Chapter 7, Problem 7.66P

To determine

Derive a differential equation for velocity $V\left(t\right)$ and show that,

$V={\left[\frac{4gD\left(S-1\right)}{3C_{d_0}}\right]}^{1/2}\tanh Ct$, $C={\left[\frac{3g{C}_{d_0}\left(S-1\right)}{3S^{2}D}\right]}^{1/2}$.

Answer to Problem 7.66P

The differential equation is,

$\frac{dV}{dt}=g\left(1-\frac{1}{S}\right)-\frac{C_D\rho gA}{2W}{V}^2$

Therefore, we can say

$V={\left[\frac{4gD\left(S-1\right)}{3C_{d_0}}\right]}^{1/2}\tanh Ct$

$C={\left[\frac{3g{C}_{d_0}\left(S-1\right)}{3S^{2}D}\right]}^{1/2}$

Explanation of Solution

Given information:

Diameter of the sphere is $D$

Density of the sphere ${\rho }_s$

The density of the fluid is $\rho$ and the viscosity is $\mu$

Drag co-efficient is $C_{d_0}$

The specific gravity S of the sphere material is defined as,

$S=\frac{{\rho }_s}{\rho }$

The Newton’s law of downward motion is defined as,

${\displaystyle \sum F=ma}$

Where,

$F$ - Downward force

$m$ - Mass

$a$ - Downward acceleration

The buoyancy force is defined as,

$F_{bouyancy}=V\rho g$

The drag force is defined as,

$F_{drag}=C_D\frac{\rho U^{2}A}{2}$

In above equation,

$C_D$ - Drag co-efficient

$A$ - Characteristic area

Calculation:

Apply Newton’s law of downward motion,

$\downarrow {\displaystyle \sum F=ma}$

The weight of the sphere $W$ is acting downwards and the drag force, buoyancy forces are acting upwards.

Therefore,

$W-F_{bouyancy}-F_{drag}=\frac{W}{g}\left(\frac{dV}{dt}\right)\to \left(1\right)$

Calculate the weight of the sphere,

$\begin{array}{l}W=mg={\rho }_{s}Vg\\ =S\rho \left(\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3\right)g\\ =\rho gS\left(\pi \frac{d^3}{6}\right)\end{array}$

Calculate the Buoyancy force exerted on sphere,

$F_{bouyancy}=V\rho g=\rho g\left(\pi \frac{d^3}{6}\right)$

Substitute in equation (1),

$\begin{array}{l}W-F_{bouyancy}-F_{drag}=\frac{W}{g}\left(\frac{dV}{dt}\right)\\ \rho gS\left(\pi \frac{d^3}{6}\right)-\rho g\left(\pi \frac{d^3}{6}\right)-C_D\frac{\rho V^{2}A}{2}=\frac{W}{g}\left(\frac{dV}{dt}\right)\\ \rho g\left(\pi \frac{d^3}{6}\right)\left(S-1\right)-C_D\frac{\rho V^{2}A}{2}=\frac{W}{g}\left(\frac{dV}{dt}\right)\end{array}$

But, according to explanation we can rewrite above equation as,

$W-\left(Vol\right)\rho g-F_{drag}=\frac{W}{g}\left(\frac{dV}{dt}\right)$

Replace buoyancy force in terms of $W$ and $S$

$W-\frac{W}{S}-F_{drag}=\frac{W}{g}\left(\frac{dV}{dt}\right)$

Rearrange,

$\frac{dV}{dt}=g\left(1-\frac{1}{S}\right)-\frac{C_D\rho gA}{2W}{V}^2$

Assume,

$\begin{array}{l}a=g\left(1-\frac{1}{S}\right)\\ b=\frac{C_D\rho gA}{2W}\end{array}$

Therefore,

$\frac{dV}{dt}=a-b{V}^2$

Integrate above equation to find $V$

${\displaystyle \int dt={\displaystyle \int \frac{1}{a-b{V}^2}dV}}$

Therefore, we get,

$V=\sqrt{\frac{a}{b}}\tanh \left(t\sqrt{ab}\right)$

Substitute,

$V={\left[\frac{4gD\left(S-1\right)}{3C_{d_0}}\right]}^{1/2}\tanh Ct$

Where,

$C={\left[\frac{3g{C}_{d_0}\left(S-1\right)}{3S^{2}D}\right]}^{1/2}$

Conclusion:

The differential equation is,

$\frac{dV}{dt}=g\left(1-\frac{1}{S}\right)-\frac{C_D\rho gA}{2W}{V}^2$

Therefore, we can say

$V={\left[\frac{4gD\left(S-1\right)}{3C_{d_0}}\right]}^{1/2}\tanh Ct$

$C={\left[\frac{3g{C}_{d_0}\left(S-1\right)}{3S^{2}D}\right]}^{1/2}$.