Pipe Diameter Formula for Various Throughputs

Author
(not verified)
Posted in: , on 25. Apr. 2004 - 21:55

I am looking for a comprehensive formula to calculate the diameter of

pipes required to handle various throughputs of different materials in

free fall. I realise this is dependent upon drop height, product

density, etc. Any advise would be greatly appreciated. "

David Harverson

Eriez Magnetics Europe Ltd

Tel : 02920 868501

Formulae

Posted on 25. Apr. 2004 - 11:29

Dear sir,

either fortunately or unfortunately you have answered most of your puzzlement. I know only of gas and fluid formulas as they relate to pipe.

The only way I or anyone else could help is if you narrowed your search on material fed through said drop pipe- the problem being of course bridging and the resultant stoppage of flow-then the problem of opening the clog or coned pile in the pipe rears its ugly head as you are dealing with material that is continualy dropped upon by said material until it plugs all the way to the surface of the delivery pipe, or the weight of the mass plugged in the pipe falls with the force of one -g to the bottom of the pipe which will no doubt create havoc unless the opening is larger at the bottom to allow said material to fall away from the resultant cone of material that it is falling upon.

I guess you could treat it as an enigma surronded by a mystery surronded by a question.

puzzles are fun :^)

PS if you were blowing silage into silo it would be easier to figure out.

lzaharis "Semper Excellsior"

Re: Pipe Diameter Formula For Various Throughputs

Posted on 26. Apr. 2004 - 01:39

Dear Mr. David Harverson,

In case of bulk material handling equipment / system, in general, one rarely comes across a situation where material is allowed to fall freely from great height, because in such situation the material will pick up very high speed and would be degrading itself on impact. It will also cause chute wear and impact on receiving surface, etc. Only equipment where such situation is present, is for equipment like ship loader etc. However, in such case material size would be comparatively small.

Pipe chute of great height are used for stockpile formation, but in such chutes the material is allowed to fall from step to step, so that it is expected to travel at nearly constant speed. The fall height and step is chosen in such a manner that energy gained in fall is dissipated at material impact on material at each step.

I suggest you to mention the name of material, lump / particle size, flow rate, application etc. This will help some respondent to give you some suggestion.

I do not have specific formula but would like to mention the following about the phenomenon :

1) Velocity at inlet will be minimum, so pipe dia at inlet will be of maximum size (often pipe diameter would be same through out)

2) Velocity increases as the material falls.

3) The material coming in contact with side-walls will tend to slow down, creating certain turbulence and certain magnitude of slow down. This slow down will depend upon the ratio of material cross section and pipe cross sectional area.

4) Air resistance will become significant when velocity becomes high (particularly if material is of smaller size i.e. grain / powdery).

5) Material will acquire terminal velocity when gravity acceleration force becomes equal to resistance force, (if the material reaches that stage).

6) If the material is powdery, its flow will be mainly governed by air resistance.

The article by Z Korzen, Poland published in “I / 94 Belt Conveyor Technology (Trans Tech Publication)” provides good information and formulae etc. for air drag resistance related to material trajectory. The information in this article may possibly be of some use to you (subject to the nature of your application, which is not known).

Regards,

Ishwar G Mulani.

Author of Book : Engineering Science and Application Design for Belt Conveyors.

Advisor / Consultant for Bulk Material Handling System & Issues.

Email : parimul@pn2.vsnl.net.in

Tel.: 0091 (0)20 25882916

Pipe Diameter Formula

Posted on 3. Jan. 2005 - 12:56

Dear Mr.David,

Why you want to calculate pipe diameter? You will get directly from standard chart. Once Mr. Vaka,of Krupp Robins,Canada published in the Bulk Solid journal. He published many tables to find out many things.

Other-wise you may go for very simple formula as given below:

3.14/4 x dia. sq.x belt speed x.7 = conveying capacity

Regards.

A.Banerjee

Re: Pipe Diameter Formula For Various Throughputs

Posted on 5. Jan. 2005 - 08:53

Hello,

There are references to standpipe capacities in pneumatic conveying textbooks. I once had a book "Principles of Pneumatic Conveying" by Marcus & 2 other experts. There was a chapter concerned with standpipes & holding vacuum legs etc.

I suggest you search along those lines. As far as I remember the Maths could be reasonably interesting.

Good Luck on your quest.

John Gateley.

John Gateley johngateley@hotmail.com www.the-credible-bulk.com

Chute Sizing

Posted on 6. Jan. 2005 - 04:57

I have attached a small program that I did that calculates chute sizes for me. You simply plug in the TPH and the Bulk Density. I use it for rock salt and I don't see why it wouldn't work for any other product.

Regrads,

Gary Blenkhorn

Gary Blenkhorn
President - Bulk Handlng Technology Inc.
Email: garyblenkhorn@gmail.com
Linkedin Profile: http://www.linkedin.com/in/gary-blenkhorn-6286954b

Offering Conveyor Design Services, Conveyor Transfer Design Services and SolidWorks Design Services for equipment layouts.

Chute Sizer

Posted on 6. Jan. 2005 - 04:58

File didn't attach first time so lets try it again.

Gary

Attachments

chutsize (ZIP)

Gary Blenkhorn
President - Bulk Handlng Technology Inc.
Email: garyblenkhorn@gmail.com
Linkedin Profile: http://www.linkedin.com/in/gary-blenkhorn-6286954b

Offering Conveyor Design Services, Conveyor Transfer Design Services and SolidWorks Design Services for equipment layouts.

Chute Size

Posted on 7. Jan. 2005 - 11:38

Dear Mr.Gary,

I tried one calculation with your programe. But how this programe is taking care of lump size with small tonnage capacity?

Regards.

A.Banerjee

Gravity Flow Through An Orifice Or Tube

Posted on 7. Apr. 2008 - 07:23

David

There is a simple calculation for determining the flow rate of material under gravity through an orifice, as follows:

Wa=(0.0038 x Pb x Do 2.5) ÷ (tan ßa)0.5

Wa=Discharge rate -Te/hour

Pb=Bulk Density (lbs/ft3)-50 lbs/ft3

Do=Orifice Diameter (Ins)-4 Inches

ßa=Angle of Repose-35degrees

0.0038 is a constant.

Wa=(0.0038 x 50 x 42.5) ÷ (tan 35)0.5 = 7265kgs/hour

I have been using this formula for years and it is extremely correct, I have checked it against actual flow a number of times. Hope this helps. Stewardson Process Design (07876 165211)

Re: Gravity Flow Through An Orifice Or Tube

Posted on 7. Apr. 2008 - 08:27

Originally posted by Tim Stewardson

I have been using this formula for years and it is extremely correct, I have checked it against actual flow a number of times. Hope this helps.

Interesting. What bulk materials have you checked the validity against?

Re: Pipe Diameter Formula For Various Throughputs

Posted on 8. Apr. 2008 - 09:18

Please note

Do (orifice diameter) is to the power of 2.5

(tan ßa) is to the power of 0.5

When I posted the text earlier this has shown as "multiplied by" and I did not notice.

I have checked this formula on free flowing products such as;

Sugars, flours, salt, silica sand, PVC powder, PVC granule saccharine, cereal grits, calcium carbonate, sodium bicarbonate, sodium alginate.

surrisri_bo
(not verified)

Not Yet

Posted on 10. May. 2008 - 01:49

Dear Tim Stewardson,



I had calculated the flow rate by using the formula given by you. I am not getting the answer.

Wa = (0.0038 x Pb x Do 2.5) ÷ (tan ßa)0.5

Wa = Discharge rate - Te/hour

Pb = Bulk Density (lbs/ft3)- 50 lbs/ft3

Do = Orifice Diameter (Ins)- 4 Inches

ßa = Angle of Repose - 35degrees

0.0038 is a constant.

In the above formula Do 2.5 is Do power 2.5 or multiplied

(tan ßa) 0.5 is tan ßa power 0.5 or multiplied

Wa = (0.0038 x 50 x 42.5) ÷ (tan 35)0.5 = 7265kgs/hour

Please expline how you got the answer 42.5 in the above calculation and final answer 7265 kgs/hr

I am seaching for Throught output formula through a pipe.

Thanking you

-Surri

Re: Pipe Diameter Formula For Various Throughputs

Posted on 12. May. 2008 - 11:08

Dear surrisribo

The formula is not correct as it appears in the text of this web site. Please send me your Email address and I will send you the formula in a Word document and also an Excell spreadsheet programme. Please send you Email to tim.stewardson@sky.com.

Tim

surrisri_bo
(not verified)

Thanks

Posted on 13. May. 2008 - 12:55

Sir,

Your Explaination for the formula is given in the second page. But did't saw that. Now i got the idea and answer for your example. I am in Cement Industry, i will do some trail calculation, then i will revert you.

Thanking you

Surri