The theoretical settling velocity is of no practical use in actual situations. You may very well be able to suspend a 40 micron particle of iron ore at 0.36 m/s but if you want a functioning de-dusting system you will design the ducting for 20 to 22 m/s and end up with economicaly sized ducts with minimal potential for blockage and minimal abrasive wear.

Michael Reid ■

In practial terms, if you are handling iron ore dust in a typical de-dusting system, the maximum particle size will be much larger than 40 microns. Every change in direction or enlargement in the ducting will cause a drop in velocity. There is a marked velocity profile across the duct, even in ideal conditions (again, see ACGIH illustrations).

Do not try to design a system from first principles but follow the guidelines. They are well proven from long practical experience.

Michael Reid. ■

Dear Pavan,

The velocity profile in the ducts has to be turbulent, whereby a minimum of dead zones occur.

That is the reason that the average air velocity needs to be higher than the suspension velocity of the particles.

To create an air velocity along the duct wall (thin boundary layer), the average velocity needs to be higher.

As Re=(density*Velocity*D)/(viscosity), small ducts require higher velocities, to reach the necessary Re number.

The wall velocity for a round duct is:

wall velocity = (-0.0165 + 0.0369 * ln(Re)) * average velocity

Dust settling problems are therefore related to the design, in which dead velocity zones must have been prevented. (Just as in pneumatic conveying)

Have a nice day

Teus ■

Do you understand that the system airflows need to be "balanced" using a damper in each branch? This is so that the design velocity is maintained in all parts of the system. Blockages will inevitably happen if the dampers are not adjusted properly.

Michael Reid. ■

Originally Posted by Teus Tuinenburg

Dear Pavan,

The wall velocity for a round duct is:

wall velocity = (-0.0165 + 0.0369 * ln(Re)) * average velocity

Have a nice day

Teus

Hi Tues, may i know this wall velocity Reynolds number that you are using, is it pipe reynolds number or particle reynolds number ? ■

The Re-number for a pipe is used here.

Have a nice day

Teus ■

Hi Tues,

I tried calculating for the duct velocity using, wall velocity = (-0.0165 + 0.0369 * ln(Re)) * average velocity

In this case the wall velocity will be the one calculated Pavan to be 0.366 m/s.

Assuming pipe diameter of 200 mm, a duct velocity of 2 m/s will have a wall velocity of 0.72 m/s which is still 1.5 times higher ( taken from your quote in Pneumatic Transport with different grain size ) than the suspension velocity ( far from the 20 m/s suggested by ACGIH ). Your kind assistance is deeply appreciated, thanks in advance

PS : May i know from what book do you made this " wall velocity = (-0.0165 + 0.0369 * ln(Re)) * average velocity " formula ? ■

Dear SeventhEro,

Calculating your example:

Re=(density*Velocity*D)/(viscosity)

Re = 1.12*2*0.2/(2.10^-5) = 22400

ln(Re) = 10.01

wall velocity = (-0.0165 + 0.0369 * ln(Re)) * average velocity

wall velocity = (-0.0165 + 0.0369 *10.01) * 2 =0.353 m/sec

The wall velocity in a pneumatic conveying pipe (or dust duct) needs to stay above the particle local suspension velocity (for small particles approx. 1.5 to 2 times).

In this example, the average velocity/wall velocity = 2/0.353 = 5.66 times.

Hence the average velocity in this case should be 5.66 * 1.5 = 8.49 times.

However, then the calculation has to be repeated for the higher average velocity, as this higher average velocity influences the Reynolds number and thereby the wall velocity. (Iteration)

The formula is a regression formula based on velocity profile graphs and must be considered as a general approximation. Influence of pipe roughness and presence of material neglected.

Have a nice day ■