Re: Primary Resistance Of The Conveying Flight Fh

Erstellt am 7. Aug. 2007 - 03:29

I believe it is the length of the flight to be analyzed.

Lawrence Nordell Conveyor Dynamics, Inc. website, email & phone contacts: www.conveyor-dynamics.com nordell@conveyor-dynamics.com phone: USA 360-671-2200 fax: USA 360-671-8450
Lyle Brown
(not verified)

Re: Primary Resistance Of The Conveying Flight Fh

Erstellt am 7. Aug. 2007 - 09:41

Typically these equations are saying

f=uN

Where f (FH) if the friction force

u (f) = coefficient of friction (or as I and some others like to call it "artificial coefficient of friction"

The other terms represent N (mass (idler, belt material etc) x gravity). I haven’t got DIN in front of me at the moment, however suspect the cos term is to resolve forces on angled components.

Regards,

Lyle

Axel Witt
(not verified)

Primary Resistance Per Din

Erstellt am 7. Aug. 2007 - 11:52

As far as I remember you have to consider the mass of the belt twice (carrying AND return run)!

Best Regards from Germany

Thanks

Erstellt am 7. Aug. 2007 - 04:57

Good day colleagues

Thank you very much for your help.

In fact cos () refers to the conveying/inclination angle,

I read that the use of these of the equation.

FH = f · L · g · [ m’R + (m’G + m’L ) · cos ä ] , is only justifiable

in a case where there are no exacting requirements.

On the other hand, if measurements for the running resistance of the idlers FR and the identation rolling resistance

of the conveyor belt FE are available, primary resistances

of the (loaded) top run and (unloaded) return run to be

determined more exactly as follows, rather than with

the use of the value f:

Top run FHo = · (FRo + FEo )/ qo

Return run FHu = · (FRu + FEu ) /qu

q is the coefficient to determine the primary resistances:

Top run: 0.5 ≤ qo ≤ 0.85; medium value: 0.7. Return run: qu = 0.9

But how can I get FRo and FEo

Fernando Diaz fernando.diaz@mecanika-sms.com

Re: Primary Resistance Of The Conveying Flight Fh

Erstellt am 19. Aug. 2007 - 02:44

Dear Mr. Fernando Diaz,

The main resistance for carrying run, the simple formula is = mu x (moving mass of carrying run supported by the inner race of the bearing) x 9.81 = mu x (carrying idler rotating mass per meter length of conveyor + belt mass per meter length of conveyor + material mass per meter length of conveyor) x conveyor length x 9.81

The answer will be in Newton when unit of mass is kg and unit of dimension is meter.

Similarly, main resistance for return run formula is = mu x (moving mass of return run supported by the inner race of the bearing) x 9.81 = mu x (return idler rotating mass per meter length of conveyor + belt mass per meter length of conveyor) x conveyor length x 9.81

I have not mentioned the effect of cos(delta) in above formula / explanation, because the question doesn’t warrant the degree of precision here. Generally conveyors are inclined up to 15 degrees and in that situation, it will affect the result up to 1.5%.

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.

Email : parimul@pn2.vsnl.net.in

Tel.: 0091 (0)20 25871916