1. Mass Transfer for 4 th Year Chemical Engineering Department Faculty of Engineering Cairo University

2. Today’s Agenda Adiabatic Absorption Stripping Multi-component Absorbtion

3. Adiabatic Absorption Absorption is an exothermic process. If absorber is not cooled, the temperature of the liquid will increase leading to changing the equilibrium relation used. Remember: y=mx & For the same system m=f(T,P) Our task now is to get an equilibrium curve that includes the changes in temperature occurring.

4. Adiabatic Absorption We will assume that the heat liberated by absorption will be gained ONLY by liquid. Overall M.B.: G’(Y in -Y out )=L’(X out -X in ) E.B. on loop: Heat liberated by absorption =heat gained by liquid

5. Adiabatic Absorption E.B. on loop: Heat liberated by absorption =heat gained by liquid L’(X out(i) -X in )×q abs =L’×C PL (T out(i) -T in ) So we can get X out(i) by knowing T out(i)

6. Problem 3: 470 lb/hr of an air-NH 3 mixture containing 5% by weight NH 3 are scrubbed with water at 70 o C in an adiabatic tower in order to remove 95% of the NH 3 present. The heat of solution of NH 3 in water is 300 BTU/lb, and the isothermal equilibrium data at different temperatures may be approximated into straight line as below: Where Y and X are mass ratios Calculate: L’ min. Required number of actual plates if stage efficiency is 35% using 1.5 L’ min. The number of theoretical stages using the same liquid flow rate at constant temperature of 70 o C and Y out Temperature, o C70758085 Y/X1/61/32/31

7. Givens: G= 470 lb/hry in =0.05  =0.95 q=300 BTU/lbx in =0 Cp W =1 BTU/lb.R G’=470*(1-0.05)=446.5 lb/hr Y in =0.05/(1-0.05)=0.0526 X in =0 Y out =0.0526*(1-0.95)=0.00263

12. To get the points on the curves to get the new equilibrium curve, use the relation: T in =70 o C=158 o F T out T out ( o F)X out 701580 751670.03 801760.06 851850.09

16. Operating line: G’=446.5 lb/hr Y in =0.0526 X in =0 Y out =0.00263

20. i- L’ min /G’ =0.735 L’ min =0.735*446.5=328.1 lb/hr ii- L’ op =1.5L’ min =492.16 lb/hr L’ op /G’=1.1 X out =0.0453

24. NAS=7 iii- L’/G’=1.1 Y in =0.0526 X in =0 Y out =0.00263

26. iii- NTS=1.6

27. STRIPPING

28. The process opposite to absorption, the solvent is stripped to be recovered. The operating line equation will not change, but it will be located BELOW the equilibrium curve. The main objective will no longer be Y out, it will be X out, so Recovery (  ) will be used as follows: X out =X in (1-  ) Same steps will be done: getting G’ min, calculating G’ op and determining the number of ideal stages.

29. Y in X Out X in GETTING G min Y out

30. Y in X Out X in GETTING G min Y out FALSE Minimum G’ operating line

31. Y in X Out X in GETTING G min Y out TRUE Minimum G’ operating line (tangent)

32. Y in X Out X in GETTING G min Y out TRUE Minimum G’ operating line (may not be tangent)

33. Y in X Out X in Operating Line Y out

34. Problem 5: As a result of an absorption process, there is a solution which must be stripped of absorbed solute. 500 Kmole/hr of this solution of benzene in a non-volatile oil containing 0.1 mole fraction benzene. The solution will be preheated to 250 o F and will be stripped at 1 atm absolute with superheated steam at 250 o F. The liquid effluent from the stripper is to contain no more than 0.005 mole fraction benzene. Assume isothermal operation and Raoult’s law is applied. Vapour pressure of benzene at 250 o F is 2400 mmHg. Determine the minimum steam rate, the number of ideal trays required for 1.25 times the minimum rate.

35. Problem 5: x in =0.1X in =0.1/(1-0.1)=0.111 L= 500 Kmole/hr L’=500*(1-0.1)=450 Kmole/hr T=250 o FP=1 atm=760 mmHg x out =0.005X out =0.005/(1-0.005)=0.00503 y in =0Y in =0 Isothermal operation and Raoult’s law is applied. P o B = 2400 mmHg.m=2400/760=3.16 y=3.16x G’ op =1.25 G’ min

38. i- L’/G’ min =3.96 G’ min =450/3.96=113.66 Kmol/hr ii- G’ op =1.25 G’ min =142.08 Kmol/hr L’/G’ op =3.17

40. i- L’/G’ min =3.96 G’ min =450/3.96=113.66 Kmol/hr ii- G’ op =1.25 G’ min =142.08 Kmol/hr L’/G’ op =3.17 NTS=10.5

41. Multi-Component Absorption - This is the general case, and the most abundant. - The solvent will have the ability to dissolve more than one component from the gas stream. - Different recovery percentages will be achieved for each component. -Here is the only case where we can work by mole fractions not ratios.

42. Design Steps Information available: 1- L/G or n 2- Feed gas composition (purity) y inA, y inB, y inC,.. 3- Available solvent purity x inA, x inB, x inC,….. 4- Equilibrium of each component with solvent. 5- Key component Recovery (   ) Usually he wants you to calculate the recovery of other components.

43. Multi-Component Absorption

44. Graphical method

45. Analytical method: