A water droplet nucleates in a fog, then falls downward due to the gravitational
ID: 2076403 • Letter: A
Question
A water droplet nucleates in a fog, then falls downward due to the gravitational field, g. As it falls, it sweeps up the fog in its path and grows in size. Assume that the droplet remains spherical, that the fog is stationary, and that the droplet experiences no drag forces due to air resistance. Call the density of the water in the droplet 1 and the density of the fog 2. Find a relationship between the speed of the droplet and d(r)/dt, where r is the radius of the droplet at time t. A water droplet nucleates in a fog, then falls downward due to the gravitational field, g. As it falls, it sweeps up the fog in its path and grows in size. Assume that the droplet remains spherical, that the fog is stationary, and that the droplet experiences no drag forces due to air resistance. Call the density of the water in the droplet 1 and the density of the fog 2. Find a relationship between the speed of the droplet and d(r)/dt, where r is the radius of the droplet at time t. Find a relationship between the speed of the droplet and d(r)/dt, where r is the radius of the droplet at time t.Explanation / Answer
A falling drop when falling feels a frictional force from the air's viscosity that opposes gravity (the drop has to push its way through the air), the drop reaches a constant terminal velocity, and this terminal velocity is smaller the smaller the radius of a fluid particle.
we can say the following:
The weight of a drop of air of radius R is
4{pi}R3{rho1}g/3
where {rho}1 is the density of the water (=103 kg/m3, and g (=9.8 m/sec2) free-fall acceleration.
If the drop is falling through air with velocity v, the viscosity of the air n (=1.75*10-5 N*sec/m2) will cause friction (Stokes law) 6{pi}nRv.
The terminal velocity of the falling drop is obtained by equating the weight and the friction force, and is given by
v=2{rho1}gR2/9n .
Thus 1 micron drop will have velocity of 0.13 mm/sec, or 11 m/day. Larger, 10 micron drops will still fall slowly, - 1.1km/day. Such fall rates can be neglected, especially since the motion of the air itself can be faster than that.
Drops significantly smaller than 1 micron are not visible and will not be percieved as clouds, while 0.1 mm drops will fall with velocity of about 1 m/sec, i.e. it will rain.
Larger drops will fall with even larger velocities; however, the air-friction starts increasing faster than than v, i.e. the velocity will not increase as fast with increasing drop size.
At such large speeds the weight of the drop is balanced by the drag force
(1/2){rho}2C{pi}R2v2,
where {rho}air is the density of the air (=1.2kg/m3) and C=1.2 is the drag coefficient for sphere. According to this equation 1 mm drop will fall with velocity of 4.3 m/sec and 10 mm drop will fall with velocity of 13.6 m/sec. However, drops larger than 5 mm are usually broken into smaller drops that fall more slowly.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.