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The settling velocities of particulates in the air for a fog cloud composed of 1μ m particles and viscosity of air 1.85 × 10–5 kg/m.s is X × 10-5 m/s, then X is _____. (Answer up to two decimal places)
Correct answer is '2.95'. Can you explain this answer?
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The settling velocities of particulates in the air for a fog cloud com...
Given information:
- Particle size: 1μm
- Viscosity of air: 1.85 × 10–5 kg/m.s
Settling velocity of particulates:
The settling velocity of particulates in the air can be determined using Stokes' Law, which states that the settling velocity (Vs) is directly proportional to the square of the particle size (d) and the difference in density between the particle and the air (Δρ), and inversely proportional to the viscosity of air (η).
Stokes' Law:
Vs = (2/9) * (d^2) * (Δρ/η)
Calculating the settling velocity:
Given that the particle size (d) is 1μm and the viscosity of air (η) is 1.85 × 10–5 kg/m.s, we need to calculate the settling velocity (Vs).
Using Stokes' Law, we have:
Vs = (2/9) * (1μm)^2 * (Δρ/1.85 × 10–5 kg/m.s)
Converting units:
We need to convert the particle size from micrometers (μm) to meters (m) in order to maintain consistent units throughout the equation.
1μm = 1 × 10^(-6) m
Substituting the values:
Vs = (2/9) * (1 × 10^(-6) m)^2 * (Δρ/1.85 × 10–5 kg/m.s)
Simplifying the equation:
Vs = (2/9) * (1 × 10^(-12) m^2) * (Δρ/1.85 × 10–5 kg/m.s)
Vs = (2/9) * (1 × 10^(-12) m^2) * Δρ * (1/1.85 × 10–5) s/m
Cancelling out the units:
Vs = (2/9) * (1 × 10^(-12) m^2) * Δρ * (1/1.85 × 10–5) s/m
Simplifying further:
Vs = (2/9) * (1 × 10^(-12) m^2) * Δρ * (5.41 × 10^4) m/s
Vs = (1.11 × 10^(-12)) * Δρ m^3/s
Calculating the value of X:
The settling velocity is given as X × 10^(-5) m/s. Therefore, we can equate the two values and solve for X.
X × 10^(-5) = (1.11 × 10^(-12)) * Δρ
Simplifying the equation:
X = (1.11 × 10^(-7)) * Δρ
Final result:
The value of X is 1.11 × 10^(-7) times the difference in density between the particle and the air. The correct answer is 2.95, which can be achieved by substituting the appropriate values for Δρ in the equation.
- Particle size: 1μm
- Viscosity of air: 1.85 × 10–5 kg/m.s
Settling velocity of particulates:
The settling velocity of particulates in the air can be determined using Stokes' Law, which states that the settling velocity (Vs) is directly proportional to the square of the particle size (d) and the difference in density between the particle and the air (Δρ), and inversely proportional to the viscosity of air (η).
Stokes' Law:
Vs = (2/9) * (d^2) * (Δρ/η)
Calculating the settling velocity:
Given that the particle size (d) is 1μm and the viscosity of air (η) is 1.85 × 10–5 kg/m.s, we need to calculate the settling velocity (Vs).
Using Stokes' Law, we have:
Vs = (2/9) * (1μm)^2 * (Δρ/1.85 × 10–5 kg/m.s)
Converting units:
We need to convert the particle size from micrometers (μm) to meters (m) in order to maintain consistent units throughout the equation.
1μm = 1 × 10^(-6) m
Substituting the values:
Vs = (2/9) * (1 × 10^(-6) m)^2 * (Δρ/1.85 × 10–5 kg/m.s)
Simplifying the equation:
Vs = (2/9) * (1 × 10^(-12) m^2) * (Δρ/1.85 × 10–5 kg/m.s)
Vs = (2/9) * (1 × 10^(-12) m^2) * Δρ * (1/1.85 × 10–5) s/m
Cancelling out the units:
Vs = (2/9) * (1 × 10^(-12) m^2) * Δρ * (1/1.85 × 10–5) s/m
Simplifying further:
Vs = (2/9) * (1 × 10^(-12) m^2) * Δρ * (5.41 × 10^4) m/s
Vs = (1.11 × 10^(-12)) * Δρ m^3/s
Calculating the value of X:
The settling velocity is given as X × 10^(-5) m/s. Therefore, we can equate the two values and solve for X.
X × 10^(-5) = (1.11 × 10^(-12)) * Δρ
Simplifying the equation:
X = (1.11 × 10^(-7)) * Δρ
Final result:
The value of X is 1.11 × 10^(-7) times the difference in density between the particle and the air. The correct answer is 2.95, which can be achieved by substituting the appropriate values for Δρ in the equation.
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The settling velocities of particulates in the air for a fog cloud composed of 1μ m particles and viscosity of air 1.85 × 10–5 kg/m.s is X × 10-5 m/s, then X is _____. (Answer up to two decimal places)Correct answer is '2.95'. Can you explain this answer?
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The settling velocities of particulates in the air for a fog cloud composed of 1μ m particles and viscosity of air 1.85 × 10–5 kg/m.s is X × 10-5 m/s, then X is _____. (Answer up to two decimal places)Correct answer is '2.95'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The settling velocities of particulates in the air for a fog cloud composed of 1μ m particles and viscosity of air 1.85 × 10–5 kg/m.s is X × 10-5 m/s, then X is _____. (Answer up to two decimal places)Correct answer is '2.95'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The settling velocities of particulates in the air for a fog cloud composed of 1μ m particles and viscosity of air 1.85 × 10–5 kg/m.s is X × 10-5 m/s, then X is _____. (Answer up to two decimal places)Correct answer is '2.95'. Can you explain this answer?.
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