Lecture 5 – Evaporation (Part 2)
Dr. Hanee Farzana Hizaddin
KIL3007 Separation Processes 2
www.um.edu.my
Learning Outcomes
Calculations for multiple effect evaporators
2
Motivation
In evaporation of solutions in a single-effect evaporator, a
major cost is the cost of the steam used to evaporate the
water.
A single-effect evaporator is wasteful of steam costs, since
the latent heat of the vapor leaving the evaporator is usually
not used.
However, to reduce this cost, multiple-effect evaporators are
used which recover the latent heat of the vapor leaving and
reuse it.
3
Vi Va V3
F T
Yi
I
v, Is
v2 , Tsz
Ps S TsI
,
,
T2
L1 ,
x
, Le , 2 L3 X3,
In the first effect raw steam is used as the heating medium to this first effect, which is boiling at
temperature T1 and pressure P1.
The vapor removed from the first effect is used as the heating medium, condensing in the second effect
and vaporizing water at temperature T2 and pressure P2 in this effect.
To transfer heat from the condensing vapor to the boiling liquid in this second effect, the boiling
temperature T2 must be less than the condensing temperature. This means that the pressure P2 in the
second effect is lower than P1 in the first effect.
In a similar manner, vapor from the second effect is condensed in heating the third effect.
Hence, pressure P3 is less than P2 .
If the first effect is operating at 1 atm abs pressure, the second and third effects will be under vacuum.
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In the first effect, raw dilute feed is added, and it is partly concentrated.
Then this partly concentrated liquid flows to the second evaporator in series, where it is
further concentrated.
This liquid from the second effect flows to the third effect for final concentration.
When a multiple-effect evaporator is at steady-state operation, the flow rates and rate of
evaporation in each effect are constant.
The pressures, temperatures, and internal flow rates are automatically kept constant by
the steady-state operation of the process itself.
To change the concentration in the final effect, the feed rate to the first effect must be
changed.
The overall material balance made over the whole system and over each evaporator itself
must be satisfied.
If the final solution is too concentrated, the feed rate is increased, and vice versa.
Then the final solution will reach a new steady state at the desired concentration.
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Comparison between Forward Feed vs. Backward/Reverse Feed
https://www.yasa.ltd/post/multiple-
effect-evaporator-forward-feed-
backward-feed-parallel-feed-for-
multi-effect-evaporation
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Temperature Drops and Capacity of Multiple-Effect Evaporators
constant
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/Va
ATa =
EAT
10. + /on + Vs
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Calculations for Multiple-Effect Evaporators
In doing calculations for a multiple-effect evaporator system, the values to
be obtained are usually the area of the heating surface in each effect, the kg
of steam per hour to be supplied, and the amount of vapor leaving each
effect, especially the last effect.
The given or known values are usually as follows:
» 1. steam pressure to first effect,
» 2. final pressure in vapor space of the last effect,
» 3. feed conditions and flow to first effect
» 4. the final concentration in the liquid leaving the last effect
» 5. physical properties such as enthalpies and/or heat capacities of the liquid and vapors, and
» 6. the overall heat-transfer coefficients in each effect. & 10%
Usually, the areas of each effect are assumed equal. by finding but later we have to reconfirm
aug area
The calculations are done using material balances, heat balances, and the
capacity equations = for each effect.
A convenient way to solve these equations is by trial and error.
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Step-by-Step Calculation Methods for Triple-Effect Evaporators
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area value are
see if each
close to each other .
so kind average ,
if less than
then
M % difference
10
acceptable .
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Example 1: Evaporation of Sugar Solution in a Triple-Effect Evaporator
A triple-effect forward-feed evaporator is being used to evaporate a sugar
solution containing 10 wt.% solids to a concentrated solution of 50%. The boiling-
point rise of the solutions (independent of pressure) can be estimated from
BPR°C = 1.78 + 6.22 2 ,where is wt. fraction of sugar in solution. Saturated
steam at 205.5 kPa [121.1°C saturation temperature] is being used. The pressure
in the vapor space of the third effect is 13.4 kPa. The feed rate is 22 680 kg/h at
26.7°C. The heat capacity of the liquid solutions is cP = 4.19 2.35 kJ/kg· K.
The heat of solution is considered to be negligible. The coefficients of heat
transfer have been estimated as U1=3123, U2=1987, and U3=1136 W/m2.K. If each
effect has the same surface area, calculate the area, the steam rate used, and
Az A, Az
the steam economy. = =
S
Vi + v2 + Ve
steam economy
=
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Solution
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F + b =
$ + vi + vz + vz + 43
F = V
,
+ Ve + Vz + 43
V, vz
22688 = + + Vz +
13
Fxz =
L3Xz
(22680) (0 10) .
=
13 (0 50) .
13 =
4536kg/h
Assume V, =
V2 =
V.
22680 =
Vi + V2 + V3 + 4536
22680 = 3V , + 4536
Vz
Vz V= 6048 kgth
= = =
effect 1 :
FXF =
L, X
effect 2 : L , X1 :
12 X
2
effect
:
3 LeXc =
13 X
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17
T
these all letak 15 C
%
nak
are just estimates ,
bolen je
it must add tak boleh lebih
Kalar increase AT AT2 and AT3 Kena decrease to
S
, up
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(difference between steam temp and AT )
,
Cactual boiling point)
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datum °
:
0 C
temperature
BPR
20
BPR
BPR
21
22
23
24
25
26
27
different values of V
.
bila semva we get to see
solve ,
atas was just an assumption .
> davi
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29
Reflections
Multiple effect evaporator – arrangement vs calculations
Reflect on the calculation method for multiple effect
evaporator
30
