[Unit conversion] From r/min to period (msec)

AI Thread Summary
The discussion focuses on converting a motor speed of 4000 r/min to a period in milliseconds. Various calculations are presented, showing that the period derived from 4000 r/min is approximately 0.25 msec or 2.39 msec, depending on the method used. The frequency calculated from 4000 r/min is 66.67 Hz, leading to a period of 15 msec, which does not match the stated 2.5 msec. A correction is made, indicating that if T is 2.5 msec, the frequency would actually be 400 Hz, suggesting an inconsistency in the original problem statement. The final correct approach involves using the expression for angular displacement to find the time that results in zero displacement, concluding that the initial calculations were not aligned with the given parameters.
JJ91
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Homework Statement


I would like to know how a 4000 r/min is converted to period of t=2.5mSec


Homework Equations


Speed of motor = 4000 r/min
Build-up time of current = 2.5msec
Frequency = 60Hz


The Attempt at a Solution


\frac{1}{4000}=0.25msec not 2.5msec

OR
Assuming it means 4000RPM which is equivalent to 419 rad/sec we obtain:
\frac{1}{419}=2.39msec not 2.5msec

OR
If frequency is 60 Hz, and 1-phase is conducting for 60° we can take period as:
t=\frac{1}{f}=\frac{1}{60}=16.67msec
Conducting for \frac{1}{3} of the cycle (60°), t = 5.55msec

Where have I made a mistake ?
 
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tomek91 said:

Homework Statement


I would like to know how a 4000 r/min is converted to period of t=2.5mSec


Homework Equations


Speed of motor = 4000 r/min
Build-up time of current = 2.5msec
Frequency = 60Hz


The Attempt at a Solution


\frac{1}{4000}=0.25msec not 2.5msec

OR
Assuming it means 4000RPM which is equivalent to 419 rad/sec we obtain:
\frac{1}{419}=2.39msec not 2.5msec

OR
If frequency is 60 Hz, and 1-phase is conducting for 60° we can take period as:
t=\frac{1}{f}=\frac{1}{60}=16.67msec
Conducting for \frac{1}{3} of the cycle (60°), t = 5.55msec

Where have I made a mistake ?

If T = 2.5ms, then the frequency is 400Hz. There is something wrong with your initial statement. Where did it come from?
 
It comes from "Electric Machinery" by Mr. Fitzgerald

14ybs6b.jpg

At the very bottom of the screen a period, T, of build-up of current is defined as 2.5msec.
 
For N = 4000 rpm, then n = 4000/60 rps = 66.67 rev/s

1 rev = 2 pi radians

omega = 2 pi * 66.67 = 133.33 pi radians/s

T = 1/omega = .0024 s = 2.4 msec
 
Disregard previous post:

For N = 4000 rpm, then frequency f = 4000 / 60 = 66.67 cycles / s

T = 1/f = 0.015 s = 1.5 msec
 
All right, thank you very much. I will use the very first formula, still it isn't exact 2.5msec.
 
The solution gives the expression:

##\theta_m = -\frac{\pi}{3} + \omega_m t##

Plug in your value for ##\omega_m## and solve for t that makes ##\theta_m = 0##.
 
gneill said:
The solution gives the expression:

##\theta_m = -\frac{\pi}{3} + \omega_m t##

Plug in your value for ##\omega_m## and solve for t that makes ##\theta_m = 0##.

This is the final correct solution.

Thanks, thread can be now closed.
 

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