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TheBlacksmith (Mechanical) (OP)
5 Nov 03 10:52
I am involved in a generator development project.  The development process involves preliminary testing on full coils in a test jig, followed by testing the completed generator.  Obviously some time will elapse between the two.  How reliable are the results of a power factor tip up/tan delta test performed on the test coils in predicting the performance of the completed unit?  Both tests will be performed at 25% and 100% of line to ground voltage using the same (Doble) test equipment.

Thank you.

Blacksmith
electricpete (Electrical)
5 Nov 03 13:03
When properly performed, it is probably the best test out there as a qc test for coils which are not yet wound into the machine.  (There are some people who prefer partial discharge, but it is newer and not as proven).

"Properly performed" includes precuatins taken to guard out the effect of the non-linear grading coating.  That cannot be done once the coil is in the slot.

We also do it on some in-service machines.  We recognize there is some tip-up due to that non-linear coating. For these in-service machines we view it more as a trending/comparison thing than an absolute. Doble has a fairly large database of results by machine manufacturer, voltage, horsepower.  Using this you can get a pretty good idea whether a machine is an outlier.

TheBlacksmith (Mechanical) (OP)
5 Nov 03 13:39
Electricpete-

Thanks for the reply - we are doing a comprehensive battery of tests, including partial discharge and tip up.  There is a concern that the treatment necessary to meet our PD requirements may preclude meeting the tip up in the assembled machine.  These machines are new designs, so the database from either Doble or Iris may not be of much use.  I have recommended trend analysis in service of the PD to monitor the long term health.  Your answer confirms my suspicion that we cannot assume the assembled machine tip up will be the same as the test coils and we will have to wait until the final test to determine if the machine is meeting our requirements.

Blacksmith
edison123 (Electrical)
5 Nov 03 20:28
theblacksmith,

how big is your generator ? if the coils are long (say over 1.5 meter), then the stress grading system would not have much effect on the pf tip up either in coils or in winding. However, for short coils, the stress grading system drastically affects the pf tip up of coils and hence the winding.

Personally, I have seen 3 metre long roebel bars showing no significant change in tip up with stress grading coating. But in coils of about 1 meter long, both the absoulte and tip up increased by 3 times.
QCE (Electrical)
7 Nov 03 13:29
I just have a quick question - I have seen on some coils a negative tip up - what is the deal with that?
electricpete (Electrical)
7 Nov 03 14:21
I haven't heard of negative tip-up.

I have heard of negative power factor, which is quite a counter-intuitive thing. Doble published an article on that with detailed phasor vector diagrams that I could never digest.  Is that what you're talking about?
QCE (Electrical)
7 Nov 03 17:17
No the tag that comes with the coils has a tip up value and it is 99% of the time positive but it is on occasion negative.  Does this mean that the lower kV test has a higher pf than the higher voltage test?  I have been told that it is possibly something to do with hooking the test leads up backwards, however I don't undersdtand this arguement.  I believe that the test is done at 10kV and 2 kV on hydro generator coils.

Sorry to take over the discussion in someone elses thread.
edison123 (Electrical)
7 Nov 03 21:17
I have personally tested coils with negative tan delta with increasing voltage. Though IEEE std 286-2000 makes a complex explanation (in section 4) for this phenomenon, it doesn't say whether it is bad or good.
edison123 (Electrical)
7 Nov 03 21:19
correction to previous posting

" tested coils with decreasing tan delta with increasing voltage".
electricpete (Electrical)
8 Nov 03 17:53
I didn't realize that was in ieee286 as edison mentioned, but there it is.  Here is a piece of that discussion:

"Hence,on the assumption that tan ä ,which is determined by the dielectric losses,remains unchanged with
voltage,the overall tan ä value of the bar insulation will vary with the second term on the right hand side of
Equation (3),which represents the discharge power loss contribution to the dissipation factor.
As long as an increasingly larger number of voids begin to undergo discharge with rising applied voltage,
the value of tan ä will continue to increase.Once all voids become ionized and are discharging,the tan ä
value,after attaining a maximum,will commence decreasing with voltage.This behavior is manifested
when the power loss due to all the partial discharges is increasing at a lower rate than the square of the
applied voltage term,V 2 ,in the denominator of the second term of the right hand side of Equation (3).
Consequently,a negative tip-up value of tan ä (if it is caused by partial discharge losses)occurs when all the
existing voids become ionized and begin discharging at some lower voltage,and a further rise in applied
voltage does not result in any additional discharging voids."

Sorry, it loses a little bit without the equations which I can't text copy in here.

What they're saying kind of makes sense.
Each void can discharge only once.
IF the pd at the low votlage has pretty well got all of the voids discharging, then increasing the voltage results in increased capacitive current but no further resistive (arcing) component, so overall decrease in power factor.

Now trying to remember the theory... which type of void arcs at a lower voltage: small or large?  That's the one you've got (according to the theory which is a little dubious imho). I have to go back and look at some references to figure that out.

I'm sure doble can provide plenty of info on the subject.
jbartos (Electrical)
8 Nov 03 18:07
Suggestion:
Generator Windings.- The power-factor-voltage characteristic (power-factor tip-up) is used primarily as a quality-control criterion in manufacturing. It is sometimes used as an acceptance test on individual coils. Power-factor tip-up has been used as a maintenance test because a change in the tip-up value over a period of time is an indication of change of condition of the coil insulation. The sensitivity of the power-factor tip-up test decreases with the length of coil included in the measurement. Therefore, one coil, or coil side, is the preferred test unit, although coils are sometimes tested in groups of two or three to expedite testing. Refer to IEEE Standard No. 286, July 1975, "IEEE Recommended Practice for Measurement of Power-Factor Tip-Up of Rotating Machinery Stator Coil Insulation," for a description of the power-factor tip-up test.
jbartos (Electrical)
8 Nov 03 18:08
edison123 (Electrical)
9 Nov 03 20:19
pete,

On reading this IEEE std explanation, what do you conclude ? Negative tan delta is ok or it is bad ? I sure can't make it.
generatorinsulation (Electrical)
10 Nov 03 1:47
TheBlacksmith
I think that your aim of testing after assembelling is assembel process controling
thus you can compare results of individual test on parallel winding circuits to each other and results of different phases to each other.
but be care, all tests must be done in same condition especially humidity and temperature, and sterss grading rings should be used on bare connection.   
electricpete (Electrical)
10 Nov 03 22:20
I remember that Iris did a lot with a capacitive voltage divider model of a void.
There is a capacitance due to void (air) and a capacitance due to solid insulation (insul).

Symbols
Vapplied = test voltage
Vair = voltage across the void
Xair = capacitive reactance of air
Xinsul = capacitive reactance of insulation.
dair = depth of air void
dinsul = depth of insulation
d = total depth void plus insulaton
eair = permtivity of air
einsul = permitivity of insulation

Voltage divider:
Vair = Vapplied * Xair/(Xair+Xinsul)
     substitute: X ~ d/e
             assumes area A is comparable.
             simplistic geometry ignores fringing fields
Vair = Vapplied * (dair/eair) / [(dair/eair) +(dinsul/einsul) ]
     multiply top and bottom by eair
Vair = Vapplied * (dair) / [(dair) +(dinsul*eair/einsul) ]
      substitute eair/einsul = 0.25
Vair = Vapplied * (dair) / [(dair) +0.25*dinsul) ]
substitute dinsul=d-dair
Vair = Vapplied * (dair) / [(dair) +0.25*(d-dair) ]
Vair = Vapplied * (dair) / [0.75*dair +0.25*(d) ]
Eair = Vair/dair
        = Vapplied / [0.75*dair +0.25*(d) ]
As gap depth dair increases, Eair decreases.
Therefore Eair is highest in the smallest voids.
Discharge occurs when Eair exceeds ~ 3kv/mm
(assuming V>Paschen minumum).
 smallest voids will discharge at lower applied voltage).
 Larger voids will not discharge until higher applied voltage.

Negative tipup implies the majority of voids are already discharging at the low test voltage and very few additional voids discharge at the high voltage.
 the majority of the voids are small.
 Generally large voids are the ones that cause the most concern.
 This would seem to imply the negative power factor is not a concern.
 But  it is a very simple model of very complex stuff.  
 Doble can provide the best advice.
jbartos (Electrical)
10 Nov 03 22:46
Suggestion: Visit
http://www.partial-discharge.com
etc. for more info
edison123 (Electrical)
10 Nov 03 22:59
Thx pete for your time & interesting analysis.

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