Energy consumption per ton of a pneumatic conveying system.


Pneumatic conveying installations suffer from the image of being not energy efficient.

Usually, this bad image is explained by the statement that a pneumatic conveying system is based on high velocities. High velocities are usually synonymous for high energy demand.


The definition of the efficiency of a pneumatic conveying system is (in case of an electric drive):


Total efficiency = Electric energy / tons       (in kWh/ton)


This Total efficiency can be divided in 4 partial efficiencies.


1) Drive efficiency = Mechanical energy / Electric energy

 2) Compressor efficiency = Thermal compressing energy / Mechanical energy

 3) Thermo dynamic conveying efficiency = Thermal compressing energy / Thermal expansion energy

 4) Pneumatic conveying efficiency = Thermal expansion energy / tons


Ad 1)

It is obvious that the energy conversion from the power supply to the mechanical shaft energy for driving the compressor should have the highest possible efficiency.


In case of an electric drive a high efficiency electric motor must be applied.

In addition, account for cable losses have to be made.


In case of a diesel engine drive, it is important to apply a suitable diesel engine design.

As a compressor runs normally at constant rpm, the diesel engine should also be designed for constant rpm. (f.i. optimal valve- and injection timing)

The chosen operational point in the diesel curves must be chosen at the lowest specific fuel consumption. (Normally at the highest torque)

Sometimes thease requirements are conflicting, however, the diesel manufacturer can advise here.


Ad 2)

The energy demand of the compressor depends on the termo dynamic compressing principle.

          Isochoric compression (at constant volume, blower) uses the highest amount of energy, but the absolute energy demand is proportional to the pressure.

          Adiabatic compression (without heat exchange). Adiabatic compression requires less energy than isochoric compression. Widely used in pneumatic conveying are screw compressors with internal compression to the outlet pressure. In case the outlet pressure differs from the internal compression pressure, the difference in pressure is equalized by isochoric compression.The efficiency is maximum when the compressor is operatig at an outlet pressure equal to the internal compression pressure. At lower pressures, than designed for, a blower (isochoric compression) can be more energy efficient than a screw compressorwith internal compression. The internal compression energy cannot be recuperated and is lost.This is an important issue when an installation is operating at partial pressure for a longer period.

Low pressure pneumatic conveying installations can be more energy efficient with a blower than with a screwcompressor. In addition, the internal leakage of blowers and screw compressors decrease the energy efficiency. A turbo compressor is compressing adiabatically. For an extensive description of a turbo application in pneumatic conveying, see

          Isothermal compression (piston compressors) uses the lowest specific energy.

For the application in pneumatic conveying installations, piston compressors are not often applied.

Ad 3)

The compressed air expands along the pipeline and the internal energy of the air is transferred to the material being conveyed. ( see also: and )

The external energy, developed by the expanding air, is not used for material conveying.


Ad 4)

 The total air expansion energy per conveyed ton is :

    E(total) = E(intake) + E(suspension) + E(kinetic)  + E(potential)   + E(product) + E(air)


Each partial energy can be derived from the pneumatic conveying formulas and are given below:

 E(intake)         = F1 * dpdoes / (mu * (1+dp))

E(suspension) = F2 * L

E(kinetic)        = F3 * vp^2

E(potential)     = F4 * H

E(product)     = F5 * mu^(a) * v(air)^2 * vp * L

E(air)               = F6 * v(air)^2 / mu * L/D


mu = SLR (Solid Loading Ratio)


           In case “dp = dp(intake)”  ,the loading ratio must be equal to zero.

            Then the items E1 and E6 will be infinite.

            E(total) will be infinite as well.


          In case “dp(intake)< dp” , then the loading ratio has a real value.

            All items of E(total) have also a real value.

            The value and plot of a minimum for E(total) is then

            depending on the combination of loading ratio ,vacuum ,product properties

            and chosen installation parameters.


           E(intake) and E(air) are decreasing with increasing mu, from infinite to 0

           E(product) is increasing with increasing mu from 0 to infinite.


Between mu=0 and mu=infinite, a lowest value of the Total energy consumption per conveyed ton exists.


As an example, the following 2 installations are calculated:


Material cement.

Horizontal conveying length = 100 m

Vertiacal conveying length = 25 m

Bends = 5


INSTALLATION 1 (Usual design)

Pipe diameter = 12” (300 mm)

Screw compressor with internal compression 1.183 m3/sec (70 m3/min)

Capacity = 287 tons/hr

Pressure 2.5 bar



Energy consumption = 0.98 kWh/ton

SLR = 56.9




INSTALLATION 2 (Alternative design)

 Pipe diameter = 18” (450 mm)

Blower 2.72 m3/sec (163 m3/min)

Capacity = 287 tons/hr



Pressure = 0.545 bar

Energy consumption = 0.66 kWh/ton

SLR = 26.3


As installation 1 is the usual design, it must be concluded that a pneumatic conveying design can be much more energy efficient (in this example approx 33%) than commonly built.

However, it must also be noted that the capital investment for a low pressure installation is higher, due to the bigger pipes, valves, filters and pumps.

In combination with a screwfeeder, the gain in energy consumption will be even more.

The reason for this is that the energy consumption for the screwfeeder is almost proportional to the conveying pressure at the same feeder rate.


Think of the wasted energy over the past 100 years in pneumatic conveying!

The bad energy image of pneumatic conveying is now reduced by at least 30%.



2 thoughts on “Energy consumption per ton of a pneumatic conveying system.”

  1. Hi Teus,
    Man you are a wealth of information. Are you sure you want to stay retired? There is so much to learn out there.

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