How do I find the constants in this differential equation that describes the time taken for a bubble to rise to the surface.

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I'm trying to model the behaviour of bubbles in a water column as accurately as I'm able to, and I modelled a Differential equation using F = ma and a free body diagram. The equation of motion I started with was

Upthrust - Weight - Drag = ma

(the hyphens are minus signs). And I replaced drag as D = 6πμrv where v is velocity. Weight is obviously the weight of the gas and Upthrust is the weight of the volume of water displaced, also called Buoyancy. I then rewrote v and a as $\frac{dh}{dt}$ and $\frac{d^2h}{dt^2}$ where h is height and got a differential equation of

$\rho_w - \rho_g = \rho_g \frac{d^2h}{dt^2} + \frac{6\pi\mu r v}{V}\frac{dh}{dt}$

where V is the volume of the gas bubble. I put the equation in a more simpler form into Wolfram Alpha (I grouped the ρ_w and ρ_g as one constant A, coefficient of d^2h/dt^2 as Q and coeff of dh/dt as R) and I got the following answer.

$h(t)= \frac{At}{R} + \frac{C_1Qe^{-\frac{R}{Q}t}}{R}+C_2$

I found C_2 in terms of C_1 by giving it the inital condition of @ t=0 , h(t)=0, so the new equation I have is

$h(t)= \frac{At}{R} + \frac{C_1Q}{R}(e^{-\frac{R}{Q}t}-1)$

The question is how do I get rid of $C_1$ and get a completely analytical solution?? Do I find initial conditions/boundary conditions from the problem? The only others I can think of are velocity is 0 at t=0 and velocity and acceleration never < 0, but I don't even know how to implement them in either.