This set of Mass Transfer Multiple Choice Questions & Answers (MCQs) focuses on “Non Dimensionless Correlation for Convective Mass Transport”.
1. Assume the following thermos-physical properties of air
D = 0.256 * 10 -4 m2/s
ยต = 1.86 * 10 -5 kg/m s
c p = 1.005 k J/kg degree Celsius
p r = 0.701
p =1.165 kg/m3
Find the value of Schmidt number
a) 0.423
b) 0.523
c) 0.623
d) 0.723
View Answer
Explanation: Sc = ยต/p D.
2. Air at 30ยฐC flows at 45 m/s past a wet flat plate of length 0.5 m. Estimate the value of the mass transfer coefficient, assuming the water vapor content of air is initially negligible. The thermophysical properties of air are: diffusion coefficient D = 0.256 ร 10-4 m2/s, dynamic viscosity ยต = 1.86 ร 10-5 kg/mยทs, specific heat cp = 1.005 kJ/kgยทยฐC, Prandtl number Pr = 0.701, and density ฯ = 1.165 kg/m3.
a) 0.1076 m/s
b) 0.2076 m/s
c) 0.3076 m/s
d) 0.4076 m/s
View Answer
Explanation: h m = 0.0296 (Re) -0.2 (V)/ (Sc) 0.667.
3. Consider air at 30ยฐC flowing past a wet flat plate at a velocity of 45 m/s and a plate length of 0.5 m. Given that the dynamic viscosity ยต = 1.86 ร 10-5 kg/mยทs and density ฯ = 1.165 kg/m3, determine the Reynolds number for the flow.
a) 14.09 * 10 8
b) 14.09 * 10 7
c) 14.09 * 10 6
d) 14.09 * 10 5
View Answer
Explanation: Re = V l p/ ยต.
4. Assume the following thermos-physical properties of air
D =0.82 * 10 -6 m2/s
v = 15.5 * 10 -6 m2/s
Find the value of Schmidt number
a) 2.89
b) 1.89
c) 3.89
d) 0.89
View Answer
Explanation: Schmidt number = v/D = 1.89.
5. Air at 1 atm and 25 degree Celsius temperature, flows inside a 3 cm diameter tube with a velocity of 5.25 m/s. Determine mass transfer coefficient for iodine transfer from the air stream to the weak surface. Assume the following thermos-physical properties of air
D =0.82 * 10 -6 m2/s
v = 15.5 * 10 -6 m2/s
a) 0.0176 m/s
b) 1.0176 m/s
c) 2.0176 m/s
d) 3.0176 m/s
View Answer
Explanation: h m = (Sherwood number) D/d =0.0176 m/s.
6. Given the following parameters for iodine transfer from an air stream to a weak surface:
Air velocity = 5.25 m/s
Tube diameter = 3 cm
Diffusion coefficient (D) = 0.82 ร 10-6 m2/s
Kinematic viscosity (v) = 15.5 ร 10-6 m2/s
Find the value of the Sherwood number for this mass transfer process.
a) 84.44
b) 74.44
c) 64.44
d) 54.44
View Answer
Explanation: Sherwood number = -0.023 (Re) 0.83 (Sc) 0.44 = 64.44.
7. The empirical correlation for local mass transfer coefficient for laminar boundary layer flow past a flat plate is given by
a) Sh X = 0.332 (Re) 0.5 (Sc) 0.23
b) Sh X = 0.332 (Re) 0.5 (Sc) 0.33
c) Sh X = 0.332 (Re) 0.5 (Sc) 0.43
d) Sh X = 0.332 (Re) 0.5 (Sc) 0.53
View Answer
Explanation: Sh = h ml/D = .332 (Re) 0.5 (Sc) 0.33.
8. The empirical correlation for local mass transfer coefficient for turbulent boundary layer flow past a flat plate is given by
a) Sh X = 0.332 (Re) 0.5 (Sc) 0.33
b) Sh X = 0.332 (Re) 0.6 (Sc) 0.33
c) Sh X = 0.0298 (Re) 0.7 (Sc) 0.33
d) Sh X = 0.0298 (Re) 0.8 (Sc) 0.33
View Answer
Explanation: Sh = h ml/D.
9. The expression for average mass transfer coefficient is
a) Sh = 0.664 (Re) 0.5 (Sc) 0.33
b) Sh = 0.664 (Re) 0.6 (Sc) 0.33
c) Sh = 0.332 (Re) 0.7 (Sc) 0.33
d) Sh = 0.332 (Re) 0.8 (Sc) 0.33
View Answer
Explanation: Sh = h ml/D = 0.664 (Re) 0.5 (Sc) 0.33.
10. Sherwood number is a function of
a) Lewis number and Reynolds number
b) Prandtl number and Lewis number
c) Reynolds number and Schmidt number
d) Schmidt number and Lewis number
View Answer
Explanation: It is a function of Reynolds number and Schmidt number.
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