Dear Mr Nogueras,

Actually, you are asking for a preliminary design, rather than posing a technical or technological question.

But using your information, this case can be used to illustrate the way how such a pneumatic system is working and how it can be designed.

To be complete, I would like to know the make and type of the roots blower and if possible the name plate data.

This enables me to calculate the volumetric efficiency.

The rotary lock, I can calculate, unless you already have one. Then I need the data.

Regarding your questions:

a)30 m/sec is far too high

b)Friction coefficient is related to the calculation algorithm used and therefore no answer is possible.

c)pipe diameter (0.128 m instead of 0.128mm) and airlock losses will follow from the calculation for 12 tons/hr.

hear from you

teus ■

This sounds like a typical CKD scenario.

1. You will be fine with 15-16 m/s inlet velocity

2. 5” (130mm id) pipe. 12tph will work fine in this size pipe.

3. 20 m3/min FAD and 800 mbar differential will be sufficient for your application. This flow rate is in the middle of your blower curve so you will be fine with this blower and don’t forget to add RV air leakage to this value.

4. As far as frictional coefficients are concerned they are material dependent and are normally determined in pilot scale plants. ■

Dear Rene Laserna Nogueras,

The first step in this preliminary design was to calculate the suspension velocity of the

material.

I gave it the name limestone, although it is a mixture of 75% limestone and 25% clay.

Based on 80 microns average size and a particle density of 2500 kg/m3 the floating velocity of one particle is 2.04 m/sec

Then the minimum wall velocity is assumed to be 1.5 times the local floating velocity.

(Locations in the pipe line where the pressure is higher, the floating velocity is lower)

In the mean time the pipe line is modeled into a matrix, describing the pipe diameters, -lengths and angles as well as the curvature and orientation of the bends.

As I am very familiar with Aerzen equipment, ( I will be visiting them tomorrow) I just have to fill in the displaced volume per revolution and the leakage per 100 mbar op pressure difference and together with the blower rpm, the blower performance is calculated for the

set pressure.

After that, the capacity of the rotary lock is set at 20 tons/hour,

resulting in a rotary lock of 0.040 m3/rev. and 15 rpm at a filling efficiency of 0.7.

The following step was to calculate a capacity – pressure curve for the lowest acceptable velocities or airflows (blower rpm)

.

I made 2 calculations, one for 1200 rpm and the second for 1000 rpm.

The results of the calculations are given below.

Rene Laserna Nogueras rene d 09-06-2006

Pressure discharge Limest

Blower Aerzen GMa 13f7 – 1200 rpm

Convey length = 62 m

Nu of bends= 3

Pump vol = 0 .16 + 0.18 m^3/s

q-convey = 0.18 m^3/s

Dia begin = 128 mm Dia end = 128 mm

Pipevolume =0 .79 m^3

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

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

1.000------- 29--------- 44-----rot.lock ------- 7.0----------13.5-------0.82-----------8.46

0.750------- 24----------35-----rot.lock -------8.3----------14.1-------0.74-----------7.36

0.500------- 18---------25------ rot.lock -------10.2---------14.9------0.68-----------6.24

0.250------- 10----------12------ rot.lock-------13.2--------16.2-------0.71---------- 5.10

Rene Laserna Nogueras rene d 09-06-2006

Pressure discharge Limest

Blower Aerzen GMa 13f7 – 1000 rpm

Convey length = 62 m

Nu of bends= 3

Pump vol = 0.12 + 0.14 m^3/s

q-convey = 0.15 m^3/s

Dia begin = 128 mm Dia end = 128 mm

Pipevolume = 0.79 m^3

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

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

1.000----------20-------40-------rot.lock-------7.0 ------10.2---------1.00-------11.10---sediment

0.750----------20-------37------- rot.lock-------7.5------10.8---------0.77-------- 9.75---sediment

0.500----------15-------27-------rot.lock-------7.9-------11.7 --------0.68--------8.34

0.250----------9--------14-------rot.lock -------10.5----12.9----------0.66--------6.62

The product collision losses are considered to be similar to cement, by using the same

friction factor.

The product is also considered dry and non cohesive and therefore fluidizable and conveyable, which I think is justified.

As I do not have field data of this limestone product, the made calculations are to be considered as indicative.

If you already have pneumatic conveying installations for limestone in your plant, I could use the field data of these installations to increase the accuracy of the predicted results.

If you equip the rotary lock with a Volt/Hertz regulated drive, you will be able to control the capacity and thereby the pressure.

Take care for a high precision rotary lock, because the blow back of convey air through the rotary valve can influence the filling degree in such a way that the pneumatic conveying starts oscillating, whereby the pressure varies between f.i. 800 mbar and 150 mbar.

Also the rpm regulation can be coupled to the conveying pressure, in order to obtain a constant conveying pressure and capacity.

Do not forget to consult your regular suppliers with experience in this branch.

hope to have informed you sufficiently

best regards

teus ■

Dear Mr Nogueras,

Darcy equation is normally used for calculating air only losses. A number of modified forms of this equation are available to calculate solid contribution to the total system pressure loss which require these friction factors. Again these are only estimations and give a rough pressure drop estimate. Since there a number of equations available these coefficients also vary depending upon the equation form you are using. For better algorithms, pressure drop is broken into horizontal vertical straight pipe, bend and acceleration pressure drops they give better accuracies.

As I suggested in my previous post 800 mbar differential will be enough for your system. From my experience with this product I think your system pressure drop will be around 500mbar for the lengths involved.

Dear Mr Teus in my experience this type of dust is not fluidizable so your assumption for using cement conveying characteristic might not be correct. Couple of years ago we carried out conveying characteristics trials on different grades of this dust and found it cannot be conveyed in dense phase / low velocities

Regards ■

Dear Rene

From the data given for the vertical lift the pick up velocity is approx 14.75 m/s,

which is sufficient to convey the material at an SLR of 12-13 In a vertical pipe

I see no reason to be fearful for horizontal.

Furthermore as you can see this velocity is sufficient in a 500mm dia pipe and

saltation velocity will be lower for a 128mm pipe as slatation velocity is a function

of Froude Number. This has been discussed in the thread

“Froude Number”. Hope it is helpful in overcoming your fear.

Just out of curiosity what is the height of your vertical lift? ■

Dear Rene,

Geldart C gives < 30 microns and cohesive and difficult to fluidize.

I looked up the chemical notation of Limestone and found that it is mainle CaCO3.

This product is also used in the desulphurization of coal powered power plants.

It is called Cacium Carbonate and with this product I have recent experience in pneumatic conveying, both in vacuum and pressure mode.

It is correct that this product is difficult to fluidize, but it was also conveyable.

As this product is cohesive, there will be agglomerations of particles,forming a kind of virtual particles that are bigger.

Compared to cement (particle size and material density), I expect that a floating velocity of 1.35 m/sec can be used.

With your info from the air lift, I calculated the product friction factor.

The results of that calculation are presented as:

5/5 230.0 tons/hr

Press: 4500 Press

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

1 intake-----1.0---------17.7-------15.5------------458.------------ 7.3-------------- no

2 pipe-------74.0--------24.4-------22.2-----------4363. ----------- 8.6---------------no

3 outlet------------------24.4--------22.2------------4398.

4 filter-------------------------------------------------4500.

v-filter 0.50 m/min

No booster

Length 75 M

Residence time in pipe line = 4.03 sec

Power 267.kW

Energy consumption 1.16 kWh/ton (0.47 ltr gasoil/ton)

Re = 5.68 * 10^5

Rene Limestone ald



Notice that v-wall/v-susp has a value of 7.3 to 8.6.

This ratio represents the air velocity along the wall of the pipeline (which is lower than the average air velocity) divided by the local suspension velocity of the (virtual) particle.

A ratio of approx 2 is sufficient to keep particle afloat (no sedimentation)

The conveying curve is:

Rene al d 09-12-2006

Pressure discharge Limest

Convey length = 75 m

Nu of bends= 0

Pump vol = 4.19 + 4.19 m^3/s

q-convey = 4.20 m^3/s

Dia begin = 500 mm Dia end = 500 mm

Pipevolume = 14.72 m^3

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

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

0.450--------229------12 ------17.7-----------24.4-----------1.16-----------4.03

0.400--------214------11 ------18.3-----------24.4-----------1.17----------- 3.93

0.350--------196------10-------18.9-----------24.3-----------1.20-----------3.84

0.300--------176-------9--------19.7-----------24.3-----------1.26-----------3.74

0.250--------153-------8--------20.4-----------24.3-----------1.36-----------3.65

0.200---------125------7--------21.3-----------24.3-----------1.54-----------3.56

0.150---------92-------5---------22.2-----------24.2-----------1.95-----------3.47

0.100---------52-------3---------23.2-----------24.2-----------3.15-----------3.39

Then I recalculated the dia 128 mm installation, using the derived product resistance factor from the air lift installation.

Rene Laserna Nogueras rene d 09-12-2006

Pressure discharge Limest

GMa 13f7 1200 rpm

Convey length = 62 m

Nu of bends= 3

Pump vol = .16 + .18 m^3/s

q-convey = 0.18 m^3/s

Dia begin = 128 mm Dia end = 128 mm

Pipevolume = .79 m^3

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

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

1.000--------21--------31------ rot.lock ------ 6.9-------13.5-------1.14-----------8.40

0.750--------18--------26------- rot.lock-------8.3-------14.1------1.00----------- 7.22

0.500--------15--------20--------rot.lock------10.2-------14.9-----0.85---------- 6.04

0.250---------9--------11--------- rot.lock-----13.2-------16.2-----0.77-----------4.85

and:

Rene Laserna Nogueras rene d 09-12-2006

Pressure discharge Limest

GMa 13f7 1000 rpm

Convey length = 62 m

Nu of bends= 3

Pump vol = .12 + .14 m^3/s

q-convey = 0.15 m^3/s

Dia begin = 128 mm Dia end = 128 mm

Pipevolume = .79 m^3

----------Pipeline

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

1.000------18-------37-----rot.lock------5.2-----10.2---------1.06----- 11.42

0.750------16-------31-----rot.lock----- 6.3-----10.8---------0.93------9.67

0.500------13-------23 -----rot.lock-----7.9-----11.7---------0.79------ 7.95

0.250-------8-------13------rot.lock-----10.5-----12.9--------0.69-------6.22

No sedimentation.

If you are afraid whether the chosen velocities are too low, you can always increase the rpm of the Aerzen GMa 13f7 blower (max 3800 rpm)

A new calculation for a higher rpm is now easily performed as all the data are now stored in my computer.

Best regards

teus ■

dear Rene,

Yes, your interpretation of the calculated tables is correct.

With a pipeline length of 62 m and a residence time of 7.95 seconds, the average product velocity is 62 / 7.95 = 7.8 m/sec.

How the velocity is varying along the pipeline can be seen from the calculation below.

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

Calculation for the Aerzen blower GMa 13f7 (max rpm 3000)

GMa 13f7 1000 rpm

5/5 13.6 tons/hr

Press: 5000 mmWC

Press.drop : 5000 mm WC

Q-blower 0.137 m3/sec

Q-pipeline 0.130 m3/sec

rotary lock loss 0.07 m3/sec

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

1 intake----1.0------8.0---------7.0-----------231.---------------2.8-----------------NO

2 pipe------4.0------8.0---------7.3----------324.----------------2.8-----------------NO

3 bend---------------8.0--------4.3-----------325.

4 pipe------5.0------8.1--------7.4-----------499.---------------2.9-----------------NO

5 bend--------------8.2---------4.0-----------501.

6 pipe------47.0----11.5-------9.6----------4707.---------------3.4-----------------NO

7 bend--------------11.5--------5.7---------4709.

8 pipe------5.0-----11.7------10.4---------4930.----------------3.4 -----------------NO

9 outlet-------------11.7------10.4---------4938.

10 filter--------------------------------------5000

v-filter 0.39 m/min

No booster

Length 62 M

residence time 7.95 sec

power 11.kW

Energy consumption 0.82 kWh/ton

Re = 0.69

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

The calculation for the Aerzen blower GMa 12.5 (max rpm 3600)

GMa 12.5 1800 rpm

5/5 13.4 tons/hr

Press: 5000 mmWC

Press.drop : 5000 mmWC

Q-blower 0.134 m3/sec

Q-pipeline 0.127 m3/sec

rotary lock loss 0.07 m3/sec

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

1 intake----1.0---------7.7---------6.8--------------226.----------2.7---------------NO

2 pipe------4.0---------7.8---------7.1--------------318.----------2.8 ---------------NO

3 bend-----------------7.8----------4.2--------------319.

4 pipe------5.0--------7.9----------7.2--------------490.----------2.8 ---------------NO

5 bend-----------------7.9---------3.9---------------491.

6 pipe-----47.0-------11.2--------9.3--------------4717.---------3.3---------------NO

7 bend-----------------11.2-------5.5---------------4719.

8 pipe-----5.0---------11.4------10.2--------------4934.---------3.3---------------NO

9 outlet----------------11.4------10.2--------------4941.

10 filter ----------------------------------------------5000.

v-filter 0.38 m/min

No booster

Length 62 M

Residence time 8.19 sec

power 10.kW

Energy consumption 0.74 kWh/ton

Re = 0.67

To obtain the desired rpm values it is necessary to have a v-belt drive, as the required

speeds deviate from the synchronous speed of the electric power grid.

All for now

teus ■