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 = \(\frac {Cd \times ρ \times Ap}{2}\) u2
We can substitute Fe, Fb and Fd in equation 1 to get:
m\(\frac {du}{dt}\) = ma – \(\frac {Cd \times ρ \times Ap}{2}\) u2 – \(\frac {m \times ρ \times a}{ρP}\)
Dividing by m on both sides and rearranging terms we get:
\(\frac {du}{dt} = \frac {a(ρP – ρ)}{ρP} – \frac {Cd \times ρ \times Ap}{2 \times m}\) u2
When \(\frac {du}{dt}\) = 0 we get maximum settling velocity or terminal velocity
\(\frac {a(ρP – ρ)}{ρP} = \frac {Cd \times ρ \times Ap}{2 \times m}\) u2
u = \(\sqrt {\frac {2ma(ρP – ρ)}{Cd \times ρ \times Ap \times ρP}}\)
When we substitute Ap = πr2 (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/s2?
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/s2 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/m3 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/s2, 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/m3
Since the fluid is water, ρ = 1000 kg/m3
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 = \(\frac {g \times (ρP – ρ)}{18 µ}\) × D2
u = 1.34 × 10-7 m/s
4. What is the diameter of a particle that has a density of 2600 kg/m3 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/m3
ρP = 2600 kg/m3
0.005 = \(\frac {9.8 \times (2600-1000)}{18 \times 8.11}\) × 104 × D2
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/m3 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/m3.
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/m3
ρP = 1300 kg/m3
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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