Flash Distillation in Chemical and Process Engineering (Part 2 of 3)

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1. Drum Design
 Vertical vs. Horizontal
 Hold-up times
 Equipment & Auxiliary Pieces
2. Control Systems

 Vertical preferred when:
 small liquid load
 limited plot space
 ease of level control is desired 12
 Prefered when L/D < 5
 Horizontal preferred when:
 large liquid loads are involved
 consequently hold-up will set the size
 three phases are present
 Preferred when L/D > 5

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 These are some common dimensions
 More will be seen in:
 Flash sizing
 Exercise:
 Find L/D Ratio

 Hold Up Time (at half full)
 2 to 32 minutes depending on quality of
control for each outgoing stream
 5 to 10 minutes is sufficient with modern
control systems to handle minor upsets
 30 minutes provides a 99% separation in
most cases
 Engineering Judgement !

 The Mesh
 Nozzles
 Drains & Manholes
 Vortex Breakers & Foam Breakers
 Straightening Vanes
 Distributors
 Level Gauges & Thermowells

 The mist extractor is the final gas cleaning device in a conventional separator.
 The selection, and design to a large degree, determine the amount of liquid
carryover remaining in the gas phase.
 The most common types include:
 Wire mesh pads (“mesh pads”)
 Vane-type (vane “packs”)
 Axial flow demisting cyclones.
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 The formation of mists in gas streams often results in process inefficiencies
and/or product loss in equipment such as:
 knockout drums, distillation columns, evaporators, and environmental scrubbers.

 Mists can also cause serious damage to downstream equipment.
 Compressors may fail if liquid droplets are not separated from the gaseous input
stream

 Typical nozzles are shown next
 The main purpose of an inlet device is to improve separation performance.
 This is achieved by maximizing the amount of gas-liquid separation occurring in the feed
pipe
 This minimizes droplet shearing, and
optimizing the downstream velocity
distributions of the separated phases
into the separator.
 Schematics for inlet devices are shown
in Figure 6.

 In large capacity, more critical separation
applications, the vane-type and cyclonic
inlet devices are commonly used.
 The simpler, and less expensive, impact
(or diverter plates) are often used where
the separation performance is less
critical.

 Ideal vs Vortex Formation due to Drain / outlets
 Excessive Pressure Drops
 Erosion of nozzle
 Poor Separation Steps
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 There are several types of breakers:

 Foam can be formed, recall that foam is a colloidal system
between liq-vap.
 This must be avoided. We use Foam breakers

 Disfavors turbulence
 Flow improves

 Baffles avoid direct flow, is used to avoid certain profiles

 Typically you will find:
 Level Controllers
 Pressure Controllers
 Temperature Controllers
 Flow Rate controllers
 Most of the variables can be controlled with
valves
 Pressure  open Gas Vent
 Liquid Level  Open Liquid Outlet
 Flow Rate  Open valve on inlet
 Temperature  Heating Jacket & Flow Rate
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 Shown above is a simple flash vessel.
 The aim of this is to separate a stream into vapour and
liquid by altering the pressure (and hence temperature).
 There are three control loops.
 F  Flow control on feed line
 L  Inventory control on liquid line
 V  Pressure control on vapour line

 Flow control on feed line
 On a flash vessel it is necessary to have the flow of one stream
controlled.
 This is to ensure that the throughput is known.
 This flow control operates the same as that shown above i.e if the
flow decreased then the valve is opened and vice versa.

 Pressure control on vapour line
 The composition of the resulting streams depends on the
pressure in the vessel.
 If the pressure:
 increases then the valve is opened to allow more vapour through.
 decreases then the valve is closed to allow the pressure to build back
up again.
 Notice, however, that both these actions will have an effect
on the level of liquid in the tank (WHY?)
 If the pressure decreases then more liquid will vapourise and the
level will decrease.
 If the pressure increases then the level will rise.

 Inventory control on liquid line
 The shaded valve denotes an inventory loop.
 This is to ensure that the mass balance around the vessel balances
i.e. what goes in must come out.
 In this case it is simply a feedback liquid level control loop.
 If the level increases then the valve is opened and if the level
decreases then the valve is closed.
 Thus the vessel should not overflow or run dry.
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 Other variables you might encounter:
 CV = Control Valve
 RC = Ratio Controller
 SV = Set Variable
 FT = Flow Transmitter
 PV = Process variable
 PC = Pressure Controller
 LC = Level Controller

 Typical Arrangements for:
Non-Adiabatic
Isothermal
Adiabatic 0Q 0Q 

 Identify the controllers
 Verify why are they required
 Why are there more controllers in this diagram?
 Tip  RC:
 What relates Flow of Vapor and Liquid?

 Sequential Method:
 Flash Distillation  Operation Line (FOL)
 Analysis of Operating Line  Changes in Fraction Vapourised
 Exercises!

 Degrees of Freedom  6
 Variables 
 Feed Flow Rate, F
 Feed Composition, xF
 Temperature, T1
 Pressure, p1
 Remaining:
 Drum Pressure, Pdrum
 X  this is the interesting part!
 Vapor mole fraction, y
 Liquid mole fraction, x
 Fraction feed vaporized, f = V/F
 Fraction of feed remaining in liquid, q = L/F
 Temperature of the flash drum Tdrum
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 There are several ways we can solve a Flash System:
 Sequential
 Equation by equation, variable by variable
 More intuitive
 Simultaneous
 All equations are solved at the same time
 Convergence Methods
 Note that in both cases, we calculate:
 Flow rates (F, V, L)
 Ccompositions of vapor, liquid and feed in A and B
 Temperatures, Pressures, etc..

 We will first solve the mass balance and equilibrium relationships
 Then we will solve the energy balance & enthalpy equations (if required)
 We can do this in a binary system if at least one of these is given:
 Vapor mole fraction, y
 Liquid mole fraction, x
 Fraction feed vaporized, f = V/F
 Fraction of feed remaining in liquid, q = L/F
 Temperature of the flash drum Tdrum
 Recall that due to equilibrium we just need a single variable to “fix” the system
 We will need an Operation Line

 Operation Lines are very commonly used in Separation Processed & Mass Trnasfer
Operations
 This is a line in which the operation will take place
 Initial spot
 Final spot
 Typically, there will be a change regarding the position in the EQUILIBRIUM
 The Driving Force of the process ( change in concentrations) is given by this
“change”
 The Operating Line is also based on a material balance
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 Example of Operation Lines
 Gas Absorption
 Flash
 Distillation (stripping, enriching sections)
 Batch Distillation
 Liquid-Liquid

 Operation Line of a Flash Distillation
 0 < q < 1
 Slope = -(1-f)/f
 f  quality inlet

 Operation Line of a Batch Rectifier Distillation
 Line varies with time
 Depends:
 Constant Reflux
 Constant Distillate Composition
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 Operation Line of a Gas Absorber
 No intersection
 Quasi-parallel lines
 Min. Operating Points

 Operation Lines of a Binary Fracitonal Distillation
 Rectifying/Enriching Section
 Feeding
 Stripping Section

 Recall that we have three main lines:
 45° line
 Relates x=y
 Equilibrium Curve
 Relates mol x = mol y
 Operation Line
 Our actual operation

 If no vapourization takes place:
 the liquid leaving the separator will have the same
composition as the feed
 If total vapourization occurs:
 the vapour will also have the same composition as the feed.
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 Clearly, for effective separation, there must be
some vapour-liquid mixture present.
 This means that there exists a certain relationship
between:
 the extent of heating
 the concentration in the vapour and liquid streams.

 Define ”f”
 molar fraction of the feed that is vaporized and withdrawn continuously as vapour
 Therefore:
 for 1 mole of binary feed mixture (basis F = 1 mol)
 (1- f) is the molar fraction of the feed that leaves continuously as liquid
 (Also total mol flow of B)
 And, by definition, f will be the total mol flow of D

 Assume:
 yD = mole fraction of A in vapor leaving
 xB = mole fraction of A in liquid leaving
 xF = mole fraction of A in feed entering.
 Streams:
 F = Feed
 D = Distillate
 B = Bottoms

 Based on the definition for “f”:
 the greater the heating is:
 the larger the value of “f”
 If the feed is completely vaporized
 then f = 1.0
 Thus, the value of f can vary:
 From 0 (no vaporization) to 1 (total vaporization).
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 To see how yD and xB changes when f change:
 We need the material balance to obtain a so-called operating line equation
 It should relate the variables:
 yD
 xB
 xf

 From total Mass Balance:
 F = D + B
 From material balance for the more volatile component (A):
 
1
( ) (1 )
( ) (1 )
F D B
F D B
F D B
F D B
D F B
F D B
Fx Dy Bx
Fx Dy Bx
F
D B
x y x
F F
x f y f x
f y x f x
 
 
 
 
  
  

 Re-arrange into the form y = f(x):
( ) (1 )
(1 )
(1 )
(1 )
D F B
F B
D
F B
D
B F
D
f y x f x
x f x
y
f
x f x
y
f f
f x x
y
f f
  
 


 

  

 We see that for a given value of “f”, we will obtain certain values for yD and
xB respectively.
 The fraction “f” depends on:
 Enthalpies going “in”
 Enthalpies going “out” (Vapor and Liquid)
(1 ) B F
D
f x x
y
f f

  
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 For simplification:
 Let us drop the subscript 'D' and 'B' gave the general operating line equation:
 yD = y
 xB = x
(1 ) B F
D
f x x
y
f f

  
(1 ) 1
D F
f x
y x
f f

  

 The equation is a straight line equation…
 Slope = -(1-f)/f
 Y-intercept = xF / f

 The equation is a straight line equation…
 Slope = -(1-f)/f
 Y-intercept = xF / f
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 The operating line crosses the point
 (xF, xF) for all values of f
 WHY?
 This provides one point on the straight line
 The 45° Line, or x=y line
 The other point can be obtained from the
intercept of from (XF/f)

 With xF known, we can construct the
operating line on the equilibrium curve.
 Intersection between the:
 operating line
 equilibrium curve
 Will yield the values for:
 yD
 xB
 WHAT ARE THESE?

 Steps:
1. Identify XF  Known
2. Identify the x=y (Xf) point
3. Identify y-intercept  xf/f
 Slope can be used as well
4. Write the Operation Line ( to )
5. Find point in which:
 Operation Line crosses Equilibrium Line
6. yD and xB are obtained from this point

 Recall that the operating line equation for flash distillation is:
 For a given feed composition  xF is fixed.
 When the fraction of feed vapourised is changed:
 the mole fraction MVC in the vapour and liquid products changes accordingly.
(1 ) 1
D F
f x
y x
f f

  

 If f = 0
 no vaporization; vertical
 If f = 1
 total vaporization; horizontal
 If 0< f <1
 90° < angle < 180°
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Counter-Clockwise  x/y in equilibrium decreases

 Summary
 Verify:
 “f” values (0, x, 1)
 Diagram type
 Type of solution
 45°
 If vapor increases  rotates clockwise
 If vapor decrases  rotates counter-clockwise

 A simple algebraic manipulation of the two equations gives the following quadratic
equation, which can be solved for x:
http://demonstrations.wolfram.com/FlashDistillationOfAConstantRelativeVolatilityMixture/

 Use the Sequential Method:
 From Wankat Ex 3-1
 A flash distillation chamber operating at 1 atm is going to separate ethanol-water
mix.
 The feed mix is 40% ethanol.
 A) What is the max. vapor composition
 B) What is the min. liquid composition
 C) if V/F = f = 0.4, find L and V compositions
 D) Repeat c) given that F = 1000 kmol/h

 Step 1 - Define Requirements
 Step 2 – Draw the x-y diagram
 Step 3 – Get Equilibrium data for ethanol-water system at P = 1 atm
 Step 4 – Solve for the given cases
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 Step 1. Define Requirements
 For a) find the highest y (vapor composition) possible
 This is only possible at V/F = (WHY)
 For b) find the lowest x (liquid composition) possible
 This is at V/ = 1.0, why?
 For c) given V/F = 0.4
 use operation line for x and y
 For d) given F = 1000, re-solve c
 The answer should be the same, since equilibrium is based on composition, not mass amount
 Sizing will change
 Energy requirements will change

 Step 2 – Draw the x-y diagram
 Step 3 – Get Equilibrium data for:
 ethanol-water system
 P = 1 atm

 Step 4 – Solve for a, b, c and d using the
Operation Line:

 A) ymax given xf = 0.40
 In order to maximize ymax, we need to
have a vertical line
 WHY?
 The red dot will be the composition
 y = 0.61, x = 0.4

 B) xmin given xf = 0.40
 In order to minimze xmin, we need to
have a horizontal line
 WHY?
 The red dot will be the composition
 y = 0.40, x = 0.08

 C) Given V/F = f = 0.40
 Substitute in:
 yD = -(1-0.4)/0.4x + 1/0.4x(0.4)
 yD= -1.5x+1
 Graph either:
 By slope + x-y point (blue point)
(1 ) 1
D F
f x
y x
f f

  
Slope = for every 0.5 dx there is -0.75 of dy
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 C) Given V/F = f = 0.40
 Substitute in:
 yD = -(1-0.4)/0.4x + 1/0.4x(0.4)
 yD= -1.5x+1
 Graph either:
 By y-intercept (red point) + x-y point (blue
point)
(1 ) 1
D F
f x
y x
f f

  

 In both cases… the cross in the equilibrium
line is the same point
 (x = 0.29, y = 0.57)

 D) Given F = 1000 kmol/h
 No effect… since slope and intercepts do not
depend on the value…
 Given V/F = f = 0.40
 Substitute in:
 Note that V/F = 0.4
 V = 0.4xF = 0.4x1000 = 400
 F = L+V  L = F-V = 1000-400 = 600
 The line remains the same, as well as the
slope and y-intercept
(1 ) 1
D F
f x
y x
f f

  
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 Typically we will continue with an Energy balance to determine enthalpies and Heat
exchange
 Finally, we continue with Flash Drum Design & Sizing…

 This is a problem that follows the common steps!
 The sequential method!
http://demonstrations.wolfram.com/ConstructAnXYDiagramForFlashDistillation/

 An equimolar mixture of benzene and toluene is subjected to flash distillation at a
pressure of 1 bar in the separator.
 Determine the compositions (in mole fraction benzene) of the liquid and vapour
leaving the separator when the feed is X% vaporized.
 A) Estimate the temperature in the separator for 25%
 B) Calculate the composition
 Proposed Homework:
 Bi) Rpeat (A) for cases (0%, 50%, 75%, 100%)
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 Equilibrium data for benzene-toluene system at 1 bar is given in this table:

 Solution (A) Temperature for a 25%
 The operating line equation is given as follows:
 f = 0.25 (25% of feed is vapourised)
 xF = 0.50 (equimolar mixture, MVC = benzene)
 Flash distillation operating line:
 Locate the first point ( x = xF = 0.50, y = xF = 0.50 ) on the 45o diagonal.
 Locate the second point using the operating line equation.
 Plot the operating line by joining the 2 points.
 Intersection between the operating line and equilibrium curve gives the solution for xB and yD.

 From graph: xB = 0.44, yD = 0.66
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 The separator temperature can be estimate by interpolation or determined from the
equilibrium phase diagram.
 T is approx.  92.4 oC.

 Solution (B) for 0, 50, 75, 100%
 The same calculations can be repeated for the other scenarios (0,50,75,100) and the results
are tabulated in the following Table:
 Conclusion:
 We can see from the results that there is a trade-off between quantity and concentration:
 the higher the fraction of feed vapourised (i.e. the larger the vapour product) the lower the mole
fraction MVC in the vapour.

 Conclusion:
 We can see from the results that there is a trade-off between quantity and
concentration:
 the higher the fraction of feed vapourised (i.e. the larger the vapour product) the lower the
mole fraction MVC in the vapour.
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 Simulate the previous problem
 Components  Benzene, Toluene
 Use  Peng Robinson Model
 Use this Blocks:
 HEATER
 VALVE
 FLASH2
 Set up accordingly
 HEATER
 Vapor fraction = X
 VALVE
 P = 1 bar
 FLASH2
 Q = 0, P = 1 bar

 A liquid mixture containing 50%-50% n-heptane and n-octane mix at 30°C is to be
flash vaporized at 1 atm. The idea is to vaporize 60% of the feed (40% liquid,
60%vapor)
 A) Calculate the vapor’s final composition

 Solution
 Assume 
 F = 100 mol
 zF = 0.50 (equimolar)
 B = 40, D = 60  -W/D = -40/60 = -0.667
 From the plot:
 Identify FEED (point P, in x=y line)
 Draw Operation line:
 With the slope  -0.667
 Intersect in equilibrium curve (Point “T”)
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 Point T:
 yD* = 0.575 approx.
 Therefore, xD = 0.387

 A vapor at its dew point and 1 atm contain: 40%
Benzene and 60% Toluene.
 Total mix is approx. 100 kmol.
 It is contacted with a 110 kmol mix of liquid at its
boiling point
 This mix contain 30%B, 70%T
 They are sent to a flash unit (single stage)
 The outlet stream leave in equilibrium.
 A) Identify final compositions and stream flows
 B) Identify stream flows distillate and bottoms

 Solution:
 Mass balance:
 Total:
 F = L+D = 100+110 = 210 kmol
 On Benzene:
 Lxb + Dyb = Fxb
 (110)(0.3) + (100)(40) = Fxb
 Benzene flow = 73 kmol B (in both, liquid & vapor)
 In equilibrium:
 (110)(0.3) + (100)(40) = (Bvap + Bliq)
 (110)(0.3) + (100)(40) = 110x + 100y*
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 Now, we need equilibrium data
 We can’t guess T-B in liquid/vapor…
 Option 1)
 Trial & Error (no analytical equation)
 Option 2)
 Fit an equation to Equilibrium line

 Option 1)
 Iteration 1
 Balance on Benzene (before equilibrium  after equilibrium)
 (110)(0.3) + (100)(40) = 110x + 100y*
 Assume x = 0.20
 (110)(0.3) + (100)(40) = 110(0.20) + 100y*
 y = (73-22)/(100) = 0.51
 Verify if true…
 x = 0.2, y = 0.51  NOT TRUE
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 Option 1)
 Iteration 2
 (110)(0.3) + (100)(40) = 110x + 100y*
 Assume x = 0.40
 (110)(0.3) + (100)(40) = 110(0.40) + 100y*
 Y = (73-44)/(100) = 0.29
 X = 0.4, y = 0.29  NOT TRUE

 Option 1)
 Iteration 2
 (110)(0.3) + (100)(40) = 110x + 100y*
 Assume x = 0.25
 (110)(0.3) + (100)(40) = 110(0.25) + 100y*
 Y = (73-27.5)/(100) = 0.45
 X = 0.25, y = 0.45  TRUE!

 Final vapor composition:
 Based on Benzene:
 X = 0.25, y = 0.45  TRUE!
 Yb = 0.45
 Therefore, for toluene:
 Yt = 1-0.45 = 0.55
 Calculate for Liquid as well…
 Xb = 0.25
 Xt = 1-xb = Xt = 0.75

 For the system methanol-water; we have an azeotrope..
 Try to obtain the best separation!
http://demonstrations.wolfram.com/AdiabaticFlashDrumWithBinaryLiquidFeed/
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 Previously:
 Material Balance (Total)
 Material Balance (MVC)
 More data can be related for our system… typically:
 Energy balance
 Enthalpies
 Temperatures
 Heatloads
F V L 
i i iF y xFz V L 

 Energy Balance
 Isothermal:
 Adiabatic:
F f lFh Q Vh Lh  
F f lFh Vh Lh 

 The variables hF, hL, and hV are the enthalpies of:
 Feed
 Liquid
 Vapor streams
 These are all functions of pressure, temperature, and compositions:
( , , )
( , , )
( , , )
F F F F
L L L L
V V V V
h f P T z
h f P T z
h f P T z




 The unit is generally assumed to be under:
 Mechanical equilibrium 
 Thermal equilibrium
 Phase equilibrium 
 Ki is generally a function of (P,T)
F L VP P P 
F L VT T T T  
i i iy K x
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 With the compositions, temperature, and pressure known, the enthalpies of the
outlet streams, hL and hV, can be found from:
 The feed enthalpy, hF, can be found from the enthalpy balance and also the feed
temperature TF.
 The amount of energy required in the heater, Q, can be determined from an energy
balance around the heater:
F f lFh Q Vh Lh  
( , ,x)
( , ,y)
L L L
V V V
h f P T
h f P T



 Finally, the heat balance:
( , , ) ( , , )
( , , ) ( , , )
[ ( , , ) ( , , )]
hx in in F F F F
hx F F F in in F
hx F F F in in F
Q Fh P T z Fh P T z
Q Fh P T z Fh P T z
Q F h P T z h P T z
 
 
 

 Regarding Energy balances we will need to use T-xy diagram..
http://demonstrations.wolfram.com/TXYAndXYDiagramsForBinaryVaporLiquidEquilibriumVLEInFlashDru/

 A vertical Flash Drum is used to separate 100 kmol/h of (40-60%) water-ethylene
glycol mix.
 Feed conditions: P = 300 kPA, T = 150°C
 Flash Drum Conditions: P = 30.4 kPa, 0.33 V/F ratio
 Get the required equilibrium data from Aspen Plus or HYSYS
 A) Find the compositions in the outlet
 If Adiabatic:
 Bi) Temperature of the drum
 Ci) Temperature at the outlet (vapor/liquid)
 Di) Is the Heat-load (added or removed)

 Using HYSYS:
 First, let us setup the components
 Then, use NRTL
 Then, go to simulation Environment
 Select the Equilibrium Operator (OP-100)
 Select the XY Plot
 Select both Species
 Select 30.4 Kpa

 From the diagram…
 The Operation Line
1
(0.33 1)
0.33 0.33
2 1.2
y mx b
V
zFy
V V
F F
z
y x
y x
 

 

 
  
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 Operation Line:
 Adjust to x,y= feed = 0.4
 A) Compositions
 X-water = 0.1073
 Y-water = 0.9756
1
(0.33 1)
0.33 0.33
2 1.2
y mx b
V
zFy
V V
F F
z
y x
y x
 

 

 
  

 If Adiabatic…
 Using the same Block E, then change XY to TXY
 For:
 X = 0.1073
 Y = 0.9756
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 Bi) The Temperature of the Flash Drum
 Using the same Block E, then change XY to TXY
 For:
 X = 0.1073
 T  118.6°C
 Ci) The Temperature of the Vapor/Liquid
 Since they are in equilibrium,
 Di) By definition, this is Adiabatic Operation:
 Q = 0
drum outletT T

 Conclusion…
 A good follow up question…
 If the system is to be Isothermal, calculate Q required

 A binary system of Ethanol-Water is to be flashed
 Conditions:
 P = 98kPa, T = 180°F
 q = 1 (saturated vapor)
 L:V ratio is  3V = L
 Basis may be assumed 100 kg/h (L = 75, V = 25)
 A) Determine composition of exiting streams (V/L)
 B) Determine composition of INLET (feed)
 C) Find the Energy required to add/remove
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 For this, we will need an enthalpy-composition (H-
xy) diagram
 WHY? we do not have inlet compositions, but we
DO have saturation and T/P
 According to Phase rule, we can get compositions in
equilibrium as well 
 Use that of 180°F (orange)
 Red dots  Saturation points

 Liquid composition (ethanol) = 0.45
 Vapor Composition (ethanol) = 0.75

 B) Determine composition of INLET (feed)
 Perform Mass balance for compositions:
 Mass Balance
 Mass Balance over Ethanol:
100 3
100 / 4 25; 3 25 75
F V L
V V
V L x
 
 
   
     
   
 0.75 25 0.45 75
100
0.525
Feed Vapor Liquid
z F y V x L
y V x L
Z
F
x x
Z
Z
 
 






 For the inlet:
0.525
1 0.525 0.475
Z ethanol
Z water
 
   

 C) If Isothermal Operation, Find the Energy
required to add/remove
 Getting Enthalpies:
 Substitute:
     ( ) –( )
Feed Vapor Liquid
Vapor Liquid Feed
vap liq feed
H Q H H
Q H H H
Q V h L h F h
  
  
 
0.75 640
0
( )
( ).45 110
( 0.52 800)
vap axis
liq axis
feed axis
h x
h x
h x
 
 
 
     55 640 165 110 – 220 800
122,650 /Q
Q x x x
BTU h
  



:
25 / 25 / 0.454 / 55lb
75 / 75 / 0.454 / 165lb
100 / 100 / 0.454 / 220lb
Mass
kg h kg lb
kg h kg lb
kg h kg lb
 
 
 

 Final notes…
 What will be the inlet temperature? Is it the
same as the outlet?
 NO  WHY?
 This is not adiabatic, i.e. Q=/0, therefore there
was a temperature change…
 Between:
 200°F
 Let it be approx. 200F  92°C
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 We need to setup:
 Components: Water, Ethanol
 Method : NRTL
 Simulation: Heater + Flash Drum
 Streams  Feed, Feed2, Vapor, Liquid

 There are several variables to consider whenever
sizing a Flash Drum Vessel
 Basic:
 Dv  Diameter of Drum
 L/H  Length of Drum
 Mesh/DeMister:
 Design Height (Nozzle to Mesh)
 Mesh/Mist Eliminator Height (6” = 150 mm)
 Mesh to Top Height
 Detail
 Nozzle Level
 LLL  Low Liquid Level
 H  Holdup Level
 S  Surge Level
 HLL  High Liquid Level
 HLIN  (HLL to Nozzle)

 There are several variables to consider whenever
sizing a Flash Drum Vessel
 Basic:
 ID  Diameter of Drum
 L  Length of Drum
 Mesh/DeMister:
 Hv
 Detail
 LLL  Low Liquid Level
 NLL  Normal Liquid Level
 HLL  High Liquid Level
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 Once vapor, liquid composition and flow rates have been determined…
 We can set the Size now!
 The following steps are used
 Step 1. Calculate Permesible Vapor Velocity (u perm)
 Step 2. Convert uperm and Vapor rate “V” to horizontal area
 Step 3. Set Diameter:Length Ratio
 Step 4. Verify Calculations

 Step 1. Calculate Permissible Vapor Velocity (uperm)
 As the name implies, it is the maximum permissible vapor velocity in ft/s at the
maximum cross-sectional area.
 The densities of liquid and vapor are crucial
 Typically, Kdrum is an empirical constant which is correlated from experimental data.
 For a 75-85% flooding + no demister
L V
perm drum
V
u K
 




 Demister Notes:
 No use of demister
 Approx. 5% of liquid will be entrained in the vapor
 Use of demister
 Approx- 1% of liquid
 Typical sizing depens on manufacturer
 Generally, assume 6 in or 150 mm
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 The following equation is used to calculate “K” of drum, the constant for the drum
sizing
 Where:
 and are the flow rates of liquid and vapor in lb/h
0 1 2 3 4
lv lv lv lv(lnF ) (ln ) C(lnF ) (lnF ) (lnF )
e lvA B F D E
drumK    

VL
lv
V
W
F
W L



1.8774
0.8146
0.1871
0.0145
0.0010
A
B
C
D
E
 
 
 
 
 LW VW

 Typical Kdrum-values
0 1 2 3 4
lv lv lv lv(lnF ) (ln ) C(lnF ) (lnF ) (lnF )
e lvA B F D E
drumK    


 Step 2. Convert uperm and Vapor Rate V to horizontal area:
 Solving for Cross-Sectional Area (Ac)
   2
3
3600
1
( / )
perm c V
vapor
s lbmftu A ft
s h ft
V lbmol h
lbm
MW
lbmol

  
   
   
( )
3600
v
c
perm V
V MW
A
u 

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 Step 3. Set Diameter:length ratio
 Option 1: Rule of thumb
 Vertical  3.0-5.0
 Option 2: As a surge tank
 Typically:
1
236 ; 48
12
0.54
v feed line
f
l
h in D in
h in
h m
  


hV  HD
hf  HLIN
hL  Sump
hV
hf
hL

 Step 4. Verify Calculations
 The depth of the liquid pool, hL, can be determined from the desired surge volume,
Vsurge:
 The Geometry can now be checked as follows:
 IF:
  large liquid surge volume possible
 L/D > not recommended, change for an horizontal flash drum
2
4
surge
L D
V
h


V f Lh h hL
D D
 

3L
D 

 A Vertical Flash drum is to flash a feed of F = 1500 lbmol/h at 40% of n-hexane and
60% of n-octane. The system is at atmospheric conditions
 Requirements if 60% mol of n-hexane on the top product
 A) Get the required flow rates and compositions
 B) Calculate the uperm permissibly velocity
 C) The tank specifications
 D) L/D calculation
378
/ 0.51
/ 3 5
durmT K
q V F
L D between and

 


 For A), Get all streams & compositions
 Verify given data
 For material balance:


 We can get V, L from here (Material Balance)
 From compositions:


 We can get from a material balance OR equilibrium line
1500 /
0.51V
F
F lbmol h

0.4
0.6
F
D
x
y



 For A)
 Perform Material balances for flow rates and compositions
 Given
 Now, all compositions and flow rates are calculated
0.51 0.51(1500) 765 /
1500 765 735 /
V
F V lbmol h
L F V lbmol h
   
    
      0.4 1500 735 0.6 765
0.1918
F l D
l
B l
x F x L x V
x
x x
 
 
 

 For B) Calculate u-perm.
 From these data, calculate properties for flow rates:
 Recall that we will need Mass Flow Rates & Densities
 Molar weights are also required for pure and mixed materials
l
v
W v
Wlv
l
F




 For the LIQUID:
 Molar Weight for Liquid:
 Density of the liquid
     
6 6 8 8
0.19 86 0.81 114
108.9
liq c c c c
liq
liq
MW x MW x MW
MW
MW
 
 

6 6 8 8
(0.19)(0.659) (0.81)(0.703) /
0.696
mix C C C C
mix
g
mLmix
x x
g mL
  


 
 

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 Now, for the VAPOR
 Molar Weight for Vapor:
 Density of the Vapr (use Idela Gas LaW)
     
6 6 8 8
0.60 86 0.40 114
97.4
vap c c c c
vap
vap
MW x MW x MW
MW
MW
 
 

3
(1 )(97.4 )
(82.06 )(378 )
3.14 10
g
mol
mLatm
molK
g
ml
PV nRT
PMW
D
RT
atm
D
K
D x 





 Now, Calculate Kdrum (Via Equation)
 Now, Calculate Flv:
 Use Equation for Kdrum:
97.4 (765 ) 74,500
108.9 (735 ) 80,034
lb lbmol lb
lbmol h hvap
lb lbmol lb
lbmol h hliq
Mass Flow of V MW V
Mass Flow of L MW L
  
  
3
74500 3.14 10
80034 0.696
0.0722
l
v
g
mLW v
Wlv g
mLl
lv
x
F
F



 
 
 
 

0 1 2 3 4
lv lv lv lv
0 1 2 3 4
0 1
0.0722 0.0722) 0.0722) 0.0722)
(lnF ) (ln ) C(lnF ) (lnF ) (lnF )
(ln( ) (ln( ) C(ln( ) (ln( ) (ln( )
( 1.8774)(ln
0.0722)
0.0722 0.07( ) ( 0.8146)(ln( ) ( 0.1871)(ln22)
e
e
e
lvA B F D E
drum
A B D E
drum
drum
K
K
K
   
   
    



2 3 4
( ) ( 0.0145)(ln( ) ( 0.0010)(ln(0.0722) 0.0722) 0 ).0722)
0.443drumK
   


 Now, Calculate Kdrum (Via Graph)
 Using Flv:
 Use Equation for Kdrum:
 Note that both cases are pretty
similar (0.410-0.443)
 Use the one of the equation
0.0722lvF 
0.41drumK 

 Still on… B) Calculate uperm:
0.6960 0.00314
(0.433)
0.00314
6.5849
l v
perm drum
v
perm
ft
sperm
u K
u
u
 






3
3.14 10 g
mlvap
PV nRT
PMW
D
RT
D x 


 
0.696g
mLmix l  
0.433drumK 

 Now, for C) get the Tank specifications…
 Get D:
3
g
2
(3600 )
(765 / )(97.39)(454 )
(6.5849 / )(3600 / )(0.00314 )(28316.85 )
16.05
( )v
e
perm v
g
lb
e mL
mL ft
e
V MW
A
u
lbmol h
A
ft s s h
A ft




4
4(16.05)
4
eA
D
D
D ft





97.4vapMW 
3
3.14 10 g
mlvap
PV nRT
PMW
D
RT
D x 


 
0.51 (1500 / )(0.51)
V 765 /
0.696g
mLmix l
V F lb h
lb h
 
 

 
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 Let us get L/D equation:
 For the following values of D:
 Note that the common standard sizing is 4.50 ft, so choose that one
4
4
L
D
L D


4(4.00 ) 16
4(4.25 ) 17
4(4.50 ) 18
L ft ft
L ft ft
L ft ft
 
 
 

 Vertical:
 https://checalc.com/calc/vertsep.html
 Horizontal:
 https://checalc.com/calc/horizsep.html

 Size the Drum from Ex 1.
 https://checalc.com/calc/horizsep.html
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 Intuitive Approach
 Flash Cascades Analysis of Case 1
 Task – Flash Cascades - Simulation of Case 1
 Animation! Flash Distillation Cascade for an Acetone-Chloroform Mixture
 Animation! Flash Distillation Cascade in a Constant Relative Volatility Mixture

 Assume that you want to purify a stream containing A and B
 You can use as many flashes, sizes, temperature changes…
 Purity Required was 95% for A
 What will happen?

 Assume that you want to purify a stream containing A and B
 You can use as many flashes, sizes, temperature changes…
 Purity Required was 95% for A
 What will happen?
F = 100 mol
z = 0.5
V = 50 mol
x = 0.75
50 mol of A  37.5 mol of A

 Analysis:
 Too many Coolers
 Too many Heaters
 Loss of Streams

 If we wanted to purify A and B…
 We will require a lot of flashes
 Different Temperature each

 A single-stage flash operation can rarely produce the
required purity or fractional recoveries.
 An obvious but inefficient approach is to apply a series
of flash separators condensing part of the vapour and
boiling part of the liquid products from successive
stages.
 The top section of the cascade produces vapours from
subsequent stages that are enriched in the MVC
 The purity of the LVC increases in the liquids in the
bottom section of the cascade.

 The main problem:
 Flash Equipment (Drum) is expensive
 There is loss of “mid” streams
 Final product stream is lower in yield

 ASPEN PLUS SIMULATION
 Case 1  Different temperatures
 Components:
 Toluene (50%)
 Benzene (50%)
 Package:
 Peng Robinson
 Unit: Flash2
 Adiabatic operation
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 Vary:
 Composition
 Number of Flashes
http://demonstrations.wolfram.com/FlashDistillationCascadeForAnEthylAcetateEthanolMixture/

 Case 2  Different Temperatures & Recycle of streams
 What if we recycle the equilibriums streams to avoid mass loss
 What if we tried to interconnect the heat-cool streams to avoid heating systems?

 In the previous setup we have:
 A large number of intermediate process vapour and liquid
streams have no destinations
 Lots of heaters and coolers
 One alternative arrangement is to recycle all
 Most of the intermediate heaters and coolers will be
eliminated by direct contact of the intermediate vapour
streams with the liquid streams.

 In the top section of the cascade, the intermediate
vapour streams are contacted with the recycled liquid
streams.
 Some of the vapour will condense while some of the
liquid will boil.
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 If:
 heat of vapourization and heat of condensation is
approximately constant
 the system is adiabatic
 Then:
 vapour condensed = liquid vapourised
 A similar analysis can be made for the bottom section
of the cascade, where the intermediate liquid stream
is directly contacted with the recycled vapour
streams.

 Case 2 (A)  Recycle Streams
 Case 2 (B)  Recycle Streams + Remove Chillers & Heaters

 Verify:
 Changes in Q
 Feed Compositions Z
 Relative Volatility (WHY)
 Number of Flashes
http://demonstrations.wolfram.com/FlashDistillationCascadeInAConstantRelativeVolatilityMixture/

Flash Distillation in Chemical and Process Engineering (Part 2 of 3)

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Flash Distillation in Chemical and Process Engineering (Part 2 of 3)