Dear Mr Kaufman,

Am I right that the italic typed text is describing an existing installation and the normal typed text is the to be designed installation?

What do you mean by “from camera”?

Which installation do you want to compare with a double tank system versus a screw feeder?

Is the tankvolume given as total volume or as cement content?

Required are tankvolume and cement volume, in order to calculate the pressurizing time and the unloading time.

The answer to your main question of which installation is more energy efficient is easy to answer.

A double tanksystem serves, just as the screwfeeder, as a feeder system.

The double tanksystem uses almost no energy/ (any pressurizing energy is later on used for pneumatic conveying)

A screwfeeder uses a considerable amount of energy, forcing the cement plug as a seal through the screw feeder barrel. More energy than needed for a tank system.

This is the case when the pneumatic conveying system for both feeders is the same.

When the pneumatic system is designed for each particular technology, the difference in energy consumption per ton changes, but will be in favor of the double tank system.

To calculate your installations, I need the additional information.

best regards

Teus ■

Originally Posted by Kaufman

Hello

For my training work I have to calculate what is better Double Pressure Tank System or Screw-Feeder (FK-pump). Which one is less energy consumption for existing pipe line with Double Pressure Tank System and optimize the pneumatic conveying of cement with air (main target less kwt per tn) .

Some info:

Cement transporting amount 40-45tn/h; 3200cm2/gm blaine

Existing Double Pressure Tank System:

Productivity of an unloading of cement from camera, t/h61.5

Length of transport pipeline, m1000

Also vertical, m, not longer than 50

air-quantity flow, m3/mun 58

Diameter of transport pipeline (internal diameter), mm200

Overall dimension, mm: -

Length 3770

Width 3350

Height 4340

Capacity, m3:

of 1 vessel 6.3

of 2 vessels 12.6

Control system remote

Type of uploading upper/top

Pipeline stainless steel with 7 90° bend total horizontal flow 800m; vertical flow 30m pipeline diameter 219mm

Compressor 250m3/min 7 bar but using bypass on 4 bar (also other users of air are in the system).

Thanks in advance

==========================================================

My dear friends Teus Tuinenberg or Mr. Jack Hilbert can surely solve this directly.

but it is your assignment to do this!

You have done most of the work already with the known factors of the system in use.

the twin tanks system allows for continued batching and dilute phase conveying.

Any elbow is a restriction and requires more energy to move the material, any long distance requires more energy. any elbow becomes place where sedimentation/build up can occur.

A solid auger is linear in design and has either a hollow auger or a solid flighting

For long distances you will need several augers with transition hoppers and for simplicities sake an incline auger instead of a vertical auger for the transition to avoid plugging.

Either way (ENERGY) is used with numerous electric motors to power the auger drives or the blower system. The six basic laws of gasses are used you should already understand that.

lzaharis ■

Dear Mr Kaufman,

I calculated your existing cement conveying system with a length of 1000 m.

The results were:

-Pipeline capacity 69.5 tons/hr at 2.5 bar and a pipeline energy consumption of 3.33 kWh/ton

-System capacity 57 tons/hr at 2.5 bar and a system energy consumption of 3.67 kWh/ton

Then I calculated the same pipeline for a screwfeeder, whereby the conveying pressure is reduced to 1.8 bar, which is the maximum pressure a screwfeeder can handle.

Resulting in:

-Pipeline capacity 56.6 tons/hr at 1.8 bar and a pipeline energy consumption of 3.22 kWh/ton

-System capacity 56.6 tons/hr at 1.8 bar and a system energy consumption of 3.92 kWh/ton

-The energy consumption of the screwfeeder is taken as 0.7 kWh/ton at 1.8 bar from an existing installation I once built. A better design of the screwfeeder with a lower energy demand makes the 2 systems more equal.

Manufacturer’s input is here important.

The pipeline energy consumption at 1.8 bar is less than the pipeline energy consumption at 2.5 bar (better pneumatic conveying efficiency)

When applying a screwfeeder, the pressurizing energy and purging energy in a vessel system is avoided and in a comparison used for the screwfeeder.

The long pipeline with a 2 vessel system has an extra issue to attend to.

The rest pressure in an empty vessel is about 0.8 bar that has to be vented from the kettle before a refill can take place. At that moment there is no airflow in the pipeline until the next vessel is released. Therefore, it is necessary that the pipeline is properly purged, to prevent sediment and possible blockage.

The application of a booster, taking care of continuous purging, is a solution.

When applying a screwfeeder, the constant purging is inherent to the system.

A screwfeeder system is simpler and easier to maintain.

Comparing the 2 systems properly, requires the design of two optimal systems.

If you need more information or the output of the calculations, let me know.

Have a nice day

Teus ■

Dear Mr Kaufman,

I will calculate the existing 8” pipeline of 800m + 30m.

Your energy calculations are based on a 100% full load of 4bar for the double tank system.

This is in reality not the case as the pressure rises from 0 to opening pressure of the kettle and decreases when purging the pipeline after the kettle is emptied.

Furthermore, I doubt whether the existing system is really reaching 4 bar. (Please check)

A double tank system also requires maintenance, especially on the valves.

And valve failures always occur at inconvenient moments.

A double tank system requires quite a complicated control.

On long pipelines, the kettles need to be bigger, in order to minimize the time lost in changing kettles. To reach a certain overall capacity, the installation needs to be designed for a higher pipeline capacity.

A screwfeeder is simple and reliable for a longer service period (250.000 tons or more).

Maintenance stops can be scheduled and replacing a screwfeeder is easy. Replacing the wearing parts of a screwfeeder is also not a big deal.

Where a double tank system is discontinuous, a screwfeeder is continuous.

When you state 2.2 kWh/t for the FK pump at 63 t/hr, means a motor of 138 kW.

Which type of FK pump is this? I once used FK pumps of 110 kW for 150 tons/hr, which is 0.7 kwh/ton.

Please check the dimension "kwt/min".

Just saw your las reply.

A pneumatic conveying system should work better at lower conveying rates (lower pressures).

I assume that you are taking air from a main system (7 bar) by using sonic chokes or flow regulators. Maybe this equipment is delivering less air at lower pressures. please check.

All for now

Teus ■

Dear Mr Kaufman,

Attached are the promised calculations.

Reconstructing your table is not easy.

Maybe, you can separate the columns with slashes.

The table gives also conveying lengths, diameters and air volumes.

However, these data are not relevant for the capacity of the screwfeeder.

A screwfeeder injects a certain amount of cement into the pipeline against a certain pressure.

How the pipeline is conveying that amount of cement at the same pressure is not known by the screwfeeder.

A 30% higher power consumption for a screwfeeder pneumatic conveying installation, without giving any secondary references, does not mean anything.

The longer the pipeline is, the less this % will be.

have a nice day

Teus

Attachments

cement pneumatic conveying pipeline 800m_1000m (PDF)

■

Dear Mr Kaufman,

The table is now clear.

I put the table in a spreadsheet and calculated the “kWh/ton” and “m/sec” values for the respective models.

(see attachment)

The “kWh/ton” for the 2 bar installations are ranging from 1.07 kWh/ton to 2.1 kWh/ton

The “kWh/ton” for the 3 bar installations are ranging from 0.87 kWh/ton to 1.29 kWh/ton

The terminal air velocities are ranging from 12.9 m/sec to 31.8 m/sec.

The product is not mentioned but let us assumes it is cement.

My first comment would be that the conveying pressure should not exceed approx. 1.8 bar.

At higher pressures, there is a chance of stalling the screw and excessive wear.

The velocity range cannot be explained as logical.

The figures are at least a bit confusing.

After you have studied the calculations and commented them, we proceed in this interesting case.

Are you in contact with Fuller?

best regards

Teus

Attachments

screwfeedertable (PDF)

■

Dear Mr Kaufman,

1)In most installations, a screwcompressor with internal compression is used above 1 bar. Therefore, my program uses this type of compressor as a default. The only influence on the pneumatic conveying calculation is the consumed power and the outlet temperature. Another type of compressor, incorporated in the program is a blower and a turbine. Other types can manually be corrected.

2)I used 41 m3/min in one of your threads. In the calculations, this choice leads to a working system and is therefore Ok for the time being.

3)The used compressor is for maximum 3.5 bar, but the calculation is not influenced by that figure.

4)If the cement temperature is up to 150 degrC, I can calculate that. Using a double kettle system with rubber-seated valves will be a problem then.

5)You refer to the bulk density in settled condition. In a tank it is in fluidized condition, hence the bulk density is lower.

6)If you have the actual filter data, those values should be used as input. The influence, if properly selected should be minimal.

7)My mistake, I will change it. The influence on the total calculation will be there, but I do not expect a drastic change in the conclusions.

8)That is referring to the chosen compressor. At other pressures, the internal leakage of the compressor is accounted for in the calculation. Internal leakage as a function of the pressure has an influence on the capacity/pressure curve.

9)This value reflects the ratio between the wall velocity of the conveying air in relation to the local suspension velocity. The wall velocity must be higher than the local suspension velocity in order to maintain the particle in suspension. If this ratio becomes lower than a preset value, sedimentation is detected and accounted for in the calculations.

10) This figure represents the time that a particle is travelling in the pipeline from the beginning to the end.

11) This figure reflects the amount of cement in kg that is present in the respective part of the pipeline.

12) The pipe diameter for the calculation must be chosen as the inner diameter. You refer to the outer diameter of 219 mm

13) This figure represents the power used in each respective pipeline section

14)This is a factor representing the unload efficiency of a kettle. If the kettle releases not enough cement into a pipeline, then the designed pressure cannot be reached. In this case no problem.

15) This is also a figure used for the same issue as under 13). Also, no problem for this calculation.

16)The volume in the tank must be less than the total volume of the tank. I guessed 5 m3.

17)The program calculates the 237 kW because of the airflow and the pressure for a screwcompressor with internal compression. In this case, a 2 stage screwcompressor should be calculated or if a sonic choke is used a constant mass flow of air should be calculated.

18)As the screwcompressor is designed for 3.5 bar, the outlet temperature at 4 bar is calculated too high. The influence on the calculation is existing, but is not big, because the mixture temperature is close to the cement temperature and along the pipeline the cooling effect against the environment is accounted for.

Hopefully, the set up and the possibilities of the program becomes a bit more clear now.

If would like to have other calculations executed, it is now only a matter of manipulating the input.

Best regards

Teus

Visit also:

Pneumatic conveying, Performance and Calculations:

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

Dense phase- or dilute phase pneumatic conveying:

https://news.bulk-online.com/?p=238

Pneumatic conveying, turbo- or positive displacement air mover:

https://news.bulk-online.com/?p=309

Energy consumption per ton of a pneumatic conveying system:

https://news.bulk-online.com/?p=331

Pneumatic conveying, an unexpected relationship.

https://news.bulk-online.com/?p=445 ■

Dear Mike,

11) The amount of cement in a respective part of the pipeline in kg:

If you take a picture of the pipeline (time period = 0), then you will see that there is at that moment x kg cement underway.

Total energy consumption is the totally consumed energy during a conveying cycle divided by the conveyed tons per cycle in kWh/ton for the calculated system.

Consumed energy is, the time weighed, compressor energy + (booster energy) + (feeder energy)

The 273 kW is the consumed power of the compressor.

For your 400 m system, I have the following question:

“3)Gas pump: centrifugal compressor 250m3/min at 1640kwt/h pressure 4 bar (it used for several system)”

How much air is used for the 400 m pneumatic conveying system? ( I need that figure for a the right calculation).

In addition, I would like to know how the amount of conveying air is taken from the compressed air system (Which technology is used. (Flow regulator, sonic choke?)

The centrifugal compressor is probably a turbine, but that is not really important here, because just a part of the air capacity is used through an auxiliary device.

The comparison double tank versus screwfeeder is already done (Cement pneumatic conveying pipeline 800m 1000m.pdf)

The screwfeeder is working independently from the pneumatic conveying system as long as the created conveying pressure can be handled by the screwfeeder.

The TA-41A screwfeeder will work up to 3 bar conveying pressure on any size of pipeline (6”-8”- 10” – 12”etc)

It is better to consult Fuller (Bethlehem USA) for the application of a certain type.

I cannot find any type data on their internet site, which is a pity.

cheers

Teus ■

Dear Mike,

1)An air volume of 60 m3/min means 1 m3/sec. In combination with an 8” pipeline, the resulting air velocities are between 10 m/sec to 35 m/sec, which is for cement unnecessarily and far too high.

2)You do NOT need a pressure regulator but a flow regulator. To have a constant flow of air from the flow regulator, a pressure regulator before the flow regulator makes the airflow independent from the pressure fluctuations in the main air supply.

3)OK.

4)Keep trying

5)See attachment

Double tank system: System capacity 92 tons/hr at 2.5 bar – 2.37 kWh/ton

Screw feeder system: System capacity 92 tons/hr at 1.8 bar – 3.29 kWh/ton

Screw feeder system: System capacity 82 tons/hr at 1.8 bar – 3.47 kWh/ton

Velocities far too high for cement.

Best regards

Teus

Attachments

cement pneumatic conveying pipeline400m kaufman (PDF)

■

Dear Mike

A 400 m long pipeline can convey a much higher rate of cement than a 800 m pipeline

see:

Pneumatic conveying, an unexpected relationship.

href="https://news.bulk-online.com/?p=445" target="blank">https://news.bulk-online.com/?p=445

Attached the 4 bar calculation with 60 m3/min.

The pressure/capacity gradient becomes very small at the higher pressures, indicating that the risk of plugging is increasing.

A small rise in feeding and the pressure exceeds the available pressure (safety valve opens) and the conveying halts

Whether the 60 m3/min is a mistake, I do not know. I just calculate based on your information.

I cannot judge from here, whether the information is right or reflecting the actual situation.

The “missing” pages do not give necessary extra information.

The program calculates 239 kW and if the programming is correct, it should be OK.

have a nice day

teus

Attachments

cement pneumatic conveying pipeline400m kaufman4ba (PDF)

■

Dear Mike,

1)Pressure/capacity gradient is defined as the tangent of the capacity=function(pressure) curve.

At 4 bar it is calculated as (139.5-138.6)/(4-3.95) =18

At 2.5 bar = 23

At 1.5 bar = 35

At 1 bar = 57

At 4 bar conveying pressure, there is less capacity increase required for the same resulting pressure increase, which increases the chance of choking.

2)If your main supply compressor is operating at 7 bar, which pressure is reduced to 4 bar, you are correct. In the mean time you have destroyed approx 400 – 250 = 150 kW in throttling. A more energy efficient system needs to be designed, using dedicated compressors. In addition, the 60 m3/min airflow is not optimal.

3)The attached file, which you refer to, is not included.

4)My program is not available on the market. Mr Agarwal of this forum shares his software (excel sheet) with forum members.

I am interested in your double tank models and will have a look at them for you.

take care

Teus ■

Dear Mike,

Calculation results of your systems:

TA-29A 1000m:

system capacity 60 tons/hr at 4 bar

TA-28A 1000m:

system capacity 109 tons/hr at 4 bar

TA-23B 230m:

system capacity 40 tons/hr at 2.5 bar

Maximum pressure 2.5 bar because of sedimentation at higher pressures

TA-29B 410m:

system capacity 47 tons/hr at 4 bar

Regarding the 40 t/hr installation:

Please indicate:

- conveying length horizontal

- conveying length vertical

- number of bends

- double tank system?

- screwfeeder ?

best regards

Teus ■

Dear Mike,

Double tank system 40 tons/hr:

Horizontal length 430 m

Vertical 35 m

Stepped pipediameter 154mm / 203mm

Tank volume 6.3 m3

Material volume 5m3

Screwcompressor: Aerzen VM21R at 20737 rpm (0.394 m3/sec)

System capacity:

48 tons/hr at 2.5 bar – 1.82 kWh/ton

40 tons/hr at 1.5 bar – 1.58 kWh/ton

40 tons/hr on a 1000m long pipeline corresponds with 107 tons/hr on a 400m long pipeline.

have a nice day

Teus ■

Dear Mike,

The calculations you asked for are attached.

You did not mention the air volume of the installed compressor.

Therefore, I assumed 0.512 m3/sec.

A system capacity of 40 tons/hr generates a “steady” conveying pressure of 0.7 bar.

In practice, the pressure fluctuates in time from 0 to the opening pressure of the tank.

In addition, during purging the pipeline, the pressure drops gradually until the set closing pressure of the tank.

success

Teus

Attachments

cement pneumatic conveying pipeline430m kaufman (PDF)

■

Good Day

While the input of Teus is very comprehensive, I must let you know that his comments about conveying line pressure of 1.8 bar for the screw pump are misleading to a slight degree as he is limiting his analysis to just one screw pump supplier ( he refers to Fuller)

There are other screw pump manufacturers such as Claudius Peters and IBAU and even FLS ( Fuller) themselves, have an older pump design as well, which can pump against higher line pressures then the 1.8 bar limit for the more recent design "M" type pumps.

Regardless of the pump design - the end result will be the same -- the dual pressure tank will be more energy efficient but the "complete" evaluation can swing either way depending on what specifically you include such as maintenance costs, installation costs, spare parts, man-hours, reliability factors, etc etc

Regards ■

Hello Jack,

Thank you for the addition to my replies.

According to the websites of Claudius Peters and Ibau, they feed up to resp. 3 bar and 2.5 bar.

The required power for the screwpumps is not given for each of the suppliers nor can any claim of low energy consumption be found.

As the power demand is almost proportional to the conveying pressure, a 3 bar feeder system uses twice as much as a 1.5 bar feedersystem.

Of course, this choice influences the pneumatic conveying design a great deal.

For a good project comparison of both systems, it is absolutely necessary to design both systems in various configurations.

Then the evaluation can take place with all the items you mentioned included.

Thanks again

Have a nice day

Teus ■

Dear Mike,

1)I am not sure what you mean by “Specific erosion = (Ratio of velocity) involution 2.65”. Increasing the conveying speed will cause more wear, not really on the pipes as well in the bends. However, why increase the velocity, as this will cost also more energy per conveyed ton?

2)I do not know the meaning of that relationship.

3)The reply of Jack covers the whole issue. Energy demand is clearly a pro for the double tank system. All the other parameters depend on the specific installation.

A double tank system is discontinuous and more sensible to choking and a screwfeeder system is 100% continuous.

If you asked for a quotation of both alternative systems, the differences become clear.

An experienced consultant can do the job for you.

4)All 4 screenshots of the installation you asked for are attached.

Take care

Teus

Attachments

cement pneumatic conveying pipeline400m kaufmancom (PDF)

■

Dear Mike,

I assume 370m HORIZONTAL

and

30m VERTICAL

Correct?

Stay tuned

Teus ■

Dear Mike,

Attached the calculation results.

At 1.6 bar(g):

Pick-up air velocity = 13.62 m/sec

Terminal air velocity = 33.62 m/sec

Material to air ratio = 19.1

Air mass of 1 m3 at 4 bar(g) at 90 degrC = 1.293 * 5 * 1 * 273/(273+90) = 4.86 kg

Considering the high velocities for cement, it is most certainly dilute phase conveying.

The compressor displacement = 232/4*60= 3480 m3/h3

The pipeline capacity = 81.4 tons/hr at 1.6 bar(g)

Resulting in 3480/81.4 = 42.75 m3/ton

I suppose it is a little bit high, because of the high air velocities.

Keep in mind that the conveying length is 400 m, limiting the maximum SLR.

Increasing the conveying pressure (by increasing the SLR) to 2.5 bar(g) will increase the system capacity from 72 tons/hr at 1.6 bar(g) to 86 tons/hr.

The energy consumption increases from 2.17 kWh/ton at 1.6 bar(g) to 2.45 kWh/ton at 2.5 bar(g)

At your service

Teus

Attachments

pninstportlandcement400m (PDF)

■

Dear Mike,

The 232 m3 is atmospheric air displacement (NOT at 4 bar(g)) in 4 minutes.

Therefore, the air mass flow = 232 * 1.293 * 1 * 273/(273+25) = 274.8 kg/4min.

(assuming an intake air temperature of 25 degrC.)

As you calculate the SLR = (3500 kg/4min)/(274.8 m3/4min) = 12.7

3500 kg/4min = 3.5*60/4 = 52.5 tons/hr.

In my calculation, the SLR is calculated as the ratio between the pipeline capacity (excluding pressurizing, purging and filling) and the calculated air mass flow.

In my program, the internal leakage of an (assumed) screwcompressor with internal compression is accounted for at different pressures.

If you have an Aerzen compressor, I can calculate the air flow automatically.

In the screenshots, you can trace the following figures:

1.0038 * 1.293 * 1 *273/(273+26) * 3600 = 4250 kg/hr

SLR = 81400/4250 = 19.1

Accurate definitions of the used figures are essential.

regards

Teus ■

Dear Mike,

I executed the calculations of the pneumatic conveying installation of 400m for portland cement, in order to verify the conveying regime, dense phase or dilute phase.

The results are presented in the attachment.

My initial guess was far from the reality.

The system is dense phase and the installation working point is in the diagram located just left of the lowest pressure drop, which is the separation between dense- and dilute phase conveying.

It seems that velocity is not such a good indication whether a system is dilute- or dense phase.

The longer length of a conveying installation shifts the difference between dense- and dilute phase conveying towards higher velocities.

Again something learned.

Have a nice day

Teus

Attachments

zenzdiagram400mportlandcement (PDF)

■

Dear Mike,

In my program, vertical pipelines are calculated as vertical pipelines and horizontal pipelines are calculated as horizontal pipelines.

Consequently, there is no need to convert vertical pipelines into equivalent pipelines.

Moreover, this way of approximations is inaccurate, as they depend on loading ratio.

In bends, only the velocity loss is calculated. The related pressure drop is calculated as subsequent acceleration.

The discharge time for 5 m3 cement is calculated as:

5/6.2 * 242.9 = 195.9 sec. (see calculation pdf)

47 seconds less.

Have a nice day

Teus ■

Dear Mike,

Air volume through a pipe is given by:

Volume at p1 and T1 = velocity * pi/4*D^2

and

Volume at standard conditions p2=1 bar(abs) (Atmospheric) and T2=0 degrC:

Volume in nm3/dt = Volume at p1 and T1 * p1/p2 * T2/T1

or

Volume in nm3/dt = velocity * pi/4*D^2 * p1/1* 273/(273+temp)

Recalculated for every location in a pipeline.

Have a nice day

Teus ■