VFD & Energy Saving-4 : Energy Saving in Belt Conveyor Application

You might be able to reduce the 14.8 kW part of the operation to about 7.4 kW (probably not a full 60% reduction).

Here is my home work. Your help is required to make it correct (in case of error) and more precise.

To avoid one long Post, I broke it into three posts

1. BELT CONVEYOR USING DIRECT ON LINE STARTER

2. BELT CONVEYOR USING VARIABLE FREQUENCY DRIVE

3. AIR CONDITIONING COSTS AND OTHER FACTORS

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1 BELT CONVEYOR USING DIRECT ON LINE STARTER

1.1 Motor data at Full Load

Rated Power (Mechanical) : 26 kW

Rated Power (Electrical) : 29.54 kW

Rated Current : 49.6 A

Power Factor : 0.87

Efficiency : 88 %

Rated Speed : 1485 RPM

Rated Torque : 167.2 N.m

1.2 Power consumption during loaded condition of Belt conveyor

Power delivered (Mechanical) : 18.48 kW

Power consumed (Electrical) : 21 kW (Motor Efficiency is 88 %)

kWh consumed in 35 minutes : 12.25 kWh

1.3 Power consumption during un-loaded condition of Belt conveyor

Power delivered (Mechanical) : 13.02 kW

Torque : 83.8 N.m

Power consumed (Electrical) : 14.8 kW

kWh consumed in 25 minutes : 6.16 kWh

1.4 Power consumption for an hour operation

Total Consumption (35 min +25 min) : 18.41 kWh (12.25 +6.16)

2 BELT CONVEYOR USING VARIBLE FREQUENCY DRIVE

2.1 Additional Cost associated with VFD

a. There are losses in the VFD. VFD is around 92 % Efficient

b. Induction Motor Efficiency drops by 1 to 2 %

c. Air Conditioning expenditures

2.2 Power consumption during loaded condition of Belt conveyor

Power delivered (Mechanical) : 18.48 kW

Power consumed (Electrical) : 21.24 kW (Motor Efficiency is 87 %)

[Note: 21.24 kW is the power delivered by VFD to the Motor. Instead of 88 %, now efficiency of Industion motor will be 87 %]

Power consumed by VFD : 23.08 kW (VFD Efficiency is 92 %)

[Note:To deliver Power of 21.24 kW, Drive will take 23.08 kW]

kWh consumed in 35 minutes : 13.46 kWh

2.3 Power consumption during un-loaded condition at 40 % of the rated speed of Belt conveyor

Power requirement at 40 % of the rated speed

Power (Mechanical) = Torque x Speed

= 83.8 x (2x 2.31415/60) x 0.40

Power (Mechanical) = 5.21 kW

Note: 5.21 kW is the 20 % of the rated power 26 kW. At 20 % of the rated load, efficiency of motor will be around 65 % instead of 87 %.

Power delivered (Mechanical) : 5.21 kW

Power consumed (Electrical) : 8.015 kW (Motor Efficiency is 65 %)

Power consumed by VFD : 8.711 kW (VFD Efficiency is 92 %)

kWh consumed in 25 minutes : 3.62 kWh

2.4 Power consumption for an hour operation by using VFD

Total Consumption (35 min +25 min) : 17.08 kWh (3.62+13.46)

3. AIR CONDITIONING COSTS AND OTHER FACTORS

3.1 Air Conditioning Costs

Air conditioning cost = Input Power x (100 - Drive Efficiency) x 2

As 60 % of the year is warm then =17.08 X 0.08 x 2 x 0.6 = 1.64 kW

3.2 Energy Saving or Not

kWh consumption with D.O.L = 18.41 kWh

kWh consumption with VFD (Excluding Air condition costs)= 17.08 kWh (Saving 1.33 kWh)

kWh consumption with VFD (with Air condition costs)= 17.08 + 1.64 = 18.72 kWh (Saving - 0.30 kWh)

3.3 Other Factors

i. There will be drop (No Idea how much) in VFD efficiency when operated at 20 % of its rated capacity.

ii. At Unloaded condition at 40 % of the rated speed, there will be problems of motor over heating.

I have one non related query.. how you calculated the motor output power as 18.48 kW.? ...just to know how...

Power (Mechanical) = Power (Electrical) x Efficiency

Power (Electrical) = 21 kW

"i. Belt conveyor runs 35 minutes in loaded condition consumes 21 kW."

Power delivered (Mechanical) = 21 x 0.88 = 18.48 kW

The easiest way to save energy is, while the belt is unloaded, switch the silly thing off.

"iii. Belt conveyor is not being stopped in unloaded condition to avoid breakage of belt due to frequent starts and stops."

This Belt Conveyor has approximate length of 350 meter. Longer belts are not stopped for short intervals to avoid belt from breaking.

Use heavy duty soft starter.......

Without seeing the material flow diagram it is difficult to comment. But I think in a material handling system it is not possible to stop a conveyor even if it is not loaded. Besides the conveyor damage as the OP has pointed out, stopping of one equipment in a chain will stop all other conveyors, feeders etc. before the conveyor as per the stopping sequence envisaged in the application software. Hence during restarting not only this conveyor but also other equipment which had stopped, need to be started again.

If a VFD were installed with a start-ramp and a stop-ramp, never mind the energy savings, the belt breakage savings and loss of production time for repairs to it would easily pay for the drive. Go get 'em, Floyd.

"Belt conveyor is not being stopped in unloaded condition to avoid breakage of belt due to frequent starts and stops."

Have worked on many conveyors. Stopping and starting should not be a problem. Sounds more like you need a new belt. The money you will spent on the VFD applied to a new belt and turning the conveyor off during times of non use. Would be more cost effective.

Not knowing what type of conveyor you have. If a slider bed. Slowing the belt down may increase the frictional force between the bed and back of the belt. Which will throw all these computations off.

If a roller bed I would be inspecting the rollers that they turn free and are not the source of the additional drag at start up that's causing the belt to break.

Also if being over loaded in a stopped position then started could be the problem with it breaking.

ADD LOAD OF VFD TRANSFORMERS, IN BELT CONVEYOR BEING CONSTANT TORQUE APPLICATION, THERE ARE HARDLY ANY POWER SAVING, YES, BENIFIT IS SMOOTH STARTING , WHICH BENIFITS LIGHT STRUCTURE, MORE LIFE FOR BELT , BUT FOR VFD DRIVE WHEN BREALDOWN OCCURED, IT IS EXPREINCED , THE REPAIR HAS TO DEPENDANCE FROM SUPPLIER FOR INTGRATED CIRCUITS AND USUALLY SUCH PLANT DOES NOT EMPLY HIGH PAID EXPERT FOR RARE USE.

Stop SHOUTING!

In your workout you forgot some other losses:

When going from low speed to higher speed, you will spend some extra energy to overcome the inertia of the system (this is a one off each time you re-accelerate). The kWhs you calculated are the friction losses to maintain the conveyor speed.

When slowing down, you lose this energy and have to re-spend it again.

Therefore, a more in depth study of the operations involved in your case is necessary before getting to the real gain or saving in energy.

It is better to avoid having the conveyor belt running unloaded in the first place! Maybe balancing the outputs of the different processes in sequence so that you could reduce the conveyor speed and keep it loaded most of the time (?).