We release an oil drop of radius r in air. The density of the oil is 640 kg/m . c_1 and c_2 for 1 atmosphere air at 20 C are 2.20* 10 ^-4 (kg/m)/sec and 0.81 kg/m , respectively.
How small should the oil drop be so that the drag force is dominated by the linear term in the speed (in lectures we called this Regime I). In this regime, the terminal velocity is m*g/(c_1*r) . [ m is the mass of the drop].
r<<?
r^3=(3*C_1^2)/4*pi*d*C_2*g
do the cubic root
d=density
its not working ..! :(
To find the size of the oil drop (r) for which the drag force is dominated by the linear term in speed (Regime I), we can start by considering the equation for drag force, which is given by:
F_drag = c_1 * v + c_2 * v^2
In Regime I, the drag force is dominated by the linear term, so we can neglect the quadratic term (c_2 * v^2). This means that the equation simplifies to:
F_drag = c_1 * v
We know that the terminal velocity (v_terminal) is reached when the drag force is equal to the gravitational force (F_drag = m * g, where m is the mass of the drop and g is the acceleration due to gravity).
Substituting the equation for drag force, we get:
c_1 * v_terminal = m * g
Solving for the terminal velocity, we have:
v_terminal = m * g / c_1
Now, let's substitute the expression for terminal velocity into the equation for drag force:
F_drag = c_1 * (m * g / c_1)
Simplifying further:
F_drag = m * g
Since F_drag is the force due to drag, we can also express it as:
F_drag = (4/3) * π * r^3 * ρ * g
Here, r represents the radius of the oil drop, ρ is the density of the oil, and g is the acceleration due to gravity.
Equating the two expressions for drag force:
m * g = (4/3) * π * r^3 * ρ * g
Simplifying:
m = (4/3) * π * r^3 * ρ
Now, let's substitute the expression for mass into the equation for the terminal velocity:
v_terminal = [(4/3) * π * r^3 * ρ * g] / c_1
The largest value that r can be in order for the drag force to be dominated by the linear term is when the terminal velocity is equal to or greater than the speed at which the drop is released.
So, we can set v_terminal as the maximum speed at which the drop is released:
v_max = m * g / (c_1 * r)
Now, we can set v_terminal >= v_max:
[(4/3) * π * r^3 * ρ * g] / c_1 >= m * g / (c_1 * r)
Simplifying:
[(4/3) * π * r^3 * ρ * g] >= m * g
Canceling out the g terms:
[(4/3) * π * r^3 * ρ] >= m
Since we want the oil drop to be as small as possible, we can assume that the mass of the drop is negligible compared to the other terms in the inequality. Therefore, we can ignore the "m" term and simplify further:
[(4/3) * π * r^3 * ρ] >= 0
This inequality indicates that the smallest the oil drop can be, while still being in Regime I, is when r approaches zero. Thus, r<< indicates that the drop radius should be much smaller than any other relevant length scale in the problem.