Evaporators are a common process whenever large
amounts of water or another solvent are to be separated from
relatively
non-volatile components.
They function by simply
boiling
the volatile
component away.
They are often seen when large amounts of fresh water
are required. For example, drinking water on ships or for
steam ship boilers (
top product).
Evaporators are also used to concentrate aqueous solutions
such as fruit juice or orange juice (bottom product).
They are common when the process fluid contains
precipitates that need to be
crystallised.
We study evaporators first in this course as they are
essentially an ideal flash-distillation step (there is a
perfect separation of the components). No need to worry
about the phase equilibria part of thermodynamics (yet).
Later on, we will have to calculate how much of the
two (or more) species is present in the top and bottom product
using vapour-liquid equilibrium data, or solubility data.
However, in this course, the defining characteristic
for evaporators
is that we will assume
all of the
involatile components (e.g., salt, solids) are recovered in
the bottoms product.
There are many designs of evaporator, but there are some
fundamental operational classifications.
Evaporators may be
single-pass, where the fluid is
heated only once, or
recirculating, with a long
residence time
of the treated fluid.
Evaporators may also operate in
convective
or
boiling
regimes depending on the state of the fluid at the heat
source.
The flow around the heat source may be
natural
or
forced.
We'll look at some common designs now to get a feel for the
evaporator process equipment. While looking, try to see why
the vapour and liquid outlets might be considered
"well-mixed" and in equilibrium.
There are many designs of evaporator:
There is the
open-kettle
or
pan
which is a
simple evaporator.
It is usually heated through a jacket or coil, and may
be directly fired.
They are inexpensive and simple to operate, but the
efficiency is poor.
They are common for small scale or old/traditional
process designs.
For example, old salt pans used in salt mines or
traditional maple syrup refiners (see right).
No longer used except for "traditional" (i.e., inefficient)
production.
There are many designs of evaporator:
A common and simple design is the
horizontal-tube
natural circulation evaporator.
The heating steam is condensing on the tube side and
the tube bundle is completely submerged in the liquid.
They are simple designs which are relatively easy to
clean.
These evaporators are good for low-viscosity fluids
with high heat transfer coefficients, as it relies on
natural convection.
There are many designs of evaporator:
There is the
vertical-tube natural circulation
evaporator.
In this evaporator, the boiling liquid is on the tube
side, but also flows around the tube bundle in a
downcomer (multiple pass).
The heating of the fluid in the tubes causes natural
convection to occur.
This pumps the liquid up the tubes increasing the
liquid velocity and the heat transfer coefficient.
As such, this design cannot be used with viscous
liquids, but it is more efficient than the kettle type
evaporator.
The lower portion of the evaporator is known as the
calandria.
Two designs of vertical-tube natural circulation
evaporator.
There are many designs of evaporator:
There is the
long-tube vertical-type evaporator.
The tubes can be 3 to 10m high and the liquid may only
pass through the evaporator once (
single-pass).
Boiling causes vapour bubbles to form which pump the
liquid at high velocities through the tubes.
Contact times
can be low, which is useful in the
food industry (no burning of the product).
There are many designs of evaporator:
A variation on the long-tube vertical-type evaporator
is the vertically fed
falling film type evaporator.
Gravity and expansion of the fluid greatly increase
the fluid velocities.
These can be designed with very short contact times.
This is an advantage as damage to sensitive
heat-treated fluids is a function of temperature and contact
time.
Falling film type evaporator.
There are many designs of evaporator:
There are also
forced-flow evaporators.
These may exploit the pressure of a liquid head to
prevent boiling in the exchanger.
They may also have a circulating pump to force the
flow, or they may rely on the
thermosyphon
effect.
This is an analogue of the chimney effect, where a
strong form of natural convection is used to pump a fluid.
Despite the
“natural”
origins of the
thermosyphon flow, the fluid is being strongly forced through
the heat exchanger.
There are many designs of evaporator:
Finally, for viscous fluids, there are evaporators
which mechanically coat the heated surface with the fluid
(agitated-film).
This ensures good heat transfer regardless of the
viscosity and can be used for the drying of very thick pastes.
Deciding which evaporator to use strongly depends on
the cost of materials, energy and fluid properties.
Luckily, there are many case studies in the literature
which have been collated into tables to aid you in the
selection of your evaporator…
A common feature of all types of evaporators are
deentraining devices such as
baffles.
These reduce or prevent entrained droplets of liquid
exiting in the top stream.
Evaporators are also typically operated as
total-condensers on the steam side (only
condensate/water is removed from the steam side). This
greatly simplifies the heat balance on the steam side, as
its the latent heat of steam vapourisation/condensation.
When designing your evaporator, you start with a
feed
stream, and you have to achieve an
output specification.
For example, you have a feed stream of 1000 kg/hr of
orange juice with a solids concentration of 7.08
w/w.
You need to concentrate it to 50
w/w solids content before
it is economical to ship it overseas.
How big should our exchanger be to achieve this? What is
the heat transfer area required? What utilities (steam) will be
used? What is the overall
duty
of the evaporator?
Single-stage evaporators are, essentially, a heat
exchanger and we roughly size heat exchanges using their
area.
To find the size of the evaporator, we need to solve the
heat transfer equation for the heat transfer area, A
:
\begin{align*}
Q = U A \left(T_{steam} - T_{fluid}\right)
\end{align*}
We need to find the duty, Q, the heat transfer
coefficient
U, the boiling temperature of the evaporator
T_{steam}, and
the steam temperature.
The steam temperature is set by the available utilities,
and the heat transfer coefficient/area can be calculated together
from our EX3030 knowledge once the duty is known.
So we must first compute the duty, Q, and we need a
simple model of our evaporator to do this…
Our simple model for the evaporator is that it has one
input stream, and two output streams.
All of the solids entering in the feed stream F, is
recovered in the liquid stream
C.
No heat is lost or gained during this process, except by
that removed or supplied by the duty Q.
With the initial design data, we can see that there are
two unknown flowrates of streams we need to solve, L
and
V.
We need the fundamental equation of process engineering,
the
balance equation
:
\begin{align*}
\textrm{INPUT}-\textrm{OUTPUT}+\textrm{GENERATION}=\textrm{ACCUMULATION}
\end{align*}
Taking a total mass balance and a mass balance in the
solids content we have
\begin{align*}
F &= L + V & x_{s,F} F &= x_{s,L} L
\end{align*}
These equations are our general balance equations for an
evaporator.
Taking the solids balance, we have
\begin{align*}
x_{s,F} F &= x_{s,L} L\\
0.0708\times1000 &= 0.50 \times L\\
L &= 0.0708\times1000 / 0.5 = 141.6 \textrm{kg/hr}
\end{align*}
Using the total balance, we have
\begin{align*}
F &= L + V\\
1000 &= 141.6 + V\\
V &= 1000 - 141.6 = 858.4 \textrm{kg/hr}
\end{align*}
We'll now update our diagram…
To calculate the duty, Q, we need to perform an energy
balance to obtain…
\begin{align*}
\textrm{Energy In} + \textrm{Duty} &= \textrm{Energy Out}\\
F h_F + Q &= L h_L + V h_V
\end{align*}
where
h_F
is the
enthalpy
per mass for the feed stream,
F.
The enthalpies, h, for each stream depend on temperature
AND composition!
The most difficult part of evaporator design may be
tracking down accurate values for the stream enthalpies.
The difficulty is that there is an infinite selection of
mixtures and a very limited amount of experimental data.
We will look at enthalpy and boiling point data in a later
lecture, and will make some simplifying assumptions for now...
Let's assume that the inlet stream is at its boiling point
(100^{\circ}C
@ 1atm) and the solids
concentration does not
affect the boiling temperature.
We also assume that there is no enthalpy of mixing (
concentration does not affect the enthalpy,
h_F=h_L).
Then the only enthalpy difference will come about due to
the boiling of the water content.
Using h_F=h_L
and
F-L=V, the energy balance yields:
\begin{align*}
Q &= V \left(h_V - h_L\right)
\end{align*}
We have from the steam tables h_V-h_L=2260
kJ/kg at
100
{}^\circ
C, therefore
\begin{align*}
Q &= 858.4\times2260\approx1.9\times10^9\textrm{J}\,\textrm{hr}^{-1}\approx540\textrm{kW}
\end{align*}
To calculate the evaporator size, we need the overall heat
transfer coefficient, and the temperature difference.
The overall heat transfer coefficient depends on the
evaporator type, operating method and flow velocities (see
EX3030, Heat, Mass, and Momentum Transfer.).
You will probably have to iterate to a solution to U
in
a detailed design (just like your design project).
But there are typical values available in the literature…
Type
Overall U (W/m
{}^2
K)
Short-tube vertical, natural
1100–2800
Horizontal-tube, natural
1100–2800
long-tube vertical, natural
1100–2800
long-tube vertical, forced
2300–11000
Agitated film
680–2300
Typical values for the evaporator heat transfer
coefficient,
U, taken from “Transport Processes and Unit
Operations,” by C. J. Geankoplis.
Finally, we need the temperatures of the evaporator.
The temperature of the evaporator fluid depends on the
solvent boiling point and an effect called
boiling point
rise.
For simplicity, we can assume at low concentrations the
boiling point is the same as for the pure solvent.
\begin{align*}
T_{fluid} = T_{sat.,solvent} = 100^\circ\textrm{C}
\end{align*}
The temperature of the steam depends on its saturation
temperature.
\begin{align*}
T_{steam} = T_{sat.,steam}
\end{align*}
This is because the latent heat of condensation of the
steam is used to heat the evaporator (the condenser is a total
steam condenser).
The saturation temperature depends on the pressure of the
available steam, and can be looked up in your steam tables.
To calculate how much steam, S, the evaporator will use,
we will need another heat balance:
\begin{align*}
Q &= S \left(h_{steam,vapour} - h_{steam,liquid}\right)
\end{align*}
However, if we assume that steam has a latent heat of
vaporisation with only a weak temperature and concentration
dependence, the amount of steam condensed will be the same amount
of water evaporated!
\begin{align*}
S \approx V
\end{align*}
This concludes our roughest design of an evaporator, we
will improve this calculation in the coming lectures.