Distillation

Presented By
Miss A Mukahanana
 Introduction
• Distillation is a separation process that utilizes differences in boiling point to separate
components of a liquid mixture.
• It involves heating, vaporisation, cooling and condensation to obtain purified substances.
• Application: used in industries such as petroleum refining, chemical production and beverage
manufacturing.
 Basic Principles of Distillation
• Raoult’s law: the partial vapor pressure of a component is proportional to its mole fraction in the
liquid.
• Dalton’s Law: the total pressure of a gas mixture is the sum of the partial pressure of its
components.
• Vapor-liquid equilibrium (VLE): the condition where the rate of evaporation equals the rae of
condensation for each component in a mixture.
 Types of Distillation
• Simple distillation: used for separating components with large boiling points.
• Fractional distillation: uses a fractionating column for better separation of close-boiling
components.
• Steam distillation: used for temperature-sensitive materials like essential oils.
• Vacuum distillation: lowers the boiling point by reducing pressure, useful for heat-sensitive
substances.
• Azeotropic and extractive distillation: used for breaking azeotropes (mixtures that cannot be
separated bby simple distillation).
 Components of Distillation system
• Distillation column: the main structure where
separation occurs.
• Reboiler: heats the liquid to generate vapors.
• Condenser: cools and condenses vapors back
into liquid.
• Trays/packing: increases surface area for
better separation.
• Reflux system: returns part of the condensed
liquid to imprve efficiency
 Operation and Terminology
• The liquid mixture that is to be processed is known as the feed and this is introduced usually
somewhere near the middle of the column to a tray known as the feed tray

• The feed tray divides the column into a top (enriching or rectification) section and a bottom
(stripping) section

• The feed flows down the column where it is collected at the bottom in the reboiler

• The Rectifying Section in a distillation column is where the lighter distillate is being enriched
and the heavier components removed

• The Stripping Section where the lighter components are being "stripped" out of the bottom
product and the heavier components consequently concentrated
 Stripping Secction
• Heat is supplied to the reboiler to generate vapour
• The source of heat input can be any suitable fluid, although in most chemical plants this is
normally steam
• In refineries, the heating source may be the output streams of other columns
• The vapour raised in the reboiler is re- introduced into the unit at the bottom of the column
• The liquid removed from the reboiler is known as the bottoms product or simply, bottoms
 Rectifying Sector
• The vapour moves up the column, and as it exits the top of the unit, it is cooled by a
condenser

• The condensed liquid is stored in a holding vessel known as the reflux drum

• Some of this liquid is recycled back to the top of the column and this is called the
reflux

• The condensed liquid that is removed from the system is known as the distillate
or top product
 Types of distillation columns
• We will consider 2 types used in distillation,
- Packed Bed Columns
- Tray Columns
 Packed bed distillation column
• Used more often for absorption and distillation of vapor-liquid mixtures
• Packing provides a large surface area for vapour-liquid contact
• The liquid flows downward through the packing and the vapor flows upward
through the column
• Advantages
✓ Cost efficient for diameters less than 0.6m
✓ Inert packing materials can be used to handle corrosive materials
✓ Lower pressure drop than in plate columns
✓ Good for thermally sensitive liquids
• Disadvantages
✓ Packing can break during installation or due to thermal expansion
✓ Not cost efficient for high liquid flow rates
✓ Contact efficiencies are decreased when the liquid flow rate is too low
 Tray Distillation Column
• Most widely used type of column

• The number of trays or stages is dependent on the application

➢ Advantages

• Cost efficient for diameters greater than 0.6m

• Cooling coils can easily be added to the plate column
(cryogenic applications)

• Can handle high liquid flow rates cost-effectively

➢ Disadvantages

• Higher pressure drops than packed columns

• Foaming can occur due to agitation of the liquid by vapour

flowing up through it
 Factors affecting distillation
• Reflux ratio: higher reflux improves separation but increases energy use.
• Pressure: lowering pressure reduces boiling points
• Tray/packing type: determines mass transfer in the column.
• Feed composition and location: affacts the number of required stages for separation, state of
feed
• Weather conditions
 Relative volatility or separation factor
• Assume that the vapour liquid equilibrium of two species i and j at a specific T, P is
given by,
𝑦𝑖 𝑦𝑖
𝐾𝑖 = 𝐾𝑗 = 𝑖𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝐾𝑖 = 𝑓(𝑇, 𝑃, 𝑥, 𝑦)
𝑥𝑖 𝑥𝑖
• This is a simplification. Ki is found from thermodynamic calculations which can include all the non-
ideal behaviour such as fugacity coefficients, activity coefficients using activity models or equations of
state
• For ideal solutions
𝑃𝑖0
𝐾𝑖 =
𝑃𝑇
• Ki is thus inversely proportional to pressure and a strong function of temperature
through 𝑃𝑖0
 Separation factor 𝛼𝑖𝑗
𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑖 𝑖𝑛 𝑣𝑎𝑝𝑜𝑟 𝑇(𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑖 𝑖𝑛 𝑙𝑖𝑞𝑢𝑖𝑑) 𝑦𝑖 𝑇𝑥𝑖 𝐾𝑖
𝛼𝑖𝑗 = = =
𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑗 𝑖𝑛 𝑣𝑎𝑝𝑜𝑟 𝑇(𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑗 𝑖𝑛 𝑙𝑖𝑞𝑢𝑖𝑑) 𝑦𝑗 𝑇𝑥𝑗 𝐾𝑗
• For a binary mixture
𝑦𝐴 𝑇𝑥𝐴 𝑦𝐴 𝑇(1 − 𝑦𝐴 )
𝛼𝐴𝐵 = =
𝑦𝐵 𝑇𝑥𝐵 𝑥𝐴 𝑇(1 − 𝑥𝐴 )
• And for binary ideal mixture

𝑃𝐴0 𝑃𝐴 𝑥𝐵
𝛼𝐴𝐵 = =
𝑃𝐵0 𝑥𝐴 𝑃𝐵
 Separation factor 𝛼𝑖𝑗
➢ For ideal binary mixtures ij is a function only of T
• ij: represents the degree of separation that has occured between two species in the two
phases. The larger ij the better the separation. Typically ij should be greater than 1.3
for reasonable distillation column operation
• ij provides a relationship between the compositions between 2 phases in
equilibrium
•  is the ratio of 𝑃𝐴0 to 𝑃𝐵0 , which are both T dependent. The ratio, however, varies very
slowly with changing T and thus ij can often be taken as approximately constant over
small temperature ranges (generally those over which the McCabe-Thiele method
apply). Assuming ij constant does not mean that the temperature in the distillation
column is constant
 For binary mix in which 𝛼𝑖𝑗 is approx. constant
𝑦𝐴 𝑇𝑥𝐴 𝑦𝐴 𝑇(1 − 𝑦𝐴 )
𝛼𝐴𝐵 = =
𝑦𝐵 𝑇𝑥𝐵 𝑥𝐴 𝑇(1 − 𝑥𝐴 )
• Re-arrange
𝑎𝑥
𝑦=
1+ 𝛼−1 𝑥
• Or
𝑎𝑦
𝑥=
𝑎− 𝛼−1 𝑦

xy diagram for constant 𝛼𝑖𝑗

 McCabe-Thiele theory
• The McCabe-Thiele approach is a graphical one, and uses the Vapour Liquid Equilibrium (VLE) plot
to determine the theoretical number of stages required for effective separation of a binary mixture

• It uses the following assumptions,

− equal molar overflow and vaporisation

− no heat losses, no heat of mixing, constant ΔHvap

− constant pressure, negligible pressure drop

• The molar flows of V, L are constant in the enriching and stripping sections, this is known as the
constant molar overflow assumption
 McCabe-Thiele analysis
• Above the feed
✓ (enriching, absorption, rectifying) Liquid reflux returned to the
column to remove (absorb) less volatile components from the
vapour

• Below the feed

✓ (stripping, exhausting) vapour from the reboiler is returned to the
column to remove the more volatile component from the liquid

• Distillate: rich in the more volatile component (low T)

• Bottoms: rich in the less volatile components (high T)

 McCabe-Thiele analysis: Ideal gas
• Vapour leaving the stage is in equilibrium with the liquid leaving
the stage (good mixing, sufficient contact time)

• The number of moles crossing the gas-liquid interface from gas to

liquid is approximately equal to the moles crossing from liquid to gas

• The heat released by 1 mol of vapour condensing is approximately the

heat needed to vaporise 1 mol of liquid
• Constant molar heats of vaporisation means that the flow rate of
liquid and gas is approximately constant from tray to tray in each
section (unequal molar heats of vaporization lead to varying flows of
gas and liquid down the column)
 Mass Balances
• Overall: F=L+V
• Components: Fz=Lx+Vy
• Solving for y to get the operating line
𝐿 𝐹
𝑦=− 𝑥+ 𝑧
𝑉 𝑉
• The slope of the operating line is –L/V, rearranging to get the q-line (V/F is the fraction of feed
vaporized)
𝑉𝑇𝐹 − 1 𝐹
𝑦= 𝑥+ 𝑧
𝑉𝑇𝐹 𝑉
 M-T overall balances
• Overall: F=D+B
• Component: 𝐹𝑥𝐹 = 𝐷𝑥𝐷 + 𝐵𝑥𝐵
• Energy: 𝐹𝐻𝐹 + 𝑄𝐵 = 𝐷𝐻𝐷 + 𝐵𝐻𝐵 + 𝑄𝐵
• Heat losses have been neglected, given the feed, distillate and bottoms
specification, the energy between 𝑄𝐶 𝑎𝑛𝑑 𝑄𝐵 .
• Also by definition, 𝑥𝐵 < 𝑥𝐹 < 𝑥𝐷 for distillation to occur.
• These represent the mole fraction of the more volatile component.
• Re-arrangement of the mass balance gives D as a function of the split
between 𝑥𝐷 and 𝑥𝐵
𝑥𝐹 − 𝑥𝐵
𝐷=𝐹
𝑥𝐷 − 𝑥𝐵
 BALANCES
Rectifying Section Balances

• Assumptions:

• Condenser (Qc) removes only latent heat and

produces
liquid at its bubble point.

− V, L are constant in the section

− Ideal equilibrium stages: yn in equilibrium with xn

𝐿 𝑙𝑖𝑞𝑢𝑖𝑑 𝑓𝑙𝑜𝑤 𝑟𝑒𝑡𝑢𝑟𝑛𝑒𝑑 𝑡𝑜 𝑐𝑙𝑢𝑚𝑛
• Reflux ratio: 𝑅 = =
𝐷 𝑑𝑖𝑠𝑡𝑖𝑙𝑙𝑎𝑡𝑒 𝑡𝑎𝑘𝑒𝑛 𝑜𝑓𝑓

• A balance around the split point below the condenser

(note, total condenser the composition does not
change) Top MB (above feed)
𝑉 = 𝐿 + 𝐷 =(𝑅 + 1)𝐷
 RECTIFYNG SECTION BALANCES
➢ Mass balance around the complete column section,

𝑉 =𝐿+𝐷

𝑉𝑦𝑛+1 = 𝐿𝑥𝑛 + 𝐷𝑥𝐷

𝐿 𝐿
𝑦𝑛+1 = 𝑥𝑛 + 𝑥𝐷
𝑉 𝐷

𝑹 𝒙𝑫
𝒚𝒏+𝟏 = 𝒙𝒏 +
𝑹+𝟏 𝑹+𝟏

➢ This is the Top Operating line for the column and slope = R/(R+1) and
intercept = xD/(R+1)
 Rectifying Section Balances
➢ Graphical Interpretation: Set xn = xD in the mass balance and then gives
yn+1 = xD, thus the operating line starts at y = x = xD (see figure) Then the
stepping down procedure can be calculated as follows, starting from xD
𝛼𝑥
➢ Assuming equilibrium is given by, 𝑦 =
1+ 𝛼−1 𝑥
Stage x y
𝑥 = 𝑥𝐷
1 𝑦1 𝑦1 = 𝑥𝐷
𝑥1 =
𝛼− 𝛼−1 𝑦
1
2 𝑦2 𝑅 𝑥𝐷
𝑥2 = 𝑦2 = 𝑥 +
𝛼− 𝛼−1 𝑦 𝑅+1 1 𝑅+1
2
3 𝑦3 𝑅 𝑥𝐷
𝑥3 = 𝑦3 = 𝑥 +
𝛼− 𝛼−1 𝑦 𝑅+1 3 𝑅+1
3
…and so on
 Rectifying Section Balances

➢ The figure shows the

systematic method of moving

from the distillate mole fraction

(xD) down towards the stripping

section

McCabe Thiele enriching section

 Stripping Section Balances
➢ Assume constant liquid (L’) and vapour
(V’) flow rates in the Bottom section
➢ These are not the same as the flow rates
(L,V) in the Top section, because of the feed
tray

➢ Mass balances
𝐿′ = 𝑉 ′ + 𝐵
𝐿 ′ 𝑥 𝑚 = 𝑉 ′ 𝑦 𝑚+1 + 𝐵𝑥𝐵
𝑳′ 𝑩
𝒚𝒎+𝟏 = 𝒙𝒎 − ′ 𝒙𝑩
𝑽 ′ 𝑽
Bottom MB (below feed)
 Stripping Section Balances
′
𝑉
➢ Boil up ratio: 𝑅𝐵 =
𝐵

➢ The Bottom Operating line is,

𝑹𝑩 + 𝟏 𝒙𝑩
𝒚𝒎+𝟏 = 𝒙𝒎 −
𝑹𝑩 𝑹𝑩
➢ Graphically: The starting point can be
found by substituting xm = xB into the
operating line to find that y= xB, thus the
bottom point must lie on the y = x line

➢ The Figure shows the stepping up

process given xB and RB McCabe Thiele stripping section
 Stripping Section Balances
➢ Notice that the counting of stages goes backwards, make sure that the equations and
mass balance sketch correspond
➢ Also the last stage +1 (RB) is the PARTIAL reboiler
➢ This so because a separation takes place, as the liquid is partially vaporised by QB
➢ Suppose a distillation column has 10 stages, starting at the reboiler the calculation is,
Stage x y
R 𝑥𝑁 = 𝑥𝐵 𝛼𝑥𝐵
𝑦 =𝑦 =
𝑁+1 𝐵
1 + 𝛼 − 1 𝑥𝐵
10 𝑅𝐵 𝑥𝐵 𝛼𝑥10
𝑥10 = 𝑦𝐵 − 𝑦10 =
𝑅𝐵 + 1 𝑅𝐵 1 + 𝛼 − 1 𝑥10
9 𝑅𝐵 𝑥𝐵 𝛼𝑥10
𝑥9 = 𝑦10 − 𝑦10 =
𝑅𝐵 + 1 𝑅𝐵 1 + 𝛼 − 1 𝑥10
8 … and so on
 The Feed Tray
➢ Feed condition influences the choice of operating conditions of the
distillation column. The choice of a feed condition can cause a distillation
operation to fail - i.e. the system will "pinch"

➢ At the feed tray, the compositions and flows change abruptly

➢ The feed can be liquid, vapour or a mixture of both (super cooled liquid to
super heated vapour)

➢ Mass and enthalpy balances across the feed tray can be used to determine the
effect that the feed condition will have on the vapour and liquid flows in each
section, which in turn influences the L/V ratios, and thus the column
operation
 The Feed Tray
➢ The mass balance is
𝐹 + 𝐿 + 𝑉 ′ = 𝑉 + 𝐿′
𝐿′ − 𝐿 = 𝑉 − 𝑉 ′ + 𝐹
➢ The energy balance is
𝐹𝐻𝐹 + 𝐿𝐻𝐿 + 𝑉 ′ 𝐻 𝑉 ′ = 𝑉𝐻𝑉 + 𝐿′𝐻𝐿′
➢ Assumptions,
− vapour and liquid inside the column all saturated
(i.e. Dew pt and Bubble pt)
− molar enthalpies of saturated vapour and liquid Feed tray enthalpy balance
approximately the same

− small compositions changes

 The Feed Tray
➢ We then get 𝐻 𝑉 ′ ≈ 𝐻𝑉 and 𝐻𝐿 ≈ 𝐻𝐿′

➢ The energy balance becomes

𝐿′ − 𝐿 𝐻𝐿 = 𝑉 ′ − 𝑉 𝐻𝑉 + 𝐹𝐻𝐹

➢ combining with the mass balance,

𝐿′ − 𝐿 𝐻𝑉 − 𝐻𝐹
= =𝑞
𝐹 𝐻𝑉 − 𝐻𝐿

𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑡𝑜 𝑏𝑟𝑖𝑛𝑔 𝑡ℎ𝑒 𝑓𝑒𝑒𝑑 𝑡𝑜 𝑑𝑒𝑤 𝑝𝑜𝑖𝑛𝑡 𝑣𝑎𝑝𝑜𝑢𝑟

𝑞=
𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 𝑜𝑓 𝑣𝑎𝑝𝑜𝑟𝑖𝑠𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑓𝑒𝑒𝑑

𝐻𝐹 𝑇𝑑𝑒𝑤 − 𝐻𝐹(𝑇𝐹 )
𝑞=
𝐻𝐹 𝑇𝑑𝑒𝑤 − 𝐻𝐹(𝑇𝑏𝑢𝑏𝑏𝑙𝑒)
 Intersection Of Operating Lines
𝐿′ − 𝐿 = 𝑞𝐹 𝑎𝑛𝑑 𝑉′ − 𝑉 = 𝑞 − 1 𝐹
➢ This gives the flow rates in each section and the hence connects the operating lines
above and below the feed tray
➢ Let (xq, yq) be the point of intersection, at the intersection the operating lines
become
𝑇𝑂𝑃: 𝑦𝑞𝑉 = 𝐿𝑥𝑞 + 𝐷𝑥𝐷
𝐵𝑂𝑇𝑇𝑂𝑀: 𝑦 𝑞 𝑉 ′ = 𝐿′𝑥𝑞 − 𝐵𝑥𝐵
➢ subtracting the equations: 𝑉 ′ − 𝑉 𝑦𝑞 = 𝐿′ − 𝐿 𝑥𝑞 − (𝐷𝑥𝐷 + 𝐵𝑥𝐵)
➢ Overall: 𝐹𝑥𝐹 = 𝐷𝑥𝐷 + 𝐵𝑥𝐵
➢ solving the equations yields the q-line
𝑞 𝑥𝐹
𝑦= 𝑥−
𝑞−1 𝑞−1
 Q-LINE
➢ The intersection of the Top and Bottom
operating line must lie on this line

➢ NB: given the condition of the feed (q) and

fixing the reflux ratio (R) determines boil-up
rate (RB) and the reboiler heat load

➢ This provides a connection between QC and

QB via the feed condition
➢ Choosing xD or xB will then define the
operation of the column
McCabe Thiele q-line
➢ Graphically: the 5 possible feed conditions
of the q-line
 Boil-up And Reflux
➢ Cases (b), (c), (d) the following applies

➢ balance around the top: 𝑉 ′ = 𝐿 + 𝐷 − 𝑉𝐹

𝑉 ′ 𝐿 + 𝐷 − 𝑉𝐹
𝑅𝐵 = =
𝐵 𝐵
➢ balance around the bottom: 𝐿 = 𝑉 ′ + 𝐵 − 𝐿𝐹

➢ (e) require the use of the energy balance to get actual flow of L’, V’ ....

➢ NB: q will affect the number of stages required

➢ Cold feed will increase the liquid flow below the feed, increases the reboiler
duty
OVERALL OPERATION

McCabe Thiele operating lines

 FEED TRAY LOCATION
➢ This is the best, results in the
minimum number of stages (tray 3)

➢ Calculations proceed until the

composition (xq, xq) of the intersection
of the operating lines is reached, when
stepping down from the top

➢ Beyond the q-line O/L intersection the

lower operating line must be used until
xB is reached or exceeded
 FEED TRAY LOCATION
➢ The feed stage is below the q-line intersection
(tray 5)
➢ Calculations from the top down would only cross
over onto the lower operating line once the feed
tray has been reached
➢ The q-line does not represent a O/L cross over
point and this scheme results in more stages
➢ The top O/L starts pinching, and the composition
cannot proceed beyond point K where a pinch
occurs
➢ Point K represents the smallest composition that
can be achieved with the top section only
➢ Using the lower O/L can of course go further
down
 FEED TRAY LOCATION
➢ Feed stage is above the q-line
intersection

➢ This results in more stages by the

same arguments as for feed on tray 5

➢ Here point R is the maximum

composition achievable in the bottom
section only, a pinch occurs at R
 THE TOP DOWN CALCULATION
Stage x y
𝑥 = 𝑥𝐷
1 𝑦1 = 𝑥𝐷
1
𝛼 − 𝛼 − 1 𝑦1
2 𝑅
2 𝑦2 =
𝛼 − 𝛼 − 1 𝑦2 𝑅+1 1 𝑅+1
3 𝑦3 𝑅 𝑥𝐷
3 < 𝑥 𝑦3 = 𝑥 +
𝛼− 𝛼−1 𝑦 𝑅+1 3 𝑅+1
After x < xq change and use bottom Operating line to get y4
4 𝑅𝐵 + 1 𝑥
4 𝑦4 = 𝑥3 − 𝐵
𝛼 − 𝛼 − 1 𝑦4 𝑅𝐵 𝑅𝐵
5 𝑅𝐵 + 1 𝑥
5 𝑦5 = 𝑥4 − 𝐵
𝛼 − 𝛼 − 1 𝑦5 𝑅𝐵 𝑅𝐵
Stop calculation after x<xB and determine number of stages
 REFLUX CONDITIONS
➢ As the reflux ratio is increased, the gradient of operating line for the rectification
section moves towards a maximum value of 1

➢ Physically, what this means is that more and more liquid that is rich in the more
volatile components are being recycled back into the column

➢ Separation then becomes better and thus less trays are needed to achieve the
same degree of separation

➢ Minimum trays are required under total reflux conditions, i.e. there is no
withdrawal of distillate
 REFLUX CONDITIONS
➢ On the other hand, as reflux is decreased, the operating line for the
rectification section moves towards the equilibrium line

➢ More trays are required to achieve the same degree of separation

➢ So there is an inverse relationship between the reflux ratio and the number
of theoretical plates

➢ The limiting condition occurs at minimum reflux ratio, when an infinite

number of trays will be required to effect separation

➢ Most columns are designed to operate between 1.2 to 1.5 times the minimum
reflux ratio because this is approximately the region of minimum operating
costs (more reflux means higher reboiler duty)
 MINIMUM NUMBER OF STAGES

McCabe Thiele minimum number of stages

 MINIMUM NUMBER OF STAGES
➢ Operation Under Total Reflux: This means that all of the liquid from the condenser is
returned to the column (R = L/D → ∞), also all of the liquid entering the reboiler is
returned to the column

➢ By a mass balance, the feed must be zero i.e. the column is brought to steady state and
the flow rates turned off

➢ When R→∞, then R/(R+1)→1, and the slope of the operating line L/V= 1
➢ The operating lines thus co-inside with the y = x line and the liquid and vapour flows up
and down the column in both sections are equal

➢ The size of the steps is a maximum as the distance between the equilibrium line and the
operating line is a maximum, thus the number of stages are a minimum Nmin

➢ For a system with approximately constant, and analytical solution for Nmin is possible
 MINIMUM NUMBER OF STAGES
➢ The Fenske equation is used to calculate the minimum number of stages,

𝑥𝐷 𝑁+1
𝑥𝐵
= 𝛼
1 − 𝑥𝐷 1 − 𝑥𝐵

➢ N excludes the reboiler i.e. with the reboiler = N+1. Therefore this becomes,

𝑥𝐷 1 − 𝑥𝐵
log
1 − 𝑥𝐷 𝑥𝐵
𝑁+1 =
log 𝛼
 MINIMUM REFLUX RATIO
➢ Minimum reflux: This
represents the smallest
reflux ratio that can be
used to still achieve the
desired separation to xD
and xB
➢ It also represents the
smallest condenser (Qc)
and reboiler(QB) heat
loads
➢ This requires the largest
possible column, or an
infinite number of stages McCabe Thiele for minimum reflux (ideal case on the right)
 MINIMUM REFLUX RATIO
➢ For a almost ideal distillation, this condition is met when the operating lines, q-line
and equilibrium line all meet at point P
➢ The stepping procedure from either the top or bottom then produces an infinite
number of stages, point P thus represents the pinch point of the operation, or
physically this can be viewed as a point of constant composition, i.e. composition
and T do not vary with increased number of stages

➢ This is very important concept in multi-component distillation

➢ Since Rmin represents the intersection of the Top and Bottom operating lines with
the q-line, then we can calculate it as follows,

𝑥𝐷 − 𝑦𝑞
𝑅𝑚𝑖𝑛 =
𝑦𝑞 − 𝑥𝑞
 EXAMPLE
A continuous fractionating column is to be designed for separating a saturated liquid
mixture containing 40 mole percent methanol and 60 mole percent water into an overhead
product containing 97 mole percent methanol and a bottom product having 98 mole
percent water. The flow rate of the liquid mixture is 10,000 kg per hour. A mole reflux ratio
of 3 is used. The equilibrium relationship for the mixture is given in the Table below
x 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

y 0.0 0.417 0.579 0.669 0.729 0.780 0.825 0.871 0.915 0.959 1.0

x = mole fraction of methanol in liquid and y = mole fraction of methanol in vapor

Calculate (i) the moles of overhead product obtained per hour and (ii) the number of
plates; (iii) determine the location of the feed plate at its bubble point.
Mr (CH3OH) = 32, Mr (H2O) = 18
SOLUTION
 SOLUTION
➢ Calculate the molar flowrate of feed.

➢ Average Mr = 0.4×32 + 0.6×18 = 23.6 g/mol

➢ Average molar flowrate of feed = 1000 kg.hr-1/23.6 g.mol-1 = 423.7 kmol/hr

➢ Mass balances
𝐹 =𝐷+𝐵 → 𝐹𝑥𝐹 = 𝐷𝑥𝐷 + 𝐵𝑥𝐵

➢ using these 2 equations we can substitute the knowns

423.7 = 𝐷 + 𝐵
423.7 × 0.4 = 0.97 𝐷 + 0.02 𝐵

➢ We can now solve simultaneously for D and B i.e. D = 169.5 kmol/hr and B =
254.2 kmol/hr
 SOLUTION: RECTIFYING SECTION
𝑅 𝑥𝐷
𝑦𝑛+1 = 𝑥𝑛 +
𝑅+1 𝑅+1
𝑦𝑛+1 = 0.75𝑥𝑛 + 0.2425

➢ x0 = xD = 0.97. Therefore we can calculate y1

𝑦1 = 0.75 0.97 + 0.2425 = 0.97

➢ therefore the top OL passes through the points (0.97, 0.97) and (0, 0.2425)
𝐿
𝑅= → 𝐿 = 3𝐷 = 3 × 169.5 = 508.5 𝑘𝑚𝑜𝑙Τℎ𝑟
𝐷
𝑉 = 𝐿 + 𝐷 = 508.5 + 169.5 = 678 𝑘𝑚𝑜𝑙Τℎ𝑟
➢ Since its saturated liquid, which implies it is at its bubble point, this means there is no
SOLUTION:
vapour in the feed (q=1). Therefore,STRIPPING SECTION
𝑉 ′ = 𝐿 + 𝐷 − 𝑉𝐹 = 508.5 + 169.5 − 0 = 678 𝑘𝑚𝑜𝑙Τℎ𝑟
𝑉′ 678
𝑅𝐵 = = = 2.67
𝐵 254.2
𝐿′ = 𝑉 ′ + 𝐵 = 678 + 254.2 = 932.2 𝑘𝑚𝑜𝑙Τℎ𝑟

𝑅𝐵 + 1 𝑥𝐵
𝑦𝑚+1 = 𝑥𝑚 −
𝑅𝐵 𝑅𝐵
𝑦𝑚+1 = 1.375𝑥𝑚 − 0.0075
➢ If we use xm=xB the yB = 0.02. Therefore the stripping line passes through (0.02, 0.02)
➢ The bottom OL is drawn by connecting (xB, yB) with the intersection point of the q-line
and the top OL.
 SOLUTION: MCCABE-THIELE

➢ The McCabe-Thiele
graphical method estimates
7 theoretical plates

➢ The feed is located on plate

number 5