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DM7.ENG

Hi there

I wonder if anyone can shed any light on this. I was at ajob last week when I came across a pump with an integrated VSD and display. Onthe name plate I noticed it said Delta – 380V Star 660V.
It was always my understanding that star configuration resultedin less voltage across the windings as Vphase = Vline/√3 hence its advantageouscharacteristics for handling starting current in motors.:mad2:

The only sense I can make of this is that 380V – Delta isthe line voltage applied to the Delta connected windings and hence 660V is theline voltage when connected in star, But my question really is where does this660V come from? Any help is appreciated:smile5:

 
You looking at it the wrong way ... but kind of there.


Draw out the winding connections for both star and delta ...what you will notice is in star between any 2 phases you have 2 windings but in delta only one so for example if you have a Star/Delta starter it will put 400v across a set-up rated for 660v thus limiting current as the windings are undervoltaged, when it changes to delta then it has 400v across a set-up rated for 400v thus full voltage.

EU Dual voltage motors are usually 230vdelta - 400vstar and the same principle exists although you cannot start these with a star/delta starters and other methods may be required for soft starting, the 230v option is common for wiring up to VSD's that convert 1ph to 3ph giving 3phase 230v.

Also some countries have 3ph this high like sweden for example who 1ph is 230/400 and 3ph 400/690 so it suits certain arrangements elsewhere.
 
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380V is no different to 433V.

You apply 433V Ph→Ph to a star winding you will get 250V Ph→Y point and 250V ∆ is correct.

Likewise 660V Ph→Ph to a star winding you will get 380V Ph→Y point. Therefore 380V ∆ is correct.

I doubt your working on 660V but if you are be very careful about the meter you use.
 
380V is no different to 433V.

You apply 433V Ph→Ph to a star winding you will get 250V Ph→Y point and 250V ∆ is correct.

Likewise 660V Ph→Ph to a star winding you will get 380V Ph→Y point. Therefore 380V ∆ is correct.

I doubt your working on 660V but if you are be very careful about the meter you use.
a lot of testers dont like v that high, as i recall the megger 1720 i used in my am2 had 550v then a red cross through it.

some of the martindale indicators are rated to 600v and some 1kv.

i believe its the grey that are rated to 1kv
 
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In one of my many moves I lost the two adapter boxes that clipped on to an AVO. One was for 5KV the other for 300A, I’ll be honest I never had the nerve to use either.
 
Hi There thanks for the reply

Just to clarify, in other words although the windings are rated for 660V technically there will never be 660V present as the max voltage typically applied is 400V between phases. Also is the starting current reduced for star because there is 2 windings between phases instead of 1 with delta hence double the impedance?
 
Hi There thanks for the reply

Just to clarify, in other words although the windings are rated for 660V technically there will never be 660V present as the max voltage typically applied is 400V between phases. Also is the starting current reduced for star because there is 2 windings between phases instead of 1 with delta hence double the impedance?

In our voltage set-up you are correct and we call these star/delta motors by undervoltaging the motor by having 2 field coils in series the current is limited but as I said earlier 660/680v is actually used by some countries so serves a few purposes, putting a lower voltage through the field coils at 50hz is only advisable for starting a motor otherwise too much heat would be generated on loading ... a VSD gets around this by speed controlling the motor in a way that the frequency is reduced in line with voltage so a 400v motor at 50hz set to 50% speed would be injected with 200v at 25% hertz ... this is just a simplified example as its not as simple as that but give you an idea but in reality torque is effected also so the drives may require programming for each application.
 
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