Belt Tensions

Posted in: , on 10. Dec. 2008 - 14:58

Please assume that I am having a conveyor with centre to centre distance as 100 metres.

It has vertical take up with 3 tonnes. Belt is 800mm , EP 800/4. Drive pulley dia is 500 mm.

Motor is consuming 20 amps in no load and 30 amperes with load. Voltage is 415 with 3 phase and Power factor angle is 0.9.

Is it possible to find T1 and T2 of the conveyor ?

Can I use the above values of T1 and T2 when replacing the full belt, so that the conveyor elongation will be "beneficially" achieved?

Do we need any further details ?

Please note that the above values are given as examples.

Thanks & Regards

Re: Belt Tensions

Posted on 10. Dec. 2008 - 03:49

I thought it is normal to start with the duty of the belt, i.e. te/hr, m3/hr, length, inclination, etc and then design the belt conveyor accordingly i.e. belt width, belt speed, tensions and power.

To me your approach of working backwards is somewhat novel.

PS, you haven't given the motor power (kW)

Re: Belt Tensions

Posted on 10. Dec. 2008 - 03:49

I thought it is normal to start with the duty of the belt, i.e. te/hr, m3/hr, length, inclination, etc and then design the belt conveyor accordingly i.e. belt width, belt speed, tensions and power.

To me your approach of working backwards is somewhat novel.

PS, you haven't given the motor power (kW)

Lyle Brown
(not verified)

Re: Belt Tensions

Posted on 10. Dec. 2008 - 09:11

Yes it is, though it requires a few assumptions:

1. P=sqrt(3)*v*i*cos(theta)*efficiency

2. P=Te*s

Rearrange 2.

3. Te=P/s (power / belt speed)

Using:

4. T1/T2 = exp(u*theta)

5. T1=Te+T2

Rearrange: 5. and 6.

6. T2= Te/(exp (u*theta)-1)

You have Te from 3.

Assuming u and observing theta, you can calculate T2, from 6.

You can calculate T1 using 5.

Note you would have to apply some "starting factor" for Te when starting (and use the largest calculated T2).

Also note the actual T2 may be larger (or less..) than calcualted due to: subjective selection of u / theta, construction ambiguity, required for brakes that we are not currently aware of, control of transtion or curves etc, etc.

We would also suggest if using this approach you would use installed power rather than demand power. If it is to be used for critical applications (i.e. where there are significant safety or cost implications: belt selection, structural / civil design) you really need to validate the assumptions.

If this requires further clarification we would suggest it may be best not to use this approach.

Regards,

Lyle

Lyle Brown
(not verified)

Re: Belt Tensions

Posted on 10. Dec. 2008 - 09:11

Yes it is, though it requires a few assumptions:

1. P=sqrt(3)*v*i*cos(theta)*efficiency

2. P=Te*s

Rearrange 2.

3. Te=P/s (power / belt speed)

Using:

4. T1/T2 = exp(u*theta)

5. T1=Te+T2

Rearrange: 5. and 6.

6. T2= Te/(exp (u*theta)-1)

You have Te from 3.

Assuming u and observing theta, you can calculate T2, from 6.

You can calculate T1 using 5.

Note you would have to apply some "starting factor" for Te when starting (and use the largest calculated T2).

Also note the actual T2 may be larger (or less..) than calcualted due to: subjective selection of u / theta, construction ambiguity, required for brakes that we are not currently aware of, control of transtion or curves etc, etc.

We would also suggest if using this approach you would use installed power rather than demand power. If it is to be used for critical applications (i.e. where there are significant safety or cost implications: belt selection, structural / civil design) you really need to validate the assumptions.

If this requires further clarification we would suggest it may be best not to use this approach.

Regards,

Lyle

Re: Belt Tensions

Posted on 11. Dec. 2008 - 05:46

Thank you for your replies.

Many times the take up margin available are found shorter or neck to neck. That means when I replace the full length of the belt, if elongation becomes more, I need to trim the belt after some time. If elongation becomes nil, I will not have any margin in take up for having future joint.

The elongation of the belt may have close relation with the tensions, T1 and T2 of the belt.

So when I am replacing the belt if my final tightening goes beyond T1 , there may not be belt elongation. If It is very less, belt may elongate more leading to trimming of belt in near future.

Hence I need to go for adding piece belt, mean longer shutdown.

I need to have a tight rope walk.

Though the new belt elongation at 10% of it's strength is known to me when new belt is purchased along with it's test certificate, the take up margin length is found shorter in many places.

Hence the thread was started in maintenance point of view only and not for designing new conveyors and their structures.

I need to find the relation between T1, T2 & the belts elongation.

I find installed power is found atleast 1.5 to 1.7 times higher than that of demand power.

Hence I have given the amperes and power could be calculated.

Thank you & regards,

Re: Belt Tensions

Posted on 11. Dec. 2008 - 05:46

Thank you for your replies.

Many times the take up margin available are found shorter or neck to neck. That means when I replace the full length of the belt, if elongation becomes more, I need to trim the belt after some time. If elongation becomes nil, I will not have any margin in take up for having future joint.

The elongation of the belt may have close relation with the tensions, T1 and T2 of the belt.

So when I am replacing the belt if my final tightening goes beyond T1 , there may not be belt elongation. If It is very less, belt may elongate more leading to trimming of belt in near future.

Hence I need to go for adding piece belt, mean longer shutdown.

I need to have a tight rope walk.

Though the new belt elongation at 10% of it's strength is known to me when new belt is purchased along with it's test certificate, the take up margin length is found shorter in many places.

Hence the thread was started in maintenance point of view only and not for designing new conveyors and their structures.

I need to find the relation between T1, T2 & the belts elongation.

I find installed power is found atleast 1.5 to 1.7 times higher than that of demand power.

Hence I have given the amperes and power could be calculated.

Thank you & regards,

Lyle Brown
(not verified)

Re: Belt Tensions

Posted on 11. Dec. 2008 - 10:21

We note:

Becareful applying desing loads in the manner described here - i.e. assuming a tower mounted drive with the winch on the ground, the drive pulley and associated support may see T1+T1 (no torque as line pull is provided by other means such as a winch etc) in lieu of normal T2+T1 (+ torque). Hence the load question is still an issue.

You may be trying to achieve "average tension" in lieu of T1/T2.

It seems as though maybe the take length up may not be optimal. Designers generally make some allowance for elastic and permenant stretch (amongst other things).

The factor mentioned earlier referred to the increased torque relative to normal operation associated with starting, not the "spare capacity of installed power relative to demand".

Regards,

Lyle

Lyle Brown
(not verified)

Re: Belt Tensions

Posted on 11. Dec. 2008 - 10:21

We note:

Becareful applying desing loads in the manner described here - i.e. assuming a tower mounted drive with the winch on the ground, the drive pulley and associated support may see T1+T1 (no torque as line pull is provided by other means such as a winch etc) in lieu of normal T2+T1 (+ torque). Hence the load question is still an issue.

You may be trying to achieve "average tension" in lieu of T1/T2.

It seems as though maybe the take length up may not be optimal. Designers generally make some allowance for elastic and permenant stretch (amongst other things).

The factor mentioned earlier referred to the increased torque relative to normal operation associated with starting, not the "spare capacity of installed power relative to demand".

Regards,

Lyle

Counterweight Mass

Posted on 3. Dec. 2009 - 10:10

Dear Sir:

Kindly help me to find mass of the counterweight (vertical) of the transport belt after we change the trough angle of carrier idler supports from 45 degree to 35 degree the characters and type of this belt are:

1.Belt width 1600 mm.

2.Distance between centers 118.99 m

3.Warping 30.8 m.

4.Inclination 15º

5.Capacity 2500 ton/hr.

6.Speed 2.3m/s.

7.Material caliza/arcil.

8.Size 0-115 mm.

9. Power 2160 KW (50HZ).

10.Temperature 25.

11.Tension 380 V 50 HZ.

And I ask if you can give me the formula or method to calculate the mass of counterweight of rubber belts.

Thanks,







Ammar Thamer

Mechanical Engineer

Bazian Cement Company

Mob. NO. 00967702239160

BCC

Counterweight Mass

Posted on 3. Dec. 2009 - 10:10

Dear Sir:

Kindly help me to find mass of the counterweight (vertical) of the transport belt after we change the trough angle of carrier idler supports from 45 degree to 35 degree the characters and type of this belt are:

1.Belt width 1600 mm.

2.Distance between centers 118.99 m

3.Warping 30.8 m.

4.Inclination 15º

5.Capacity 2500 ton/hr.

6.Speed 2.3m/s.

7.Material caliza/arcil.

8.Size 0-115 mm.

9. Power 2160 KW (50HZ).

10.Temperature 25.

11.Tension 380 V 50 HZ.

And I ask if you can give me the formula or method to calculate the mass of counterweight of rubber belts.

Thanks,







Ammar Thamer

Mechanical Engineer

Bazian Cement Company

Mob. NO. 00967702239160

BCC