Chapter 7, Problem 7.72P

To determine

(a)

Estimate the minimum length of the tank if the particle diameter is equal to $1mm$ ?

Answer to Problem 7.72P

$L=6.162m$

Explanation of Solution

Given information:

The depth of the water tank is $2.5m$

The velocity of water flow is $35cm/s$

The specific gravity of tank material is $SG=2.55$

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:

The horizontal velocity $U$ of the particles is equal to $35cm/s$

Assume vertical velocity as $V$

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,

$W-F_{bouyancy}-F_{drag}=0\to \left(1\right)$

Calculate the weight of the sphere,

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

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),

$\begin{array}{l}W-F_{bouyancy}-F_{drag}=0\\ \rho gS\left(\pi \frac{d^3}{6}\right)-\rho g\left(\pi \frac{d^3}{6}\right)-C_D\frac{\rho V^{2}A}{2}=0\\ \rho g\left(\pi \frac{d^3}{6}\right)\left(SG-1\right)-C_D\frac{\rho V^2}{2}\left(\frac{\pi }{4}{D}^2\right)=0\end{array}$

Solve for $V$

$V={\left[\frac{4gD\left(SG-1\right)}{3C_D}\right]}^{1/2}$

The length of the tank can be defined as,

$L=\frac{Uh}{V}$

Where,

$h$ - Height

According to the definitions, we can assume,

$C_D=1.0$

Therefore,

Find vertical velocity,

$V={\left[\frac{4gD\left(SG-1\right)}{3C_D}\right]}^{1/2}={\left[\frac{4\left(9.81m/s^2\right)\left(0.001m\right)\left(2.55-1\right)}{3\left(1.0\right)}\right]}^{1/2}=0.142m/s$

Calculate the minimum length of the tank,

$L=\frac{Uh}{V}=\frac{\left(0.35m/s\right)\left(2.5m\right)}{0.142m/s}=6.162m$

Conclusion:

The minimum length of the tank if $D=1mm$ is equal to,

$L=6.162m$.

To determine

(b)

Estimate the minimum length of the tank if the particle diameter is equal to $100\mu m$ ?

Answer to Problem 7.72P

$L=116.67m$

Explanation of Solution

Given information:

The depth of the water tank is $2.5m$

The velocity of water flow is $35cm/s$

The specific gravity of tank material is $SG=2.55$

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:

According to sub-part a,

We can say,

$V={\left[\frac{4gD\left(SG-1\right)}{3C_D}\right]}^{1/2}$

The length of the tank can be defined as,

$L=\frac{Uh}{V}$

Where,

$h$ - Height

According to the definitions, we can assume,

$C_D=36$

Therefore,

Find vertical velocity,

$V={\left[\frac{4gD\left(SG-1\right)}{3C_D}\right]}^{1/2}={\left[\frac{4\left(9.81m/s^2\right)\left(100\times {10}^{-6}m\right)\left(2.55-1\right)}{3\left(36\right)}\right]}^{1/2}=0.0075m/s$

Calculate the minimum length of the tank,

$L=\frac{Uh}{V}=\frac{\left(0.35m/s\right)\left(2.5m\right)}{0.0075m/s}=116.67m$

Conclusion:

The minimum length of the tank if $D=1mm$ is equal to,

$L=116.67m$.