friction models conveyor ppt
TRANSCRIPT
9th ICBMH Newcastle NSW 2007. Keynote Address
Summary
Running friction and profile determine conveyor power and dynamics.
Numerous design method exist (e.g. DIN 22101, CEMA, In‐house).
Computer programs incorporate design simulation algorithms.
For long conveyors, design models need to accommodate :‐
1. Temperature effects.2. Vertical and horizontal curves.3. Troughed, pipe and cable belt systems.4. Idler, belt and material properties.
1.0 IntroductionConveyor Belt Rolling Resistance Methods – References (1950 – 2007)
1950 – 1970
1954 – P. Lachmann (Clouth) > Resistance of Full Conveyors (Germany)
1956 – 1960. A. Vierling & H.H. Oehman > Travel resistance, measurements (Germany)
1960 – 1970. U. Behrens, F Schwartz > Conveyor Resistance Theories (Germany)
1970 – 1980
1976 – CEMA and Goodyear, Conveyor Equipment Manufactures , handbooks (USA)
1978 – 1991. C. Spaans >Indentation Resistance and Belt Flexure (Netherlands)
1979 – ISO 5048 > Calculations for Belt Conveyors (and DIN 22101) (Germany/Europe)
1980 – C. Jonkers > Indentation Rolling Resistance (Netherlands)
1980 ‐2007
1987 – 1998. A. Harrison, et al : various > Rolling Loss Test Rigs, Alt. Models (Australia/USA)
1987 – 1997. L. Nordell, et al. : various > Power/Rubber losses, Applications (USA)
1993 – 2005. M. Hager/A. Hintz, et al > Energy Saving, Long Conveyors (Germany)
1995 – 2003. G. Lodewijks : various > Rolling Resistance of Rubber Types (USA/Netherlands)
2003 – 2006. C. Wheeler, et al. : various > Indentation, FEA, Material Flexure (Australia)
General :Te = L g [R + B + V] + Q v + P + O : T1 = (Te + T2) (in N)
T1
T2V
L
Q (t/h)v (m/s)
R = Resistance of Idlers (carry and return)‐(CEMA Kx, DIN has separated values)B = Belt and Material Flexure (carry and return)‐(CEMA Ky, DIN n/a)
Variable CEMA DIN 22101 (or ISO 5048)R Kt Kx = Kt (0.0068(Mb+Mm) + Ai/Sc f ct C Ai* (1 < ct < 1.7)B Kt Ky Mb + 0.015 Mb Kt + Mm Ky f ct C (2 Mb + Mm) cos δ
(cos δ ~1 if slope δ<18 º)(0.01 < f < 0.04) DIN factor
V Mm H/L Mm H/L
Comparison of coefficients for CEMA and DIN methods.
General Components of “Te” at the Element Level
F j F j+1
D
MF(j)= f(v, Sag, Type)
BI(j)BF(j)
R(j)
j j+1
Fn = (Mm+Mb) g
MmMb
Fn
BI = f (T, D, Fn )1.33
T = TemperatureD = Idler DiameterS = Idler Spacing
S j
‐40 C 0 40 C
R(j)SBR
KNNNSC
Small D
Large D
Coal
Lig
Cem
MF(j)
Ore
3.0 Hybrid Non‐Lin Models
for (int j = 0; j < lsn; j++) //
{ T = g L [R + {BI + BF} + MF + V] + Qv + P + O)}
Material Flex: MF = M Sz (p + q v)(1.66 Sc/Sr))s
(0.2 < p < 3), (0.02 < q < 0.6), (1 < s < 3)M =(Mb+Mm), z = carry or return
Belt Flexure : BF = mu Kt1/2 M Sz3/2 / L (0.003 < mu < .008)
Indentation : BI =(d tanδ v1/4 / g f(Dz)) . (g M Sz)4/3
tanδ = Rubber Loss(T) = f(v, Sz(j), τ) + f(T) τ = Viscoelastic time constant at a frequency = v/Sz (Hz).f(Dz) = (k1. idz + k2)
Idlers : R = Kt1/2 [Ks + Az) (v/2)2/3exp(- c T) / Sz) ]
RUN >
Input Selections
Model Values
Example 1: 16.8 km belt, 5% slope
Example 1. : Model Values
Inputs
Example 1. – Long Conveyor, 5% Slope
SBR
Nat./SBR
Tensions
Example 1. – Natural Rubber
Example 2. Horizontal and Vertically Curved Pipe Belt
Example 2 : Pipe Conveyor (Cement & Limestone
Example 2 : Dual Carry, Natural Rubber
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