In this video, we will review how to calculate required belt pull and mechanical power to move discrete packages on a horizontal roller-supported conveyor belt. For more conveyor information, go to [ Ссылка ]
0:00 Start
0:17 Related Videos
1:05 Belt Pull Equation
5:33 Convert Belt Pull to Required Power
6:47 Make Drive Power Selection
7:08 Check Design Assumptions
7:38 Verify Drive Power Selection
Use this video in conjunction with our other videos presenting how to calculate required belt pull and mechanical power to move:
• Discrete packages on a horizontal slider bed supported conveyor belt.
• Discrete packages on an inclined slider bed supported conveyor belt.
• Continuously flowing bulk materials on a inclined roller-supported conveyor belt.
To watch those videos click the link in the upper right-hand corner of your screen.
The power required to move a load on a belt conveyor may be calculated as follows:
Required Power = Force x Velocity, or
Required Power = Belt pull x belt speed
The belt pull required to move packages on a roller-supported conveyor belt equals the total weight of all packages plus the weight of the belt plus the weight of all rotating parts times the coefficient of bearing friction of the supporting rollers, as summarized in this equation:
F = 0.04 x L x (2Pn +Ppr + Pm)
Where,
L = conveyor length
Pn = weight per foot of conveyor belt
Ppr = weight per foot of all rotating parts
Pm = weight per foot of product to be handled
If a roller brand and belt supplier have not been selected, then we use:
0.04 as the roller bearing frictional coefficient
5 lbs per foot as the weight per foot of all rotating parts
5 lbs per foot of conveyor belt
If the roller is especially long or has a thick wall, then you should consult your roller supplier for specific details.
For example, consider a 100-foot-long, horizontal, roller-supported, belt conveyor continuously moving 25 evenly-spaced boxes, with an average weight of 50 lbs/box, at a belt speed of 100 fpm during an 8 hour shift. The belt pull is calculated as follows.
• Pn = 2 x 5 lbs/ft of belt = 10 lbs/ft of belt
• Ppr = 5 lbs/ft of rotating parts
• Pm = (25 packages x 50 lbs/package) / 100 feet = 12.5 lbs/foot of product
• Total weight on rollers = 27.5 lbs/ft
Belt pull = 0.04 x 100 ft x 27.5 lbs/ft = 110 lbs
To determine required power use this equation:
Required Power = Force x Velocity
Required Power = Belt pull x belt speed
A belt pull of 110 lbs times a belt speed of 100 feet per minute equates to a required power of 11,000-foot-pounds per minute.
Required power = 110 lbs x 100 fpm
= 11,000 ft-lbs/min
In imperial units, one horsepower equals 33,000-foot-pounds per minute.
Therefore,
Required power = (11,000 ft-lbs/minute) / (33,000 ft-lbs/ minute / HP)
= 0.33 HP
Now we can make a selection of 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.
Let’s select a conveyor drive of 0.40 HP 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 weight handled during 8 hour shift = 600,000 lbs
• Total number of packages handled during 8 hour shift = 12,000 packages
• Average package weight = (600,000 lbs/shift) / 12,000 packages/shift) = 50 lbs/package
However, suppose while double-checking the design parameters you learn that heavy packages (120 lbs/box) are handled from 8:00am to 10:00 am and light packages (26.67 lbs/box) are handled from 10:00am to 4:00pm on every shift.
Check required conveyor drive power from 8:00 am to 10:00am.
• Pn = 2 x 5 lbs/ft of belt = 10 lbs/ft of belt
• Ppr = 5 lbs/ft of rotating parts
• Pm = (25 packages x 120 lbs/package) / 100 feet = 30 lbs/foot
• Total weight on rollers = 45 lbs/ft
Belt pull = 0.04 x 100 ft x 45 lbs/ft = 180 lbs
Convert this required belt pull to required power.
Required Power = Belt pull x belt speed
Required power = 180 lbs x 100 fpm
= 18,000 ft-lbs/min
Therefore,
Required power = (18,000 ft-lbs/minute) / (33,000 ft-lbs/ minute / HP)
= 0.55 HP
If we installed 0.4 HP the drive we would not have enough power to move the heavy boxes from 8:00 am to 10:00am.
The correct drive selection would be 0.75 HP.
