How to Calculate Belt Pull on RollerSupported Conveyor Belts metric

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Rulmeca Corporation explains how to calculate required belt pull and power to move packages on a roller-supported belt conveyor, using metric units. For more conveyor info, go to https://rulmecacorp.com/motorized-pul... • 0:00 Introduction • 0:23 Roller Conveyor Definitions • 0:43 Related Videos • 1:19 Power Calculation Equation • 1:37 Mass to Weight Conversion • 2:25 Belt Pull Equation • 2:52 Definitions of Terms • 3:18 How to Use the Equation • 7:39 Selecting Drive Power • 7:54 Sensitivity Analysis • The power required to move a load on a belt conveyor may be calculated as follows: • Required Power = Force x Velocity • Required Power = Belt pull x belt speed • When using metric units cargo is usually specified in units of mass (e.g. kg), but belt pull is expressed in units of force (e.g. N). Therefore, it is always necessary to convert units of mass to units of weight (measured at the earth’s surface). • 1 Newton = 1 kg x 1m/sec2 • At the earth's surface, a 1 kg mass has a weight of 9.81 N. • 9.81 N = 1 kg x 9.81 m/sec2 • For example, a package with a mass of 25 kg has a weight of 245 Newtons. • Use this equation to make the conversion. • Weight = mass x acceleration due to gravity • Weight = 25 kg/pkg x 9.81 m/sec2 • Weight = 245 N/pkg • The belt pull required to move packages on a roller-supported conveyor belt equals • F = (0.04 x L x (2Pn +Ppr + Pm)) + (H x Pm) • Where, • L = conveyor length • H = material lift height • Pn = (weight/meter of conveyor belt) • Ppr = (weight/meter of all rotating parts) • Pm = (weight/meter of product to be handled) • If a roller brand and belt supplier have not been selected, then use: • 0.04 as the roller bearing frictional coefficient • 73 N/m of conveyor belt • 73 N/m as the weight per foot of all rotating parts • If the roller is especially long or if the roller has a thick wall, then you should consult your roller supplier for specific details. • Consider a 30-meter-long, inclined, roller-supported, belt conveyor continuously moving 25 evenly spaced boxes, with an average weight of 245 N/pkg, at a belt speed of 0.5 m/s on a conveyor belt during an 8 hour shift. • The belt pull is calculated as follows. • Belt pull required to overcome roller bearing friction: • • L = 30 meters • • Pn = 2 x 73 N/m of belt = 146 N/m of belt • • Ppr = 73 N/m of rotating parts • • Pm = (25 packages x 245 N/package) / 30 meters = 204 N/m of product • • Total weight on rollers = 423 N/m • Belt pull to overcome roller bearing friction = 0.04 x 30 m x 423 N/m = 508 N • Belt pull required to overcome gravity: • • H = 10 meters • • Pm = (25 packages x 245 N/package) / 30 meters = 204 N/m of product • Belt pull required to overcome gravity = 10 meters x 204 N/m = 2,040 N • Total belt pull required to move packages = 508 N + 2,040 N = 2,548 N • Required Power = Belt pull x belt speed • Required power = 2,548 N x 0.5 m/s • = 1,274 Nm/sec • Now convert that to a useful unit of measure. • In metric units, 1 kW = 1,000 Nm/sec. • Required power = (1,274 Nm/sec) / (1,000 Nm/sec)/ kW • = 1.27 kW • Now select an appropriate conveyor drive system. • Here’s a tip to selecting an adequate amount of drive power for a conveyor. Select a power and then do a sensitivity analysis, verifying that all design assumptions are correct. • Select a 2 kW drive and check the validity of the design rate. • Suppose the design rate was based on a daily average handling rate calculated as follows: • • Total package mass handled during 8-hour shift = 300,000 kg • • Total number of packages handled during 8-hour shift = 12,000 packages • • Avg. pkg mass = (300,000 kg/shift) / 12,000 packages/shift) = 25 kg/package • However, suppose while double-checking the design parameters you learn that heavy packages (60 kg/box) are handled from 8:00am to 10:00 am and light packages (13.3 kg/box) are handled from 10:00am to 4:00pm on every shift. • Weight = mass x acceleration due to gravity • Weight = 60 kg/pkg x 9.81 m/sec2 • Weight = 589 N/pkg • Check required conveyor drive power from 8:00 am to 10:00am. • Belt pull required to overcome roller bearing friction: • • L = 30 m • • Pn = 2 x 73 N/m of belt = 146 N/m of belt • • Ppr = 73 N/m of rotating parts • • Pm = (25 packages x 589 N/package) / 30 m = 491 N/m • • Total weight on rollers = 710 N/m • Belt pull = 0.04 x 30 m x 710 N/m = 852 N • Belt pull required to overcome gravity: • • H = 10 m • • Pm = (25 packages x 589 N/package) / 30 m = 491 N/m • Belt pull required to overcome gravity = 10 m x 491 N/m = 4,910 N • Total belt pull required to move packages = 852 N + 4,910 N = 5,762 N • Convert this required belt pull to required power. • Required Power = Belt pull x belt speed • Required power = 5,762 N x 0.50 m/s • = 2,881 Nm/sec • Required power = (2,881 Nm/sec) / (1,000 Nm/sec/kW) • = 2.88 kW • If we installed 2 kW, the drive would not have enough power to move the heavy boxes from 8:00 am to 10:00am. • The correct drive selection would be 3 kW or 4 kW.

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