A fuel droplet with a time constant K of 3 times 107 m2s eva

A fuel droplet with a time constant K of 3 times 10^-7 m^2/s evaporates in 1ms. What was the initial droplet diameter, D_i? Sketch what D^2 vs. time will look like. For the same initial droplet, if the constant were increased to K = 6 times 10^-7 m^2/s, roughly sketch what the new D^2 vs time curve will look like on the same axes as your previous sketch.

Solution

Using Knudsen droplet equations, The simplified approach suggests that the radius follows

Rs² -R₀² = -Kt, where R₀ is the inital droplet size, and Rs that at time t, K the rate constant 3 * 10 ^-7   m²/s

Given that Rs =0 at time 1 ms (millisecond), initial R0 = sqrt( 3* 10^-10 )m = 1.732 * 10 ^-5 m

The shape of the curve is decreasing parabolic with time going to zero at large t. K chnages the latus rectum ie the width of the curve.