3. Decide on the Best Possible Operating Conditions.

The most common rule is to decide based on Capital-Energy Trade Off.

Process Variables/Parameters

CostCapital

TOTAL

Energy

Optimum Parameter Value

The study is conducted during design phase.

Simple parametric study

Select the optimum value for operation.

Can you give examples ?

SUGGESTED PROCEDURE (cont...)

Lecture 11 : SYNTHESIS OF SEPARATION SYSTEM – OPERATING PARAMETER

COST

REFLUX RATIO

Capital

TOTAL

Energy

optimumrefluxratio

1. DISTILLATION

Some quick guideline basedon past operational experience

RRopt = (1.1-1.5) * RRmin

However, this serve as a starting pointwhich can be optimised later once thepotential of integrating the columnoperation with the process has beenexplored!

What about other parameters ?

Pressure/Temperature;

Decide by the vaporisation & condensing T of the mixture.Decide by degradation T ofthe distilled material.

eg. Separation of Gas Mixture

P set in the column should allow for condensationof lighter component atcooling media T in condenser.

Let us look at the distillation case more closer ….

Temperature of the cooling media will set the constraint for the condensing temperature in condenser. In turn will set the required column pressure for operation.

Setting the pressure of the column will affect the degree of difficulty for separation as it has an impact on the relative volatility between the components. Generally, higher pressure will lead to more difficult separation requiring higher reflux ratio for a fixed no. of stages.

Higher pressure causes higher density for liquid and vapour flow leading to higher vapour/liquid flow traffic, more difficult disengagement between vapour and liquid leading to flooding.

Setting the pressure of the column will affect the boiling temperature and latent heat of vaporisation of the mixture. Though the latent heat tend to decrease but the boiling point will increase as the pressure is increased.

Setting the pressure of the column will more or less determined the thickness of the column wall.

Occasionally, you will confront mixtures which are hard to separate due to very close relative volatility or due to formation of azeotrope. A slightly different configuration of distillation column is required.

i. Use of two columns with different operational pressure

Temp.

AB

maximum boiling azeotrope

P1

P2

pure A

pure B

P1

P2

ii. Use of entrainer which forms binaryor ternary azeotrope with the top product components but upon condensation, forms a two phase liquid layer which can be separated in the decanter.

Decanterentrainer

A

B

entrainer is a volatilecomponent which formsa low boiling azeotropewith the products.

iii. Use of solvent to break the azeotrope by increasing the relative volatility between the component to be separated.

pure A

pure B

P1

P2

solvent + B

solvent

Note1. The amount of entrainer or solvent used has an impact on the separation capability and the energy required by the distillation column in undertaking the separation. This can be optimised on stand alone basis but if integration with the process can provide the energy required, optimisation should be left at later stage.

2. ABSORPTION

Some quick guideline basedon past operational experience

Absorption Factor (A) for component i (L/VKi) is within the range of 1.2 - 2.0.

In the absorption process, solvent is used to dissolve certain preferential component termed as the solute, which normally appears as minor component. The mixture of solvent and solute is then separated in the stripping process to recover the solvent.

Temp.

P1

P2

solvent + B

solvent

solute

product free of solute

Stripping Factor for component i (VKi/L) is within the range of 1.2 - 2.0.

The typical value is about 1.4

Absorber Stripper

High P Low P

Low T High T

Why?

i. Decide by the equilibrium property between the solvent and the system which gives the best extraction performance.

ii. Decide by the condition of which the solvent can be vaporised and condensed to recover it.

In the case of physical absorbtion, the vapour liquid equilibrium can be approximated by Henry’s Law. iii Hxp

Partial pressure of component iAssuming ideal gas behaviour, Pyp ii

The K value could be calculated using equationP

H

x

yK i

i

ii

A straight line would be expected in the plot of y i versus xi.

yi

xi

Equilibrium lineSlope = Ki

A material balance around the absorber;

Lin, xin Vout, yout

Lout, xout Vin, yin

outin x

V

Lyx

V

Ly

L/V – slope of operating line

yin

yout

xoutxin

Minimum solvent flowrate

In the case of desorption (stripping), the vapour liquid equilibrium can be approximated using the same approach.

However the operating line for the stripping will be situated underneath the equilibrium line

yi

xi

Equilibrium lineSlope = Ki

A material balance around the stripper;

Lin, xin

Vout, yout

Lout, xout

Vin, yin

outin x

V

Lyx

V

Ly

L/V – slope of operating line

yout

yin

xinxout

The number of stages could be determined either through the graphical construction or using Kremser equations as below;

KVL

KVLKxy

Kxy

KVL

KVL

N

outout

inin

log

11

log

Number of stages for absorption

If L/KV =1

inout

outin

Kxy

yyN

KVL

KVL

K

yx

K

yx

KVL

N

inout

inin

1log

1log

Number of stages for stripping

If L/KV =1

Ky

x

xxN

inout

outin

Assume that L, V and K remain constant.

Example on Absorption Process.

A hydrocarbon gas stream containing benzene is to be stripped in an absorption column using a heavy liquid hydrocarbon stream with an average molar mass of 200 kg/kmol. The concentration of benzene in the gas stream is 2% by volume and the liquid contains 0.2% benzene by mass. The gas stream flowrate is 850 m3/hr, its pressure and temperature is 1.07 bar and 25 C. It is required to remove 95% of the benzene from the vapour flow.Data : At standard atm. condition, 1 kgmol of gas occupies 22.4 m3 volume ; Molar mass of benzene 78 kg/kmol ; Assume the equilibrium obeys Raoult’s Law and at the temp. of separation, the saturated liquid vapour pressure of benzene is 0.128 bar.

Lin, xin Vout, yout

Lout, xout Vin, yin

outin x

V

Lyx

V

Ly

From Raoult’s Law : K = Psat / P = 0.128 / 1.07 = 0.12

hrkmolV /6.36

07.1013.1

278298

4.22

1850

= Vin = Vout

Earlier, the estimates for A = L/KV from practice is within 1.2 – 2.0 for optimum operation. Let’s assume a value of 1.4

Calculate L based on the A factor = 1.4

L = 1.4 x 0.12 x 36.6 = 6.15 kmol/hr = 6.15 kmol/hr x 200 kg/kmol = 1230 kg/hr

Calculate xin = ( 0.2 / 100 ) x (200/78) = 0.0051

Calculate yout = (2/100) x (1 – 0.95) , assuming V is constant = 0.001

KVL

KVLKxy

Kxy

KVL

KVL

N

outout

inin

log

11

log

Applying the equation

N = 8 theoretical stages.

COST

SOLVENT FLOWRATE

Capital

TOTAL

Operating(Solvent + Energy)

optimumsolventflow

So what sort of optimisation are we dealing with?

The amount of solvent used have an impact on :

i. The separation capabilityii. The amount of energy required to recover the solvent

But how does the two link?

ABSORBER

solvent

feed

solvent + dissolved component(require separation to recover- use desorber)

eg. CO2 absorption using amine soln.

Example of application.

3. LIQUID – LIQUID EXTRACTION

Similar to the absorption process, liquid-liquid extraction separates a homogenous mixture by addition of another phase (solvent) – in this case an immiscible liquid with the process liquid. The separation occurs as a results of components (solute) in the feed distributing themselves differently between the two liquid phases. The use of sovent is to extract solute from the feed.

FXF,i

RXR,i

SXS,i

E XE,i

iR

iE

iR

iEi x

xK

,

,

,

,

Distribution coefficient reflected by the composition ratio or activity coefficient ratio between extract and raffinate

Comparing components I and j ;

jR

jE

iR

iE

jR

jE

iR

iE

j

iji

x

x

x

x

K

K

,

,

,

,

,

,

,

,

,

Separation factor reflects the tendency of component I to be extracted from the raffinate to the extract compared to component j

When choosing the proper solvent for an extraction process, the issues to be considered are;

1. Distribution coefficientThe solvent should give a large K (distribution coefficient) value to minimise solvent amount.Select solvent that is chemically similar to the solute – “like dissolves like” . Eg. A polar liquid such as water would be good to dissolves ionic and polar compounds while Non polar compound such as hexane would be a better choice for dissolving non polar compounds such as hydrocarbon in general.

2. Separation FactorThe separation factor should be greater than unity but preferably as large as possible.

3. Insolubility of the SolventThe solubility of the solvent in the raffinate and vice versa should be as low as possible.

4. Ease of RecoveryThe solvent should be easily recoverable (by distillation), thermally and chemically stable, does not formed azeotropes with solute, much lower relative volatility from the solute and small latent heat of vaporisation.

5. Significant Density Difference between the solvent (extract phase) and the feed (raffinate)The density difference has to be significant enough to enable the two liquid phases to coalesce more readily.

6. Difference in Interfacial TensionThe larger the difference in interfacial tension between the two phases, the more readily coalescence will occur. However, higher interfacial tension will lead to more difficult dispersion in the extraction.

The vapour pressure at working condition should be kept low (low T) especially when organic solvent is used to avoid emitting VOCs. In addition, the solvent should be non toxic , non flammable and low viscosity.

For liquid-liquid extraction, the liquid-liquid equilibrium could be treated similar to the vapour liquid equilibrium in absorption.

The K value could be determined from equation

iR

iE

iR

iEi x

xK

,

,

,

,

Eout, xi,E,out

Ein, xi,E,in

Rin, xi,R,in

Rout, xi,R,out

L/V – slope of operating line

Xi,E

Xi,R

Equilibrium line is normally a curve

xE,in

xR,inxR,out

xE,out

If the equilibrium curve is linearised, then the Kremser equation could be used to determine number of stages.

Eout, xi,E,out

Ein, xi,E,in

Rin, xi,R,in

Rout, xi,R,out

L/V – slope of operating line

Xi,E

Xi,R

Equilibrium line is a linear line

xE,in

xR,inxR,out

xE,out

Minimum solvent

rate

log

11log

outout

inin

Kxy

Kxy

N e = KE/R

4. ADSORPTION

Adsorption is a process in which molecules of adsorbate become attached to the surface of a solid adsorbent.There are 2 broad classes of adsorption namely;

i. Physical Adsorption – physical bonds form between the adsorbent and the adsorbate.ii. Chemical Adsorption - chemical bonds form between the adsorbent and the adsorbate.

L or Vout, x or yout

L or Vin, x or yin

Adsorbent bed

During adsorption process, the adsorbate (normally gases or liquid components) were removed from the bulk gas/liquid flow.

The process continues until the bed gets saturated with the adsorbate.

L or Vout

L or Vin

Adsorbent bed

During desorption process, the adsorbent bed is regenerated by removing the adsorbate from the bed using hot fluid flushing, temperature swing or pressure swing.

While the bed is being regenerated, the adsorption process continues using a parallel bed.

The capacity of an adsorption bed can be represented by the adsorption isotherms.

P

Volume adsorbed

Increasing temperature

T

Volume adsorbed

Increasing pressure

Adsorption of gases/vapours on solid

The concentration profile in an adsorption bed changes according to the pattern below;

Gas Flow + Adsorbate

t1 t2 t3 t4t6

At t6, the bed become saturated

Gas Flow Adsorbate free

The adsorption can be represented using eqn.

npkV '

V – volume absorbedP – partial pressure of adsorbate in the systemk’ , n – constants determined from experiment

5. EVAPORATORS

What about other parameters ?

Pressure/Temperature;

Decide by vaporisation ability at the giventemperature of the heating media. Also the product degradation temperature helps to set the operating temperature and pressure.

COST

AREA OF EVAPORATOR UNIT

Capital

TOTAL

Energy

optimumarea of evaporator

Similar to distillation, only usesheat for separation of material.

4 different arrangement ;

1. Forward feed operation2. Backward feed operation3. Parallel feed operation4. Mixed feed operation

No. of Evaporator UnitDecide by amount of feed materialand the area requirement of the evaporator.

Difficult to decide !

For a single evaporator !

COST

AREA OF EVAPORATOR UNIT

CapitalTOTAL

Energy

optimumarea of evaporator

T

H

2 evaporator

3 evaporator

T

H

T

H

4 evaporator

What can you say the relation between No. of evaporator unit and the two cost components ?Do you expect the overall cost curve to be smooth ?

Remember Q = UADT

Questions.

6. DRYERS

Drying refers to the removal of moisture from solid component.There are many forms of dryers namely ;

Tunnel dryers Rotary dryers

Drum dryers Spray dryers

Fluidised bed dryers

Selection depends on the nature of application.

Production of powdered milk - spray dryingDrying of solid slab - tunnel dryersDrying of grains - Fluidised bed dryers

examples

Dryer efficiency is a measure that can also be used for comparisonand selection particularly when external heating source is required.

Dryer efficiency = heat of vaporisation / total heat consumed.

Conclusions

So far, we have explored a number of methods to separate homogenous mixture.

The setting of operating parameters (T, P, flowrate of solvent etc.) have to take into account of several factors such as the physical properties of the components in the mixture, the separation ability reflected by the K value etc.

The capital trade off effect in setting the operating parameter have to be considered.