1. 3. Decide on the Best Possible Operating Conditions. The most common rule is to decide based on Capital-Energy Trade Off. Process Variables/Parameters Cost Capital 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

2. COST REFLUX RATIO Capital TOTAL Energy optimum reflux ratio 1. DISTILLATION Some quick guideline based on past operational experience RR opt = (1.1-1.5) * RR min However, this serve as a starting point which can be optimised later once the potential of integrating the column operation with the process has been explored! What about other parameters ? Pressure/Temperature; Decide by the vaporisation & condensing T of the mixture. Decide by degradation T of the distilled material. eg. Separation of Gas Mixture P set in the column should allow for condensation of lighter component at cooling media T in condenser.

3. 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.

4. 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. A B maximum boiling azeotrope P1P1 P2P2 pure A pure B P1P1 P2P2 ii. Use of entrainer which forms binary or ternary azeotrope with the top product components but upon condensation, forms a two phase liquid layer which can be separated in the decanter. Decanter entrainer A B entrainer is a volatile component which forms a low boiling azeotrope with the products.

5. iii. Use of solvent to break the azeotrope by increasing the relative volatility between the component to be separated. pure A pure B P1P1 P2P2 solvent + B solvent Note 1. 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.

6. 2. ABSORPTION Some quick guideline based on past operational experience Absorption Factor (A) for component i (L/VK i ) 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. P1P1 P2P2 solvent + B solvent solute product free of solute Stripping Factor for component i (VK i /L) is within the range of 1.2 - 2.0. The typical value is about 1.4 Absorber Stripper High P Low P Low THigh 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.

7. In the case of physical absorbtion, the vapour liquid equilibrium can be approximated by Henry’s Law. Partial pressure of component i Assuming ideal gas behaviour, The K value could be calculated using equation A straight line would be expected in the plot of y i versus x i. yiyi xixi Equilibrium line Slope = K i A material balance around the absorber; L in, x in V out, y out L out, x out V in, y in L/V – slope of operating line y in y out x out x in Minimum solvent flowrate

8. 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 yiyi xixi Equilibrium line Slope = K i A material balance around the stripper; L in, x in V out, y out L out, x out V in, y in L/V – slope of operating line y out y in x in x out

9. The number of stages could be determined either through the graphical construction or using Kremser equations as below; Number of stages for absorption If L/KV =1 Number of stages for stripping If L/KV =1 Assume that L, V and K remain constant.

10. 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 m 3 /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 m 3 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. L in, x in V out, y out L out, x out V in, y in From Raoult’s Law : K = P sat / P = 0.128 / 1.07 = 0.12 = V in = V out 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 x in = ( 0.2 / 100 ) x (200/78) = 0.0051 Calculate y out = (2/100) x (1 – 0.95), assuming V is constant = 0.001 Applying the equation N = 8 theoretical stages.

11. COST SOLVENT FLOWRATE Capital TOTAL Operating (Solvent + Energy) optimum solvent flow So what sort of optimisation are we dealing with? The amount of solvent used have an impact on : i. The separation capability ii. 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. CO 2 absorption using amine soln. Example of application.

12. 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. F X F,i R X R,i S X S,i E X E,i Distribution coefficient reflected by the composition ratio or activity coefficient ratio between extract and raffinate Comparing components I and j ; Separation factor reflects the tendency of component I to be extracted from the raffinate to the extract compared to component j

13. When choosing the proper solvent for an extraction process, the issues to be considered are; 1. Distribution coefficient The 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 Factor The separation factor should be greater than unity but preferably as large as possible. 3. Insolubility of the Solvent The solubility of the solvent in the raffinate and vice versa should be as low as possible. 4. Ease of Recovery The 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 Tension The 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.

14. 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 E out, x i,E,out E in, x i,E,in R in, x i,R,in R out, x i,R,out L/V – slope of operating line X i,E X i,R Equilibrium line is normally a curve x E,in x R,in x R,out x E,out

15. If the equilibrium curve is linearised, then the Kremser equation could be used to determine number of stages. E out, x i,E,out E in, x i,E,in R in, x i,R,in R out, x i,R,out L/V – slope of operating line X i,E X i,R Equilibrium line is a linear line x E,in x R,in x R,out x E,out Minimum solvent rate  = KE/R

16. 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 V out, x or y out L or V in, x or y in 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 V out L or V in 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.

17. 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 t1t1 t2t2 t3t3 t4t4 t6t6 At t 6, the bed become saturated Gas Flow Adsorbate free The adsorption can be represented using eqn. V – volume absorbed P – partial pressure of adsorbate in the system k’, n – constants determined from experiment

18. 5. EVAPORATORS What about other parameters ? Pressure/Temperature; Decide by vaporisation ability at the given temperature 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 optimum area of evaporator Similar to distillation, only uses heat for separation of material. 4 different arrangement ; 1. Forward feed operation 2. Backward feed operation 3. Parallel feed operation 4. Mixed feed operation No. of Evaporator Unit Decide by amount of feed material and the area requirement of the evaporator. Difficult to decide ! For a single evaporator !

19. COST AREA OF EVAPORATOR UNIT Capital TOTAL Energy optimum area 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 = UA  T Questions.

20. 6. DRYERS Drying refers to the removal of moisture from solid component. There are many forms of dryers namely ; Tunnel dryers Rotary dryers Drum dryersSpray dryers Fluidised bed dryers Selection depends on the nature of application. Production of powdered milk - spray drying Drying of solid slab - tunnel dryers Drying of grains - Fluidised bed dryers examples Dryer efficiency is a measure that can also be used for comparison and selection particularly when external heating source is required. Dryer efficiency = heat of vaporisation / total heat consumed.

21. 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.