Pressure Drop in Vacuum Conveying

Posted in: , on 12. Jun. 2006 - 05:04

Hey Guys,

When doing a pressure drop calculation for vacuum conveying for design purposes, do you include the pressure drop from air only between the air inlet point and feed point and also from the cyclone to the blower? The system I'm looking at feeds from the silo into the system using a rotory airlock, drops into the end point using a cyclone, then the air exiting goes through a filter receiver, then to the blower.

How many pressure drop (from a rule of thumb), do you add for the cyclone and filter receiver (FR only has air mostly going to it). Any equivalent length addition or extra pressure drop from slide gate valves in the line?

Thanks

Re: Pressure Drop In Vacuum Conveying

Posted on 12. Jun. 2006 - 07:56

Pneumatic conveying pressure losses are broken down into air only losses and solids contributions. Air only losses will always occur even if there are no solids present in the line and they are relatively easier to calculate. Don’t forget to add the pipe length after the cyclone to the exhauster for air only losses. Pressure losses across the filter can be taken as 20 mbar. For a suitable size filter it is always in a single figure. Losses in the cyclone will depend on the cyclone design. Since you have a filter why are you using a cyclone? A suitable shape receiving hopper with filter on top works fine. Last but not least do take into account the rotary air lock air leakage.

Mantoo

Re: Pressure Drop In Vacuum Conveying

Posted on 12. Jun. 2006 - 05:17

Thanks for the reply.

The polystyrene pellets that we airvey gets collected in the cyclone and put into a packaging bin. The filter receiver after the cyclone is in case there are fine materials that don't get captured in the cyclone so they don't get exhausted into the atmosphere (this is a company standard). Would this drop the pressure drop in the filter receiver after the cyclone?

The RA is currently the same used. We are adding extra piping and bends to the system and I wanted to calculate the new delta P to see if this will give us close to the same capacity as we currently have. So I'm not taking RA leakage into account.

Thanks again.

Re: Pressure Drop In Vacuum Conveying

Posted on 12. Jun. 2006 - 06:36

Dear RJB324

If you want to know the pressuredrop after you have modified the installation, proceed as follows:

-model your existing installtion

-collect all the operational parameters (by measuring)

-Make the pressure drop calculation inputs consistent with the measured outputs.

Now you have your calculation data.

-Remodel your new installation

-Run the calculation again, using the obtained calculation data from the measurement.

Then you have your new pressure drop.

(Use the existing installation as a 1 to 1 scale test facility)

Success

Teus

Re: Pressure Drop In Vacuum Conveying

Posted on 12. Jun. 2006 - 07:41

The additional pressure drop will depend on how much conveying pipe and bends are added. Since it is a modification to a running system and you would know current system pressure drop it should not be difficult to just use a linear scale for extension. Since polystyrene pellets have very low bulk density it will work.

The pressure drop across the filter should be less then 20 mbar for correctly sized a reverse jet filter with cloth area large enough to give an upward velocity of 1m/min.

Mantoo

Re: Pressure Drop In Vacuum Conveying

Posted on 13. Jun. 2006 - 03:31

Thanks Again. One more thing that I could not find in literature. Does slide gate valve have any equivalent length in conveying systems (if they are fully open all the time)? Also, how can you estimate the pressure drop through a manual slide gate partially open?

Thanks.

Re: Pressure Drop In Vacuum Conveying

Posted on 13. Jun. 2006 - 05:38

Dear RJB324 and Mr Mantoo,

The general formal for a flow is:

p = * cw * rho * v^2

For a pneumatic conveying installation, this formal can be written as

(and this is a VERY ROUGH estimate):

p = function (constant * v^2 * mu^(alpha) * Re * L)

In which :

v = velocity

mu = loading ratio

Re = Reynolds number

L = conveying length

By changing the length and maintaining the capacity (rotary lock feeding),

the terms constant, mu^(alpha) and Re do not change.

This results in :

p1 = function (constant * v1^2 * mu^(alpha) * Re * L1)

p2 = function (constant * v2^2 * mu^(alpha) * Re * L2)

further :

v1 = Q/(A*p1) and v2 = Q/(A*p2)

in which:

Q = airvolume

A = cross sectional area of pipeline

resulting in :

v1/v2 = p2/p1

Substituted :

p1 = function (constant * 1/p1^2 * mu^(alpha) * Re * L1)

p2 = function (constant * 1/p2^2 * mu^(alpha) * Re * L2)

p1/p2 = L1/L2 * p2^2/p1^2

p2^3 = p1^3 * L2/L1

p2 = pressure drop after modification = p1 * (L2/L1)^0.3333

The increase in pressure drop is certainly not proportional and therefore I still prefer the previous method I mentioned.

Then the pressure drop over a partially opened slide gate is difficult to predict, as this depends

on the velocity increase in the slide gate opening and the resulting product velocity losses and how much energy is regained afterwards.

But why should you close a slide gate in a pneumatic conveying flow ????.

have good day

Teus