Vibration

[magneticted]

Can one of you kind gents help me out here. We have specification from a client stating "" that the vibration of the motor shall not exceed 0.5 mil double amplitude in any direction at full or no load "". The motor in question is a IEC 400 frame with sleeve bearings. The motor spec states vibration according to IEC60034-14.and is expressed in vibration velocity as 2.3mm/s rms. So how do you convert 2.3mm/s to mil double amplitude value.

 

[GTstartup]

At what RPM?

 

[magneticted]

It is a 4 pole motor 50 hz 1500 rpm

 

[Skogsgurra]

The general relation is that speed v = ds/dt = (A/2)*d(sin(wt))/dt = A/2*w*cos(wt).

Where A is double amplitude, v is velocity and w is angular speed in rad/second.

Since cos(wt) cannot be more than 1, you get:

v = A/2*w
so, v = A/2*6.28*RPM/60

This is true for SI units, but if you express amplitude and speed in mils and mils/s, you can use it without any further unit conversions.

Convert mm/s to mils/s:

2.3 mm/s = 2.3*1000/25.4 = 91 mils/s

Which results in

v = A/2*6.28*RPM/60

or, in your case:

2.3*91 = A/2*6.28*RPM/60

A= 2*60*2.3*91/(6.28*RPM)

If your machine does 1500 PRM, you get

A = 2.7 mils double amplitude

And, at 3600 RPM, 1.1 mils double amplitude.




Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Skogsgurra]

Sorry, I see now that your velocity is rms. Multiply by sqrt(2) to get your values.

As your machine is 1500 RPM, the double amplitude is 3.8 mils.

Either I am wrong - or the spec calls for a specially balanced machine.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...