Simplified method for estimating the acceleration time of centrifugal drives
Simplified method for estimating the acceleration time of centrifugal drives
(OP)
Greetings:
If anyone has the book "Industrial Power Systems" by Khan the subject refers to the section (9.7.7) in that book.
There have been many helpful posts on motor acceleration time on eng-tips, but none which I have found which might explain why:
average motor torque, Tm = (3*Tlr + Tbdn)/4 (Tlr = locked-rotor torque, Tbdn = breakdown torque)
average load torque, Tl = Tl/3
Can you explain how these averages are obtained?
These go into the equation for acceleration time,
Acceleration time in seconds = (WR2 × rpm change)/ (308×TA) , where TA = Tm - Tl
Thanks for your comments.
If anyone has the book "Industrial Power Systems" by Khan the subject refers to the section (9.7.7) in that book.
There have been many helpful posts on motor acceleration time on eng-tips, but none which I have found which might explain why:
average motor torque, Tm = (3*Tlr + Tbdn)/4 (Tlr = locked-rotor torque, Tbdn = breakdown torque)
average load torque, Tl = Tl/3
Can you explain how these averages are obtained?
These go into the equation for acceleration time,
Acceleration time in seconds = (WR2 × rpm change)/ (308×TA) , where TA = Tm - Tl
Thanks for your comments.





RE: Simplified method for estimating the acceleration time of centrifugal drives
ASSUMPTION 1: torque ~ speed^2 (reasonable for centrifugal load)
ASSUMPTION 2 speed increases linearly vs time (SIMPLIFYING assumption, not particularly accurate)
Combine assumption 1 and 2 => torque ~ t^2
i.e.
torque ~ Tmax * (t/tfinal)^2 as t goes from 0 to tfinal
Find average value by integrating over the interval and dividing by length of interval:
<torque> = (1/tfinal) * int{torque(t)}dt
<torque> = (1/tfinal) * int{Tmax * (t/tfinal)^2}dt, t=0..tfinal
<torque> = (1/tfinal) * int{Tmax * (t^2/tfinal^2)}dt, t=0..tfinal
<torque> = (Tmax/tfinal^3) * int{ (t^2}dt, t=0..tfinal
<torque> = (Tmax/tfinal^3) * { [t^3/3]@t=tfinal - [t^3/3]@t=0]}
<torque> = (Tmax/tfinal^3) * { tfinal^3/3 - 0}
<torque> = Tmax/3
=====================================
(2B)+(2B)' ?
RE: Simplified method for estimating the acceleration time of centrifugal drives
Actually the average of the torque is not particularly relevant. The inverse of the average of the inverse of the accelerating torque is relevant. If you have no load and want to find an equivalent constant motor torque that exactly determines accelerating time when plugged into your formula, T = (WR2 × rpm change)/ (308×Tm), then you would invert the torque vs speed curve, find the average of the inverse of torque, and invert the result. It's not the same as the average, but it's close and it's easy. Likewise subtracting average of load torque from average of motor torque is not really what we're after (we're after inverse of average of inverse of accelerating torque) but it's close and it's easy.
=====================================
(2B)+(2B)' ?
RE: Simplified method for estimating the acceleration time of centrifugal drives
Look towards the end of this thread for some more info why and for other approaches that can give better results.
http://www.eng-tips.com/viewthread.cfm?qid=277213