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ENGR5002 / Week 10
ENGR263 2022 Q A2(a)-(b)Past paperOld spec ENGR2632:111 marks13 min

ENGR263 Process of Mass Transfer 2021-22 Q A2[PARTIAL] official answers Left out: (c) (14 marks), oxygen through a stagnant CH4/H2/CO2 mixture, is word for word the ENGR271 2026 Q3 already in the bank (engr271-2026-q3), so it is not converted twice.

Answer ALL parts (a) - (c). [Part (c), oxygen diffusing through a stagnant CH4_4/H2_2/CO2_2 mixture, is the same question as ENGR271 2026 Q3 (already in the bank) and is omitted here.]

Formulas you may need
  • Mass transfer coefficient NA=kc(cA,s−cA,∞)N_A = k_c(c_{A,s} - c_{A,\infty}) (on the formula sheet)
  • Diffusion through a stagnant gas: NA=PDABRTδln⁡P−pA2P−pA1N_A = \dfrac{PD_{AB}}{RT\delta}\ln\dfrac{P - p_{A2}}{P - p_{A1}} (on the formula sheet)
  • 1 atm=760 mm Hg1\ \mathrm{atm} = 760\ \mathrm{mm\,Hg}; 1 dm3=1000 cm31\ \mathrm{dm^3} = 1000\ \mathrm{cm^3} (learn this)
  1. (a)
    Starch in the human diet is digested in the stomach and small intestine into monosaccharides such as glucose, which are the only forms that can be absorbed. For one set of experiments, the rate of intestine uptake of glucose is 1.6×10−10 mol/(cm2 s)1.6 \times 10^{-10}\ \mathrm{mol/(cm^2\,s)} from a solution containing 2.65×10−4 mol/dm32.65 \times 10^{-4}\ \mathrm{mol/dm^3} glucose. Calculate the mass transfer coefficient if the amount of glucose in blood is very small.
    [3]Third
  2. (b)
    The rate of evaporation of water from a flat surface maintained at a temperature of 60∘C60^\circ\mathrm{C} is 2.91×10−4 kg/(s m2)2.91 \times 10^{-4}\ \mathrm{kg/(s\,m^2)}. What will be the rate of evaporation of benzene from a similar flat surface, but maintained at 26∘C26^\circ\mathrm{C}, if the effective film thicknesses are the same in both cases? It is given: molecular weight of water 18 g/mol; vapour pressure of water at 60 C = 149 mm Hg; vapour pressure of benzene at 26 C = 99.5 mm Hg; diffusivity of air-water vapour at 60 C = 2.6×10−5 m2/s2.6 \times 10^{-5}\ \mathrm{m^2/s}; diffusivity of air-benzene vapour at 26 C = 0.98×10−5 m2/s0.98 \times 10^{-5}\ \mathrm{m^2/s}; atmospheric pressure in both cases = 1.013×105 N/m21.013 \times 10^5\ \mathrm{N/m^2}; the ideal gas constant is equal to 8.314 m3 Pa K−1 mol−18.314\ \mathrm{m^3\,Pa\,K^{-1}\,mol^{-1}}.
    [8]