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ENGR5002 / Week 10
ENGR262 2025 Q3(b)Past paperOld spec ENGR2622:29 marks11 min

ENGR262 Particle Technology and Separation Summer 2025 Q3[PARTIAL] Left out: (a) minimum water-to-air ratio (8 marks) and (c) outlet liquid composition at a given water rate (8 marks) are absorption-column design and not on the current ENGR5002 syllabus.

[Parts (a) and (c), absorption-column design, are omitted.]

An exhaust gas stream (100 kmol/h) from a chemical plant contains 15 mol% of a pollutant while the rest is air. An absorption column using pure water as the absorbent operates isothermally at 30∘C30^\circ\mathrm{C}. The measured vapour-liquid equilibrium data at 30 C are:

Mole fraction of pollutant in liquid, xx0.0200.0600.1000.1400.1800.2200.260
Mole fraction of pollutant in vapour, yy0.0100.0310.0530.0750.0990.1240.149
Formulas you may need
  • Two-film theory: NA=ky(y−yi)=kx(xi−x)N_A = k_y(y - y_i) = k_x(x_i - x), with (xi,yi)(x_i, y_i) on the equilibrium curve (learn this)
  • Graphical construction: line of slope −kx/ky-k_x/k_y from the bulk point (x,y)(x, y) to the equilibrium curve (learn this)
  1. (b)
    After taking samples at a certain point in the column, the technician measured the mole fraction of pollutant in bulk gas and bulk liquid: y=0.11y = 0.11 and x=0.15x = 0.15, respectively. Find the interface mole fraction of pollutant in liquid and gas phases at that point in the column if the mass transfer coefficient for pollutant in the gas phase is 3×10−3 kmolA s−1 m−2(mole fraction)−13 \times 10^{-3}\ \mathrm{kmol_A\,s^{-1}\,m^{-2}(mole\ fraction)^{-1}} and in the liquid phase is 4×10−1 kmolA s−1 m−2(mole fraction)−14 \times 10^{-1}\ \mathrm{kmol_A\,s^{-1}\,m^{-2}(mole\ fraction)^{-1}}. (Present your results on a y-x graph.)
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