Simulation of Double-Effect Evaporator for Concentrating Orange Juice

Australian Journal of Engineering Research

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Australian Society for Commerce Industry & Engineering

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Simulation of Double-Effect Evaporator for Concentrating

Orange Juice

Sara Fatima (Corresponding author)

Department of Chemical Engineering, Aligarh Muslim University

Aligarh 202002, Uttar Pradesh, India

Tel: 91-976-048- 8956 E-mail: [email protected]

Aishatul Bushra



E-mail: [email protected]

Raunaq Hasib



E-mail: [email protected]

The research was part of MTech dissertation topic of author Sara Fatima.

Abstract

The aim of this work is to carry out a steady state simulation of double effect evaporator for

concentrating orange juice. Besides empirical correlations for overall heat transfer coefficient and

boiling point rise of orange juice are developed. A new arrangement of model equations (Reduced

Equations model) is also developed to facilitate the solution. The model equations are solved using

fsolve, Newton Raphson, Relaxation Strategy and Reduced Equations techniques in MATLAB and

results so obtained compared with experimental data available in Moresi and Angletti [1]

. First linear

technique and then non-linear technique has been used to solve the simulation problem.

Keywords: Double-effect evaporator, Orange juice, Simulation, Newton Raphson, Non-linear Model,

Linear Model, Relaxation Strategy

1. Introduction

Fruits are perishable. Hence they need to be processed in time to increase their shelf-life, preserve

nutritional value and hence reduce storage, packaging and shipping costs. Oranges are the most

produced fruit in the world, comprising 22.5% of global fruit production. Oranges for orange juice

production are primarily grown in US and Brazil, which together produce 87% of the frozen juice

concentrate. Other major producing countries include Mexico, China, India and Spain. Orange juice is

marketed for its minimal use of additives and preservatives.

Considerable amount of work has been present in open literature on multiple effect evaporators but due

to their non-linearity, they are a difficult system to simulate. Not much work has been reported in open

literature on orange juice (except Moresi and Angletti [1]

but model proposed by them is complex).

Hence in present study, a simpler method of simulation of orange juice production has been proposed.

The evaporator used in this study for orange juice concentration is a TASTE (Thermally Accelerated

Short Residence Time) Falling Film evaporator. Figure 1 shows a double effect evaporator in forward

feed arrangement for concentrating orange juice. Orange juice manufacturers use falling film

evaporators. They are especially popular in the food industry where many substances are heat sensitive.

A thin film of the product to be concentrated trickles down inside of heat exchanging tubes. Steam

condenses on the outside of the tubes supplying the required energy to the inside of the tubes. Juices


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are heat sensitive and their viscosities increase significantly as they are concentrated. Small solids in

the juices tend to cling to the wall and cause fouling. Juice evaporations are usually performed in

vacuum to decrease boiling temperature. High flow circulation rates avoid build-up on the tube walls.

Vacuuming the product side decreases boiling point of the liquid, thereby, the product boils at relatively

low temperatures, thus minimizing the heat exchange.

LINEAR TECHNIQUE

1.0 Mathematical Model

Steady state model equations for the system considered (Fig. 1) are as follows (Koko and Joye[5]

).

First Effect

Mass Balance: F- V1- L1=0 (1)

Heat Balance: F[h(TF,XF)-h(τ 1,x1)]+V0λ0-V1[H(τ 1)-h(τ 1,x1)]=0.0 (2)

Heat Transfer Rate:U1R1 A(T0- τ 1)-V0λ0=0.0 (3)

Second Effect

Mass Balance: L1+ (- 1)L2 + (-1) V2=0 (4)

Heat Balance: L1[h(τ 1,x1)-h(τ 2,x2)]+V1[H(τ 1)-h(T1)]-V2[ H(τ 2)-h(τ 2,x2)]=0.0 (5)

Heat Transfer Rate: U2R2 A(T1- τ 2)- )]-V1[H(τ 1)-h(T1)]=0.0 (6)

The variable area Ai has been written as ARi to allow for variable area situation, where Ri is the ratio of

area of the ith

stage to A.

Some of the assumptions are Forward feed steady state operation, unequal areas of the two effects,

equilibrium conditions achieved in each effect, temperature of boiling liquor is equal to temperature of

solvent vaporised, vapour leaving each effect is superheated, BPR is present, variation in heat

capacities with temperature and composition, variation in latent heat of vaporization of steam with

temperature, no reaction taking place in each effect, homogeneous temperature and composition inside

each effect and negligible heat losses to the surroundings.

Linear method is an iterative method proposed in the literature (Koko and Joye[5]

and Lambert, Koko

and Joye[6]

) for the design of multiple-effect evaporators. This method is based on the Linearization of

set of non-linear equations. The non-linearity in Eqn. (2), (3), (5) and (6) arise from two sources: the

first source in the cross product of A and T in Eqn. (3) and (6) and the other is the non-linear nature of

enthalpy functions, U’s and BPR’s with respect to temperature and concentration. The hurdle of non-

linearity can be eliminated by redefining the cross product term (i.e. AT) as a new variable. The second

source can be easily handled by iteration. In each iteration the U’s and BPR’s are assumed constant ( i.e.

at any Ti the coefficients are defined), thereby yielding linear equations; which can be put in matrix

form and solved by any Linear technique such as Gauss elimination. Of course, iterations will be

required if the new values of Ti are significantly different from old. Because the coefficients are not

strong functions of T, the equations don‟t change very much from iteration to iteration and convergence

is very rapid.

1.2 Degrees of Freedom

Total number of independent equations describing the double effect evaporator is 6. Total number of

variables are 10 (F, V1, L1, V0, AT0, Aτ1, L2, V2, Aτ2, A). Thus four boundary conditions which can be

specified are:


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B.C. 1 Given F (7)

B.C. 2 L2=F*X/x2 (8)

B.C. 3 A τ 0 = G1A (9)

B.C. 4 A τ2 = G2A (10)

where, G1 and G2 are constants.

This result is a set of 10 algebraic equations with constant coefficients. The resulting equations can be

put into matrix form as shown below:

[ ]

𝐵

[ 𝐹𝐿2 ]

𝑋

[ 0𝑉0𝐹 1𝑉1𝐿1 2𝑉2𝐿2 ]

The above matrix and vector can be put as AX = B, and solved by Gauss Elimination as X=B\A.

1.3 Algorithm

The algorithm for solving the evaporator series problem consists of the following steps:

1) The governing equations are linearized as described as above. The resulting equations are:

First Effect

Mass Balance: F+ (- 1)V1+ (- 1) L1=0 (11)

Heat Balance: aF+ bV0-cV1=0.0 (12)

Heat Transfer Rate: dAT0+eAτ 1-fV0=0.0 (13)

Second Effect


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Mass Balance: L1+ (- 1)L2 + (-1) V2=0 (14)

Heat Balance: gL1+ hV1- iV2=0.0 (15)

Heat Transfer Rate: hV1- jA τ 1+ lA τ 2 + kA=0.0 (16)

Where,

a = [h(TF,X)-h(τ 1,x1)]

b= λ0

c=[H(τ 1)-h(τ 1,x1)]

d= U1R1

e= - d= -U1R1

f= λ0

g=[h(τ 1,x1)-h(τ 2,x2)]

h=[H(τ 1)-h(T1)]

i=[ H(τ 2)-h(τ 2,x2)]

j=U2R2

k= U2R2 BPR1

l=U2R2

2) Product composition, x2, is assumed (it is unknown for the simulation problem) and then

temperature and composition in each effect is initialized by apportioning the vapor flows and

temperature drops. Since x2 is bounded by xf and 1, a reasonable starting value should be higher than

xf. The remaining variables can be initialized as in traditional trial-and error procedure also used by

Lambert and Koko[6]

. If U is a constant or some function of T, then equation (3.9.19) may be used to

initialize the temperature in each effect.

∆Ti= ∆Ttot / N (17)

Where,


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∆Ttot = T0-T2-BPR1-BPR2 (18)

∆Ttot = Temperature difference between externally supplied steam and final effect solution.

∆Ti = Temperature difference between externally supplied steam and ith

effect solution

N = Effect number

T0 = Inlet saturated steam temperature

T2 = Outlet orange juice temperature

BPR1 = Boiling point rise in Effect I

BPR2 = Boiling point rise in Effect II

3) Further, it is assumed that each effect produces the same amount of vapor, Vi. Hence

∆Vi= ∆Vtot / N (19)

∆Vi = Vapour flow rate across ith

effect

∆Vtot = Total vapour flow rate across the evaporator

4) The temperature and composition (x, T) are initialized and the coefficients of the variables are

found. These coefficients are the enthalpy values, boiling point rise, and overall heat transfer

coefficients, all of which are functions of temperature and composition, or specified as input

parameters.

5) These coefficients are filled in the 10x10 matrix as below. In matrix, A, first six rows are

coefficients from the model equations and the last four rows are the four boundary conditions. These

are nothing but input variables.

6) The coefficient matrix is solved by any Linear technique (such that AX=B) to obtain the

values of the variables (L, V, AT and A) for each effect.

7) If computed Area > A2, product concentration, x2 is incremented by 0.1 and steps (2) through

(7) is repeated.

8) Do step (7) until x2>= 0.6, i.e. till it reaches the expected product concentration.

1.4 Simulation model

The linear simulation model presented above has been solved using MATLAB. Linear method is an

alternative to the traditional trial-and-error non-linear method used in solving MEE problems. This

linear system has been solved using GUASS ELIMINATION method. Also, Koko and Joye[5]

have

solved the problem as a design problem but in this study, it has been solved as a simulation problem.

1.5 Constitutive Relationships

1. Specific Heat:

The equation for specific heat of orange juice, Cp is obtained from Moresi and Angletti[1]

:-

Cp= 4.186-2.679 x (J/g.K) (20)

where, x = weight fraction of orange juice

2. Enthalphy and Latent Heat of Vaporization:

Enthalpy of saturated and superheated steam, H(τ) and Latent heat of vaporization of steam, λ0 were


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obtained from work of Zain and Kumar[4]

:-

H(τ)=4.154T+2.0125x10-4

T2+1.62(τ-T) +2.085*10

-4 (τ

2-T

2 )-0.3747*10

-4 (τ

3-T

3)+λ0 (21)

where,

H(τ)=Enthalpy of Vapor in an Effect

λ0=Latent Heat of Vaporisation of Steam

λ0=-80.345 T -21025.8/ T +2049.123/( T)0.5

-4213.519ln( T )+0.0918 T2 -1.04*10

-4 T

3 +8597.953 (22)

3. Solution Temperatures:

Solution temperatures in Effect I and II, τ 1, and τ 2 are calculated using Boiling point rise data available

in Moresi and Spinosi[3]

. A fifth order Polynomial is developed using MS-Excel Scatter Plot. First ∆Tb,

i.e. BPR is calculated at various concentrations of orange juice using following formula. This formula

uses first principles (Moore, 1972):

∆Tb = (RTo2/∆Hv) . (MA/ MB ).(yB/(1-yB)) (23)

Here solvent is water and solute is orange juice. Boiling point rise is fit in the following equation:

BPR = ax + bx2 + cx

3 + dx

4 + ex

5(°C) (24)

BPR so obtained is:-

BPR = - 90.53x5 + 55.76x

4 - 11.81x

3 + 4.5x

2 + 2.583x(°C), (25)

τ = T + BPR (26)

Data for calculation of BPR is given in Table 2.

4. Heat Transfer Coefficients:

Heat transfer coefficient, U, is calculated using temperature and composition data available in Moresi

and Spinosi[3]

using Multiple Linear Regression in MS-Excel. Overall heat transfer coefficient is fit in

the following exponential equation:

U=τ

bx

c (27)

Equation for U so obtained is:-

U=12.21*103τ

-1.304x

-2.06543 (28)


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Data for calculation of U is given in Table 3.

NON - LINEAR TECHNIQUE

2.0 Mathematical Model for Non-linear Technique

Steady state model equations for this system are (Holland, 1975[11]

):

First Effect

f1 =F[h(TF,XF)-h(τ 1,x1)]+V0λ0-(F-L1)[H(τ 1)-h(τ 1,x1)] (29)

f2 =U1A1(T0- τ 1)-V0λ0 (30)

f3 =m(x1)T1+b(x1)- τ 1 (31)

f4 =FX-L1x1 (32)

Second Effect

f5 =L1[h(τ 1,x1)-h(τ 2,x2)]+(F-L1)[H(τ 1)-h(T1)]-(L1 –L2 )[ H(τ 2)-h(τ 2,x2)] (33)

f6 =U2A2(T1- τ 2)- )]-(F-L1)[H(τ 1)-h(T1)] (34)

f7 =m(x2)T2+b(x2)- τ 2 (35)

f8 =FX-L2x2 (36)

The assumptions are same as for linear technique.

2.1 Degrees of freedom

Specification: F, X, Tf, T0 (or P0), x2, T2 (or P2), A (or A1, A2)

To find: V0, τ 1, T1, x1, L1, τ 2, x2, L2

Number of equations, E = 8

Number of varaibles, V = 8

Degrees of Freedom, F = V-E=0

Hence, in simulation problem, there are 8 equations in 8 unknowns for the double effect evaporator.

2.2 Scaling

A scaling procedure is used to reduce the magnitude of the terms appearing in the functional equations

and matrices. For computational purposes, it is desirable to have terms with magnitudes near unity.

Scaling has been implemented using the same steps as given in Holland, 1975[11]

for multiple effect

evaporators.

2.3 Algorithm

The developed model equations have been solved using following procedure:


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1) Read input data. The values of the eight input variables (F, X, Tf, T0, x2, T2, A and A2) are set using

experimental data.

2) Scaling is implemented.

3) Iteration number is set, n=0.

4) Create vector, x0 of eight unknown variables as:-

X0= [v0, u1*, u1, x1, l1, u2*, x2, l2]

Set the initial guess, X0 of the eight unknown variables.

5) Calculate the eight scaled functional expressions (g1, g2, g3, g4, g5, g6, g7 and g8) using above values.

Fill vector, fn :

fn= [g1, g2, g, g4, g5, g6, g, g8]

6) Calculate the Jacobean matrix, Jn as follows:

J

[ ∂g1∂v0

∂g1∂u1

∗

∂g1∂u1

∂g1∂x1

∂g1∂l1

∂g1∂u2

∗

∂g1∂x2

∂g1∂l2

∂g2∂v0

∂g2∂u1

∗

∂g2∂u1

∂g2∂x1

∂g2∂l1

∂g2∂u2

∗

∂g2∂x2

∂g2∂l2

∂g3∂v0

∂g3∂u1

∗

∂g3∂u1

∂g3∂x1

∂g3∂l1

∂g3∂u2

∗

∂g3∂x2

∂g3∂l2

∂g4∂v0

∂g4∂u1

∗

∂g4∂u1

∂g4∂x1

∂g4∂l1

∂g4∂u2

∗

∂g4∂x2

∂g4∂l2

∂g5∂v0

∂g5∂u1

∗

∂g5∂u1

∂g5∂x1

∂g5∂l1

∂g5∂u2

∗

∂g5∂x2

∂g5∂l2

∂g6∂v0

∂g6∂u1

∗

∂g6∂u1

∂g6∂x1

∂g6∂l1

∂g6∂u2

∗

∂g6∂x2

∂g6∂l2

∂g7∂v0

∂g7∂u1

∗

∂g7∂u1

∂g7∂x1

∂g7∂l1

∂g7∂u2

∗

∂g7∂x2

∂g7∂l2

∂g8∂v0

∂g8∂u1

∗

∂g8∂u1

∂g8∂x1

∂g8∂l1

∂g8∂u2

∗

∂g8∂x2

∂g8∂l2 ]

7) Calculate ΔXn = -fn / Jn

Jn ΔXn=-fn

where,

Jn= Jacobean Matrix

ΔXn = Xn+1 – Xn


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ΔXn= [ Δv0 Δu1*Δu1 Δ x1 Δl1 Δu2* Δx2 Δl2]

fn=[ g1 g2 g3 g4 g5 g6 g7 g8]

8) Check the convergence criteria of the above Jacobean matrix

i.e. |ΔXn | = |Xn+1 - Xn| < e.

If |ΔXn | > e, convergence has not been met. Assign Xn+1 = Xn + ΔXn. Repeat steps (6) to (7) till

convergence is met for the given initial guess. If the convergence criterion is not met even after

1000 iterations, try another initial guess and repeat steps (4) through (8). Else if the convergence

criteria has been met in step (8), read the values of scaled variables from Xn+1

9) Calculate the un-scaled variables as:

V0= v0 F, τ1= u1*T0, T1=u1*T0, τ2=u2* T0,

L1=l1*F, L2=l2 F

10) Calculate Boiling point rise and Overall Heat Transfer coefficient in Effect I and II.

11) The above procedure is repeated until product concentration, x2 is near experimental product

concentration, i.e., the values are in good agreement with the experimental values reported in literature

(Moresi and Angletti [1]

).

2.4 Constitutive Relationships

The constitutive relationships used are same as in linear technique.

2.5 Reduced Equations Model

Careful study of Eqn. (29) to (36) reveals that the eight equations may be combined together into three

equations, namely Eqn. (29), (33) and (34). Correspondingly, the three independent variables are T1, L1

and L2. This is because that if these three variables are known, then the remaining five variables, x1, τ1,

V0, x2 and τ2 can be easily computed in this order using Eqn. (32), (31), (30), (36) and (35) respectively.

Consequently in this new model, these five equations have been eliminated from the set of non-linear

equations. This idea of new arrangement of model equations was taken from Zain and Kumar [4]

. This

reduced model is easy to solve.

2.6 Simulation model

The non-Linear simulation model presented above has been solved using MATLAB. First, fsolve

function was used to solve the non-linear set of equations. Fsolve is a built-in function available in

MATLAB to solve set of non-linear equations. It uses TRUST REGION DOGLEG method of

optimization (based on sum of least squares).

Then Newton Raphson technique was used to solve the same non-linear set of equations. First, a

convergence criterion was applied to check whether the method is converging for particular initial

guess or not. Convergence Criteria used was that the absolute value of the function of each of the

equations is less than 1.0e-15 than expected value (i.e. tolerance<1.0e-15).

Then Relaxation Strategy was used to solve the same non-linear set of equations; using under-relaxed

model of relaxation:

Xj+1

= α Xj-1

+ (1-α) Xj


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where,

α = relaxation factor (0.0< α <1.0)

j = jth

iteration of Newton Raphson

Then, Reduced Equations model was used to solve the same non-linear set of equations.

3.1 Results and Discussion for Linear and Non-linear technique

Results for simulation are shown in Table 4 and 5 for linear technique. Results for simulation are

shown in Table 6 and 15 for non-linear technique.

In linear-simulation, the area of the two effects had to be constantly monitored to get near experimental

conversion. In non-linear-simulation, different parameters affect simulation for concentrating of orange

juice:

Selection of relaxation parameter, α: The value of α at which convergence occurred has been found

to be 1.0e-07. This was obtained by trial and error. Values lower than this give complex roots and

values higher than this give negative result.

Constitutive Relationships: Two different plant sizes (Feed rate=1200kg/h and 2400kg/h) are used as

input. The BPR correlation given in Moresi and Angletti [1]

did not give reliable results. So a BPR

correlation was derived using Moore formula from Moresi and Spinosi [3]

. Also constitutive

relationships for specific heat and latent heat of vaporization of steam obtained from Zain and Kumar[4]

are helpful in giving reliable results.

Order of Variables: As many as forty possible different permutations were tried. In some cases, it was

observed that the Jacobean becomes ill-conditioned and hence no solution obtained. Hence changing of

“Order of Variables” did not yield valuable results. Finally the “Order of Variables” proposed by Zain

and Kumar[4]

was used and the results so obtained have been found to be in good agreement with

experimental data. Therefore attention must be given to the order of variables.

Single vs. Double Precision: Initially, the simulation was done using single precision arithmetic but

Newton Raphson did not converge. Then simulation was carried out using double precision arithmetic

and the algorithm converged.

Derivatives: Zain and Kumar [4]

have used the following method to calculate the derivatives:

∂f/∂X=(f(X+ε)-f(X))/ε

where,

ε = step size

But in this study, the Jacobean has been calculated using Jacobean () function given in MATLAB. This

function saves tedious manual differentiation of model equations and is also fast and accurate.

Initial Guess: Zain and Kumar[4]

have used the initial guess of 1.0 for all the unknown scaled variables.

But other initial estimates also work for caustic soda. This technique was extended to the system of

orange juice. The initial guess was kept as 1.0 but good results were not achieved. Hence in simulation

using Newton Raphson and Relaxation Strategy, other initial guesses (between 0 and 1) were tried and

they worked. For these two cases, initial guess was found to be between 0.7-0.8.

No. of Iterations: Newton Raphson and Relaxation Strategy gave similar results but latter required less

number of iterations.

Reduced Equations: The following observations were made during this numerical simulation:

i. Number of iterations was significantly reduced and hence the solution required less

computation time.

ii. The value of initial guess was 0.45-0.48 at which convergence occurred.

The advantages of linear method over non-linear method are:


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1) The algorithm can be extended to more complex systems (4 to 5 effects) for orange juice

concentration.

2) The algorithm is unusually efficient and stable, with very favorable convergence

characteristics shown even where non-linear method fails to converge.

3) It handles non-linearity simply and efficiently.

Comparing linear method with non-linear method, following major differences were found:

1) The Linear method converged in 3-4 iterations. However, the non-linear method converges in

6-11 iterations. Hence convergence requires less time in linear method than non-linear method.

2) In non-linear method, first tolerance is specified and then Jacobean matrix non-singular,

convergence test is performed. Here, epsilon (ε=X2-X1) is calculated using Newton Raphson formula

and then epsilon compared with tolerance (set as 1.0e-15). System is said to be converged if

epsilon<tolerance and the dot product ((Xn-X).(X0-X))<1. But in linear method, epsilon and tolerance

comparison is not required to achieve convergence.

3) The linear method is solved by Gauss elimination method only. While the non-linear method

can be solved by following ways:

i) fsolve

ii) Newton Raphson

iii) Relaxation Strategy

iv) Reduced Equations

4) Scaling is not required in linear method as in non-linear method.

Therefore, it was more difficult to achieve same convergence in non-linear method than linear method.

But results are closer in non-linear method to experimental data.

3.2Conclusion for Linear and Non-linear technique

Simulation model for linear as well as non-linear technique have been developed for concentrating

orange juice in a double effect evaporator using Koko and Joye[5]

and Moresi and Angletti[1]

,

respectively. Empirical correlations for Boiling point rise and overall heat transfer coefficient for

orange juice have also been developed using Moresi and Angletti[1]

and Moresi and Spinosi[2]

experimental data, respectively. Computational aspects associated with simulation of a double effect

evaporator for concentrating orange juice have been studied. In view of the results discussed, it may be

concluded that the proposed simulation model is suitable for studying the behavior of a double effect

evaporator under operating conditions.

3.3 Recommendations for Future Work

1) Simulation was done for two effect evaporator. Now a day‟s several effects, usually 4-7 are

used for the Orange Juice concentration, while a few years ago there might have been two or four

effects. Energy has become more expensive. As the number of effects is increased, capital cost

increases and operating cost decreases. Hence more number of effects can be added for concentrating

orange juice to save on operating cost.

2) Here vapor and liquid leaving each effect are assumed to be in equilibrium. However, in most

cases, vapor and liquid are not in equilibrium.

3) Change in pressure with height can also be considered.

3.4Acknowledgement

The authors are grateful to the Department of Chemical Engineering, AMU, India for providing all

necessary facilities and support for this research work.

3.5 Nomenclature

A Heat transfer area [m2]

F Feed flow rate [kg/h]

f General mathematical function

g Scaled function

H Vapor enthalpy [kJ/kg]


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h Liquid enthalpy [kJ/kg]

L Liquid flow rate [kg/h]

T Saturation temperature [°C]

τ Saturation temperature [°C]

U Overall heat transfer coefficient [kJ/h. m2°C]

V Steam rate [kg/h]

x Weight fraction of orange juice

λ Latent heat of vaporization [kJ/kg]

BPR Boiling point rise [°C]

∆Ti Temperature difference between externally supplied steam and ith

effect solution

∆Ttot Temperature difference between externally supplied steam and final effect solution

α Relaxation Factor

Xi Value of ith

iteration in Newton Raphson

ε Epsilon to compare value of subsequent iterations in Newton Raphson

R Universal gas constant [kJ/kmole.°C]

MA Molecular weight of the solvent A, water

MB Molecular weight of the solute B, orange juice

∆Hv Enthalpy of vaporization of the solvent A [kJ/kmole]

yB Weight fraction of the sucrose in the juice

<Subscript>

i Iteration number

3.6 References

[1] Angletti, S. and Moresi, M. (1983), “Modeling of multiple effect falling film evaporators”, Journal

of Food Technology, 18, 539-563.

[2] Moresi, M. (1980), “Economic Study of concentrated citrus juice production”, Department of

Chemical Engineering, University of Rome, 41, 975-991

[3] Moresi, M. and Spinosi, M. (1980), “Engineering factors in the production of concentrated fruit

juices, 1. Fluid physical properties of orange juices”, Journal of Food Technology, 15, 265-276

[4] Zain, Omer. S and Kumar, S (1996), “Simulation of a Multiple Effect evaporator for concentrating

caustic soda solution-Computational aspects”, Journal of Chemical Engineering of Japan, 29: 5, 889-

893

[5] Joye, D and Koko, F (1988), “A simpler way to tame multiple-effect evaporators”, Chemical

Engineering Education, 1, 52-56

[6] Lambert, Richard. N, Joye, Donald. D and Koko, Frank. W (1986), “Design Calculations for

Multiple-Effect Evaporators. 1. Linear Method”, American Chemical Society, 26, 104-107

[7] Joye, Donald. D and Koko, Frank. W (1988), “Design Calculations for Multiple-Effect Evaporators.

2. Comparison of Linear and Non-Linear Methods”, American Chemical Society, 26, 108-110

[8] Urgin, E. and Urbician, M. J (1999), “Design and Simulation of Multi-effect Evaporators”, Heat

Transfer Engineering, 20: 4, 38-44

[9] Geankoplis, C. (1983), Transport Processes and Unit Operations, 3rd

Ed, Allyn and Bacon, Boston,

MA

[10] Perry, R.H, D.W. Green and J.O. Maloney (1984), Perry‟s Chemical Engineer‟s Handbook, 6th Ed,

McGraw-Hill, NY

[11] Holland, C.D. (1975). Fundamentals and Modeling of Separation Processes. Prentice Hall Inc.,

Englewood cliffs, NJ


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Table 1: Operating conditions for orange juice (Moresi and Angletti[1]

)

Water removal capacity 1000 2000 kg/h

Feed flow-rate 1200 2400 kg/h

Feed temperature 18 18 °C

Feed concentration 10 10 °Brix

Output concentration 60 60 °Brix

Live steam temperature 95 95 °C

Live steam pressure 83.8 83.8 kPa

Second effect pressure 6 6 kPa

Effect I

Heat transfer surface 38 65 m2

Effect II

Heat transfer surface 71 95 m2

Table 2: Data for calculation of BPR (Moresi and Spinosi[3]

)

S. No Boiling point rise, BPR(°C) Concentration, x

(weight fraction)

1 0.2832 0.096

2 0.5462 0.170

3 0.9223 0.257

4 1.2433 0.318

5 2.0115 0.430

6 2.7477 0.5075

7 3.6076 0.575

8 4.9515 0.650

Table 3: Data for calculation of U (Moresi and Angletti [1]

)

S. No Overall heat transfer

coefficient, U(W/m2°C)

Solution

temperature,

τ(°C)

Concentration, x(weight

fraction)

1 1544 73.65 0.1709

2 1900 74.95 0.1708

3 1308 74.65 0.1708

4 1235 67.55 0.1713

5 2763 61.65 0.1718

6 171 44.5 0.6

7 162 44.5 0.6

8 158 44.5 0.6

9 254 44.5 0.6


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10 655 44.5 0.6

11 1752 79.84 0.1704

12 2159 80.84 0.1703

13 1491 80.85 0.1703

14 1352 71.95 0.1710

15 2862 65.85 0.1715

16 192 44.5 0.6

17 187 44.5 0.6

18 179 44.5 0.6

19 293 44.5 0.6

20 657 44.5 0.6

Table 4: Results for orange juice concentration using linear technique (Input 1)

VARIABLES EXPRIMENTAL

DATA BY

MORESI AND

ANGLETTI[1]

SIMULATION

RESULT

UNITS

F=1200 kg/hr - assume x2=0.2 -

No. of Iterations - 3 -

V0 600-650 593 kg/h

L1 - 791 kg/h

V1 - 469 kg/h

T1 55-75 50.03 °C

τ 1 - 50.57 °C

L2 - 240 kg/h

V2 - 491 kg/h

T2 40.4 36.15 °C

τ 2 - 40.85 °C

BPR1 0.55 0.54 °C

U 1 5558-9946 11022 kJ/hr. m2°C

BPR2 4.1 4.7 °C

U2 615-2358 1347 kJ/hr. m2°C

A1 38 38 m2

A 2 71 67.62 m2

x1 0.16-0.18 0.164 Weight fraction

x2 0.6 0.59 Weight fraction

T0 95 94.99 °C


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Table 5: Results for orange juice concentration using linear technique (Input 2)


DATA BY MORESI

AND ANGLETTI[1]

SIMULATION

RESULT

UNITS

F=2400 kg/hr - assume x2=0.2 -

No. of Iterations - 4 -

V0 1200-1300 1207 kg/h

L1 - 710 kg/h

V1 - 489 kg/h

T1 58-80 62.35 °C

τ 1 - 62.91 °C

L2 - 200 kg/h

V2 - 510 kg/h

T2 40.4 36.15 °C

τ 2 - 41.55 °C

BPR1 0.55 0.56 °C

U 1 6307-10303 7801 kJ/hr. m2°C

BPR2 4.1 5.4 °C

U2 691-2365 748 kJ/hr. m2°C

A1 65 65 m2

A 2 95 94.54 m2

x1 0.16-0.18 0.169 Weight fraction

x2 0.6 0.6 Weight fraction

T0 95 94.55 °C

Results for non-linear method are given below:

Table 6: fsolve results (Input 1)


DATA BY MORESI

AND ANGLETTI[1]

INITIAL

GUESS

SIMULATION

RESULT

UNITS

F=1200 kg/hr - - - -


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No. of Iterations - - 11 -

v0 0.5-0.54 1 0.5371 -

u1* 0.58-0.795 1 0.9189 -

u1 0.578-0.79 1 0.9128 -

x1 0.16-0.18 1 0.1706 -

l1 0.55-0.625 1 0.5860 -

u2* 0.468 1 0.6058 -

x2 0.6 1 0.6544 -

l2 0.1667 1 0.1528 -

UNSCALED RESULTS

V0 600-650 - 644 kg/h

τ 1 55.1-75.53 - 87.29 °C

T1 55-75.05 - 86.72 °C

BPR1 0.55 - 0.57 °C

L1 660-750 - 703 kg/h

τ 2 44.46 - 41.25 °C

T2 40.4 - 36.15 °C

BPR2 4.1 - 5.1 °C

L2 200 - 183.37 kg/h

U1 4446-9947 - 4989 kJ/hr. m2°C

U2 616-2358 - 534.86 kJ/hr. m2°C

Table 7: fsolve results (Input 2)


DATA BY MORESI

AND ANGLETTI[1]

INITIAL

GUESS

SIMULATION

RESULT

UNITS

F=2400 kg/hr - - - -


v0 0.5-0.542 0.8 0.5265 -

u1* 0.62-0.86 0.8 0.9111 -

u1 0.61-0.853 0.8 0.9052 -

x1 0.16-0.18 0.8 0.1679 -

l1 0.55-0.625 0.8 0.5956 -

u2* 0.468 0.8 0.5306 -

x2 0.6 0.8 0.5937 -

l2 0.1667 0.8 0.1684 -

UNSCALED RESULTS


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V0 1200-1300 - 1263 kg/h

τ 1 58.9-81.7 - 86.55 °C

T1 57.95-81.04 - 85.99 °C

BPR1 0.55 - 0.56 °C

L1 660-750 - 1429 kg/h

τ 2 44.46 - 40.55 °C

T2 40.4 - 36.15 °C

BPR2 4.1 - 4.4 °C

L2 200 - 404.24 kg/h

U1 4867-10303 - 5218 kJ/hr. m2°C

U2 691-2365 - 777.47 kJ/hr. m2°C

Table 8: Newton Raphson results (Input 1)


DATA BY MORESI

AND ANGLETTI[1]

INITIAL

GUESS

SIMULATION RESULT UNITS

F=1200 kg/hr - - - -


v0 0.5-0.54 0.8 0.536538228129474 -

u1* 0.58-0.795 0.8 0.919087645043830 -

u1 0.578-0.79 0.8 0.913059132520497 -

x1 0.16-0.18 0.8 0.170469690041145 -

l1 0.55-0.625 0.8 0.586614546995797 -

u2* 0.468 0.8 0.600381313152053 -

x2 0.6 0.8 0.650675610750239 -

l2 0.1667 0.8 0.153686412012121 -

UNSCALED RESULTS

V0 600-650 - 644 kg/h

τ 1 55.1-75.53 - 87.31 °C

T1 55-75.05 - 86.74 °C

BPR1 0.55 - 0.57 °C

L1 660-750 - 703 kg/h

τ 2 44.46 - 41.15 °C

T2 40.4 - 36.15 °C

BPR2 4.1 - 5.0 °C

L2 200 - 184 kg/h

U1 4446-9947 - 4998 kJ/hr.

m2°C


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U2 616-2358 - 548 kJ/hr.

m2°C

Table 9: Newton Raphson results (Input 2)


DATA BY MORESI

AND ANGLETTI[1]

INITIAL GUESS SIMULATION RESULT UNITS

F=2400 kg/hr - - - -


v0 0.5-0.542 0.7 0.529033284893321 -

u1* 0.62-0.86 0.7 0.909949773444520 -

u1 0.61-0.853 0.7 0.904008695559832 -

x1 0.16-0.18 0.7 0.168644873472966 -

l1 0.55-0.625 0.7 0.592961991317396 -

u2* 0.468 0.7 0.547499478259759 -

x2 0.6 0.7 0.609318779484449 -

l2 0.1667 0.7 0.164117705488432 -

UNSCALED RESULTS

V0 1200-1300 - 1269 kg/h

τ 1 58.9-81.7 - 86.45 °C

T1 57.95-81.04 - 85.88 °C

BPR1 0.55 - 0.56 °C

L1 660-750 - 1423 kg/h

τ 2 44.46 - 40.65 °C

T2 40.4 - 36.15 °C

BPR2 4.1 - 4.5 °C

L2 200 - 394 kg/h

U1 4867-10303 - 5178 kJ/hr.

m2°C

U2 691-2365 - 707 kJ/hr.

m2°C

Table 10: Relaxation Strategy results (Input 1)


DATA BY MORESI

AND ANGLETTI[1]



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F=1200 kg/hr - - - -


v0 0.5-0.54 0.8 0.536538228129474 -

u1* 0.58-0.795 0.8 0.919087645043830 -

u1 0.578-0.79 0.8 0.913059132520497 -

x1 0.16-0.18 0.8 0.170469690041145 -

l1 0.55-0.625 0.8 0.586614546995797 -

u2* 0.468 0.8 0.600381313152054 -

x2 0.6 0.8 0.650675610750239 -

l2 0.1667 0.8 0.153686412012121 -

UNSCALED RESULTS

V0 600-650 - 643 kg/h

τ 1 55.1-75.53 - 87.31 °C

T1 55-75.05 - 86.74 °C

BPR1 0.55 - 0.57 °C

L1 660-750 - 704 kg/h

τ 2 44.46 - 41.15 °C

T2 40.4 - 36.15 °C

BPR2 4.1 - 5.0 °C

L2 200 - 184 kg/h

U1 4446-9947 - 4998 kJ/hr.

m2°C

U2 616-2358 - 548 kJ/hr.

m2°C

Table 11: Relaxation Strategy results (Input 2)


DATA BY MORESI

AND ANGLETTI[1]


F=2400 kg/hr - - - -


v0 0.5-0.542 0.6 0.529033284893321 -

u1* 0.62-0.86 0.6 0.909949773444520 -

u1 0.61-0.853 0.6 0.904008695559831 -

x1 0.16-0.18 0.6 0.168644873472966 -

l1 0.55-0.625 0.6 0.592961991317396 -

u2* 0.468 0.6 0.547499478259759 -

x2 0.6 0.6 0.609318779484452 -


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l2 0.1667 0.6 0.164117705488432 -

UNSCALED RESULTS

V0 1200-1300 - 1269 kg/h

τ 1 58.9-81.7 - 86.44 °C

T1 57.95-81.04 - 85.88 °C

BPR1 0.55 - 0.56 °C

L1 660-750 - 1423 kg/h

τ 2 44.46 - 40.65 °C

T2 40.4 - 36.15 °C

BPR2 4.1 - 4.5 °C

L2 200 - 393 kg/h

U1 4867-10303 - 5178 kJ/hr.

m2°C

U2 691-2365 - 707 kJ/hr.

m2°C

Table 12: Reduced Equation results using Newton Raphson (Input 1)


DATA BY MORESI

AND ANGLETTI[1]

INITIAL

GUESS


F=1200 kg/hr - - - -


v0 0.5-0.54 0.48 -

u1* 0.58-0.795 0.48 -

u1 0.578-0.79 0.48 0.610951932243284 -

x1 0.16-0.18 0.48 0.1721* -

l1 0.55-0.625 0.48 0.580873083855312 -

u2* 0.468 0.48 -

x2 0.6 0.48 0.5872* -

l2 0.1667 0.48 0.170292418657319 -

UNSCALED RESULTS

V0 600-650 - 602 kg/h

τ 1 55.1-75.53 - 58.62 °C

T1 55-75.05 - 58.04 °C

BPR1 0.55 - 0.58 °C

L1 660-750 - 697 kg/h

τ 2 44.46 - 44.95 °C

T2 40.4 - 36.15 °C


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BPR2 4.1 - 4.3 °C

L2 200 - 204 kg/h

U1 4446-9947 - 8234 kJ/hr. m2°C

U2 616-2358 - 2019 kJ/hr. m2°C

*Calculated Results

Table 13: Reduced Equation results using Newton Raphson (Input 2)

VARIABLES EXPRIMENTAL DATA

BY MORESI AND

ANGLETTI[1]

INITIAL

GUESS


F=2400 kg/hr - - - -


v0 0.5-0.542 0.48 - -

u1* 0.62-0.86 0.48 - -

u1 0.61-0.853 0.48 0.674322640474179 -

x1 0.16-0.18 0.48 0.1722* -

l1 0.55-0.625 0.48 0.580858966440164 -

u2* 0.468 0.48 - -

x2 0.6 0.48 0.5772* -

l2 0.1667 0.48 0.173248575182311 -

UNSCALED RESULTS

V0 1200-1300 - 1306 kg/h

τ 1 58.9-81.7 - 64.64 °C

T1 57.95-81.04 - 64.06 °C

BPR1 0.55 - 0.58 °C

L1 660-750 - 1394 kg/h

τ 2 44.46 - 40.35 °C

T2 40.4 - 36.15 °C

BPR2 4.1 - 4.2 °C

L2 200 - 416 kg/h

U1 4867-10303 - 7248 kJ/hr.

m2°C

U2 691-2365 - 858 kJ/hr.

m2°C

*Calculated Results


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Table 14: Reduced Equation results using Relaxation Strategy (Input 1)


DATA BY MORESI

AND ANGLETTI[1]

INITIAL

GUESS


F=1200 kg/hr - - - -


v0 0.5-0.54 0.48 - -

u1* 0.58-0.795 0.48 - -

u1 0.578-0.79 0.48 0.610951932243284 -

x1 0.16-0.18 0.48 0.1721* -

l1 0.55-0.625 0.48 0.580873083855312 -

u2* 0.468 0.48 - -

x2 0.6 0.48 0.5872* -

l2 0.1667 0.48 0.170292418657319 -

UNSCALED RESULTS

V0 600-650 - 602 kg/h

τ 1 55.1-75.53 - 58.62 °C

T1 55-75.05 - 58.04 °C

BPR1 0.55 - 0.58 °C

L1 660-750 - 697 kg/h

τ 2 44.46 - 40.45 °C

T2 40.4 - 36.15 °C

BPR2 4.1 - 4.3 °C

L2 200 - 204 kg/h

U1 4446-9947 - 8234 kJ/hr. m2°C

U2 616-2358 - 2019 kJ/hr. m2°C

*Calculated Results

Table 15: Reduced Equation results using Relaxation Strategy (Input 2)


DATA BY MORESI

AND ANGLETTI[1]


F=2400 kg/hr - - - -


v0 0.5-0.542 0.45 - -

u1* 0.62-0.86 0.45 -

u1 0.61-0.853 0.45 0.674322640474179 -


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x1 0.16-0.18 0.45 0.1722* -

l1 0.55-0.625 0.45 0.580858966440164 -

u2* 0.468 0.45 - -

x2 0.6 0.45 0.5772* -

l2 0.1667 0.45 0.173248575182311 -

UNSCALED RESULTS

V0 1200-1300 - 1306 kg/h

τ 1 58.9-81.7 - 64.64 °C

T1 57.95-81.04 - 64.06 °C

BPR1 0.55 - 0.58 °C

L1 660-750 - 1394 kg/h

τ 2 44.46 - 40.35 °C

T2 40.4 - 36.15 °C

BPR2 4.1 - 4.2 °C

L2 200 - 416 kg/h

U1 4867-10303 - 7248 kJ/hr. m2°C

U2 691-2365 - 858 kJ/hr. m2°C

*Calculated Results

Figure 1: Double Effect Evaporator System with forward feed arrangement

F, Feed

Tf

V0, Steam

mSteaFeed

P1, τ 1

P0, T0

P2, τ2

P1, T1

Drips Condensate L1 L2

V1 V2

Documents

Simulation of Double-Effect Evaporator for Concentrating Orange Juice