### Stokes Law Calculator

A Stokes’ law calculator is a tool for analyzing the motion of a spherical particle in a falling ball viscometer.

This device is just a vertical tube filled with a viscous liquid. When dropped in the tube, a small particle is subjected to a drag force as a result of the fluid’s resistance. By measuring the velocity at the end of the pipe, you can calculate the viscosity of the liquid.

This article will explain how to use the terminal velocity equation to determine viscosity, Also let’s elaborate a bit on the definition of viscosity.

The viscosity of a fluid (gas or liquid) describes its resistance to shear stresses. For example, honey, which is “thicker” than water, has a higher viscosity and is more resistant to shear stresses. The units of viscosity are Pascal times per second (Pa·s).

Read on to know what Stokes’ law is, the definition of viscosity, and how to find terminal velocity using Stokes’ law.

### Stokes Law Terminal Velocity Formula

Stokes law terminal velocity formula V_{(m/s)} in meter per second is equal to the gravitational acceleration g_{(m/s2)} in meter per second and multiply square of the diameter of the particle d^{2}_{(m)} in meter and multiply the is the density of the particle pp_{(kg/m3) }in kilogram per meter^{3} and minus the density of the fluid pm_{(kg/m3)} in kilogram per meter^{3} and multiply the eighteen and multiply the dynamic viscosity of the fluid 18μ_{(pa)} in pascal.Hence the stokes law formula can be written as

V_{(m/s)}=g_{(m/s2)}*d^{2}_{(m)}(pp_{(kg/m3)}-pm_{(kg/m3)})(18μ_{(pa)})

Where

V=is the terminal velocity of a spherical particle in m/s

G=is the gravitational acceleration and is equal to 9.80665 m/s^{2}

D= is the diameter of the particle in m

P_{p}=is the density of the particle in kg /m^{3}

P_{m} = is the density of the fluid in kg /m^{3}

μ =is the dynamic viscosity of the fluid in pa.s

### Example:1

Calculate the diameter of the particle 7m,and density of the particle 13 kg/m^{3}, the density of the fluid 11 kg/m^{3}, dynamic viscosity of the fluid 9 pa.s.Find the terminal velocity?

### Answer:

G=9.8 m/s^{2}

D=7m

P_{p}=13 kg /m^{3}

P_{m}=11 kg /m^{3}

μ =9 pa.s

formula using

V_{(m/s)}=g_{(m/s2)}*d^{2}_{(m)}(pp_{(kg/m3)}-pm_{(kg/m3)})(18μ_{(pa)})

V=9.8*7^{2}(13-11)/(18/9)

Terminal velocity=5.92m/s

### Example:2

Calculate the diameter of the particle 9m,and density of the particle 30 kg/m^{3}, the density of the fluid 28 kg/m^{3}, dynamic viscosity of the fluid 20pa.s.Find the terminal velocity?

### Answer:

G=9.8 m/s^{2}

D=9m

P_{p}=30 kg /m^{3}

P_{m}=28 kg /m^{3}

μ =20 pa.s

formula using

V_{(m/s)}=g_{(m/s2)}*d^{2}_{(m)}(pp_{(kg/m3)}-pm_{(kg/m3)})(18μ_{(pa)})

V=9.8*9^{2}(30-28)/(18/20)

Terminal velocity=4.41m/s