Question 8.11: Bubble Nucleation and Growth A steady stream of bubbles of ......
Bubble Nucleation and Growth
A steady stream of bubbles of carbon dioxide gas is found to form on the surface of a glass and to detach and rise in a carbonated beverage of density 1030 kgm^{–3} and viscosity 0.001 Nsm^{–2} for which the following observations and measurements were made:
Determine the diameter and age of the bubbles at the point they detach from the nucleation sites on the surface of the glass, and the rate at which carbon dioxide enters the rising bubbles in terms of volume of carbon dioxide (m³ ) transferred per unit area (m² ) of bubble if the rate of carbon dioxide transfer is proportional to the surface area of the bubble.
\begin{matrix} \text{Time (s)} & 1.0 & 2.0 & 3.0 & 4.0 \\ \text{Velocity (ms)}^{-1} & 0.0225 & 0.04 & 0.0625 & 0.3 \end{matrix}
For liquids that have a high dissolved gas content such as carbon dioxide in water, the formation of bubbles is due to the extent of saturation and sufficient nucleation. Bubbles of gas are formed and grow to a size that is sufficient for them to detach and rise up through the liquid due to buoyancy effects. Their enlargement in size as seen in a stream of nucleating bubbles rising in a champagne glass may initially be thought to be due to the decrease in hydrostatic head. Their enlargement, however, is due largely to the diffusion of carbon dioxide.
If the rate of carbon dioxide gas that transfers into the bubbles changing the volume with time is proportional to the surface area of the bubble, then
{\frac{d V}{d t}}=f a (8.44)
The rate of change of bubble radius with time is therefore a constant. That is,
{\frac{d r}{dt}}=f (8.45)
From the formation of the bubble, the radius at any time t can be found by integrating Equation 8.45 to give
r=f(t+t_{o}) (8.46)
where t_{\mathrm{o}} is the time that the bubble first forms. If the density of the liquid is taken to be considerably greater than that of the gas, then the terminal velocity of the rising bubble of gas rising is approximately given by Stokes’ law (see Equation 8.1) expressed using radius rather than diameter as
U_{t}={\frac{g\left(\rho_{p}-\rho\right)d_{p}^{2}}{18\mu}} (8.1)
U_{t}={\frac{2\rho g r^{2}}{9\mu}} (8.47)
Therefore,
\sqrt{U_{t}}=f\sqrt{\frac{2\rho g}{9\mu}}(t+t_{o}) = f{\sqrt{\frac{2\times1030\times9.81}{9\times0.001}}}(t+t_{o}) (8.48)
Using the data provided gives a straight line (Figure 8.8):
The bubbles therefore form 2 seconds before detachment at which time the terminal velocity is 0.01 ms^{–1} . This corresponds to a bubble radius of
r={\sqrt{\frac{\mathrm{{9}}\mathrm{\mu}U}{2\mathrm{{\rho}}g}}}= {\sqrt{\frac{9\times0.001\times0.01}{2\times1030\times9.81}}}=6.7\times10^{-5}\ \mathrm{m} (8.49)
That is, a bubble diameter of 0.13 mm. The rate of carbon dioxide transfer into the bubble is therefore
f={\frac{r}{t+t_{o}}}={\frac{6.7\times10^{-5}}{2}} =3.35\times10^{-5} \mathrm{m^{3}}(C O_{2}) \mathrm{m^{2}s^{-1}} (8.50)
\begin{matrix} \text{Time (s)} & 1.0 & 2.0 & 3.0 & 4.0 \\ \sqrt{} \text{ Velocity }(m^{1/2}s^{-1/2}) & 0.15 & 0.2 & 0.25 & 0.3 \end{matrix}
