Find the terminal velocity of a raindrop of diameter 3mm when it fall through atmospheric air
(Appointment and is 249.7m/s)
To find the terminal velocity of a raindrop falling through atmospheric air, we can use Stoke's law. Stoke's law states that the drag force experienced by an object moving through a fluid is directly proportional to its velocity.
The formula to calculate the drag force of an object moving through a fluid is:
Drag Force (F) = 6πηrv
Where:
- F is the drag force
- η is the viscosity of the fluid
- r is the radius of the raindrop
- v is the velocity of the raindrop
Given values:
Diameter of the raindrop = 3 mm, which means the radius (r) would be 1.5 mm = 0.0015 m
Viscosity of air (η) = 1.81 x 10^-5 Pa.s (at 20°C and 1 atm)
Since we want to find the terminal velocity, we know that at terminal velocity, the drag force experienced by the raindrop is equal to the gravitational force acting on it:
Drag Force (F) = Gravitational Force (mg)
Where:
- m is the mass of the raindrop
- g is the acceleration due to gravity
The mass of the raindrop can be calculated using its density:
Density (ρ) = mass/volume
Mass = ρ * volume
The density of water droplets is around 1000 kg/m³.
The volume of a sphere can be calculated using the formula:
Volume (V) = (4/3) * π * r³
Now we can calculate the mass of the raindrop using its density:
mass = ρ * V = 1000 kg/m³ * [(4/3) * π * (0.0015)^3] m³
Once we have the mass, we can calculate the gravitational force:
Gravitational Force (mg) = mass * gravitational acceleration
Gravitational Force (mg) = mass * 9.8 m/s²
Now, equating the drag force and gravitational force, we can solve for v:
6πηrv = mass * 9.8
Solving for v:
v = [mass * 9.8] / [6πηr]
Plugging in the given values and solving the equation, we can find the terminal velocity of the raindrop.
google is your friend:
https://en.wikipedia.org/wiki/Drop_(liquid)#Speed