MULTICOMPONENT DISTILLATION

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Multi-Component Distillation (MCD) – The Problem

• For multi-component systems, C > 2, no. of equations obtained from mass and energy balances with the equilibrium relationship will always be one less than the number of unknowns.

• A complete analytical solution for multi-component distillation is difficult to attain. trial and error with the additional unknown assumed to be known

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EQUILIBRIUM DATA IN MCD

• Raoult’s Law can be used to detrmine equilibrium compositions.

• In hydrocarbon systems, because of non idealities, equilibrium data are represented as

yi = Kixi

where: Ki is the vapor-liquid equilibrium constant or distribution coefficient of I

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EQUILIBRIUM DATA IN MCD

• Relative volatility for each individual component can be defined as:

i = Ki /Kj

Where j denotes the base component• Values of Ki will be a stronger function of

temperature than the i

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BOILING PT., DEW PT. AND FLASH DISTILLATION

• BOILING POINT -For multicomponent mixture, BP must

satisfy yi = 1.0– For a mixture of A,B,C,D with C as base

component yi = Kixi = Kc ixi =1.0 Calculation is a trial and error process.-When final temperature is known, vapor

composition is calculated fromyi = ixi / (ixi )

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BOILING PT., DEW PT. AND FLASH DISTILLATION

• DEW POINT - trial and error

xi = (yi/Ki)= (1/Kc) (yi/i)=1.0 Kc = yi/i

xi = (yi/i)/(yi/i)

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FLASH DISTILLATION OF MCD

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Pb Pc PdPa >>> >

pump

heater

throttlevalve

Tb Tc TdTa ~ <

Liquid productL, xi, hL

Vapor productV, yi, Hv

Tdrum

Pdrum

Q

TF, PF, hF

FeedF, zi,T1, P1

FLASH DISTILLATION OF MCD

• f = V/F fraction of feed vaporized• 1 – f = L/F fraction of the feed remaining as liquid• Composition of i in the vapor yi = Kixi = [(f – 1) xi/f] + xiF/f• xi = xiF /[f(Kci – 1) + 1] = 1.0 - solved by trial and error by assuming T - when xi = 1.0, proper T has been chosen

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MCD – Some Additional Terminology

• Fractional recoveries• Key components• Non-key components• Splits – distributing and Non-distributing systems

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MCD – Fractional Recoveries

11What are fractional recoveries?

MCD – Fractional Recoveries• Fractional recoveries are often specified in MCD.

• A fractional recovery, FRi, is the amount or flow rate of component i in the distillate or bottoms stream with respect to the amount or flow rate of component i in the feed stream:

• It is the simple relationships expressed by the right-hand-side equations that make the use of fractional recoveries useful.

• These are also often specified simply as % recovery.

i

i

i

ii

F,

B,

F,

botbot Fz

Bx

Fz

BxFR

botdist FR1FR ii

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i

i

i

ii

F,

D,

F,

distdist Fz

Dx

Fz

DxFR

distbot FR1FR ii

MCD – Key Components

• The components that have their distillate and bottoms composition specified are known as the key components.

• The most volatile of the key components is termed the light key (LK).

• The least volatile of the key components is termed the heavy key (HK).

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MCD – Non-Key Components

• All other components not specified in the distillate or botoms are termed non-key components (NK’s).

• If a non-key component is more volatile than the light key, then it is termed a light non-key (LNK).

• If a non-key component is less volatile than the heavy key, it is a heavy non-key (HNK).

• If a non-key component is neither a heavy non-key nor a light non-key, then it is an intermediate non-key (INK) or simply NK.

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MCD – Non-Key Component Splits

• The split of the non–key components is generally defined as to where the non–key components are obtained with respect to the distillate or bottoms stream.

• One can have two types of situations concerning the split of the non–key components:

– Sharp split – Non-distribution of non-keys– Split – Distribution of non-keys

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MCD – Non-distribution of NK’s

• Non–distribution of non–keys means that essentially all of the non–keys are obtained in either the distillate stream or the bottoms stream. We obtain a sharp split of the NK’s.

• Non–distribution of non–keys can be assumed when:– All of the non-keys are either HNK’s or LNK’s– The fractional recoveries of the LK in the distillate and HK in

the bottoms are relatively large.

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MCD – Distribution of NK’s

• Distribution of non–keys means that the non-keys are not sharply split between the distillate stream or the bottoms stream. We obtain a split of the NK’s.

• Distribution of non–keys occurs when:– Not all of the non-keys are either HNK’s or LNK’s – we have

NK’s.– The fractional recoveries of the LK in the distillate and HK in

the bottoms are not relatively large.

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How do we determine the keys (LK and HK) and the non–keys (LNK’s, HNK’s and NK’s) in MCD?

• The classification of components in MCD can be determined from their relative volatilities.

• Relative volatility is defined as the ratio of the K values for two components, which is trivial for a binary system.

• In order to use relative volatilities in MCD, we choose a reference component and define all other component volatilities with respect to the reference component.

• The relative volatility for the reference component, of course, will be 1.

• We can then define relative volatilities using equilibrium coefficient K values for each component, e.g., from the DePriester charts for hydrocarbon systems.

• The choice of the reference component depends upon the problem, but in general it will be the HK component since it is less volatile than the LK component.

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Key and Non-Key Example• Consider a distillation column with the following feed components:

– propane– n–butane– n–pentane– n–hexane

• The recoveries for n–butane and n–pentane are specified for the distillation.

• What are the key and non–key designations for this separation?

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Key and Non-Key Example

• Component volatilities can be determined from the K values.

• From the DePriester charts, the order of volatility is: propane > n–butane > n–pentane > n–hexane

• Since the recoveries of n–butane and n–pentane are specified…designate LK, HK and NK

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Key and Non-Key Example

• We have: Volatilities

propane > n–butane > n–pentane > n–hexane

Component Designation Propane Light Non–Keyn–butane Light Keyn–pentane Heavy Keyn–hexane Heavy Non–Key

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Key and Non-Key Example

• If the recoveries of n-butane and n-hexane are specified:

Volatilitiespropane > n-butane > n-pentane > n-hexane

Component Designation Propane Light Non-Keyn-butane Light Keyn-pentane Non-Keyn-hexane Heavy Key

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Key and Non-Key Example

• If only the recovery of n–butane is specified:Volatilities

propane > n–butane > n–pentane > n–hexane

Component Designation Propane Light Non–Keyn–butane Keyn–pentane Non–Keyn–hexane Non–Key

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Missing Keys

• In typical MCD problems, one specifies the LK and the HK recoveries.

• If only the LK or the HK recovery is specified, one typically chooses one of the non-key components to be the HK or LK, respectively, – usually the non-key component with the greatest feed composition.

• The fractional recovery of the missing key needs to be determined, but it must be estimated since not enough information is typically given in the problem to determine it directly.

• One way to estimate the fractional recovery of the missing key component is to do an external mass balance based upon a binary system comprised of the LK and HK.

• This fractional recovery is then used in the MCD solution. A trial–and–error solution may be required to determine the actual fractional recovery.

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Key and Non-Key Example• Consider a distillation column with the following feed

components:– Methane– Ethane– Ethylene– Propylene– Propane

• It is specified that a distillate concentration, xD, for ethylene is required.

• What are the key and non-key designations for this separation?

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Key and Non-Key Example

• One source for determining the order of the component volatilities can be determined from the K values, which can be found from the DePriester charts, for example.

• The order of volatility is: methane > ethylene > ethane >

propylene > propane• Since xD for ethylene is specified it is a key component.

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Key and Non-Key Example Component Designation

Methane Light Non-KeyEthylene Light KeyEthane Non-KeyPropylene Non-KeyPropane Non-Key

• There is no heavy key specified for this problem.

• What if an xD for ethylene and an xB for propylene are specified?

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Key and Non-Key Example

Component Designation Methane Light Non-KeyEthylene Light KeyEthane Non-KeyPropylene Heavy KeyPropane Heavy Non-Key

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TOTAL REFLUX FOR MCD(FENSKE EQ.)

• Calculates Nmin for total reflux Nm =

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log[(xLDD/xHDD)(xHWW/xLWW)]

log(L,av)

Where: xLD is mole fraction of LK in Distillate

xHD is mole fraction of HK in Distillate

xLW is mole fraction of LK in Bottoms

xHW is mole fraction of HK in Bottoms

L,av = (LD LW)^0.5

TOTAL REFLUX FOR MCD

• Distribution/concentration of other components in the distillate and bottoms at total reflux can be determined by:

xiDD xiWW

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=(L,av)xHDD

xHWW

NM

Minimum Reflux Ratio for MCDUNDERWOOD EQUATIONS

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1-q = i - ixiF

Rm + 1 = i - ixiD

i is based on average T of top and bottom tower

-Trial and error process:Solve check if bet. of LK and HK =1.0 Rm

CLASS 1 SEPARATIONS

• All components in the feed distribute to both the distillate and bottoms products. Single pinch point bridges the feed stage

• Occurs with narrow boiling range mixtures or when the degree of separation between key components is not sharp.

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CLASS 2 SEPARATIONS

• One or more components appear in only one of the products

• If neither distillate nor bottoms products contains all feed components, two pinch points occur away from the feed stage.

• If all feed components appear in the bottoms, the stripping section pinch point moves to the feed stage.

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NUMBER OF STAGES AT OPERATING REFLUX RATIO FOR MCD

• EMPIRICAL CORRELATION BY ERBAR AND MADDOX (Fig. 11.7-3, p.749,Geankoplis)– Graph of R/R+1 vs Nm/N (Rm based on

Underwood method)

• Estimate of feed plate location(Kirkbride)

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Log Ne

Ns= 0.206 log

xHF

xLF

xLW

xHD

W

D

2

Ne is the number of theoretical stages above the feed plateNs is the number of theoretical stages below the feed plate

The following feed of 100 mol/h at the boiling point and 405.3 kPa pressure is fed to a fractionating tower: n-butane(xA=0.40), n-pentane(xB= 0.25), n-hexane(xc=0.20), n-heptane(xD = 0.15). This feed is distilled so that 95% of n-heptane is recovered in the distillate and 95%of n-hexane in the bottoms. Calculate the ff:

a. Moles/hr and composition of distillate and bottomsb. Top and bottom temperature of towerc. Minimum stages for total reflux and distribution of trace

componentsd. Minimum reflux using Underwood methode. No of theoretical stages if R=1.3Rm using Erbar-Maddoxf. Location of feed tray using Kirkbride

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