AVP - Distillation

Chemical Engineering Operations

Distillation

Dr. Anand V. PatwardhanProfessor of Chemical EngineeringInstitute of Chemical Technology

Nathalal M. Parekh RoadMatunga (East), Mumbai−400019

Email: anand@udct.org; avpuict@gmail.com, avpiitkgp@gmail.com

REVIEW of the following BASICS of BINARY distillation:1. Vapour liquid equilibrium2. Relative volatility3. Enthalpy − Concentration Diagram4. Application of enthalpy − concentration diagram to FLASH

distillation of a BINARY mixture5. Steam distillation6. Differential distillation / Rayleigh distillation7. Continuous multistage fractionation of binary mixtures

(McCabe−Thiele method)a) Material and energy balancesb) Determination of number of stagesc) Total reflux – Fenske’s equation – Minimum number of stagesd) minimum reflux ratio

8. Continuous multistage fractionation of binary mixtures (Ponchon Savarit Method)a) Determination of number of stagesb) Minimum reflux ratioc) Total reflux

Distillation: Technique of preferential separation of MORE VOLATILE COMPONENT(S) from LESS VOLATILE COMPONENT(S) by partial vaporisation, followed by condensation.

Distribution of components in two phases = f(vapour−liquid equilibrium).Distillation column consists of:

(1) Column having trays or packing and suitable internals

(2) Reboiler(3) Condenser

Functioning of distillation column:(1) Feed enters at a suitable point.(2) Reboiler partially vaporises the liquid received from

the column bottom.(3) Vapour flows UP through the trays (or packing),

leaves at the column top and enters into overhead condenser.

(4) A part of the condensate is withdrawn as TOP PRODUCT, and the rest is fed back to the column as REFLUX that flows down the trays (or packing).

(5) Intimate contact between UPFLOWING vapour and DOWNFLOWING liquid occurs on the trays (or packing).

(6) EXCHANGE of mass takes place between the liquid and vapour phases: M.V.C. from liquid → vapour; L.V.C. from vapour → liquid.

CondensateW

Vapour

Reflux

F: Feed (L, V, or a mixture of L and V)

D: Distillate or top productW: Bottom product

1: Distillation column2: Feed preheater3: Condenser4: Reflux drum5: Reboiler

Schematic of a typical

distillation column

Vapour−Liquid Equilibrium

Under a given set of conditions, the equilibrium vapour composition is related to the liquid composition.

F = C - P + 2

where C = number of components

P = number of phases

F = de

Gibb's P

grees of

Example

Aqueous solution of C2H5OH in a closed vessel fitted with a valve

Boiled for some time to expel the air ⇒ contains only H2O + C2H5OH

Vessel put in a constant temperature bath for sufficient time

System reaches equilibrium ⇒ L and V compositions, and PT in the vapour space attain unique constant values

⇒ Number of Components = 2 (C2H5OH and H2O)

⇒ Number of Phases = 2 (L and V)

⇒ Degrees of Freedom = 2

Total number of parameters = 4 (T, PT, L composition, V composition)

2 of these 4 parameters can be fixed to define system COMPLETELY

Accurate Vapour−Liquid Equilibrium (VLE) data are essential for reliable design of a distillation column.

If the experimental data are not available (multicomponent systems), a suitable predictive method can be adopted, for example, UNIFAC method.

If data for prediction is also not available, conduct experiments to determine the VLE.

Constant T & Constant P Binary V−L EquilibriaTB

LIQUID

VAPOURT−y

(dew point)

T−x(bubble point)

Diagonaly = x

Equilibrium (x−y) diagram

x →(0, 0)

At constant P: Boiling Temperature versus liquidcomposition ⇒ T−x (bubble point) curveAt constant P: Boiling Temperature versus vapourcomposition ⇒ T−y (dew point) curveLiquid at Point G → heated gradually → Point M→ heating continued → Liquid becomes progressively POORER in A→ Liquid Boiling Point (bubble point) increases→ Last liquid droplet at M1

→ Final VAPOUR at N1

→ further heating gives SUPERHEATED vapourSame description can be given for LV, L1V1, G1H1

F = amount of two-phase mixture, kmol (zF = mole fraction)

L = amount of liquid, kmol (x = mole fraction)

V = amount of vapour, kmol (y = mole fraction)

Overall mole balance: F = L + V

Component A balance: F zF = L x + V y

length of section Elimating : = length of section

y* - zL V

FFFV z -

er Arm

x FLFRule

Raoult’s Law: vapour-liquid equilibria for Ideal Solutions:

( )( )

v v vp = x P and p = x P = 1 - x PA A A B B B A Bv vTotal pressure = p + p = x P + 1 - x PA B A A A B

Mole fraction of A in vapour phase = yAv= p P = x P PA T A A T

The above equation can is used to compute V−L equilibrium data for IDEAL BINARY mixtures.

Deviation from Ideality and Formation of Azeotropes

POSITIVE deviation: a liquid mixture exerting an equilibrium total vapour pressure more than that computed by ideal equation.

NEGATIVE deviation: a liquid mixture exerting an equilibrium total vapour pressure less than that computed by ideal equation.

AZEOTROPES

If there is LARGE POSITIVE deviation from ideality, and the vapour pressures of A and B are not much different, the TOTAL PRESSURE CURVE may have a MAXIMUM at a certain liquid composition ⇒ constant boiling mixture ⇒ minimum boiling AZEOTROPE.

The x−PT and y−PT curves touch at Azeotropic Composition.

The x−T and y−T curves pass through a common minimum.

The equilibrium curve crosses the diagonal line at the azeotropic composition.

V−L Equilibria of a Minimum−Boiling Azeotrope (ethanol − benzene)

Azeotropic point, x = y

x →(0, 0)

x − PT curve

Ideal behaviour

x − PA curve

x − PB curve

If there is LARGE NEGATIVE deviation from ideality, the partial pressures of individual components are <ideal values, the TOTAL PRESSURE CURVE may have a MINIMUM at a certain liquid composition ⇒constant boiling mixture ⇒ MAXIMUM boiling AZEOTROPE.

The x−PT and y−PT curves touch at Azeotropic Composition.

The x−T and y−T curves pass through a common minimum.

The equilibrium curve crosses the diagonal line at the azeotropic composition.

RELATIVE VOLATILITY

Relative volatility of a component A in a mixture indicates the EASE of its SEPARATION from another component B.

( )( )

Relative volatility of A =concentration of A concentration of B phaseconcentration of A conc

vapour

liquidy 1 - y y 1 - x

α = = AB x 1 - x x 1 -

entration of B phase

yα xy =

1 + α

- 1 x⇒

For IDEAL binary solutions, α can be expressed in terms of vapour pressures of the components:

( )( )

( ) ( )( ) ( )

p P p Py 1 - y A T B Tα = = v vx 1 - x p P p PA A B B

vP vapour pressure of AA= = v vapour pressure of BPB

Equilibrium in a Multicomponent SystemHydrocarbons of a homologous series are nearly ideal solutions.For an IDEAL solution, we can write: vx Ppi = n j jv iy P = p = x P P = p yj T j j j T i j i = nP vi = 1 T x Pi ii

and = =

where = vapour pressure of pure component at the given t

emperatu=

vP jj e

∑ ⇒

Noting that , the above equation can be rewritten as:v vα = P Pij i jx xj jyj i = n

i = n x1 v i ijx P i = 1i ivP i =

⎛ ⎞

α∑⎜ ⎟∑⎜ ⎟

⎜ ⎟⎝ ⎠

For hydrocarbon mixtures, a quantity called “equilibrium vaporisation ratio” is extensively used: vy Pi iKi x Pi

Equilibrium in a nonideal system

At equilibrium, the FUGACILITIES of the component i in the vapour phase and in the liquid phase are EQUAL.

V Lf = fi iV V L L 0f = Φ y P and f = Φ x P = x γ fi i i T i i i T i i i

Vwhere Φ = fugacity coefficient of component in vapour phaseiγ = activity coefficient of component in liquid i

i phase

0f = fugacity of component at standard statei i

For low−to−moderate pressures, the “fugacity at standard state”can be approximated to the “vapour pressure” at the given temperature, that is: L vf = x γ Pi i i i

V vThus, at equilibrium: y P = x Pi i T i i iΦ γ

ENTHALPY − CONCENTRATION DIAGRAM

Change in “composition” is accompanied by a change in “enthalpy”. For a given T and concentration x of a liquid, the molar enthalpy HL can be calculated using the equation:

HL = cPS Mav (T − T0) + ΔHS

HL = molar enthalpy of the solution at temperature T, kJ/kmol

CPS = specific heat of the solution, kJ/(kg K)

Mav = average molecular weight of the solution

T0 = reference temperature, K

ΔHS = heat of solution at the reference temperature T0, kJ/kmol

ENTHALPY − CONCENTRATION DIAGRAM

Since the heat of mixing of the vapour is negligible, we can use the following equation to compute the molar enthalpy of the saturated vapour HV at a given T and y:

HV = y MA {cPA (T − T0) + λA} + (1 − y) MB {cPB (T − T0) + λB}

cPA, cPB = specific heats of pure LIQUIDS A and B, kJ/(kg K)

λA, λB = heats of vaporisation of A and B at temperature T, kJ/kg

ENTHALPY − CONCENTRATION DATA (Example)Aceone (A) − Water (B) system at 1 atm total P. Integral heat of solution (15 0C) at different concentrations, specific heat of solutions and x−y data are given:

10001001−22.20.2530.0191.7

0.98−106.80.6240.0575.70.96−159.70.7550.166.60.9−187.70.8150.262.2

0.85−171.70.830.3610.8−121.50.8390.460.4

0.75−83.560.8490.5600.7−60.30.8590.659.5

0.66−41.110.8740.758.9

0.61−23.880.8980.858.20.56−10.550.9350.957.50.5600.9630.95570.5401156.5

cPSΔHSyXT, 0C Average specific heat of liquid acetone = 0.57 kcal/(kg K)

Specific heat of water = 1 kcal/(kg K)

Heat of vaporisation of acetone = λA = 128.8 − (T 0C − 50), kcal/kg

Average heat of vaporisation of water = λB = 550 kcal/kg

x−y−H (Enthalpy−Concentration Diagram)

x →(0, 0) 1

15000 kcal/mol

y − HV curve

x − HL curve

Δ(HV)1

Δ(HV)2

Tie Lines

x, y →

ENTHALPY − CONCENTRATION DIAGRAMThe amount of vapour and liquid phases SEPARATED can be determined by “Lever Arm Rule”.Point R on the x−y diagram corresponds to Tie Line WD.Point Q on the x−y diagram corresponds to Tie Line IJ.

Consider 2 solutions whose states (H and x or y) and amounts are given by M and N. After MIXING, the resultant solution is P. Hence:

Total material balance: M + N = PComponent A balance: M zM + N zN = PzPEnthalpy balance: M HM + N HN = PHP

Eliminating P: M/N = (zN − zP)/(zP − zM)Eliminating P: M/N = (HN − HP)/(HP − HM)

⇒ (HN − HP)/(zN − zP) = (HP − HM)/(zP − zM)⇒ Slope of section NP = Slope of section MP⇒ Points M, N, and P are collinear.

FeedF, zF , HF

VapourSchematic of a

FLASH distillation unit

Top productD, xD = yD , HD

Bottom productW, xW , HW

+ QFlash drum

baffles

Flash vaporisation of a BINARY MIXTUREIf a sufficiently hot liquid is THROTTLED into a vessel, a part of it vaporises ⇒ vapour will be richer in MVC ⇒ partial separation.The liquid is heated under Pressure, and throttled into a vesselunder Reduced pressure.Let the Flow Rate, Composition, Enthalpy of FEED, DISTILLALATE (TOP), and BOTTOM products be (F, zF, HF), (D, xD, HD), (W, xW, HW), and Q be the rate of supply of heat exchanger:Total material balance: F = D + WComponent A balance: F zF = D xD + W xW

⇒ (D + W) zF = D xD + W xW

Enthalpy balance : F HF + Q = D HD + W HW

QH H x z D FW FD F ElimQD

inatix zW F H H W F F

⎛ ⎞− +⎜ ⎟− ⎝ ⎠− = =− ⎛ ⎞− +⎜ ⎟

⎝ ⎠

Flash Vaporisation of a Binary Mixture

x →(0, 0) 1

15000 kcal/mol

y − HV

F (zF, HF + Q/F)

x, y →

x − HL

F′ (zF, zF)

x zW D FSlope = x zD W F

−− = −

Flash vaporisation of a BINARY MIXTURECold feed at F1 ⇒ passes through preheater ⇒ receives Q heat energy ⇒ hot feed F ⇒ FLASHES into vapour (D) and liquid (W) upon throttling into the flash drum.The enthalpy and composition of vapour (D) and liquid (W) streams are obtained by drawing TIE LINE through F.The point F′ (zF, zF) is located on the diagonal of x−y plot.Point P is located on the equilibrium curve such that the slope of the line F′P = −W/D. The line F′P is the ‘operating line’ for flash vaporisation process.By using the H−x−y and x−y curves, the amounts and compositions of D and W can be calculated for a given feed for given Q.Alternatively, if the fraction of feed to be vaporised is specified, the required Q can be computed.The analysis can be extended to the case of a real stage with a given value of stage efficiency (How ?).

VapourSchematic of a STEAM DISTILLATION unit

Residue

Steam sparger

Receiver A

B(water)

Open steam

Steam coil

Steam Distillation

For an ideal binary solution, yA = pA/PT = (xA PAV)/PT

However, if A and B are IMMISCIBLE, their mixture exerts a vapour pressure = Σ (vapour pressures of individual components).

Hence the BUBBLE POINT of such a mixture < boiling point of either A and B.

Process: Live steam is passed through a liquid ⇒ A vaporises ⇒ leaves WITH steam ⇒ condenser ⇒ 2 layers (A and B) ⇒ separated by decantation.

Applications of steam distillation:

Separation of high boiling materials (decolourisation, deodourisaion of vegetable oils, purification of essential oils).

Separation and purification of hazardous materials like turpentine

Separation of thermally unstable materials

Separation of volatile impurities from waste water (removal of ammonia, VOCs)

Separation of A immiscible with waterPT = PA

V + PBV ⇒ PB

V = PT − PAV

If mA moles of A are volatilised by putting in mB moles of steam, and if the system operates at equilibrium,

mA/mB = PAV/PB

V = PAV/(PT − PA

V)⇒ mA = mB PA

V/(PT − PAV)

However, if the system does not operate at equilibrium, pA< PA

⇒ A factor called “vaporisation efficiency E” is defined = pA/PA

⇒mA = mB E PAV/(PT − PA

V)The above equation can be used to compute the steam requirement.The vaporisation efficiency usually ranges from 60% to 90%.

BATCH DISTILLATION of a Binary Mixture(Differential Distillation / Rayleigh Distillation)

Equilibrium is assumed in the still.Feed is charged to a still pot.heat is continuously supplied.As boiling continues, MVC in liquid decreases with timeVapour is led to condenserThe condensate (top product) is collected in a receiver.At the beginning, the condensate is very rich in MVC.MVC in the condensate decreases with time.Applicable when the components greatly differ in volatility and when ultra−pure products are NOT required.

BATCH DISTILLATION of a Binary Mixture

Theoretical analysis is based on “differential mass balance”.

Let, at instance of time,

L = moles of liquid, having mole fraction x

D = moles of accumulated condensate

y = mole fraction of equilibrium vapour

dL = change in amount of liquid over differential time

dD = change in amount of vapour withdrawn over differential time

Differential mass balance:

Total material balance: − dL = dD

Component A balance: − d(L x) = y dD

⇒ − L dx = y dD + x dL = y dD − x dD = (y − x) dD

Putting dD = − dL, and rearranging: dL dx L y x

If distillation starts with F moles of feed (xF), and continues till the liquid reduces to W moles (xW), above equation is integrated as:

x xW W WdL dx F dx ln L y x W y

(Rayle

xF x xFigh equatio

= ⇒ =∫ ∫ ∫− −

If x−y data is available, the RHS can be solved graphically.If y = f(x) is available, RHS can be solved analytically / numerically.If α = αAVERAGE, then RHS can be solved analytically, as follows:

( )( )

x x 1 x 1 xWF dx 1 F W Wln ln lnxW 1 x 1 x 1 xx - x W

In a more co

nvenient form :

FF 1 1 x

F 1 xF x WFln lnW x W 1 xW F

− −= = +∫ α α − − −

+ α −

−= α

xWF dxln (Rayleigh W

quatxF

ion)= ∫−

The above equation involves 4 quantities: F, W, xF, xW.

The AVERAGE composition of the ACCUMULATEDdistillate is given be TOTAL and COMPONENT material balances:

Total material balance: F = D + W

Component material balance: F xF = D yD,Average + W xW

Continuous MultistageFractional Distillation of

Binary Mixtures

FeedF, zF , HF

Vapour

Top productD, xD, HD

Vapour

Vn+1yn+1

HV,n+1

L’mxm

V’m+1ym+1

HV,m+1

−QCEnvelope 1

Envelope 2

Envelope 3

Envelope 4

MATERIAL AND ENERGY BALANCE EQUATIONSThe determination of number of stages is based on steady state M & E B equations over the ENVELOPES as shown in the Figure.L & V: liquid and vapour flows ABOVE the feed location.L’ & V’: liquid and vapour flows BELOW the feed location.Ln & Vn = molar liquid and vapour flow rate LEAVING nth

stage.HL,n & HV,n = molar liquid and vapour enthalpy LEAVINGnth stage.L0 = flow rate of reflux to the top stage (suffix ‘0’ signifies that the stream is ‘as if’ coming from a hypothetical 0th

stage above top stage.

QC = heat removal rate from overhead condenser.

QB = heat supply rate to reboiler (to vaporise a part of the liquid leaving the bottom stage).

F = feed rate to the column.

zF = mole fraction of MVC in feed.

D = rate of distillate removal from reflux drum.

xD = mole fraction of MVC in distillate.

W = rate of bottom product removal from reboiler.

xW = mole fraction of MVC in bottom product.

L0/D = R = reflux ratio.

Top product:D, xD, HD

Envelope 1

x 0, H

L0Stage 1

Vapour: V1, y1, HV1

ENVELOPE 1(condenser + reflux drum)

Overall balance:V1 = L0 + D = R D + D = D (R + 1)

Component “A” balance:V1 y1 = L0 x0 + D xD

Enthalpy balance:V1 HV1 = L0 HL0 + D HD + QC

⇒ QC = D { (R + 1) HV1 − R HL0 − HD }

Please note: y1 = xD = x0 (for total condenser)

Vapour

Vn+1yn+1

HV,n+1

Stage n

Rectifying section

Envelope 2

ENVELOPE 2(part of rectifying section + condenser)

Overall balance:Vn+1 = Ln + D

Component “A” balance:Vn+1 yn+1 = Ln xn + D xD

Enthalpy balance:Vn+1 HV,n+1 = Ln HL,n + D HD + QC

If you put n = 0, equations for Envelope 1 are obtained !

Vapour

L’mxm

V’m+1ym+1

HV,m+1

Envelope 3

Bottom productW, xW, HW

Stage m

ENVELOPE 3(part of stripping section + reboiler)

Overall balance:L’m = V’m+1 + W

Component “A” balance:L’m xm = V’m+1 ym+1 + W xW

Enthalpy balance:L’m HL,m + QB = V’m+1 HV,m+1 + W HW

Vapour

Envelope 4

Feed:F, zF, HF

Feed stage

Rectifyingsection

Strippingsection

ENVELOPE 4(entire column + condenser + reboiler)

Overall balance:F = D + W

Component “A” balance:F zF = D xD + W xW

Enthalpy balance:F HF + QB = D HD + W HW + QC

Above the FEED POINT: Vapour is enriched OR purified by discarding the LVC into the downflowing liquid ⇒ MVC concentration in vapour is more than that in the feed ⇒ Rectifying OR enriching section.

Below the FEED POINT: MVC is removed or stripped out of the downflowing liquid ⇒ MVC concentration in vapour is less than that in the feed ⇒ stripping section.

Flow rate, composition, state of FEED: T, P, phase, etc. Feed may be L, V, or V+L (two-phase).Required degree of separation: requirement of QUALITY or PURITY.Reflux ratio and the condition of the reflux: The ratio of reflux FED BACK to the top and the top product WITHDRAWN. Reflux may be SATURATED liquid or SUB−COOLED liquid.Operating pressure and allowable pressure drop across the column: operating pressure also determines the TEMPERATURE of column. The ΔP = f (type, number of hydraulic design of trays OR type and size of packings). ΔP is very important if the column is run at LOWER PRESSURES.Tray / packings type: determine the efficiency of separation.

Determination of Number of Trays: McCabe−Thiele MethodDeveloped in 1925: graphical solution of material balanceequations together with the equilibrium relation x−y (or equilibrium data).Assumptions:

1. CONSTANT MOLAR OVERFLOW of liquid (and vapour) from one tray to another over any section of the distillation column, that is:

L0 = L1 = L2 = ... = Ln = L (RECTIFYING section);L’m = L’m+1 = ... = L’N = L’ (STRIPPING section), AND

V1 = V2 = V3 = ... = Vn+1 = V (RECTIFYING section);V’m+1 = V’m+2 = ... = V’N+1 = V’ (STRIPPING section);

2. Heat loss from distillation column is negligible. (If there are heat losses, there will be NET condensation within the column, leading to a corresponding variation in the flow rates along the column.

Steps:

(1) Draw equilibrium curve using available x−y data

(2) Draw operating lines for the rectifying and the stripping sections

(3) Draw the STEPS between the equilibrium curve and the operating lines to find out the number of IDEAL stages.

The RECTIFYING Section

ENVELOPE 2 (part of rectifying section + condenser)

Component “A” balance: Vn+1 yn+1 = Ln xn + D xD

( )R = L D = L D0xL D L D Dy = x + x =

x + n+1 n D nV V V

e CMO: constant mol

D V DxR Dy = x + n+1 nR

ar overflow

+ 1 R + 1

The above equation is a STRAIGHT line, having slope = R/(R+1), and Y−intercept = xD/(R+1). It is satisfied by xn = xD; yn+1 = xD. This equation is called the “operating line for RECTIFYING section”.

The STRIPPING Section

ENVELOPE 3 (part of stripping section + reboiler)

Component “A” balance: L’ xm = V’ ym+1 + W xW

V' = L' - WL' Wy = x - xm+1 m WV' V'

L' Wy = x - xm+1 m WL'

- W L' -

The above equation is a STRAIGHT line, having slope = L’/(L’ − W), It is satisfied by the point (xW, xW). This equation is called the “operating line for STRIPPING section”.

The FEED line

Consider the plate ON WHICH the feed is introduced (feed plate).

F, zF, HF

L, HL, f−1

L’, HL, f V’, HV, f+1

V, HV, f

Total material balance:F + L + V’ = L’ + VEnthalpy balance:F HF + L HL, f−1 + V’ HV, f+1 =

L’ HL, f + V HV, fAssuming Uniform Enthalpy:

HL, f−1 ≈ HL, f ≈ HLand HV, f+1 ≈ HV, f ≈ HV

⇒ F HF + L HL + V’ HV = L’ HL + V HV⇒ (L’ − L) HL = (V’ − V) HV + F HF

H - H'L - L V F =

= (say) F H - HV

iminating V and V :

L’ − L = increase in the liquid flow rate across the feed stage as a result of introduction of the feed

= Rate of LIQUID INPUT with the feed

⇒ LIQUID FRACTION in the feed

heat required to convert 1 mole feed to saturated vapourq = molar heat of vaporisation of the saturated lAlso, iquid

If the feed is at its bubble point (saturated liquid), q = 1

If the feed is at its dew point (saturated vapour), q = 0

If the feed is a two-phase mixture, q = fraction of liquid in the feed

⇒ (1 − q) = quality of the feed

If the point of intersection of RECTIFICATION operating line and STRIPPING operating line is (x, y), then

V y = L x + D xD and V’ y = L’ x − W xW

Subtracting the above equations, and using the equation for component balance for the complete column (F zF = D xD + W xW), we get:(V − V’) y = (L − L’) x + (D xD + W xW) = (L − L’) x + F zF

Total material balance on feed plate:F + L + V’ = L’ + V

Using the above two equations, and the definition of q, we get the equation of FEED LINE:

( )zq Fy x z , z, passing through and

the intersection of two op

F Fq er

q ing lines

1= −

− −

The number of IDEAL TRAYS are determined by step (staircase) construction between the equilibrium curve and operating lines.

(xn, yn): on equilibrium curve.

(xn, yn+1): on operating line.

Construction may start from either D (xD) or W (xW).

When the FEED LINE is crossed, a CHANGEOVER from one operating line to the other is done (that is, a transition from RECTIFICATION to STRIPPING section, or vice versa).

1. If feed is Superheated Vapour: feed line has +VE slope

(q < 0): Explain.

2. If feed is Saturated Vapour: feed line is horizontal

(q = 0): Explain.

3. If feed is Liquid + Vapour: feed line has −VE slope

(0 < q < 1): Explain.

4. If feed is Saturated Liquid: feed line is vertical

(q = 1): Explain.

5. If feed is Subcooled Liquid: feed line has +VE slope

(q > 1): Explain.

Various Types of Feed Lines

Superheated vapour(+, +)

Saturated vapour

Vapour + Liquid(−, +)

Saturated Liquid

Subcooled Liquid(+, −)

Feed l

L L V F = = F H HV Lzq F q 1 q - 1

−−−

= −−

Feed Tray Location and Number of Ideal Trays

S’’

Feed = Saturated Liquid

TOTAL REFLUXIf the liquid from the overhead condenser is TOTALLY recycled to the column (D = 0), the column is said to run at “total reflux”. Also no product is drawn from the reboiler either ⇒ there should not be any Feed.

R = L0/D = L0/0 = ∞Operating line for Rectifying section:

xR Dy = x + n+1 nR + 1 R + 1Slope = 1,hen R ,

Operating line coinc Y-

ideintercep

s with Dt = 0

IAGONAL→ ∞

Naturally, the number of stages corresponding to infinite Rare minimum.Total reflux is very often used during the startup of a distillation column till the steady state is attained. After this, continuous flow of feed, top product, and bottom product withdrawal are SIMULTANEOUSLY started.

Fenske’s equation

Used to compute theoretically the minimum number of trays if the relative volatility remains reasonably (more or less) constant. It also assumes a total reboiler (analogous to total condenser).

If αw is relative volatility of A at the reboiler temperature, and xW, yW are Liquid and Vapour concentrations in the reboiler, then:

y xW W = α W1 - y 1 - xW W

The Vapour leaving the reboiler and entering the lowest tray (Nm) has a mole fraction yW of component A. The Liquid leaving this tray has a composition xNm ⇒ (xNm, yW) lies on the Operating Line ⇒ xNm = yW. Therefore, the above equation can be rewritten as:

x xNm W = α W1 - x 1 - xNm WApplying the same procedure to the case of tray number Nm, we get:

y x xNm Nm W = α = α α Nm Nm W1 - y 1 - x 1 - xNm Nm W

Fenske’s equation (continued … )

Continuing the procedure up to the top tray (where y1 = xD), we get:

( )( )

xx y WD 1 = = α α . . . α α α 1 2 Nm-1 Nm W1 - x 1 - y 1 - xD 1 WN + 1 xx m WD = α average1 - x 1 - xD W

x 1 - xD Wlog x 1 - xW D N + 1 = m

log αaverage

⎛ ⎞⎜ ⎟⎝ ⎠

⎧ ⎫⎪ ⎪⎨

⎬⎪ ⎪⎩ ⎭⎛

⇒⎞

⎜ ⎟⎝ ⎠

The above equation is called the “Fenske’s equation”, which is useful for the calculation of MINIMUM number of trays.

Minimum Reflux Ratio

This is exactly analogous to the (L/G)min for gas absorption. This is based on the identification of “pinch point”.

The following procedure is followed to determine Rm:

Locate the points D (xD, xD) and W (xW, xW) on the DIAGONAL of the x−y diagram. These are the terminal points of operating lines.

Draw the feed line through F from the known feed composition zF, and feed quality (given by q); locate the point of intersection of the feed line and the equilibrium curve (x−y diagram). Call this point M.

Join DM (operating line for rectifying section), and extend this line till the Y-axis.

Y−intercept = E = xD/(Rm + 1). Get Rm from the intercept.

At Reflux Ratio = Rm, the number of stages for the given separation are infinity (driving force becomes ZERO at the PINCH POINT).

Minimum Reflux Ratio

EY−intercept = xD/(Rm + 1)

Slope = Rm/(Rm + 1)

Sieve Tray

Tray Columns

Tray Deck

Reboilers

Circulating Pump

Heating Medium

Forced Circulation

Bottom Tray

Heating Medium

Vertical Thermosiphon

Bottom Tray

Bottoms Product

Reboilers

Bottoms Product

Heating Medium

Kettle

Heating Medium

Horizontal Thermosiphon

Reboilers

Bottom Tray

Bottoms ProductBottoms

Product

Feed Distributors

The “Real” World

Fouled Structured Packing Damaged Valve Tray

The “Real” World

Plugged DistributorTray “Blanking” Strips

Valve Tray Deck

Major Tray Damage

Fouled Bubble Cap Tray

Structured Packings

Fair, J.R., Seibert, A.F., Behrens, M., Saraber, P.P., and Olujic, Z. “Structured Packing Performance − Experimental Evaluation of Two Predictive Models”, Industrial and Engineering Chemistry Research 2000, 39, 1788−1796.

Structured Packing Wetted Area

Ponchon Savarit Method (Calculation of Number of Stages)

Calculations on an Enthalpy−Concentration (H−x−y) diagram.

If the dew point and bubble point lines are more or less straight and roughly parallel ⇒ Latent heat of vaporisation is basically constant with respect to composition ⇒ Prerequisite for constant molar overflow, energy balances may be neglected.

If the saturation curves show significant changes in curvature ⇒constant molar overflow cannot be assumed.

H−x−y diagram is more general than a McCabe−Thiele construction, because it takes direct account of the thermal effects and does not require an assumption of constant equimolar overflow.

Feed and product points (F, D, W) can be located on H−x−y diagram. The coordinates are their composition and enthalpy.

FeedF, zF , HF

Vapour

Vn+1yn+1

HV,n+1

L’mxm

V’m+1ym+1

HV,m+1

−QCEnvelope 1

Envelope 2

Envelope 3

Envelope 4

MATERIAL AND ENERGY BALANCE EQUATIONSThe determination of number of stages is based on steady state M & E B equations over the ENVELOPES as shown in the Figure.L & V: liquid and vapour flows ABOVE the feed location.L’ & V’: liquid and vapour flows BELOW the feed location.Ln & Vn = molar liquid and vapour flow rate LEAVING nth

stage.HL,n & HV,n = molar liquid and vapour enthalpy LEAVINGnth stage.L0 = flow rate of reflux to the top stage (suffix ‘0’ signifies that the stream is ‘as if’ coming from a hypothetical 0th

stage above top stage.

QC = heat removal rate from overhead condenser.

QB = heat supply rate to reboiler (to vaporise a part of the liquid leaving the bottom stage).

F = feed rate to the column.

zF = mole fraction of MVC in feed.

D = rate of distillate removal from reflux drum.

xD = mole fraction of MVC in distillate.

W = rate of bottom product removal from reboiler.

xW = mole fraction of MVC in bottom product.

L0/D = R = reflux ratio.

Envelope 1

x 0, H

L0Stage 1

Vapour: V1, y1, HV1

ENVELOPE 1(condenser + reflux drum)

Overall balance:V1 = L0 + D = R D + D = D (R + 1)

Component “A” balance:V1 y1 = L0 x0 + D xD

Enthalpy balance:V1 HV1 = L0 HL0 + D HD + QC

⇒ QC = D { (R + 1) HV1 − R HL0 − HD }

Please note: y1 = xD = x0 (for total condenser)

Vapour

Vn+1yn+1

HV,n+1

Stage n

Rectifying section

Envelope 2

ENVELOPE 2(part of rectifying section + condenser)

Overall balance:Vn+1 = Ln + D

Component “A” balance:Vn+1 yn+1 = Ln xn + D xD

Enthalpy balance:Vn+1 HV,n+1 = Ln HL,n + D HD + QC

If you put n = 0, equations for Envelope 1 are obtained !

Vapour

L’mxm

V’m+1ym+1

HV,m+1

Envelope 3

Stage m

ENVELOPE 3(part of stripping section + reboiler)

Overall balance:L’m = V’m+1 + W

Component “A” balance:L’m xm = V’m+1 ym+1 + W xW

Enthalpy balance:L’m HL,m + QB = V’m+1 HV,m+1 + W HW

Vapour

Envelope 4

Feed:F, zF, HF

Feed stage

Rectifyingsection

Strippingsection

ENVELOPE 4(entire column + condenser + reboiler)

Overall balance:F = D + W

Component “A” balance:F zF = D xD + W xW

Enthalpy balance:F HF + QB = D HD + W HW + QC

Rectifying SectionENVELOPE 2 (part of rectifying section + condenser)

Total material balance: Vn+1 = Ln + DComponent A balance: Vn+1 yn+1 = Ln xn + (Vn+1 − Ln) zD

⇒ Ln / Vn+1 = (zD − yn+1) / (zD − xn)Note: Reflux may be a sub−cooled liquid and hence the composition is denoted by zD.Energy balance:

Vn+1 HV, n+1 = Ln HL,n + D (HD + QC/D) = Ln HL, n + D Q’D

where Q’D = Enthalpy removed from the TOP section of the column per mole of distillate. Substituting D = Vn+1 − Ln :

Vn+1 HV, n+1 = Ln HL, n + (Vn+1 − Ln) Q’D

⇒ Ln / Vn+1 = (Q’D − HV, n+1) / (Q’D − HL, n)The quantity Ln/Vn+1 is called “internal reflux ratio”.

From the above two equations:

' ' 'Q H Q H Q Hz y D V, n 1 D L, n D V, n 1D n 1'z x z x z yQ HD n D n D n 1D L, n

'z , Q x , HD D n L, n

y , Hn 1 V, n

The 3 points , ,

and , are COLLINEAR.

that is, the points , , and

'L V are COLQn n+1 D

⎛ ⎞⇒ ⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟

− − −− + ++ = =− − −− +

+ +⎝ ⎠

LINEAR.

⇒The point D’ (zD, Q’D) can be considered as a “phase obtained by subtracting Ln from Vn+1”.The flow rate of phase D’ is also denoted by D’ = Vn+1 − Ln.However, NONE of the phases can have the composition zD, and enthalpy Q’D = HD + QC/D.The stream D’ is a “fictitious stream” defined for the purpose of graphical construction.D’ lies vertically above D, because they have same abscissa = zD.If D lies below H−x curve, D is sub−cooled liquid.Put n = 0 ⇒ D’, D, V1 are collinear.Point L1 (x1, HL1) is located by drawing a tie line from V1.L1D’ is joined to get point V2, and so on …

Consider:

Putting n = 0

' 'Q - H Q - HL VD V, n+1 D L, nn n+1 = = ' 'V LQ - H Q - Hn+1 nD L, n D V, n+1' 'Q - H Q - HL + DV D L, 0 D L, 001 = = ' 'L LQ - H Q -

H0 0D V, 1 D V, 1'Q - HD D D1 + = 'L Q -

⇒ ⇒

' 'Q - HL D VD V, 10 1 R = = =vertical dista

D H - H V DH V, 1 D 1V,

vertical distance 1

If the “reflux ratio” is given, the above relation is used to locate D’.

First, D (zD, HD) is located from the given “state” of the distillate.

A vertical line through zD intersects the y−HV curve at V1.

Then, obtain D’ so that the above relation for R is satisfied.

V1V2V3VnVn+1

Ln−1Ln

0.0 1.0

y−HV curve

x−HL curve

zDx, y

Tie lines

Operating line SEGMENTS

Rectifying Section

Stripping SectionENVELOPE 3 (part of stripping section + reboiler)

Total material balance: L’m = V’m+1 + WComponent A balance: L’m xm = V’m+1 ym+1 + (L’m −V’m+1) xW

⇒ L’n / V’m+1 = (ym+1 − xW) / (xm − xW)Energy balance:

L’m HL, m − V’m+1 HV, m+1 = W HW − QB = W (HW − QB/W)⇒ L’m HL, m − V’m+1 HV, m+1 = W Q’W

where Q’W = Enthalpy supplied to the BOTTOM section of the column per mole of bottom product. Substituting D = L’m− V’m+1 :

L’m HL, m − V’m+1 HV, m+1 = (L’m − V’m+1) Q’W

⇒ L’m / V’m+1 = (HV, m+1 − Q’W) / (HL, m − Q’W)

From the above two equations:

' ' 'H Q H Q H Qy x V, m 1 W V, m 1 W L, m Wm 1 W'x x y x x xH Qm W m 1 W m WL, m W

'x , Q x , HW W m L, m

y , Hm 1 V, m

The 3 points , ,

and , are COLLINEAR.

that is, the points , , and

are COL'L V Qm m+1 W

⎛ ⎞⇒ ⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟

− − −− + ++ = =− − −− +

+ +⎝ ⎠

LINEAR.

The flow rate of phase W’ is also denoted by W’ = L’m −V’m+1.

The stream W’ is a “fictitious stream” defined for the purpose of graphical construction.

W’ lies vertically below W, because they have same abscissa = xW.

V’N+1 (reboiler vapour) should lie on the y−HV curve on the tie line drawn from W.

The line connecting V’N+1 and W’ intersects x−HL curve at L’N (collinearity condition).

The tie line through L’N intersects the y−HV curve at V’N, and so on…

0.0 1.0

y−HV curve

x−HL curve

zDx, y

Stripping Section

VN−1

Tie linesOperating line SEGMENTS

Feed PlateA changeover from one section to the other is necessary at the feed plate, and hence feed line is required.Consider the material and enthalpy balances over “envelope 4”, that is, entire column + condenser + reboiler.Component A balance:

(D + W) zF = D zD + W xW ⇒ (D/W) = (zF − xW) / (zD − zF)Enthalpy balance:

(D + W) HF + QB = D HD + W HW + QC

⇒ D HF + W HF = D (HD + QC/D) + W (HW − QB/W)QCH H ' ' ' 'F D H Q z x H Q H Q Q HDD F W F W F W F W D F ' 'QW z z z x z zQ H Q HB D F F W D FD F D FH HW FW

⎛ ⎞⎜ ⎟− +⎜ ⎟ − − − − −⎝ ⎠= = = =

− − −⎛ ⎞ − −− −⎜ ⎟⎜ ⎟⎝

⇒ The 3 points W’(xW, Q’W), F(zF, HF), and D’(zD, Q’D) are collinear.

The intermediate point F(zF, HF) denotes the state of feed in terms of composition and enthalpy.F(zF, HF) lies on the line joining D’ and W’.W’ represents a fictitious stream of flow rate W’ = L’m − V’m+1.D’ represents a fictitious stream of flow rate D’ = V’n+1 − L’n.

⇒ F = D + W = D’ + W’Therefore, F, a real stream can be viewed as a stream obtained by mixing the 2 fictitious streams D’ and W’.The line D’W’ is called “feed line”.

The steps (graphical) for obtaining the number of ideal stages on H−x−y diagram:

1. Data supplied (directly or indirectly): Concentrations, Enthalpies, Flow Rates of Distillate, Bottom Product, as well as Reflux Ratio.

2. Points D, D’, and F are located.

3. D’F is joined, and extended to meet the vertical line {drawn through W(xW, HW)} at point W’.

4. Stages can be constructed either from D’ or from W’.

5. A changeover has to be made after the feed line is crossed.

6. The stage at which the changeover is made is “feed stage”.

7. Construction continues till the other end point is reached.

The steps (graphical) for obtaining the number of ideal stages on x−ydiagram:

1. Draw the equilibrium curve (x−y diagram) using separate axes below the H−x−y diagram using matching scale.

2. Locate (zD, zD) and (xW, xW) on the diagonal line.3. Since any line starting from D’ and W’ intersects the H−x and H−y

curves at xn and yn+1 respectively, the point (xn, yn+1) is point on the operating line in the x−y diagram.

4. Several lines can be drawn from D’ and W’, and for each of these lines, the point (xn, yn+1) is located on the x−y diagram.

5. All these points are joined, giving rise to two curves; these are the 2 operating curves, namely, rectifying section operating curve and stripping section operating curve.

6. Since the flow rates in both these sections are not constant (constant molar overflow is not applicable), the operating lines are not straight.

7. Staircase construction gives the number of ideal stages.

Minimum Reflux RatioIf a line through D’ or W’ coincides with tie line, the corresponding (x,y) values give a pinch point.Reflux ratio is given by: vertical d' 'Q - HL D VD V, 10 1R = = =

istance

vertical dist- H V DV ance, D 1 1

The above relation indicates that the smaller the distance D’V1 (or the nearer the point D’ to the H−x curve), the smaller is the reflux ratio.Most commonly, when the line D’W’ coincides with the tie line through the feed point, minimum reflux ratio occurs.However, for some highly nonideal mixtures, it may happen that atie line through a point other than F intersects the vertical line (drawn through zD) at the smallest distance. In that case, the same should be taken as the true pinch point.Once (D’)min has been located on the H−x−y diagram, then the minimum reflux ratio is computed using the above relation.

Total RefluxIn this case, the reflux ratio is infinite.Naturally, the points D’ and W’ are located at infinity.Therefore, all the lines through D’ and W’ are vertical lines.

The graphical construction involves:Locate V1 vertically above zD, on the H−y curve.Draw a tie line from V1, which will intersect H−x curve at L1.Draw a vertical line from L1, which will intersect H−ycurve at V2.Continue the above construction till xW is reached.This (obviously) gives the minimum number of ideal stages.

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