Chapter 13, Problem 76P

To determine

The terminal speed of ascent for a spherical $1.00\text{ mm}$ diameter carbon dioxide bubble rising in a carbonated beverage.

The time required for this bubble to rise $20\text{ cm}$ .

Whether this length of time is consistent with one’s observation.

Answer to Problem 76P

  $v_t=0.33\text{ m/s}$

  $\Delta t=0.61\text{ s}$

Explanation of Solution

Given:

The drag force on a moving sphere at a very low Reynolds number is $F_d=6\pi \eta av$

where, $\eta$ is the viscosity of the surrounding fluid and $a$ is the radius of the sphere.

Density of carbonated beverage, $\rho =1.1\text{ kg/L}$

Viscosity of carbonated beverage, $\eta =1.8\text{mPa}\cdot \text{s}$

Diameter of the carbon dioxide bubble $=1.0\text{mm}=1.0\times {10}^{-3}\text{m}$

Radius of the carbon dioxide bubble, $a=0.5\times {10}^{-3}\text{ m}$

Formula used:

  

FIGURE: 1

The forces acting on the bubble before it reaches to its terminal speed are shown in the free body diagram (figure 1).

Applying Newton’s second law, $\sum F_y=m{a}_y$ to the bubble,

  $B-m_{g}g-F_d=m{a}_y$  $\left(1\right)$

Where, $B$ is the buoyant force acting on the bubble,

  $m_g$ is mass of the gas in the bubble,

  $g$ is acceleration due to gravity,

  $F_d$ is drag force on a moving bubble,

  $m$ is mass of the bubble,

  $a_y$ is the acceleration of the bubble in Y-direction

Under terminal speed conditions, acceleration $a_y=0$ and hence equation $\left(1\right)$ becomes,

  $B-m_{g}g-F_d=0$  $\left(2\right)$

By using Archimedes principle, the buoyant force $\left(B\right)$ acting on the bubble can be written as,

  $B=w_f=m_{f}g$  $\left(3\right)$

Where, $w_f$ is weight of the displaced fluid, and

  $m_f$ is mass of the displaced fluid

Since $m_f={\rho }_f{V}_f$ , equation $\left(3\right)$ becomes,

  $B={\rho }_f{V}_{f}g={\rho }_s{V}_{bubble}g$

Where,

  ${\rho }_f$ is density of the displaced fluid,

  $V_f$ is volume of the displaced fluid,

  ${\rho }_s$ is density of soda (carbonated beverage),

  $V_{bubble}$ is volume of the bubble

Mass of the gas bubble can be written as,

  $m_g={\rho }_g{V}_g={\rho }_g{V}_{bubble}$

Substituting for $B$ , $m_g$ and $F_d$ in equation $\left(2\right)$ ,

  ${\rho }_s{V}_{bubble}g-{\rho }_g{V}_{bubble}g-6\pi \eta a{v}_t=0$

Where, $v_t$ is terminal speed

  $\Rightarrow {\rho }_s{V}_{bubble}g-{\rho }_g{V}_{bubble}g=6\pi \eta a{v}_t$

  $\Rightarrow v_t=\frac{V_{bubble}g\left({\rho }_s-{\rho }_g\right)}{6\pi \eta a}$  $\left(4\right)$

Volume of the carbon dioxide bubble which is spherical in shape can be written as,

  $V_{bubble}=\frac{4}{3}\pi a^3$

Substituting for $V_{bubble}$ in equation $\left(4\right)$ ,

  $\begin{array}{l}{v}_t=\frac{\frac{4}{3}\pi a^{3}g\left({\rho }_s-{\rho }_g\right)}{6\pi \eta a}\\ v_t=\frac{2a^{2}g\left({\rho }_s-{\rho }_g\right)}{9\eta }\end{array}$

Since ${\rho }_s𑨀{\rho }_g$ , above equation can be written as,

  $v_t\approx \frac{2a^{2}g{\rho }_s}{9\eta }$  $\left(5\right)$

The rise time $\left(\Delta t\right)$ of the bubble can be expressed as,

  $\Delta t=\frac{h}{v_t}$  $\left(6\right)$

Where, $h$ is the height of the soda glass

Calculation:

Substituting the numerical values in equation $\left(5\right)$ ,

  $\begin{array}{l}{v}_t\approx \frac{2{\left(0.5\times {10}^{-3}\text{ m}\right)}^2\left(9.81{\text{ m/s}}^2\right)\left(1.1\times {10}^3{\text{ kg/m}}^3\right)}{9\left(1.8\times {10}^{-3}\text{ Pa}\text{.s}\right)}\\ v_t=0.33\text{ m/s}\end{array}$ .

Substituting numerical values in equation $\left(6\right)$ ,

  $\Delta t=\frac{0.20\text{ m}}{0.33\text{ m/s}}=0.61\text{ s}$ .

The time is less than a second. It seems reasonable.

Conclusion:

The terminal speed of ascent for a spherical $1.00\text{ mm}$ diameter carbon dioxide bubble rising in a carbonated beverage is $0.33\text{ m/s}$ .

The time required for this bubble to rise $20\text{ cm}$ is $0.61\text{ s}$ .