Ask Question
27 February, 07:55

A particle is emitted from a smoke stack with diameter of 0.05 mm. In order to determine how far downstream it travels it is important to find its terminal downward velocity. If it has a density of 1200 kg/m3, its terminal downward velocity (cm) is: (assume the drag coefficient is 24/Re and the volume of a sphere is 4/3 pi R3)

+2
Answers (1)
  1. 27 February, 10:57
    0
    Answer: downward velocity = 6.9*10^-4 cm/s

    Explanation: Given that the

    Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 * 10^-5 m

    Where radius r = 2.5 * 10^-5 m

    Density = 1200 kg/m^3

    Area of a sphere = 4πr^2

    A = 4 * π * (2.5 * 10^-5) ^2

    A = 7.8 * 10^-9 m^2

    Volume V = 4/3πr^3

    V = 4/3 * π * (2.5 * 10^-5) ^3

    V = 6.5 * 10^-14 m^3

    Since density = mass / volume

    Make mass the subject of formula

    Mass = density * volume

    Mass = 1200 * 6.5 * 10^-14

    Mass M = 7.9 * 10^-11 kg

    Using the formula

    V = sqrt (2Mg / pCA)

    Where

    g = 9.81 m/s^2

    M = mass = 7.9 * 10^-11 kg

    p = density = 1200 kg/m3

    C = drag coefficient = 24

    A = area = 7.8 * 10^-9m^2

    V = terminal velocity

    Substitute all the parameters into the formula

    V = sqrt[ (2 * 7.9*10^-11 * 9.8) / (1200 * 24 * 7.8*10^-9) ]

    V = sqrt[ 1.54 * 10^-9/2.25*10-4]

    V = 6.9*10^-6 m/s

    V = 6.9 * 10^-4 cm/s
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A particle is emitted from a smoke stack with diameter of 0.05 mm. In order to determine how far downstream it travels it is important to ...” in 📙 Engineering if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers