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21-Multi-Stage-Distillation-04.html
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<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8"/>
<title>
Separation Processes 1
</title>
<meta content="Separation processes 1 notes" name="description"/>
<meta content="Marcus Bannerman <[email protected]" name="author"/>
<meta content="yes" name="apple-mobile-web-app-capable"/>
<meta content="black-translucent" name="apple-mobile-web-app-status-bar-style"/>
<meta content="width=device-width, initial-scale=1.0, maximum-scale=1.0, user-scalable=no, minimal-ui" name="viewport"/>
<script src="header.js"></script>
</head>
<body>
<div class="reveal">
<div class="slides">
<section>
<section>
<h2>
Multi-Stage Distillation: Part 04
</h2>
<div class="center">
</div>
</section>
</section>
<section>
<section data-menu-title="Reflux Ratio">
<div style="display:flex;align-items:center;">
<div style="flex: 1 1 calc(100% * 0.651);">
<ul>
<li>
When designing a distillation column from scratch, the
only design specifications we will have are the feed and
outlet concentrations.
</li>
<li class="fragment" data-fragment-index="1">
We will need to decide upon a
<b>column operating
pressure</b>
and use this to calculate the
$q$
-value of the
feed stream (see the last lecture's slides).
</li>
<li class="fragment" data-fragment-index="2">
We will also need to choose a
<b>reflux ratio</b>.
</li>
<li class="fragment" data-fragment-index="3">
Selecting the
<b>reflux ratio</b>
and the
<b>column
operating pressure</b>
requires examining the economic
trade-off between
<b>capital</b>
and
<b>operating costs</b>.
</li>
<li class="fragment" data-fragment-index="4">
The
<b>column operating pressure</b>
will require a
full optimisation study. However, we can make some educated
estimates when it comes to the
<b>reflux ratio</b>
…
</li>
</ul>
</div>
<div style="flex: 1 1 calc(100% * 0.351);">
<figure>
<div class="center">
<img src="img/Benzene_Tolulene_McCabe_Thiele_05.svg" style="width:100%"/>
</div>
</figure>
</div>
</div>
</section>
<section data-menu-title="Total Reflux/Minimum Number of Trays">
<div style="display:flex;align-items:center;">
<div style="flex: 1 1 calc(100% * 0.651);">
<ul>
<li>
To understand the values of the reflux ratio, we need
to understand its limits and what they physically correspond
to.
\begin{align*}
R&=\frac{L_{n}}{D}
\end{align*}
</li>
<li class="fragment" data-fragment-index="1">
The first limit to consider is when we have the
maximum reflux ratio $R\to\infty$.
</li>
<li class="fragment" data-fragment-index="2">
This condition is known as
<b>total reflux</b>, and
occurs when
<b>no distillate is collected</b>
(
$D=0$), but all
rising vapour is refluxed back down the column
(
$L_{N+1}=V_N$).
</li>
<li class="fragment" data-fragment-index="3">
But what effect does this have on the column design?
</li>
<li class="fragment" data-fragment-index="4">
The intercept of the enrichment
<span style="color:teal">
operating line</span>
with the
$y$
-axis (
$x=0$)
goes to zero!
\begin{align*}
y(x=0)=\lim_{R\to\infty}\left(\frac{x_D}{R+1}\right)=0
\end{align*}
</li>
</ul>
</div>
<div style="flex: 1 1 calc(100% * 0.351);">
<figure>
<div class="center">
<img src="img/fractionation_column_01_top.svg" style="width:100%"/>
</div>
</figure>
</div>
</div>
</section>
<section>
<figure>
<div class="center">
<img src="img/partial_reflux.svg" style="width:50%;"/>
</div>
<figcaption>
The last class example on the McCabe-Thiele method for a
Benzene-Toluene mixture. Here we have a typical value of $R=3$
for the operating reflux ratio.
</figcaption>
</figure>
</section>
<section>
<figure>
<div class="center">
<img src="img/total_reflux.svg" style="width:50%;"/>
</div>
<figcaption>
At a reflux ratio of
$R\to\infty$, the enrichment
operating line coincides with the
$45^\circ$
line.
</figcaption>
</figure>
</section>
<section>
<figure>
<div class="center">
<img src="img/total_reflux_stepped.svg" style="width:50%;"/>
</div>
<figcaption>
When performing distillation at total reflux, the number
of trays required to reach a certain separation is at a
minimum! (compare
<span style="color:green!40!black">
$R=3$</span>
and
<span style="color:teal">
$R=\infty$</span>).
</figcaption>
</figure>
</section>
<section>
<div style="display:flex;align-items:center;">
<div style="flex: 1 1 calc(100% * 0.651);">
<ul>
<li>
We cannot actually operate at a total reflux ratio.
</li>
<li class="fragment" data-fragment-index="1">
Without producing any distillate, if any feed is added
it must leave in the bottoms product.
</li>
<li class="fragment" data-fragment-index="2">
What goes in, must come out, so we achieve nothing in
this case.
</li>
<li class="fragment" data-fragment-index="3">
However, we can use this to estimate the minimum
number of trays needed for a given separation!
</li>
<li class="fragment" data-fragment-index="4">
Also, during the start-up of a column, it is operated
under total reflux until sufficient column vapour and liquid
flow-rates have been obtained.
</li>
<li class="fragment" data-fragment-index="5">
Thus total reflux helps us understand what is
achievable, and is industrially relevant during
<b>plant
startup</b>.
</li>
</ul>
</div>
<div style="flex: 1 1 calc(100% * 0.351);">
<figure>
<div class="center">
<img src="img/fractionation_column_01_top.svg" style="width:100%"/>
</div>
</figure>
</div>
</div>
</section>
<section data-menu-title="Minimum Reflux Ratio">
<div style="display:flex;align-items:center;">
<div style="flex: 1 1 calc(100% * 0.651);">
<ul>
<li>
Just as we had a maximum reflux ratio, we can also
have a
<b>minimum reflux ratio</b>.
</li>
<li class="fragment" data-fragment-index="1">
<b>The reflux ratio should not reach zero</b>, as some
liquid must be returned to the column to allow a multi-stage
separation to take place. For
$R=0$, we will have only one
(partial reboiler) or zero (total reboiler) stages.
</li>
<li class="fragment" data-fragment-index="2">
So what actually is the minimum (but non-zero) reflux
ratio to perform a distillation at?
</li>
<li class="fragment" data-fragment-index="3">
Lets consider a distillation of an equimolar
($x_F=0.5$) Benzene-Toluene feed stream which, upon entering
the column, flashes to equal amounts of vapour and liquid
(
$q=0.5$).
</li>
<li class="fragment" data-fragment-index="4">
If the top and bottoms product have concentrations of
90% ($x_D=0.9$) and 10% (
$x_W=0.1$) benzene respectively,
what is the minimum possible reflux ratio?
</li>
</ul>
</div>
<div style="flex: 1 1 calc(100% * 0.351);">
<figure>
<div class="center">
<img src="img/fractionation_column_01_top.svg" style="width:100%"/>
</div>
</figure>
</div>
</div>
</section>
<section>
<figure>
<div class="center">
<img src="img/min_reflux_01.svg" style="width:50%;"/>
</div>
<figcaption>
From the data given already, we can draw most of the
McCabe-Thiele constructs including the
<span style="color:red">
$q$
-line</span>. We're only missing the
<span style="color:teal">
operating lines</span>.
</figcaption>
</figure>
</section>
<section>
<figure>
<div class="center">
<img src="img/min_reflux_02.svg" style="width:50%;"/>
</div>
<figcaption>
The only restriction on the placement of the enrichment
<span style="color:teal">
operating line</span>
is that it must intersect the
<span style="color:red">
$q$
-line</span>
somewhere below the
<span style="color:blue">
VLE</span>
line.
</figcaption>
</figure>
</section>
<section>
<figure>
<div class="center">
<img src="img/min_reflux_03.svg" style="width:50%;"/>
</div>
<figcaption>
The higher the interception of the enrichment
<span style="color:teal">
operating line</span>
with the
$y$
-axis, the lower
the reflux ratio. So the minimum reflux ratio is where the
<span style="color:red">
$q$
-line</span>,
<span style="color:blue">
VLE</span>
line, and the
<span style="color:teal">
operating line</span>
coincide (
$R_{min}\approx1.31$)!
</figcaption>
</figure>
</section>
<section>
<figure>
<div class="center">
<img src="img/min_reflux_04.svg" style="width:50%;"/>
</div>
<figcaption>
At
$R=R_{min}\approx1.31$, it is still theoretically
possible to achieve a separation. However, it will take
<b>an
infinite number of plates to overcome the pinch</b>.
</figcaption>
</figure>
</section>
</section>
<section>
<section data-menu-title="Pinches">
<div style="display:flex;align-items:center;">
<div style="flex: 1 1 calc(100% * 0.6);">
<ul>
<li>
The minimum reflux ratio corresponds to the first
(highest) $R$
where a
<b>pinch</b>
is formed.
</li>
<li class="fragment" data-fragment-index="1">
We can then just read off the $y$
-intercept of the
enrichment
<span style="color:teal">
operating line</span>
to find the
minimum reflux ratio.
\begin{align*}
R_{min}=\frac{x_D}{y(x=0)}-1
\end{align*}
</li>
<li class="fragment" data-fragment-index="2">
The
<b>optimum reflux ratio</b>
(see C&R vol. 2,
pg. 575 for details) is generally around 1.1–1.5 times the
<b>minimum reflux ratio</b>.
</li>
<li class="fragment" data-fragment-index="3">
This is where the running costs of the column (heating
steam for the boiler, cooling water for the condenser, pumps)
roughly balance with the capital costs of the column (number
of trays, cost of reboiler and condenser).
</li>
<li class="fragment" data-fragment-index="4">
So we can now calculate a rough reflux ratio, given
just the outlet conditions and the $q$
value! We can design
binary distillation columns…
</li>
</ul>
</div>
<div style="flex: 1 1 calc(100% * 0.4);">
<figure>
<div class="center">
<img src="img/min_reflux_04.svg" style="width:100%"/>
</div>
</figure>
</div>
</div>
</section>
<section data-menu-title="Non-Ideal liquids">
<div style="display:flex;align-items:center;">
<div style="flex: 1 1 calc(100% * 0.6);">
<ul>
<li>
<b>Pinches</b>
don't just appear at the intersection of
the
<span style="color:red">
$q$
-line</span>,
<span style="color:blue">
VLE</span>
line, and
the
<span style="color:teal">
operating line</span>.
</li>
<li class="fragment" data-fragment-index="1">
If the fluid is a non-ideal mixture, the pinch can
occur at some other point along the
<i>enrichment</i>
<span style="color:teal">
operating line</span>
and the
<span style="color:blue">
VLE</span>
line.
</li>
<li class="fragment" data-fragment-index="2">
A pinch can also occur somewhere along the
<i>stripping</i>
<span style="color:teal">
operating line</span>
and the
<span style="color:blue">
VLE</span>
line.
</li>
<li class="fragment" data-fragment-index="3">
Lets take a look at two examples…
</li>
</ul>
</div>
<div style="flex: 1 1 calc(100% * 0.4);">
<figure>
<div class="center">
<img src="img/min_reflux_04.svg" style="width:100%"/>
</div>
</figure>
</div>
</div>
</section>
<section>
<figure>
<div class="center">
<img src="img/enrichment_pinch.svg" style="width:50%;"/>
</div>
<figcaption>
This graph illustrates a pinch occurring on the
enrichment
<span style="color:teal">
operating line</span>
for a fictitious non-ideal
mixture's VLE curve.
</figcaption>
</figure>
</section>
<section>
<figure>
<div class="center">
<img src="img/stripping_pinch.svg" style="width:50%;"/>
</div>
<figcaption>
This graph illustrates a pinch occurring on the
stripping
<span style="color:teal">
operating line</span>
for a fictitious non-ideal
mixture's VLE curve.
</figcaption>
</figure>
</section>
<section data-menu-title="Azeotropes">
<div style="display:flex;align-items:center;">
<div style="flex: 1 1 calc(100% * 0.6);">
<ul>
<li>
There is a particular form of pinch which cannot be
overcome by changing the reflux ratio.
</li>
<li class="fragment" data-fragment-index="1">
This type of pinch was introduced to you in EG3029
Chemical Thermodynamics, and is called an
<b>azeotrope</b>.
</li>
<li class="fragment" data-fragment-index="2">
The most famous azeotrope exists in Ethanol-Water
mixtures, and prevents the distillation of more than 95.6% at
atmospheric pressure.
</li>
<li class="fragment" data-fragment-index="3">
We cannot place the top and bottoms product
concentrations on different sides of the
<b>azeotrope</b>, and
enrichment beyond the
<b>azeotrope</b>
concentration in a
single column is impossible.
</li>
</ul>
</div>
<div style="flex: 1 1 calc(100% * 0.4);">
<figure>
<div class="center">
<img src="img/min_reflux_04.svg" style="width:100%"/>
</div>
</figure>
</div>
</div>
</section>
<section>
<figure>
<div class="center">
<img src="img/azeotrope_pinch.svg" style="width:50%;"/>
</div>
<figcaption>
This graph illustrates the limitations placed on
distillation by the occurrence of an azeotrope. The top product
concentration cannot be moved above the azeotrope
concentration.
</figcaption>
</figure>
</section>
<section data-menu-title="Pressure-Swing Distillation">
<figure>
<div class="center">
<img src="img/azeotrope_swing.svg" style="width:50%;"/>
</div>
<figcaption>
One way of overcoming this azeotrope is to utilise
<b>pressure swing distillation</b>. At different pressures the
position of the azeotrope shifts (dotted blue line).
</figcaption>
</figure>
</section>
<section>
<figure>
<div class="center">
<img src="img/azeotrope_swing_distill.svg" style="width:50%;"/>
</div>
<figcaption>
If we first distill using the high pressure, we can use
this top product as the feed into a low-pressure distillation
column. Then we can concentrate up to arbitrary strength.
</figcaption>
</figure>
</section>
</section>
<section>
<section data-menu-title="Tray Efficiencies">
<div style="display:flex;align-items:center;">
<div style="flex: 1 1 calc(100% * 0.651);">
<ul>
<li>
The next subject on distillation design is
<b>plate
efficiencies</b>.
</li>
<li class="fragment" data-fragment-index="1">
So far we've only concerned ourselves with an
<b>overall plate efficiency</b>,
$E_O$.
</li>
<li class="fragment" data-fragment-index="2">
We can convert from theoretical stages ($N$) to real
stages (
$N_{real}$) by just dividing by the efficiency.
\begin{align*}
N_{real} = N / E_O
\end{align*}
</li>
<li class="fragment" data-fragment-index="3">
There are expressions for the overall efficiency
available in the literature for certain types of distillations
(See C&R Vol. 2, Eq. 11.126).
</li>
<li class="fragment" data-fragment-index="4">
However, in general an overall efficiency is a very
crude approximation and can only be used for rough
calculations.
</li>
<li class="fragment" data-fragment-index="5">
But there are better definitions of the efficiency
available…
</li>
</ul>
</div>
<div style="flex: 1 1 calc(100% * 0.351);">
<figure>
<div class="center">
<img src="img/fractionation_column_01_top.svg" style="width:100%"/>
</div>
</figure>
</div>
</div>
</section>
<section data-menu-title="Murphree Tray Efficiency">
<div style="display:flex;align-items:center;">
<div style="flex: 1 1 calc(100% * 0.651);">
<ul>
<li>
The
<b>Murphree
<u>Tray</u>
Efficiency</b>
is a
parameter that appears to fit reality a little better.
</li>
<li class="fragment" data-fragment-index="1">
The expression for the Murphree tray efficiency is
\begin{align*}
E_M = \frac{y_n-y_{n-1}}{y_{n}^{*}-y_{n-1}}
\end{align*}
where
$y_n$
is the real concentration of vapour coming off a
stage, and
$y_{n}^{*}$
is the equilibrium vapour
concentration.
</li>
<li class="fragment" data-fragment-index="2">
If the tray efficiency is one, these two values are
the same. At efficiencies less than one, the outlet vapour
will not reach equilibrium with the average tray
concentration.
</li>
<li class="fragment" data-fragment-index="3">
The easiest way to use such an efficiency is to apply it as
you step. Here we'll draw the
<i>curved</i>
<span style="color:green!40!black">
Murphree line</span>
on
the almost complete McCabe-Thiele chart, just for illustration.
</li>
</ul>
</div>
<div style="flex: 1 1 calc(100% * 0.351);">
<figure>
<div class="center">
<img src="img/fractionation_column_01_top.svg" style="width:100%"/>
</div>
</figure>
</div>
</div>
</section>
<section>
<figure>
<div class="center">
<img src="img/murphy_tray_efficiency.svg" style="width:50%;"/>
</div>
<figcaption>
To plot the
<span style="color:green!40!black">
Murphree line</span>, we
just put points at a fraction of the distance between the
<span style="color:teal">
operating line</span>
and the
<span style="color:blue">
VLE
line</span>
equal to the Murphree tray efficiency. The example above
is calculated at
$E_M=0.5$.
</figcaption>
</figure>
</section>
<section>
<figure>
<div class="center">
<img src="img/murphy_stepping.svg" style="width:50%;"/>
</div>
<figcaption>
To determine the number of real trays, we just perform
the stepping as usual, but use the
<span style="color:green!40!black">
Murphree line</span>
in place of the
<span style="color:blue">
VLE line</span>.
<b>We can also calculate the Murphree line as we step, which
is more accurate, so its highly recommended</b>. This distillation requires
at least 5 real trays and an ideal reboiler stage.
</figcaption>
</figure>
</section>
<section data-menu-title="Murphree Point Efficiency">
<div style="display:flex;align-items:center;">
<div style="flex: 1 1 calc(100% * 0.651);">
<ul>
<li>
We can extend the Murphree tray efficiency to the
<b>Murphree point efficiency</b>.
</li>
<li class="fragment" data-fragment-index="1">
The expression for the Murphree point efficiency is
\begin{align*}
E_P = \frac{y'_p-y'_{n-1}}{y_{n}^{*}-y'_{n-1}}
\end{align*}
where the primes denote the concentration at a certain point
on the plate.
</li>
<li class="fragment" data-fragment-index="2">
This is used to take into account that the
concentration varies along the plate as the tray fluid is not
well mixed.
</li>
<li class="fragment" data-fragment-index="3">
Use of this efficiency requires us to consider the
hydrodynamics within a plate, and this is too detailed for
this course.
</li>
<li class="fragment" data-fragment-index="4">
But, as always, you need to be aware of the
approximations of your solutions.
</li>
</ul>
</div>
<div style="flex: 1 1 calc(100% * 0.351);">
<figure>
<div class="center">
<img src="img/murphy_point_efficiency.svg" style="width:100%"/>
</div>
</figure>
</div>
</div>
</section>
</section>
</div>
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