Chapter 15, Problem 99RQ

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,

  $\begin{array}{l}{F}_D=W-F_B\\ 3\pi \mu VD={\rho }_{s}g\text{V}-{\rho }_{f}g\text{V}\\ 3\pi \mu VD=\left({\rho }_s-{\rho }_f\right)g\frac{\pi D^3}{6}\\ V=\frac{\left({\rho }_s-{\rho }_f\right)g{D}^2}{18\mu }\end{array}$

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

The terminal velocity is,

  $\begin{array}{l}V=\frac{\left(2600{\text{ kg/m}}^3-1274{\text{ kg/m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right){\left(0.002\text{ m}\right)}^2}{18\left(1\text{ kg/m}\cdot \text{s}\right)}\\ V=0.00311\text{m}/\text{s}\\ V=3.11\text{mm}/\text{s}\end{array}$

The error in terminal velocity is,

  $\begin{array}{c}\text{Error}=\frac{V_{\exp }-V}{V_{\exp }}\times 100\%\\ =\frac{3.2-3.11}{3.2}\times 100\%\\ =2.8\%\end{array}$

Thus, the error in the terminal velocity is $2.8\%$.

The Reynolds number is,

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

Hence, Stokes law is accurate.

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

The terminal velocity is,

  $\begin{array}{l}V=\frac{\left(2600{\text{ kg/m}}^3-1274{\text{ kg/m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right){\left(0.004\text{ m}\right)}^2}{18\left(1\text{ kg/m}\cdot \text{s}\right)}\\ V=0.0124\text{m}/\text{s}\\ V=12.4\text{mm}/\text{s}\end{array}$

The error in terminal velocity is,

  $\begin{array}{c}\text{Error}=\frac{V_{\exp }-V}{V_{\exp }}\times 100\%\\ =\frac{12.8-12.4}{12.8}\times 100\%\\ =3.1\%\end{array}$

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

The Reynolds number is,

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

Hence, Stokes law is accurate.

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

The terminal velocity is,

  $\begin{array}{l}V=\frac{\left(2600{\text{ kg/m}}^3-1274{\text{ kg/m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right){\left(0.010\text{ m}\right)}^2}{18\left(1\text{ kg/m}\cdot \text{s}\right)}\\ V=0.0777\text{m}/\text{s}\\ V=77.7\text{mm}/\text{s}\end{array}$

The error in terminal velocity is,

  $\begin{array}{c}\text{Error}=\frac{V_{\exp }-V}{V_{\exp }}\times 100\%\\ =\frac{60.4-77.7}{60.4}\times 100\%\\ =-28.6\%\end{array}$

Thus, the error in the terminal velocity is $-28.6\%$.

The Reynolds number is,

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

Hence, Stokes law is inaccurate.