In this screen cast I am going to describe the operation of a single effect evaporator.
Which is used to remove water from an aqueous solution in order to concentrate the solution
this may be a solution that contains sugar, sodium chloride, orange juice, and the final
objective is this concentrated solution.
So what we are going to look at is a schematic representation of this single effect evaporator.
So in this schematic representation we have a feed coming in which is a liquid and we
are going to remove vapor and therefore end up with a concentrated liquid leaving the
system.
So we will label this feed, F in kg/hr coming in.
And it is going to be coming in with a concentration of the solute, the solid that are dissolved,
a mass fraction xF.
Vapor is going to be pure solvent, which in general most of the systems we are looking
at,this is water.
So that means that he mole fraction of the solute is zero there.
And then we have the liquid.
So this is the concentrated liquid.
This is our objective is to create this concentrated liquid.
And so we have mass fraction of the liquid.
The mass fraction of the liquid is greater than the mass fraction in the feed.
A couple of things to note, remember this is a dilute liquid, feed temperature is less
than the temperature in the evaporator.
We are going to call this temperature T of the evaporator (TE) and we are going to assume
this is well mixed.
And this is important.
Assuming this is well mixed, the temperature here, the temperature here, are also the evaporator
temperature and this is because this is mixed in the evaporator sufficiently such that we
have no gradients in temperature.
So the pressure in the evaporator is going to determine what the evaporation temperature
is for the system.
The other aspect of this single effect evaporator is that we have steam coming in to cause the
evaporations so we are going to have steam, which is causing the evaporation so we will
have steam flow, comes into the evaporator and moves through the same flow rate as steam
then leaves the evaporator.
So, also kg/hr.
Temperature of the steam (TS) and we are going to assume this is saturated steam.
Quality of 1 and here we are assuming all the steam is condense so then the liquid is
dripping out of the exchanger tube.
So this is at the same temperature.
So let's just look at some of the governing equations for this system.
One would be a mass balance and we will do the mass balance on the overall feed.
So this feed coming in is split into two streams.
Liquid plus vapor.
And then we can do a mass balance on the solute, the solid.
So the mass fraction in the feed times the flow rate.
The only place these solids leave are in the liquid phase, the mass fraction of the liquid
times the mass flow rate of the liquid that is leaving.
So this is our mass balances for the system.
One of the energy balances that is important for this single effect evaporator is the heat
transfer.
So the heat transfer rate, this is kJ per hr, equal to an overall heat transfer coefficient.
So this heat transfer coefficient is taking into account all the heat transfer both sides
of the heat transfer tubing.
So this is kJ/m^2 per hour per degree K. Heat transfer area which would be in square meters.
And then the temperature difference so the TS, which of course is higher than the TE.
Bigger this temperature difference of course the faster the heat transfer.
And so these are the basic equations.
We now also have to do enthalpy balances on the streams that we will look at in a subsequent
screen cast.
There are more efficient ways to carry out evaporation for example a triple effect evaporator
to have three of these in series, that saves a significant cost on the steam but adds a
cost of the construction and so it is a relative tradeoff, it is a tradeoff of the cost of,
if steam is cheap, then this might be the better approach.