ENGR263 2025 Q2Past paperOld spec ENGR2632:152 marks62 min

ENGR263 Mass Transfer Summer 2025 Q2[VALID] official answers

Answer ALL parts (a) - (b).

Formulas you may need
  • Flux with bulk flow NA,z=−cDABdyAdz+yA(NA,z+NB,z)N_{A,z} = -cD_{AB}\dfrac{dy_A}{dz} + y_A(N_{A,z} + N_{B,z}) (on the formula sheet)
  • Diffusion through a stagnant gas: NA=DABPRTzln⁡P−pA2P−pA1=DABPRTz pA1−pA2pBMN_A = \dfrac{D_{AB}P}{RTz}\ln\dfrac{P - p_{A2}}{P - p_{A1}} = \dfrac{D_{AB}P}{RTz}\,\dfrac{p_{A1} - p_{A2}}{p_{BM}} (on the formula sheet in the current ENGR271 sheet; on the 2025 ENGR263 sheet it had to be derived from the flux equation)
  • Multicomponent stagnant mixture 1DA,m=∑j≠Ayj′DAj\dfrac{1}{D_{A,m}} = \sum_{j \ne A}\dfrac{y'_j}{D_{Aj}}, yj′y'_j on an A-free basis (on the formula sheet)
  1. (a)
    Oxygen (A) is diffusing through carbon monoxide (B) via a rectangular plane surface between positions 1 and 2 under steady-state conditions, with the carbon monoxide non-diffusing. The total pressure is 1×105 N/m21 \times 10^5\ \mathrm{N/m^2} and the temperature 0∘C0^\circ\mathrm{C}. The partial pressure of oxygen at planes 1 and 2 is 13000 and 6500 N/m2\mathrm{N/m^2}, respectively. The two planes are 2 mm apart. The diffusivity for this mixture is 1.87×10−5 m2/s1.87 \times 10^{-5}\ \mathrm{m^2/s}. Calculate the rate of diffusion of oxygen in kmol/s through each square meter of the two planes. The following data are provided: R=8314 N m/(kmol K)R = 8314\ \mathrm{N\,m/(kmol\,K)}.
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  2. (b)
    Calculate the rate of diffusion of oxygen (A) under steady-state conditions and the same experimental conditions as the above part (a), but this time assuming that the non-diffusing gas is a mixture of methane (B) and hydrogen (C) under a volume ratio 2:1. Comment on this result with respect to those in part (a). The diffusivities of oxygen to hydrogen and methane are estimated to be DO2−H2=6.99×10−5D_{O_2-H_2} = 6.99 \times 10^{-5} and DO2−CH4=1.86×10−5 m2/sD_{O_2-CH_4} = 1.86 \times 10^{-5}\ \mathrm{m^2/s}, respectively.
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