Dear Mersam,

Can you clarify the meaning of these parameters?

have a nice day

Teus ■

mersam

I merged all of your posts into one thread. Posting multiple posts with the same question will not only frustrate those that are trying to help you but will confuse those that want to help you. Please keep your question in here.

You also need to answer the question of someone that is trying to help you. See post #4 above from Teus.

Thank you.

Gary Blenkhorn

forum moderator ■

Thank you Gary.

Dear mersam,

Explain the meaning of:

Explanation : ------------------------------

Explanation : ------------------------------

Explanation : ------------------------------

Explanation : ------------------------------

Explanation : ------------------------------

Explanation : ------------------------------

In addition, it would be helpful if you describe your installation more accurately.

BR

Teus ■

Dear mersam,

I have noticed the information, you have now supplied.

I am now studying the data in Table 1.

The calculation method whereby the presence of solid is accounted for by a factor on top of the air pressure drop is not used by me.

It seems that you must have all the information available in your assignment to do the design.

However, I am not familiar with the method and it is not doable for me to dig in to that.

I will see what I can come up with, but that will not help you with your task.

Success

Teus ■

Dear mersem,

I used table 1 (pneumatic conveying test of flour) to calculate the Solid Friction Factor for each test.

Then I plotted the found SFF against the SLR and the Reynolds number.

The SFF shows no correlation with the SLR, which should be the case.

However, the SFF shows a rather strong correlation with the Reynolds number.

However, one would expect a lower friction factor at lower velocities (low Re) but the friction factor increases with a lower Re.

Another problem is that the test runs are all in the laminar flow regime (Re << 2*10^5) and that the tests do not indicate the SFF curve under turbulent circumstances. (Re > 2*10^5)

I leave the test information for what it is right now and focus more on previous experience.

Test run number 40 seems to be corrupted, because the SFF even becomes negative.

Attached the test based calculations.

Do not follow me, but stay on your own course.

All for now

Teus

Attachments

testresultsflour (ZIP)

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Dear mersem,

I made a preliminary calculation for the 175 m long installation.

Have alook at it and in case of questions come back to the forum.

Good luck

Teus

Attachments

flour preliminary pneumatic conveying calculation (PDF)

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Dear mersem,

The calculation is for a 2-vessel (double tank) system as indicated on the screen shot “Calculation results pressure conveying” .

Valves are not considered in the calculation.

pressure drops, pressures, air velocities, material velocities are displayed in the calculation sheets.

Your design sheet of the test results is an approach that I am not familiar with (using the Froude number is f.i. unclear to me) and therefore cannot comment on that adequately.

Did you calculate in your spreadsheet the SLR somewhere?

I will calculate the 175m long system now for a smaller pipe and a higher pressure.

Take care

Teus ■

Dear Mersim,

The promised calculation.

Notice that, due to the high pressure, water condensation occurs just after the flour/air mixture is formed.

This condensation is strongly depending on temperatures and the RH humidity of the intake air.

Using a rotary valve for high pressure systems is not really recommended. Rotary valve leakeges increase with pressure and especially for very fine products, the volumetric efficiency of the rotary lock is influenced negatively.

Low pressure systems function better with a rotary lock.

A screw feeder will probably not work, due to the compressability of the flour forming lumps.

A low pressure system can use a simple and not expensive blower, opposed to a higher presure system, that requires a more expensive oil free screw compressor with internal compression.

(Oil free compressor, because of a food application)

A low pressure, rotary valve fed, installation could be the best solution.

Designing the 125 m and 150 m installations (can be executed easily) will probably not change much to this design.

A stepped pipeline design will lead to a somewhat lower pressure, but still need an oilfree screwcompressor.

As I am not aware of the calculation algorithm that you use, the outcome should be more or less the same if the installations are comparable.

Hope to have helped you a little bit

Success

Teus

Attachments

flour preliminary pneumatic conveying calculation (PDF)

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Dear mersem,

No problem

It is not slug flow, because in slug flow completely other phenomena and formulas are ruling.

Dense phase fluidized is something like an airslide?

The computer program covers dense – to dilute phase.

I calculated the Zenz diagram for the 4” installation.

The calculation is in dense phase.

The same amount of flour can be conveyed at 0.4 m3/sec against a pressure of 2.503 bar, but requires approx 3 times more power.

Air requirement is a displacement volume of the compressor of 0.12 m3/sec against a pressure of 2.85 bar.

The pipe size was a choice for calculation purposes.

Yes, it is a bit small, but that explains the high pressure.

A bigger pipe requires more air and less pressure.

2 blow tanks, each 2 m3 volume with a filled volume of 1.5 m3

see page 2, lower sheet

31 kW shaft power (motor approx. 35 kW)

see page 2.

The velocities at the end of each pipe section are presented on the upper sheet on page 2, together with other calculated parameters.

I calculated 3 bends more than required.

On page 2 upper sheet you will find:

Bend losses 0.1 kW (for 11 bends)

Conveying energy 16 kW

For 9 bends the energy loss is approx. 9/11 * 0.1 = 0.082 kW

Difference is 0.1-0.082 = 0.18 kW = 0.18/16*100=0.11%

This difference is negligible.

This is probably a scaling method by Dr David Mills.

I am not familiar with the scaling technique, as I always tested and calculate scale 1:1

Success

Teus ■

Sorry,

I forgot the Zenz diagram attachement.

Attachments

zenz diagram flour 175m 4 inch (PDF)

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Dear mersam,

Dilute phase is defined as the operating region in a Zenz diagram on the right side of the lowest point.

If an increase of air volume at a constant capacity (Ms) causes an increased pressure drop, then the pneumatic conveying is in dilute phase.

Dense phase is defined as the operating region in a Zenz diagram on the left side of the lowest point.

If an decrease of air volume at a constant capacity (Ms) causes an increased pressure drop, then the pneumatic conveying is in dense phase.

This has no relationship with the absolute SLR, as the SLR for a long conveying pipeline is lower than for a shorter pipeline.

Also, the particle size and particle density are not relevant in this definition.

I indicated the dense- and dilute region in the previously attached file.

Read:

Dense phase- or dilute phase pneumatic conveying:

https://news.bulk-online.com/?p=238

have a nice day

Teus ■

Dear mersam,

I calculated the installation for the 3” pipeline (75mm)

Notice the difference between your and my calculation of the air only pressure drop.

I calculate in a time domain of 0.001 sec along the pipe, adjusting the gas conditions every 0.001th of a second, where you use the arithmic average velocity along the whole pipe.

The formula for lambda(s) = (7.0197 * rhoair^1.379)/(D^0.5 * 100) is unknown to me, but also a key value for the design.

You probably derived this formula from the test runs.

If I calculate the installation as if it was capable of 4.6 tons/hr, then the solid friction factor must be a factor 100 times less than derived from the test runs and the pressure drop for acceleration increases because of the increased mass flow rate.

Have you tried the spreadsheet of Mr Agarwal?

Comparing different calculation algorithms, without really investigating the underlying theory is almost undoable and shows how much research in pneumatic conveying still has to be streamlined.

Have a nice day

Teus

Attachments

pneumatic conveying calculation for flour 175m 3in (PDF)

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Dear Mersam,

A pneumatic conveying system works as follows:

Imaging, having a physical system as your design.

-When you switch on the compressor and there is no material feed into the pipeline, then the conveying pressure equals the air only pressure.

-If you mix a little bit of material into the airflow, there will be more pressure generated to force the material through the pipe.

-The air velocity at the beginning of the pipeline drops because of the higher pressure. Therefore, the air pressure drop decreases a little bit due to the lower air velocity.

-Introducing more material into the pipeline increases the conveying pressure again.

-When so much material is fed int the airflow that the conveying pressure reaches the maximum pressure that the compressor can deliver and the conveying is still stable, then the maximum capacity of the system is reached. (ms is controlled from minimum to maximum)

-Changing the airflow (ma), changes the physical installation. F.i. the velocities will increase and therefore the pressure drop will increase. However, the SLR will decrease at the same capacity and therefore the pressure drop will decrease.

-Which of the two changed pressure drops is dominant determines whether the new pressure drop is lower or higher than the original pressure drop (dilute- or dense phase)

-If a system does not reach the required capacity at the maximum allowed conveying pressure, then it is too small and a bigger pipeline and compressor is required.

If you try to increases the air mass flow in your design, you will see the air velocity Ci also increase.

If you change ma then you must also change the pipe diameter to keep the velocity within acceptable values.

Pneumatic conveying is very complex in the sense that all the parameters are inter dependant in sometimes a positive and sometimes in a negative relation.

Have a nice day

Teus ■

Dear mersam,

look at:

Dense phase- or dilute phase pneumatic conveying:

https://news.bulk-online.com/?p=238

The Zenz diagram is the pressure drop curve as a function of the airflow for a defined pneumatic conveying system for a defined material and for a constant capacity.

The lowest pressure drop in that curve is defined as the boundary between dense- and dilute phase pneumatic conveying.

Probably the 6” pipeline version with double tank feeder and low pressure blower.

No problems with airlocks.

Low energy consumption.

The pipe line must probably executed in stainless steel because of food grade.

Also use your own knowledge and intelligence together with research of literature and contacting manufacturers to complete your assignment and make your thesis.

Success

Teus ■

Dear Mersam,

For the Mills scaling technique, visit this link:

http://www.teora.hit.no/dspace/bitst...Thesis-PhD.pdf

full text of link: teora.hit.no/dspace/bitstream/2282/297/1/Thesis-PhD.pdf

Success

Teus ■

Dear Mersam,

I anticpated this problem, that is why I also included the full text of the ling.

Try that text, preceeded by www.

Or start at:

www.teora.hit.no

Success

Teus ■

Dear mersam,

I googled mills scaling technique and found this link.

It is a pdf file that can be downloaded.

The titel is A Comprehensive Scaling Up Technique for Pneumatic Transport Systems.

You can aslo perform a search on the web with this text.

success

teus ■

Dear mersam,

It takes a complete study of the theory, to understand the emaning of the given formulas.

The formula, you are referring to says:

material massflow1 times Eq.Length1 = material massflow2 times Eq.Length2

or in words:

If you double the eq.length of a system, then the material massflow becomes half the original massflow.

However, I have serious doubts that this relation is valid.

Furthermore, all the curves in the document are presented for fly ash and your system is for flour

I found for a cement installation approx.:

Installation 1:

horizontal length = 50 m

vertical length = 15 m

Number of bends = 3

pipe dia = 254 mm

air flow = 0.9612 m3/sec

Installation 1:

horizontal length = 100 m

vertical length = 30 m

Number of bends = 6

pipe dia = 254 mm

air flow = 0.9612 m3/sec

Eq.length2/eq.length1 = SQRT(massflow1/massflow2)

I cannot help you further here, because in my calculation algorithm, the scaling problem is bypassed as I calculate in real size.

Have a nice day

Teus ■