using low speed motors instead of high speed motors with constant torque 1

[eng2waleed2mouhammed]

Hi, All
i have a question ...
i have a conveyor moving with a (15 kw / 1500 rpm) with 71 N.M and i wish to know if i can replace it with a motor (11 kw / 1000 rpm) with the same torque...
waiting for your help

 

[waross]

On the face of it it should drive the load.
It may be well to consider starting. NEMA motors have a design letter. This describes the curve of the starting torque. If the original motor is a special, high slip motor to aid starting and handle torque surges better, the replacement by any motor with normal torque characteristics may fail or be problematic.
You may not have NEMA spec motors, but the point is to check the starting torque curve of the replacement motor against the starting torque curve of the original motor if possible. The curves may not be the same.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jraef]

Torque is usually the determining factor in a conveyor application, but it is application specific and we dont know the application.

For example, we cannot address the effects of a slower speed and what that may mean to other equipment in series with this conveyor. Let's say this conveyor is on the input of a rock crusher. Slower speed means less material feeding in, which may be OK, or it may affect the output of the crusher if it is autogeneous (rock on rock fracture crushing). If it is on the output of the crusher feeding a vibrating screen, slowing it down means more material builds up on the belt because it moves slower beneath the output chute, which means you would need MORE torque to move it. The same might be true of moving boxes. Something is putting boxes ONTO that conveyor, so if that something is still putting them on at the same rate, you end up with more boxes per unit of belt length, more weight, so more torque required.

Nor can we tell if the motor frame will fit. Slower speed means more poles, which often means a larger frame size (although maybe not in that power rating).

"Will work for (the memory of) salami"

 

[racookpe1978]

What is the difference in design impact (of this comparison between old (high speed) and new (low speed) motor torques) on an "instantaneous start" for the conveyor? You mentioned secondary effects during long-turn operations.

(After a power failure, for example, when the conveyor line or the screen is already loaded, but needs to be restarted under load?)

 

[eng2waleed2mouhammed]

thanks for help and i am already try to get the Torque - Speed Curve for the old(high speed) and the new (low speed) and compare both of them to determine if i can replace
i will tell you about it soon.

 

[waross]

There is a good explanation of design letters here in the Cowern Papers. These are the motor characteristics.

http://www.baldor.com/pdf/manuals/PR2525.pdf

Page 7

jraef gave an excellent explanation of the some of the load factors associated with changing the speed of a conveyor.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[LionelHutz]

Maybe

Why slow it down? Some more information would allow a better answer then "maybe".

Just to note - energy savings don't correlate well with the speed of most conveyers but rather correlate well with the rate of material transfer.

 

[eng2waleed2mouhammed]

ok LionelHutz (Electrical) i'll give some info...
my application is a conveyor to transfer corn from the storing silos, 2 conveyor before it and one conveyor after it.
the power transmission is made using pully and v-belt to transfer the power from the motor to a Gearbox and then to the shaft of the conveyor.
the current final speed is 39 rpm and the planned final speed is 31 rpm (the design is made by the mechanical engineers on the site and the point for me is to save the electrical energy so i made the second choice).
the current motor data is 15 kw /1500 rpm / 98.3 N.M
the planned motor is 11 kw / 1000 rpm / 108 N.M

i hope that the info is enough...
waiting for opinions

 

[desertfox]

Hi

I am no motor expert but you mention a new planned final speed of 31 rpm from the current 39 rpm, would this be using the same gearbox?
If so you won't get to 31 rpm because the current gearbox has a ratio of 1500/39 that gives a ratio of 38.46, so if you put the new motor on the same gearbox you will only get 26 rpm.

My ten penny worth

Desertfox

 

[eng2waleed2mouhammed]

Hi desertfox (Mechanical)
i forget to tell about the pully size for the motor and the GB
the motor pully size is 10 cm and the GB pully is 27 cm
and the planned pully size for the motor is 13 cm and for the GB is 20 cm (all sizes for the outer diam.) so the input speed for the GB remain constant with the ratio remain constant the final speed will be reduced to 31 rpm instead of 39 rpm
waiting for your response.

 

[LionelHutz]

Well, you're changing the drive ratio so assuming constant torque between motors is wrong.

I calculate the 1000rpm motor needs to be 117.2Nm to get the same torque at the conveyer.

To move X kg of material per hour requires Y amount of energy. You can slow the conveyer down so that Y is less but that just means X will also be less. You could also just load the conveyer less and Y will be less, but X will still also be less. To save energy, you have to figure out how to move X kg of material while using fewer kWh of energy. In other words, you need to figure out how to make the conveyer more efficient.

When evaluating energy savings you always have to evaluate work done per unit of energy used. Just to give a few examples;
- kg of material per kWh
- liters of liquid per kWh
- cubic meters of air per kWh

Just sticking a watt meter on the motor and seeing a lower kW reading can be very misleading.

Assuming a 11kW motor will save energy compared to a 15kW motor can be very misleading.

 

[eng2waleed2mouhammed]

hi LionelHutz (Electrical)
i am sorry to say that we will reduce the final speed of the conveyor from 39 rpm to 31 rpm ...
the final speed will remain constant so that the transferred quantity sill constant hence we will reduce the energy by 4 kw and transfer the same quantity (Energy saving is achieved)...
waiting your response

 

[jraef]

That's right.
Energy = power consumed (work) x time
Production = work performed x time

So it appears you are trying to lower the power that the motor consumes, but must retain torque to make sure it works. That means slowing down the motor, which will increase the time it takes to perform the task, so the motor is doing less work over time. You then will either reduce the total production, in your case the movement of corn, or if you are needing the same amount of total production, your time will be longer and that means your net energy remains the same.

In reality there are always losses in the transfer of energy from electical to kinetic in the form of heat from motor inefficiency, conveyor friction etc. So (arguably) the longer it takes to perform a task, the longer the components waste energy as heat and you may even slightly INCREASE your energy consumption.

"Will work for (the memory of) salami"

 

[LionelHutz]

i am sorry to say that we will reduce the final speed of the conveyor from 39 rpm to 31 rpm ...
the final speed will remain constant so that the transferred quantity sill constant hence we will reduce the energy by 4 kw and transfer the same quantity (Energy saving is achieved)...
waiting your response

You lost me with the 2 contradicting bold parts I have quoted.

Did you mean to say the conveyer pulley will still run at 39rpm even though you've changed the motor speed? You won't save the energy you think by doing that. You'll likely just break the conveyer so it doesn't work for you. It's simply wrong to believe you can keep moving the same amount of material yet change from a 15kW motor to a 11kW motor and lower the power drawn by 4kW or save 4kWh of electricity per hour the motor operates.

 

[squeeky]

If your initial drive is vee belts, watch out for slipping belts because of higher torque. The larger pulley diameter will give a larger wrap contact surface but it should be checked. It appears that the conveyor speed will remain the same as the input speed to the gearbox is the same. The reduced speed in the motor has been corrected with the pulley sizes. Go back to previous comment on the mass moved over time as this will have influenced the initial motor kW selection. A motor running light will generally run less efficiently and at a worse power factor. The closer you run a motor to it's design parameters, the better for the motor. Sometimes conveyor motors are purposely oversized to clear chokes. Correct interlocking for stopping and starting is critical.

 

[waross]

The KW rating of a motor is the maximum load that it may drive without overheating. If load is less, the KW draw will be less. Many motors have best efficiency at a little less than full load. I have seen the efficiency sweet spot at 75% or 80% of full load. Motor efficiency generally drops off when the motor is overloaded.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[desertfox]

Hi just looking at the pulley ratio's that you posted, with the original motor the GB pulley was doing 555 rpm and the proposed pulley with give 650 rpm with the new motor, so unless some changes are made in the gearbox the conveyor cannot run the same speed.
Also based the motor torque's given by the OP then the torque into the gearbox will be about 100Nm less with the new motor than with the old set up.
There is some serious mis information somewhere down the line.

 

[LionelHutz]

Yes, most motors are very similar efficiency wise from at least 75% to 100% load. There are even motors that are more efficient at 75% load compared to 100% load. So, it would be better for most motors to run around 75% load range. You still get good efficient and the motor is running cooler.

As for the misinformation? Apparently the mechanical engineer should be fired. With the unsure ratio changes he's proposing, it appears the new motor may need to be capable of producing 175% of the torque of the old motor just to keep the conveyer working.

 

[waross]

You will be safe replacing the 15kW/1500 RPM motor with a 15 kW/1000 RPM motor and changing the gearing so the conveyor runs the same speed.
As motors of a given kW go down in speed they often get physically larger, more expensive, and less efficient. Let us know how it turns out.
Buying for a new installation is one thing, you want the correct size as over-sized equipment is an increased capital expense.
However buying a new motor to replace an existing motor that may be a little over-sized will still use the same energy to do the same work. no saving there. The new smaller motor may be less efficient so there is a net loss in efficiency. Even if it is a little more efficient the payback time may be extremely long. There may be exceptions for grossly over-sized motors but not many.

Bill
--------------------
"Why not the best?"
Jimmy Carter