Dear Ehab

I believe this has been discussed in the forum in the past. You will receive a better response if you define specifice problem you are trying to solve. As you see in the following, many conditions must be reviewed depending on the need for risk control and consequences of installing a improper design.

I do not know about your 2.5 times factor - product of direct-on-line start torque. Usually the standard steady-state gravity takeup type drive friction factor ( ratio of T1/T2 noted below) is taken as f=0.35 (CEMA,DIN...)for typical in-plant conv. Starting and stopping dynamic drive friction is taken as f=0.40. This can vary on many factors. TO go further on this point requires too many qualifications. In wet and slime conditions, the steady-state friction is reduced to 0.30 and in some cases 0.25 with rubber lagging.

The counterweight (takeup force) criteria has many parts to satisfy a proper design:

1. Drive tension ratio of high (T1) to low (T2) is dependent on wrap angle, steady-state contact friction (belt & lagging) and dynamic contact friction. These can vary with:

a) wrap angle

b) belt stiffness and ability to conform to pulley surface

c) type of lagging such as rubber or ceramic.

d) lagging geometry and degree of wear on grooving

e) material contamination such as water or wet slippery slime

f)

2. Account for counterweight internal hysteresis of:

a) rope on sheaves (bending of rope on sheave),

b) sheave bearings (roller or bushing),

c) quantity of sheaves,

d) sheave diameter,

e) rope type,

f) wrap angles over sheaves

g) type of carraige or sled and vertical, inclined of vertical orient.

3. Type of take-up (TUP): gravity and its hysteresis, semi-auto winch or fixed TUP and account for temperature swings due to belt thermal growth with a fixed TUP

4. Location of TUP from drive station -- remote location can require special analysis depending on starting and stopping system

5. Type of starting / stopping controls and their response to delivery exagerated high T1 tension and low T2 tension

6. Degree of reliability and maintenance free operation

7. Belt sag control to eliminate spillage and unacceptable takeup dynamic response to all major "what if events"

8. Load sharing conditions with multiple drives

I could go on, but I hope the point is made -- provide a little more detail without asking for an engineering treatise. Others will have more to say and may reference the past articles on this point. ■

Ehab

To answer your question about slippage. I am assumming that you mean belt slippage on the drive pulley.

You would never want to have slippage between the drive pulley and the belt for many reasons.

1. Loss of starting torque.

2. Glazing of any rubber lagging.

3. Possible fire due to overheating belting.

4. Cover wear to belting.

5. Nusance zero speed trips.

To name a few.

As Mr. Nordell has stated your question regarding take-up cannot be answered on this forum. Your system would required a complete engineered analysis in order to determine the amount of take-up weight and the belt sag required.

You need to look at all conditions of the system to see which creates the worst case.

Regards,

Gary Blenkhorn ■

Dear Mr. Hassan,

I normally dont do engineering in the forum. I will comment on your selections this once:

1. Power require 10kW (~7.5 kW demand)

2. Assuming 180% start torq DOL = 0.8 sec. start cycle @ 10 kW

3. Gravity CWT mass =1.4 tons = 2x belt tension @ 7000 N

4. CWT mass set by belt sag (1.5% running) and accel slippage on drive with average DOL @ 200%..

Note the potential 250% start torque will last about .2 seconds and some slip will occur. You might hear it but will not notice it.

If you have

If you apply your 15 kW drive, then you will need a 2 ton gravity counterweight per the above criteria. The conveyor will accelerate to full speed in about 0.50 seconds. It will apply unnecessary forces on all components and structures.

I do not get your point regarding ISO quotation. The required CWT mass (assuming auto- tension by gravity) is related on starting and steady-state operating conditions as I stated and you applied. It is also related to lagging.

Enough said. ■

Dear Mr.Hasan,

I understand that you have calculated the required CWT properly. In my openion you use 1.0 Ton & place the counter weight at 4.0M below the carrying belt.I do not know why the staering torque is 250%.What type of coupling you are using?Use fluid coupling or any type resilent coupling restricting the starting torque as 1.5.Please do not over load the belt.

Regards.

A.Banerjee ■

Dear all

Fluid coupling is generally not provided in small horse power conveyors . Fluid coupling is provided for conveyors having 30 or higher kw rating.

As per calculation 7.8 kw is actual power requirement of this conveyor 9.3 kw motor is adequate to run the conveyor .

However client is having 15 kw motor and intends to use the same . With 15 kw motor as per calculations starting torque is 250 percent as motor is directly coupled to conveyor without any device which increases the starting torque/ reduces the starting time

Under the circumstances it is recommended to use 9.3 kw motor only so that starting torque can be limited to reasonable value. or in case he uses 15 kw motor with direct coupling he has to provide higher take up weight and other components

A R SINGH ■

Dear Hassan,

THe friction factor "f= 0.35" is from the equation:



T1/T2 = natural log (e) raised to the power (theta x "f")

where T1 is the high incomming tension to the drive pulley

T2 is the slack or low tension leaving the drive pulley

theta is the wrap angle in radians

THere is an acceptance to increase the allowable friction "f" during starting to f=0.40, while running f=0.35. THese values are often quoted. However, in cases of heavy rains, power pulley lagging maintenance, slime like products such as iron ore, bausxite, nickel, lignite, may require much lower values of f.

Ceramic lagging ususally allows an increase in f =.40 and .45 respectively.

Dry arrid conveyor operations can also appply a higher value.

I tend not to read too much into ISO. The world of conveyor technology has pasted the out dated standards. Much more is known than the standards can cover.

There is a reasonable body of engineers that apply design factors based on what they can observe by measurements and not soley rely on ancient emperical generalized parameters. ■

Mr.Singh Shab,

If you want to restrict yourself in IS:11592-2000, yes, you are correct.For your information TATA STEEL ,INDIA never accept worm gear box whatsoever the rating. Some claint wants fluid coupling for soft start. I do not understand the requirement of 250% starting torque in 15KW motor in belt conveyor system.Is it special requirement?

As regards CWT Mr. Hasan can use 1.0 Ton by suitably locating the same i.e allowing it to move up during starting.

Regards.

A.Banerjee ■

Dear Mr Banerjee

Dear Mr Banerjee

I never restrict to IS 11592-2000

I have not recommended worm gear boxes

We always use fluid coupling for motors 30 kw and above. I have not seen / provided fluid coupling for 9.3/15 kw conveyor rating. I would like to know the client who insists fluid coupling for these small motors.

250 percent is not requirement of conveyor.

In case 15 KW motor is used it will be having much more stating torque than 9,3 kw motor such torque with higher size motor transmitted to conveyor would result in stating tension 250 percent of allowable tension may be for a short time.

If you provide 9.3 kw motor stating tension will be less compared to 15 kw rating.

For controlling belt stating tension flywheel/ fluid coupling / inverter is required

As regards CWT it should be sufficient to avoid slippage even at stating period

Regards.

A R SINGH ■

Mr. Banerjee,

The 250% starting torque is not a special requirement. It is produced by the motor. Most AC motor speed torque curves have a "pull- up" torque (electrical designation) that range from 225 -300% of nameplate rated torque. THis occurs at about 80-90% of full speed.

The more effiecient the motor the higher the pull-up peak as a rule.

I suggest you obtain such a curve from any motor manufacturer to become informed. ■

Respected MR.Nordell,

As per CEMA (page102 of 5th edn.)the starting torque could be more than 2.5 times. You are correct. But it is also written that for conveyor application it should be ristricted to 150 to180 %.In INDIA we keep those motor under special series like KL.. so I have written this.

With regards.

A.Banerjee ■

Dear Mr Banerjee

Motor starting torque is torque developed by the motor at zero speed when it is switched on. Starting torque in most of the efficient motors is 250-300 percent of rated motor nameplate torque.

If we reduce the starting torque then motor will require higher starting time. Excessive long starting period causes temperature rise in motor as Starting current of induction motor is 5-7 times the rated current. Longer starting time also develops electromehanical stresses

Now in case of direct coupling of motor with the load starting time or starting torque depends upon

1.0Total inertia of the system

2.0Torque speed curve of motor

3.0Torque speed curve of conveyor

For belt conveyors it is generally recommended to have starting belt tension 150-180 percent of allowable starting belt working tension (not motor starting torque)

In case of motors of small conveyors where there is no other means to increase starting time of conveyor ( like fluid coupling, fly wheel, inverter etc) providing higher KW motor means higher staring belt tension.

In this particular case when conveyor is directly coupled if you provide 15 kw instead of 9.3 kw motor belt starting tension will increase up to 250 percent for shorter period..

Cheers

A R SINGH ■

Dear Mr. Ehab A. Hassan:

We select belt based on maximum working tension T1 (not starting tensin) should be less than 85 percent of belt allowable working tension

Factor of safty of belt should not be less than 6 during starting and 10 during normal running

Best regards,

A R SINGH ■

Dear Mr. Singh,

Factor of safety refered by you here for NN belting only.

Regards.

A.Banerjee ■

Dear Mr Banerjee

Agreed you are right factor of safty for fabric belts we consider 10 however for steel chord belting is much lower .

As in fabric belting at spilce joint joint efficiency is reduced as 1 ply out of total ( 3,/4/5 as may be applicable) is not effective.

As in this thread we were discussing very small conveyor having .3/15 kw motor steel chord is not applicable.

Thanks for correction

A R SINGH ■

It may be that this issue is being over-analyzed. Extrapolation of the argument would follow that we would size the take-up counterweigt for 500% of the steady state running Te if we were to oversize the drive by a factor of 2. It is a matter of inertia and duration of the max torque. A low power, short conveyor will come up to speed quickly with minimal time duration of any spike.

The writer has always been adamant about structural design, at the terminals (drive, take-Up, etc) that accomodates variability, that is the likeliness that the actual loads may vary widely from the predicted. In the case of the take-up the variability may be due to the dynamics and the (subject of this thread) need to adjust (up) the take-up tension. The DSI (Dos Santos International) structural standards make ample allowance for the variabilty.

Joseph A. Dos Santos ■