Dear Xavier,

I recalculated Belt 1 and got the 6000 tons/hr with a freeboard of the ore from the belt edge of 170 mm

By using the same formula, I recalculated Belt 2 and got 2500 tons/hr with the same freeboard.

After increasing the freeboard to 205 mm, the capacity reduced to 1800 tons/hr.

As you said, “Seems simple enough”

Continuity equations never fail.

Equal calculation algorithms give equal results. Only the input influences the output.

Take care

Teus ■

I always think it important when learning to be able to work from first principles.

In the case of mechanical conveyors you have a requirement to move X te/hr (the gravimetric capacity) and Y m3/hr the volumetric capacity. The relationship between the two being the bulk material density in te/m3.

The volumetric capacity determines the physical size of the machine, and the gravimetric capacity determines the required power to drive the machine.

So establish the volumetric capacity you need and then use this to establish the machine size. Then establish the required power according to the machine size selected.

This is where experience come in. There are many design guides you can use, but most established designers will have developed their own techniques based on feedback from machines they will have installed and have factors not necessarily in accordance with those published guides or national standards. If you make a living selling equipment you must keep some knowledge commercially confidential or else you will be out of business, undercut by some dirt cheap company living on other peoples knowledge.

You say you are a student. So, speak to your professor and ask him to obtain copies of the CEMA, MHEA, ISO etc guides for your university library. They will always come in useful for those who follow you. ■

Dear Xavier,

The “freeboard” is taken from the edge of the belt.

I choose the freeboards in such a way, that the outcome matched the mentioned values.

The input variables are based on experience (which every engineer must have or every new engineer must experience).

The calculation spreadsheet is attached for you only (NO guarantee)

success

Teus

Attachments

beltconveyor_1 (ZIP)

■

Originally Posted by Xavier Balayer

Hello,

"Belt conveying of minerals" by E. D. Yardley and L. R. Stace. .....

I've been using the formula described in the book above (which I verified myself multiple times), which calculates the area between the troughing idlers and the semi-circle above it using the surcharge angle. From there I simply multiply it by the density of the iron ore being conveyed and the speed of the belt. Seems simple enough. .....

The book mentioned earlier is unfortunately too basic for the conveyor belts I have to work on.

It is simple enough: We select the belt width, centre roll length & trough angle; then we give a 50mm edge distance and add the area of the trapezium to the area of the arc sector formed between the side rolls at the surcharge angle. That is the basic geometry & hasn't changed with time.

Of course there will be those who insist that the roll has bent under the load, that the belt thickness has deformed, that the belt corner curvature detracts from the simple trapezium area, that there might be visco plastic densification of the material as it passes from centre span to roller.

I won't be listening & neither should you.■

Well: maybe I'll keep one ear open.■

Important points not discussed:

1. Allowance for belt tracking error specified by CEMA and DIN and ISO: about 5.5% * Belt Width + .9 inches

2. Lump size containment: needs about 0.5 * Lump Size + belt tracking allowance

Further, trough angle has an effect when lumps size is considered. In addition, dry vs wet bulk density, center roll length, trough angle, how material is loaded.

Take not of SME paper from KPC, authored by Mr. Poltak Sinaga, showing a much smaller edge distance is possible while operating at 8.5 m/s.

Other issues include idler spacing. Two points with wider idler spacing:

3. Idler Junction Stress - pressure on center roll at its outer edge must be limited to prevent pitting or divoting due to excessive shear stress

4. Shape loss and resulting material trampling loss (increase in power) due to exaggeration on sag shape

Standards are about 5 years or more behind common sense engineering with today's analytic tools.

Specifically, proper chute loading, perserving kinetic energy from discharge velocity (belt speed) and from fall distance and increase in particle stream velocity. This savings can amount to more than 100 kW per transfer on high tonnage and fast moving belts. This does not include the belt life wear loss factor.

Too not include these points is to through client money away. ■

That I can't spell is a mystery of old age. ■

Small Corrections to Postings:

1. 6000 t/h running at 3.95 m/s, 20 deg surcharge, 1200mm wide belt will have a clear edge distance of 83mm. Not enough to contain the belt against spillage and not recommended by any notable belt designer. Thus, what is the necessary speed to maintain belt against spillage?

Assume the maximum lump size is 152 mm. The belt speed will be 4.94 m/s to contain the lump with and edge distance of 137 mm.

2. Problem 2: Belt speed would need to equal 2.65 m/s to contain 150mm lump size at all give condtions.

------------------------------------------------------------------------------------------------------

Change the lump size, accept spillage, fiddle a little with speed and you can do almost anything.

What is the point of providing meaningless knowledge to designs that will not be built? At least I hope that it will not get that far.

Maybe it is better not to respond when the answer has no bearing on bettering the design communitee. Any answer will do. ■