Dear guga24,

Choosing the gas velocity (I asume it is air) is related to the suspension velocity of the particles to be conveyed.

100 mesh corresponds with approx 150 micron.

With a particle density of approx 2000 kg/m3 this will result in a suspension velocity of approx. 2.5 to 3.0 m/sec.

The end velocity should then be approx 10 to 12 m/sec for a pressure system.

For a vacuum system this should be the beginning velocity.

Conveying 2500 tons/day (125 tons/hr) is possible.

Whether this should be a positive pressure or a vacuum system, depends on the conveying distance and the installation requirements.

Up to approx 50 meters of conveying lengths, both systems are possible.

Success

Teus ■

Dear guga24,

I think that the 1272 kg/m3 is the BULK density of the coal.

For the suspension velocity, the PARTICLE density is important

Particle density = particle mass / particle volume

Using the particle density for the estimation of the suspension velocity is an approximation by calculation.

Measuring the suspension velocity is a better approach, but for now it is OK.

28 m/sec is used for grain unloaders with suspension velocities of grain up to 10 to 12 m/sec

have a nice day

Teus ■

Dear guga24,

If the particle density = 1272 kg/m3 , then the suspension velocity will be approx 2.0 m/sec.

The respective airvelocities then will be approx. 10 m/sec.

The suspension velocity can be calculated from:

particle density (1772)

particle size (0.000150)

drag factor (approx.0.05)

gas density. (1.293)

V-suspension = SQRT(4/3 * product density/dragfactor * particlesize/1.293)

The drag factor is an educated guess, as this factor is very much depending of the particle shape.

Therefore a suspension velocity measurement is preffered.

All for now

Teus ■

Dear guga24,

Sedimentation starts when the air velocity along the wall drops below approx 2 times the suspension velocity at the specific location.

This condition has to be checked for every location in the pipe line.

If the wall velocity needs to be approx. 2 times the local suspension velocity, then the average velocity can be taken an extra 2 times higher.

Resulting in air velocity = approx. 4 x v-suspension.

These are the figures, which can be used in the first preliminary design stage.

Later on the velocities can be chosen more appropriate.

Can you enlighten how you derived the saltation veloclty?

have a nice day

teus ■

Dear guga24,

The Rizk correlation assumes a solid loading ratio SLR

In your eqaution calculated from Mp , d and gas density.

The SLR is depending on the conveying length at a certain pressure drop and velocity.

If the SLR and the capacity is known, then the mass flow of air can be determined.

Using a gas velocity, the pipe diameter can be calculated.

If the Rizk correlation ,used for checking the saltation velocity, shows that the chosen conveying velocity is to low, then the mass flow aof gas has to be increased.

But then the SLR changes and the procedure has to be done again.

Hopefully this converges to a solution.

Also the gas density varies along the pipeline, which means that the check has to be done for each location in the pipeline.

(Especially when stepped pipelines are applied)

And then : How reliable is this equation and how is it derived and what does it really tell us ???????

SQRT(g*D) indicates the involvement of the Froude number

I never uses(d) it.

BR

Teus ■

Dear guga24,

Apparently you have a system to calculate.

That means that you also know the pipe routing, the pick-up system and the receiving hopper configuration.

The length of the pipe line and the number of bends, together with the chosen velocities and pressure drop determine the SLR. (As I explained earlier)

This counts for pressure- as well as for vacuum systems.

Therefore it is important to have that information too.

A pipe line size of 13” seems to be an off sized standard.

12”or 14” is easier available.

Not knowing your calculation program, it could be that the product factor is chosen too low, or there is a bug in the software, causing to ignore the pressure drop due to solids loading.

I think that pneumatic conveying of 150 micron coal should not be that difficult, because that is often done in coal fired power plants to transport the coal powder to the burners with the burning air.

Using the performance data of those installations would make the design more reliable, because the product loss factor can be derived from those installations.

Waiting for your reaction.

best regards

Teus ■

Dear Martin,

From the data, you supplied, the SLR you use can be calculated as follows.

13” pipe Area= 0.0855 m2

10 m/sec gives 10 * 0.0855 = 0.855 m3/sec # 1.2 * 0.855 = 1.026 kg/sec

SLR = mass flow /air mass flow = 28.95 / 1.026 = 28.2

To convey 104.23 tons/hr of coal over a distance of 40 km will only be possible with a very,very low SLR. Much lower than 28.2

40 km of pipeline might already consume the total pressure drop for air only, leaving almost 0 for the coal conveying (As your calculation already suggests)

The consequence of the lower SLR is a much higher air volume and thus a much bigger pipe and much more energy.

Based on field experience, I can sometimes guess the size of a new installation and then verify by calculation. The next step is then to fine tune the design.

But this is beyond guessing. This will be a case of trial and error until the solution is found.

Do want me to give it a shot?

If feasible a pneumatic rail car system or a closed dump car system seems more realistic.

hear from you

Teus ■

Dear Martin,

Preliminary pressure pneumatic conveying design for 100 mesh (150 micron) coal.

Coal:

100 mesh (150 micron)

material density 1272 kg/m3

bulk density 680 kg/m3

suspension velocity 2 m/sec

Pipe line:

horizontal length 40000 m

vertical length 10 m

Number of bends 3

Diameter pipe 1800/2100 mm

Compressor 46 m3/sec

Filterarea 5520 m2

Pressure 2.5 bar

Capacity 129 tons/hr

residence time 4950 sec (1 hr 22 min)

Empty pipeline pressure drop 0.72 bar

gas velocity begin 8.42 m/sec

gas velocity end 12.94 m/sec

pressure drops:

intake 110 mmWC

acceleration 19 mmWC

product 4246 mmWC

elevation 8 mmWC

suspension 16764 mmWC

gas 12434 mmWC

filter 86 mmWC

Compressor power 12434 kW

Energy consumption 96.2 kWh/ton

It takes 48 kg of coal to generate the electric energy, required to transport 1000 kg coal over a distance of 40 km.

Hardly an economic enterprise.

Figures are indicative.

Best regards

Teus ■

Dear Martin,

You are welcome.

A slurry system as you state is certainly cheaper in energy.

I assume that there also additional issues to solve.

I am using a calculation program, which I developed myself and of which a description is given on the Bulkblog.

https://news.bulk-online.com/?p=65

I modeled your project and run the program and (I must say) I was a liitle bit proud that it gave a reallistic solution right away.

It also showed that the majority of the energy is spent on keeping the coal in suspension and that could be expected over 40 km and 1hr22min.

Success

Teus ■

Dear Martin,

I started to develop the theory and the program in 1981 and still fine tuning it.

Indeed, the program is not available on the market.

have a nice day

Teus ■

■

Wuestion for TEUS,

In which book can one find the stated formula for the suspension velocity. Is this formula applicable for the particles that are rather large like 10mm and above.

Enej ■

Dear Enej

The formula, you are referring to is this one?

V-suspension = SQRT(4/3 * product density/dragfactor * particlesize/1.293)

This formula results from the equilibrium of forces (gravity and drag force) of a free falling particle in air, reaching its terminal velocity.

If the particle is not falling, but instead the air is moving upwards, we call the air velocity the suspension velocity, whereby the particle stays at the same position.

Common physics

Teus ■