At a temperature of 20ºC, what would be the terminal velocities and Reynolds Numbers of water droplets with the followin

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answerhappygod
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At a temperature of 20ºC, what would be the terminal velocities and Reynolds Numbers of water droplets with the followin

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At a temperature of 20ºC, what would be the terminal velocities
and Reynolds Numbers of water droplets with the following
diameters: a) 1 micron (10-6 m); b) 10 microns (10-5 m); c) 0.1 mm
(10-4 m); d) 1 mm (10-3 m); and e) 10 mm (10-2 m)? (Hint: You will
need to iterate to solve this problem. A spreadsheet will be very
helpful)
Equations are given bellow.
At A Temperature Of 20oc What Would Be The Terminal Velocities And Reynolds Numbers Of Water Droplets With The Followin 1
At A Temperature Of 20oc What Would Be The Terminal Velocities And Reynolds Numbers Of Water Droplets With The Followin 1 (13.51 KiB) Viewed 229 times
At A Temperature Of 20oc What Would Be The Terminal Velocities And Reynolds Numbers Of Water Droplets With The Followin 2
At A Temperature Of 20oc What Would Be The Terminal Velocities And Reynolds Numbers Of Water Droplets With The Followin 2 (53.25 KiB) Viewed 229 times
24/Re CD = 24(1+0.15 Re 0.687) / Re 0.44 Pu-u₂d₁ Re= Ha Re < 1 1 < Re<1000 Re > 1000 (3-11) (3-12)
1 mag = = = P₂C.₁A² This can be rewritten as: 1 P.( ² ) x = / P. C. (1²7) ² Pwl Pa Ca 6 4 (3-13) (3-14)
Equation 3-14 is then solved for the droplet velocity: 1/2 [4pgd Ud 3p CD (3-15) For the case of Re <1, the drag coefficient is CD = 24/Re and the terminal velocity can be expressed as: - Pwgd² นส 18μ (3-16) For the case of Re> 1000, Cp = 0.44 and the terminal velocity can be expressed as: 1/2 71/2 -- [30, 8d]"^* - [3.036 + 8d] ² 4Pgd = -gd [3PaCD- Pa (3-17) For Reynolds numbers between 1 and 1000, an iterative approach is needed to solve for the terminal velocity based on the drag coefficient, which is a function of the velocity.
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