Most of the tankers are 2 barg rated so max. conveying pressure in a tanker is about 1.8 barg. In Europe most of the tankers have a tractor based screw compressor capable of 2 barg. They come in three sizes approx. 400, 600, 900 m^3/hr depending on the product. Since roots blower has a limit of 1 barg it will not be able to achieve 30 min target but can be used if time is not critical. Reciprocating compressors are can give 8 -10 bar and will have to be regulated to 2 bar. They can work but then you will use extra power.

Type of tanker is also very important if it’s a tipping tanker it usually discharges faster then a belly tanker. Last but not least the conveying pipe distance and number of bends also have massive influence. Typically 30 ton cement tanker unloading into a silo takes less then 30 min. ■

dear mr bharat

please can you give the following additional information

-pipeline geometry , s.a. length horizontal, length vertical, number of bends.

optionally

-if there is an existing pipeline also the diameters.

-if you have a bulktruck in mind, also the compressor characteristics (volume and pressure)

These data will enable me to calculate the installation and the expected performance.

best regards ■

Assuming it is a 4" pipe line with 19m^3/ min the exit velocities will be 35-36 m/s. These are too high velocities to convey at high SLR + don't forget fly ash is highly abrasive. You should be looking at 8 -10 m^3/min range to give you approx 15-18 m/s

exit velocity. ■

600 m3/hr and 2 barg pressure will be a good point to unload a tanker. Do not worry about the system pressure it is reasonably easy to control on tankers and general rule of thumb is higher pressure gives u shorter unloading times. But then there is a limit of max. phase density for every product.

At 30 min time and these conditions the phase density will be 85. In my experience not every type flyash can do this phase density so with out testing it is difficult to say yes or no. ■

dear Mr bharat

based on your info, I made the following calculation :

As you did not specify any pipe diameters, I assumed that the pipe diameter is still free to choose.

FORUM bharat d 10-26-2005

Pressure discharge Fly-ash

Convey length = 25 m

Nu of bends= 3

Pump vol = .31 + .31 m^3/s

q-convey = 0.32 m^3/s

Dia begin = 193 mm (8”)

Dia end = 193 mm (8”)

Kettle volume = 40.0 m^3

Fly ash volume.= 32.0 m^3

C-Vessel = 600 tons/hr

Kettle cement content = 31.0 tons

Pipevolume = .73 m^3

Tabel 1

Press.PipelineCap.Syst. Cap. mu Nr ofbulktrucksReq.kettle factv-beginv-end kWh/ton .res.time

2.5001961381474.76874.4 12.4 0.34 4.96

2.2501831291404.5 5954.6 12.3 0.34 4.91

2.0001661171304.1 5004.7 12.1 0.33 4.80

1.750 1531081223.9 4224.9 12.0 0.32 4.68

1.500 141991143.6 3535.2 11.9 0.31 4.55

1.250 125891043.3 2835.5 11.7 0.30 4.35

1.000 10977912.9 2206.0 11.6 0.29 4.09

0.750 9265792.5 1616.8 11.4 0.29 3.79

0.500 7150642.0 1087.7 11.2 0.29 3.46

0.250 4431411.3 559.0 11.0 0.35 3.10

Maximum system capacity : 140 t/hr 2.26 bar(o)

From this table you can read that 4 bulk trucks of 30 tons each can be unloaded in 1 hr.

May some time has to be spend on changing bulk trucks and necessary connecting.

(Or use 2 pipelines or 2 inlets with valves on one pipeline)

The question at this moment is, will the bulk truck release enough fly ash per unit of time

to cope with the pipeline capacity of approx 175 tons/hr at approx. 2,25 bar(o)

On a 6” pipeline (143 mm inner dia) I calculated approx 90 tons/hr # approx 3 bulk trucks per hour

I assumed a “normal” type of fly ash of 30 micron, which can deviate from your fly ash.

Take care of the maximum allowable pressure in the bulk truck tank.

Did I help you ?

best regards ■

dear Mr bharat

Your compressor is OK

You can read from the table the 5th figure in each row, which represents the number of trucks that can be unloaded in 1 hour (time loss for swapping trucks NOT included)

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

For good understanding, I assumed that you will use an external compressor to the bulk truck, because regular bulk trucks have smaller compressors (may be up to 12 m3/min)

The outlet of the bulk truck should be at least 6” (which is normal).

A pipe adaptor from 6” to 8” should be used right after the bulk truck

Do not forget a 40 m2 filter + fan of 80 m3/min at 450 mmWC + safety over- and under pressure valve on the silo roof.

Also leveltesters should be installed, which will cause a butterfly valve to close the pipeline to prevent overfilling.

Note:

if you are interested in a full calculation for a 4” pipeline I will be more than willing to do that for you

success ■

dear mr Bharat.

You are welcome. The time spent on a calculation is only a few minutes of personal computer time.

The first calculation I made is for :

compressor 19 m3/min at an 8” pipeline giving the capacity at x bar(o)

This new calculation is for :

compressor 19 m3/min at a 4” pipeline giving the capacity at x bar(o)

FORUM bharat d 10-28-2005

Pressure discharge Fly-ash

Convey length = 25 m

Nu of bends= 3

Pump vol = .31 + .31 m^3/s

q-convey = 0.32 m^3/s

Dia begin = 96 mm (4”)

Dia end = 96 mm (4”)

Kettle volume = 40.0 m^3

Cement vol.= 32.0 m^3

C-Vessel = 500 tons/hr

Kettle cement content = 31.0 tons

Pipevolume = .18 m^3

Pipeline Syst. /Nr of Req.kettle

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

2.500 - 47 - 33 - 44 / 1.4 - 167 - 14.0 - 44.6 - 1.44 - 1.51

2.250 - 45 - 32 - 41 / 1.3 - 146 - 15.0 - 44.4 - 1.38 - 1.41

2.000 - 42 - 30 - 39 / 1.2 - 126 - 16.2 - 44.3 - 1.33 - 1.33

1.750 - 39 - 28 - 36 / 1.1 - 108 - 17.6 - 44.3 - 1.28 - 1.24

1.500 - 35 - 25 - 33 / 1.0 - 90 - 19.3 - 44.0 - 1.23 - 1.15

1.250 - 32 - 23 - 30 / 0.9 - 73 - 21.5 - 43.9 - 1.18 - 1.05

1.000 - 27 - 20 - 26 / 0.8 - 56 - 24.1 - 43.7 - 1.16 - 0.96

0.750 - 23 - 16 - 22 / 0.7 - 40 - 27.6 - 43.6 - 1.16 - 0.86

0.500 - 16 - 11 - 15 / 0.4 - 24 - 31.9 - 43.3 - 1.31 - 0.77

0.250 - 6 - 4 - 6 / 0.1 - 8 - 37.1 - 42.9 - 2.47 - 0.69

Possible choking of pipeline Print-out table ? y/

Maximum system capacity : 44 t/hr 2.5 bar(o)

Aproxx. 1,2 bulk trucks can be unloaded per hour with this installation.

A 19m3/min compressor is a good match with an 8” pipeline

A 19m3/min compressor is a NOT good match with an 4” pipeline (air speed is too high)

Notic the air velocities at the beginning of the pipe line and at the end.

I also made the calculation for a compressor of 9 m3/min at a 4” pipeline with the following result :

FORUM bharat d 10-28-2005

Pressure discharge Fly-ash

Convey length = 25 m

Nu of bends= 3

Pump vol = .15 + .15 m^3/s

q-convey = 0.15 m^3/s

Dia begin = 96 mm

Dia end = 96 mm

Kettle volume = 40.0 m^3

Cement vol.= 32.0 m^3

C-Vessel = 500 tons/hr

Kettle cement content = 31.0 tons

Pipevolume = .18 m^3

Pipeline Syst. /Nr of Req.kettle

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

2.500 - 47 - 70 - 40 / 1.2 - 165 - 7.0 - 22.4 - 0.68 - 3.03

2.250 - 44 - 66 - 38 / 1.2 - 145 - 7.4 - 22.3 - 0.66 - 2.89

2.000 - 41 - 62 - 36 / 1.1- 126 - 7.9 - 22.2 - 0.63 - 2.75

1.750 - 39 - 58 - 34 / 1.0 - 107 - 8.5 - 22.0 - 0.60 - 2.60

1.500 - 36 - 54 - 32 / 1.0 - 90 - 9.3 - 21.9 - 0.58 - 2.43

1.250 - 32 - 49 - 29 / 0.9 - 74 - 10.2 - 21.7 - 0.55 - 2.26

1.000 - 29 - 43 - 26 / 0.8 - 58 - 11.3 - 21.6 - 0.52 - 2.08

0.750 - 24 - 37 - 22 / 0.7 - 44 - 12.8 - 21.4 - 0.50 - 1.89

0.500 - 19 - 29 - 18 / 0.5 - 30 - 14.8 - 21.2 - 0.50 - 1.71

0.250 - 12 - 18 -11 / 0.3 - 15 - 7.5 - 21.0 - 0.60 - 1.52

Print-out table ? y/

Maximum system capacity : 40 t/hr 2.5 bar(o)

Conclusion is that 19 m3/min on a 4” pipeline results in approx. the same capacity/hr as a compressor of 9 m3/min on a 4” pipeline.

The smaller compressor is favorite on a smaller pipeline

Pneumatic conveying at proper velocities ( approx 4-8 times the local suspension velocity)

do not wear straight pipelines and even not so much on elbows.

High velocities propel particles through the boundary laye of the flow profile and can hit the wall of the pipe, causing wear as a result.

May be I am generating to much figures now, so that a proper overview of the subject becomes a little vague now.

all for now■

Dear Mr bharat,

Using a Roots blower of 900 m3/hr ( # 15m3/min # .25 m3/sec ) is no problem at all.

I recalculated your installation, but now with a diameter of 6” (150 mm)

The outlet of the bulk tanker should also be 6”.

Use a proper spring relieve valve to protect your Roots blower.

The result is for 1 bar(o) # 10000 mmWC:

FORUM REMARK : 11-17-2005

***************

PRODUKT :Fly-ash

Pipeline capacity ........= 68.39 ton/hr

System-pressure...........= 10000. mmwC

Q-pump....................= 0.26 m^3/s

Q-convey-pipe ............= 0.26 m^3/s

loading-ratio ............= 58.56 - -

Convey Length = 25 m

Number of Bends = 3 -

D-begin =150 mm

D-end =150 mm

Outlet force .. = 1481 N (dynamic)

NO BOOSTER

Ambient temperature ......= 40.0 °C

Reynoldsnumber ..[ Re ]...= 1.28 --

spec.energy-consumption...= 0.51 kWh/ton

backpressure at pipe-end..= 0. mmwC

p-accell.excl prod.rest..= 1561. mmwC

p-suspension.............= 2284. mmwC

p-lifting................= 1695. mmwC

p-airfriction............= 50. mmwC

p-productresistance......= 4233. mmwC

p-intake productcolumn...= 100. mmwC

p-intake ................= 19. mmwC

p-nozzle ................= 657. mmwC

p-filter.................= 54. mmwC

density product/air mix ..= 69.3 kg/m^3

Compr power .....= 35. kW

Mass in pipeline.= 67 kg

For different pressures the capacitycurve is :

FORUM bharat d 11-17-2005

Pressure discharge Fly-ash

Convey length = 25 m

Nu of bends= 3

Pump vol = .26 + .27 m^3/s

q-convey = 0.27 m^3/s

Dia begin = 150 mm

Dia end = 150 mm

Kettle volume = 50.0 m^3

Cement vol.= 37.3 m^3

C-Vessel = 500 tons/hr

Kettle Fly-ash content = 25.0 tons

Pipevolume = .44 m^3

Pipeline Syst. /Nr of Req.kettle

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

1.000 - 68 - 59 - 54 / 2.1 - 137 - 8.1 - 15.5 - 0.51 - 2.99

0.750 - 58 - 49 - 48 / 1.9 - 102 - 9.2 - 15.5 - 0.47 - 2.70

0.500 - 46 - 38 - 39 / 1.5 - 69 - 10.8 - 15.6 - 0.43 - 2.41

0.250 - 29 - 24 - 26 / 1.0 - 36 - 13.0 - 15.8 - 0.43 - 2.08

Maximum system capacity : 54 t/hr 1 bar(o) or 2,1 bulk trucks / hr

Using this blower on a 8” pipeline will brin an ever higher capacity, but there might be some sedimentation in the pipeline due to the lower velocity.

The pipeline has then to be purged, after the unloading, with an extra tank of air, which should be released at 1 bar(o)

Using 2 blowers makes it possible to unload 2 bulk tankers at the same time, providing you make a second pipeline.

Using 2 blowers on 1 tanker requires a bigger pipeline and also a bigger outlet of the tanker.

In both cases the rate will be higher and the energy consumption per ton will be less.

Also keep in mind that higher air volumes require also bigger dedusting filters on the silo.

So far

best regards ■

Dear Mr Bharat

A pneumatic conveying system requires a certain amount of air at a certain pressure.

This pneumatic conveying system does not care where the air comes from.

However, the type of compressor determines its characteristics and makes them more- or

less suitable for the chosen application.

For your installation an oil free roots type blower or an oil free screwcompresseor are very suitable (Due to their positive displacement characteristics).

Also look at www.aerzener.de

A blower will require (isochoric compression principle) a bit more energy than a screwcompressor ( adiabatic internal compression principle).

Of course, the compressor installation must comply with the manufacturers instructions.

I made you a list of compressor types and there properties :

(I do not claim full accuracy)

Types of convey air compressors

Roots type blower

-high volumes-/-900 m^3/min

-pressure0,5-/-1,0 bar

-vacuum0,5-/-0,6 bar

-/-,85 bar (with pre inlets)

-oil free

-isochoric compression(high power demand for high

pressure ratio)

-mainly used in low vacuum installations(low energy consumption per ton

due to the efficiency of the

pneumatic system and the low

power demand at partial load)

-used in high vacuum installations with pre inlets

Screwcompressor

-volumes-/-250 m^3/min

-pressure-/-3,5 bar

-vacuum-/-0,7 bar

-/-,85 bar (with pre inlets)

-oil free

-internal adiabatic compression followed by isochoric expansion or compression

-used in high vacuum installations with pre inlets

-mainly used in pressure discharge systems

Oil-filled screw compressor

-volumes-/-100 m^3/min

-pressure-/-6,0 bar -/- 10 bar

-vacuum-/-not used

-not oil free

-oil separators

-inlet closes at set minimum and maximum discharge pressure

-internal adiabatic compression followed by throttled expansion or isochoric compression

-used in pressure discharge systems with pressure reducer.

-rental units available as replacement for a broken compressor or as additional booster.

Vane compressor

-volumes-/-100 m^3/min

-pressure-/-2,5 bar

-vacuum-/-0,6 bar

-not oil free

-oil lubricated vanes

-inlet closes at set maximum discharge pressure

-internal adiabatic/isothermic compression followed by isochoric expansion or compression

-used in vacuum systems

-used in pressure discharge systems.

Piston compressors

-volumes-/-80 m^3/min (double acting)

-pressure-/-4,0 bar (single stage)

-vacuum-/-0,65 bar

-not oil free

-lubricated pistons

-inlet closes at set maximum discharge pressure

-internal adiabatic/isothermic compression to delivery pressure

-used in pressure discharge systems.

-low power demand.

Turbo compressors

-volumes-/-750 m^3/min

-pressure-/-5,0 bar

-vacuum-/-0,5 bar

-oil free

-diffuser vane control

-internal adiabatic compression

-used in low vacuum systems

-used in pressure discharge systems.

-complicated installation

-high energy demand

-expensive.

Water-ring compressor

-volumes-/-200 m^3/min

-vacuum-/-0,75 bar

-moisture to atmosphere

-internal adiabatic compression

-used in vacuum systems

-auxiliary equipment : water pump

-very high energy demand over full range. ■

Dear sir,

First of all thanks for your valuable knowledge sharing..

I am also having same kind of problem, In my case, for pneumatic unloading of Coarse flyash from bulker (40t) i choosen the spec. 27m3/min , 3 bar pressure, 6" dia pipe & steel silo(1000t).

please guide me whether this spec. is sufficient for unloading or shell i change volume 20m3/min @4 bar pressure?

and Can I get the excel which you are refering, for my future work.

Regards,

damu ■

Dear Teus,

thanks for your knowledge sharing.

In pneumatic conveying system how we select volume, pressure & pipe dia. etc.

example: I want to unload flyash from 40t bulker to 1000t steel silo with the help of pneumatic unloading system.

Can i get any calculation sheet to design the suitable spec. for this system.

Regards,

DAMU

Originally Posted by Teus Tuinenburg

dear mr Bharat.

You are welcome. The time spent on a calculation is only a few minutes of personal computer time.

The first calculation I made is for :

compressor 19 m3/min at an 8” pipeline giving the capacity at x bar(o)

This new calculation is for :

compressor 19 m3/min at a 4” pipeline giving the capacity at x bar(o)

FORUM bharat d 10-28-2005

Pressure discharge Fly-ash

Convey length = 25 m

Nu of bends= 3

Pump vol = .31 + .31 m^3/s

q-convey = 0.32 m^3/s

Dia begin = 96 mm (4”)

Dia end = 96 mm (4”)

Kettle volume = 40.0 m^3

Cement vol.= 32.0 m^3

C-Vessel = 500 tons/hr

Kettle cement content = 31.0 tons

Pipevolume = .18 m^3

Pipeline Syst. /Nr of Req.kettle

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

2.500 - 47 - 33 - 44 / 1.4 - 167 - 14.0 - 44.6 - 1.44 - 1.51

2.250 - 45 - 32 - 41 / 1.3 - 146 - 15.0 - 44.4 - 1.38 - 1.41

2.000 - 42 - 30 - 39 / 1.2 - 126 - 16.2 - 44.3 - 1.33 - 1.33

1.750 - 39 - 28 - 36 / 1.1 - 108 - 17.6 - 44.3 - 1.28 - 1.24

1.500 - 35 - 25 - 33 / 1.0 - 90 - 19.3 - 44.0 - 1.23 - 1.15

1.250 - 32 - 23 - 30 / 0.9 - 73 - 21.5 - 43.9 - 1.18 - 1.05

1.000 - 27 - 20 - 26 / 0.8 - 56 - 24.1 - 43.7 - 1.16 - 0.96

0.750 - 23 - 16 - 22 / 0.7 - 40 - 27.6 - 43.6 - 1.16 - 0.86

0.500 - 16 - 11 - 15 / 0.4 - 24 - 31.9 - 43.3 - 1.31 - 0.77

0.250 - 6 - 4 - 6 / 0.1 - 8 - 37.1 - 42.9 - 2.47 - 0.69

Possible choking of pipeline Print-out table ? y/

Maximum system capacity : 44 t/hr 2.5 bar(o)

Aproxx. 1,2 bulk trucks can be unloaded per hour with this installation.

A 19m3/min compressor is a good match with an 8” pipeline

A 19m3/min compressor is a NOT good match with an 4” pipeline (air speed is too high)

Notic the air velocities at the beginning of the pipe line and at the end.

I also made the calculation for a compressor of 9 m3/min at a 4” pipeline with the following result :

FORUM bharat d 10-28-2005

Pressure discharge Fly-ash

Convey length = 25 m

Nu of bends= 3

Pump vol = .15 + .15 m^3/s

q-convey = 0.15 m^3/s

Dia begin = 96 mm

Dia end = 96 mm

Kettle volume = 40.0 m^3

Cement vol.= 32.0 m^3

C-Vessel = 500 tons/hr

Kettle cement content = 31.0 tons

Pipevolume = .18 m^3

Pipeline Syst. /Nr of Req.kettle

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

2.500 - 47 - 70 - 40 / 1.2 - 165 - 7.0 - 22.4 - 0.68 - 3.03

2.250 - 44 - 66 - 38 / 1.2 - 145 - 7.4 - 22.3 - 0.66 - 2.89

2.000 - 41 - 62 - 36 / 1.1- 126 - 7.9 - 22.2 - 0.63 - 2.75

1.750 - 39 - 58 - 34 / 1.0 - 107 - 8.5 - 22.0 - 0.60 - 2.60

1.500 - 36 - 54 - 32 / 1.0 - 90 - 9.3 - 21.9 - 0.58 - 2.43

1.250 - 32 - 49 - 29 / 0.9 - 74 - 10.2 - 21.7 - 0.55 - 2.26

1.000 - 29 - 43 - 26 / 0.8 - 58 - 11.3 - 21.6 - 0.52 - 2.08

0.750 - 24 - 37 - 22 / 0.7 - 44 - 12.8 - 21.4 - 0.50 - 1.89

0.500 - 19 - 29 - 18 / 0.5 - 30 - 14.8 - 21.2 - 0.50 - 1.71

0.250 - 12 - 18 -11 / 0.3 - 15 - 7.5 - 21.0 - 0.60 - 1.52

Print-out table ? y/

Maximum system capacity : 40 t/hr 2.5 bar(o)

Conclusion is that 19 m3/min on a 4” pipeline results in approx. the same capacity/hr as a compressor of 9 m3/min on a 4” pipeline.

The smaller compressor is favorite on a smaller pipeline

Pneumatic conveying at proper velocities ( approx 4-8 times the local suspension velocity)

do not wear straight pipelines and even not so much on elbows.

High velocities propel particles through the boundary laye of the flow profile and can hit the wall of the pipe, causing wear as a result.

May be I am generating to much figures now, so that a proper overview of the subject becomes a little vague now.

all for now

■

Dear damu,

-a standard bulker is for maximum 2.0 bar working pressure. Usual operating pressure is approx. 1.8 bar.

-Your component's choice must be based on material properties, pipeline geometry and ambient conditions. (not mentioned in your thread)

-a standard bulker is for maximum 2.0 bar working pressure. Usual operating pressure is approx. 1.8 bar. (4.0 bar is too high)

By calculating the velocities, pressure drops, heat exchanges, sedimentation, etc. along the pipeline for a certain capacity.

This requires full understanding of the physics of pneumatic conveying and an extensive computer program.

Have a nice day ■

Hi sir,

very glad to meet you,

In one of my project am dealing with bagfilter vent ducts & process ducts design, as per my knowledge i designed the duct sizing and dead load of same.

Now i want to know the Live load calculation for my sturcture design. Please guide me the general norms for calculating the live loads for vertical, horizontal and inclined ducts in cement industry and if you have any velocity norms for material flow inside the ducts please share for my reference.

Thanks.

Damu ■

The thread starter is now in its 8th year and was a fairly usual straightforward request for a free design of a pneumatic conveyor.

Now pipe support is involved, assuming circular section ducting. In such a case you must design the piping in accordance with the appropriate B31 family of ASME Standard dealing with 2 phase flows. Nothing less is legally acceptable, except maybe some European metricised imitations.

PS. If the live loads are an issue how can you be sure that even your dead loads are accurate. You could also invest in a piping handbook. I use McGraw Hill's tome on occasions. Most handbooks give copious reference to the appropriate codes and contain a plethora of worked examples. ■

Originally Posted by Mantoo

Assuming it is a 4" pipe line with 19m^3/ min the exit velocities will be 35-36 m/s. These are too high velocities to convey at high SLR + don't forget fly ash is highly abrasive. You should be looking at 8 -10 m^3/min range to give you approx 15-18 m/s

exit velocity.

PVC Compounds in Powder Form available at Cost-Effective Prices -any SPECS possible; manufacturing done as per your formulation and your formula remains confidential.... PVC Soft and Rigid all are possible. ■

Originally Posted by bharat

I want to unload fly ash bulk tanker of 30ton with the help of compressor/roots blower.

Now maximum time which i required is 30min to unload tanker.

Now what pressure and what capacity of Compressor /roots blower i will required to unload tanker.

And which is convenient from roots blower and reciprocating compressor.

Why?

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

Dear Bharat

Could you be kind enough to provide me contact details for sending flyash bulker to our brick site at Bansa, Barabanki UP 225204? We have red bricks Field and started flyash bricks plant just now. But getting fly ash & stone dust in trucks is a big headache. So I need it in bulkers whose suppliers I am not getting in our area. We are nearby Lucknow (55km in the east) or any smaller cost effective machine which we can attach to truck for unloading it i.e. dependency on labourers can be reduced. You can also call me on 9648570886. Thanks. ■

...and still rolling along.

The latest question/reply is not very clear. If you don't have tubs (pneumatic discharge tankers) then covered (sheeted) tippers are your best alternative. These could be tipped into a ground hopper or tailgate nozzle and blown into the silo for longer term storage. Using tubs is a bit of an overkill. If you can tip the material out why do you need a pressurised tank? Don't say dust! ■

It seems like I have decimated the pneumatically discharged road tanker business.

Vehicle prices will come down, payloads will increase and unloading times will reduce.

My problem is why they were ever built in the first place?

Of course you need more details. Just ask. ■