How to Calculate Belt Tensions in Bulk Handling Belt Conveyors
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- Опубліковано 7 лип 2024
- This tutorial is for bulk handling conveyor belt engineers and technicians. For a free copy of our power calculation go to rulmecacorp.com/motorized-pul...
0:00 Introduction
0:28 Agenda
0:54 Definitions
1:31 Methodology
2:08 CEMA Equation
2:39 Effective Tension Components
3:44 Calculating Required Power
4:40 Other Belt Tensions
5:03 Slack Side Tension (slip)
6:03 Slack Side Tension (sag)
7:51 Conclusion
This tutorial:
• explains basic principles involved in calculating bulk conveyor belt tensions.
• introduces terms: effective belt tension, slack side tension, and maximum belt tension.
• serves as an introduction to our advanced tutorial on how to reduce belt tension by switching from a single drive to a dual drive system.
For a free copy of the company's power calculation program use this link:
rulmecacorp.com/motorized-pul...
For complete text, screen shots, and equations go to:
rulmecacorp.com/how-to-calcul...
Common components on a typical bulk handling belt conveyor include head pulley, snub pulley, flat belt, tail pulley, carrying idlers, return idlers, belt plow, counterweighted take-up, belt cleaner, loading skirts, hopper feeder, and slider bed.
The CEMA Conveyor Design Manual provides this (historical method) equation to calculate effective belt tension.
Te = LKt (Kx + KyWb + 0.015Wb) + Wm(LKy + H) + Tp + Tam + Tac
These parameters enable the designer to calculate belt tension required to overcome:
• Friction
• Gravity
• Momentum
Some Friction components are:
Tbc, the tension required to overcome belt cleaner drag
Tsb, the tension required overcome skirt board drag
Tyr, the tension required to overcome friction in the bearings of the return rollers as the empty belt travels over the return strand.
Gravity components include:
Tb, the tension required to lift or lower the conveyor belt;
Tm, the tension required to lift or lower material.
The Momentum component, Tam, is the tension required to accelerate the material on the belt from the initial velocity, as it hits the conveyor, to the terminal velocity, which is defined as the conveyor belt speed.
Required power equals effective tension times belt speed:
Required Power = Te x V
In imperial units, Te is expressed in pounds (lbs), and V is expressed in feet per minute (fpm). The product of the two factors is expressed in foot-pounds per minute (ft-lbs/min). Since one horsepower (HP) = 33,000 ft-lbs/min, required conveyor drive power may be expressed in HP as follows,
(Te in lbs) x (V in fpm)/((33,000 ft-lbs/min)/HP) = HP
After calculating Te, it is important to calculate T2slip (slack side tension required to resist slippage of the belt on the pulley.) The historical method that CEMA provides to calculate T2slip is as follows:
T2slip = Te x Cw, where Cw is the CEMA wrap factor for a rubber surfaced belt.
The wrap factor can be as small as 0.08 for dual drive systems with rubber-lagged drive pulleys, an automatic take-up, and 420° of wrap angle, or as large as 1.2 for a single drive system, an unlagged pulley, manual take-up and 180° of belt wrap.
Next it is important to calculate T2sag (slack side tension required to prevent belt sag.) Sag is a phenomenon which can occur at the point of minimum tension in the carrying strand of a conveyor belt. On an inclined conveyor, it is usually in the vicinity of the loading zone.
The CEMA historical method allows the designer to select an appropriate percentage of sag between the carrying idlers, to prevent lumps or material from coming out of the conveyor belt. The three values CEMA provides are 3%, 2%, and 1.5% sag Those percentages are based on material lump size, the proportion of lumps vs. fines, and the idler troughing angle. Note that T0, the minimum tension to prevent sag, may be reduced if the belt is less than fully loaded.
The three pertinent equations are:
T0 = 4.20 Si (Wb + Wm) for 3% sag
T0 = 6.25 Si (Wb + Wm) for 2% sag
T0 = 8.40 Si (Wb + Wm) for 1.5% sag
In these equations, Si is idler spacing in feet, Wb is a weight per foot of the belt, and Wm is a weight per foot of the material.
After calculating T0 it is essential to add or subtract the weight of the carrying and return strands of the belt for a sloped conveyor and add or subtract Tyr, which is the tension required for the empty belt to overcome idler friction.
After calculating T2slip and T2sag , it is essential select the larger of the two values.
Once T2 has been determined, maximum belt tension may be calculated using this equation:
T1 = Te + T2
The T1 value is required to select a belt. - Наука та технологія
We are happy so many conveyor designers find our free power calculation program useful. We appreciate all of your comments.
I am a huge fan of rumelca material. Thank you guys.
This is absolutely great shares!
You are very welcome. If you have any specific questions, please ask us.
I appreciate your help in providing me a brief equation or rule of thumb to calculate the specific energy consumption as Kwh/Mt/Km , belt speed = 2.5 meter per second, trough angle = 30 , surcharge = 25, material = excavated soil (sand) , lump size less than 150 mm , density =1,550 kg/M3 , belt length = 5 Km (multiple flights ) , Flow = 3500 to 4,000 Mt/Hr , Route inclination= zero inclination, Width = 1,200 mm , considering these inputs , how much electricity will be consumed in electrical motor as Kwh per metric ton (Weight) per Km (distance)
Hi there. What's the average cost of a 50-60ft mobile belt conveyor for hauling dirt away?
GREAT INFORMATION thank you!!!
I have a question.
L the leght of the conveyor
Is it the totally leght such the load side as the return side or is it just the leght from center of the drive pulley to the center of the tail pulley?
Dear Norwin, Length should be considered to be the distance from the centerline of the tail pulley to the centerline of the head pulley, assuming material is loaded near the tail and discharged near the head. Mike
@@rulmecacorporation4596 a lot of thanks!!
@@norwingradisrivera1368 It's our pleasure to help in any way we can.
Thanks for this information, it will be great in metric units also...
You are welcome. Our International Group is finalizing a metric version. It should be generally available within one year.
The metric version is now available to the public. Please provide your email address and I will send a link to the program
@@rulmecacorporation4596 rajsekcae012@gmail.com
@@rulmecacorporation4596 nguyendangthelisemco@gmail.com
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Mr. Bear Whip. It's good to see you , and, that your expertise is still valuable. Please let me know how to use the program.
MIKE LOCHE! There's a name I haven't heard in a while. I hope you are doing well Mike. We posted a one hour UA-cam video on how to use our bulk calculation program, entitled Webinar #102
ua-cam.com/video/0-Vz5GrFYs4/v-deo.html
@@rulmecacorporation4596 Wonderful to hear from you. So, I sold half of my business to Rema in 2013 and the other half in 2019. It would be great for you guys to get a hold of Ramon regarding your product. I'm out here in Columbia Tn. now. Following my youngest son at a church thebelongingco.com
Hoping to get the rest of the grandkids out here.
Is there an email that we can correspond through? I'd like to communicate with you regarding further distribution, etc.?
HOW TO CHOOSE WHAT BELT TYPE
Great video! just one question: lbs units in mass or force?
Dear Carlos, We're happy you like the video. In our power calculation program, we assume that conveyors are near the surface of the Earth. So, pound-force and pound-mass may be considered equal for our purposes. Our power calculation software, to which you are welcome, uses slugs when we define mass in our equations. I hope that helps.
Thanks for information ❤ but I have few doubt
How may we help you resolve your doubts?
I had go threw the rulmecca problem solving section but I haven't found any details about skirt board in equation, how friction act on the conveyor
@@rizanmd9334 ua-cam.com/video/5QJjy67Z-NM/v-deo.html please watch this video
@@rizanmd9334 ua-cam.com/video/T7mgqj3bXSA/v-deo.html please watch this video too
Outstanding Explanation, have couple of questions. Can you please share your email I'd.!!
Hello , are you come here
Sir hindi milenga kya
I appreciate your help in providing me a brief equation or rule of thumb to calculate the specific energy consumption as Kwh/Mt/Km , belt speed = 2.5 meter per second, trough angle = 30 , surcharge = 25, material = excavated soil (sand) , lump size less than 150 mm , density =1,550 kg/M3 , belt length = 5 Km (multiple flights ) , Flow = 3500 to 4,000 Mt/Hr , Route inclination= zero inclination, Width = 1,200 mm , considering these inputs , how much electricity will be consumed in electrical motor as Kwh per metric ton (Weight) per Km (distance)
Dear sir, We will be happy to help you. Please provide your name and email address.
Thanks,
Mike