Chapter 15, Problem 100RQ

To determine

The error involved in each case and the accuracy of Stokes law.

Explanation of Solution

Given:

The density of aluminum balls is, ${\rho }_s=2600{\text{ kg/m}}^3$.

Temperature of glycerin is, $T=22°\text{C}$.

The density of glycerin is, ${\rho }_f=1274{\text{ kg/m}}^3$.

The viscosity of glycerin is, $\mu =1\text{ kg/m}\cdot \text{s}$.

The diameter of the aluminum balls are, $D=2,4,10\text{ mm}$.

The terminal velocities are, $V_{\exp }=3.2,12.8,60.4\text{mm}/\text{s}$.

Calculation:

From the force balance,

  $\begin{array}{l}{F}_D=W-F_B\\ 3\pi \mu VD+\frac{9\pi }{16}{\rho }_s{V}^2{D}^2={\rho }_{s}g\text{V}-{\rho }_{f}g\text{V}\\ 3\pi \mu VD+\frac{9\pi }{16}{\rho }_s{V}^2{D}^2=\left({\rho }_s-{\rho }_f\right)g\frac{\pi D^3}{6}\\ V=\frac{-3\pi \mu D+\sqrt{{\left(3\pi \mu D\right)}^2-4\left(\frac{9\pi }{16}{\rho }_s{D}^2\right)\left[-\left({\rho }_s-{\rho }_f\right)g\frac{\pi }{6}{D}^3\right]}}{2\left(\frac{9\pi }{16}{\rho }_s{D}^2\right)}\end{array}$

Case 1: $D=2\text{ mm}$ and $V_{\exp }=3.2\text{mm}/\text{s}$.

The terminal velocity is,

  $\begin{array}{c}V=\frac{-3\pi \mu D+\sqrt{{\left(3\pi \mu D\right)}^2-4\left(\frac{9\pi }{16}{\rho }_s{D}^2\right)\left[-\left({\rho }_s-{\rho }_f\right)g\frac{\pi }{6}{D}^3\right]}}{2\left(\frac{9\pi }{16}{\rho }_s{D}^2\right)}\\ =\frac{\begin{array}{l}-3\pi \left(1\text{ kg/m}\cdot \text{s}\right)\left(0.002\text{ m}\right)\\ +\sqrt{{\left[3\pi \left(1\text{ kg/m}\cdot \text{s}\right)\left(0.002\text{ m}\right)\right]}^2-\left\{\begin{array}{l}4\left[\frac{9\pi }{16}\left(2600{\text{ kg/m}}^3\right){\left(0.002\text{ m}\right)}^2\right]\\ \left[\begin{array}{l}-\left(2600{\text{ kg/m}}^3-1274{\text{ kg/m}}^3\right)\\ \times \left(9.81{\text{ m/s}}^2\right)\frac{\pi }{6}{\left(0.002\text{ m}\right)}^3\end{array}\right]\end{array}\right\}}\end{array}}{2\left[\frac{9\pi }{16}\left(2600{\text{ kg/m}}^3\right){\left(0.002\text{ m}\right)}^2\right]}\\ =0.00310\text{ m/s}\\ =\text{3}\text{.10}\text{m}/\text{s}\end{array}$

The error in terminal velocity is,

  $\begin{array}{c}\text{Error}=\frac{V_{\exp }-V}{V_{\exp }}\times 100\%\\ =\frac{3.2-3.10}{3.2}\times 100\%\\ =3.1\%\end{array}$

Thus, the error in the terminal velocity is $3.1\%$.

The Reynolds number is,

  $\begin{array}{c}\mathrm{Re}=\frac{{\rho }_{f}VD}{\mu }\\ =\frac{\left(1274{\text{ kg/m}}^3\right)\left(0.0032\text{m}/\text{s}\right)\left(0.002\text{m}\right)}{1\text{ kg/m}\cdot \text{s}}\\ =0.008\ll 1\end{array}$

Case 2: $D=4\text{ mm}$ and $V_{\exp }=12.8\text{mm}/\text{s}$.

The terminal velocity is,

  $\begin{array}{c}V=\frac{-3\pi \mu D+\sqrt{{\left(3\pi \mu D\right)}^2-4\left(\frac{9\pi }{16}{\rho }_s{D}^2\right)\left[-\left({\rho }_s-{\rho }_f\right)g\frac{\pi }{6}{D}^3\right]}}{2\left(\frac{9\pi }{16}{\rho }_s{D}^2\right)}\\ =\frac{\begin{array}{l}-3\pi \left(1\text{ kg/m}\cdot \text{s}\right)\left(0.004\text{ m}\right)\\ +\sqrt{{\left[3\pi \left(1\text{ kg/m}\cdot \text{s}\right)\left(0.004\text{ m}\right)\right]}^2-\left\{\begin{array}{l}4\left[\frac{9\pi }{16}\left(2600{\text{ kg/m}}^3\right){\left(0.004\text{ m}\right)}^2\right]\\ \left[\begin{array}{l}-\left(2600{\text{ kg/m}}^3-1274{\text{ kg/m}}^3\right)\\ \times \left(9.81{\text{ m/s}}^2\right)\frac{\pi }{6}{\left(0.004\text{ m}\right)}^3\end{array}\right]\end{array}\right\}}\end{array}}{2\left[\frac{9\pi }{16}\left(2600{\text{ kg/m}}^3\right){\left(0.004\text{ m}\right)}^2\right]}\\ =0.0121\text{ m/s}\\ =12.1\text{m}/\text{s}\end{array}$

The error in terminal velocity is,

  $\begin{array}{c}\text{Error}=\frac{V_{\exp }-V}{V_{\exp }}\times 100\%\\ =\frac{12.8-12.1}{12.8}\times 100\%\\ =5.5\%\end{array}$

Thus, the error in the terminal velocity is $5.5\%$.

The Reynolds number is,

  $\begin{array}{c}\mathrm{Re}=\frac{{\rho }_{f}VD}{\mu }\\ =\frac{\left(1274{\text{ kg/m}}^3\right)\left(0.0128\text{m}/\text{s}\right)\left(0.004\text{m}\right)}{1\text{ kg/m}\cdot \text{s}}\\ =0.065\ll 1\end{array}$

Case 3: $D=10\text{ mm}$ and $V_{\exp }=60.4\text{mm}/\text{s}$.

The terminal velocity is,

  $\begin{array}{c}V=\frac{-3\pi \mu D+\sqrt{{\left(3\pi \mu D\right)}^2-4\left(\frac{9\pi }{16}{\rho }_s{D}^2\right)\left[-\left({\rho }_s-{\rho }_f\right)g\frac{\pi }{6}{D}^3\right]}}{2\left(\frac{9\pi }{16}{\rho }_s{D}^2\right)}\\ =\frac{\begin{array}{l}-3\pi \left(1\text{ kg/m}\cdot \text{s}\right)\left(0.010\text{ m}\right)\\ +\sqrt{{\left[3\pi \left(1\text{ kg/m}\cdot \text{s}\right)\left(0.010\text{ m}\right)\right]}^2-\left\{\begin{array}{l}4\left[\frac{9\pi }{16}\left(2600{\text{ kg/m}}^3\right){\left(0.010\text{ m}\right)}^2\right]\\ \left[\begin{array}{l}-\left(2600{\text{ kg/m}}^3-1274{\text{ kg/m}}^3\right)\\ \times \left(9.81{\text{ m/s}}^2\right)\frac{\pi }{6}{\left(0.010\text{ m}\right)}^3\end{array}\right]\end{array}\right\}}\end{array}}{2\left[\frac{9\pi }{16}\left(2600{\text{ kg/m}}^3\right){\left(0.010\text{ m}\right)}^2\right]}\\ =0.0597\text{ m/s}\\ =59.7\text{m}/\text{s}\end{array}$

The error in terminal velocity is,

  $\begin{array}{c}\text{Error}=\frac{V_{\exp }-V}{V_{\exp }}\times 100\%\\ =\frac{60.4-59.7}{60.4}\times 100\%\\ =1.2\%\end{array}$

Thus, the error in the terminal velocity is $1.2\%$.

The Reynolds number is,

  $\begin{array}{c}\mathrm{Re}=\frac{{\rho }_{f}VD}{\mu }\\ =\frac{\left(1274{\text{ kg/m}}^3\right)\left(0.0604\text{m}/\text{s}\right)\left(0.010\text{m}\right)}{1\text{ kg/m}\cdot \text{s}}\\ =0.770<1\end{array}$