Fluid Mechanics, Renoylds Number

[simonre7]

A particle falls in water and reaches its terminal velocity. if the density of the particle is 1100kg/m3, what is the velocity and diameter is Re = 1.0


Again guys bit lost with this one, there are so many different equations relating to this i don't know where to start. hopefully some one with a bit more experience can help me.

Thank you in advance.

 

[Astronuc]

A particle falls under gravity. What is the condition at terminal velocity, i.e. the particle no longer accelerates?

the Reynolds number is the ratio of inertial forces (v_sρ) to viscous forces (μ/L) and consequently it quantifies the relative importance of these two types of forces for given flow conditions.

http://en.wikipedia.org/wiki/Reynolds_number

v_s is mean velocity of fluid or velocity of object in fluid.

 

[piercas]

I can provide a response to this content by utilizing the principles of fluid mechanics and the Reynolds number. The Reynolds number is a dimensionless quantity that relates the inertial forces to the viscous forces in a fluid flow and is often used to determine the type of flow (laminar or turbulent). In this scenario, we are dealing with a particle falling in water, which can be considered a fluid.

Firstly, we must determine the terminal velocity of the particle. This is the maximum velocity that the particle will reach as it falls through the water and is dependent on various factors such as the density and size of the particle, as well as the viscosity and density of the fluid.

Using the equation for terminal velocity in a fluid (Vt = (2mg)/(ρACd)), where m is the mass of the particle, ρ is the density of the fluid, A is the cross-sectional area of the particle, and Cd is the drag coefficient, we can calculate the velocity of the particle at terminal velocity.

Given the density of the particle (1100 kg/m3), we can assume a spherical shape for simplicity and calculate the diameter (d) of the particle using the equation for volume of a sphere (V = (4/3)πr3) and solving for r. This gives a diameter of approximately 0.082 meters or 82 mm.

Substituting the values into the equation for terminal velocity, we get:

Vt = (2 * (1100 kg/m3) * 9.8 m/s2) / ((1000 kg/m3) * (π * (0.082 m)2) * 0.5)

Vt = 2.62 m/s

This means that the terminal velocity of the particle is approximately 2.62 m/s. Now, to determine the Reynolds number, we can use the equation Re = (ρVD)/μ, where ρ is the density of the fluid, V is the velocity, D is the diameter of the particle, and μ is the dynamic viscosity of the fluid.

Substituting the values, we get:

Re = ((1000 kg/m3) * (2.62 m/s) * (0.082 m)) / (0.001 kg/(m*s))

Re = 214,360

Since the Reynolds number is greater than 1, we can conclude that the flow around the particle is turbulent.

In