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# Mechanical Operations MCQ’s – Terminal Falling Velocities of Spherical Particles

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 }}

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Answer: b

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

mdudt = Fe – Fb – Fd (Equation 1)

Fe = ma

Fb = m×ρ×aρP

Fd = Cd×ρ×Ap2 u^{2}

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

mdudt = ma – Cd×ρ×Ap2 u^{2} – m×ρ×aρP

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

dudt=a(ρP–ρ)ρP–Cd×ρ×Ap2×m u^{2}

When dudt = 0 we get maximum settling velocity or terminal velocity

a(ρP–ρ)ρP=Cd×ρ×Ap2×m u^{2}

u = 2ma(ρP–ρ)Cd×ρ×Ap×ρP−−−−−−−−−√

When we substitute Ap = πr^{2} (projected area of sphere) in the equation of u, we get:

u = 8×r×a(ρP–ρ)3×Cd×ρ−−−−−−−−−√

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 = 4x5×r×(ρP–ρ)3×Cd×ρ−−−−−−−−√

b) u = 4xr×(ρP–ρ)Cd×ρ−−−−−−√

c) u = 5x5×(ρP–ρ)3×Cd×ρ−−−−−−−√

d) u = 8×(ρP–ρ)3×Cd×ρ−−−−−−−√

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Answer: a

Explanation: We know that u = 8×r×a(ρP–ρ)3×Cd×ρ−−−−−−−−−√ for a spherical body.

When a = 10 m/s^{2} we get u = 80×r×(ρP–ρ)3×Cd×ρ−−−−−−−−−√, which can also be written as

u = 4x5×r×(ρP–ρ)3×Cd×ρ−−−−−−−−√

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

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Answer: a

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 = ρvDµ

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 = g×(ρP–ρ)18µ × D^{2}

u = 1.34 × 10^{-7} m/s

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

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Answer: d

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

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 = 9.8×(2600−1000)18×8.11 × 10^{4} × D^{2}

D = 6.83 × 10^{-5} m

D ≈ 0.7 × 10^{-6} m

D ≈ 0.7 µm

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

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Answer: a

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(gD(ρP–ρ)ρ−−−−−−−√

u = 2.336 m/s

u ≈ 2.3 m/s