Dear Mark,

Testting a pneumatic conveying installation should comply with the design specification.

Run the installation first without convey air and without product, after that with convey air and without product and finally with product,

starting at lower pressures (capacity) and then increasing the pressures gradually, until the design situation is reached.

With a better description of your installation a testing protocol can be defined.

Do not be surprised, when your testing procedure has to be adapted in the testing proces.

Also the circumstances and the available equipment can influence the testing possibilities.

success ■

A PFD or P&ID would enable a test protocol to be defined and explained.

We do this all the time for companies like yourself.

Please don't hesitate to contact us for detailed assistance once you have more information to share

Regards ■

dear Myuan62,

I calculated your installation as

Cement

conveying length 213 m (700 ft)

pipe diameter 0.145 (6 inch)

Pumpvolume 0.5 m3/sec (based on 100 ft/sec at atmospheric conditions and a 6" pipe)

pressure 0.9140 bar (13 psi)

The resulting capacity is then approx 37 tons/hr at 9140 mmWC

In your post it is not clear what the 65 psi means.

To calculate your installation as built, you should provide the real geometry of the pipe line

and the used convey air compressor properties.

(Preferably in metric units)

best regards ■

Dear Mark,

I calculated a positive pressure system with 0.9140 bar(o) or 13 psi pressuredrop.

If the pipeline pressure of 65 psig is the conveying pressure drop, then the capacity in cement will be approx 105 tons of cement per hr.

But the conveying is very unstable due to the fact that the

Capacity = function(pressuredrop) curve is almost horizontal at 4.5 bar(o).

A small increase in feeding will result in a significant pressure increase and possibly a choking pipe.

Now I cannot understand what the pressure drop of 13 psi means.

If it is a pressure system, should the 65 psig not be the air outlet gauge of the compressor (or are you referring to the air inlet gauge of the pipe?)

greetings

teus ■

What K factor did u use for these calculations ??? ■

For 8000 lb /hr in the given system gives SLR of approx 1.7 and calculated 0.9 bar pressure drop is very high. Kindly describe the system in detail to get any meaningful explanation.

Just for curiosity how did you come up with this K value of 1.2 ? ■

dear Mark,

There seems to be a discrepancy between your calculation and my calculation.

If you can share the description of the real installation and the field observations made so far,

it will be possible to solve those discrepancies.

awaiting your information,

have a nice day

teus ■

Dear Mark,

gas density = 0.37 lbs/ft3 # 6 kg/m3

This density occurs at 4.5 bar and 50 degrC

This represents the measured situation and not your calculated pressuredrop

(The bulk density of the cement is not important for the pneumatic conveying calculation)

95 lbs/m3 # 1522 kg/m3, which is settled, not fluidized cement.

The pneumatic conveying calculation results in :

Cement

Capacity 105.2 tons/hr

Press: 45000

Press.drop :45000

prod.loss.fact 0.02140

Velocities and pressure drop along the pipeline

Part-------------length ----------v-air-------v-product-----press.drop-----v-wall/v-susp

1 intake --------1.0 -------------7.2-----------6.7------------882.------------ 4.1

2 pipe ---------141.1----------11.4---------10.5 ----------22140------------ 5.0

3 d.tr --------------------------11.4----------10.5-----------22140.

4 pipe ---------48.1-----------15.7 ---------14.2 -----------31547.-----------5.9

5 bend-------------------------15.4----------8.0--------------31556.

6 pipe -----------0.0----------16.2----------8.8--------------31645.-----------6.1

7 d.tr ---------------------------16.2-------- 8.8-------------31645.

8 pipe ----------13.1-----------22.6--------19.6-------------39019-----------7.0

9 bend------------------------- 22.7---------11.3----------- 39031.

10 pipe-----------10.1---------34.9---------29.8 -----------44838.------------8.7

11 bend ------------------------35.9---------16.9-----------44858.

12 outlet------------------------35.9---------16.9-----------44932.

13 filter--------------------------0.5 ---------------------- 45034.

v-filter 0.49 m/min

No booster Length 213 M

Residence time = 22.73 sec

Power = 194.kW 1.84 kWh/ton

Re = 2.31

Overall data and partial pressure drops

PRODUKT :Cement

Pipeline capacity ........= 105.16 ton/hr

System-pressure...........= 45034. mmwC

Convey Length = 213 m

Number of Bends = 3 -

D-begin =145 D-end =145

T-Cement 54.5/ 46.5 Deg.C

Q-pump....................= 0.500 m^3/s

Q-convey-pipe ............= 0.500 m^3/s

loading-ratio ............= 46.84 - - (dynamic)

Outlet force .. = 5245 N

Compr power .....= 194. kW

spec.energy-consumption...= 1.84 kWh/ton

NO BOOSTER

Ambient temperature ......= 40.0 °C

Reynoldsnumber ..[ Re ]...= 2.31 --

backpressure at pipe-end..= 0. mmwC

p-accell.excl prod.rest..= 6236. mmwC

p-suspension.............= 6019. mmwC

p-lifting................= 1329. mmwC

p-airfriction............= 1739. mmwC

p-productresistance......= 29463. mmwC

p-intake productcolumn...= 100. mmwC

p-intake ................= 35. mmwC

p-nozzle ................= 882. mmwC

p-filter.................= 102. mmwC

density product/air mix ..= 56.4 kg/m^3

Mass in pipeline.= 803 kg

Empty pipe dp....= 14767 mmWC

Due to the high pressure (4.5 bar(o) the air pressure drop is only 1739 mmWC # 3.8%

The air pressure drop increases to 14767 mmWC # 1.47 bar(o) for an empty pipe line

It could also be, that K does not play a big role because K is so small.

To prevent that we are getting mixed up in a forest of numbers, it should be better to focus on the as built installation first (I understand it is already existing) and then start calculating.

Let us know

teus ■

Good day Mark,

The given capacity calculation is for 4.5 bar(o) # 45000 mmWC (mm Water Column) # 65 psi

In the mean time I discovered an error in the previous reply:

The empty pipe line pressure drop (The blower only blowing air through the pipe line without any product) is NOT 14767 mmWC but only 4100 mmWC.

I apologize for that mistake.

The used computer algorithm is developed by myself over the past 27 years and is based on the calculation of energies acquired by the particles and friction losses.

That energy is delivered by the expansion of the carrying gas.

The calculation develops in the time domain (dt=0.001 sec) and through an iteration the formulas are solved.

The input values are then corresponding with the output values.

F.i. A capacity of x tons/hr at a set pressure of 2.5 bar is iterated towards 2.5 bar by solving the value for x.

One result is calculated at one time

Many calculations can form a table or a graph.

take care

teus ■