Flash Distillation in Chemical and Process Engineering (Part 3 of 3)

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1. Multicomponent VLE
 Calculation of Bubble & Dew Points
 K-Values for Hydrocarbon Systems (dePriester)
2. Multicomponent Flash Distillation
 Methodology: Rachford-Rice
 Multicomponent Flashing Exercises
 Multicomponent Flashing Simulations

 Multicomponent VLE Theory
 Calculation of Bubble & Dew Points
 K-Values for Hydrocarbon Systems (dePriester)

 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…

 In this section, we will cover only multiple-alkane systems

 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

 Verify K-Values of several Hydrocarbons
 Pressure – Temperature Relationship
http://demonstrations.wolfram.com/KValueOfSeveralHydrocarbonsVersusTemperatureAndPressure/

 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

 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/

 For multicomponent, we are interested on calculating dew and bubble points

 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

 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 

 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


 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
   
   
   

 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
  
  
  

 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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 Solution:
 Step 1. Guess any T:
 ; nearest to nC63 6C nCT T T 
200 392
700 101
T C F
P kPa psi
   
 

 Step 2. Read K values

3
4
5
6
390
nC
nC
nC
nC
K
K
K
K
T F





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3
4
5
6
9.8
5.7
3.1
1.8
390
nC
nC
nC
nC
K
K
K
K
T F






 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  

3
4
5
6
2.66
?
?
?
?
nC
nC
nC
nC
K
K
K
K
T






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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 Normalize (divide by K function value)
( ) 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
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  

3
4
5
6
4.36
1.80
0.85
0.36
200
nC
nC
nC
nC
K
K
K
K
T F




 

( ) 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
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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3
4
5
6
3.70
0.78
0.65
0.28
180
nC
nC
nC
nC
K
K
K
K
T F




 

( ) 0.93i iF x K Z 
3.68
3.94
0.93
K  

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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 At this point, K = 1 approx.
 Accept T = 188°F or 190°F
 Note that Real Value of mix is  T = 87°C  188.6°F

 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

 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


 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  

 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

 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?

 Rachford-Rice Equation
 Derivation
 Procedure – Newton’s Method
 Worked Example
 Multicomponent Flash Distillation:
 Exercises
 Simulations

 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

 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
 

 
  
 

  

 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


  


 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


 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
 
   
        

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)]
benzene toluene xylenei
z Kz K z K
f
K K K
 
   
        
2
2
(1 )
'( )
[1 ( 1)]
i i
i
z K
f
K

 
  


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
 

 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
  



 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




 Step 3. Assume phi
 Recommended is:
 WHY?
0.5 

 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
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


  
  

 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
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 6. Calculate the new “phi”
( )
'( )
0.0823
0.50
0.4172
0.6972
old
new old
old
new
new
f
f

   


  
 

 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






 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
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
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

  

  

 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?

 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

 Step 3. Assume phi-values
0.50 
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 Step 4. Calculate RRE
 Step 5. Calculate Derivative of RRE:
Ethane  12.5
isobutane  2.80
n-pentane  0.95
n-hexane  0.45
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

 Step 7. Verify if error is acceptable.
 V/F = 0.7597
 A) Percentage of Vapor = 75.97%
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?
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

  

  

 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?

 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 3: Assume a phi value
 Phi = 0.5
 Step 4&5: Get RRE and its derivative

 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

 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

 Phi = 0.7597
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 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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1. Introduction
2. Review of Mass Transfer
3. Flash Distillation Concepts
4. Multicomponent Flashing
5. Conclusion

1. Introduction
1. Why Flash Distillation?
2. Objectives & Goals
3. Resources & Downloads
4. Additional Notes
2. Review of Mass Transfer
1. Ideal Solution & Gas, Equilibrium, Vapor & Partial Pressures, VLE
2. Volatility / Relative Volatility
3. Phase Diagrams (Txy, Pxy, XY)
4. Ideal Solution – Ideal Gas: Raoult’s Law (Vapor-Liquid)
5. Deviations: Azeotropes
3. Flash Distillation Concepts
1. Process Technology Overview
2. Equipment
3. Operation Line
4. Flash Cascades
4. Multiple Components
1. Introduction
2. Rachford-Rice
3. Alkane System
5. Conclusion

 More Engineering Courses
 Process Design & Simulation
 Aspen Plus
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 Unit Operations
 Gas Absorption & Stripping
 Flash Distillation
 Binary Distillation
 Oil & Gas
 Petrochemical Industries
 Petroleum Refining

Flash Distillation in Chemical and Process Engineering (Part 3 of 3)

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