11 : Optimum number of effects in a multiple-effect evaporation system
Vapour Recompression
Thermal energy in the vapour generated from a boiling solution can be utilised to
vaporise more water if there is a temperature drop for heat transfer in the desired direction.
In multiple-effect evaporation systems, this temperature drop is created by gradually
lowering the boiling point of the solution in a series of evaporators by operating them
successively under lower absolute pressures.
The desired driving force (i.e., temperature drop) can also be created by increasing the
pressure (therefore, the condensing temperature) of the vapour generated by (a) mechanical
recompression or (b) thermal recompression.
The compressed vapour having a higher condensing temperature is fed to the steam chest
of the evaporator from which it is generated. Therefore, the economy of an evaporator is also
increased by recompressing the vapour from the evaporator and condensing it in the steam
chest of the same evaporator.
In this method, the vapours from the evaporator are compressed to a saturation pressure
of steam in order to upgrade the vapours to the condition of the original steam to allow their
use as the heating medium. The cost of compression is usually smaller than the value of
latent heat in the vapour. By this technique we can obtain the multiple effect economy in a
single effect.
Mechanical recompression :
In this method, the vapour generated from an evaporator is compressed to a certain
higher pressure by a positive displacement or centrifugal compressor and fed to a heater. As
the saturation temperature of the compressed vapour is higher than the boiling point of the
solution, heat flows from the vapour to the solution and more vapours are generated. The
principle of mechanical vapour recompression is shown in Fig. 6.12.
It is used for the concentrations of very dilute radioative solutions and production of
distilled water.
Unit Operations – II 6.17 Evaporation

Evaporator body

Makeup
steam Vent

Heater
Compressor (Shell and
tube type)
Thick liquor

Feed
Condensate

Fig. 6.12 : Mechanical recompression applied to forced circulation evaporator

Thermal recompression :
In this method, the vapour is compressed by means of a steam jet ejector. Here the high
pressure steam is used to draw and compress the major part of vapours from the evaporator,
while the remaining part of vapours is separately condensed for compensating motive steam
added.
H.P.
Steam
To condenser

Ejector Evaporator body

Vent

Heater
(Shell and Thick liquor
tube type)
Feed
Drips

Fig. 6.13 : Thermal recompression

Thermal recompression is better than mechanical recompression as a substitute to
vacuum operation as steam jets can handle large volumes of vapour. Jets are cheap and easy
to maintain compared to compressors/blowers. Disadvantages of thermal recompression
include : (i) low mechanical efficiency of the jets and (ii) lack of flexibility in the system to
meet changes in the operating conditions.
By this method also, we can obtain the multiple-effect economy in a single-effect.
Unit Operations – II 6.18 Evaporation

Choice of steam pressure :

In any evaporation operation, generally, a low pressure steam is used. We cannot use a
high pressure steam that gives a larger temperature drop that consequently decrease the size
and therefore the cost of the evaporator, since it is much more valuable as a source of power
than as a source of heat. High pressure steam has a lower latent heat than a low pressure
steam and hence the low pressure steam delivers more latent heat than the high pressure
steam for evaporation of the solvent. Also the construction of the evaporator to hold the high
pressure steam would be much more expensive (since it demands heavy thickness) than the
construction to hold the low pressure steam.
Pressure in the vapour space :
It is always desirable to have a larger temperature drop between the condensing steam
and the boiling point of a solution because as the temperature drop increases, the heating
surface and in turn the cost of the evaporator decreases. By making use of a condenser and a
vacuum pump, the pressure in the vapour space of the evaporator is made less than
atmospheric. So by operating the evaporator at a pressure less than atmospheric, i.e.,
operating under a vacuum results in decrease in the boiling point of the solution which in
turn leads to a larger value of temperature drop that decreases the required heat transfer
surface.
It is not necessary to operate the evaporators under vacuum, but as it is economical to
feed them with steam at a relatively moderate pressure, a vacuum is necessary in order to get
an economical temperature drop (∆T). There are certain cases where the operation under
vacuum is necessary, for example, while dealing with heat sensitive materials (they may get
decomposed/degraded if boiled at higher temperatures). Evaporators operating at high
pressures requires much more expensive construction than when operating at lower
pressures. Basically evaporators are operated under vacuum to obtain a larger temperature
drop (∆T).
Evaporator accessories :
These are devices that must be supplied with every evaporator.
(i) condensers (ii) vacuum pump / steam jet ejector.
There are two types of condensers :
(i) Surface condensers and (ii) Contact condensers : parallel current, counter current etc.
In a surface condenser, the vapour to be condensed and the cooling medium are separated
by a metal wall, whereas in a contact condenser the vapour and the cooling medium/liquid
are mixed directly.
Parallel-current condenser : It is the one in which the non-condensed gases leave at a
temperature of the exit cooling water.
Counter-current condenser : It is the one in which the non-condensed gases leave at a
temperature of the entering cooling water.
Wet condenser : It is the one in which the non-condensed gases and cooling water are
removed by the same pump.
Dry condenser : It is the one in which the non-condensed gases and cooling water are
removed by separate pumps.
Barometric condenser : It is the one that is placed high enough so that water escape
from it by a barometric leg.
Low level condenser : It is the one in which water is removed by a pump.
Unit Operations – II 6.19 Evaporation

In practice parallel current condensers are almost always wet condensers, while counter-
current condensers are always dry.
Materials of construction for evaporators : Evaporator bodies are generally fabricated
from mild steel as it is least expensive and easy to fabricate. For corrosive solutions,
materials like monel, inconel and stainless steel are used. The tubes are made of copper,
stainless steel and the tube sheet may be of cast bronze, nickel clad steel or stainless steel.
SOLVED EXAMPLES
Example 6.1 : Calculate the boiling point elevation of a solution and the driving force
for heat transfer using the following data :
Data : Solution boils at a temperature of 380 K (107 oC) and the boiling point of water
at a pressure in the vapour space is 373 K (100 oC)
Temperature of condensing steam is 399 K (126o C).
Solution : Boiling point of the solution = T = 380 K
Boiling point of water = T' = 373 K
Boiling point elevation = T – T' = 380 – 373 K
= 7 K (or oC) … Ans.
Saturation temperature of the condensing steam = Ts = 399 K
Driving force for heat transfer = Ts – T = 399 – 380
= 19 K … Ans.
Note : Whenever the pressure at which steam is available and pressure prevailing in the
vapour space of an evaporator are given, then the steam table should be referred to find the
temperature of steam, latent heat of condensation of steam based on a given steam pressure
and the latent heat of evaporation of water, i.e., latent heat of vapour based on a pressure
prevailing in the vapour space of the evaporator.
Example 6.2 : An evaporator operating at atmospheric pressure (101.325 kPa) is fed at
the rate of 10000 kg/h of weak liquor containing 4 % caustic soda. Thick liquor leaving the
evaporator contains 25% caustic soda. Find the capacity of the evaporator.
Solution :
Basis : 10,000 kg/h of weak liquor entering the evaporator.
.
Let m' be the kg/h of thick liquor leaving the evaporator.
Material balance of caustic soda :
Caustic soda in the feed = Caustic soda in the thick liquor
.
0.04 × 10000 = 0.25 × m'
.
m' = 1600 kg/h
Overall material balance :
kg/h of feed = kg/h water evaporated + kg/h of thick liquor
10000 = kg/h water evaporated + 1600
water evaporated = 10000 – 1600 = 8400 kg/h
∴ Capacity of the evaporator = 8400 kg/h … Ans.
Unit Operations – II 6.20 Evaporation

Example 6.3 : An evaporator is operating at atmospheric pressure. It is desired to

concentrate a feed from 5 % solute to 20 % solute (by weight) at a rate of 5000 kg/h. Dry
saturated steam at a pressure corresponding to the saturation temperature of 399 K (126o C)
is used. The feed is at 298 K (25o C) and the boiling point rise (elevation), i.e., B.P.E.
(B.P.R.) is 5 K. The overall heat transfer coefficient is 2350 W/(m2·K). Calculate the
economy of the evaporator and the area of heat transfer to be provided.
Data : Treating the solution as a pure water and neglecting the B.P.R., the latent heat of
condensation of steam at 399 K is 2185 kJ/kg.
Latent heat of vaporisation of/evaporation of water at 101.325 kPa and 373 K = 2257 kJ/kg.
Specific heat of feed = 4.187 kJ/(kg·K)
Solution : Basis : 5000 kg/h of feed to the evaporator.
. . ⋅
Let mf, m' and mv be the kg/h of feed, thick liquor and water vapour / water evaporated.
Material balance of solute :
Solute in the feed = Solute in the thick liquor
.
0.05 × 5000 = 0.20 × m'
.
m' = 1250 kg/h
Overall material balance (i.e., material balance over evaporator) :
kg/h feed = kg/h water evaporated + kg/h thick liquor
.
Water evaporated = mv = 5000 – 1250 = 3750 kg/h
⋅
Let ms be kg/h of steam required (steam consumption)
λs = latent heat of condensation of steam at 399 K = 2185 kJ/kg
λv = λ = latent heat of vaporisation of water at 373 K = 2257 kJ/kg
T = temperature of thick liquor
= T' + B.P.E. = 373 + 5 = 378 K
(λv is taken as equal to λ for the calculation purpose, since the effect of B.P.E. is to be
neglected)
Tf = temperature of feed = 298 K
⋅
mf = 5000 kg/h
Cp = 4.187 kJ/(kg·K)
f
Let us calculate the economy.
Heat balance over evaporator :
Heat given out/lost by condensing steam (latent heat) =
= Heat gained by the solution to increase its temperature to boiling point
+ Heat required to vaporise / evaporate water
Unit Operations – II 6.21 Evaporation

⋅ ⋅ · λ
ms λs = mf Cp (T – Tf) + m v
f
⋅
ms × 2185 = 5000 × 4.187 (378 – 298) + 3750 × 2257
⋅
ms = 4640.1 kg/h
Steam consumption = 4640.1 kg/h
kg/h water evaporated
Economy of the evaporator = kg/h steam consumed
3750 kg evaporation
= 4640.1 = 0.808 kg steam … Ans.
Let us calculate the heat transfer area.
⋅
Rate of heat transfer = Q = ms λs
= 4640.1 × 2185
= 10138619 kJ/kg
10138619 × 1000
= 3600
= 2816283 J/s ≡ 2816283 W
Ts = saturation temperature of steam = 399 K
Temperature driving force = ∆T = Ts – T
= 399 – 378
= 21 K (21o C)
U = 2350 W/(m2·K)
We have : Q = UA ∆T
∴ A = Q/U ∆T = 2816283 / (2350 × 21)
= 57.07 m2
Heat transfer area to be provided = 57.07 m2 … Ans.
Example 6.4 : A solution containing 10 % solids is to be concentrated to a level of
50 % solids. Steam is available at a pressure of 0.20 MPa [saturation temperature of
393 K (120o C)]. Feed rate to the evaporator is 30000 kg/h. The evaporator is working at
reduced pressure such that boiling point is 323 K (50o C). The overall heat transfer
coefficient is 2.9 kW/(m2·K). Estimate the steam economy and heat transfer surface for :
(i) Feed introduced at 293 K (20o C)
(ii) Feed introduced at 308 K (35o C).
Data : Specific heat of feed = 3.98 kJ/(kg·K)
Latent heat of condensation of steam at 0.20 MPa = 2202 kJ/kg
Latent heat of vaporisation of water at 323 K (i.e. at pressure in the vapour space
= 2383 kJ/kg).