18
Vaporizing Exchangers
Key Learning Objectives
Specific objectives and topics covered
are:
A
Understanding the working principle of a vaporizer
A Formulating
the model of a vaporizer
A Developing the
simulation algorithm to solve the vaporizer model
A Revising the
concept of evaporator system
A Deriving the model
of an evaporator used for tomato juice
A Studying the
evaporator dynamics in open- and closed-loop mode
18.1
VAPORIZER
A typical layout of avaporizer is shown in Figure 18.1. It is a jacketed vessel, in which a
liquid
(feed) enters at the bottom with a volumetric flow rate F; and temperature T, and the produced
vapour is taken out from the top. Steam is supplied from the external source as a heating
medium. Due to the continuous input of heat (O) to the vaporizer, the liquid
temperature
gradually rises and finally reaches the boiling point. The boiling of a liquid starts at some
temperature (and composition if more than one comnponent is present) when the vapour
pressure of the liquid (P) tends to exceed the pressure in the vapour phase above it (P). An
infinitesimal difference between these two pressures is sufficient to provide the vapour flow.
At this point it is worthy to mention that the vapour boil-up rate (vg) is proportional to the
driving force P- Py that is, Vg k(P- P). At equilibrium, P = Py, which indicates very large
gain (mass transfer coefficient), k. Recall the well-known fact that the system temperature does
not rise beyond the boiling point when there is no resistance to the departure of the produced
vapour.
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Chemical Process Modelling and Computer Simulation
394
18.1.1
Model Development
The model for a vaporizer is constructed based on the following assumptions:
The liquid is pure, incompressible and perfectly mixed in the vessel
The heat losses are assumed negligible.
The density (p) and heat capacity (C) remain constant.
The liquid and vapour phases are in equilibrium. It implies that
(a) the temperatures of liquid and vapour are identical (that is, T, =Ts
energy balance is required for the vapour phase.
(b) the vapour pressure
of liquid (pure) and the pressure in the vapour nhae
identical (that is, P = Pp.
The temperature everywhere in the jacket is the same (T).
The perfect gas law is used.
Po
Fo
PC
Py
Vy
my
VB
Py
Steamn
Vapour
LC
Liquid
T
Ts
PL
F, T;
Figure 18.1
Ts
P
Schematic representation of a vaporizer.
The balance equations for the representative system are presented in the
following:
Mass balance (liquid phase)
dmL
dt
d(pVL) = FiPL
dt
VB
(18.1)
Fo Py
(18.2)
Mass balance (vapour phase)
dmy
dt
dpyVv) = VBdt
where V, denotes the volume of liquid, Vy the volume of vapour, Fo the volumetric
outflow through exit valve, m, the mass in liquid phase and my the mass in vapour vapour
phase.
Vaporizing Exchangers
395
Energy balance (liquid phase)
d(pLVLCT)
dt
= FPLC,T; - VB (À + C,T) + Q
(18.3)
with
Q= UA(Ts - T)
(18.4)
Obviously, the reference temperature (Tre) is considered here zero. In the above
equations,
Ais the latent heat of vaporization at the system temperature, Uthe overall heat
transfer coefficient, A the heat transfer area and O the heat supplied by the steam. Using the
following cquation, the steam temperature (Ts) in degree Celsius can be determined knowing
the pressure (Psteam) in bara:
Psteam =exp 11.6859
3822.3186
227.47+ Ts
[from Equation (2.82)]
Energy balance (vapour phase)
T; = Ty = T
(18.5)
myRT
MW
(18.6)
Equation of state (vapou phase)
PvVy =
with
(18.7)
Here, V represents the total volume of the system and MW the molecular weight.
Valve equation (gas flow)
Considering V = Py0:
vO = Ky yR (R - P)
(18.8)
where Ky is the valve constant and Po the exit pressure.
Vapour pressure (boiling relation)
In P=A -
B
T+C
[Antoine equation, that is Equation (9.28)]
Note that it is straightforward to calculate the boiling point temperature, T and pressure,
P(= Pr by solving simultaneously Equation (18.6) and the Antoine equation. As shown in
Figure 18.1, two controllers can be employed for the example system. One is for maintaining
the liquid level in the vessel by the manipulation of feed inflow rate and the other one can
Control the pressure by adjusting the steam flow rate.
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ChemicalProcess Modelling and Computer Simulation
18.1.2
Simulation Algorithm with a Boiler Example
In the preceding discussion, we came to know how to model a vaporizer. It
should be pointed
out that this process model can also be used to represent a boiling unit. Now,
we would like
to simulate this developed model for a boiler.
Consider a boiler that aims to produce steam by receiving heat (0) from
firing.
In order to simulate the boiler model, required data are documented in Tableexternal
18.1.
In
this
problem, neglect the left hand side of Equation (18.3) because this differential term is
with respect to Q. The computer-assisted simulation algorithm in step-wise fashion is negligible
the following. The development of computer code and the dynamics
given
are left for inh
students. Investigate the closed-loop performance as well implementing,study
say the proportional
integral (PI) controller, for maintaining both the level and pressure.
Table 18.1
Given data for a boiler
Q= 5000 kcal/h
2 = 475 cal/g
Cp = 1 cal/g-°c
Iref = 0°C
T; = 20°C
F;
=1000 Vh
PL = 1g/cc
V= 5500 1
VL (at t =0) 2800 1
my (at t= 0) = 9 kg
Ky = 40 kg/h-bara
Po = 10 bara
MW = 18
R= 1.987 cal/gmol-K
Simulation steps
Step 1: Compute Vy from Equation (18.7) knowing V and V (at t = 0).
Step 2: Knowing my (at t = 0), R, MW and V, calculate T and P by
simultaneously solving
the boiling relation (Antoine) and Equation (18.6).
Step 3: Solve Equation (18.3) with left-hand side equals zero for finding va
based on the
known information of F; PL, Cp T;, d, T and Q.
Step 4: Compute Vo from Equation (18.8) at given Ky and Po- P is
available.
Step 5: Update my (at t + l) using Equation (18.2). V; and vo (
PyFo) are obtained
previously.
Step 6: Update V (at t + 1) using Equation (18.1).
Step 7.: Go to Step 1 until the steady state has reached.
Vaporizing Exchangers
397
coMMERCIAL DOUBLE-EFFECT EVAPORATOR: TOMATO JUICE
18.2
Introduction
18.2.1
Evaporation technique is used for removing the solvent as vapour from a solution or slurry. For
most of the evaporation systems, the _olvent is water. The purpose of an evaporation process
is to
1
produce
a concentrated
solution (thick liquor or slurry) by vaporizing a portion of the
In many situations, evaporators operate under a vacuum. For this, a vacuum pump or
jet ejector
and
paper
vacuum system
is needed on the last effect. Evaporators are widely used in food.
pulp, and chemical industries.
In evaporation, no attempt is made to separate components of the vapour and this
iefinguises evaporation from distillation method. Evaporation differs from drying in that the
due is always a liquid, which may be highly viscous or aslurry. Again it is distinguised from
talization in that evaporation technique is employed for concentrating a solution rather than
producing or building crystals.
Evaporator may have a single unit, called single-effect, or multiple units, called multiple
effect. Most important advantage of multiple-effect over the single-effect evaporator is the
oronomy!l,Multiple-effect scheme evaporates morewater per kilogram of steam fed to the unit
wreusing the vapour from one effect as the heating medium for the next. In contrast, the
multiple-efect
scheme provides lower capacity² than the single-effect evaporator. To know
more about evaporators, the reader can consult any standard textbook on heat transfer
(for example, McCabe et al., 1993).
Here our learning objective includes a systematic simulation study on a commercial
double-effect tomato paste evaporator. The dynamic process model, consisting of mass
differential
balance, energy balance and empirical correlations, is derived in the form of
algebraic equations. The validation of this model is reported in literature (Runyon et al., 1991).
The simulation of the model structure is performed here for open-loop process dynamics. To
investigate
the closed-loop behaviour, the three single-loop proportional integral (PI) control
strategies along with twvo level controllers (P-only) have been employed around the interactive
evaporation system.
18.2.2
The Process
The
example process shown in Figure 18.2 is a double-effect evaporator with backward
feeding arrangement used for tomato concentrate. The two effects are numbered from left to
right as Tank 1 and Tank 2, respectively. The raw juice (feed) having flow rate F, composition
Xf and temperature T, enters Tank 2, and the steam with flow rate S and temperature Ts enters
Tank 1.The mass hoidups in the two tanks are defined as M, and M,. VËand V; are the vapour
flow rates from the overhead of two tanks with temperature TË and T2, respectively. Pi and
Pz are the product flow rates from the two effects with composition Xp and 3. and
perature 7| and T2, respectively. The steady state and parameter values are listed in
Table 18.2.
2
3
Eeonomy is the number of kilograms vaporized per kilogram of steam fed to the evaporator.
Capacity is the number of kilograms of water vaporized per nour.
Eitects are always nu1nbered according to decreasing pressure (steam flow),irespective of the feeding pattern.
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Chemical Process Modelling and Conputer Simulation
CC= Composition controller
rc
Vapour
LC= Level controller
TC =Temperature controller
Tank I
Steam
M
S, T
Vapour
Tank 2
LC)
M,
L)
rC)
Feed
F. T, X,
Product
Condensate
Product
P, T, X,
CË, T,
Condensate
C, T,
P, T, Xp
Figure 18.2 Schematic of a double-effect evaporator.
Table 18.2 Steady state* and parameter values
Term
Tank 1 mass holdup
Tank 2 mass holdup
Input feed flow rate
Input steam flow rate
Tank 1 liquid product flow rate
Tank 2 liquid product flow rate
Vapour flow rate from Tank I
Vapour flow rate from Tank 2
Feed composition
Tank 1 composition
Tank 2 composition
Steam temperature
Feed temperature
Temperature in Tank 1
Temperature in Tank 2
Heat transfer area in Tank 1
Heat transfer area in Tank 2
Overall heat transfer coefficient for Tank 1
Overall heat transfer coefficient for Tank 2
suffix 0.
Abbreviation (unit)
Mo (kg)
M20 (kg)
Value
2268
So (kg/h)
2268
26,103
11,023
P1o (kg/h)
5006
P2o (kg/h)
Vio (kg/h)
V20 (kg/h)
9932
Fo(kg/h)
Xo (kgkg)
Xpo (kgkg)
14,887
11,165
0.05
0.2607
X20 (kgkg)
Tso (°C)
To (°C)
T1o (°C)
74.7
T20 (°C)
52.0
0.0874
115.7
85.0
A, (m²)
102
A, (m²)
U, (kJ/h-m².°C)
412
U, (kJ/h-m2.°C)
2453
5826
Vaporizing Exchangers
399
Model development
Rvanoration process involves mass and heat transfer. The tomato juice is considered as a
Linary solution of water and soluble solids (Miranda and Simpson, 2005), both considered inert
constructed based on the
in a chemical sense. The macroscopical evaporator model has been
phase is
conservation laws and empirical relationships. It should be noted that only the juice
mathematical
of
the
formulation
considered for modelling. The assumptions involved in the
end of the discussion to
model are listed below. A separate Notation' section is given at the
define the terms used.
Negligible heat losses to the surroundings.
each effect.
Homogeneous composition and tenmperature inside
Variable liquid holdup and negligible vapour holdup.
Overhead vapours considered as pure steam.
" Latent heat of vaporization varied with temperature.
Total mass balance
First effect:
dM1 = P2 -
Pi -
(18.9)
VË
dt
Second effect:
M 2=F -
Pa
CENTRAL LIBRARY
V2
dt
(18.10)
INSTITUTE OF ECHNOL00
Component (solids) nass balance
First effect:
(18.11)
d (M 1X p) = P2X 2 - PiX p
dt
dX p + Xp dM i
M1
dt
dt
dX
M1
’
(18.12)
P»X2 - PAp
PiXp - Xp
= P2X 2
(18.13)
dM1
dt
dt
Substituting Equation (18.9), we obtain
M1
M
**P= PTX>- PXn -
Pi
V)
(18.14)
dt
dt
dX
’
Xp (P2 -
dt
= P2X2-
PX p -
XpP: + Xp P1+Xp Vi
P2(X2- X,) + X,Vi
Mi
(18.15)
(18.16)
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Chemical Process Modelling and Computer Simulation
Second effect:
d(M2X 2) = FX f dt
P2X2
Simplifying and rearranging, finally we get
F(Xf - X2)+X 2V2
dX 2
dt
M2
Energy balance
The steam flow rate to the first effect is
heat exchanger as:
(18.18)
obtained through the energy
SA(T) = UjA|(Ts - T)
That means,
(18.17
balance on the
first effect
(18.19)
S= UlA|(Ts - T)
Similarly, the vapour flow rate to the second effect is
derived from the
second effect heat exchanger as:
Vja(Ti) = UyAz(T - T)
(18.20)
energy balance on the
(18.21)
V-U2A2(T1 - T2)
A(T)
=
In the following, the energy
balance equations are derived.
First effect: d[M h(T 1 X)
= Pzh(T2» X2) + Sa(Ts)
dt
- Pih(T1, X p) ’
M1
dh(T, Xp)
+ h(T,, Xp) dM1 =
dt
’
M1
dt
(18.22)
ViH (T)
(18.23)
Ph(T2, X2) + SA(T) - Pih(T, Xp) ViH(T)
dh(T1, Xp)
(18.24)
= Pzh(T2, X2) + SA(Ts) P1h(T, X )- VjH(T)- h(T1 Xp) dM1
d
Substituting Equation (18.9), we
(18.25)
obtain
dt
M1
dh(T1, X)
dt
= Plh(T2 X2) - h(T1, Xp)l +
Using Equation (18.19), we
MI
dh(Ti, X)
dt
obtain
Sa(Ts) - V[H(T) h(T; X!
(18.26)
= P[h(T2> X2)h(T, Xp)] + UjA(Ts - T) - V, [H(T) - h(TI, Xp))
(18.27)
Vaporizing Exchangers
Ph(T, X)
dh(T, A )
p
401
h(T: X)l+ UjA{Ts - T)- (H(T)- h(T, X)!
MI
dt
(18.28 )
d|M
Second effect:
T; X ]
dt
=Fh(T, X;) + V,2(T) - Ph(T2, X2) - V,H(T)
(18.29)
Simplifying and rearranging,
dh(T. X,).
finally we have
F[h(T,, Xf)- h(T2 X)]+U2A2(Ti -T) -
V[H (T;) - h(T 2x X >))
M2
dt
(18.30)
Empirical correlations
The enthalpy of the product (tomato juice) is
represented as (Heldman and Singh, 1981):
(18.31 )
n(T, A) = (4.177 - 2.506X)T
The pure solvent vapour (steam) enthalpy is computed
of values fromn the steam
tables as:
H() = 2495.0 + 1.958T
For the condensate streams, the pure solvent liquid
tables as:
using a polynomial regression equation
(18.32)
0.002128T²
enthalpy is also obtained from the steam
(18.33 )
h(T) = 4.1777T
follovwing form:
So, the latent heat of vaporization has the
(18.34)
h(T) = 2495.0 - 2.219T - 0.002 128T?
a(T) = H()
[(18.28) and (18.30)] have the
correlations, the energy balance equations
Using the above
following finalI forms:
dTi
P:(4.177 -
dt
dT2
T)
2.506X )(T: -T) -U2A2(Ti -T2) + UiA|(Ts M(4.177 - 2.506X )
l4.177 T2
F(4.177- 2.506 X,) (T, -T)+U2A;(Ti -T) + V
dt
- H(T))
M (4.177 - 2.506X )
(18.36)
In the following, the open-loop as well as the
18.2.3
(16.35)
closed-loop dynamics
are discussed.
Application of Control Algorithm
Control objectives
product
system are selected taking into account the
the
process,
Ihe control objectives for an evaporation
concerned
the
and cost considerations. For
Specifications, operational constraints
Primary objective is to maintain the product
solids concentration (or product viscosity) at its
402
Chemical Process Modelling and Computer Simulation
desired value. In addition to the product specifications, the following aspects need to
care of:
To prevent the overflow or drying out of evaporator tubes, liquid
required to control.
mass
holdup is
To avoid the product degradation or damage, temperature is required to maintain at
desired value.
the
Steam economy should be maximized.
Degrees of freedom analysis
The evaporator model presented earlier includes fifteen independent variables () [F, Pi,
M1, Mz, T, Ts Ii, T2, S, V, V, X, Xp and X,] and eight independent equations (E) (8oPz,
(18.10), (18.16), (18.18), (18.20), (18.22), (18.35) and (18.36)]. We know that the
of degrees of freedom () is given as
f= V- E
(18.37)
The process is said to be:
Case A: exactly specified when f = 0
Case B: underspecified when f> 0
Case C: overspecified when f< 0.
It should be noted that for an underspecified system having f degrees of freedom, we
require
fnumber of additional equations to make the system completely specified. On the other hand.
f number of equations has to be removed for the overspecified case.
Underspecified systems are quite common. To make them exactly specified, there are.
however, two possibilities:
(i) Specify the values of disturbance variables mostly through direct
(iü) Include control equations selecting best possible control pairs.
measurements.
At this point, it is important to mention that the implementation of the above
two strategies
must be done judiciously and care must be taken to avoid the
overspecification
of the system.
For the representative evaporation system, the number of degrees of
freedom is seven
(= 15
8), that is, f> 0 (underspecified). In order to have a completely
as stated, the number of its degrees of freedom should
the control objectives, five control pairs have been selected and
and T% are to be treated as known disturbances.
determined process,
be zero. For this purpose and to meet
two input variables, namely Xf
Selection of control pairs
A control pair is selected taking into consideration the following:
(i) the manipulation has a direct and rapid effect on the
controlled variable,
(ii) the dead-time is as small as possible between the
manipulated
input and controlled
variable, and
(iii) for multi-input/multi-output system, the loop interaction is
minimal.
4 A control pair is a pair between a controlled variable and a
manipulated input.