Cap 18 Evaporation

© 2003 by CRC Press LLC

625

18

Evaporation

18.1 Introduction

Evaporation is a unit operation that consists of the elimination of water ofa fluid food by means of vaporization or boiling. Several foods are obtainedas aqueous solutions and, in order to facilitate their preservation and trans-port, they are concentrated during a water elimination stage. This elimina-tion can be performed in different ways, although evaporation is one of themost used methods. The equipment used to remove this water from the foodproduct is called an evaporator.

An evaporator consists mainly of two chambers, one for condensation andanother for evaporation. Steam condenses in the condensation chamber, giv-ing off the latent heat of condensation, which is contained in the evaporationchamber. The evaporated water leaves the evaporation chamber at boilingtemperature, obtaining at the same time a stream of concentrated solution.

V

while that of food is

w

A

, obtaining a stream of vapor

V

and another of concen-trated solution (or liquid)

w

C

. The removed vapor

V

is driven to the condenserwhere it condenses. It is important to note that many food solutions are heatsensitive and can be adversely affected if exposed to high temperatures. Forthis reason it is convenient to operate under vacuum conditions in the evapo-ration chamber, which causes the boiling temperature of the aqueous solutionto decrease, and the fluid to be affected by heat to a lesser extent. If it is desiredto operate under vacuum, a vacuum pump is needed. Also, a barometric col-umn to compensate for the pressure difference with the exterior is needed inthe condenser to condense the vapor released in the evaporation chamber.

The capacity of the evaporator (

V

) is defined as the amount of waterevaporated from the food per unit time. The consumption (

w

V

) is the amountof heating steam consumed per unit time. The economy (

E

) is the amountof solvent evaporated per unit of heating steam:

(18.1) E

VwV

= = capacity

consumption

TX69299 ch01 frame.book Page 625 Wednesday, September 4, 2002 2:13 PM

Figure 18.1 shows a scheme of an evaporator. The mass flow of steam is w ,


626

Unit Operations in Food Engineering

18.2 Heat Transfer in Evaporators

Figure 18.2 presents a scheme of a single-effect evaporator including thedifferent variables of each stream. The condensation chamber is fed with asaturated vapor stream

w

V

that has a temperature

T

and an enthalpy ˆ

H

w

.Vapor condenses and the only heat given off is that of condensation, so astream

w

V

of liquid water leaves this chamber at the condensation temper-ature

T

, and with enthalpy ˆ

h

w

, which corresponds to the enthalpy of waterat the boiling point. The condensation heat flow

•

Q

is transferred through theexchange area of the evaporator to the food stream in the evaporation chamber.

FIGURE 18.1

Scheme of the installation of an evaporator.

FIGURE 18.2

Simple evaporator.

Vapor

Barometric column

Condenser

Evaporator

wa

ww

wV

wc

Pt

Vapor

wA

wV

wCtA.hA

T, HW

wV

T, hW

Q

t, HV

tC, hC

.



Evaporation

627

A stream

w

A

is fed into the evaporation chamber at a temperature

t

A

withenthalpy ˆ

h

A

. Due to the heat released by the condensed vapor (

•

Q

), a concen-trated stream

w

C

is obtained, with temperature

t

C

and enthalpy ˆ

h

C

. Also, avapor stream

V

is obtained, at a temperature

T

V

and with enthalpy ˆ

H

V

. Notethat the temperatures of the concentrated and vapor streams are equal andcorrespond to the boiling temperature of the concentrated solution thatleaves this chamber.

The energy balances that should be performed are:

Condensation chamber: (18.2)

Evaporation chamber: (18.3)

Exchange area: (18.4)

where

U

is the overall heat transfer coefficient and

A

is the area of theevaporator.

18.2.1 Enthalpies of Vapors and Liquids

In the notation used here, the enthalpies per unit mass of vapor streams willbe designated by ˆ

H

, and those of liquid by ˆ

h

.The enthalpy per unit mass of vapor at a temperature

T

can be expressedas the summation of the enthalpy at saturation ( ˆ

H

sat

) plus the integral betweenthe boiling temperature

T

b

and the enthalpy at

T

of the specific heat times

dT

:

(18.5)

The term ˆ

H

SAT

is the enthalpy of the vapor at its condensation temperature.The specific heat of the water vapor ( ˆ

C

P

)

V

depends on the pressure, althoughits value is close to 2.1 kJ/(kg ·°C).

Since enthalpy is a function, the state of the enthalpy of a liquid shouldbe expressed as a function of a reference temperature. If this temperature is

t

*

and the liquid is at a temperature

t

, it is obtained that:

(18.6)

Tables used for the calculation of these enthalpies can be found in theliterature. Generally, the reference temperature is the freezing temperatureof water (0°C).

w H w h Q V w V wˆ ˆ ˙= +

w h Q w h V HA A C C Vˆ ˙ ˆ ˆ+ = +

Q U A T U A T t˙ = = −( )∆

ˆ ˆ ˆH H C dTSAT P VTe

T

= + ( )∫

h C dT C t tP P*

t*

t

ˆ ˆ ˆ= = −( )∫



628


The enthalpy of the liquid at its boiling temperature is called ˆ

h

SAT

. Thelatent heat of condensation or evaporation (

λ

) will be the difference betweenthe saturation enthalpies of the vapor and the liquid, since the evaporationand condensation temperatures are the same.

(18.7)

The numerical values of the enthalpies of saturated vapor and of the liquidcan be obtained from saturated water vapor tables, and the latent heat ofcondensation can be calculated. However, this value can be obtained in anapproximate way from the equation of Regnault as follows:

(18.8)

where

T

is in °C.The enthalpies of the liquid streams, food ( ˆ

h

A

) and concentrated ( ˆ

h

C

), thatappear in Equation 18.3 are expressed as:

(18.9)

(18.10)

The enthalpy of the vapor in Equation 18.3 will be different if the solutionbeing concentrated presents or does not present a boiling point rise. In caseof no increase in the boiling point of the concentrated solution, the enthalpyof the vapor will be the sum of the saturated liquid plus the latent heat:

(18.11)

where

t

b

is the boiling temperature of the solution.In case of an increase in the boiling point, the boiling temperature of the

solution (

t

e

) will be greater than that of pure water (

t

), so the vapor enthalpywill be:

(18.12)

In order to simplify the calculations, the reference temperature usuallyselected is the boiling point of pure water,

t

* =

t

b

, which makes the enthalpyof the vapor leaving the evaporation chamber coincide with the latent heat

λ = ˆ ˆH hSAT SAT−

λ = −2538 2 91. T kJ kg

h C dT C t tA P A P A A*

t

tA

ˆ ˆ ˆ

*

= ( ) = ( ) −( )∫

ˆ ˆ ˆh C dT C t tC P C P C C*

t*

tC

= ( ) = ( ) −( )∫

ˆ ˆH C t tV SAT P b

*( ) = −( )+ λ

ˆ ˆ ˆH C t t C t tV P b

*P V b= −( ) ( ) −( )+ +λ



Evaporation

629

of condensation if there is no increase in the boiling point. Also, the enthalpyof the concentrated stream will be annulled, since

t

C

=

t

b

=

t

.

18.2.2 Boiling Point Rise

Water boils at a fixed temperature whenever the pressure remains constant.If the pressure varies, the boiling point varies too. For aqueous solutions,the boiling temperature depends not only on pressure, but also on theamount of solute contained, in such a way that the presence of the solutecauses the boiling temperature to increase. The determination of the boilingpoint rise presented by food solutions is very important for the calculationof evaporators. For this reason, expressions and means to calculate theincrease in boiling temperature will be given next.

For diluted solutions that comply with the law of Raoult, the boiling pointincrement can be calculated by the expression:

(18.13)

where

M

S

is the molecular weight of the solute,

X

is the ratio of kg solute/kgsolvent, and

Ke

is the so-called boiling constant of the solvent.For aqueous solutions the following equation can be used:

(18.14)

where

C

is the molal concentration of the solute.A general expression that allows the calculation of the boiling point rise

considering an ideal solution is the equation:

(18.15)

If the solutions are diluted, the following equation can be used:

(18.16)

In these two last equations,

X

w

is the mass fraction of water,

λ

is the latentheat of evaporation,

R

is the gas constant, and

t

b

is the boiling temperatureof pure water.

For real solutions, the boiling point rise can be calculated by the empiricalrule of Dühring, which states that the boiling point of the solution is a linear

∆T Kb X

MbS

=1000

∆T Cb = 0 52.

∆Tt

R t X

bb

b

=1+

w

−λln

∆T

RtXb

b-

w

2

1λ

−( )



630


function of the boiling point of the pure solvent at the same pressure. For agiven concentration of solute, the plots of the boiling temperatures of thesolution against those corresponding to the pure solvent yield straight lines.

In the case of sugar solutions, empirical correlations exist that allow theboiling point increase of the solutions to be obtained. One of these expres-sions is (Crapiste and Lozano, 1988):

(18.17)

in which

C

is the concentration of the solution in °Brix,

P

is the pressure inmbar, and

α

,

β

,

δ

, and

γ

are empirical constants whose values depend on the

sugars, and fruit juice solutions.This equation has been modified for juices by adding a new term to the

exponential term in such a way that the final expression is (Ilangantilekeet al., 1991):

(18.18)

FIGURE 18.3

Diagram of Dühring for tamarind juices. (Adapted from Manohar, B. et al.,

J. Food Eng.,

13,241–258, 1991.)

45 55 65 7540

50

60

70

80

62.1%

54.8%43.1%

30.7%

20.2%

Boiling temperature of pure water (°C)

Bo

ilin

g t

em

pe

ratu

re o

f ju

ice

s (°

C)

∆T C P Cb = expα γβ δ ( )

∆T C P C Cb = − + ×( )−0 04904 0 03889 6 52 100 029 0 113 4 2. . .. . exp

TX69299 ch01 frame.book 30 Wednesday, September 4, 2002 2:13 PM

solute. Table 18.1 shows the values of these parameters for sucrose, reducing

Figures 18.3 and 18.4 present the diagram of Dühring for two aqueous systems.


Evaporation

631

18.2.3 Heat Transfer Coefficients

The calculation of the overall heat transfer coefficient can be obtained fromthe expression:

(18.19)

FIGURE 18.4

Diagram of Dühring for aqueous solutions of sucrose.

TABLE 18.1

Parameters

α

,

β

,

δ

, and

γ

Sample

�

×

10

2

� � �

×

10

2

Sucrose 3.061 0.094 0.136 5.328Reducing sugars 2.227 0.588 0.119 3.593Juices 1.360 0.749 0.106 3.390

Source:

Crapiste, G.H. and Lozano, J.E.,

J. Food Sci.,

53(3), 865–868, 1988.

50

75

100

125B

oili

ng

te

mp

era

ture

of

the

so

lutio

n (

°C)

50 75 100

Boiling temperature of water (°C)

1000800600700

200

g sucrose100y water

1 1 1U A h A

ek A h Aw

P

P m c

=′

+ +′′



632


in which

h

w

is the individual convective heat transfer coefficient for thecondensing vapor, while

h

c

is the coefficient corresponding to the boilingsolution. The parameters

e

P

and

k

P

are the thickness of the solid throughwhich heat transfer occurs and its thermal conductivity, respectively. In thistype of operation, it is assumed that the areas are the same, so the expressionis simplified to:

(18.20)

In case there is deposition on the heat transfer surface, the resistance offered(

R

D

) should be taken into account. Hence, the real overall coefficient

U

D

is:

(18.21)

Although the calculation of the theoretical overall coefficient should bemade by Equation 18.20, values for this coefficient can be found in theliterature, depending on the type of evaporator. Table 18.2 shows the valuesfor this coefficient.

18.3 Single Effect Evaporators

ables. In order to perform the calculation of this type of evaporator, massand energy balances should be conducted.

TABLE 18.2

Global Heat Transfer Coefficients for Different Types of Evaporators

UEvaporator (W/m2·°C)

Long tube verticalNatural circulation 1000–3500Forced circulation 2300–12000

Short tubeHorizontal tube 1000–2300Calandria type (propeller calandria) 800–3000

Coiled tubes 1000–2300Agitated film (Newtonian liquids)

Viscosity 1 mPa ·s 2300100 mPa ·s 1800104 mPa ·s 700

Source: McCabe, W.L. and Smith, J.C., Operaciones Bósi-cas de Ingeniería Quínica, Reverté, Barcelona, Spain, 1968.

1 1 1U h

ek hw

P

P c

= + +

1 1U U

RD

D= +


Figure 18.5 shows a single-effect evaporator with all the streams and vari-


Evaporation 633

Mass balances: A global balance and a component balance are carriedout.

(18.22)

(18.23)

where XA and XC are the mass fractions of solute in the food and inthe concentrated solution streams, respectively.

Energy balances: Balances around the condensation and evaporationchambers are performed, in addition to the equation of heat trans-fer rate through the exchange area. These balances are similar tothose performed previously and are given in Equations 18.2, 18.3,and 18.4. If the expressions for the enthalpies of the liquid and thevapor given in Section 18.2.1 are taken into account, then:

Condensation chamber:

(18.24)

Evaporation chamber:

(18.25)

Exchange area:

(18.4)

FIGURE 18.5Single effect evaporator.

Pt

Wa, XA

Wv, PC

Wv, PC

wC, XC

V, P

tA,hA

T, hw

T, HW

t, H V

tC, h C

w w V A C= +

w X w X A A C C=

w H h w QV w w V w

ˆ ˆ ˙−( ) == λ

w C t t Q w C t t V C t tA P A A b C P C C b V P V b

ˆ ˙ ˆ ˆ( ) −( ) + = ( ) −( ) + + ( ) −( )

λ

Q U A T U A T t˙ = = −( )∆



634 Unit Operations in Food Engineering

Note that tC = t; that is, the temperatures of the streams leaving the evap-oration chamber are equal, and the boiling point rise of the solution ∆Tb =t – tb. When combining Equations 18.24 and 18.25, it is obtained that:

(18.26)

In case there is no boiling point rise (∆Te = 0), the previous equationbecomes:

(18.27)

18.4 Use of Released Vapor

The vapor released in the evaporation chamber contains energy that may beused for other industrial purposes. Such vapor has a temperature lower thanthat of steam, so its latent heat of condensation is greater; for this reason, itis very important to take advantage of this latent heat. There are differentmethods for using this energy, among which are vapor recompression, ther-mal pump, and multiple effect. The first and second methods mentionedwill be briefly studied here, while in the following section the multiple-effectmethod will be studied in detail.

18.4.1 Recompression of Released Vapor

One way to use the energy contained in released vapor is to compress anduse it as heating steam. There are two methods in practice to compress thisvapor: mechanical compression and thermocompression.

18.4.1.1 Mechanical Compression

This type of operation consists of compressing the vapor released in theevaporation chamber using a mechanical compressor. The vapor leaving theevaporation chamber at a temperature t1 and a pressure P1 is compressed toa pressure P2, corresponding to the pressure of the steam used in the con-

sented by point 1. Mechanical compression, generally, is an isoentropic pro-cess, so a perpendicular straight line is followed until reaching the isobarcorresponding to pressure P2. The condition of this steam can be found inMollier’s diagram, where it is obtained that the outlet temperature of the

w w C T V C T w C t tV V C P C b V P V b A P A A bλ λ= ( ) + + ( )

− ( ) −( )ˆ ˆ ˆ∆ ∆

w V w C t t V V V A P A A bλ λ= − ( ) −( )ˆ


densation chamber (Figure 18.6). In the enthalpy–entropy Mollier diagramfor steam (Figure 18.7), the condition of the released vapor can be repre-


Evaporation 635

FIGURE 18.6Simple evaporator with mechanical compression of released vapor.

FIGURE 18.7Evolution of released vapor during mechanical compression.

1

3

Wr

Wr

Wc

V

Wv

Wv*+V

Wv*

Wv

Compressor

Entropy

1

3

2

Enthalpy

H2

H3

H1

P2

P1




compressor is t2, its pressure is P2, and its enthalpy is H2. It can be observedthat the vapor obtained after compression is a reheated steam, so beforemixing it with the saturated steam coming from the boiler, its temperatureis lowered by recirculating a stream wR. In this way, the condensation cham-ber can be fed with saturated steam.

around the evaporation chamber are not affected. However, additional bal-ances should be performed in the condensation chamber.

Energy and mass balances:

(18.28)

(18.29)

(18.30)

The calculation of the evaporator is similar to that described in the singleeffect, although in this case these additional balances should be taken intoaccount.

18.4.1.2 Thermocompression

Another way to use the energy of the released vapor is to use a jet whichcarries away part of the vapor and joins the steam coming from the boiler.

device that functions, due to the effect of Venturi, in such a way that a steamjet carries away part of the vapor released in the evaporation chamber. Thevapor entering the condensation chamber is saturated, although its pressureis intermediate between steam and the released vapor. If the jet is fed witha steam flow wV

* that has a pressure PW and an enthalpy H*, then this steamcarries away a fraction of vapor V, which is at pressure P1, temperature t1,and has an enthalpy H1. The vapor that leaves the jet will have a pressurePC and an enthalpy Hw, with a mass flow rate wV. Similarly to mechanicalcompression, the balances around the evaporation and condensation cham-bers are unaltered. However, new balances should be carried out in the jet.

Balances around the jet:

(18.31)

(18.32)

w V w wV R V*= + +

w H V H w h w HV w R w V*

wˆ ˆ ˆ ˆ= + +1

Q wV W= λ

w aV wV*

V+ =

w H* aV H w H V*

Vˆ ˆ ˆ+ =1 w


As can be observed in Figure 18.6, in this type of operation the balances

Figure 18.8 represents a diagram of the evaporator jet system. The jet is a


Evaporation 637

An empirical equation correlates the different variables and allows calcu-lations of this type of compression. Thus, the expression to be used is (Viánand Ocón, 1967):

(18.33)

in which R is the thermal performance of the jet.

18.4.2 Thermal Pump

The so-called thermal pump is usually employed in heat sensitive productsin which high temperature can adversely affect the product. Low boiling

for the heating fluid.This installation has two evaporators. The evaporation chamber of the first

is fed with the fluid to be concentrated, obtaining a vapor stream V that isused as heating fluid for the second evaporator. The condensation chamberof the first evaporator is fed with a vapor, which could be NH3, that con-denses. This liquid exits the condensation chamber (point 1), expands in avalve (point 2), and is employed to feed the evaporation chamber of the

FIGURE 18.8Simple evaporator with thermocompression of released vapor.

P1

t

(1 – a) VV

aV

JetWv*

Wv

Wc

Wv

Wa

aVw

R

PP

P

PV*

C

+ =

1 1

1

log

log

w


temperatures can be achieved with this device. Figure 18.9 presents a schemeof this installation, while Figure 18.10 shows a temperature–entropy diagram



other evaporator, thus yielding a vapor stream (point 3) fed to a mechanicalcompressor, with the objective of raising its pressure and obtaining a moreenergetic vapor (point 4). This vapor is used as heating vapor for the firstevaporator. It should be pointed out that the circuit followed by the heatingvapor is a closed circuit.

This type of installation is usually employed to concentrate juices, forexample, orange juice, that are affected by high temperatures.

18.4.3 Multiple Effect

One of the most usual ways of employing the vapor released in the evapo-ration chamber is by using it as heating fluid in another evaporator.

observed that the vapor released in the first evaporator is used as heatingfluid in the second one, while the vapor liberated in this effect is used to

FIGURE 18.9Installation with thermal pump.

Pt

2

1

4

To vacuum

P1T1

3

Wv

Wa

Wv

Wc

hV

V

HV

V


Figure 18.11 shows a diagram of a triple-effect evaporation system. It can be


Evaporation 639

heat the third effect. Finally, the vapor released in the last effect is driven tothe condenser.

For notation, the different streams have subscripts corresponding to theeffect that they are leaving. The vapor released during the different effectsis, each time, at a lower temperature and a lower pressure:

FIGURE 18.10Evolution of the heating fluid in the temperature–entropy diagram for the thermal pump.

FIGURE 18.11Scheme of released vapor use in a triple effect evaporator.

Entropy

1

32

4

Temperature

P2

P1

TC

T1

V2

P1

t1P2

t2P3

t3V1

WVPC T

V1P1 tb1

V2P2 tb2

Wv

PC T

V1

P1 t1

V2

P2 t2V3

P3 t3

V3P3 tb3

T t t t t t t

P P P P

b b b

C

> > > > > >

> > >

1 1 2 2 3 3

1 2 3




where t1, t2, and t3 are the boiling temperatures of the solutions leaving theevaporation chambers of the first, second, and third effects, respectively. Thetemperatures tb1, tb2, and tb3 are the boiling temperatures of pure water atpressures P1, P2, and P3, respectively.

The evaporation chamber of the first effect is at a pressure P1 and temper-ature t1. The vapor that leaves this effect V1 does so at these conditions andis used as heating fluid in the second effect, where it is supposed to enterin saturated conditions, i.e., at its boiling temperature tb1. The pressure inthe condensation chamber of the second effect is still P1, while the temper-ature is tb1. The evaporation chamber of the second effect is at pressure P2

and temperature t2, the same as those of the vapor V2 leaving this chamber.This vapor condenses in the condensation chamber of the second effect at atemperature tb2 and a pressure P2. The evaporation chamber of the third effectis at a pressure P3 and a temperature t3; thus, the vapor V3 leaving this effecthas the same characteristics. This vapor is driven to a condenser where itcondenses at the temperature tb3, which corresponds to that of pressure P3.

When there is no boiling point rise of the solutions that pass through theevaporator, the temperatures of the evaporation chamber of one effect andthat of condensation of the following effect will be equal (ti = tbi).

Note that, in this type of installation, vacuum pumps are required to reachadequate temperatures in the chambers of each effect. A more detailed study,which allows the calculation of multiple-effect evaporators, is performed inthe following section.

18.5 Multiple-Effect Evaporators

Only the case of a triple-effect evaporator will be studied; however, themathematical treatment in other cases for multiple effects is similar.

18.5.1 Circulation of Streams

As previously mentioned, the vapor released in the evaporation chamber ofone effect is employed as heating fluid for the next effect. However, depend-ing on the circulation system of the solutions to be concentrated, differentpassing systems are obtained.

18.5.1.1 Parallel Feed

The food is distributed in different streams fed into each of the effects

gathered in only one stream, which will be the final concentrate.


(Figure 18.12A), while the concentrated solution stream of each effect is


Evaporation 641

FIGURE 18.12Circulation systems of fluid streams for triple effect evaporators.

V 1 V 2

Wv

Wv V1 V2

V3

Wc

WA

V1 V2

I II IIIP1

t1P2

t2P3

t3

A) Parallel System

V1

V1 V2

V2

V3

V1 V2WV

WA

WV

WC

III I I IP2

t2P1

t1P3

t3

B) Forward Feed

P1

t1P2

t2P3

t3

wV

V1 V2

V1 V2

V1 V2

V3

WA

I I I III

WC

WA

C) Backward Feed




18.5.1.2 Forward Feed

The diluted stream is fed into the first effect, while the concentrate stream

observed that the vapor and concentrate streams of each effect are parallelflows. This pass system is frequently used for solutions that can be thermallyaffected, since the most concentrated solution is in contact with the vaporat the lowest temperature.

18.5.1.3 Backward Feed

The flows of the solutions to concentrate and of vapor are countercurrent

vapor has less energy, and the concentrated solution leaving this effect isused to feed the previous effect and so on. This type of arrangement shouldbe carefully used in the case of food solutions, since the solution with thehighest concentration gains heat from the vapor at the highest temperature,which may affect the food.

18.5.1.4 Mixed Feed

In this type of arrangement, the diluted solution can be fed into any of theeffects, while the concentrated solutions can be used to feed a previous orfollowing effect. Figure 18.2D shows a mixed-feed arrangement in which thediluted solution is fed into the third effect, while the solution leaving thiseffect is used to feed the first one. The food stream feeding the second effectis the concentrated solution that leaves the first effect, obtaining the finalconcentrated solution at the second effect.

FIGURE 18.12 (continued)

P1

t1P2

t2P3

t3

WV

WV

V1 V2

V1 V2

V1 V2

W A

I I I III

WC

V3

D) Mixed Feed


leaving each effect is fed into the following effect (Figure 18.12B). It can be

(Figure 18.12C). The diluted solution is fed into the last effect, where the


Evaporation 643

18.5.2 Mathematical Model

Of the different cases that can be studied, this section will examine only thetriple-effect evaporator in which the pass system is a backward feed(Figure 18.13). The mathematical model here and its solution are similar forany type of circulation and number of effects.

In order to set up the mathematical model, the global and component massbalances should be performed, as well as the enthalpy balances and equa-tions for rate of heat transfer through the exchange area in each effect.

The reference temperatures for calculating the enthalpies of the differentstreams are the boiling temperatures of pure water at the pressure in theevaporation chamber of each effect tbi. The boiling point rise of each effectis the difference between the boiling temperature of the solution leaving thechamber of each effect and the boiling temperature of pure water at thepressure in this chamber ∆Tbi = ti – tbi .

The notation used here implies that the streams leaving an effect will havethe subscript corresponding to each effect.

Mass balances:

(18.34)

(18.35)

FIGURE18.13Backward feed triple effect evaporator.

P1

t1

P2

t 2

P3

t3

W V

C

W V

C

V1 V2

V1

P1 t1

V2

P2 t 2

V1

P1 tb 1

V2

P2 tb 2

V3

P3 t 3

V3

P3 tb 3

wC

IIII II

P T

P T

w w V V VA C= + + +1 2 3

w X w X A A C C=




(18.36)

(18.37)

Enthalpy balances:The enthalpy balances performed around each effect lead to the following

equations:

Substitution of the expressions of the enthalpies of each stream and rear-rangement yields:

(18.38)

(18.39)

(18.40)

These are general equations, which means that it is assumed that there isa boiling point rise. However, when there is no boiling point rise, the equa-tions are simplified.

Even when the vapor leaving the evaporation chambers is reheated, it isassumed that, when it enters the condensation chamber of the next effect, itenters as saturated vapor. This simplifies the latter equations to:

(18.41)

(18.42)

w w V VA2 2 3= − −

w w VA3 3= −

w H w h w h w h V HV V W C C Vˆ ˆ ˆ ˆ ˆ

W + = + +2 2 1 1

V H w h V h w h V HV V V1 1 3 3 1 1 2 2 2 2ˆ ˆ ˆ ˆ ˆ+ = + +

V H w h V h w h V H V A A V V2 2 2 2 3 3 3 3ˆ ˆ ˆ ˆ ˆ+ = + +

w H h w C t t V C t t

w C t t

V w w C P C C b V P V b

P b

ˆ ˆ ˆ ˆ

ˆ

−( ) = ( ) −( ) + + ( ) −( )

− ( ) −( )1 1 1 1 1

2 2 2 1

λ

V C t t w C t t V C t t

w C t t

V P V b P b V P V b

P b

1 1 1 1 2 2 2 2 2 2 2 2

3 3 3 2

λ λ+ ( ) −( )

= ( ) −( ) + + ( ) −( )

− ( ) −( )

ˆ ˆ ˆ

ˆ

V C t t w C t t V C t t

w C t t

V P V b P b V P V b

A P A A b

2 2 2 2 3 3 3 3 3 3 3 3

3

λ λ+ ( ) −( )

= ( ) −( ) + + ( ) −( )

− ( ) −( )

ˆ ˆ ˆ

ˆ

w w C t t V C T w C t tV w C P C C b V P V b P bλ λ= ( ) −( ) + + ( )

− ( ) −( )ˆ ˆ ˆ

1 1 1 1 2 2 2 1∆

V w C T V C T w C t tP b V P V b P b1 2 2 2 2 2 2 3 3 3 2λ λV1 = ( ) + + ( )

− ( ) −( )ˆ ˆ ˆ∆ ∆



Evaporation 645

(18.43)

Rate of heat transfer equations:The heat transferred through the exchange area of each effect is obtained

from the following equations:

(18.44)

(18.45)

(18.46)

It is assumed that the vapor entering the condensation chambers is satu-rated and the only heat that dissipates is the condensation heat.

18.5.3 Resolution of the Mathematical Model

Generally, the data available in evaporator problems are the flow rate of foodto concentrate, as well as its composition and temperature. Also, the com-position of the final concentrated solution is known. The characteristics ofthe vapor of the steam boiler are known; generally the pressure and, sinceit is saturated vapor, its temperature and latent heat can be obtained fromthermodynamics tables. The pressure of the evaporation chamber of the thirdeffect is usually known; therefore, its characteristics are also known. It ispossible to obtain the boiling point rise by Dühring diagrams or adequateequations once the composition of the streams leaving the evaporation cham-bers is known.

A ten-equation system is obtained from the mass and enthalpy balancesas well as from the rate equations:

(18.34)

(18.35)

(18.36)

(18.37)

(18.41)

V w C T V C T w C t tP b V P V b A P A A b2 3 3 3 3 3 3 3λ λV2 = ( ) + + ( )

− ( ) −( )ˆ ˆ ˆ∆ ∆

Q w U A T tV w1 1 1 1= = −( )λ

Q V U A t tV b2 1 1 2 2 1 2= = −( )λ

Q V U A t tV b3 2 2 3 3 2 3= = −( )λ

w w V V VA C= + + +1 2 3

w X w X A A C C=

w w V VA2 2 3= − −

w w VA3 3= −

w w C t t V C T w C t tV w C P C C b V P V b P bλ λ= ( ) −( ) + + ( )

− ( ) −( )ˆ ˆ ˆ

1 1 1 1 2 2 2 1∆




(18.42)

(18.43)

(18.44)

(18.45)

(18.46)

Since the number of unknowns is larger than the number of equations,there are infinite solutions. It is assumed that the area of each effect is equalin order to solve the problem. Also, the heat flow rate transferred througheach effect is similar, and can be assumed to be equal (

•

Q1 =•

Q2 =•

Q3).Since it is complied that

•

Qi/Ai = constant, then:

(18.47)

Due to one property of the ratios, they will be equal to the sum of thenumerators divided by the sum of the denominators:

(18.48)

where ∆Tbi is the boiling point rise experienced by the solution in the itheffect.

These assumptions allow solution of a system of ten equations, althoughan iterative process is needed to solve the mathematical model.

18.5.4 Calculation Procedure

The calculation procedure requires an iterative method, which is simplerwhen there is no boiling point rise.

V w C T V C T w C t tP b V P V b P b1 2 2 2 2 2 2 3 3 3 2λ λV1 = ( ) + + ( )

− ( ) −( )ˆ ˆ ˆ∆ ∆

V w C T V C T w C t tP b V P V b A P A A b2 3 3 3 3 3 3 3λ λV2 = ( ) + + ( )

− ( ) −( )ˆ ˆ ˆ∆ ∆

Q w U A T tV w1 1 1 1= = −( )λ

Q V U A t tV b2 1 1 2 2 1 2= = −( )λ

Q V U A t tV b3 2 2 3 3 2 3= = −( )λ

T t U

t t U

t t U

b b−=

−=

−1

1

1 2

2

2 3

3

1 1 1

T t

U

t t

U

t t

U

T t T

U

b b b bi

i

−=

−=

−=

− −

∑∑

1

1

1 2

2

2 3

3

3

1 1 1 1



Evaporation 647

18.5.4.1 Iterative Method when there is Boiling Point Rise

The calculation steps are:

1. Assume that the flow of heat transfer at each stage is the same.Also, assume that the exchange areas of the different stages areequal.

2. Determine wC by Equations 18.34 and 18.35 as the total flow ofreleased vapor (V1 + V2 + V3).

3. Assume that the flows of eliminated vapor in each effect are equal:V1 = V2 = V3.

4. Calculate the concentrations X2 and X3.5. Use the concentration of each solution to determine their corre-

spondent specific heats: (CP)i.6. Calculate the boiling point rise in each effect. Such rises are calcu-

lated with the concentration of the solution leaving the effect.7. Calculate the unknown temperatures of the evaporation and con-

densation chambers using Equation 18.48.8. Calculate the latent heat of condensation λw, λV1, λV2, and λV3 of the

saturated steams at temperatures T, tb1, tb2 and tb3.9. Solve the equation systems obtained from enthalpy and mass bal-

ances, Equations 18.34 through 18.43. This operation allows deter-mination of ww, w2, w3, V1, V2, and V3.

10. Obtain the areas of each effect A1, A2, and A3 from the rate equations(Equations 18.44, 18.45, and 18.46).

11. Check whether the areas obtained are different by less than 2%with respect to the mean value Am, in which case the iterativeprocess is finished.

12. If the areas are different, recalculate X2 and X3 using the values ofV2 and V3 obtained in step 9.

13. Recalculate the boiling point rise with the new concentrations.14. Determine the new temperatures of the different evaporation and

condensation chambers using the following expressions:

(18.49)

(18.50)

T t T tAAj j

m

−( ) = −( )

−1 1 1

1

t t t tAAb j b j

m1 2 1 2 1

2−( ) = −( )

−




(18.51)

These equations point out that the temperature rise between thecondensation and evaporation chambers of each effect in an itera-tive stage j is equal to the rise existing between such chambers inthe previous calculation stage j – 1, multiplied by the ratio betweenthe area of each effect and the mean area. Two pairs of values areobtained for each unknown temperature, so the value taken is thearithmetic mean.

15. Continue the calculations beginning with step 8 until the values ofthe areas of each effect coincide.

18.5.4.2 Iterative Method when there is No Boiling Point Rise

The calculation steps are listed below:

1. Assume that the heat flows transferred in each stage are the same.Also, assume that the exchange areas of the different stages areequal.

2. Determine wC and the total flow of released vapor (V1 + V2 + V3)by Equations 18.34 and 18.35.

3. Assume that the flows of vapor eliminated in each effect are equal:V1 = V2 = V3.

4. Calculate the concentrations X2 and X3.5. Use the concentration of each solution to determine their corre-

spondent specific heats: (CP)i.6. Calculate the unknown temperatures of all evaporation and con-

densation chambers using Equation 18.48.7. Use temperatures T, tb1, tb2, and tb3 to calculate the latent heat of

condensation of saturated steams λw, λV1, λV2, and λV3 at thesetemperatures.

8. Solve the equation systems obtained from the enthalpy and massbalances, Equations 18.34 through 18.43, which will allow calcula-tion of ww, w2, w3, V1, V2, and V3.

9. Obtain the areas of each effect A1, A2, and A3 from the rate equations(Equations 18.44, 18.45, and 18.46).

10. Check whether the areas obtained are different by less than 2%with respect to the mean value Am, in which case the iterativeprocess is finished.

11. If the areas are different, recalculate X2 and X3 using the values ofV2 and V3 obtained in step 9.

t t t tAAb j b j

m2 3 2 3 1

3−( ) = −( )

−



Evaporation 649

12. Calculate the new temperatures of the different evaporation andcondensation chambers using Equations 18.49, 18.50, and 18.51.Two pairs of values are obtained for each unknown temperature,so the value taken is the arithmetic mean.

13. Continue from step 7 until the values of the areas of each effectcoincide.

18.6 Evaporation Equipment

Different types of equipment are used for evaporation processes: those inwhich the fluid circulates by pumps and those that do not need these devices.The former are called forced circulation evaporators, and the latter are callednatural circulation evaporators. Also, there are long tube evaporators ofascending and descending film, as well as plate and expanded flow evapo-rators. A brief description of each type is given next.

18.6.1 Natural Circulation Evaporators

There are different types of evaporators based on the natural circulation offluids, with the simplest being the open evaporator. There are also tubeevaporators, usually short, based on this principle.

18.6.1.1 Open Evaporator

These evaporators are the simplest, consisting mainly of a container open tothe atmosphere in which fluid is heated directly, or by a heating coil orexternal jacket. Often, evaporators have a low evaporation rate, hence show-ing a poor thermal economy. Sometimes, to allow operation under vacuum,the container may have a hermetic seal.

The main advantage of these evaporators is that they are useful when lowcapacity units are required. However, heating is not effective in large capacityunits, since the ratio of heat transfer surface to volume of liquid is low. Also,heat transfer is reduced in units that have internal coils, since they make thecirculation of liquid difficult.

This type of evaporator is used in the food industry to concentrate tomatopulp, prepare soups and sauces, and boil marmalades and confectioneryproducts.

18.6.1.2 Short Tube Horizontal Evaporator

These evaporators are formed by a chamber, the bottom of which is crossed

acting as heating fluid. Above the tubes is a space that allows the separationby gravity of drops carried away by the vapor released in the base.


by a bundle of interior horizontal tubes (Figure 18.14) that circulate steam



Impact slabs are arranged in order to facilitate the separation and carryingaway of drops. Since the tube bundle makes the circulation of liquid difficult,such evaporators present poor global heat transfer coefficients. They areusually employed for the concentration of low viscosity liquids.

18.6.1.3 Short Tube Vertical Evaporator

steam condenses outside tubes vertically arranged inside the evaporationchamber. The tube sheet, called the calandria, has a large central return tubethrough which a liquid colder than the liquid that circulates in the heatingascending tubes, thus forming natural circulation streams. The length of thetubes usually ranges between 0.5 and 2 m, with a diameter of 2.5 to 7.5 cm,while the central tube presents a transversal section between 25 and 40% ofthe total section occupied by the tubes.

These evaporators show adequate evaporation velocities for noncorrosiveliquids with moderate viscosity. The units can be equipped with basketcalandria that facilitate cleaning, since they can easily be dismounted.

FIGURE 18.14Short tube horizontal evaporator. (Adapted from Brennan, J.G. et al., Los Operaciones de laIngeniería de los Alimentos, Acribia, Zaragoza, Spain, 1980.)

Steam

Feed

Product Condensate

Vapor


Figure 18.15 shows a scheme of this type of evaporator in which the heating


Evaporation 651

Short tube vertical evaporators are usually employed for the concentrationof sugar cane and beet juices, as well as in the concentration of fruit juices,malt extracts, glucose, and salt.

18.6.1.4 Evaporator with External Calandria

the vapor separator. The evaporator usually operates at reduced pressuresand has easy access to the tube bundle. Also, a plate heat exchanger cansubstitute for the calandria, which is useful in case crusts form, since theplates are easy to dismount and clean.

Since these evaporators can operate under vacuum, they are used to con-centrate heat sensitive foods such as milk, meat extracts, and fruit juices.

18.6.2 Forced Circulation Evaporators

Circulation in these evaporators is achieved by a pump that impels the foodthrough the calandria into a separation chamber, where vapor and concen-

a velocity between 2 and 6 m/s; when it passes through the tube bundle,

FIGURE 18.15Short tube vertical evaporator. (Adapted from Brennan, J.G. et al., Los Operaciones de la Ingenieríade los Alimentos, Acribia, Zaragoza, Spain, 1980.)

Vapor

Steam

FeedCondensate

Product


In this type of evaporator (Figure 18.16), the tube bundle is located outside

trate are separated (Figure 18.17). The pump causes the fluid to circulate at



the fluid gains enough heat to reheat it, but the liquid is subjected to a staticcharge that prevents boiling inside the tubes. However, when the fluidreaches the chamber, there is a sudden evaporation, and the impact platefacilitates the separation of the liquid phase from vapor.

These evaporators are capable of concentrating viscous liquids when thepump impels the liquid at an adequate velocity. For this reason, centrifugalpumps are used if the liquids present low viscosity. If the liquids have ahigher viscosity, then positive displacement pumps should be used.

18.6.3 Long Tube Evaporators

Some evaporators consist of a vertical chamber made of a tubular exchangerand a separation chamber. The diluted liquid is preheated to almost boilingtemperature before entering the tubes. Once inside the tubes, the liquidbegins to boil, and the expansion due to vaporization yields the formationof vapor bubbles that circulate at high velocity and carry away the liquid,which continues concentrating as it moves forward. The liquid–vapor mixture

FIGURE 18.16Evaporator with exterior calandria: A, vapor inlet; B, liquid feed inlet; C, concentrated liquidoutlet; D, vapor outlet; E, condensate outlet; F, noncondensable gases outlet. (Adapted fromBrennan, J.G. et al., Los Operaciones de la Ingeniería de los Alimentos, Acribia, Zaragoza, Spain, 1980.)

Cyclon separator

Heating chamber

A

D

B

F

C

E



Evaporation 653

then enters the separation chamber, where baffle plates facilitate vapor sep-aration. The concentrated liquid obtained can be directly extracted or mixedwith nonconcentrated liquid and recirculated, or it can go into anotherevaporator where concentration can be increased.

Long tube evaporators can be of ascending film, falling film, or ascend-ing–falling film. In ascending film evaporators, the liquid enters the bottomof the tubes; vapor bubbles that ascend through the center of the tube beginto form, creating a thin film on the tube wall that ascends at great velocity.In falling film evaporators, the feed is performed at the top of the tubes, sothe vapor formed descends through the center of the tubes as a jet at greatvelocity. When high evaporation velocities are desired, ascending–fallingfilm evaporators are used, where ascending film evaporation is used toobtain an intermediate concentration liquid with a high viscosity. This liquidis then further evaporated in tubes, where it circulates as falling film.

Generally, overall heat transfer coefficients are high. In film evaporators, theresidence time of the liquid treated in the heating zone is short since it circu-lates at great velocity. The product is thus not greatly affected by heat; there-fore, these evaporators are useful for evaporation of heat-sensitive liquids.Descendent film evaporators are widely used to concentrate milk products.

FIGURE 18.17Forced circulation evaporator. (Adapted from Brennan, J.G. et al., Los Operaciones de la Ingenieríade los Alimentos, Acribia, Zaragoza, Spain, 1980.)

Steam

Condensate

Feed Product

Pump

Impact plate

Vapor




18.6.4 Plate Evaporators

Plate evaporators consist of a set of plates distributed in units in which vaporcondenses in the channels formed between plates. The heated liquid boilson the surface of the plates, ascending and descending as a film. The liquidand vapor mixture formed goes to a centrifugal evaporator.

These evaporators are useful to concentrate heat sensitive products, sincehigh treatment velocities are achieved, allowing good heat transfer and shortresidence times of the product in the evaporator. Also, plate evaporatorsoccupy little space on the floor and are easily manipulated for cleaning, sincesetup and dismount are easy and quick. Plate evaporators are usuallyemployed to concentrate coffee, soup broth, light marmalades, and citric juices.

Besides the evaporators described here, there are other types, such asexpanded flow, scrape surface and those based on the functioning of thethermal pump used for evaporation of products very sensitive to heat.

Problems

18.1

A salt solution is concentrated from 5 to 40% in weight of salt. For this reason,15,000 kg/h of the diluted solution are fed to a double-effect evaporator thatoperates under backward feed. The steam used in the first effect is saturatedat 2.5 atm, maintaining the evaporation chamber of the second effect at apressure of 0.20 atm. If feed is at 22°C, calculate: a) steam flow rate neededand economy of the system; b) heating area of each effect; c) temperaturesand pressures of the different evaporation and condensation chambers.

Data: consider that only the 40% salt solution produces a boiling point riseof 7°C. The specific heat of the salt solutions can be calculated by theexpression: CP = 4.18 – 3.34 X kJ/(kg ·°C), where X is the mass fraction ofsalt in the solution. The global heat transfer coefficients of the first and secondeffect are, respectively, 1860 and 1280 W/(m2 °C). Specific heat of water vaporis 2.1 kJ/(kg ·°C).

Properties of the saturated steam:

P T

h

H

w

w

w w

= = = °

=

= =

2 5 2452 126 8

533

2716 2183

. .

ˆ

ˆ

at mbar C

kJ kg

kJ kg kJ kgλ

P t

h

H

b

V

V V

2 2

2

2 2

0 2 196 59 7

250

2609 2359

= = = °

=

= =

. .

ˆ

ˆ

atm mbar C

kJ kg

kJ kg kJ kgλ


The diagram of the double-effect evaporator is represented in Figure 18.P1.


Evaporation 655

Global and component mass balances:

obtaining:

Initially, it is assumed that V1 = V2 = 6562.5 kg/h, supposing that thecomposition of the stream w2 is X2 = 0.09.

The specific heats of each stream are obtained from the equation given inthe problem statement:

For XA = 0.05, CPA = 4.01 kJ/(kg ·°C).For XC = 0.40, CPC = 2.84 kJ/(kg ·°C).For X1 = 0.09, CP1 = 3.88 kJ/(kg ·°C).

According to the statement of the problem, boiling point rise is only in thefirst effect, while it can be neglected in the second one, yielding that t2 = tb2 =59.7°C.

FIGURE 18.P1Backward feed double effect evaporator.

P1

t1

P2

t2

Wv

WV

PC T

C

V1V2

V1

P1 t1

V2

P2 t 2

V1

P1 tb 1V2

P2 tb 2

I II

P T

Wc Wa

15 000

15 000 0 05 0 40

1 2,

, . .

= + +

( )( ) = ( )w V V

w

C

C

w V V C = + =1875 13 1251 2kg h and kg h,




In order to perform the calculation process, it is supposed that the areasand the heat flows transferred through these exchange areas are equal forthe two effects, complying with (Equation 18.48):

Hence,•

Q/A = 45,569 W/m2.The temperature te1 is obtained from the heat transfer rate equation in the

second effect:

The boiling temperature in the first effect is:

It is possible to find the properties of the saturated steam from the tem-perature tb1 = 95.3°C and the saturated steam tables:

Enthalpy balances applied to both effects yield:

1st effect:

2nd effect:

Together with the following balance equations:

˙ . .QA

T t T

U U

b b=

− −

+=

− −( )+

2 1

1 2

1 1127 2 59 7 7

11860

11280

∆ º C

m º C W

2

˙ .Q A U t t tb b= −( ) = °2 1 2 1 95 3 C

t t Tb b1 1 1 95 3 7 102 3= + = + =∆ . . º C

t P

h

H

b

V

V V

1 1

1

1 1

95 3 855

399 3

2668 2268 7

= ° =

=

= =

.

ˆ .

ˆ .

C mbar

kJ kg

kJ kg kJ kgλ

2183 = 2268.7 + 2.1 7 2.84 1875 102.3 95.3

3.88 59.7 95.32

w V

w

V ( ) ⋅ ( )( ) + ( )( ) −( )− −( )

1

2268 7 2359 4 4 01 15 000 22 59 71 2. . . , .V V= − ( )( ) −( )

w w V

V V

A2 2

1 2 13 125

= −

+ = ,



Evaporation 657

A four-equation system with four unknowns is obtained, which, whensolved, yields:

The value of the areas through which heat is transferred can be obtainedby means of the equations of heat transfer rate through such areas:

1st effect:

2nd effect:

The mean area is Am = 105.56 m2. Since these areas differ by more than2%, the calculation procedure should begin again, rectifying the intermediatetemperatures t1 and te1, since the other temperatures remain the same:

Hence, tb1 = 101.3 – 7 = 94.3°C.Since they are different, the mean value of each temperature is taken: tb1 =

94°C and t1 = 101°C.The new enthalpies for 94°C can be found in the saturated steam tables:

The four-equation system stated above is solved again using the new valueof λV1, yielding:

w w

V V

V = =

= =

8102 9056 5

7181 5 5943 5

2

1 2

kg h kg h

kg h kg h

.

. .

8102 2183 3600 1 86 126 8 102 31( ) = −( ). . . A

7181 5 2268 7 3600 1 28 95 3 59 7

107 81 99 32

2

1 2

. . . . .

. .

( ) = −( )= =

A

A A m m2 2

t A A t

t A A t

b m b

m

1 2 1

1 1 1

59 7 95 3 59 7 93 8

126 8 126 8 102 3 101 3

− = −( )( ) = °

− = −( )( ) = °

. . . .

. . . .

C

C

t P

h

H

b

V

V V

1 1

1

1 1

94 815

393 8

2666 2272 2

= ° =

=

= =

C mbar

kJ kg

kJ kg kJ kg

ˆ .

ˆ .λ

w w

V V

V = =

= =

8090 9051

7176 5949

2

1 2

kg h kg h

kg h kg h




The areas are recalculated from the velocity equations:

(a) Economy of the system:

(b) Area per effect:

(c) 1st effect:

2nd effect:

18.2

A double-effect evaporator, operating under forward feed, is used to con-centrate clarified fruit juice from 15 to 72 °Brix. The steam available from theboiler is saturated at 2.4 atm, and the vacuum pressure in the evaporationchamber of the second effect is 460 mm Hg. The diluted juice is fed into theevaporation chamber at a temperature of 50°C and a mass flow rate of 3480kg/h. If the overall heat transfer coefficients for the first and second effectsare 1625 and 1280 W/(m2 ·°C), respectively, determine: (a) steam flow ratefrom the boiler and economy of the system; (b) heating surface for eacheffect; and (c) temperatures and pressures in the condensation and evapo-ration chambers for each effect.

Data: properties of the fruit juices: the boiling point rise can be calculatedaccording to the expression ∆Tb = 0.014 C0.75 P0.1exp(0.034 C)°C, where C isthe soluble solids content in °Brix and P is the pressure in mbar. The specificheat is a function of the mass fraction of water according to the equation:

A A1 2102 22 103 16= =. . m m2 2

E

V VwV

=+

= =1 2 13 1258090

1 62,

.

A m = 102 7. m2

P T

P t

C = = °

= = °

2452 126 8

815 101 01 1

mbar C

mbar C

.

.

P t

P t

b1 1

2 2

815 94 0

196 59 7

= = °

= = °

mbar C

mbar C

.

.

ˆ . .C XP WATER= + ⋅°( )0 84 3 34 kJ kg C



Evaporation 659

There is a vacuum pressure of 460 mm Hg in the evaporation chamber ofthe second effect, so its pressure is P2 = 300 mm Hg.

Properties of the saturated steam:

Figure 18.P2 shows a diagram of the double-effect evaporator that worksunder forward feed.

Global and component mass balances:

FIGURE 18.P2Forward feed double-effect evaporator.

Wa

P1

t 1

P2

t 2

Wv

PC T

Wv

C

V1V 2

V2

P2 t 2

V1

P1 tb 1 V2

P2 tb 2

W C

II

P t 11

V1

P T

P T

h

Hw

w

w

w

= = = °

=

= =

2 4 2353 125 5

527

2713 2186

. .

ˆ

ˆ

atm mbar C

kJ kg

kJ kg kJ kgλ

P t

h

H

b

V

V V

2 2

2

2 2

300 400 75 8

317

2637 2320

= = = °

=

= =

mmHg mbar C

kJ kg

kJ kg kJ kg

.

ˆ

ˆ λ

3480 1 2= + +w V VC

3480 0 15 0 72( )( ) = ( ). . wC




Yielding:

Initially it is supposed that V1 = V2 = 1377.5 kg/h, which makes one assumethat the composition of the stream w2 is X2 = 0.248, corresponding to a contentof 24.8 °Brix.

The boiling point rises are calculated by the equation given in the datasection. The concentration used is that of the stream leaving each effect:

1st effect:

2nd effect:

obtaining:

The boiling temperature of the second effect is:

P1 should be known in order to obtain ∆Te1, but it can be estimated. In caseP1 = 1000 mbar, the boiling point rise is ∆Te1 = 0.7°C.

The specific heat of each stream is obtained from the equation given in thestatement of the problem:

For CA = 15°Brix:

For CC = 75°Brix:

For C1 = 24.8°Brix:

w V V C = + =725 27551 2kg h and kg h

C P1 124 8= ° =. ?Brix

C P2 272 0 400= ° =. Brix mbar

∆ ∆T P Tb b1 1

0 1

20 36 7 3= ( ) = °. ..

C

t t Tb b2 2 2 75 8 7 3 83 1= + = + =∆ . . . º C

ˆ .C PA = ⋅°( )3 68 kJ kg C

ˆ .C PC = ⋅°( )1 78 kJ kg C

ˆ .C P1 3 35= ⋅°( )kJ kg C



Evaporation 661

In order to perform the calculation procedure, it is supposed that the areasand the heat flows transferred through these exchange areas are equal forboth effects, complying with:

Hence,•

Q/A = 29,858 W/m2.Temperature te1 is obtained from the heat transfer rate equation for the

second effect:

and the boiling point in the first effect is:

The properties of saturated steam can be found using the temperature tb1 =106.4°C from the saturated steam tables:

If the boiling point rise in the first effect is recalculated using pressure P1 =1271 mbar, a slightly different value is obtained, so the same boiling pointrise obtained before is taken.

Application of enthalpy balances to both effects yields:

1st effect:

2nd effect:

˙ . . . .QA

T t T T

U +

U

+

b b b=− − −

=− − −( )

2 1 2

1 2

1 1125 5 75 8 0 7 7 3

11625

11280

∆ ∆ º C

m º C W

2

˙

.

Q A=U t t

t

e

b

2 1 2

1 106 4

−( )= °C

t t Tb b1 1 1 106 4 0 7 107 1= + = + = °∆ . . . C

t P

h

H

b

V

V V

1 1

1

1 1

106 4 1271

447

2685 2238

= ° =

=

= =

.

ˆ

ˆ

C mbar

kJ kg

kJ kg kJ kgλ

2186 2238 2 1 0 7

3 35 107 1 106 4 3480 50 106 4

1

1

( ) = + ( )( )(+ ( ) −( ) − ( ) −( )

w V

w

V . .

. . . .

2238 2319 2 1 0 7

725 1 78 83 1 75 8 3 35 107 1 75 8

1 2

1

( ) = + ( )( )(+ ( )( ) −( ) − ( ) −( )

V V

w

. .

. . . . . .




With the equations of the balances:

a four-equation system with four unknowns is obtained that, when solved,yields:

These new values are used to calculate the soluble solids content in thestream w1, obtaining 24.6°Brix, which allows calculation of the specific heatand the boiling point rise: CP1 = 3.36 kJ/(kg ·°C) and ∆Tb1 = 0.7°C.

The value of the areas through which heat is transferred can be obtainedby the equations of heat transfer rate through such areas:

1st effect:

2nd effect:

The mean area is Am = 31.70 m2. Since these areas are different by morethan 2%, the calculation procedure should begin again, rectifying the inter-mediate temperatures t1 and tb1 (Equations 18.49 to 18.51), as the other tem-peratures do not vary:

With this temperature, it is obtained that:

Since they are different, the mean value of each one is taken, so:

w w V

V V

A1 1

1 2 2755

= −

+ =

w w

V V

V = =

= =

1726 2120

1360 1395

1

1 2

kg h kg h

kg h kg h

1725 2186 3600 1 625 125 5 107 11( ) × ( ) ( ) = ( ) −( ). . .A

1360 2238 3600 1 28 106 4 83 2

35 05 28 35

2

1 2

( ) × ( ) ( ) = ( ) −( )= =

. . .

. .

A

A A m m2 2

t A A t

t A A t

b m b

m

1 2 1

1 1 1

83 1 106 4 83 1 103 9

125 5 125 5 107 1 105 2

− = −( )( ) = °

− = −( )( ) = °

. . . .

. . . .

C

C

tb1 105 2 0 7 104 6= − = °. . . C

t tb1 1104 2 104 9= ° = °. .C and C



Evaporation 663

The new enthalpies for 104.2°C can be found in the saturated steam tables:

The four-equation system stated above is solved again, using the new valueof λV1, yielding:

The areas are recalculated from the rate equations:

(a) Mass flow rate of steam from the boiler:

Economy of the system:

b) Area per effect:

c) 1st effect:

2nd effect:

t P

h

H

b

V

V V

1 1

1

1 1

104 2 1177

437

2682 2245

= ° =

=

= =

.

ˆ

ˆ

C mbar

kJ kg

kJ kg kJ kgλ

w w

V V

v = =

= =

1720 8 2118 6

1361 4 1393 6

2

1 2

. .

. .

kg h kg h

kg h kg h

A A1 231 22 31 43= =. . m m2 2

wV = 1720 7. kg h

EV V

wV

=+

= =1 2 27551720 8

1 6.

.

Am = 31 33. m2

P T

P t

C = = °

= = °

2353 125 5

1177 104 91 1

mbar C

mbar C

.

.

P t

P t

b1 1

2 2

1177 104 2

400 83 1

= = °

= = °

mbar C

mbar C

.

.




18.3

Tamarind is an important culinary condiment used as an acidifying ingre-dient. Due to the cost of transport, it is convenient to obtain tamarind as aconcentrated juice through an evaporation stage. An Indian industry desiresto obtain 1000 kg/h of a 62°Brix of concentrated juice beginning with 10°Brix.For this reason, the possibility of installing a single effect with mechanicalcompression of steam or a double effect operating under forward feed isstudied. The global heat transfer coefficients of the first and second effectare 2100 and 1750 W/(m2 °C), respectively. The food is at 22°C, while the62°Brix juice cannot withstand temperatures higher than 70°C. The industryhas a saturated steam stream at 1.8 kg/cm2 used to carry out the juiceconcentration. Calculate: (a) the 10 °Brix juice flow that can be concentratedand the consumption of steam at 1.8 kg/cm for both options; (b) the com-pression power, for the first option, if the isentropic performance of thecompressor is 88%; (c) the more profitable option, if the cost of each m2 ofevaporator is $22 U.S., of each kW of compression power is $4 U.S., of eachkW.h is $0.08 U.S., and of each kg of steam at 1.8 kg/cm2 is 1 cent. Considerthat the amortization of the equipment is estimated in 1 year.

Data and notes: juices with soluble solids content lower than 18°Brix donot present an appreciable boiling point rise. The plant functions 300 daysa year, 16 hours daily. Specific heat of the tamarind juices:

where XS is the percentage of soluble solids and T is the temperature inKelvin.

Single effect evaporation with mechanical compression:

temperature in the evaporation chamber is t1 = 70°C, while the boilingtemperature of pure water is obtained from the graph of Dühring

e1 = 66°C (∆Tb1 = 4°C).The following conditions can be obtained from saturated steam tables:

ˆ . . .C T XP S= + × −( ) ⋅( )−4 18 6 84 10 0 05035 kJ kg K

P T

h

H

C C

w

w

= = = °

=

= =

1 8 116 3

488

2700 2212

. .

ˆ

ˆ

atm 1765 mbar C

kJ kg

kJ kg kJ kgw λ

t

h

H

b

V

V V

1

1

1 1

66 262 0 27

276

2619 2343

= ° = =

=

= =

C P mbar atm

kJ kg

kJ kg kJ kg

1 .

ˆ

ˆ λ


The diagram of this type of installation corresponds to Figure18.6. The

(Figure 18.3): t


Evaporation 665

The vapor that leaves the evaporation chamber is reheated to a tempera-ture of t1 = 70°C, with an enthalpy:

The compression of the vapor that leaves the evaporation chamber is an

to point 2 at the 1765 mbar isobar. The conditions of point 2 correspondingto the vapor outlet at the compressor can be obtained by means of the graph:

Vapor (V) with these conditions is reheated and mixed with the vaporcoming out of the boiler (w′V ) that is saturated at the same pressure, yieldinga reheated vapor mixture. In order to avoid this, part of the condensate (wR)is recirculated to obtain a saturated vapor at point 3 (wV), which is fed to thecondensation chamber of the evaporator (Figures 18.6 and 18.7).

The specific heats of the food and concentrate streams are calculated attheir respective concentrations:

Food:

Concentrate:

Mass balances:

obtaining:

If an enthalpy balance is performed in the evaporator, it is possible toobtain the amount of the vapor wV that gets into the evaporation chamber:

ˆ ˆ ˆ . .H H C T V PV b1 1 1 2619 2 1 4 2627 4= + ⋅ = + ( )( ) =∆ kJ kg

P P t

H

C2 2

2

1765 270

3009

= = = °

=

mbar C

kJ kg

ˆ

C C A PA= ° = ⋅°( )10 3 88Brix kJ kg Cˆ .

C C C PC= ° = ⋅°( )62 2 52Brix kJ kg Cˆ .

w w V w V

w X w X w w

A C A

A A C C A C

= + = +

= =

1000

0 1 0 62. .

w V A = =6200 5200kg h kg h


isentropic process (Figures 18.6 and 18.7) from point 1 at a 262 mbar isobar



When performing the mass and enthalpy balances around point 3, wherethe streams of steam from the boiler (w′V), compressed vapor (V2 = V), andrecirculated condensate (wR) get together, it can be obtained that:

Solution of this system yields:

The area of the evaporator is obtained from the heat transfer rate equation:

Hence, the area is:

The theoretical compression power is obtained from the expression:

The real power is obtained by dividing the theoretical power by the isen-tropic performance:

2212 5200 2343 2 1 4 100 2 52 4 6200 3 88 22 66

6010 2

w

w

V

V

= ( )( ) + ( )( ) + ( )( )( ) − ( )( ) −( )=

. . .

. kg h

6010 2 5200

6010 2 2700 2700 488 5200 3009

.

.

= ′ + +

( )( ) = ′ ( ) + ( ) + ( )( )w w

w w

V R

V R

′ =

=

w

w

V

R

83 8

726 4

.

.

kg h

kg h

w U A T tV Cλw = −( )1 1

6010.2 2212 3600 2.1 116.5 70( )( ) ( ) = ( ) −( )A

A = 37 82. m2

Pow V H H

Pow s

Pow

T

T

T

( ) = −( )( ) =( )( ) −( )( ) =

ˆ ˆ

.

.

2 1

5200 3600 3009 2627 4

551 2

kg kJ kg

kJ kg

PowR

( ) = ( ) =551 2 0 88 62 4. . . kJ kg



Evaporation 667

The annual operation cost is calculated from the expression:

where CA, CP, CV, and CPOWxh are the cost per m2 of evaporator area, cost per

compression power installed, cost of the waste of steam coming from theboiler, and operation cost of the compressor, respectively, while hT denotesthe annual operation hours.

Operation hours:

The annual cost is:

Hence: C = U.S. $250,713.

Forward feed double-effect evaporator:

vapor at a pressure PC = 1.8 atm = 1765 mbar enters the condensationchamber of the first effect, while in the second effect the boiling temperatureof the 62°Brix juice is t2 = 70°C, so the boiling temperature of pure water istb2 = 66°C (∆Tb2 = 4°C).

The following is obtained from the saturated steam tables:

The global and component mass balances yield:

C C A C Pow C h C w hA P R POWxh T V V T= + ( ) + ′

hT = ( )( ) =16 300 4800 h

C = ( )( ) + ( )( ) + ( )( )( ) + ( )( )( )22 37 82 4 626 4 0 08 626 4 4800 0 017 83 8 4800. . . . . .

P T

h

H

C C

w

= = = °

=

= =

1 8 1765 116 3

488

2700 2212

. .

ˆ

ˆ

atm mbar C

kJ kg

kJ kg kJ kg

w

w λ

t P

h

H

b

V

V V

2 2

2

2 2

66 262 0 27

276

2619 2343

= ° = =

=

= =

C mbar atm

kJ kg

kJ kg kJ kg

.

ˆ

ˆ λ

w V V A = + =6200 52001 2kg h kg h


This type of evaporator is similar to the one presented in Figure 18.P1. A



Initially it is supposed that V1 = V2 = 2600 kg/h, allowing the concentrationof stream w1 to be obtained, leaving the first effect: C1 = 17.2°Brix, pointingout that there will not be an appreciable boiling point rise (∆Tb1 = 0).

The specific heats of the different juice streams are:

Food:

Stream:

Concentrate:

To perform the calculation process, it is supposed that the areas and heatflows transferred through such exchange areas are equal in both effects,complying with:

Thus,•

Q/A = 44,196 W/m2.Temperature te1 is obtained from the heat transfer rate equation in the

second effect:

and the boiling temperature in the first effect is:

It is possible to find the properties of saturated steam, using the temper-ature tb1 = 95.3°C, from saturated steam tables:

C C A PA= ° = ⋅°( )10 3 88Brix kJ kg Cˆ .

C C P1 117 2 3 75= ° = ⋅°( ). ˆ .Brix kJ kg C

C Brix kJ kg CC = ° = ⋅°( )62 2 52ˆ .C PC

˙ .QA

T t T

U +

U

=

+

b b=− − − −( )

2 2

1 2

1 1116 3 66 4

12100

11750

∆ º C

m º C W

2

˙ .Q A U t t tb e= −( ) = °2 1 2 1 95 3 C

t t Tb b1 1 1 95 3 0 95 3= + = + = °∆ . . C

t P

h

H

b

V

V V

1 1

1

1 1

95 3 855

399 3

2668 2268 7

= ° =

=

= =

.

ˆ .

ˆ .

C mbar

kJ kg

kJ kg kJ kgλ



Evaporation 669

Thus, applying enthalpy balances to both effects:

1st effect:

2nd effect:

besides the equations of the balances:

A four-equation system with four unknowns is obtained, which, whensolved, yields:

The composition of stream w1 is recalculated, obtaining C1 = 17°Brix, sothere is no boiling point rise in the first effect (∆Tb1 = 0), and its specific heatis almost the same as the one calculated before.

The value of the areas through which heat is transferred can be obtainedfrom the heat transfer rate equations through such areas:

1st effect:

2nd effect:

and a mean area Am = 42.10 m2. Since these areas are different by more than2%, the calculation procedure should begin again, rectifying the intermediatetemperatures t1 and tb1, since the other temperatures remain the same:

2212 = 2268.7 3.75 0 6200 3.88 22 95.3( ) ( ) + ( )( ) − ( )( ) −( )w V wV 1 1

V V w1 2268 7 2343 2 1 4 1000 2 52 70 66 3 75 95 3 662 1. . . . .( ) = + ( )( ) + ( )( ) −( ) − ( ) −( )(

w w V V

V V

A1 1 1

1 2

6200

2600

= − = −

+ =

w w

V V

V = =

= =

3425 3637 8

2562 2 2637 8

1

1 2

kg h kg h

kg h kg h

.

. .

3425 2212 3600 2 16 116 3 95 31( ) ( ) ( ) = ( ) −( ). . .A

2562 2 2268 7 3600 1 28 95 3 702. . . .( ) ( ) ( ) = ( ) −( )A

A A1 247 72 36 47= =. . m m2 2




Since ∆Tb1 = 0, it is complied that te1 = t1.Since they are different, the mean value of both is taken, thus: te1 = t1 =

92.3°C.The new enthalpies for the temperature 92.3°C can be found in the satu-

rated steam tables:

The four-equation system stated above is solved using the new value ofλV1, obtaining:

The areas are recalculated from the rate equations:

Thus, the mean area by effect is:

The annual cost is obtained from the expression:

Hence:

According to the result obtained, it is better to install one effect with vaporrecompression, since the annual cost is lower.

t A A t

t A A t

b m b

m

1 2 1

1 1 1

70 95 3 70 92 0

116 3 116 3 95 3 92 5

− = −( )( ) = °

− = −( )( ) = °

. .

. . . .

C

C

t P

h

H

b

V

V V

1 1

1

1 1

92 3 770

387

2662 2275

= ° =

=

= =

.

ˆ

ˆ

C mbar

kJ kg

kJ kg kJ kgλ

w w

V V

V = =

= =

3405 4 3632 3

2567 7 2632 3

1

1 2

. .

. .

kg h kg h

kg h kg h

A A 1 241 52 41 58= =. .m m2 2

A m = 41 55 2. m

C C A w C hA m V V T= +2

C

C

= ( ) ( ) + ( )( )( )=

22 2 41 55 3405 4 0 017 4800

279 709

. . .

$ ,U.S.


Cap 18 Evaporation

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