Engineering Questions with Answers - Multiple Choice Questions
Mechanical Operations MCQ’s – Terminal Falling Velocities of Spherical Particles
1 - Question
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
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 u2
We can substitute Fe, Fb and Fd in equation 1 to get:
mdudt = ma – Cd×ρ×Ap2 u2 – m×ρ×aρP
Dividing by m on both sides and rearranging terms we get:
dudt=a(ρP–ρ)ρP–Cd×ρ×Ap2×m u2
When dudt = 0 we get maximum settling velocity or terminal velocity
a(ρP–ρ)ρP=Cd×ρ×Ap2×m u2
u = 2ma(ρP–ρ)Cd×ρ×Ap×ρP−−−−−−−−−√
When we substitute Ap = πr2 (projected area of sphere) in the equation of u, we get:
u = 8×r×a(ρP–ρ)3×Cd×ρ−−−−−−−−−√
2 - Question
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 = 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−−−−−−−√
View Answer
Answer: a
Explanation: We know that u = 8×r×a(ρP–ρ)3×Cd×ρ−−−−−−−−−√ for a spherical body.
When a = 10 m/s2 we get u = 80×r×(ρP–ρ)3×Cd×ρ−−−−−−−−−√, which can also be written as
u = 4x5×r×(ρP–ρ)3×Cd×ρ−−−−−−−−√
3 - Question
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
Answer: a
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 = 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 × D2
u = 1.34 × 10-7 m/s
4 - Question
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
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/m3
ρP = 2600 kg/m3
0.005 = 9.8×(2600−1000)18×8.11 × 104 × D2
D = 6.83 × 10-5 m
D ≈ 0.7 × 10-6 m
D ≈ 0.7 µm
5 - Question
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
Answer: a
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(gD(ρP–ρ)ρ−−−−−−−√
u = 2.336 m/s
u ≈ 2.3 m/s