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

1. Nov 24, 2012

### JJ91

1. The problem statement, all variables and given/known data
I would like to know how a 4000 r/min is converted to period of t=2.5mSec

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

3. 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 ?

2. Nov 24, 2012

### Staff: Mentor

If T = 2.5ms, then the frequency is 400Hz. There is something wrong with your initial statement. Where did it come from?

3. Nov 24, 2012

### JJ91

It comes from "Electric Machinery" by Mr. Fitzgerald

At the very bottom of the screen a period, T, of build-up of current is defined as 2.5msec.

4. Nov 24, 2012

### SteamKing

Staff Emeritus
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

5. Nov 24, 2012

### SteamKing

Staff Emeritus
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

6. Nov 25, 2012

### JJ91

All right, thank you very much. I will use the very first formula, still it isn't exact 2.5msec.

7. Nov 25, 2012

### Staff: Mentor

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 $\theta_m = 0$.

8. Nov 25, 2012

### JJ91

This is the final correct solution.

Thanks, thread can be now closed.