This set of Mechanical Operations Questions & Answers for Exams focuses on “Terminal Falling Velocities of Spherical Particles”.

1. What is the equation for drag force on a spherical body, when external force acts with acceleration a? Here ρ is the density of fluid, ρP is the density of particle, m is mass of the particle, Cd is the drag coefficient and r is the radius of the sphere.

a) u = \(\sqrt {\frac {8 \times r \times a(ρ – ρP)}{3 \times Cd \times ρP}}\)

b) u = \(\sqrt {\frac {8 \times r \times a(ρP – ρ)}{3 \times Cd \times ρ }}\)

c) u = \(\sqrt {\frac {3 \times r \times a(ρP – ρ)}{Cd \times ρ }}\)

d) u = \(\sqrt {\frac {4 \times r \times a(ρP – ρ)}{3 \times Cd \times ρ }}\)

View Answer

Explanation: When the forces on a spherical particle falling through a fluid are balanced, we get:

m\(\frac {du}{dt}\) = Fe – Fb – Fd (Equation 1)

Fe = ma

Fb = \(\frac {m \times ρ \times a}{ρP}\)

Fd = \(\frac {Cd \times ρ \times Ap}{2}\) u

^{2}

We can substitute Fe, Fb and Fd in equation 1 to get:

m\(\frac {du}{dt}\) = ma – \(\frac {Cd \times ρ \times Ap}{2}\) u

^{2}– \(\frac {m \times ρ \times a}{ρP}\)

Dividing by m on both sides and rearranging terms we get:

\(\frac {du}{dt} = \frac {a(ρP – ρ)}{ρP} – \frac {Cd \times ρ \times Ap}{2 \times m}\) u

^{2}

When \(\frac {du}{dt}\) = 0 we get maximum settling velocity or terminal velocity

\(\frac {a(ρP – ρ)}{ρP} = \frac {Cd \times ρ \times Ap}{2 \times m}\) u

^{2}

u = \(\sqrt {\frac {2ma(ρP – ρ)}{Cd \times ρ \times Ap \times ρP}}\)

When we substitute Ap = πr

^{2}(projected area of sphere) in the equation of u, we get:

u = \(\sqrt {\frac {8 \times r \times a(ρP – ρ)}{3 \times Cd \times ρ }}\)

2. Which of the following is the equation of a spherical ball of radius r, settling under gravity through a fluid, if acceleration due to gravity is considered as 10 m/s^{2}?

a) u = 4x\(\sqrt {\frac {5 \times r \times (ρP – ρ)}{3 \times Cd \times ρ}}\)

b) u = 4x\(\sqrt {\frac {r \times (ρP – ρ)}{Cd \times ρ}}\)

c) u = 5x\(\sqrt {\frac {5 \times (ρP – ρ)}{3 \times Cd \times ρ}}\)

d) u = \(\sqrt {\frac {8 \times (ρP – ρ)}{3 \times Cd \times ρ}}\)

View Answer

Explanation: We know that u = \(\sqrt {\frac {8 \times r \times a(ρP – ρ)}{3 \times Cd \times ρ}}\) for a spherical body.

When a = 10 m/s

^{2}we get u = \(\sqrt {\frac {80 \times r \times (ρP – ρ)}{3 \times Cd \times ρ}}\), which can also be written as

u = 4x\(\sqrt {\frac {5 \times r \times (ρP – ρ)}{3 \times Cd \times ρ}}\)

3. A spherical particle of diameter 1 µm and density 1200kg/m^{3} is falling freely through a column of water with a velocity of 0.7 m/s. Considering the viscosity of water to be 8.11 × 10^{-4} kg/ms and acceleration due to gravity as 9.8 m/s^{2}, calculate the terminal velocity of the particle.

a) 0.34 × 10^{-6} m/s

b) 1.34 × 10^{-7} m/s

c) 1.34 × 10^{-8} m/s

d) 2.34 × 10^{-7} m/s

View Answer

Explanation: Given,

µ = 8.11 × 10

^{-4}kg/ms

v = 0.7 m/s

D = 1 × 10

^{-6}m

ρP = 1200 kg/m

^{3}

Since the fluid is water, ρ = 1000 kg/m

^{3}

Re = \(\frac {ρvD}{µ}\)

Re = 0.863

Since Re < 1, the flow is laminar.

Drag coefficient Cd can be given as, Cd = 24/Re

Terminal velocity for laminar flow of spherical particle is given by, u = \(\frac {g \times (ρP – ρ)}{18 µ}\) × D

^{2}

u = 1.34 × 10

^{-7}m/s

4. What is the diameter of a particle that has a density of 2600 kg/m^{3} and flows through water under laminar conditions. Assume viscosity of water as 8.11 × 10^{-4} kg/ms. The terminal velocity is found to be 0.005 m/s.

a) 0.9 µm

b) 1 µm

c) 0.4 µm

d) 0.7 µm

View Answer

Explanation: Terminal velocity for laminar flow of spherical particle is given by, u = \(\frac {g \times (ρP – ρ)}{18 µ}\)

Given,

u = 0.005 m/s

µ = 8.11 × 10

^{-4}kg/ms

ρ = 1000 kg/m

^{3}

ρP = 2600 kg/m

^{3}

0.005 = \(\frac {9.8 \times (2600-1000)}{18 \times 8.11}\) × 10

^{4}× D

^{2}

D = 6.83 × 10

^{-5}m

D ≈ 0.7 × 10

^{-6}m

D ≈ 0.7 µm

5. What is the terminal settling velocity of spherical particle flowing through a liquid of density 1100 kg/m^{3} and Reynolds number 1500 with a velocity 0.04 m/s? The diameter of the particle is 10 m a its density it 1300 kg/m^{3}.

a) 2.3 m/s

b) 2.5 m/s

c) 2 m/s

d) 2.7 m/s

View Answer

Explanation: Given,

ρ = 1100 kg/m

^{3}

ρP = 1300 kg/m

^{3}

D = 10 m

V = 0.04 m/s

Since Re > 1000 the flow is turbulent in nature

Therefore, u = 1.75\(\sqrt {\frac {(g D (ρP – ρ)}{ρ}}\)

u = 2.336 m/s

u ≈ 2.3 m/s

**Sanfoundry Global Education & Learning Series – Mechanical Operations.**

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