You will find plenty of data on both cement (OPC and oil well) and barite in my Pneumatic Conveying Design Guide - 2nd Edition, published in 2004 by Elsevier Butterworth-Heinemann. There is data on both materials conveyed through rubber hose as well as steel pipeline. If you prefer the information in US engineering units try my Handbook of Pneumatic Conveying Engineering published by Marcel Dekker (now CRC Press). ■

Dear Matthew,

I can calculate the relevant parameters in a cement/barite conveying pipeline at any

desirable location as you request.

(for cement I have accurate data, for barite I do have also some data, but limited)

To perform such a calculation I need the piping geometry (diameter, length, angle and bends) and the data of the chosen compressor (type, displacement, maximum pressure).

Let me know those data together with the results of your theoretical calculations.

To check your calculations with an experiment would require building that installation, which would be a bit costly.

hear from you

teus ■

Dear Matthew,

After modeling your system, the following result was calculated:

Matthew mat d 03-12-2007

Pressure discharge Cement

Convey length = 54 m

Nu of bends= 3

Pump vol = .14 + .14 m^3/s

q-convey = 0.14 m^3/s

Dia begin = 127 mm Dia end = 127 mm

Pipevolume = 0.68 m^3

------------Pipeline

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

4.000 ------129----------203 ------------- 4.5---------14.0 ----------0.37--------------10.56

3.500-------121----------190 ------------ 4.7 ----------13.9---------0.35--------------10.22

3.000-------111----------175 ------------- 5.0----------13.8---------0.32------------- 9.88

2.500-------101----------159 ------------- 5.4----------13.7---------0.30-------------9.46

2.000 --------90----------141 ------------ 5.8-----------13.6--------0.27 -------------8.98

1.500---------77----------120 ---------------6.3 ----------13.4--------0.25 -------------8.42

1.000---------59-----------92 ----------------7.1----------13.3 --------0.24 -------------7.81

0.500---------35----------54 ----------------9.0 -----------13.0-------- 0.27 ------------6.78

Empty pipeline pressure ...= 267.1 mmWC

The detailed data for the calculation for 3 bar(o) or 30000 mmWC are as follows:

5/5 111.4 tons/hr

Press: 30000

Press.drop :30000

Part--------length ----v-air------v-product-----press.drop

1 intake ---1.0 -----5.0 ------- 4.6 --------------1194

2 pipe-----24.0-----5.7---------5.2---------------8715.

3 bend --------------4.7--------2.0-------------- 8716.

4 pipe-----12.0-----9.4--------7.5--------------24440.

5 bend--------------9.4---------4.0 -------------24442.

6 pipe ----12.1-----11.6------10.2--------------27801.

7 bend-------------11.8--------5.5--------------27804.

8 pipe------5.1-----13.9-------12.0-------------29971

9 outlet------------13.9 -------12.0------------29982.

10 filter-------------0.2 -------------------------30000

v-filter 0.21 m/min

No booster > Length 54 M

Residence time 9.89 sec

36.kW

0.33 kWh/ton

Matthew Cement matd Re = 0.75

Matthew

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

PRODUKT :Cement

Convey Length = 54 m

Number of Bends = 3 -

D-begin =127 D-end =127

Outlet force .. = 2145 N (dynamic)

Q-pump....................= 0.142 m^3/s

Q-convey-pipe ............= 0.142 m^3/s

Ambient temperature ......= 40.0 °C

backpressure at pipe-end..= 0. mmwC

NO BOOSTER

Pipeline capacity ........= 111.40 ton/hr

System-pressure...........= 30000. mmwC

loading-ratio ............= 175.11 -

T-Cement--------=41.9/ 42.1 Deg.C

Reynoldsnumber ..[ Re ]...= 0.75 --

Compr power .....= 36. kW

T-out compressor = 319. Deg.C

Mass in pipeline.= 569 kg

density product/air mix ..= 197.9 kg/m^3

spec.energy-consumption...= 0.33 kWh/ton

dp-accell.excl prod.rest..= 4753. mmwC

dp-suspension.............= 12625. mmwC

dp-lifting................= 5296. mmwC

dp-airfriction............= 125. mmwC

dp-productresistance......= 7075. mmwC

dp-intake productcolumn...= 100. mmwC

dp-intake ................= 7. mmwC

dp-nozzle ................= 1194. mmwC

dp-filter.................= 18. mmwC

Empty pipe dp....= 265 mmWC

The above calculation(s) now can be made for any chosen situation for air volume and pressure.

That will generate so many data, that it would become a problem to monitor them all.

In case you want to have calculated special situations, please let me know.

Note that the given capacities are pipeline capacities and are not corrected for the timelosses caused by a double kettle system.

Have fun

teus ■

Dear matthew,

it is not that amazing. The computer does all the work.

1)

cap =capacity in tons/hr

mu = Solid Loading Ratio in (kg material/sec) / (kg air/sec)

residence time = time that a particle travels from the beginning of the pipeline to the end.

dp = (delta pressure) pressure drop (increment)

2)

The inflow of material to the intake requires a force that is generated by a pressure drop and air has to be accelerated.

Turbulence at the outlet consumes also some pressuredrop

3)

filter is for dust/air separation, so that the convey air can be released to the atmosphere clean.

4)

Total set pressuredrop is 3 bar(g) # 30000 mmWC

velocity in m/sec

5)

Mixture density is the weight of 1 m3 of product/air mixture

When you indicate the conditions you want to have calculated, the program will solve that.

best regards

teus ■

dear Matthew,

I can calculate some more conveying situations for you, no problem.

But if you indicate the situations you want to calculate, to meet your requirements, please let me know.

The you can get the most benefit of it.

best regards

teus ■

Dear matthew,

The calculation of 60 tons/hr at 60 psi (4.14 bar) will be made.

Low pressure drop over bend is explained as follows:

In a bend there will occur complete air-product segregation.

The only pressure drop is because of air friction.

The material slides against the outer wall of the bend, where it is decelerated.

After the bend, the material is entrapped in the air stream again and re-accelerated,

causing an increased pressure drop, due to the loss of velocity in the bend.

best regards

teus ■

Sliding flow in bends only happens in lean phase and is pretty much as described by Mr Teus. The actual bend pressure drop occurs in the straight section immediately after the bends known as “acceleration zone” and it is normally attributed to bend pressure losses. This is pretty much text book stuff.

But at the above mentioned solid loading ratios / velocities, which by all definitions is dense phase material will convey in plug flow. In plug flow there is no significant velocity loss in the bend because of low velocities and the plug already sliding on the pipe wall. For approximation in dense phase bend pressure losses are almost equal to the equivalent straight length.

For above mentioned distance for cement you will get a constant conveying rate of 60 tph in 5” line with exit velocity of 14m/s blow tank pressure will be between 1.8 - 2 barg. ■

Moving bed or dune flow depends on material properties and conveying velocities (2-6 m/s range). Granular material like sugar salt etc. are conveyed in this mode. Dune flow is limited to first horizontal and as soon as flow moves form horizontal to vertical full bore plugs are formed which convey to the end conveying line end if the line is not stepped.

If you know your air flow rate at FAD exit velocity is calculated by dividing it with the cross sectional area of the conveying pipe. As far as conveying pressure drops are concerned they are from experience. I haven’t seen a reliable pressure drop predicting model for dense phase. Scaling from test plant data is used for designing purposes and is very reliable.

Your original post asking about the actual data was in the right direction. You can only get this data either from books as mentioned above or from some published papers (academic sources) only. Commercially all the good PC manufactures have test plants and have plenty conveying data but is never shared. ■

Dear Matthew,

The original calculation are for 1.5D bends (R-out = 0.191 m)

The result was 111.4 tons/hr at 3 bar

The recalculation for 5 D bends (R-out=0.7m) results in:

11.6 tons/hr at 3 bar.

Not much difference though.

The velocity losses in the 1.5D and 5 D bends are almost the same.

More over the bends form just a small part of the installation and therefore their influence is limited.

take care

teus ■

Correction:

111.6 tons/hr at 3 bar

sorry

teus ■

Dear Matthew

Attached the promised calculations.

I calculated for 4 different air volumes for the same pipeline.

you will find the capacity = function (pressure) tables

and the velocity calculations for each air volume at 3 bar.

A 60 ton/hr installation can be designed with a smaller installation.

The chosen high pressure installation can be improved with stepped pipelines.

best regards

teus ■

the file

teus

Attachments

dear matthe2 (ZIP)

■

dear Matthew,

You are correct.

The figures in tons/hr are the cement rates (capacity) at 3 bar.

At the end of the conveying line thers is the filter pressure left (which is almost atmospheric)

The pressure of the compressor is completely used after the filter.

When I calculate a pneumatic conveying system, I want to know which capacity belongs to which pressure.

That requires that I choose one of the 2 as a set value.

In this case I chose the pressure to be set at 3 bar and calculated the corresponding capacity.

If you want it the other way around (pressure at a set capacity), that is also possible.

The installation is now modeled in the computerfiles and I can calculate in any setup now as you like.

success

teus ■

Dear Matthew,

I cakculated your installation for :

60 tons/hr

5 inch pipeline

various airflows until the pressure was approx 4 bar

Actually this is the Zenz diagram

Air flow---------pressure----------kwh/ton---------SLR------mix density

m3/sec----------mmWC----------------------------kg/kg----------kg/m3

0.06-------------40686---------------0.35----------220.7--------222.7

0.07-------------32133--------------0.32---------190.9-----------214.4

0.08-------------25229--------------0.29-----------167.0---------189.4

0.10--------------15319--------------0.24-----------133.6---------153.6

0.12--------------11337--------------0.22----------111.4----------129.2

0.14167---------10240--------------0.25-----------94.3-----------110.4

0.18--------------10277--------------0.31----------74.2------------87.8

0.24--------------11270--------------0.45----------55.7-------------66.6

0.28--------------11999--------------0.55-----------47.7------------57.4

0.32---------------12795-------------0.66-----------41.7-------------50.5

0.38---------------13917-------------0.83-----------35.2------------42.8

0.45---------------14633-------------1.03-----------29.7-------------36.4

If you plot this table into a graph you will see the Zenz diagram

success

teus ■

Dear Matthew,

You are welcome

teus ■