In this video I am going to show an example where we calculate the heat transfer area
for a single effect evaporator that is being used to concentrate a salt solution.
the feed salt solution is 1.5 % salt in water and the exit is 4% salt in water.
And the feed rate is given and the temperature of the evaporator is at 1 bar and we are using
saturated steam at 170 kPa to provide the heat for vaporization.
And the question is, if we know the heat transfer coefficient 2500 W/m^2 per degree K, what
is the heat transfer area and what assumptions do we have to make to do this calculation
if this is the only information that we have available to us.
So the first thing we do in a problem like this is to draw a diagram and define our variables.
So here is a diagram and the single effect evaporator is explained in more detail in
another video.
Feeding in saturated steam to the evaporator through some sort of heat transfer coil, so
the steam leaving, is condensed liquid at saturated conditions.
Feed is 1.5% and the feed flow rate is 7500 kg/hr.
One of the assumptions we make is there is no salt in the vapor which is certainly what
we would expect.
So the other assumptions, one is that the evaporating liquid is well mixed.
And so that means the temperature for the vapor, temperature inside the temperature
liquid, all the same temperature.
it also means the concentration, this weight percent leaving is the same as the weight
percent inside our evaporator.
The second assumption we are going to make is that the heat of vaporization is going
to be calculated based on the heat of vaporization of water.
The fact that it is a solution, we could still use the steam tables to get the heat of vaporization.
We also assume there is no heat of mixing or unmixing, and the salt is added to the
water, water is removed from the salt.
This may or may not be a good assumption based on information we have not knowing the salt.
This is a reasonable first approximation.
We are also going to assume that the boiling point temperature increase, namely the fact
that we add salt, lowers the water fugacity if you like, and therefore raises the boiling
point.
We are going to assume for now that that is zero.
That the concentration is low on a molar basis such that we don't get much of a temperature
increase.
This may not be a good assumption depending on the system and this can have a significant
effect on the heat transfer area because looking at the temperature difference between this
steam and the temperature TE.Our next step now is to write down mass balances for this
system because we need to know the flow rates of the liquid and the vapor leaving in order
to calculate the energies.
And there is the overall mass balance, the flow coming in, the molar flow rate is leaving
as liquid or as vapor.
And then there is a balance on the salt.
So if we take the mole fraction of the salt and the feed, times the molar flow rate, all
of the salt appears in the liquid leaving.
so this is 7500 kg/hr and we don't know L, but of course we can solve this to determine
L. So this is the total liquid flow rate.
Which means the vapor flow rate is 7500-2812.
So the vapor flow rate 4688 kg/hr.
Now that we have the mass balances, the next step is to do energy balance on the stream
that we are evaporating.
Apply the first law, steady state system, a flow system.
No accumulation.
The energy coming in is the mass flow rate times the enthalpy per kg entering the system.
And then the energy leaving is the mass flow rate of the vapor times the enthalpy of the
vapor per kg, minus the mass flow rate of the liquid times the enthalpy of the liquid
per kg.
And then the other term that we have is the heat transfer.
So it is a heat transfer coefficient times the heat transfer area times the temperature
difference, the temperature of the steam, saturated steam coming in, saturated liquid
leaving.
So everything is at water vapor and liquid TS- minus the temperature in the evaporator.
so this is the first law.
We can use steam tables to look up values.
So what I have listed is flow rate, steam coming in 85 degrees we will assume that the
1.5 wt% solution can be modeled as pure water at 85 decrees C.
We will use the enthalpy of saturated steam at 85 degrees C, that is a pretty good approximation
to the steam at 1 bar and 85 degrees C. Vapor at 1 bar, so 100 degrees C, and then liquid
also at 1 bar, 100 degrees C. And then we look up for 170 kPa the saturation temperatures
around 115 degrees C. So we can substitute the values for mass flow rates and our enthalpies
into this equation.
We are also given the heat transfer coefficients so our only unknown is the heat transfer area
and that is what we are going to solve for.
So let's substitute the numbers in.
So here are the values substituted in that we looked up in the steam table and we calculated
from the mass balances.
And then we have to be aware of a Watt is a J/s.
We have to keep units consistent and that is why it is valuable to write the units with
each of the numbers.
So we convert time to hours, convert J to kJ for the heat transfer coefficient.
And then the temperature difference is the same as the temperature difference in K.
Now certainly two significant figures is probably more than we can justify since we made a number
of assumptions.
And one of the big assumptions that affects things is the fact that the temperature difference
here is going to affect the heat transfer area a lot is temperature is going to be higher
for a 4% solution than the temperature difference is less the heat transfer area would have
to be more.