in this example, calculate the number of ideal plates by Kremser Equation in Y form.

EXAMPLE 2 2.2. Ammonia is stripped from a dilute aqueous solution by countercurrent contact with air in a column containing seven sieve trays. The equilibrium rela. tionship is ye=0.8xe, and when the molar flow of air is 1.5 times that of the solution, 90 percent of the ammonia is removed. (a) How many ideal stages does the column have, and what is the stage efficiency? (b) What percentage removal would be obtained if the air rate were increased to 2.0 times the solution rate? Solution (a) For a dilute solution and a dilute gas, L and V are assumed constant, and the stripping factor is S=LmV=0.81.5=1.2 All concentrations can be expressed in terms of xa, the mole fraction of NH3 in the entering solution: xb=0.1xaxb=0sinceyb=0 From an ammonta balance, vy=vysL Also, From Bq. (20.28). ya=VL(0.9x0)=1.50.9xi=0.6xixn=my0=0.80.6x0=0.75xc0N=losln[(x0.75xa)/(0.1x20)]=ln1.2ln(0.25x0/0.1xi)=5.02 The separation correuponds to 5.02 ideal stages, so the stage efficitacy is 502/7=72 percent: (b) If V/L is increased to 2.0 and the number of ideal stager N does not chunge (same stage efficiency), S=0.82.0=1.6. Then from Eq(20,30) lnxbxpxa=5.02ln1.6=2.36xbxuxe=10.59 Let f be the fraction of NH3 removed. Then x2=(1fxn. By a material bodance. yaxa=VL(xaxb)=21[xa(1f)xa)=21fxa=mya=0.80.5fxi=0.625fxa Thus: xaxa=(10.625f)xdxaxa=10.59xh=10.59(1f)xc From these, f=0.962, or 96.2 percent is removed. The conditions for the original case and the new case are sketchad in Fig. 20.10 . FIGURE 20.10 Diagram for Example 20.2