Chapter 11, Problem 109P

To determine

The error involved in the given case.

Answer to Problem 109P

The error involved between experimental and Stokes law is $9.69\%$ in first two cases of ball diameter of 2 mm and 4mm, and -19.65% for the third case of ball diameter 10 mm. The Stokes law is valid for all three cases.

Explanation of Solution

Given information:

  $\text{Diameter}=2\text{mm}$

  $\begin{array}{l}{\rho }_s=2600\frac{\text{kg}}{{\text{m}}^{\text{3}}}\\ {\rho }_f=1274\frac{\text{kg}}{{\text{m}}^{\text{3}}}\end{array}$

  $V_{\text{experimental}}=3.2\frac{\text{mm}}{\text{s}}$

  $\mu =1\frac{\text{kg}}{\text{m}\cdot \text{s}}$

Concept used:

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

  $\text{Error}=\frac{V_{\text{experimental}}-V_{\text{Stokes}}}{V_{\text{experimental}}}$

  $\mathrm{Re}=\frac{\rho VD}{\mu }$

  $\begin{array}{c}{F}_D\to \text{Drag force}\\ C_D\to \text{Drag coefficient}\\ {\rho }_s\to \text{Density of balls}\\ {\rho }_f\to \text{Density of fluid}\\ A\to \text{Area}\\ V\to \text{Velocity}\end{array}$

Calculation:

  $\begin{array}{c}{V}_{\text{Stokes}}=\frac{g{D}^2\left({\rho }_s-{\rho }_f\right)}{18\mu }\\ =\frac{9.81\times {\left(\frac{2}{1000}\right)}^2\times \left(2600-1274\right)}{18\times 1}\times \left(\frac{\text{1000}\text{mm}}{\text{1}\text{m}}\right)\\ =2.89\frac{\text{mm}}{\text{s}}\end{array}$

  $\begin{array}{c}\text{Error}=\frac{V_{\text{experimental}}-V_{\text{Stokes}}}{V_{\text{experimental}}}\times 100\\ =\frac{3.2-2.89}{3.2}\times 100\\ =9.69\%\end{array}$

  $\begin{array}{c}\mathrm{Re}=\frac{\rho VD}{\mu }\\ =\frac{1274\times 0.00289\times \frac{2}{1000}}{1}\\ =0.0074<<1\end{array}$

Hence, the Stokes law is valid.

Now let us calculate the error for the aluminium ball of 4 mm diameter.

  $\text{Diameter}=4\text{mm}$

  $\begin{array}{l}{\rho }_s=2600\frac{\text{kg}}{{\text{m}}^{\text{3}}}\\ {\rho }_f=1274\frac{\text{kg}}{{\text{m}}^{\text{3}}}\end{array}$

  $V_{\text{experimental}}=12.8\frac{\text{mm}}{\text{s}}$

  $\mu =1\frac{\text{kg}}{\text{m}\cdot \text{s}}$

Concept used

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

  $\text{Error}=\frac{V_{\text{experimental}}-V_{\text{Stokes}}}{V_{\text{experimental}}}$

  $\mathrm{Re}=\frac{\rho VD}{\mu }$

  $\begin{array}{c}{F}_D\to \text{Drag force}\\ C_D\to \text{Drag coefficient}\\ {\rho }_s\to \text{Density of balls}\\ {\rho }_f\to \text{Density of fluid}\\ A\to \text{Area}\\ V\to \text{Velocity}\end{array}$

Calculation:

  $\begin{array}{c}{V}_{\text{Stokes}}=\frac{g{D}^2\left({\rho }_s-{\rho }_f\right)}{18\mu }\\ =\frac{9.81\times {\left(\frac{4}{1000}\right)}^2\times \left(2600-1274\right)}{18\times 1}\times \left(\frac{\text{1000}\text{mm}}{\text{1}\text{m}}\right)\\ =11.56\frac{\text{mm}}{\text{s}}\end{array}$

  $\begin{array}{c}\text{Error}=\frac{V_{\text{experimental}}-V_{\text{Stokes}}}{V_{\text{experimental}}}\times 100\\ =\frac{12.8-11.56}{12.8}\times 100\\ =9.69\%\end{array}$

  $\begin{array}{c}\mathrm{Re}=\frac{\rho VD}{\mu }\\ =\frac{1274\times 0.01156\times \frac{4}{1000}}{1}\\ =0.06<<1\end{array}$

Hence, the Stokes law is valid.

The error involved between experimental and Stokes law is $9.69\%$.

Now let us calculate the error for the aluminium ball of 10 mm diameter.

  $\text{Diameter}=10\text{mm}$

  $\begin{array}{l}{\rho }_s=2600\frac{\text{kg}}{{\text{m}}^{\text{3}}}\\ {\rho }_f=1274\frac{\text{kg}}{{\text{m}}^{\text{3}}}\end{array}$

  $V_{\text{experimental}}=60.4\frac{\text{mm}}{\text{s}}$

  $\mu =1\frac{\text{kg}}{\text{m}\cdot \text{s}}$

Concept used:

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

  $\text{Error}=\frac{V_{\text{experimental}}-V_{\text{Stokes}}}{V_{\text{experimental}}}$

  $\mathrm{Re}=\frac{\rho VD}{\mu }$

  $\begin{array}{c}{F}_D\to \text{Drag force}\\ C_D\to \text{Drag coefficient}\\ {\rho }_s\to \text{Density of balls}\\ {\rho }_f\to \text{Density of fluid}\\ A\to \text{Area}\\ V\to \text{Velocity}\end{array}$

Calculation:

  $\begin{array}{c}{V}_{\text{Stokes}}=\frac{g{D}^2\left({\rho }_s-{\rho }_f\right)}{18\mu }\\ =\frac{9.81\times {\left(\frac{10}{1000}\right)}^2\times \left(2600-1274\right)}{18\times 1}\times \left(\frac{\text{1000}\text{mm}}{\text{1}\text{m}}\right)\\ =72.27\frac{\text{mm}}{\text{s}}\end{array}$

  $\begin{array}{c}\text{Error}=\frac{V_{\text{experimental}}-V_{\text{Stokes}}}{V_{\text{experimental}}}\times 100\\ =\frac{60.4-72.27}{60.4}\times 100\\ =-19.65\%\end{array}$

  $\begin{array}{c}\mathrm{Re}=\frac{\rho VD}{\mu }\\ =\frac{1274\times 0.07227\times \frac{10}{1000}}{1}\\ =0.92<1\end{array}$

Hence, the Stokes law is valid.

The error involved between experimental and Stokes law is $-19.65\%$.

Conclusion:

The error involved between experimental and Stokes law is $9.69\%$ in first two cases and -19.65% for the third case. The Stokes law is valid for all three cases.