ENGR262 2018 Q B1(a)-(b)Past paperOld spec ENGR2622:212 marks14 min

ENGR262 Particle Technology and Separation 2018 Section B Q B1[PARTIAL] Left out: (c) minimum liquid flow rate (7 marks) and (d) number of theoretical stages (6 marks) are absorption-column design (operating lines, stage stepping), not on the current ENGR5002 syllabus.

Answer ALL parts (a) - (d). [Parts (c)-(d), absorption-column design, are omitted.]

The equilibrium distribution of a solute A between gas and water at a particular temperature is given as yA=1.2xAy_A = 1.2x_A, where xAx_A and yAy_A stand for mole fractions of solute A in the liquid and in the gas phases, respectively. At a certain point in a mass transfer device, the concentration of solute A in the bulk air is 0.04 mole fraction and that in the bulk aqueous phase is 0.025. At the same point, the local individual mass transfer coefficients in the gas and liquid phases are ky=7.2 kmol h−1 m−2(Δy)−1k_y = 7.2\ \mathrm{kmol\,h^{-1}\,m^{-2}}(\Delta y)^{-1} and kx=4.6 kmol h−1 m−2(Δx)−1k_x = 4.6\ \mathrm{kmol\,h^{-1}\,m^{-2}}(\Delta x)^{-1}.

Formulas you may need
  • Two-film theory, flux continuity at the interface: NA=ky(y−yi)=kx(xi−x)N_A = k_y(y - y_i) = k_x(x_i - x), with yi=mxiy_i = mx_i (interface at equilibrium) (learn this)
  • Overall coefficients: 1Ky=1ky+mkx\dfrac{1}{K_y} = \dfrac{1}{k_y} + \dfrac{m}{k_x}, NA=Ky(y−y∗)N_A = K_y(y - y^*), y∗=mxy^* = mx (learn this)
  1. (a)
    In which direction does the transport of solute A occur (from gas to liquid or from liquid to gas)? Explain your answer.
    [7]Third
  2. (b)
    Determine interfacial concentrations in mole fractions in both the gas-phase and the liquid-phase. (Present your results on a y-x graph.)
    [5]