Speed Reducer or Increaser, which is better ? 5

[edison123]

An ID fan running speed is 850 RPM. We have two motors – one with 720 RPM (8 pole) and other 990 RPM (6 pole). Which motor will give an overall better efficiency – the 720 RPM motor with speed increaser or the 990 RPM one with speed reducer ? Forget VFD’s.

 

[acmotorengineer]

In general, higher speed (lower pole) motors with gear reducers have the best combination of:
efficiency
initial cost
service reliablity
availability

For instance, US Motors quotes 89.5% efficiency and 85.1% Power Factor for a 10HP, General Purpose, 6Pole and 88.5% Efficiency and 74.6% Power Factor for an 8Pole.

Since gear efficiency is readily avail from manufacturers, it should be easy to check overall efficiency, at least.

What is your HP? Duty Cycle? Motor Efficiency?

 

[jbartos]

Suggestion: Typically, if pulleys are used, the motors have the higher speed than the fan.

 

[edison123]

Both the motors' HP is 250 HP, 415 V. Both the motors' efficiency is 87% (wound rotors). p.f data not available. My question is since the efficiency is same for both the motors, how will be the overall system efficiency (motor + gear) for the two scenarios?

 

[Marke]

It is usual for the motor efficiencies to fall as the number of poles increase. From you quoted figures, this is not the case in this instance. As the pulley ratios required for both cases are similar, it is reasonable to assume that the mechanical efficiency would also be similar and so there will probably be no difference between either motor. I would tend to run with the higher speed motor, but that is just based on experience.
Best regards, Mark Empson

http://www.lmphotonics.com

 

[edison123]

I thought that with lower speed motor (720 RPM, 8 pole), the losses will be lesser in the gear box as compared to the higher speed motor (990 RPM, 6 pole).

You know, friction and other rotational losses.

Do you think that I better post this thread in mechnical engineering forum ?

 

[jbartos]

Suggestion: When it comes to mechanical engineering, many electrical engineers try to do their best. How many electrical engineers are using mechanical software?

 

[lanier]

If you are using a gear reduction or increaser to create the 850 rpm that you need for the fan, then remember that you will also be affecting the torque that the motor can produce. By using a 720 rpm motor, increasing the speed of the motor will reduce the amount of torque the motor can apply to the fan and will also require a higher power consumption to operate at that speed. The opposite affect will be seen using a 990 rpm motor with a reducer. I would recommend using the 990 rpm motor as it will be able to handle the load much easier and will also provide you energy savings in the process.

 

[electricpete]

lanier - if the motors are rated at the same horsepower, then when you transform their rated torque to a common speed, you will get the same torque capability (neglecting losses).

One thing I was going to mention is the smaller motor is probably a little cheaper. Maybe not a problem if you have the motors on-hand already, but may save money if you ever have to rewind. Also the smaller motor will be a little smaller and therefore a little easier to install and if need be remove.

Given you have identified similar efficiencies on the motor, I wouldn't think that the efficiency difference in two gearbox setups would be enough to worry about. But you are certainly welcome to ask the mechanical guys. You might consider the mechanical vibration forum since there are a lot of rotating equipment guys there. Also of course the pump forum.

 

[edison123]

Thank you gentlemen for your suggestions and time.

 

[jOmega]

Edison123:

What is the Brake Horsepower of the ID Fan at 850 RPM?

 

[edison123]

jOmega

BHP at 850 RPM is 230.

 

[jOmega]

Edison123:

Thank you for your quick reply.

Based upon the information you have provided, I offer the following calculations for consideration.

The two motors are considered as CASE I and CASE II.

[blue]CASE I - 250 HP 720 RPM motor : Use gear increase to raise applied RPM to 850[/blue]

[green]CASE II - 250 HP 990 RPM Motor : Use gear reduction to reduce applied RPM to 850[/green]
---------------------------------------------------------
Common detail:

..... ID Fan BHP 250 HP @ 850 RPM

..... HP = (T x N) ÷ 5252

..... T = (5252 x HP) ÷ N

..... where: T = torque in ft-lbs
..... N = speed in RPM

ID Fan Torque requirement at 850 RPM:

..... T = (5252 x 250) ÷ 850

..... T = 1545 ft-lbs (nominal)
-----------------------------------------------------

CASE I - 250 HP 770 RPM motor

Motor Torque:

..... T = (5252 x 250) ÷ 770 = 1705 ft-lbs

Gear Increase ratio :

..... 770 ÷ 850 = 0.90588 : 1

Assuming that this ratio can be obtained with sheave diameter ratio and a timing belt... and an efficiency of .985

Delivered Torque = Motor Torque x Gear Inc. Ratio x eff.
................ = 1705 x 0.90588 x 0.985 = 1521 ft-lbs

So, the delivered torque would be short about 1.3% which means that the motor would operate just a wee bit above its rated amps to make up the difference.. However, the slip would increase some and the actual motor rpm would be slightly less than 770 ... But, you should be pretty close at normal operating air temperature.
(At cold air temp) you won't quite get up to speed because the greater air density will require more torque.. but the speed should increase as the air temp warms up and the density decreases with it.

Ok ? then on to CASE II

CASE II - 250 HP 990 RPM motor

Motor Torque:

..... T = (5252 x 250) ÷ 990 = 1326 ft-lbs

Gear Increase ratio :

..... 990 ÷ 850 = 1.1647: 1

Assuming that this ratio can be obtained with sheave diameter ratio and a timing belt... and an efficiency of .985

Delivered Torque = Motor Torque x Gear Inc. Ratio x eff.
................ = 1326 x 1.1347 x 0.985 = 1482 ft-lbs

So, the delivered torque would be short about 4% which means that the motor would operate about 5% above its rated amps to make up the difference.. However, the slip would increase some and the actual motor rpm would be something less than 990 ... maybe 940-950 RPM.

==========================================================

These are rough calculations and assume that implementing the required gear ratio is done with sheave diameter or if you use a timing belt .... the teeth ratio of the two sheaves.... the assumed efficiency would be approx. 98 - 99% (compromised value of 98.5% used for above calculations) .... Your power transmission distributor can advise actual efficiency to be obtained from the belts selected. I would not recommend V-belts for this application.

Now then, I doubt that you'll be able to get the exact ratio required, so additional compromise will need to be made. However, the foregoing can serve as a model for recalcuating.

From the foregoing, it would appear that the 8-pole motor would be the better choice..

Please let us know (with a follow-up posting) what you decide to do and how it works out.

 

[jOmega]

Edison123:

One additional consideration...

The CASE I (770 RPM) motor with the gear increase...

Will take longer to accelerate the fan.

Reason: fan inertia is reflected to the motor shaft by 1/(GR²)

This means that the inertia of the ID fan (as seen by the motor shaft) is the fan inertia lb-ft² divided by 0.90588²

or...

Fan inertia divided by 0.8206 using the numbers from t CASE I example.

This will increase the accel time .. and increase the time that the motor is pulling greater than nameplate amps while accelerating. However, as this is a Wound Rotor motor, it should have sufficient thermal capacity to accelerate the fan. However, you may have to go to a Class 30 overload relay to keep it from trippiing during accel.

 

[edison123]

jOmega

Thanks for your time and post. Your calculations were eductive. Will get back to you after our trials.

 

[edison123]

jOmega

Thanks for your time and post. Your calculations were educative. Will get back to you after our trials.

 

[electricpete]

It seems to me that the conlcusion that the faster motor accelerates the load faster is in error.

The motors both have the same horsepower. That means that the rated torque varies in inverse proportion to the speed. Assuming that they both had identical torque vs slip curves (normalized to their own rated torque), then when we transform them to the same speed we will get identical torque speed curves (in the load reference frame).

Or if you prefer to work in the reference frame of the slower motor, you find that:
#1 - The inertia is higher by factor of (1/speed)^2 (speed<1)
#2 - The rated torque is higher by factor of 1/speed (speed<1)
#3 - The speed range that we have to accelerate is lower by factor of speed (speed <1)

#2 and #3 cancel out the effects of #1. Time to accelerate is the same with either motor (nelgecting differences in inertia of the motors themselves). Right?

 

[electricpete]

I am not that familiar with using a wound rotor motor to start a large fan. That sounds like a strange application. Is the rotor to be shorted out or resistors applied during starting.

If rotor is to be shorted out, I would think that the ability of the motor to accelerate the load safely without damage will be a prime consideration. A large fan such as this can present a challenge. NEMA lists standard values of inertia which motors can safely accelerate. It does list higher inertia values for lower speed motors (of similar horsepower). I'm not sure what applies to wound rotor motors.

I was under the general impression that wound rotor motors (with rotors shorted) have less capability to accelerate large loads than comparable squirrel cage motors. The reason being that squirrel cage rotor can withstand a lot of heating, wound rotor can withstand much less due to insulation. Correct me if I'm wrong.

 

[electricpete]

j[ignore]Omega[/ignore]; - can you explain why you think one of the motors would be overloaded and one would not?
I guess I am a little skeptical of your conclusion.

 

[electricpete]

I can see where you made the error.
Case 2 grear ratio is 1.1647
You types 1.1347.
If you correct your error you calculate 1521 ft-lb for both motors. i.e. both motors by your calculation show torque delievered is 1.3% less than rated. That is very closely tied to the 98.5% assumption on the gearbox.

The exact operating point of course is determined by intersection of the load and motor torque-speed curves.