Question

What is the settling velocity of a discrete particle in a wide body of water when the relevant Reynolds number is less than 0.5? The diameter and specific gravity of the particle are 2 × 10^-3 cm and 2.65, respectively. Water temperature is 20°C. (Kinematic viscosity = 2 × 10^-2 cm²/ sec.)
1. 0.018 cm / sec
2. 0.025 cm / sec
3. 0.18 cm / sec
4. 0.25 cm / sec

Answer 1

Correct Answer - Option 1 : 0.018 cm / sec

Concept:

Settling velocity of particles when Reynolds no less than 0.5 is given by Stocks Law:

\(V = \frac{{\left( {G - 1} \right){\gamma _w}{d^2}}}{{18\;\mu }}\)

Where,

G is the Specific Gravity

d is the particle size

μ is the dynamic viscosity of water.

Calculation:

Given,

G = 2.65;  Re < 0.5;  d =  2 × 10^-3 cm

Kinematic viscosity = 2 × 10^-2 cm²/sec

Let density of water is 1000 kg/m³  

Dynamic viscosity of water, μ = 2 × 10^-2 × 10^-4 × 1000 = 2 × 10^-3 kg/m-sec

The settling velocity is

\(V = \frac{{\left( {G - 1} \right){\gamma _w}{d^2}}}{{18\;\mu }}\)

\(V = \frac{{\left( {2.65 \;- \;1} \right)\; \times \;9810\;{{\left( {2\; \times\; {{10}^{ - 3}}\; \times\; {{10}^{ - 2}}} \right)}^2}}}{{18\; \times\; 2\; \times \;{{10}^{ - 3}}}}\)

V = 0.000179 m/sec = 0.0179 cm/sec