1) when you say 50m high do you mean it's an inclined conveyor?

2) power is a function of the te/hr conveyed, which you haven't told us1 ■

As designer says we need to know your tonnage. At first glance the power demand seems excessive for the largest standard belt sizes and carrying coal which is comparatively light, however wet. Lets see your numbers.■

Dear Aspiring Engineer,

It is not possible to explain the conveyor design procedure on such forum. Please obtain the belt conveyor design standard such as DIN, ISO or your countries published standard for this purpose. Such standard describes the power calculation formulae in a very clear manner. The conveyor power calculation is relatively easy in context of mathematical formulae. Of course the complexity or crux of the issue is to have a right kind of input into these formulae which makes difference with respect to expertise, experience and ultimately quality of the result. Your calculation for conveyor power requires many important decisions such as belt width, speed, idler type and size, idler pitch and so on which become input into those formulae. So these preparatory decisions are also complex. The conveyor design beyond the power calculation is specialised issue.

Regards,

Ishwar G Mulani.

Author of Book : Engineering Science and Application Design for Belt Conveyors.

Author of Book : Belt Feeder Design and Hopper Bin Silo

Advisor / Consultant for Bulk Material Handling System & Issues.

Pune, India.

Tel.: 0091 (0)20 25871916

Email: parimul@pn2.vsnl.net.in ■

Originally Posted by I G Mulani

Please obtain the belt conveyor design standard such as DIN, ISO or your countries published standard for this purpose. Such standard describes the power calculation formulae in a very clear manner.

Not sure I can agree for an "Aspiring Engineer". I think it it good for engineering students to develop the ability to "work from first principles", not just stick some numbers in someone else's formula. While in the case of belt conveyors there are numbers of formulae that can be used, for many other bulk materials handling machines published formulae do not exist, or those that do can be very suspect (and don't get me started on the subject of "standards" )

An engineer needs to be able to think and develop his own methods, dropping numbers into a formula from a book doesn't make you an engineer ■

[QUOTE=designer;63833].....I think it it good for engineering students to develop the ability to "work from first principles", ......QUOTE]

The Egyptian, so called, engineers I'm working with here in KSA don't exhibit any principles at all; let alone first principles.

But the intent of the reply is perfectly correct. I just have a plague of frogs and locusts outside my office!

And I still think the power is too high for a young feller to be involved with. ■

Not too high if you are transporting 5000 T/hr ?? ■

Since you did not offer the tonnage, maybe we give the following:

900 kW = 5400 t/h based on 5 m/s speed, 2 m spacing, belt width according to tonnage and coefficient = 0.1667 (kW/Ton (m)/ 200m long x 50 m lift)

900 kW = 0.1667 x 5400 t/h pretty close to Mr. Holt's 5000 t/h comment. ■

I should have stated the formula is good down to about 2000 t/h.

ie. at 2000 t/h you need about 350 kW. ■