FULL COURSE:
https://courses.chemicalengineeringguy.com/p/flash-distillation-in-chemical-process-engineering/
Introduction:
Binary Distillation is one of the most important Mass Transfer Operations used extensively in the Chemical industry.
Understanding the concept behind Gas-Gas, Liquid-Liquid and the Gas-Liquid mass transfer interaction will allow you to understand and model Distillation Columns, Flashes, Batch Distillator, Tray Columns and Packed column, etc...
We will cover:
REVIEW: Of Mass Transfer Basics (Equilibrium VLE Diagrams, Volatility, Raoult's Law, Azeotropes, etc..)
Distillation Theory - Concepts and Principles
Application of Distillation in the Industry
Equipment for Flashing Systems such as Flash Drums
Design & Operation of Flash Drums
Material and Energy Balances for flash systems
Adiabatic and Isothermal Operation
Animations and Software Simulation for Flash Distillation Systems (ASPEN PLUS/HYSYS)
Theory + Solved Problem Approach:
All theory is taught and backed with exercises, solved problems, and proposed problems for homework/individual study.
At the end of the course:
You will be able to understand mass transfer mechanism and processes behind Flash Distillation.
You will be able to continue with Batch Distillation, Fractional Distillation, Continuous Distillation and further courses such as Multi-Component Distillation, Reactive Distillation and Azeotropic Distillation.
About your instructor:
I majored in Chemical Engineering with a minor in Industrial Engineering back in 2012.
I worked as a Process Design/Operation Engineer in INEOS Koln, mostly on the petrochemical area relating to naphtha treating.
There I designed and modeled several processes relating separation of isopentane/pentane mixtures, catalytic reactors and separation processes such as distillation columns, flash separation devices and transportation of tank-trucks of product.
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We have been studying binary systems, that is two species
For this case, the Phase Rule stated:
If given:
F as the number of degrees of freedom
C as the number of components
P as the number of phases
Then this is true:
F = C-P+2
For a Ternary (3 species in equilibrium) System, then we get:
F = C-P+2 = 3-2+2 = 3
For a Quaternary system… and so on..
F = C-P+2 = 4-2+2 = 4…
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Recall that K-Value is a relationship between liquid and vapor phases:
Ki = yi/xi
According to chemistry, the hydrocarbons’ boiling point depends on their size, as
they will have mostly van der waal forces, i.e. the greater the size of the HC the
greater its boling point.
It is safe to assume that:
The larger (heavier) the HC, the greater its BP, i.e. the least volatile
If this is true:
Low boiling point HC have HIGH K-values
High boiling point HC have LOW K-values
It will now be convenient for us to work with K-Values in multicomponent systems
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Verify K-Values of several Hydrocarbons
Pressure – Temperature Relationship
http://demonstrations.wolfram.com/KValueOfSeveralHydrocarbonsVersusTemperatureAndPressure/
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For light hydrocarbons, the value of Ki of each species can be obtained from the
graph (called the “K chart”) prepared by DePriester
The temperature and pressure of the system must specified
Note that each plot/graph of each hydrocarbons can be written in the form of
equation:
X values vary from substance to substance and the units being used
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Input variables:
Set a Temperature
Set a Pressure
Set a component
Output variable
K-Value
As you will see, if T/P are fixed, then, for pure substances:
There is only a SINGLE line that describes these characteristic
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If we set T/P:
There is a unique
condition for a PURE
substance
See Lines:
Orange (high P– Low T)
Yellow (high P– high T)
Blue (low P– Low T)
Red (Low P– high T)
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Use the animation to verify certain conditions.
Methane
Propane
Octane
http://demonstrations.wolfram.com/DePriesterChartForHydrocarbons/
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If the specifications that you are given for your single-stage equilibrium separations
process are not T and P alone but say:
V/F = 0 and T or P
(which is a bubble point temperature or pressure, respectively)
or V/F = 0 and T or P
(which is a dew point temperature or pressure, respectively)
Do NOT use Rachford-Rice Equation!
In this case, we will have something between 0 < V/F <1
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Let us first consider bubble point calculations
In this case the liquid-phase composition xi is given
it corresponds to the case where V is very small and
Recall:
The bubble point of a liquid is the point where the liquid just starts to evaporate (boil),
that is, when the first vapor bubble is formed.
If the temperature is given:
then we must lower the pressure until the first bubble is formed.
If the pressure is given:
then we must increase the temperature until the first bubble is formed.
0V
i ix z
1i iK x
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NOTE:In both cases, this corresponds to adjusting T or p until the computed sum of
vapor fractions is just 1, that is,
Since:
Then
1iy
i
i i i i
i
y
K y K x
x
1i iK x
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At bubble point, V/F = 0, by definition in equilibrium
From the equation:
Step 1: Guess any Bubble Point Temperature
Step 2: Determine K-values from the Chart/Equation/Table/Plot
Step 3: If the function ( ) then Bubble Point is correct
Step 4: If the function is not 1 change Bubble point accordingly:
function >1 reduce T
function < 1 increase T
Step 5: Repeat Iteration until % error is met
Tip:
Best Educated Guess is
1i iK x 1i iK z
i ix z
1i iK z
old
new
i i
K
K
K z
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Bubble point at given temperature T.
A liquid mixture contains 50% pentane (1), 30% hexane (2) and 20% cyclohexane (3)
(all in mol-%), i.e.,
At T = 400 K, the pressure is gradually decreased.
What is the bubble pressure and composition of the first vapor that is formed?
Assume ideal liquid mixture and ideal gas (Raoult’s law).
1 2 30.5; 0.3; 0.2x x x
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Solution.
The task is to find a Pressure that satisfies
Since T is given, this is trivial (not required)
We can simply calculate P from the previous equation
We start by computing the vapor pressures for the three components at T = 400K.
Using the Antoine data, we get:
At the bubble point, the liquid phase composition is given, so the partial pressure of
each component is
( )i ix P T p
1
2
3
( 400 ) 10.248
( 400 ) 4.647
( 400 ) 3.358
P T K bar
P T K bar
P T K bar
1 1 1
2 2 2
3 3 3
(0.5)(10.248 ) 5.124
(0.3)(4.647 ) 1.394
(0.2)(3.358 ) 0.672bar
p x P bar bar
p x P bar bar
p x P bar
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Thus, from the equation of the bubble pressure we get:
Finally, the vapor composition (composition of the first vapor bubble) is
1 2 3 7.189p p p p bar
1
1
2
2
3
3
5.124
0.713
7.189
1.394
0.194
7.189
0.672
0.093
7.189
p bar
y
p bar
p bar
y
p bar
p bar
y
p bar
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A hydrocarbon liquid mix with
Composition (10,20,30,40% mol of; nC3, nC4, nC5, nC6)
Find the temperature at which we will get the first bubble formation.
Do NOT Assume ideal solution/gas
/ 0, 700V F P kPa
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Step 2. Read K values
3
4
5
6
390
nC
nC
nC
nC
K
K
K
K
T F
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Step 3. Use f(x) function
Step 4. K-value is Not what we expected
Guess new T… Recommended is to: Normalize (divide by K function value)
Do this for the smallest (MVC) Propane
Repeat Iteration!
( ) 3.69i iF x K Z
3 3 4 4 5 5 6 6
3 4 5 6
(0.10) (0.20) (0.30) (0.40)
(9.8)(0.10) (5.7)(0.20) (3.1)(0.30) (1.8)(0.40)
3.70
i i nC nC nC nC nC nC nC nC
i i nC nC nC nC
i i
i i
K Z K Z K Z K Z K Z
K Z K K K K
K Z
K Z
9.8
2.66
3.7
K
27. www.ChemicalEngineeringGuy.com
Step 2. Read K values
3
4
5
6
2.66
0.80
0.30
0.12
128
nC
nC
nC
nC
K
K
K
K
T F
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Step 3. Use f(x) function
Step 4. K-value is Not what we expected
Normalize (divide by K function value)
Do this for the smallest (MVC) Propane
Repeat Iteration!
( ) 0.61i iF x K Z
3 3 4 4 5 5 6 6
3 4 5 6
(0.10) (0.20) (0.30) (0.40)
(2.66)(0.10) (0.80)(0.20) (0.3)(0.30) (0.12)(0.40)
0.61
i i nC nC nC nC nC nC nC nC
i i nC nC nC nC
i i
i i
K Z K Z K Z K Z K Z
K Z K K K K
K Z
K Z
2.66
4.36
0.61
K
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Step 3. Use f(x) function
Step 4. K-value is Not what we expected
Normalize (divide by K function value)
Do this for the smallest (MVC) Propane
Repeat Iteration!
( ) 1.18i iF x K Z
3 3 4 4 5 5 6 6
3 4 5 6
(0.10) (0.20) (0.30) (0.40)
(4.36)(0.10) (1.80)(0.20) (0.85)(0.30) (0.36)(0.40)
1.18
i i nC nC nC nC nC nC nC nC
i i nC nC nC nC
i i
i i
K Z K Z K Z K Z K Z
K Z K K K K
K Z
K Z
4.36
3.70
1.20
K
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Step 3. Use f(x) function
Step 4. K-value is Not what we expected
Normalize (divide by K function value)
Do this for the smallest (MVC) Propane
Repeat Iteration!
( ) 0.93i iF x K Z
3.68
3.94
0.93
K
33. www.ChemicalEngineeringGuy.com
Step 2. Read K values
3
4
5
6
3.94
1.40
0.60
0.27
188
nC
nC
nC
nC
K
K
K
K
T F
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Let us next consider dew point calculations.
In this case the vapor-phase composition yi is given
(it corresponds to the case where L is very small ( ) and
The dew point of a vapor (gas) is the point where the vapor just begins to condense,
that is, when the first liquid drop is formed.
If the temperature is given
then we must increase the pressure until the first liquid is formed.
If the pressure is given
then we must decrease the temperature until the first liquid is formed.
0L i iy z
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In both cases, this corresponds to adjusting T or p until
Or, more conveniently:
1ix
1i
i
y
K
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At the dew point, by definition, and from f(1) = 0, we can find:
If this is true, then, when P and T are given:
Must be changed since T is too large… The steps are as follows:
Step 1. Guess any Dew Point Temperature
Step 2: Determine the K-values based on that
Step 3. Calculate . If function is not near 1, then:
Increase T when Function is less than 1.
Decrease T when Function is greater than 1.
Step 4. Recalculate New K-values based on normalization of K-old
Step 5. Repeat Iteration until acceptable % error value.
1i
i
Z
K
1i
i
Z
K
/ 1V F
1i
i
Z
K
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Calculate the Dew point at given Temperature T.
A vapor mixture contains:
50% pentane (1), 30% hexane (2) and 20% cyclohexane (3) (all in mol-%), i.e.,
At T = 400 K, the pressure is gradually increased.
What is the dew point pressure and the composition of the first liquid that is
formed?
Assume ideal liquid mixture and ideal gas (Raoult’s law).
1 2 30.5; 0.3; 0.2y y y
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Solution.
The task is to find the value of p that satisfies
Since T is given, this is trivial; we can simply calculate 1/p from (7.48).
From previous experiments and data, we got the following regression:
and we find
The liquid phase composition is:
Then, we find
1
( )
i
i
y
P T p
11 0.5 0.3 0.2
0.1729
10.248 4.647 3.358
bar
p
5.75p bar 1
( )
i
i
i
y
x
P T p
1 2 3
0.5 5.78 0.3 5.78 0.2 5.78
0.282; 0.373; 0.345
10.248 4.647 3.749
x x x
x x x
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For the previous mixture of Bubble Point:
Calculate its Dew Point
That is, assume it is a vapor
You are looking for condensation point
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Component / Molar flow
C1 20
C2 15
C3 12
C4 15
IC4 12
NC5 15
IC5 10
C6 5
C7 3
• A) Get Dew point @ T= ? P = 50bar
• B) Get Bubble point @ T= 220°C, P = 10bar
• C) What phase do we have at 25/25
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Component / Molar flow
B 0.25
T 0.25
O-X 0.25
P-X 0.25
• A) Get Dew point @ T= 150°C, P = 50bar
• B) Get Dew point @ T= 220°C, P = 10bar
• C) What is the Critical Point & Meaning?
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Next, consider a flash where a feed F (with composition zi) is split into
A vapor product V (with composition yi)
A liquid product (with composition xi)
For each of the Nc components, we can write a material balance:
In addition, the vapor and liquid is assumed to be in equilibrium,
i i iFz Lx Vy
i i iy K x
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The K-values:
Must be computed from the VLE model.
In addition, we have the two relationships:
With a given feed (F, zi), we then have:
3Nc + 2 equations
3Nc + 4 unknowns (xi , yi , Ki , L, V, T, p).
Thus, we need two additional specifications, and with these the equation set should
be solvable
, ,( ),i i i iK K T P x y
1
1
1
i
i
i i
x
y
x y
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The simplest flash is usually to specify p and T (pT-flash)
because Ki depends mainly on p and T .
Let us show one common approach for solving the resulting equations, which has
good numerical properties.
Substituting into the mass balance:
Gives
Solving with respect to xi gives:
Simplify via L = F − L (total mass balance) to derive
i i iFz Lx Vy
i i iy K x
(K )i i i iFz Lx V x
( )
( ) (1)
( ) 1 ( 1)
1 ( 1)
i i i
i i
i V
Fi i
V
F
i
i
i
Fz x L VK
L F z z
x
L V K K
z
x
K
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Here, we cannot directly calculate xi because the vapor split V /F is not known.
To find V /F we may use:
the relationship
alternatively
OR the addition of both…
However, it has been found that the combination Σi(yi−xi) = 0
It results in an equation with good numerical properties
This is the so-called Rachford-Rice Flash Equation
1
1
1
i
i
i i
x
y
x y
isat
i
p T
K
p
( 1)
0
1 ( 1)
i i
i
z K
K
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Rachford-Rice Equation:
Is a monotonic function in V/F
It is easy to solve numerically.
A physical solution must satisfy 0 ≤ V /F ≤ 1.
If we assume that Raoult’s holds, then Ki depends on p and T only.
Then, with T and p specified, we know Ki and the Rachford-Rice equation can be
solved for V /F.
For non-ideal cases, Ki depends also on xi and yi
One approach is add an outer iteration loop on Ki .
isat
i
p T
K
p
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This will be the typical procedure for the RRE
Note that this is based on a numerical method
Newton-Raphson Method
Uses the original function, f(phi)
It also requires the derivative of the function, f’(phi)
( 1)
( )
1 ( 1)
i i
i
z K
f
K
0.50
V
F
2
2
(1 )
'( )
[1 ( 1)]
i i
i
z K
f
K
( , , , ... )i j k zF z z z z
(V,y ,y ,y ...y )i j k z
(L,x ,x ,x ...x )i j k z
( , )T P
( , , ... )i j k zK K K K
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1. Given F, zi, T and P
2. Get Ki for species (either graph, equations or experimental values)
Antoine Equation (ideal)
K-Values from DePriester Chart
3. Assume a phi value (vaporized material in the feed) (hint good start is 0.5)
4. Get the Numerical Value of Rachford Rice Equation
Example:
BTX ( Benzene, Toluene, Xylene) System:
( 1)
0
1 ( 1)
i i
i
z K
K
0.50
V
F
(1 )(1 ) (1 )
( )
1 ( 1) 1 ( 1) 1 ( 1)
xylene xylenebenzene benzene toluene toluene
benzene toluene xylene
z Kz K z K
f
K K K
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5. Get the Numerical Value of the Derivative of Rachford Rice Equation
Example:
BTX ( Benzene, Toluene, Xylene) System:
V
F
22 2
2 2 2
(1 )(1 ) (1 )
'( )
[1 ( 1)] [1 ( 1)] [1 ( 1)]
xylene xylenebenzene benzene toluene toluene
benzene toluene xylenei
z Kz K z K
f
K K K
2
2
(1 )
'( )
[1 ( 1)]
i i
i
z K
f
K
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6. Recalculate phi (Newton Raphson Method)
7. Verify Rel. Error, if < 0.0001, this is ok, otherwise go to step 3 (repeat iteration)
V
F
( )
'( )
old
new old
old
f
f
% .error 100%new old
old
rel abs x
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8. Get V, L, xi, and yi (see equations)
Note that in all cases:
You will need K and phi
V V F
L L F V
1 ( 1)
1 ( 1)
i i
i
i
i
i
i
K z
y
K
z
x
K
V
F
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Given a Flash with the following data:
Composition (zi)
(BTX, 0.60, 0.25, 0.15)
VLE (Antoine Constant) Data:
A) Get the compositions, flow rates of Vapor & Liquid streams
i A B C
B 6.879 1196.700 219.160
T 6.950 1342.000 219.190
X 7.000 1476.390 213.870
1 , 100
100
&
kmol
h
P atm T C
F
V L unknown
55. www.ChemicalEngineeringGuy.com
Step 1 – Get the given Data:
1 , 100 , 100
0.6; 0.25, 0.15
kmol
h
benzene toluene xylene
P atm T C F
Z Z Z
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Step 2 Get Ki-values
Via Raoult’s Law
Via DePriester Diagrams (not available for BTX)
If we use Raoult’s Law & Antoine’s Equation:
1.7756
0.7322
0.2611
benzene
toluene
xylene
K
K
K
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Step 4. Calculate The Value of the Rachford-Rice Equaiton (RRE)
0.5
(1 )
( )
1 ( 1)
(1 )(1 ) (1 )
( )
1 ( 1) 1 ( 1) 1 ( 1)
0.60(1 1.7756) 0.25(1 0.7322)
( )
1 0.50(1.7756 1) 1 0.50(0.732
i i
i
xylene xylenebenzene benzene toluene toluene
benzene toluene xylenei
z K
f
K
z Kz K z K
f
K K K
f
0.15(1 0.2611)
2 1) 1 0.50(0.2611 1)
( ) 0.0823f
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Step 5. Calculate The Value of the derivative of RRE
2
2
22 2
2 2 2
2
2
0.5
(1 )
'( )
[1 ( 1)]
(1 )(1 ) (1 )
'( )
[1 ( 1)] [1 ( 1)] [1 ( 1)]
0.60(1 1.7756) 0.25
'( )
1 0.50(1.7756 1)]
i i
i
xylene xylenebenzene benzene toluene toluene
benzene toluene xylenei
z K
f
K
z Kz K z K
f
K K K
f
2 2
2 2
(1 0.7322) 0.15(1 0.2611)
1 0.50(0.7322 1)] 1 0.50(0.2611 1)]
'( ) 0.4172f
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Step 7. Verify phi value (error)
Error is too large, repeat from step 3
% . 100%
0.6972
% . 100%
0.5000
% . 139%
new
old
rel error x
rel error x
rel error
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The most convenient way to do this is via a Spreadsheet… as we will need to iterate
Step 7. Verify phi value (error)
Best Case phi = 0.6806; error is acceptable
Trial phi f(1) f(2) f(3) f(phi) f'(1) f'(2) f'(3) f'(phi) New Phi %error
1 0.5 -0.33532 0.0773 0.175775 -0.08225 0.187402 0.023901 0.205979 0.417282 0.6971 39.42
2 0.6971 -0.30205 0.08232 0.228567 0.008834 0.152057 0.027105 0.348286 0.527447 0.6804 2.40
3 0.6804 -0.30462 0.08187 0.222879 0.000126 0.154654 0.026808 0.026808 0.20827 0.6797 0.09
4 0.6797 -0.30471 0.08185 0.222679 -0.00018 0.154749 0.026797 0.026797 0.208344 0.6806 0.13
5 0.6806 -0.30458 0.08187 0.222971 0.000269 0.154611 0.026813 0.026813 0.208236 0.6793 0.19
6 0.6793 -0.30478 0.08184 0.222544 -0.00039 0.154813 0.02679 0.02679 0.208394 0.6812 0.28
7 0.6812 -0.30448 0.08189 0.223167 0.000572 0.154518 0.026823 0.026823 0.208164 0.6785 0.40
2
2
22 2
2 2 2
2
2
0.5
(1 )
'( )
[1 ( 1)]
(1 )(1 ) (1 )
'( )
[1 ( 1)] [1 ( 1)] [1 ( 1)]
0.60(1 1.7756) 0.25
'( )
1 0.50(1.7756 1)]
i i
i
xylene xylenebenzene benzene toluene toluene
benzene toluene xylenei
z K
f
K
z Kz K z K
f
K K K
f
2 2
2 2
(1 0.7322) 0.15(1 0.2611)
1 0.50(0.7322 1)] 1 0.50(0.2611 1)]
'( ) 0.4172f
f’(1) f’(2) f’(3)
0.5
(1 )
( )
1 ( 1)
(1 )(1 ) (1 )
( )
1 ( 1) 1 ( 1) 1 ( 1)
0.60(1 1.7756) 0.25(1 0.7322)
( )
1 0.50(1.7756 1) 1 0.50(0.732
i i
i
xylene xylenebenzene benzene toluene toluene
benzene toluene xylenei
z K
f
K
z Kz K z K
f
K K K
f
0.15(1 0.2611)
2 1) 1 0.50(0.2611 1)
( ) 0.0823f
f(1) f(2) f(3)
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Step 8. Calculate all other Values
For compositions, use Spreadsheet
(0.6803)(100) 68.03 /
100 68.03 31.97 /
V V F V kmol h
L L F V kmol h
1 ( 1)
1 ( 1)
i i
i
i
i
i
i
K z
y
K
z
x
K
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Step 8. Calculate all other Values
1 ( 1)
1 ( 1)
i i
i
i
i
i
i
K z
y
K
z
x
K
Species i Ki zi phi yi xi
Benzene 1 1.78 0.60 0.6806 0.69728 0.392703
Toluene 2 0.73 0.25 0.6806 0.22385 0.305722
Xylene 3 0.26 0.15 0.6806 0.07879 0.301747
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A feed:
(1) 20 mol % ethane
(2) 20 mol % isobutane
(3) 20 mol % n-pentane
(4) 40 mol % n-hexane
Flash Operation is at T = 100°c, P = 600kPa
A) What mole fraction of the feed is vaporized?
B) Composition of vapor?
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Step 1: Calculate K Values
F = must be assumed, 100
Z = given, 0.2,0.20,0.20,0.40
Given, 100°C (212F), 600kPa (87 psi)
Step 2. Calculate K Values
Use DePriester Chart.
Ethane 12.5
isobutane 2.80
n-pentane 0.95
n-hexane 0.45
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Step 3. Assume phi-values
0.50
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Step 7. Verify if error is acceptable.
V/F = 0.7597
A) Percentage of Vapor = 75.97%
Trial phi f(1) f(2) f(3) f(4) f(phi) f'(1) f'(2) f'(3) f'(4) f'(phi) New Phi %error
1 0.500 -0.34074 -0.1895 0.010256 0.303448 -0.21651 0.580521 0.179501 0.000526 0.230202 0.990751 0.7185 43.71
2 0.7185 -0.2483 -0.157 0.010373 0.363752 -0.03115 0.308257 0.123206 0.000538 0.330788 0.762789 0.7594 5.68
3 0.7594 -0.23632 -0.1521 0.010395 0.37778 -0.00024 0.279228 0.115673 0.115673 0.356794 0.867368 0.7596 0.04
4 0.7596 -0.23624 -0.1521 0.010395 0.37788 -3.2E-05 0.279044 0.115624 0.115624 0.356983 0.867274 0.7597 0.00
5 0.7597 -0.23623 -0.1521 0.010395 0.377893 -4.3E-06 0.279019 0.115617 0.115617 0.357008 0.867262 0.7597 0.00
6 0.7597 -0.23623 -0.1521 0.010395 0.377895 -5.7E-07 0.279016 0.115616 0.115616 0.357011 0.86726 0.7597 0.00
7 0.7597 -0.23623 -0.1521 0.010395 0.377895 -7.5E-08 0.279016 0.115616 0.115616 0.357012 0.86726 0.7597 0.00
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B) Composition of vapor?
Species i Ki zi phi yi xi
Ethane 1 12.5 0.20 0.7597 0.256764 0.020541
i-Butane 2 2.8 0.2 0.7597 0.23654 0.084479
n-Pentane 3 0.95 0.2 0.7597 0.197502 0.207897
n-Hexane 4 0.45 0.4 0.7597 0.309191 0.68709
0.999998 1.000007
(0.7597)(100) 75.97 /
100 75.97 24.03 /
V V F V mol h
L L F V kmol h
1 ( 1)
1 ( 1)
i i
i
i
i
i
i
K z
y
K
z
x
K
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For a mix at T= 95°C and P = 700kPa
The composition of 40, 30, 20, 10 mol percent:
propane (1)
n-butane (2)
n-pentane (3)
n-hexane (4)
A) What percentage of the feed enters as liquid?
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Step 1. Get data
F, T, P, zi are given!
T= 95°C = 203°F
P = 700kPa = 101 psi
Step 2. Get Ki
From K-chart:
4.20
1.75
0.74
0.34
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Step 6: Verify, iterate
NOTE
If The Value of RRE < 0, this is above the bubble point… WHY?
If the Value of RRE derivative > RRE, this is ABOVE the dew point…
Trick question… This is all VAPOR!
phi 0.5
i zi Ki f(phi) f'(phi)
1 0.4 4.2 -1.28 4.096
2 0.3 1.75 -0.225 0.16875
3 0.2 0.74 0.052 0.01352
4 0.1 0.34 0.066 0.04356
Sum = -1.387 4.32183
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In the following animation:
Change the Flash Pressure
See the Effect of Q in temperature
Compare the volatility of species:
Butane vs. heptane
https://demonstrations.wolfram.com/FlashDistillationOfAMixtureOfFourHydrocarbons/
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Use Aspen Plus Software to Simulate Ex 1. Rachford-Rice Flashing
A) Verify Composition & Streams
B) Use Physical Property Analysis:
PV-Curve
Mixture Properties
C) Use Sensitivity Analysis for:
Effect of Temperature in Vapor
Tmin? WHY
Tmax? WHY
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Phi = 0.7597
Species i Ki zi phi yi xi
Ethane 1 12.5 0.20 0.7597 0.256764 0.020541
i-Butane 2 2.8 0.2 0.7597 0.23654 0.084479
n-Pentane 3 0.95 0.2 0.7597 0.197502 0.207897
n-Hexane 4 0.45 0.4 0.7597 0.309191 0.68709
0.999998 1.000007
(0.7597)(100) 75.97 /
100 75.97 24.03 /
V V F V mol h
L L F V kmol h
Trial phi f(1) f(2) f(3) f(4) f(phi) f'(1) f'(2) f'(3) f'(4) f'(phi) New Phi %error
1 0.500 -0.34074 -0.1895 0.010256 0.303448 -0.21651 0.580521 0.179501 0.000526 0.230202 0.990751 0.7185 43.71
2 0.7185 -0.2483 -0.157 0.010373 0.363752 -0.03115 0.308257 0.123206 0.000538 0.330788 0.762789 0.7594 5.68
3 0.7594 -0.23632 -0.1521 0.010395 0.37778 -0.00024 0.279228 0.115673 0.115673 0.356794 0.867368 0.7596 0.04
4 0.7596 -0.23624 -0.1521 0.010395 0.37788 -3.2E-05 0.279044 0.115624 0.115624 0.356983 0.867274 0.7597 0.00
5 0.7597 -0.23623 -0.1521 0.010395 0.377893 -4.3E-06 0.279019 0.115617 0.115617 0.357008 0.867262 0.7597 0.00
6 0.7597 -0.23623 -0.1521 0.010395 0.377895 -5.7E-07 0.279016 0.115616 0.115616 0.357011 0.86726 0.7597 0.00
7 0.7597 -0.23623 -0.1521 0.010395 0.377895 -7.5E-08 0.279016 0.115616 0.115616 0.357012 0.86726 0.7597 0.00
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