With 8.6 m3/min of air in a 100 mm or 4" pipe your exit velocity will be about 16.2 m/s not 20 m/s assuming no air leakage. Recheck your calcs. You should be looking at 12 – 12.5 m3/min range for this assignment but then your solid loading ratio will drop to 6 if you want to keep it at 8 use 3” pipe. ■

For a 4” pipeline doing 5 tph using 750 m3/hr of air will give you an exit velocity of 23.61 m/s assuming no air leakage through feeding device. At these conditions the system pressure drop will not exceed 0.5 bar and SLR would be 5.6. At these sort of pressure the inlet velocity would be about 16 m/s. If you want to keep SLR at 8 you will need to increase the solid flow rate to 7.2 tph. 4” line will take this increased solid flow rate but the system pressure drop will increase and slightly higher amount of air will be required due to increased system pressure drop. The other option is reducing the pipe size to keep the SLR high as I mentioned before.

If your system is running at 1 bar and your inlet velocity is 20 m/s only then the exit velocity will become 40 m/s. 20 m/s inlet is slightly higher for this product 15-17 m/s is sufficient. Try to keep exit velocity low 40 m/s is bit too high. ■

Well I think this system is a suitable candidate for vacuum conveying in a 4” pipe. 520 m3/hr of air at the product intake point is enough to do the job. 400 mbar vac. is enough. There is no optimised SLR range but general range for lean phase is between 5- 10 below SLR 5 the wear rate increase exponentially. I would try to keep the air velocity as low as possible to reduce air only losses as low as possible. ■

Dear kevinjcg

How are you ?

I will run my program for you on this installation, but I am wondering if the particle size you state in your first post of :

"over 60% less than 0.2um,"

0,2 um is extremely small, is this correct ?

It is essential to know this size, in order to calculate (estimate) the suspensionvelocity.

For that matter the material density is also important.

Please let me know those figures and I will come back to you.

best regards ■

Dear Kevinjcg

Input data :

Limestone = CaCO3

average particle size = approx. 140 micron

suspension velocity = 2,5 m/sec

bulk density = 640 kg/m3

material density = 2670 kg/m3

pipeline :

horizontal length = 38 m

vertical length = 12 m

number of bends = 4

pipe diameter = 0,1 m

Roots blower : 8,6 m3/min at 1 bar max.

Calculation results for pressure conveying :

kevinjcg lstke d

Pressure discharge CaCO3

Convey length = 50 m

Nu of bends= 4

Pump vol = .14 + .14 m^3

q-convey = 0.14 m^3/s

Dia begin = 100 mm

Dia end = 100 mm

Kettle volume = 4.0 m^3 CaCO3

vol.= 3.0 m^3

Kettle CaCO3 content = 1.9 tons

Pipevolume = .39 m^3

Pipeline

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

1.000 - 22 - 36 - 9.9 - 18.6 - 0.64 - 5.8

0.900 - 21 - 34 - 10.5 - 18.6 - 0.63 - 5.5

0.800 - 20 - 31 - 11.0 - 18.5 - 0.62 - 5.3

0.700 - 18 - 29 - 11.7 - 18.4 - 0.61 - 5.0

0.600 - 17 - 27 - 12.4 - 18.4 - 0.61 - 4.7

0.500 - 15 - 24 - 13.2 - 18.3 - 0.62 - 4.5

0.400 - 13 - 21 - 14.1 - 18.3 - 0.65 - 4.2

0.300 - 10 - 16 - 15.2 - 18.2 - 0.73 - 3.9

0.200 - 6 - 10 - 16.4 - 18.1 - 1.08 - 3.6

In my work, I have some recent experience in vacuum-unloading from coastal vessels and pressure discharging Calcium Carbonate into a silo at a distance of approx 130 m horizontal and 30 m vertical

I used the product loss factor, derived from those unloadings.

The Calcium Carbonate is used to desulphurize the exhaust gases of a coal-fired power plant.

Any more questions, let us know

greetings ■