Pneumatic Conveying, Performance and Calculations!

In many industrial processes and transport, materials have to be stored and moved from one location to another location. For long distances, e.g. from one country to another country (or continents), modalities are used e.g. ships, aircraft, trains, trucks, etc.

Where changes are made in the transport (or storage) modality, various technologies are used to move the material from one modality to the other modality.

The basic applied technologies are :

  • mechanical systems
  • grabs
  • screws
  • belt conveyers
  • buckets
  • etc.
  • Carrying medium systems
  • Hydraulic systems using liquids as carrying edium
  • Pneumatic systems using gas as carrying medium

The bulk handling sector over the world is a key player in economics as it handles all kinds of commodities such as cereals, seeds, derivatives, cement, ore, coal, etc., which are processed in the industry to other commodities, which have to be transported and handled again.

To manufacture all the necessary equipment for the bulk handling alone a whole industry exists. The magnitude of financial investment is tremendous as well as the operating cost involved.

The importance of economic handling is not only a matter of the handlers, but also to third parties such as the transport sector.

The technology of bulk handling equipment is crucial to all the involved parties and therefore it is of the upmost importance that the bulk handling industry employs the best engineers and operators, who design, develop, build, calculate, operate the installations and do research and document their achieved knowledge and experience.

One sector of bulk handling is the pneumatic unloading and conveying of cereals, seeds, derivatives and powdery products such as cement, fly ash, bentonite, etc.

The first pneumatic unloaders were built around 1900. In 1975, there were still, steam driven, floating grain unloaders operating in the ports of Rotterdam and Antwerp. Unloaders, which even dated back from 1904.

How these installations were calculated is not really known as the manufacturers did not reveal their knowledge publicly for obvious (commercial) reasons. Trial and error must have played a significant role in the beginning of this industry.

Calculating a pneumatic system was, before computers were introduced, done by applying practice parameters, based on field data from built machines.


Example 1975

Calculation grain unloader anno 1975

Set Capcity grain                                 440                  tons/hr

Bulk density grain                                0,75                tons/m3

Suction height (elevation)                     30                    m

Air displacement pump                        500                  m3/min

Vacuum air pump                                0,4                   bar

Absolute pressure vacuum pump          0,6                   bar

air density                                            1,2                   kg/m3

pressure drop nozzle                            0,16                 bar

Nozzle diameter                                  0,45                 m

Cross section nozzle                            0,1590             m2

Grain volume                                       9,778               m3/min
(Capacity/grain density/60)

Air volume at nozzle                            357,1               m3/min
(Air displ pump * abs press pump /(1-press drop nozzle))

Transport volume after nozzle               367                  m3/min
(Grain volume + Airvolume at nozzle)

Grain mass                                          7333                kg/min (capacity *1000/60)

Air mass                                              360                  kg/min
(Air displ pump * abs press pump * 1,2)

Transport mass after nozzle                  7693                kg/min (Grain mass + Air mass)

Specific density mixture                       20,97               kg/m3 (Transport mass after nozzle / Transport volume after nozzle)

Mean velocity of
mixture after nozzle                              2307                m/min
38,45              m/sec
(Transport volume after nozzle / Cross section nozzle)

Pressure drop nozzle                           1577                mmWC
11,98              cmHg
(specific density mixture * mean velocity^2 / (2 * 9,83))

pressure drop miscellaneous                  5                      cmHg

pressure drop vacuum pump                30,4                 cmHg
(Vacuum air pump * 76)

Available pressure drop elevation 1       3,416               cmHg
(press. drop vacuum pump – press, drop nozzle –
press, drop misc.)

Elevation per available press. drop        2,2361             m/cmHg (elevation / available pressure drop elevation)

Loading factor from diagram                24,45               kg/m3 (Loading factor = function (elevation per available pressure drop))

Calculated Capacity                            440,1               tons/hr
(60 * loading factor from diagram * air displ vacuum pump * (1- vacuum))

Table loading factor kg/m3 = function (elevation per available pressure drop)

Loading factor  elevation/cmHg
16                               3.6
17                               3.44
18                               3.28
19                               3.16
20                               3.0
21                               2.84
22                               2.64
23                               2.48
24                               2.26
25                               2.04
26                               1.74
27                               1.35
By changing the figures in this calculation, an iteration process is executed until the set capacity equals the calculated capacity The calculation can be started by assuming the pressure drop over the nozzle at 0.15 bar.

If a parameter is not known, assume this parameter and vary until an optimum is found.

It is clear, that this method is not really accurate, nor gives it a scientific insight how the physics of pneumatic conveying work.

Calculation of Pneumatic Systems using Gas as Carrying Media

Since computers are available, it became possible to build an algorithm that can execute calculations in a time domain, whereby the conveying length is divided in differential pipe lengths, which are derived from the elapsed time increment.

The physical principle of this technology is:
A gas flow in a pipeline will induce a force on a particle, which is present in the gas flow. This force (if of sufficient value) will accelerate and/or move that particle in the direction of the flow. (Impulse of air is transferred to particles) The particle is moved from location 1 to location 2.

Between pipe location 1 and pipe location 2 , impulse is transferred from the gas to the particles and to friction.

This transferred impulse is used for:

  • acceleration of particles
  • collisions between particles and from the particles to the wall
  • elevation of  the particles
  • keeping the particles in suspension
  • air friction

Bends are calculated only for product kinetic energy losses by friction against the outer wall and air friction pressure drop. The calculation of velocity losses in a bend are depending on the orientation of the bend in relation to the product flow.

There are 5 bend orientations to be considered:

  • vertical upwards to horizontal       (type 1)
  • horizontal to vertical downwards   (type 2)
  • vertical downwards to horizontal   (type 3)
  • horizontal to vertical upwards       (type 4)
  • horizontal to horizontal                (type 5)

All these energy transfers result in a change in the gas conditions (p,V,T) and changing velocities of the carrying gas and the particles.

All these energies, velocity changes and gas conditions can be calculated and combined into a calculation algorithm.

This algorithm calculates in the time domain (dt=0,01 sec)

The physical laws involved in this algorithm are:

  • Newton laws
  • Bernoulli laws
  • Law of conservation of energy
  • Thermo dynamic laws

From the original (start) conditions, the changes in those 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, can be calculated.

By adding those changes to the begin conditions at location 1, the 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, both these values are 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) (Notice the similarity of the iteration process with the example 1975)

This iteration can also be executed, whereby the capacity is set and the pressure drop is iterated.

Example of a computer calculation 2007



Example of a modern computer calculation 2008



The computer program is originally written in Q-basic under DOS and still operates, although some features are now lost under Windows

By changing the program from Q-basic to VisualBasic, the screens appear in a Windows form and more Windows features can be applied, but the program algorithm stays the same.

A very important feature of this algorithm is that performance data from existing installations can be used to determine the product loss factors for certain products. That opens the opportunity to build a database of various products that can be conveyed pneumatically and be calculated. As the used physics are basic, the calculations work as well as in pressure mode as in vacuum mode with the same formals, product parameters and product loss factors. (Adaptations are made for the different behavior of the gas pumps in pressure mode and vacuum mode)

As the pneumatic conveying calculation is basic, the calculation program can be extended with many other features s.a. booster application, rotary locks, high back pressure at the end of the conveying pipe line, heat exchange along the conveying pipe line, energy consumption per conveyed ton, Δp-filter control, double kettle performance, sedimentation detection, 2 pipelines feeding one pipeline, etc. Also it becomes now possible to evaluate product pneumatic conveying properties from field data and tests and also investigating operating machines for functioning. (Defects were found, just by calculating the actual situation).

Based on the properties of pneumatic conveying, derived from the above described theory, the used technology is chosen. The used technology and operational procedures are also depending on the type of application and product.

The above only describes the calculation of pneumatic conveying based on physics. The connection between theory and practice is made by measured and calculated parameters from field installations. In addition to this theory, there are many technological issues to be addressed, ranging from compressor technology to the structural integrity of a complete unloader as well as PLC controls, hydraulics, pneumatics, electric drives motors, diesel engines, filter technology, ship technology, soil mechanics (product flow), maintenance, methods of operation, etc.

The mathematical approach with the field verification (resulting in many corrections and extra features), documented description and creating the computational software is (was) a matter of many years of persistent labor but worthwhile. This approach also resulted in a better and still growing understanding of the pneumatic conveying technology. The influence of the various parameters and there effects (sometimes hidden by counter action) was revealed step by step.

July, 2008
Teus Tuinenburg
The Netherlands

78 thoughts on “Pneumatic Conveying, Performance and Calculations!”

  1. Will you please send me a copy of your computer programme? My email id is “”. Thanks and regards,

  2. Will you please send me a copy of your computer programme? My email id is “”. Thanks and regards,Anil patel

  3. Can u explain me how should i calculate flow of pipeline which having air velocity 500 milibar and 8inch line diameter with 100 meter length

  4. Very good article.
    Will you please send me a copy of your computer programme? My email id is “”. Thanks and regards,
    Ajit Gohad

  5. Dear Sir

    I’m a student. I study particle losses in 180 degree bend, compare particle deposition due to bend axises(horizon and vertical)and air flow rates(laminar regime only). I focus on inertial and gravitational forces by using Pich(1972)’s model for gravitational settling and Pui(1987)for inertial deposition in a bend. I found that my studies are compatible with Pich when the flows are low and compatible with Pui when high flow. But in moderate flows, my experiment did not go along with both models. So would you please give me some suggestions that what particle deposition mechanism that I should be consider for moderate flow.

    Best regards
    Parkpoom P.

  6. Dear Parkpoom P.

    Particles in a pneumatic conveying system, which enter a bend will hit the outer wall of the bend, while initially maintaining their velocity and direction.

    Then you can assume that the velocity component, perpendicular to the outer bend wall is lost.
    The tangential velocity of the particle causes the particle to slide along the wall, while losing velocity by friction and loosing or gaining velocity through gravity (depending on the bend type: horizontal to horizontal – vertical up to horizontal – etc.)

    By this, full segregation is occurring.
    In the bend, there is no interaction between the particles and the conveying gas, until the reduced particle velocity start to reduce the remaining gas area too much.

    Have a nice day,

  7. Teus,
    Can you help in calculating the amount of air and pressure required for following:
    Horizontal Length: 165 meters
    Vertical Height : 21 meters
    Particle Size: 10 mm
    Density: 0.38
    Velocity Required: 12 m/sec
    Flow of material: 1.3 kg/sec
    Pipe size: 6 inches.
    I sells blowers but do not know how to calculate. Kindly help out please.
    VT Ten

  8. Hi Teus,
    I have been looking through the forum and found you as knowledgable in pneumatic conveying.
    I have a pneumatic conveying problem I am looking at solving and hope you might be able to offer some guidance.
    I have a requirement of conveying ordinary portland cement – required mass flow of 80tonnes/hour, pipeline diameter of 6″. Conveying distance about 120m horizontally, and 10m vertically (I can send a drawing). Max pressure differential 2.5bar – airflow choices of 720 or 1440m3/hour.
    Will I be able to convey using 720m3/hour?
    Many thanks in advance and kind regards

  9. Hi Joby,

    Thank you for your interest.

    I assume that you want to unload a cement bulk truck.
    Normally, cement tankers discharge at approx. 1.8 barg as the tank is made for 2.0 barg.

    I calculated the 2 compressor options for:

    horizontal length = 120 m
    vertical length = 10 m
    pipe size = 6” and 8″
    material = cement

    Calculation results:

    6”pipe size
    720 m3/hr at 2.5 bar:
    approx. 60 tons/hr at 2.5 barg
    Sedimentation in the pipeline
    1 bend filled with cement over the last 9 degrees. (But flowing)

    6”pipe size
    1440 m3/hr at 2.5 bar:
    approx. 71 tons/hr at 2.5 barg
    No sedimentation in the pipeline

    8”pipe size (bulk truck outlet 6”)
    1440 m3/hr at 1.8 bar:
    approx. 90 tons/hr at 1.8 barg
    No sedimentation in the pipeline

    Take care

  10. Dear Teus,

    We have a project to convey poultry meal with bulk density 730kg/m3 from a hopper to silo. The required flow rate is 25t/h. First horizontal distance 3m, 90 degree bend to vertical, vertical 20m, 90 degrees bend to horizontal, 5m horizontal, 90 degree bend, 6m horizontal. Pipe diameter 6″ and max differential pressure 1 bar. Dilute phase preffered. Can we convey with max 1500m3/h airflow?

    Thank you.


  11. I didn’t know that the first pneumatic unloaders were built in 1900. I’d say that pneumatic conveyors have made a lot of advancements since them! In general, how much does an entire pneumatic conveyor system cost?

  12. Sir i have to design for vacuum conveying system

    i require calculation for sand transfer system.. how much capacity and vacuum require for suction of sand..

    vacuum tank , vacuum with ventury (air jet ejector) air consumtion available : 250cfm @9bar

    sand density : 1988 kg/m3
    suction distance : 600 feet
    pipe size : 4″
    collection tank size : 1500 ltr

    Please give us best reply with calaculation of capacity and vacuum

  13. Hi Sir Theus,
    Please help me do some calculation for ash conveying. Need to know the Pump capacity, Cu. M and Kw required
    Here are the datas:

    Req’d Flow rate – 70 tons/hr
    Pipe dia. 150 mm
    Pipe Horizontal Length 110 meters
    Vertical length 22 meters
    No. Bends 8 x 90
    D/d 10


  14. Dear Mr. Noel,

    Because you did not give the type of ash I assumed that it is fly ash with density 750 kg/m3. your pump power calculated is 105 kW, air flow : 36 m3/min; compressed air pressure is 2.25 bar. I hope this is useful for you.

  15. Hi Teus, we have some issues with our pneumatic conveyors systems, we have 3 system in our line production, vacuum arrangement (pull system). We are transporting kibbles (pet food). The main problem that we have is too many fines in collectors, about 30%. The rate of production is 5 to 10 ton/hr, may u give me ur email to send u more information about it please. Thanks..!

    B. Rgds.

  16. Good data
    Could you please send me a copy of your computer programme?
    My email is “”.

    Thanks and regards,

  17. Dear, sir

    We have the make new project for the pnumeric vacuum conveyor system so please you best your experiance give me the Hartley so i used witch pipe size and witch size rotary valve
    material data bellow
    1 -material B/D 980
    2-material transportation 50ft hight
    3 material transportation 2000kg/h

  18. Dear Sir,

    I need to convey Chana Dal to 4 MT / hr , to height of 60 Mtr thru Pneumatic conveying, kindly help us to know the pipe dia and air capacity and pressure require to centrifugal blower.

  19. Dear sir,
    I want to transport rock Phosphate 10 tonns per hour.
    Conveying distance.
    6 meter high and 35 mtr long.
    Using 4 inch pipe and 2 90 degree albo to send material into precipitator.
    Can you please help me calculate cfm and blower requirement.

  20. calculate how much ash conveyed by vessel have capacity 16cubic feet,available pressure 4.4kg/cm2,,conveying distance-206m,compressor capacity 14.7m3/minute..

  21. Thank you for all the information about pneumatic conveying. My husband and I visited a factory that used a system, and I got really curious about how they worked. That is really cool that the first pneumatic unloaders were built around the year 1900. I can’t believe they have been around that long. That is amazing we still use them today!

  22. Dear Sir,
    I have to select a compressor to unload the Coal Powder from truck to Silo. detailed are following:
    Unloading Capacity of Coal: 25 tons/hr
    Particle Size of Coal: very fine- 12% Residue on 90 micron
    Bulk density of Coal: 0.4 Ton/m3
    Pipe vertical Height: 38 meter
    Pipe horizontal height: 15 meter
    Velocity in pipe should be 35 m/sec
    Kindly help me in calculation to select a compressor .

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