WRAP ANGLE DEPEND ON THE FOLLOWING PARAMETERS :-

-- TYPE OF CONVEYOR.

-- LENGTH OF CONVEYOR.

-- BELT WIDTH.

-- CAPACITY OF CONVEYOR.

-- BELT TENSIONS ETC.

THANKS

BALVINDER SINGH

M/S C.M.L.ENGG PROJECTS P LTD

NEW DELHI-110014

E-MAIL:rajputeng2000@rediffmail.com ■

Dear Mr shahverdi,

The wrap angle must be sufiecient to transmit the belt conveyor forces to the driving drum.

This in combination with the other parameters, such as pre-tension and friction.

Therefore the belt conveyor has to be fully designed first and then the drive can be calculated.

The wrap angle itself is limited by the structural design of the drive.

A spanner –guide roll can increase the wrap angle a little in case of a single drive drum.

A multiple drum design increases the total wrap angle to a greater extend, but the power sharing of the drives becomes an issue.

( A mooring rope on a ship can have a wrap angle of 1440 degrees, 4 times around a boulder,

so that a manual force is capable of holding a heavy ship)

Attached a file that describes the mathematics of a drive torque transmittal.

success

Attachments

torquetransmittal (ZIP)

■

Dear Mr shahverdi,

A quick calculation :

Tangential driving force = 2160 -1073 = 1087 kg # 10870 Newton

Torque = 10870 * 0.325 = 3533 Nm

radial speed of drum = omega = V/R = 1.3 / .325 = 4

Drive power = Torque * omega = 3530 *4 = 14.13 KW

Calculation of wrap angle (alpha):

assume friction factor mu=0.4

then: T2/T1 = e^(mu*alpha)

or:

alpha = 1/mu * ln(T2/T1)

substituted :

alpha = 1/0.4 * ln(2160/1073) = 1.749 rad, # 100 degrees

In case mu=0.2 then alpha= 200 degrees

does this answers your question ?

best reards ■

Above calculation in spreadsheet

Attachments

beltconveyor (ZIP)

■

HI,

WE CAN SUPPLY THESE BELT CONVEYORS TO YOU.PLEASE ARRANGE TO SEND YOUR ENQUIRY.

THANKS

BALVINDER SINGH

CHIEF EXECUTIVE

M/S CML ENGG PROJECTS P LTD

NEW DELHI-110014

MO-09899777763

E-MAIL:rajputeng@rediffmail.com ■

Hello Shahverdi..

The answer to your question is 180 degrees

Snub pulleys are awkward horrible things that should be avoided wherever possible.

They operate on the dirty side of the belt and have to be lagged with rubber or ceramics.

They cause fines to be removed from the belt which goes on the floor next to the chute. Unless the snub is in the chute, then you need a horrible high transfer.

For head drive pulleys 800mm diameter and less I don't use snubs. Sometimes up to 1000mm depending on what I can do with the return belt line.

Regards

LSL Tekpro ■

Dear Mr. shahverdi,

Firstly, the wrap angle depends upon the distance between carrying run and return run. Suppose, this distance between these two runs is 350 mm and one is using 400 mm diameter pulley, then he has to be satisfied with the wrap angle as available in this situation which could be 185 degree or like that.

The general approach would be to go for the highest wrap angle within the constraints fixed by the pulleys arrangement, the distance between carrying run and return run, etc. In case of usual conveyors or single snub drive, the wrap angle value that can be achieved can range from 180 degree to 220 degrees. There is no hard and fast rule for the required wrap angle except as can be achieved. Having fixed the wrap angle the designer decides the belt tensions T1 and T2, which would be in conjunction with the wrap angle. Therefore, if you have a warp angle of 185 degrees and if conveyor is designed properly, then belt will not slip on drive pulley, and in this context it would be in no way inferior to properly designed conveyor having wrap angle of 210 degrees. I mean to say performance with respect to slippage between belt and drive pulley. Of course, the conveyor which has lesser wrap angle will have higher value of tensions and possibly more price, but that does not mean inferior performance.

I have given the aforesaid reply because there is a perception that if a particular conveyor is not having wrap angle of say 210 degrees, then it is not a good conveyor. This is faulty perception. Having achieved the maximum available wrap angle, the conveyor designed properly will perform equally good.

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 25882916 ■

It is also interesting to note that if you have a dual drive pulley configuration with each drive pulley having equal wrap, then the amount of power you can put on the higher tension (primary) pulley is about three times the amount of power you can put on the lower tension (secondary) one.

It was for this reason that I designed and patented a single power proportioning gearbox with two output shafts, and only one input.

It is based on an epicyclic arrangement so as to act like a differential, and always proportion the power and torque 3:1

LSL Tekpro ■

I agree with Mr. Spriggs and the 180 degrees of wrap. Most conveyors should be designed without a drive snub.

Build up of adhesive contaminant material has many detrimental conditions. First, it causes belt flutter making belt cleaning difficult. Second, it can cause premature failure of the belt splices due to high local stresses created between drive and snub. Third, it can cause fluctuations of the demand power such as by the need to accelerate or decelerate the snub pulley RPM with the changing or dialating diameter. Fourth, if there is a dual drive with a intervening snub, the drives will not properly load share. Fifth, the diameter dialation will magnify the resultant forces and reduce the effective life of pulleys and bearings.

A need for extra wrap can be offset by more TUP force. This means an increase in belt rating and its implications. High performance and high reliablility plants focus favor their spending on added belt rating, pulleys, and structural strength. This may not be the case if belt standardization, structural or civil costs become highly unfavorable.

I disagree with Mr. Spriggs about the cleaning. If your belt cleaner is not effective, and the snub is not installed, then the contaminants will be passed down to the return idlers making a multi-station mess instead of one local clean-up zone as is obvious with the use of winged pulleys.

Another note to Mr. Spriggs. Dual or multiple drives inherently load share according to belt elasticity, belt cover gauges, gearboxes ratios, motor electrical slip, and pulley diameters which control equilibrium and compatibility states of operation.

The conveyor will demand a power level to be shared between drives according to the elastic stretch of the belt as it moves from high T1 tension to lower T2 tension. The loss of belt extension across the pulley means the belt is traveling slower when leaving the first pulley ( positve power). Therefore, the second pulley will see a slower belt speed and then climb to a higher drive electric slip position which means higher power. Knowing this must occur then the primary drive sheds power. Their sum is the demand power to move the conveyor.

I claim the notion that a split gearbox solves this condition is wrong. Maybe, there is another level of physics I do not understand and am willing to be taught. Load sharing is typically solved by other means. Often load sharing can only be achieved at one belt T1 and T2 condition. Inverters, DC drives and scoop couplings can achieve load sharing by mechanical slip control with speed sensor or power feedback.

The differential can aid. However, if one applies a 1:1 or 2:1 or 1:2 power ratio would not the differential control the speed error in the same manner as a 3:1? Just because there is more or less wrap, I claim does not control load sharing.

We solved the load sharing equations between multiple drives for the design of La Caridad, in 1977, where the 20 drives on multiple conveyors, with multiple uphill and downhill configurations, are transfered to the grid once each conveyor reaches full speed.

By the way, split gearboxes have been used for generations on underground conveyor drives. They have lost favor due to their poor load sharing and lagging wear difficulties. The idea of a gear differential has only been applied to the Cable Belt drives due to the two cables having differing lengths from installation of thermal expansion. There may be other advantages implied by Mr. Spriggs, which he can explain. ■

Morning Larry et al...

Let me try and get across what my power proportioning gearbox does. Its really quite logical, and will mean all things to all long overland conveyors.

First imagine a car with a differential that has been modified in such a crazy way that the torque you get out of the Right wheel is three times that which you get out of the Left wheel.

Now drive the car on the road.

The car will have traction from both back wheels and there will be no creep or slip thanks to the differential. It will be able to be steered, but will have a preference for going left.

(Drive the car through some mud and the right wheel will spin first due to the additional 3 x torque.)

OK so far..

Now draw a Spriggs two pulley conveyor single drive in plan on a piece of paper, starting with a single vertical rectangle representing the motor at the bottom of the page.

Connect the motor vertically up the page to under the right hand side of a single horizontal rectangle which represents the gearbox.

Now draw two output shafts from the upper side gearbox one left of centre the other right of centre. These couple to two pulley rectangles drawn vertically, the left being the secondary and the other the primary.

The pulleys have equal wrap but the right hand side primary can take three times the torque of the left hand side secondary thanks to the higher belt tensions around the primary.

My simple differential gearbox produces exactly that ratio of torque, so if for example, I put on a single one Megawatt motor, through a single gearbox with one single VSD, I have a very economical single drive, albeit on two drive pulleys for the lowest belt class and which provides 333kW to the secondary pulley and 667 kW to the primary.

The two pulleys don't even have to be exactly the same diameter any more thanks to the differential effect!!

A simple invention. (Incidentally, my father was an inventor too, and invented many things including the building scaffolding that slots together. He did this in the early sixties and made a stack out of it when he sold the patent. Its the one with the four U shaped slot- in points at 90 degrees to each other on the vertical pipes..maybe keep a look out for it)

Regards

LSL Tekpro ■

Hi All,

I consider that two important issues are being neglected.

One without some form of snub, how do we stop the belt vibrating when we leave the head pulley, and therefore how do we clean the belt as the vibration will cause the belt scraper to “chatter” and be ineffective.

This talk of double output shaft gearboxes must have a detrimental effect on the distance between drive pulleys and thus splice fatigue?

Thanks ■

Hi all..

For what it is worth, I never seem to have this belt flapping scenario on my conveyors, and certainly would not encourage the use of snub pulleys for this reason.

(Maybe its like one of those urban legends like the one where they say that two conveyors in series must run at the same speed, or you must put the drive at ground level, and all the others that give me a hard time)..

Regards

LSL Tekpro ■

I am fascinated by Mnr Spriggs gearbox (if you say "give me a hard time" then Mnr it is). Can we see a drawing on the forum please? In 1968 I was introduced to differential turbocharging principles & so far imagine that there is a similar principle involved. "Drawings are the language of Engineering." (Lobel, Loughborough University 1964 & before).

I've never come across flappy belt situations either & snub pulleys are a real nightmare. ■

Originally Posted by Teus Tuinenburg

Dear Mr shahverdi,

A quick calculation :

Tangential driving force = 2160 -1073 = 1087 kg # 10870 Newton

Torque = 10870 * 0.325 = 3533 Nm

radial speed of drum = omega = V/R = 1.3 / .325 = 4

Drive power = Torque * omega = 3530 *4 = 14.13 KW

Calculation of wrap angle (alpha):

assume friction factor mu=0.4

then: T2/T1 = e^(mu*alpha)

or:

alpha = 1/mu * ln(T2/T1)

substituted :

alpha = 1/0.4 * ln(2160/1073) = 1.749 rad, # 100 degrees

In case mu=0.2 then alpha= 200 degrees

does this answers your question ?

best reards

Hello sir

Well being a new to this thread and also having to have teh concrete concept ifor belt conveyor , i would like to be advised by you experts that

1) What is te general approach to design the belt conveyor for a given input of throuput capacity and distance of conveyor. Can you enlighten teh design procedure

Any spreadsheet wil also be helpful for me

thanks in anticipation ■

Go buy yourself a copy of "BELT CONVEYORS FOR BULK MATERIALS" from CEMA

http://www.cemanet.org/publications/index.html

It explains how to design belt conveyors ■

Originally Posted by Teus Tuinenburg

Dear Mr shahverdi,

A quick calculation :

Tangential driving force = 2160 -1073 = 1087 kg # 10870 Newton

Torque = 10870 * 0.325 = 3533 Nm

radial speed of drum = omega = V/R = 1.3 / .325 = 4

Drive power = Torque * omega = 3530 *4 = 14.13 KW

Calculation of wrap angle (alpha):

assume friction factor mu=0.4

then: T2/T1 = e^(mu*alpha)

or:

alpha = 1/mu * ln(T2/T1)

substituted :

alpha = 1/0.4 * ln(2160/1073) = 1.749 rad, # 100 degrees

In case mu=0.2 then alpha= 200 degrees

does this answers your question ?

best reards

Dear Mr. Tesus,

Thank you very much, very good explanation.

Regards

Uditha ■

Originally Posted by Teus Tuinenburg

Dear Mr shahverdi,

A quick calculation :

Tangential driving force = 2160 -1073 = 1087 kg # 10870 Newton

Torque = 10870 * 0.325 = 3533 Nm

radial speed of drum = omega = V/R = 1.3 / .325 = 4

Drive power = Torque * omega = 3530 *4 = 14.13 KW

Calculation of wrap angle (alpha):

assume friction factor mu=0.4

then: T2/T1 = e^(mu*alpha)

or:

alpha = 1/mu * ln(T2/T1)

substituted :

alpha = 1/0.4 * ln(2160/1073) = 1.749 rad, # 100 degrees

In case mu=0.2 then alpha= 200 degrees

does this answers your question ?

best reards

How the radial speed is derived by dividing belt velocity by radius is not clear.

Radial velocity ( omega ) has the unit of degree per second or radian per second, I think.

Rgds, ■

omega given by:

omega= (2*pi/n)/60 in rad/sec

velocity = omega*R in m/sec

omega = velocity/R in rad/sec ■