Asked by Ajiri
Find the limiting velocity of a rain drop assuming diameter of 5.0×10^-4m.The density of air relative to water is 1.3×10^-3
Answers
Answered by
Damon
The trick here is guessing the drag coefficient of the raindrop shape.
See:
http://home2.fvcc.edu/~dhicketh/DiffEqns/Spring11projects/Ben_Lynch_Gavin_Lommatsch/DiffEqProject/DiffEqProjectGavinBen.pdf
It is a function of Reynolds number even if the shape is spherical. However using .3 as a guess:
v = sqrt [ 2 m g / (rho A C) ]
here rho = (1.3*10^-3)(1000 kg/m^3)
= 1.3 kg/m^3
m = 1000 *(4/3) pi r^3
m g = 9,800 *(4/3)pi r^3
2 m g = 26.1 *10^3 pi r^3
Rho A C = 1.3 *pi r^2 * .3 = .39 pi r^2
so
v^2 = 26.1 *10^3 pi r^3/(.39 pi r^2)
v^2 = 66.9 *10^3 *2.5*10^-4
v^2 = 16.7
v = 4.1 m/s
See:
http://home2.fvcc.edu/~dhicketh/DiffEqns/Spring11projects/Ben_Lynch_Gavin_Lommatsch/DiffEqProject/DiffEqProjectGavinBen.pdf
It is a function of Reynolds number even if the shape is spherical. However using .3 as a guess:
v = sqrt [ 2 m g / (rho A C) ]
here rho = (1.3*10^-3)(1000 kg/m^3)
= 1.3 kg/m^3
m = 1000 *(4/3) pi r^3
m g = 9,800 *(4/3)pi r^3
2 m g = 26.1 *10^3 pi r^3
Rho A C = 1.3 *pi r^2 * .3 = .39 pi r^2
so
v^2 = 26.1 *10^3 pi r^3/(.39 pi r^2)
v^2 = 66.9 *10^3 *2.5*10^-4
v^2 = 16.7
v = 4.1 m/s
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.