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This section includes 7 Mcqs, each offering curated multiple-choice questions to sharpen your Mass Transfer knowledge and support exam preparation. Choose a topic below to get started.
1. |
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 |
Answer» D. Schmidt number and Lewis number | |
2. |
The expression for average mass transfer coefficient is |
A. | Sh = 0.664 (Re)<sup> 0.5</sup> (Sc)<sup> 0.33</sup> |
B. | Sh = 0.664 (Re)<sup> 0.6</sup> (Sc)<sup> 0.33</sup> |
C. | Sh = 0.332 (Re)<sup> 0.7</sup> (Sc)<sup> 0.33</sup> |
D. | Sh = 0.332 (Re)<sup> 0.8</sup> (Sc)<sup> 0.33</sup> |
Answer» B. Sh = 0.664 (Re)<sup> 0.6</sup> (Sc)<sup> 0.33</sup> | |
3. |
The empirical correlation for local mass transfer coefficient for turbulent boundary layer flow past a flat plate is given by |
A. | Sh <sub>X</sub> = 0.332 (Re)<sup> 0.5</sup> (Sc)<sup> 0.33</sup> |
B. | Sh <sub>X</sub> = 0.332 (Re)<sup> 0.6</sup> (Sc)<sup> 0.33</sup> |
C. | Sh <sub>X</sub> = 0.0298 (Re)<sup> 0.7</sup> (Sc)<sup> 0.33</sup> |
D. | Sh <sub>X</sub> = 0.0298 (Re)<sup> 0.8</sup> (Sc)<sup> 0.33</sup> |
Answer» E. | |
4. |
The empirical correlation for local mass transfer coefficient for laminar boundary layer flow past a flat plate is given by |
A. | Sh <sub>X</sub> = 0.332 (Re)<sup> 0.5</sup> (Sc)<sup> 0.23</sup> |
B. | Sh <sub>X</sub> = 0.332 (Re)<sup> 0.5</sup> (Sc)<sup> 0.33</sup> |
C. | Sh <sub>X</sub> = 0.332 (Re)<sup> 0.5</sup> (Sc)<sup> 0.43</sup> |
D. | Sh <sub>X</sub> = 0.332 (Re)<sup> 0.5</sup> (Sc)<sup> 0.53</sup> |
Answer» C. Sh <sub>X</sub> = 0.332 (Re)<sup> 0.5</sup> (Sc)<sup> 0.43</sup> | |
5. |
Consider the above problem, find the value of Sherwood number |
A. | 84.44 |
B. | 74.44 |
C. | 64.44 |
D. | 54.44 |
Answer» D. 54.44 | |
6. |
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
|
A. | nv = 15.5 * 10 -6 m<sup>2</sup>/s |
B. | 0.0176 m/s |
C. | 1.0176 m/s |
D. | 2.0176 m/s |
E. | 3.0176 m/s |
Answer» B. 0.0176 m/s | |
7. |
Consider the above problem, find out the value of Reynolds number |
A. | 14.09 * 10 <sup>8</sup> |
B. | 14.09 * 10 <sup>7</sup> |
C. | 14.09 * 10 <sup>6</sup> |
D. | 14.09 * 10 <sup>5</sup> |
Answer» E. | |