Chapter 10, Problem 19P

To determine

The velocities with actual form of stocks theorem and compare with actual determined value.

Answer to Problem 19P

The theoretical velocities for all different cases are $3.121\text{mm/s}$, $12.12\text{mm/s}$ and $55.56\text{mm/s}$. The difference between experimental and theoretical velocities for all three cases are $0.079\text{mm/s}$, $0.68\text{mm/s}$ and $\text{4}\text{.84}\text{mm/s}$.

Explanation of Solution

Given information:

Actual stokes law is $F_D=3\pi \mu DV+\left(9\pi /16\right)\rho V^2{D}^2$. Three aluminum balls of diameters $\text{2}\text{mm}$, $4\text{mm}$ and $10\text{mm}$.

Concept used:

For constant velocity of ball under glycerin, the relation for summation of forces on the ball is expressed as follows:

  $F_D+F_B=W$...... (1)

Here, drag force is $F_D$, Buoyant force is $F_B$, Weight of the ball is $W$.

Calculation:

Substitute $3\pi \mu DV+\left(9\pi /16\right)\rho V^2{D}^2$ for $F_D$, ${\rho }_{al}g\times \frac{\pi D^3}{6}$ for $W$ and ${\rho }_{gl}g\times \frac{\pi D^3}{6}$ for $F_B$ in equation (1).

  $\begin{array}{l}3\pi \mu DV+\left(9\pi /16\right)\rho V^2{D}^2+{\rho }_{gl}g\times \frac{\pi D^3}{6}={\rho }_{al}g\times \frac{\pi D^3}{6}\\ \left(\frac{9\pi }{16}{\rho }_{gl}D\right)V^2+3\mu V+\frac{D^{2}g}{6}\left({\rho }_{gl}-{\rho }_{al}\right)=0\end{array}$

For first case:

Substitute $\text{2}\text{mm}$ for $D$, $\text{1260}\text{Kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{gl}$, $2700\text{Kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{al}$ and $1\text{Kg-m}/\text{s}$ for $\mu$ in above equation.

  $\begin{array}{l}\left(\frac{9\pi }{16}\times 1260\times \left(\text{2}\text{mm}\right)\left(\frac{\text{1}\text{m}}{1000\text{mm}}\right)\right)V^2+3\times 1\times V+\frac{{\left(\left(\text{2}\text{mm}\right)\left(\frac{\text{1}\text{m}}{1000\text{mm}}\right)\right)}^2\times 9.81}{6}\left(1260-2700\right)=0\\ 4.4532V^2+3V-0.0094176=0\end{array}$

On solving quadratic equation, the value of $V$ are $0.003121\text{m/s}$ and $-0.6768\text{m/s}$. Velocity cannot negative for this case. So, the theoretical velocity for $\text{2}\text{mm}$ diameter ball is $0.003121\text{m/s}$ or $3.121\text{mm/s}$.

Difference between actual and theoretical velocity is calculated as follows:

  $\begin{array}{l}\Delta \text{V=}{V}_{\text{experimental}}-V_{\text{theoretical}}\\ \Delta \text{V=}3.2\text{mm/s}-3.121\text{mm/s}\\ \Delta \text{V=}0.079\text{mm/s}\end{array}$

For second case:

Substitute $4\text{mm}$ for $D$, $\text{1260}\text{Kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{gl}$, $2700\text{Kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{al}$ and $1\text{Kg-m}/\text{s}$ for $\mu$ in above equation.

  $\begin{array}{l}\left(\frac{9\pi }{16}\times 1260\times \left(4\text{mm}\right)\left(\frac{\text{1}\text{m}}{1000\text{mm}}\right)\right)V^2+3\times 1\times V+\frac{{\left(\left(4\text{mm}\right)\left(\frac{\text{1}\text{m}}{1000\text{mm}}\right)\right)}^2\times 9.81}{6}\left(1260-2700\right)=0\\ 8.9064V^2+3V-0.03767=0\end{array}$

On solving quadratic equation, the value of $V$ are $0.01212\text{m/s}$ and $-0.3489\text{m/s}$. Velocity cannot negative for this case. So, the theoretical velocity for $4\text{mm}$ diameter ball is $0.01212\text{m/s}$ or $12.12\text{mm/s}$.

Difference between actual and theoretical velocity is calculated as follows:

  $\begin{array}{l}\Delta \text{V=}{V}_{\text{experimental}}-V_{\text{theoretical}}\\ \Delta \text{V=12}\text{.8}\text{mm/s}-12.12\text{mm/s}\\ \Delta \text{V=}0.68\text{mm/s}\end{array}$

For Third case:

Substitute $10\text{mm}$ for $D$, $\text{1260}\text{Kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{gl}$, $2700\text{Kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{al}$ and $1\text{Kg-m}/\text{s}$ for $\mu$ in above equation.

  $\begin{array}{l}\left(\frac{9\pi }{16}\times 1260\times \left(10\text{mm}\right)\left(\frac{\text{1}\text{m}}{1000\text{mm}}\right)\right)V^2+3\times 1\times V+\frac{{\left(\left(10\text{mm}\right)\left(\frac{\text{1}\text{m}}{1000\text{mm}}\right)\right)}^2\times 9.81}{6}\left(1260-2700\right)=0\\ 22.26V^2+3V-0.23544=0\end{array}$

On solving quadratic equation, the value of $V$ are $0.05556\text{m/s}$ and $-0.190339\text{m/s}$. Velocity cannot negative for this case. So, the theoretical velocity for $10\text{mm}$ diameter ball is $0.05556\text{m/s}$ or $55.56\text{mm/s}$.

Difference between actual and theoretical velocity is calculated as follows:

  $\begin{array}{l}\Delta \text{V=}{V}_{\text{experimental}}-V_{\text{theoretical}}\\ \Delta \text{V=60}\text{.4}\text{mm/s}-55.56\text{mm/s}\\ \Delta \text{V=4}\text{.84}\text{mm/s}\end{array}$

Thus, the theoretical velocities for all different cases are $3.121\text{mm/s}$, $12.12\text{mm/s}$ and $55.56\text{mm/s}$. The difference between experimental and theoretical velocities for all three cases are $0.079\text{mm/s}$, $0.68\text{mm/s}$ and $\text{4}\text{.84}\text{mm/s}$.

Conclusion:

The theoretical velocities for all different cases are $3.121\text{mm/s}$, $12.12\text{mm/s}$ and $55.56\text{mm/s}$. The difference between experimental and theoretical velocities for all three cases are $0.079\text{mm/s}$, $0.68\text{mm/s}$ and $\text{4}\text{.84}\text{mm/s}$.