How many bends are there in the first 50 m ??? ■

Dear kevinjcg

I calculated your test installation and found the following result:

pressure (baro) tons/hr

2,5 13

2,25 13

2,00 12

1,75 11

1,50 10

1,25 9

1,00 8

0,75 6

0,50 4

You can notice that the pressure increase significantly although the flow rate of the flyash changes very little.

This means that a little bit of fly ash extra into the air flow will increase the pressure very much.

By the preesure increase, the air velocity will eventually drop below the suspension velocity and a blockage is created.

That is presumably what happened in your installation.

OVER FEEDING THE PIPELINE

Also opening the outlet valve at 0,4 MPa (4 baro if I am correct) will overfeed the pipeline instantaniously and lead to a blockage in the first section after the outlet.

You may try to start conveying at abeginning pressure of f.i. ,5 baro and keeping the convey pressure below 2 baro, by controlling the feeding.

succes fro

Teus ■

How do you control the discharge rate of fly ash from your silo (blow tank)?

If all of your air is directed into the bottom of the silo to fluidise the fly ash you will only ever block such a long pipeline. Part of the supply air must by-pass the silo and be fed into the conveying pipeline as close to the top of the silo as possible. This by-pass, or supplementary, air will dilute the high concentration of fly ash leaving the silo to a value that will be possible to convey through the length of the pipeline with the pressure available. The discharge rate of any material from any blow tank is infinitely controllable by means of this air proportioning.

The concentration of the material discharged from a blow tank must be balanced by the conveying line pressure gradient. Too high a concentration and you will block the pipeline. Too low a concentration and it will convey, if the conveying air velocity is high enough, but you will not utilise all the air supply pressure that you have available.

If your pipline is 410 m horizontal and 45 m vertically up, with bends, and your air supply pressure is 100 psig your available pressure gradient for conveying will be about 12 mbar/m. A pressure gradient of 12 mbar/m should convey your fine fly ash at a solids loading ratio of about 50 and so the volumetric flow rate of your air supply should be satisfactory. You should achieve about 15 tonne/h through your pipeline with an air supply pressure of 0.7 Mpa. ■

Dear kevinjcg

I recalculated your installation from 7 bar(o) down and found the following results :

Convey length = 455 m Nu of bends= 8 Pump vol = .1 + .1 m^3/s

q-convey = 0.10 m^3/s Dia begin = 80 mm Dia end = 100 mm

Kettle volume = 2.0 m^3 Cement vol.= 1.5 m^3 C-Vessel = 500 tons/hr

Kettle cement content = 1.0 tons Pipevolume = 2.71 m^3

Pipeline Syst. /Nr of Req.kettle

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

7.000 21 49 15 / 10.3 6.500 21 47 15 / 10.3 6.000 20 46 15 / 10.3 5.500 19 44 15 / 10.3 5.000 19 43 15 / 10.3 4.500 18 41 14 / 9.6 4.000 17 39 14 / 9.6 3.500 16 36 13 / 8.9 3.000 15 34 13 / 8.9 2.500 13 31 11 / 10.4 2.250 13 29 11 / 10.4 2.000 12 28 10 / 9.5 1.750 11 26 10 / 9.5 1.500 10 24 9 / 8.5 1.250 9 22 8 / 7.6 1.000 8 19 7 / 6.6 0.750 6 16 6 / 5.7 0.500 4 11 4 / 3.8

Pneumatic conveying is possible at 1 bar(o).

The loading ratio is calculated as 49 kg/kg, which matches the value of Dr Mills

However this calculation assumes all the air going through the pipeline from the beginning to the end.

No air is by passed to a point further in the pipeline, as this would reduce the velocity and also increasing the local loading ratio before that injection point .

This will cause higher pressure drops and eventually choking

As only a small change in material feeding causes a big change in conveying pressure, the accuracy of the feeding control is very important

The cause of the choked pipeline must be too much material in the pipeline.

This can only be controlled by bypassing the convey air around the kettle.

This can be done by a manual valve as this is only a test installation.

In practice it should be done with an automatic regulating valve based on compressor pressure

Try to start conveying as follows.

open the manual by pass valve

pressurize the kettle to approx . 0,5 bar(o) (0,05 Mpa)

open the outlet valve

wait for approx 2 to 2,5 minutes (residence time of a particle can go up to 106 sec) until stable conveying occurs.

If the convey pressure can be increased gradually close the by pass valve.

After each new setting wait for stable conveying.

This can also be done with the opening pressure for opening the outlet valve

Also notice the difference in energy consumption/ton at different convey pressures

(this calculation is based on the use of a screw compressor)

Success

teus■

Sorry for an error in my message.

It should read " Pneumatic conveying is possible at 7 bar(o)"

------------

teus ■

dear kevinjcg

starting conveying is assumed on an empty pipeline .

As soon as conveying starts there is airflow in the pipeline mixed with fly-ash

The maximum pressure in the blow tank is reached when the complete pipeline is filled with flowing fly-ash mixed with air with the maximum, regulated, loading ratio.

To start the next cycle of discharging , the pipeline must be purged empty, until all the fly-ash is removed.

This can be monitored on the pressure in the blow tank by reaching the “empty pipeline pressure”

Discharging at low pressure means a low loading ratio, little product resistance and because of the low pressure, higher air velocities.

As this is a very long pipeline, designed for a high pressure, this installation will always be very sensitive for choking and will require a very accurate by-pass control.

Note: your installation is designed for 7 bar(o), which should require a step up in diameter of approx. 1,7x instead of 1,25x, for optimal pneumatic conveying This to keep the air velocity just above the sedimentation velocity at any location in the pipeline

Pneumatic conveying might be a simple principle, it is not a simple art.

PS I tried to insert a drawing in this message, but that failed.

succes ■

Dear kevinjcg

you can start conveying at any (low) pressure you want. If it results in a too low continuous discharge pressure at a certain setting of the by pass valve (called “extra air”) you can increase f.i the opening pressure for the next run.

May be you have to open the “extra air valve” again a little.

The stability of the pneumatic discharge-conveying is a result of the combined settings of opening pressure and extra air.

Start with a low opening pressure (f.i. 0,5 bar(o)) and an open “extra air valve”.

You are not by-passing pressure around the blow tank, but you are bypassing air-volume.

The less air goes through the blow tank, the less fly-ash is taken from the tank and the lower the loading ratio will be, resulting in a lower convey pressure.

The intention is to dilute the concentration of the mixture to prevent choking.

The trick is to find the maximum loading ratio, whereby the pneumatic conveying works without choking and keeping it that way.

An automatic “extra air valve” can do that for you

Success ■

Your blow tank with no bypass air sounds like a high pressure single plug type system but then your conveying pipe diameter is bit small and batch size bit high. You are on the right track now ,bypass air on the blow tank is essential for a this type of system. Avoid putting bypass air in conveying line it just increase the velocity / wear in the system and could also reduce solid flowrate.

In the mean time u could try conveying flyash in single plug mode but using smaller amount per batch. Start with 175 kg/ batch at 3-4 barg and keep increasing/ decreasing the batch size until the optimum point is reached. This system is normally used with a doom valve as it requires higher number of cycles + there is higher wear in the system due to blowdown. But sicne it is a test plant no harm in trying !!! ■

If you put bypass air after the blowtank it will reduce the solids flow rate which will result in lower system pressure drop. You will then not need injection points on the conveying line. In any case uncontrolled amount of air injection into the conveying line is always a big NO. ■

dear kevinjcg

please can you give the following information when you are injecting supplementary air into the pipeline at f.i. 300m after the blow tank.

-air volume through the blow tank

-air volume by passing the blow tank

(air volume through the blow tank + air volume by passing the blow tank = air volume entering the pipeline at the beginning)

-air volume injected at 300 m distance from the blow tank

Using all the data already known this enables me to calculate the effect of such a trial.

best regards ■

dear kevinjcg

As you can see from previeous calculations, I assumed 0,1 m3/sec (6 m3/min) free air delivery from the beginning of the pipeline to the end,

If you set the pressure of the by passes at 6 bar, not knowing the injected air volumes, I cannot calculate the system because the velocities, loading ratio, product resistence etc cnnot be calculated.

Airflow should be known at every location in the pipeline.

My advise is to start testing the installation with the available air used from the beginning of the pipeline.

do not give up

all for now ■

dear kevinjcg

yes you must be able to convey with only a (controllable) air by pass around the blowtank.

The total amount of air flow I already indicated in my answer dd 22nd of October 2005.

0,1 m3/sec # 6 m3/min. (based on your earlier information

The capacity is given in the table in that answer.

Maybe you should type this table out on your computer and repair the table, then it will be clear(er)

13 tons/hr at an average pressure of 5 bar is expected.

Use your common sense during trials s.a. :

lowering loading ratio decreases pressure drop and capacity and reduces the chance of choking. It also increase the velocities-

success ■

Originally Posted by Dr David Mills

How do you control the discharge rate of fly ash from your silo (blow tank)?

If all of your air is directed into the bottom of the silo to fluidise the fly ash you will only ever block such a long pipeline. Part of the supply air must by-pass the silo and be fed into the conveying pipeline as close to the top of the silo as possible. This by-pass, or supplementary, air will dilute the high concentration of fly ash leaving the silo to a value that will be possible to convey through the length of the pipeline with the pressure available. The discharge rate of any material from any blow tank is infinitely controllable by means of this air proportioning.

The concentration of the material discharged from a blow tank must be balanced by the conveying line pressure gradient. Too high a concentration and you will block the pipeline. Too low a concentration and it will convey, if the conveying air velocity is high enough, but you will not utilise all the air supply pressure that you have available.

If your pipline is 410 m horizontal and 45 m vertically up, with bends, and your air supply pressure is 100 psig your available pressure gradient for conveying will be about 12 mbar/m. A pressure gradient of 12 mbar/m should convey your fine fly ash at a solids loading ratio of about 50 and so the volumetric flow rate of your air supply should be satisfactory. You should achieve about 15 tonne/h through your pipeline with an air supply pressure of 0.7 Mpa.

KA Dr David mills

Dear sir

I am doing one practical project in dense phase conveying system for indian fly ash. I was wondering about the advantage of bypass pipe technology in dense phase conveying syste. Apparantly in indian context , in almostall power plant teh pipe-inpipe techology is gaining its importance for fly ash pneumatic conveying system . In thsi technology where in there is a small pipe (having the hole at regular interval) air flow is through this small pipe

Please comment on below and rectify in my understanding .

My understanding is , teher is energy saving comparively for a reason as teh , pressure required for a short plug is less than pressure required for long plug. In pipe in pipe technology , since the air is being inserted at certain distance which maintains teh plug length distrubution

2) Secondly in pipe in pipe the permiability and air retention will be higher for a reason i dont no . pelase explain

3) Indeed there is reduction in area of conveying pipe and incerase in wall friction , but it seems teh advantage in this case plays a stronger influence over teh disadvantages of reducing the area. Please comment

Sir other than this , please explain why this technology been adopted and what is teh major significance in regard of pressure gradiant and air flow ■

Originally Posted by Teus Tuinenburg

dear kevinjcg

starting conveying is assumed on an empty pipeline .

As soon as conveying starts there is airflow in the pipeline mixed with fly-ash

The maximum pressure in the blow tank is reached when the complete pipeline is filled with flowing fly-ash mixed with air with the maximum, regulated, loading ratio.

To start the next cycle of discharging , the pipeline must be purged empty, until all the fly-ash is removed.

This can be monitored on the pressure in the blow tank by reaching the “empty pipeline pressure”

Discharging at low pressure means a low loading ratio, little product resistance and because of the low pressure, higher air velocities.

As this is a very long pipeline, designed for a high pressure, this installation will always be very sensitive for choking and will require a very accurate by-pass control.

Note: your installation is designed for 7 bar(o), which should require a step up in diameter of approx. 1,7x instead of 1,25x, for optimal pneumatic conveying This to keep the air velocity just above the sedimentation velocity at any location in the pipeline

Pneumatic conveying might be a simple principle, it is not a simple art.

PS I tried to insert a drawing in this message, but that failed.

succes

To Mr Teus

Sir, there is a distance limitation in vaccum conveying system, what is that limitation (may be in terms of distance) I got to size the fly ash conveying system in vaccum mode system for 35TPH at distance of 250mtr 5 bends. Now if i maintain the pickup and terminal velocity (16m/s & 27m/s respectively) to acceptable level and resulting volume with pressure gradiant is x m3/hr @ 16in HG. Now since teh same volume and pressure rating of vaccum equipment is available in the market , then why shouldnt i gp for it? ■

Dear kj,

In vacuum conveying, the available pressure drop is 1 bar.

However, at 1 bar vacuum, the air mass flow becomes zero and then there is no pneumatic conveying possible.

Vacuum cement unloaders are designed for a vacuum of approx. 0.75 bar, which is close to the choking condition.

The vacuum is related to the vacuum conveying length, as the length determines the vacuum.

The maximum vacuum conveying length is therefore a function of the design vacuum.

I designed your system for 4 pipe sizes and found the following:

Pipe 8”/10”

Vacuum pump 0.75 m3/sec

32 tons/hr – 5500 mmWC vacuum

Pipe 10”

Vacuum pump 0.75 m3/sec

35 tons/hr – 2678 mmWC vacuum

Pipe 10”/12”

Vacuum pump 1.0 m3/sec

35 tons/hr – 2205 mmWC vacuum

Pipe 12”/14”

Vacuum pump 1.4 m3/sec

35 tons/hr – 2606 mmWC vacuum

If the equipment is available on the market, as you state, then there is no reason to ignore that.

Have a nice day

Teus ■

Originally Posted by Teus Tuinenburg

Dear kj,

In vacuum conveying, the available pressure drop is 1 bar.

However, at 1 bar vacuum, the air mass flow becomes zero and then there is no pneumatic conveying possible.

Vacuum cement unloaders are designed for a vacuum of approx. 0.75 bar, which is close to the choking condition.

The vacuum is related to the vacuum conveying length, as the length determines the vacuum.

The maximum vacuum conveying length is therefore a function of the design vacuum.

I designed your system for 4 pipe sizes and found the following:

Pipe 8”/10”

Vacuum pump 0.75 m3/sec

32 tons/hr – 5500 mmWC vacuum

Pipe 10”

Vacuum pump 0.75 m3/sec

35 tons/hr – 2678 mmWC vacuum

Pipe 10”/12”

Vacuum pump 1.0 m3/sec

35 tons/hr – 2205 mmWC vacuum

Pipe 12”/14”

Vacuum pump 1.4 m3/sec

35 tons/hr – 2606 mmWC vacuum

If the equipment is available on the market, as you state, then there is no reason to ignore that.

Have a nice day

Teus

Sir

1) As per your calculation vacum pump for 0.75m3/sec @5500mwc is available in market . But to get better understanding , what is the exactly disadvantage in vaccum conveying system , if the equipment is available. I know and can make a qualified guess of vaccum having disadvantages in long distance , but in what sense?

2) For a calculated suspension velocity as you taught , how to calculate pickup velocity in vaccum conveying system. Please specy the factor to be used ■

Dear kj,

1)Compare the vacuum calculations with the pressure calculation for the same project.

250 m conveying length

5 bends

Capacity 35 tons

pipe size 6” (vacuum 10”)

Air displacement pump 0.3 m3/sec at 1 bar (vacuum 0.75 m3/sec at 0.2678 bar vacuum)

power (indication) = airflow*pressure ratio 0.3*2=0.6 (Vacuum 0.75 * 1/0.2678 = 1.027)

The advantage of the pressure system is:

smaller pipe diameter

smaller sized air pump

less energy consumption per conveyed ton.

This is caused by the lower velocities in pressure conveying as a result of the denser air.

2)The pick up velocity in a vacuum system is calculated as:

Air flow/pipe area = Pump displacement*(1-vacuum)/(3.141593/4 * D^2)

The pick up velocity in a vacuum system (at the beginning of the pipe) is initially chosen as 6@7 times the suspension velocity and checked in the calculation.

Have a nice day

Teus ■