# Non-linear ODE: Use change of variable to solve the equation for initial conditions

1. Jul 12, 2007

### VinnyCee

1. The problem statement, all variables and given/known data

When an insoluble bubble rises in a deep pool of liquid, its volume increases according to the ideal gas law. However, when a soluble bubble rises from deep submersion, there is a competing action of dissolution that tends to reduce size. Under practical conditions, it has been proven that the mass transfer coefficient, $k_c$, for spherical bubbles in free-fall (or free-rise) is constant. Thus, for sparingly soluble bubbles released from rest, the applicable material balance is

$$\frac{d}{dt}\,\frac{4\,C\,\pi}{3}\,R^3\,=\,-k_c\,C^*\,4\pi\,R^2$$

where the molar density of the gas

$$C\,=\,\frac{P}{R_g\,T}$$

with $R_g$ and T as the ideal gas constant and temperature, respectively. $C^*$ is the molar solubility of the gas in the liquid, and R(t) is the bubble radius, which changes over time. The pressure P at a distance z from the top of the liquid surface is

$$P\,=\,P_A\,+\,\rho_L\,g\,z$$

where $\rho_L$is the liquid density and g is the gravitational acceleration. The rise velocity, $\frac{dz}{dt}$, follows a linear relation between speed and size

$$\frac{dz}{dt}\,=\,\beta\,R(t)$$

where $\beta$ is a constant that depends on the liquid viscosity.

SHOW that a change of variables allows the material balance equation to be written as

$$R\,\frac{dR}{dP}\,+\,\left(\frac{1}{3}\right)\,\frac{R^2}{P}\,=\,-\frac{\lambda}{P}$$

and

$$\lambda\,=\,\frac{k_c\,R_g\,T\,C^*}{\rho_L\,g\,\beta}$$

2. Relevant equations

Algebra and the Calculus.

3. The attempt at a solution

$$\frac{dz}{dt}\,=\,\beta\,R(t)$$

$$\frac{d}{dt}\,=\,\beta\,R(t)\,\frac{d}{dz}$$

Substituting into the original material balance equation:

$$\beta\,R(t)\,\frac{d}{dz}\,\left(\frac{P}{R_g\,T}\,\frac{4\pi\,R^3}{3}\right)\,=\,-k_c\,C^*\,4\pi\,R^2$$

Here I am stuck, how do I "show" that the two versions of the material balance equations are equivalent?

Last edited: Jul 12, 2007
2. Apr 30, 2011

### Pequod

Re: Non-linear ODE: Use change of variable to solve the equation for initial conditio

Hey I am doing this question at the moment, I found this thread while doing a google search. Do you remember how you solved it?

I've reached about just as far, but I've found lambda by using dP/dz = rhoL*g

(Don't really know how to use html code tbh.)

Any help is appreciated, thanks.