Mass Transfer Questions and Answers – Non Dimensionless Correlation for Convective Mass Transport

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This set of Mass Transfer Interview Questions & Answers 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
µ = 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

Explanation: Sc = µ/p D.

2. 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
c p = 1.005 k J/kg degree Celsius
p r = 0.701
p =1.165 kg/m3
a) 0.1076 m/s
b) 0.2076 m/s
c) 0.3076 m/s
d) 0.4076 m/s

Explanation: h m = 0.0296 (Re) -0.2 (V)/ (Sc) 0.667.

3. Consider the above problem, find out the value of Reynolds number
a) 14.09 * 10 8
b) 14.09 * 10 7
c) 14.09 * 10 6
d) 14.09 * 10 5

Explanation: Re = V l p/ µ.
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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

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

Explanation: h m = (Sherwood number) D/d =0.0176 m/s.

6. Consider the above problem, find the value of Sherwood number
a) 84.44
b) 74.44
c) 64.44
d) 54.44

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

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

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

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

Explanation: It is a function of Reynolds number and Schmidt number.

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