Speed Reducer or Increaser, which is better ?
Speed Reducer or Increaser, which is better ?
(OP)
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.





RE: Speed Reducer or Increaser, which is better ?
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?
RE: Speed Reducer or Increaser, which is better ?
RE: Speed Reducer or Increaser, which is better ?
RE: Speed Reducer or Increaser, which is better ?
Best regards,
Mark Empson
http://www.lmphotonics.com
RE: Speed Reducer or Increaser, which is better ?
You know, friction and other rotational losses.
Do you think that I better post this thread in mechnical engineering forum ?
RE: Speed Reducer or Increaser, which is better ?
RE: Speed Reducer or Increaser, which is better ?
RE: Speed Reducer or Increaser, which is better ?
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.
RE: Speed Reducer or Increaser, which is better ?
RE: Speed Reducer or Increaser, which is better ?
What is the Brake Horsepower of the ID Fan at 850 RPM?
RE: Speed Reducer or Increaser, which is better ?
BHP at 850 RPM is 230.
RE: Speed Reducer or Increaser, which is better ?
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]
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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)
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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.
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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.
RE: Speed Reducer or Increaser, which is better ?
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.
RE: Speed Reducer or Increaser, which is better ?
Thanks for your time and post. Your calculations were eductive. Will get back to you after our trials.
RE: Speed Reducer or Increaser, which is better ?
Thanks for your time and post. Your calculations were educative. Will get back to you after our trials.
RE: Speed Reducer or Increaser, which is better ?
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?
RE: Speed Reducer or Increaser, which is better ?
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.
RE: Speed Reducer or Increaser, which is better ?
I guess I am a little skeptical of your conclusion.
RE: Speed Reducer or Increaser, which is better ?
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.
RE: Speed Reducer or Increaser, which is better ?
Let the two speeds of interest be speed 1 (880rpm) and speed 2 (770rpm).
N = speed
Trated = Torque at rated power
Taverage = average torque over the speed range 0 to N
t = acceleration time
N = speed
J = inertia
Tload – load torque assumed zero during start for simplicity
t1 = J1*N1/(Taverage1)
Trated2 = Trated1*(N1/N2)
Assuming similar torque vs slip curves:
Taverage2 = Taverage1 * (N1/N2)
J2=J1*(N1/N2)^2
t2 = J2*N2/(Taverage2)
t2 = [J1*(N1/N2)^2] * [N2] / [Taverage1*(N1/N2)]
t2 = J1 * N1/Taverage1
t2 = t1
The quantity of interest is not likely starting time, but the ability to start the motor safely. As noted before, the lower speed motors have ability to start larger inertia's, BUT as jomega pointed out the inertia will also look bigger to the lower speed motor. I'm not sure which effect in general is more important.
RE: Speed Reducer or Increaser, which is better ?
Load WK2 =+A*hp^0.95/(rpm/1000)^2.4 - 0.0685 *hp^1.5/(rpm/1000)^1.8
If I am not mistaken, it's telling me the max load inertia is roughly propotional to 1/speed^2 for a given horsepower. Since the inertia seen will also be proportional to 1/speed^2, there doesn't seem to be a reason to prefer one over another based on motor safe starting considerations. But you should investigate starting carefully for yourself.
RE: Speed Reducer or Increaser, which is better ?
The fan torque requirement was computed from 250hp, should be computed from 230hp. That will give you a little bit of margin to keep the motors less than full load amps. If you still feel you are too close to full load amps (and the process allows), you can change the ratio to decrease the speed sligthly..... fan power will change according to speed^3.
RE: Speed Reducer or Increaser, which is better ?
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.
///The wound rotor induction motor has a lot of heat dissipated in the motor starting resistor during the motor start. In fact, the wound rotor starting resistor is sometimes used for the motor speed control. However, the starting resistor has to be properly designed with respect to the wound rotor motor rotor and stator parameters.\\\