Chapter 9, Problem 76P

To determine

The terminal speed of an air bubble rising through water.

Answer to Problem 76P

The terminal speed of an air bubble rising through water is $\underset{\_}{2.9\text{cm/s}}$.

Explanation of Solution

Given that the specific gravity of aluminium sphere is $2.7$, and its terminal speed is $5.0\text{cm/s}$.

The free-body diagram of the aluminium sphere and the air bubble is given in Figure 1.

Write the equilibrium condition on forces in $y$ direction.

${\displaystyle \sum F_y}=0$ (I)

Apply the condition in equation (I) to the aluminium sphere.

$\begin{array}{c}{F}_{\text{1D}}+F_{\text{1B}}-m_{1}g=0\\ F_{\text{1D}}=m_{1}g-F_{\text{1B}}\end{array}$ (II)

Here, $F_{\text{1D}}$ is the drag force on the aluminium sphere, $F_{\text{1B}}$ is the buoyant force on the aluminium sphere, $m_1$ is the mass of aluminium sphere, and $g$ is the acceleration due to gravity.

Apply the condition in equation (I) to the air bubble.

$\begin{array}{c}{F}_{\text{2B}}-F_{\text{2D}}-m_{2}g=0\\ F_{\text{2D}}=F_{\text{2B}}-m_{2}g\end{array}$ (III)

Here, $F_{\text{2D}}$ is the drag force on the air bubble, $F_{\text{2B}}$ is the buoyant force on the air bubble, and $m_2$ is the mass of air bubble.

Divide equation (III) by (II).

$\frac{F_{\text{2D}}}{F_{\text{1D}}}=\frac{F_{\text{2B}}-m_{2}g}{m_{1}g-F_{\text{1B}}}$ (IV)

Write the expression for the drag force on the aluminium sphere.

$F_{\text{1D}}=6\pi \eta r{v}_1$ (V)

Here, $\eta$ is the viscosity of water, $r$ is the radius of the sphere, and $v_1$ is the speed of the sphere.

Write the expression for the drag force on the air bubble.

$F_{\text{2D}}=6\pi \eta r{v}_2$ (VI)

Here, $v_2$ is the speed of the air bubble.

Divide equation (VI) by (V).

$\begin{array}{c}\frac{F_{\text{2D}}}{F_{\text{1D}}}=\frac{6\pi \eta r{v}_2}{6\pi \eta r{v}_1}\\ =\frac{v_2}{v_1}\end{array}$ (VII)

Equate the right hand sides of equations (IV) and (VII).

$\frac{F_{\text{2B}}-m_{2}g}{m_{1}g-F_{\text{1B}}}=\frac{v_2}{v_1}$ (VIII)

The buoyant force on aluminium sphere and the air bubble is equal to $m_{\text{w}}g$, where $m_{\text{w}}$ is the mass of displaced water. Modify equation (VIII) using the expressions for buoyant force.

$\begin{array}{c}\frac{m_{\text{w}}g-m_{2}g}{m_{1}g-m_{\text{w}}g}=\frac{v_2}{v_1}\\ v_2=v_1\left(\frac{m_{\text{w}}-m_2}{m_1-m_{\text{w}}}\right)\\ =v_1\left(\frac{1-\frac{m_2}{m_{\text{w}}}}{\frac{m_1}{m_{\text{w}}}-1}\right)\end{array}$ (IX)

Replace ratios of mass in equation (IX) with densities.

$v_2=v_1\left(\frac{1-\frac{{\rho }_{\text{a}}}{{\rho }_{\text{w}}}}{\frac{{\rho }_{\text{Al}}}{{\rho }_{\text{w}}}-1}\right)$ (X)

Here, ${\rho }_{\text{a}}$ is the density of air, ${\rho }_{\text{w}}$ is the density of water, and ${\rho }_{\text{Al}}$ is the density of aluminium.

Since the specific gravity of aluminium is $2.7$, the density of aluminium can be replaced as $2.7{\rho }_{\text{w}}$. Thus, equation (X) becomes,

$\begin{array}{c}{v}_2=v_1\left(\frac{1-\frac{{\rho }_{\text{a}}}{{\rho }_{\text{w}}}}{\frac{2.7{\rho }_{\text{w}}}{{\rho }_{\text{w}}}-1}\right)\\ =v_1\left(\frac{1-\frac{{\rho }_{\text{a}}}{{\rho }_{\text{w}}}}{1.7}\right)\end{array}$ (XI)

Conclusion:

Substitute $1.20{\text{kg/m}}^3$ for ${\rho }_{\text{a}}$, $1001.8{\text{kg/m}}^3$ for ${\rho }_{\text{w}}$ and $5.0\text{cm/s}$ for $v_1$ in equation (XI) to find the terminal speed the air bubble.

$\begin{array}{c}{v}_2=\left(5.0\text{cm/s}\right)\left(\frac{1-\frac{1.20{\text{kg/m}}^3}{1001.8{\text{kg/m}}^3}}{1.7}\right)\\ =\left(5.0\text{cm/s}\times \frac{1\text{m}}{100\text{cm}}\right)\left(\frac{1-\frac{1.20{\text{kg/m}}^3}{1001.8{\text{kg/m}}^3}}{1.7}\right)\\ =0.029\text{m/s}\\ =0.029\text{m/s}\times \frac{100\text{cm}}{1\text{m}}\\ =2.9\text{cm/s}\end{array}$

Therefore, the terminal speed of an air bubble rising through water is $\underset{\_}{2.9\text{cm/s}}$.