dear sea-will,

The presence of product in a cross-section, decreases the area through which air flows.

Definitions :

At (m2)= total area (/4*D^2)

Ap (m2)= area for product

Al (m2)= area for air

Ql (kg/sec)= mass air flow

rho(l) (kg/m3)= air density

rho(p) (kg/m3) = product material density

mu ((kg/sec)/(kg/sec))= loading ratio

vp (m/sec)= product velocity

vl (m/sec)= air velocity

with : mu = Qp/Ql and Qp = vp * Ap * rho(p)

Results in : Ap = (mu * Ql) / (vp * rho(p))

Further : vl = Ql / (rho(l) * (At - Ap))

After substitution :

Ql

vl = -----------------------------------

mu * Ql

rho(l) * ( At - ------------------ )

vp * rho(p)

In case the term : mu * Ql

--------------- = At

vp * rho(p)

(meaning that the complete area is filled with product), the air-velocity will be infinite.

Whether you can neglect the presence of material depends on loading ratio, product velocity and material density.

Air density determines the air volume and thereby the air velocity and the product velocity.

As the various parameters are dependent on each other it will be rather complicated to calculate the air velocity.

best regards ■

as the formals come out a little bit distorted, I will try again.



vl = Ql / (rho(l) * (At - (mu * Ql )/(vp * rho(p))





In case the term : (mu * Ql )/(vp * rho(p)) = At ■

as the formals come out a little bit distorted, I will try again.



vl = Ql / (rho(l) * (At - (mu * Ql )/(vp * rho(p)))





In case the term : (mu * Ql )/(vp * rho(p)) = At ■

Sea-Will,

When the pipeline is stepped-up, solids to gas mass flow ratio does not change, only the gas velocity decreases. This decrease must be calculated to maintain the same Froude Number before and after the step. I'm sure you know how to calculate the Froude Number. If you need my help, you may send me an email at the address given below.

Regards,

Amrit T. Agarwal

Consulting Engineer

Pneumatic Conveying Consulting LLC

Email: polypcc@aol.com

Ph and Fax: 304 346 5125 ■

dear Mr Cacing

In general, a system, properly designed for a 8” pipeline, must be capable of functioning on a 8”to 6” pipeline.

As long as the maximum system conveying pressure is not exceeded (by feeding control), it will operate.

Of course, the velocities in the 6”part of the pipeline will be higher and therefore the capacity will be lower,

The effect depends on the location of the diameter transfer.

The product (which you did not mention) must be fine enough, not to choke in the diameter transfer section. Normally, when using a high capacity installation on a smaller than designed

pipeline, must be controlled in the feeding more accurately.

The higher velocities must be acceptable for product degradation and pipe wear etc.

And: everything can be calculated (more or less)

Success. ■

dear Mr Cacing

For a pipeline counts; what goes in (air + product) must come out. (law of continuity)

If the same amount of air goes through an 8”pipeline the air velocity =

vel(8”) = airflow / (/4*82)

for a 6”pipeline :

vel(6”) = airflow / (/4*62)

This results in vel(6”) = 82 / 62 * vel(8”)

The extra air velocity losses have to be compensated by a lower loading ratio, in order to maintain the same (maximum) conveying pressure drop.

And a lower loading ratio with the same airflow equals less capacity in tons/hr

best regards ■

cacing

I presented a paper at the Chemical Engineering Conference last year which I addressed line stepping in more detail.

If you would like a copy, send me an e-mail to my address below and I'll send you a copy.

Regards ■

dear Mr Cacing

The character is actually pi=3.141593.

This came out misformed in the reply.

The influence of the loading ratio (mu) and the velocity is actually the core science of pneumatic conveying.

In short :

The total pressure drop in a conveying pipeline is the addition of partial pressure drops.

-kinetic energy = function(v2)

-potential energy = function(H)

-product resistance energy = function(mu^a,v2,Reynoldsnumber)

-suspension energy = function(residence time)

-air friction energy = function(v2)

Increasing v (by reducing the pipe diameter), increases the pressure drop for kinetic-, product resistance- and air friction- energy.

Because of the higher velocity, the residence time of the particles in the pipeline reduces, and thereby the pressure drop for keeping the particles in suspension.

Tust by decreasing the diameter (and keeping the other parameters constant), the increase of pressure drops is higher than the decrease of pressuredrop.

To compensate that in order to keep the total pressure drop constant, the loading ratio has to be decreased.

That is all

best regards ■

Hello my friends,I'm very sorry because I have something have to copewith so there was a long time for me not on line. And I can't on line next week.But thank you very much for your reply.

And jack ,can you give me the copy of your paper?I will be very glad if you give me.I need that kinds of paper very much .my E-mail is

sea-will@163.com ■

jack,i'm needing the paper too.and my e-mail:zpewljg@yahoo.com.cn ■

I will be happy to send a PDF copy of my calculations for stepped pipelines (10x12x14 inch dia). These calculations are based on my article "Theory and Design of Dilute Phase Pneumatic Conveying Systems" and are very easy to use if you know basic Excel. Please send your request to my email address given below:

Regards,

Amrit T. Agarwal

Consulting Engineer

Pneumatic Conveying Consulting

Email: polypcc@aol.com

Ph and Fax: 304 346 5125 ■

Dear Armit

Can you send me a PDF copy of your calculations for stepped pipelines

thanks alot

Best Regards,

ShooCHiang ■

Dear Amrit,

Thank you very much for the article and respond me so quickly

the article you send me is very useful for me.

Thanks again!

ShooChiang ■

Dear Sea-Will,

Please send an email to me to get a PDF copy of these calculations.

Regards,

Amrit T. Agarwal

Consulting Engineer

Pneumatic Conveying Consulting LLC

Email: polypcc@aol.com

Ph and Fax: 304 346 5125 ■