Your forgetting the fact that gravity will affect the flow of the wood dust so the more flow the better as it-(the wood dust)-will stop moving even though it is in an enclosed tube as it will settle and begin to pile upon itself.

The high volume is justified by the low weight of the wood grindings-just like herding cats with a squirt gun :^) but a bowl of warm milk works better. ■

Another point to remember is that the material to air ratios you cite from the reference material are based on using blowers and / or compressors for the air supplies.

The system you describe is a fan conveying system and due to the performance characteristics of a fan as compared to a positive displacement blower or compressor, you want to keep a very smooth and consistant flow regime within the pipeline.

It's not uncommon to have fan conveying systems designed in the range of material to air that you mention. I must admit, it is very low, even for a fan system but not beyond the range of application. Most of the systems I have been involved with are designed around a ratio of 0.5, based on air at standard conditions.

Your velocity range is okay.

Regards ■

Dear Pablo,

I do not know whether my answer is of use for you. We had many installations where the goal was to make the material flow as fast as needed to prevent congestions but to keep it as slow as possible to prevent wear and tear of pipe and material and to keep operating costs at bay. The device "ABC Even Flow" is a self regulating device that is fully maintaining that goal. If you are interested look it up under

http://www.feldhaus-technik.de/eng/i...rarbeitsweise

Regards, Thomas ■

If you use a roots type blower 2 tph can be easily be achieved in a 2 1/2 inch with about 320 m3/hr air and SLR of approx 5. If you want to push the system harder you can use 2 inch pipe with 180 m3/hr air giving a SLR of approx 9(don’t forget to add RV air leakage to these figures). For fine wood dust with low bulk density 15 m/s pick up velocity is normally sufficient.

Wood dust is not considered abrasive you don’t have to worry about wear and tear in the system. ■

Pablo,

A dilute phase system can operate at M-to-A ratios anywhere from zero up to 10, 12 or sometimes even 15 depending on the conditions (material, length, etc). Beyond that, the cat/squirt gun analogy takes over.

For a set line length, the lower the M-to-Air ratio the lower the press/vac requirements to drive the system. Thus a 0.2 could easily be driven by a fan instead of a blower. This is not a very efficient method though. Consider just the size of the filter to handle such a small amount of product.

For the application you described, a blower system would be the efficient and economical. The normal process is to select a line size, then choose an airflow to put you in the correct velocity range then verify the resulting pressure drop is within the operating range of the equipment.

Mantoo have given you a good start with the 2.5" line, but I would probably not go any smaller. In fact a larger line would be more conservative with very little impact on cost or efficiency. ■

Thank you Mr. Thorn,

The International Squirt Gun Cat Herders Association-

ISGCHA thanks you for your response.

lzaharis ■

Dear Pablo and others,

In this thread, the physics of pneumatic conveying are discussed, based on the questions “How high should the loading ratio be? What should be the velocity? What pipeline size should be applied? Which blower is necessary? What is the pressure drop? “

Consider the installation as a system consisting of a feeder, a pneumatic conveying installation and a receiver/separator section.

A feeder just brings in the material into the air flow at the beginning of the pipeline. The feeding rate can be chosen freely.

The pneumatic conveying installation, in its turn, consists of an air pump and a pipe line.

The air pump, together with the pipeline air resistance, requires a certain pressure at the air pump, resulting in a pressure drop over the pipeline in case the feeding rate is 0.

Then imagine that the feeding starts at a low rate.

Then, the material has to be accelerated and lifted to a different level. Also particles start to collide to each other and against the wall (and pipe bends) And the particles have to be kept afloat.

All these losses result in a pressure drop of the air.

The air pump wants to maintain the airflow and generates a higher pressure.

The higher pressure causes a lower specific air volume at the beginning of the pipeline, changing (lowering) the air velocity at the beginning of the pipeline and changing the air density.

The higher the feeding rate (higher loading ratio), the stronger these effects become. (higher pressure)

The pressure drop is a response of the system to the higher loading ratio.

In general form : dp = function ( airvolume, loading ratio, dp)

dp = pressure drop

As dp is a function of itself, it is not possible to capture pneumatic conveying in one formula.

The only way open, is to calculate the pressure drop through iteration.

This is also the reason that discussions about pneumatic conveying get confusing, especially when someone is not having a full mathematical overview of the physical events.

In this case of wood powder, where the floating velocity (reference in atmospheric pressure at 0 degrC ) is low, due to the small particle size (powser) and the low material density, the air velocities also can be kept low.

Choosing a high air flow, the loading ratio is kept low, the pressure drop is kept low (not much difference between air pressure drop and conveying pressure drop)

A fan can be a feasible choice then.

Choosing a low air flow, the loading ratio is kept high, the pressure drop is kept high

A high pressure blower is needed then.

As the pipe diameter is related to the floating velocity, a high air flow will require a larger diameter and a low air flow will require a smaller diameter.

Designing your installation, requires some performance data of existing installations, from which the parameters can be derived (calculated back).

For products as cement, fly ash, alumina, etc., these figures are known, as ther are many built installations.

For wood powder, I do not know those figures, but I would expect that it is easy to convey (if it is not sticky).

I heard of installations where wood dust is sucked up by a fan, passing the fan blades, where the impulse is added and then conveyed through the pipe.

I have not given the numerical answer to your question, but I tried to explain that pneumatic conveying is not as simple as it looks like.

If the wood powder is part of a combustion process, then the loading ratio could also be dictated by chemical parameters. (do not forget explosion hazards)

If ther are forum members, having operational performance data of wood powder installations, I would welcome those, to evaluate the conveying parameters.

Thanks in advance and

best regards. ■

Pick up velocity doesn’t change with Solid loading Ratio (SLR) but is a function of pipe diameter. As the solid flow rate increases so does the system pressure drop, the air flow rate is increased to keep the pick up velocity same this results in higher exit velocities. I am not sure the velocities you have given are they pick up or exit?

Wood powder is explosive and smouldering fires in woodwork factories are rather common. The SLR at which the pneumatic conveying system normally runs is way above the maximum explosive limit. These concentrations can only fall within the explosive limit only during start /stop and only for a very short time. Explosion of wood dust in pneumatic conveying line is highly unlikely, but all pipe work should be earth properly. Further more the flame front doesn’t travel in such small diameter pipes for material in Kst class 1. ■

Dear Pablo,

For a pressure of 70 kPa # 0.7 bar a Roots type blower will be a good choice.

Visit www.aerzen,de

For your fan, search the web for -high pressure centrifugal fans- or -high pressure centrifugal ventilators-.

Your MFR seems to be in (kg/sec wood powder)/(m3/sec air).

Normally the Solids Loading Ratio (SLR) in (kg /sec wood powder)/(kg/sec air).is used in calculations. The SLR is considered constant along the pipeline, while the MFR is varying along the pipe line length, due to the pressure gradient.

What are the wood powder data ?

Your end air velocity (between 24 ¡V 30 m/sec) seems far too high for wood powder (small particle size + low material density --> low suspension velocity)

Also your calculated pipeline range (1.5¡¨ to10¡¨) seems far too wide for a good judgement.

Attached, a spreadsheet evaluation of your calculations.

Happy New Year ■

Sorry,

I forgot the file

Attachments

calculationevaluation (ZIP)

■

Dear Pablo,

MFR - SLR:

If you have a look in the spreadsheet in the cells K6 to K9.

The figures in these cells are calculated as (kg/secMAT)/(m3/secAIR) and correspond to the figures you mentioned in this thread.

Regarding the air velocity, I compare this to cement conveying (my daily work), where end-velocities of 12 m/sec are sufficient.

Wood POWDER, far easier to convey, should easily be conveyed at lower speeds.

What particle size distribution and density are you talking about?

best regards

teus ■

Dear Pablo,

Thanks for your prompt reply and the graph.

I assume that the 425 kg/m3 is the bulk density.

Using :

v(suspension) = SQRT(4/3 * d/dens.air * dens.wood/cw)

assuming dens.wood = 850 kg/m3 and a cw=.05 the calculated suspension velocity becomes:

v(suspension) = SQRT(4/3 * .003/1.2 * 850/.05) = 7.5 m/sec

A factor 2.66 times on the suspension velocity gives approx 20 m/sec

It is the particle size of 0.003 m that causes the high velocity. (I assumed that powder has a much smaller particle size)

This part is now understood and solved.

In one of your calculations you mention a pipe diameter of 1.5 inches.

That means that 12.7 particles can form the diameter of that pipe, indicating that a blockage is likely to occur.

Having set the air velocity, the next step is to determine the relationship between pipediameter (which now is also airflow), SLR and pressuredrop.

There should be one diameter, SLR and pressuredrop combination that gives a solution for the lowest energy consumption at the given capacity.

All for now ■

Dear Pablo,

I modeled your pipeline and then calculated the pneumatic conveying performance

for 5 different pipe diameters.

The calculations are based on the following assumptions:

v(suspension) = 7.5 m/sec

wood density = 850 kg/m3

Product loss factor = 0.01 (constant) (comparable to hard cereals and seeds)

THESE CALCULATIONS ARE NOT SUPPORTED BY FIELDTESTS AND HAVE TO BE CONSIDERED AS A ROUGH INDICATION (EDUCATED GUESS)

However the figures show how pneumatic conveying works.

Capacity = 5 tons/hr

pipe dia------Q----------SLR--------v-begin----------v-end--------pressuredrop

inch---------m3/sec-------------------m/sec------------m/sec------------bar--------------kWh/ton

2.5----------0.057-------19.54--------12.8-------------21.7-----------0.7632-------------0.98

3------------0.0866-------12.86-------15.4--------------23.4----------0.5727-------------1.25

4------------0.162---------6.87--------18.3-------------24.1-----------0.3578-------------1.85

5------------0.253---------4.4----------19.7-------------23.9-----------0.2527-------------2.53

6-----------0.365----------3.05---------20.7------------23.8-----------0.1953-------------3.36

A rootstype blower is required for the above calculated installations

Bigger pipe diameters will result in lower pressures, where a fan can be applied, but the energy consumption will become higher and higher.

If you have other configurations to calculate, it is now just a matter of input manipulation.

Also field test data on this wood powder can improve the reliability of these calculations.

best regards

teus ■

Dear Pablo,

I also calculated an 1.5 inch pipeline.

The capacity of 5 tons/hr is not possible under 3 bar(o)

The airflow resistance with no product present is already 0.65 bar(o)

From the table you can read that the difference in capacity between 2.5 bar(o) and 3.0 bar(o) is only 300 kg/hr

If set at 2.5 bar(o), an increase of 300 kg/hr (7%) will cause a pressure rise of 0.5 bar (20%) to 3.0 bar(o)

If the screw compressor, which is needed for this pressure, starts then relieving air, a blockage is imminent.

Table

Pablo wopo d

Pressure discharge Wood Powder 3 mm

Convey length = 95 m

Nu of bends= 5

Pump vol = 0.0193 + 0.0193 m^3/s

q-convey = 0.019382 m^3/s

Dia begin = 38.1 mm Dia end = 38.1 mm

Pipevolume = 0.1 m^3

------------Pipeline

Press. -------Cap------. mu----v-begin----v-end-------kWh/ton------- res.time

3.000-------- 4.6----------54 -----7.0---------23.7---------1.06-------------13.23

2.750--------4.5---------- 53------7.4---------23.6---------1.00-------------13.59

2.500--------4.3 ----------50------7.6---------23.6---------0.96-------------13.92

2.250--------4.0---------- 46------8.0---------23.2---------0.95-------------13.98

2.000--------3.6 ----------42------8.3---------22.8---------0.94-------------13.94

1.750--------3.2---------- 37------8.8---------22.4---------0.94-------------13.81

1.500--------2.7 ----------32 -----9.3---------22.0---------0.96-------------13.62

1.250--------2.2 ----------26-----10.0--------21.5---------1.02--------------13.36

1.000--------1.7--------- -20-----10.9--------21.0---------1.14--------------13.00

Empty pipeline pressure ...= 6517.9 mmWC (0.652 bar(o))

(Figures are indicative only. You should not build an installation, based on these calculations) ■

Hi Pablo,

I've been following all posted replies and found invaluable information in all of them. I just want to point a few things regarding wood sawdust/powder conveying:

1) Plenty of people make pneumatic conveying systems in the region, but very few really know what they are doing. The majority, use general purpose data as the one you've got from a fan manufacturer. So, any contractor with poor engineering background, will always stand on the safe side of the matter. Extremely dilute phase conveying is the most common one as for SLR of 0.2 and lower, air only data is OK for pressure drop calculations and pipeline and blower sizing.

2) In sawmill facilities (and many other industries) you need to separate particle capture processes from conveying to silo stages. You will need very high air velocities (in as much as 34m/s) to change the momentum of blasted particles from the moulder knives to inside a pipeline through adequately designed hoods. Once you've got the particles inside the pipeline then you may want to switch from a high air volume demand system to an energy efficient one. There is where fun begins!

3) Plenty of care has to be taken when an energy efficient system is to be designed. I have witnessed cases where the presence of a mix of sawdust, shavings and powder have dismissed all calculations based on bulk density, particle size and conveying velocities taken from hanbbooks. Manufacturers will supply data suitable for their equipment, so it is very unlikely a fan manufacturer will show up with high SLR information.

4) Regarding your question about power unit suppliers, you may want to visit Chicago Blowers website (they are present in Argentina), Repicki for root type blowers (also in Argentina) and also the very well know: Sullair, Atlas Copco, Compair and Kaeser piston & screw compressor companies, all present in the region.

5) Green sawdust is a sticky material, you should not risk in building a too small pipeline diameter conveying system based only in others experience or in technical literature. You should use experimental data to confirm results with your own material. In Chile you have the chance to contact Eng. Francisco Cabrejos, one of the "gurus" in the matter who works for Jenike & Johanson in Via del Mar and has a test rig at the University of Via del Mar.

6) This final point goes for Teus:

Hi Teus!, would be nice if you could enlighten a little more your point about maximum particle size / pipe diameter to prevent blockages.

And, which procedure do you recommend to iterate in your calculations to get v-end in each pipe section? (as end density depends on pressure drop, which depends on velocity, etc, etc).

Thanks you all,

Eng. Walter Lusiardo

PRAXIA

Consulting Services for the Wood Industry

URUGUAY ■

dear Walter,

Before my remark about particle size/pipe diameter takes on a life of its own, do not take that too seriously.

When I saw the small pipe diameter of 1.25 inch # 31.75 mm, another remark crossed my mind.

Somewhere in the past, I read a study paper about hopper outlets, where was stated that the opening should be at least 10 to 15 times the particle size, to prevent a blocked outlet.

(Like when too many people are trying to get through a too small door)

Places where bigger particles can gather in a small opening they can tend to block (bridging or arching)

May be my associative level is a bit too high.

About the iteration of pneumatic conveying calculations, please read the following passage

from my efforts (1982-today) to build a pneumatic conveying theory description and calculation algorithm

quote

From the original (start) conditions, the changes in the particle- and gas conditions are calculated for a time period of dt.

Using the average velocity over the period dt, the covered length dLn can be calculated.

At the end of this calculation the energy acquired by the particles and spent on losses are equal to the expansion energy of the carrying gas.

By adding those changes to the begin conditions at location 1, the end conditions at location 2 can be calculated for the particles as well as for the gas.

From there, the calculation is repeated for the next interval of time dt (and length dLn+1), covering the distance from location 2 to location 3.

The output of section dLn is used as the input for section dLn+1.

This procedure is executed until the end of the whole installation is reached.

All the conditions at the intake of a pneumatic conveying system are known.

Therefore the intake is chosen as the start of the calculation.

In vacuum- and pressure pneumatic conveying calculations, the used product properties are identical.

The only difference is the mass flow generated by a compressor in vacuum mode or pressure mode.

The calculation result should be the capacity at a certain pressure drop.

However, one of these values is not known.

To calculate the capacity, the pressure drop must be set and the capacity must be iterated from a guessed value.

The calculated pressure drop from a “wrong” guess will be different from the set pressure drop.

Therefore the capacity guess is renewed in such a way that the new, to be calculated, pressure drop, approaches the set pressure drop.

This iteration ends when the calculated pressure drop equals the set pressure drop.

The capacity that resulted in this pressure drop equality is the wanted value.

(Input and output are consistent)

unquote

Thanks for the attention ■

Dear Teus,

thank you for such a prompt and detailed reply. No doubt pneumatic conveying is more than art, it's passion!

We shall be moving a little apart from Pablo's original post but for the sake of knowledge it's worthwhile doing so. This is why I will add a few more questions:

1) If, for instance, a gas conveying velocity of 10m/s is stablished which would you consider a reasonable time interval: 1 sec, 0.5 secs, other?

2) I find a similar approach in Mr. Tim Agarwal's simplified dilute phase calculation method, when he divides the pipeline in short sections. You talk about time he does about distance, at the end of the day, is it the same thing?

3) It would be great if you could fully enligthen the iteration process with an example (if it's an excel file, much better!) as you have more than 20 year experience but for legos it doesn't look that simple!.

Thanks again for your very kind cooperation,

Walter ■

dear Walter,

1)My computer program (in Q-basic) calculates in the time domain

with a dt of 0.001 sec.

This results in pipe sections of approx. 10 * 0.001 = 0.01 m

In regions of high acceleration, the program switches to 0.0001 sec, in order to prevent too high calculated velocities.

2)dL = v * dt

v is taken as the product velocity (NOT the gas velocity)

3)Calculation of previous wood powder installation for 5.5 tons/hr with a 4 inch pipeline and 0.162 m3/sec gas flow

------------------capacity----------guessed pressuredrop ------calculated pressuredrop

1st calculation-------5.5-------------------4000-------------------------3535

2nd calculation-------5.5-------------------3535-------------------------3578

3rd calculation-------5.5-------------------3574-------------------------3571

4th calculation-------5.5-------------------3572-------------------------3572

5th calculation-------5.5-------------------3572-------------------------3572

When the guessed pressure drop equals the calculated pressure drop, the calculation is finished and the output is consistent.

What does the word –legos- mean ?

have a nice day

teus ■