MCQs
Total Questions : 115
| Page 6 of 12 pages
Question 51. Estimate the diffusion coefficient of carbon monoxide through air in which the mole fractions of each constituents are
O₂ = 0.18
N₂ = 0.72
CO = 0.10
The gas mixture is at 300 K and 2 atmosphere total pressure. The diffusivity values are
D CO = 18.5 * 10⁻⁶ m²/s at 273 K, 1 atm
D CN = 19.2 * 10⁻⁶ m²/s at 288 K, 1 atm
The C is carbon monoxide, O is oxygen and N is nitrogen.
O₂ = 0.18
N₂ = 0.72
CO = 0.10
The gas mixture is at 300 K and 2 atmosphere total pressure. The diffusivity values are
D CO = 18.5 * 10⁻⁶ m²/s at 273 K, 1 atm
D CN = 19.2 * 10⁻⁶ m²/s at 288 K, 1 atm
The C is carbon monoxide, O is oxygen and N is nitrogen.
Answer: Option A. -> 10.29 * 10⁻⁶ m²/s
Answer: (a).10.29 * 10⁻⁶ m²/s
Answer: (a).10.29 * 10⁻⁶ m²/s
Answer: Option C. -> 64
Answer: (c).64
Answer: (c).64
Answer: Option A. -> Ideal
Answer: (a).Ideal
Answer: (a).Ideal
Answer: Option B. -> (Nᴀ+Nв)Cᴀ/C, DᴀвdCᴀ/dz
Answer: (b).(Nᴀ+Nв)Cᴀ/C, DᴀвdCᴀ/dz
Answer: (b).(Nᴀ+Nв)Cᴀ/C, DᴀвdCᴀ/dz
Answer: Option B. -> L² T⁻¹
Answer: (b).L² T⁻¹
Answer: (b).L² T⁻¹
Answer: Option D. -> 2630
Answer: (d).2630
Answer: (d).2630
Answer: Option A. -> Leather
Answer: (a).Leather
Answer: (a).Leather
Question 58. Air at 30 degree Celsius temperature flows at 45 m/s past a wet flat plate of length 0.5 m. Estimate the value of mass transfer coefficient. Assume that the water vapor content of air initially is negligible and take the following thermos-physical properties of air.
D = 0.256 * 10⁻⁴ m²/s
µ = 1.86 * 10⁻⁵ kg/m s
cₚ = 1.005 k J/kg degree Celsius
pᵣ = 0.701
p =1.165 kg/m³
D = 0.256 * 10⁻⁴ m²/s
µ = 1.86 * 10⁻⁵ kg/m s
cₚ = 1.005 k J/kg degree Celsius
pᵣ = 0.701
p =1.165 kg/m³
Answer: Option A. -> 0.1076 m/s
Answer: (a).0.1076 m/s
Answer: (a).0.1076 m/s
Answer: Option B. -> 545
Answer: (b).545
Answer: (b).545
Answer: Option C. -> 0.623
Answer: (c).0.623
Answer: (c).0.623