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Single-Effect Evaporator - Heat Transfer Area

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Language: en

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

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