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VinnyCee
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Homework Statement
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_Lis 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}
Homework Equations
Algebra and the Calculus.
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?
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