# Air Pollution Control Questions and Answers – Gravitational Settling Chambers – Set 2

This set of Air Pollution Control Multiple Choice Questions & Answers (MCQs) focuses on “Gravitational Settling Chambers – Set 2”.

1. Stoke’s law is used to estimate the settling velocity of particles separating in the gravitational settlers.
a) True
b) False

Explanation: Stoke’s law can be used to determine the settling velocity of the particles in the gravitational chambers. Studies have shown that this law can accurately predict the settling velocities of particles smaller than 100 microns.

2. The settling velocities of which particles sizes can be accurately predicted by Stoke’s law?
a) 200 microns
b) 75 microns
c) 150 microns
d) 400 microns

Explanation: Settling velocities of particle sizes lesser than 100 microns can be easily predicted by the Stoke’s law. Among the given options, only 75 micron falls in this range. Hence, 75 microns is the answer.

3. Which equation determines the minimum particle size that can be removed with 100% efficiency in the gravitational settler?
a) dp,min = √(nWLg(ρp – ρg) / ρgf)
b) dp,min = √(nWLg(ρp – ρg) / 18Qμg)
c) dp,min = √(18Qμg / (nWLg(ρp – ρg))
d) dp,min = 18Qμg / nWLg(ρp – ρg)

Explanation: The following equation
dp,min = √(18Qμg / (nWLg(ρp– ρg))
is used to evaluate the size of the smallest particle that is removed with 100% efficiency. Here, Q denotes volumetric gas flowrate, L represents the distance the stream has to travel, and W represents the chamber width.

4. Which term does Q represent in the following equation?

`dp,min = √(18Qμg / (nWLg(ρp- ρg))`

a) Particulate flow rate
b) Molecular flow rate
c) Flow rate
d) Volumetric flow rate

Explanation: The given equation represents the minimum particle diameter for which the settler has a collection efficiency of 100%. Universally, the volumetric flow rate is denoted by Q. It is the same for this equation. Volumetric flow rate is the product of the area and the flow velocity of the gas.

5. Why can the equations calculating the drag coefficient of particulates in the gravity settling chamber not predict actual settling?
a) Deviations from ideal conditions
b) Equations are based on trial and error
c) Because of approximations
d) Because of assumptions

Explanation: The equations used to determine the settling velocity of particles in settling chambers cannot be predicted accurately. It is because the particles do not adhere to the ideal conditions, and the settling is susceptible to deviations.

6. Which of the following is a factor causing the deviation of particulate settling from ideal conditions in the gravity settlers?
a) Uniform gas velocity
b) Turbulence
c) Laminar flow
d) High pressure drop

Explanation: Turbulence is one of the reasons the particles do not settle ideally. It is intended that the gas flow in laminar conditions, but due to various reasons this may not happen. Laminar flow is a significant condition used to predict the flow.

7. Which of the following is not a condition causing deviation from the ideal settling of particulates in settling chambers?
a) Varied gas velocity
b) Turbulence in the chamber
c) Low concentration causes deviation
d) Particle become airborne after settling

Explanation: All the given options, except for “low concentration cause deviation”, are responsible for the deviation. The equations to determine the settling velocity of the particles determined on the assumption of uniform gas velocity and laminar flow. On the other hand, high concentrations of particulates may cause deviations.

8. Which velocity should be maintained in the gravity settler to prevent the particulates from becoming airborne again?
a) 8 m / s
b) 6 m / s
c) 4 m / s
d) 2 m / s

Explanation: Particles, if they become re-entrained, may cause deviations from ideal settling. Laminar flow of lesser than 3 metres per second prohibits this situation from occurring. Only 2 m / s falls below this threshold.

9. When is laminar flow in the gravity settlers possible?
a) When particles sizes are large
b) High pressure drops
c) High fall distances
d) Large chamber size

Explanation: Laminar flow is usually only possible with large particle sizes. Gases entrained with large particles is a rarity in the industries. This is one of the reasons why the flow in settling chambers may be turbulent.

10. Even though the laminar flow is preferred in gravity settling chambers, why is it not often achieved?
a) Particle sizes are small
b) Settling velocities are high
c) High number of trays
d) Shape of chamber

Explanation: Laminar flow is not achieved because of several reasons. Particle sizes need to be large for laminar flow, but this is not the case for industries. Obstacles like the high number of trays may also cause the flow to become turbulent, along with the shape of the chamber.

11. What is the speciality of the well-mixed model for gravitational settling?
a) Assumes high pressure drop
b) Assumes laminar flow
c) Assumes turbulent flow
d) Assumes low pressure drop

Explanation: The well-mixed model assumes turbulent flow, which is ideal for gravitational settling. The assumption is that the gas flow is mixed in the vertical direction. The mass balance results in the expression of efficiency.

12. Which of the following is the expression of turbulent mixing efficiency according to the well-mixed model?
a) ηlaminar = 1 – exp(ηturb)
b) ηlaminar = 1 – exp(-ηturb)
c) ηturb = 1 – exp(ηlaminar)
d) ηturb = 1 – exp(-ηlaminar)

Explanation: The expression used to determine the efficiency of turbulent flow in the gravitational settler is :
ηturb = 1 – exp(-ηlaminar)
This equation is the result of the well-mixed model.

13. A settling chamber has 7 trays and a width of 1.2 m and a length of 5 metres. If the flow rate of the gas is 8 cubic metre per second, and the density of the particles is 1500 kg / m3, what is the minimum size of particles that will be collected with 100% efficiency? The viscosity of air is 1.81 * 10-5 kg / m s
a) 65 microns
b) 75 microns
c) 85 microns
d) 95 microns

Explanation: Given,
Q = 8 m3 / s
n – 7
W = 1.2 m
L = 5 m
ρp = 1500 kg / m3
ρg = 1.2 kg / m3
We know that,
dp,min = √(18Qμg / (nWLg(ρp– ρg))
dp,min = √ (18 * 8 * 1.81 * 10-5 / (7 * 1.2 * 9.81 * 5 * (1500-1.2)))
dp,min = 65 microns.

14. A settling chamber has 7 trays and a width of 1.2 m and a length of 5 metres. If the flow rate of the gas is 8 cubic metre per second, and the density of the particles is 1500 kg / m3. What is the efficiency of the settler, if particles of 56 microns have to be removed? The viscosity of air is 1.81 * 10-5 kg / ms.
a) 76%
b) 86%
c) 96%
d) 66%

Explanation: Given,
Q = 8 m3 / s
n – 7
W = 1.2 m
L = 5 m
ρp = 1500 kg / m3
ρg = 1.2 kg / m3
I) We know that,
dp,min = √(18Qμg / (nWLg(ρp – ρg))
dp,min = √ (18 * 8 * 1.81 * 10-5 / (7 * 1.2 * 9.81 * 5 * (1500 – 1.2)))
dp,min = 65 microns
II) Effciency = dp,rem / dp,min
Given, dp,rem = 56 micron
Therefore, Efficiency = 56 / 65 = 86.1%

15. A settling chamber has 7 trays and a width of 1.2 m and a length of 5 metres. If the flow rate of the gas is 8 cubic metre per second, and the density of the particles is 1500 kg / m3. What is the efficiency of the turbulent mixing?
a) 72%
b) 62%
c) 52%
d) 42%

Explanation: Given,
Q = 8 m3 / s
n – 7
W = 1.2 m
L = 5 m
ρp = 1500 kg / m3
ρg = 1.2 kg / m3
I) We know that,
dp,min = √(18Qμg / (nWLg(ρp– ρg))
dp,min = √ (18 * 8 * 1.81 * 10-5 / (7 * 1.2 * 9.81 * 5 * (1500 – 1.2)))
dp,min = 65 microns
II) Terminal Velocity, vt = 30000 * ρp * dp2 = 30000 * 1500 * (65 * 10-6)2
vt = 0.141
III)ηturb = 1 – exp(-nWLvt / Q) = 1 – exp(-7 * 1.2 * 5 * 0.141 / 8) = 0.523 = 52%.

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