Chapter 4, Problem 4.1P

To determine

The air velocity required for pneumatic conveyors to move wood chips

Answer to Problem 4.1P

  $3302.49\frac{\text{ft}}{\text{min}}$

Explanation of Solution

Given information:

Diameter of wood chips $=1\text{in}$

Bulk density of wood chips $=500{\text{lb/yd}}^{\text{3}}$

Concept used:

  $SG=0.1{\left(W\right)}^{2/3}$

  $v_f=3250\sqrt{\left(SG\right)d}$

  $v_m=585\sqrt{W}$

  $v_a=v_f+v_m$

  $\begin{array}{c}W\to \text{Bulk density of material}\\ v_m\to \text{Material velocity}\\ v_a\to \text{Velocity of air stream}\\ v_f\to \text{Floating velocity}\end{array}$

Calculation:

  $\begin{array}{c}W=500\frac{\text{lb}}{{\text{yd}}^{\text{3}}}\times \frac{\text{1}{\text{yd}}^{\text{3}}}{{\left(\text{3}\text{ft}\right)}^{\text{3}}}\\ =18.52\frac{\text{lb}}{{\text{ft}}^{\text{3}}}\end{array}$

  $\begin{array}{c}SG=0.1{\left(W\right)}^{2/3}\\ =0.1{\left(18.52\right)}^{2/3}\\ =0.70\end{array}$

  $\begin{array}{c}{v}_f=3250\sqrt{\left(SG\right)d}\\ =3250\sqrt{0.7\times 1\text{in}\times \frac{\text{1}\text{ft}}{\text{12}\text{in}}}\\ =784.95\frac{\text{ft}}{\text{min}}\end{array}$

  $\begin{array}{c}{v}_m=585\sqrt{W}\\ =585\sqrt{18.52}\\ =2517.54\frac{\text{ft}}{\text{min}}\end{array}$

  $\begin{array}{c}{v}_a=v_f+v_m\\ =784.95+2517.54\\ =3302.49\frac{\text{ft}}{\text{min}}\end{array}$

Conclusion:

The air velocity required for pneumatic conveyors to move wood chips is $3302.49\frac{\text{ft}}{\text{min}}$ .