I have an 11 KV, 1 MVA VPI insulated generator stator which was thoroughly overhauled and dried out. While the PI (10 min / 1 min) value of individual phases (with other two phases grounded) is more than 2, the PI value for three phases together with respect to ground is only 1.1. Can anyone explain how this could happen ? ( All values were taken at the same temperature and with the same megger at the same voltage).
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Generator PI value problem
- Thread starter edison123
- Start date
hi, edison123.
i would like to clarify this;
1.) WAS THE TEST CONDUCTED RIGHT AFTER IT WAS DRIED OUT?
2.) HAD YOU MONITORED THE INSULATION RESISTANCE STATUS WHILE IT WAS IN THE OVEN OR IN THE PROCESS OF DRYING?
IF SO, DO YOU GRAPH THE TREND?
3.)ALL WERE TAKEN IN THE SAME TEMPERATURE, HAD YOU CONSIDER THE RELATIVE HUMIDITY FROM SINGLE PHASE TESTING UNTIL THE OVERALL TEST?
4.) YOU MAY CHECK YOUR TEST EQUIPMENT.
REGARDS,
FBJAVIER
i would like to clarify this;
1.) WAS THE TEST CONDUCTED RIGHT AFTER IT WAS DRIED OUT?
2.) HAD YOU MONITORED THE INSULATION RESISTANCE STATUS WHILE IT WAS IN THE OVEN OR IN THE PROCESS OF DRYING?
IF SO, DO YOU GRAPH THE TREND?
3.)ALL WERE TAKEN IN THE SAME TEMPERATURE, HAD YOU CONSIDER THE RELATIVE HUMIDITY FROM SINGLE PHASE TESTING UNTIL THE OVERALL TEST?
4.) YOU MAY CHECK YOUR TEST EQUIPMENT.
REGARDS,
FBJAVIER
electricpete
Electrical
Tough question. I can't think of any good explanation off-hand.
Were the windings discharged adequately between test? (4x length of time that voltage was applied)
Was the test repeated for double-check with same results?
Were the windings discharged adequately between test? (4x length of time that voltage was applied)
Was the test repeated for double-check with same results?
A problem I find with a standard PI test is that it looks at two values in time, but ignores activity at all other points in time.
I have had several motors pass the PI test that were blowing fuses. However, these motors showed periodic discharges when the PI test results are continuously plotted. The periodic discharges resulted in often dramatically reduced megohm values, which would then climb to acceptable levels as the PI test continued.
Perhaps your motor is similar--you have periodic discharges that happened to be caught during some tests and not caught during others.
I have had several motors pass the PI test that were blowing fuses. However, these motors showed periodic discharges when the PI test results are continuously plotted. The periodic discharges resulted in often dramatically reduced megohm values, which would then climb to acceptable levels as the PI test continued.
Perhaps your motor is similar--you have periodic discharges that happened to be caught during some tests and not caught during others.
xabproject
Electrical
dear edison123,
That seems okay. When you measure the three phases together you have larger mass of insulation and the winding to ground capacitances add up (something like, parallel capacitances = series resistances) to give you a bigger capacitor for charging. The larger and longer duration displacement currents give you lower PI if measured on the same basis as single winding measurements.
This is true only if there has been no other mistake like, misconducted test, undischarged windings, incorrect connections, timing errors, instrument errors etc.
If you have closed the thread, please keep us posted on how you concluded your observations.
Did you conduct the PI test again on single winding basis?
Best regards,
That seems okay. When you measure the three phases together you have larger mass of insulation and the winding to ground capacitances add up (something like, parallel capacitances = series resistances) to give you a bigger capacitor for charging. The larger and longer duration displacement currents give you lower PI if measured on the same basis as single winding measurements.
This is true only if there has been no other mistake like, misconducted test, undischarged windings, incorrect connections, timing errors, instrument errors etc.
If you have closed the thread, please keep us posted on how you concluded your observations.
Did you conduct the PI test again on single winding basis?
Best regards,
The insulation resistance for the whole winding (three phases together) will result in lower Resistance as compared to a single phase, but the PI should be around the same figure as for individual phase test. The PI is a measure of the insulation polarization. You should review the Megohms readings of each test.
For the performance described it seems that you did not discharge properly your winding after the first test. (you started the final test with a partially polarized insulation).
If that is not the case then the stator got humidity which will reflect too as much reduced insulation resistance and up and downs in the Megohms readings during the 10 min test.
For the performance described it seems that you did not discharge properly your winding after the first test. (you started the final test with a partially polarized insulation).
If that is not the case then the stator got humidity which will reflect too as much reduced insulation resistance and up and downs in the Megohms readings during the 10 min test.
jbartos.
Capacitive current will decay in less than 1 minute, Absorption or Polarization current slope is mainly proportional to the PI. Consult IEEE Std 43-2000 part 5.
Remember that each phase individually tested had PI>2; which means the winding is not deteriorated ( chemically affected after washed and dried out ???)
Capacitive current will decay in less than 1 minute, Absorption or Polarization current slope is mainly proportional to the PI. Consult IEEE Std 43-2000 part 5.
Remember that each phase individually tested had PI>2; which means the winding is not deteriorated ( chemically affected after washed and dried out ???)
- Thread starter
- #11
To answer all the above questions :
1. All the PI values were taken at room temperature after thorough overhaul and dry out. (PI, in any case, is supposed to be temperature independent.)
2. Sufficient time gap was maintained between each phase test (remember, when one phase is tested, other two phases are grounded for at least ten minutes i.e. the time taken to measure the PI of the first phase + a discharge time of 10 minutes). After testing the individual phases, all the 3 phases were grounded for over 15 minutes, before measuring PI of the 3 phases together.
3. To reconfirm our test results, we did repeat the PI tests the next day with the same results (i.e, individual phases having a PI of over 2 and 3 phases together having only 1.1)
4. Since we had to ac hipot the windings at 16.5 KV (1.5 times the rated voltage), I took the above PI values. With the PI test results being a mystery, I chanced my arm and went ahead with successfully hipotting the individual phases at 16.5 KV with the same leakage current for all the 3 phases. The generator is online now.
I still haven't a clue to explain the above PI values, which I frankly admitted to my client.
Thank you guys for keeping this thread still alive and look forward to some satisfactory explanation.
1. All the PI values were taken at room temperature after thorough overhaul and dry out. (PI, in any case, is supposed to be temperature independent.)
2. Sufficient time gap was maintained between each phase test (remember, when one phase is tested, other two phases are grounded for at least ten minutes i.e. the time taken to measure the PI of the first phase + a discharge time of 10 minutes). After testing the individual phases, all the 3 phases were grounded for over 15 minutes, before measuring PI of the 3 phases together.
3. To reconfirm our test results, we did repeat the PI tests the next day with the same results (i.e, individual phases having a PI of over 2 and 3 phases together having only 1.1)
4. Since we had to ac hipot the windings at 16.5 KV (1.5 times the rated voltage), I took the above PI values. With the PI test results being a mystery, I chanced my arm and went ahead with successfully hipotting the individual phases at 16.5 KV with the same leakage current for all the 3 phases. The generator is online now.
I still haven't a clue to explain the above PI values, which I frankly admitted to my client.
Thank you guys for keeping this thread still alive and look forward to some satisfactory explanation.
- Thread starter
- #14
kantor & aolalde,
Here is the info on PI (10 min / 1 min IR):
Day 1
R10/R1 = 422/197 = 2.14 Y10/Y1 = 514/247 = 2.08 B10/B1 = 456/219 = 2.08
(R+Y+B)10/(R+Y+B)1 = 530/441 = 1.20
Day 2
R10/R1 = 443/207 = 2.14 Y10/Y1 = 506/237 = 2.13 B10/B1 = 469/221 = 2.12
(R+Y+B)10/(R+Y+B)1 = 546/452 = 1.20
Hope this helps.
Here is the info on PI (10 min / 1 min IR):
Day 1
R10/R1 = 422/197 = 2.14 Y10/Y1 = 514/247 = 2.08 B10/B1 = 456/219 = 2.08
(R+Y+B)10/(R+Y+B)1 = 530/441 = 1.20
Day 2
R10/R1 = 443/207 = 2.14 Y10/Y1 = 506/237 = 2.13 B10/B1 = 469/221 = 2.12
(R+Y+B)10/(R+Y+B)1 = 546/452 = 1.20
Hope this helps.
electricpete
Electrical
OK, just thinking out loud...
The theory as I have heard (you have probably heard as well), there are 3 components of current:
Ic = capacitive charging current
Ia = absorption current
Ir = resistive leakage current
Ic is supposed to die out before the 1 minute reading.
Ia is supposed to die out before the 10 minute reading.
Ir is supposed to be constant
PI = R10/R1 = I1/I10 = (Ia+Ir)/Ir
The individual phase measurements give the sum of phase-ground (Ipg) and phase-phase currents (Ipp).
The combined measurement gives only phase-to-ground (Ipg). Assume it is roughly 3* the individual.
The combined phase-to-ground PI measurement tells us
(3*Ia_pg+3*Ir_pg) / (3*Ir_pg) ~ 1
(Ia_pg+Ir_pg) / (Ir_pg) ~ 1
The individual measurement tells us
(Ia_pg+ Ir_pg + Ia_pp+Ir_pp) / (Ir_pp + Ir_pp) ~ 3
The logical conclusion is that we have a large phase-to-phase absorption current which gives the high (normal) PI phase-to-phase and almost zero phase-to-ground absorbption current which gives PI~1 during combined test. This seems almost backwards of what one would expect based on the fact that absorption current arises in the bulk insulation while resistive may be bulk or surface. Clearly the slot section has bulk insulation under stress which should give absorption current phase to ground. I am not sure what conclusion to draw.
The theory as I have heard (you have probably heard as well), there are 3 components of current:
Ic = capacitive charging current
Ia = absorption current
Ir = resistive leakage current
Ic is supposed to die out before the 1 minute reading.
Ia is supposed to die out before the 10 minute reading.
Ir is supposed to be constant
PI = R10/R1 = I1/I10 = (Ia+Ir)/Ir
The individual phase measurements give the sum of phase-ground (Ipg) and phase-phase currents (Ipp).
The combined measurement gives only phase-to-ground (Ipg). Assume it is roughly 3* the individual.
The combined phase-to-ground PI measurement tells us
(3*Ia_pg+3*Ir_pg) / (3*Ir_pg) ~ 1
(Ia_pg+Ir_pg) / (Ir_pg) ~ 1
The individual measurement tells us
(Ia_pg+ Ir_pg + Ia_pp+Ir_pp) / (Ir_pp + Ir_pp) ~ 3
The logical conclusion is that we have a large phase-to-phase absorption current which gives the high (normal) PI phase-to-phase and almost zero phase-to-ground absorbption current which gives PI~1 during combined test. This seems almost backwards of what one would expect based on the fact that absorption current arises in the bulk insulation while resistive may be bulk or surface. Clearly the slot section has bulk insulation under stress which should give absorption current phase to ground. I am not sure what conclusion to draw.
Hi
I like to contribute to this discussion.
Doble testing guide quotes referring to the AIEE No.62(old)as Ia, the absorbtion current as
Ia = KCV t^(-n) where C is the Capacitance, V the voltage, K a constant and n being between 0.5 to 1.0.
The new IEEE 43-2000 states Ia being equal to
Ia = K t^(-n)
K = function of the particular insulation system and the applied voltage.
Looking at the previous equation let us take the Capacitance:
All three phases joined together to earth gives the total Capacitance as 3 * C where C is the Capacitance of a single phase to earth. Reason is that the individual phase winding Capacitance forms a parallel configuration.
When you test a single phase to earth with the other two phases bonded to earth gives you a series configuration.
Now PI = One minute current / ten minute current.
This now gives you a picture of the one minute current values for the two configuration you were referring. Note the absorption current for the two configuration will be different for the one minute reading.
How to confirm:
When doing a power factor test carry out the Capacitance value test for your two configuration.
When doing a PI test record the current every minute plus the curent at 3.16 minute. Draw a graph, calculate the value of K and do an analysis.
Refer IEEE 43-2000.
My view is that PI is not merely the
10minute IR / 1 minute IR.
I hope someone agrees with me.
Thanks
Kantor
I like to contribute to this discussion.
Doble testing guide quotes referring to the AIEE No.62(old)as Ia, the absorbtion current as
Ia = KCV t^(-n) where C is the Capacitance, V the voltage, K a constant and n being between 0.5 to 1.0.
The new IEEE 43-2000 states Ia being equal to
Ia = K t^(-n)
K = function of the particular insulation system and the applied voltage.
Looking at the previous equation let us take the Capacitance:
All three phases joined together to earth gives the total Capacitance as 3 * C where C is the Capacitance of a single phase to earth. Reason is that the individual phase winding Capacitance forms a parallel configuration.
When you test a single phase to earth with the other two phases bonded to earth gives you a series configuration.
Now PI = One minute current / ten minute current.
This now gives you a picture of the one minute current values for the two configuration you were referring. Note the absorption current for the two configuration will be different for the one minute reading.
How to confirm:
When doing a power factor test carry out the Capacitance value test for your two configuration.
When doing a PI test record the current every minute plus the curent at 3.16 minute. Draw a graph, calculate the value of K and do an analysis.
Refer IEEE 43-2000.
My view is that PI is not merely the
10minute IR / 1 minute IR.
I hope someone agrees with me.
Thanks
Kantor
- Thread starter
- #17
pete,
A small correction in your second equation
"The individual measurement tells us
(Ia_pg+ Ir_pg + Ia_pp+Ir_pp)/(Ir_pp + Ir_pp)" should be
(Ia_pg+ Ir_pg + Ia_pp+Ir_pp)/(Ir_pg + Ir_pp)
= 1 + {(Ia_pg + Ia_pp)/(Ir_pp + Ir_pg)}
Normally, phase-phase IR is about two times phase-ground IR (twice insulation thickness et.al.), so that we can assume Ia_pp = 0.5*Ia_pg and Ir_pp = 0.5*Ir_pg.
Substituting these values in above equation we get
PI (Individual phase) = 1 + (1.5*Ia_pg/1.5*Ir_pg)= 1 + (Ia_pg/Ir_pg)
PI (Combined) = (Ia_pg+Ir_pg)/(Ir_pg) = 1 + (Ia_pg/Ir_pg)= PI (Individual phase).
Well, that was interesting discussion. Bit unfortunate it still does not solve the mystery.
Kumar
kantor,
IEEE 43-2000 does state the PI is normally defined as (10 min IR/1 min IR). See definitions 3.6.
It also says that with modern insulation systems in form wound coils, PI can be defined as (60 secs IR/30 secs IR) or (5 min IR/1 min IR) but the values of acceptable PI in these cases is not established (refer annex A - Variants in PI). Given this, traditional PI definition and acceptance seems safe at the moment.
Thx for all the people who posted here and made this thread interesting. I hope this thread stays alive with lively posts.
A small correction in your second equation
"The individual measurement tells us
(Ia_pg+ Ir_pg + Ia_pp+Ir_pp)/(Ir_pp + Ir_pp)" should be
(Ia_pg+ Ir_pg + Ia_pp+Ir_pp)/(Ir_pg + Ir_pp)
= 1 + {(Ia_pg + Ia_pp)/(Ir_pp + Ir_pg)}
Normally, phase-phase IR is about two times phase-ground IR (twice insulation thickness et.al.), so that we can assume Ia_pp = 0.5*Ia_pg and Ir_pp = 0.5*Ir_pg.
Substituting these values in above equation we get
PI (Individual phase) = 1 + (1.5*Ia_pg/1.5*Ir_pg)= 1 + (Ia_pg/Ir_pg)
PI (Combined) = (Ia_pg+Ir_pg)/(Ir_pg) = 1 + (Ia_pg/Ir_pg)= PI (Individual phase).
Well, that was interesting discussion. Bit unfortunate it still does not solve the mystery.
Kumar
kantor,
IEEE 43-2000 does state the PI is normally defined as (10 min IR/1 min IR). See definitions 3.6.
It also says that with modern insulation systems in form wound coils, PI can be defined as (60 secs IR/30 secs IR) or (5 min IR/1 min IR) but the values of acceptable PI in these cases is not established (refer annex A - Variants in PI). Given this, traditional PI definition and acceptance seems safe at the moment.
Thx for all the people who posted here and made this thread interesting. I hope this thread stays alive with lively posts.
Kumar (edison123)
I agree with all the IEEE 43-2000. I do not differ in your opinion anyway. However, a good value is worked out for the different classes of insulation by IEEE. Absorption current plays an important role as the time for decay of the current is important. Therefore PI depends on the time taken for the absorption current to decay and become steady (approximately)and this is different to types of insulation.
My answer is to your question and not on IEEE 43-2000. Your question of why the PI is different to the two configuration. Sketch a diagram and you could anaylse the case.
Thanks anyway.
I agree with all the IEEE 43-2000. I do not differ in your opinion anyway. However, a good value is worked out for the different classes of insulation by IEEE. Absorption current plays an important role as the time for decay of the current is important. Therefore PI depends on the time taken for the absorption current to decay and become steady (approximately)and this is different to types of insulation.
My answer is to your question and not on IEEE 43-2000. Your question of why the PI is different to the two configuration. Sketch a diagram and you could anaylse the case.
Thanks anyway.
edison123
Just to add.
During an insulation test the following components of currents are monitored.
1. Absorption current -Ia
2. Leakage current -Ii which is constant
3. Capacitive current -Ic which decays within a minute
4. Conduction current -Ig which is constant
Therefore the PI is dependent on the Absorption current. The formula gives the response of the Absorption current to an applied voltage which is dependent on the Capacitance and the constant K.
I think I can accept this fact.
Thanks
Just to add.
During an insulation test the following components of currents are monitored.
1. Absorption current -Ia
2. Leakage current -Ii which is constant
3. Capacitive current -Ic which decays within a minute
4. Conduction current -Ig which is constant
Therefore the PI is dependent on the Absorption current. The formula gives the response of the Absorption current to an applied voltage which is dependent on the Capacitance and the constant K.
I think I can accept this fact.
Thanks
jbartos
Electrical
- Jan 15, 2000
- 6,440
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When all phases are tested simultaneously, only the insulation to ground is tested. Insulation resistance measurements should be made with all external equipment (cables, capacitors, surge arresters, etc.) disconnected and grounded as these items may influence the resistance reading.
etc. for more info
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When all phases are tested simultaneously, only the insulation to ground is tested. Insulation resistance measurements should be made with all external equipment (cables, capacitors, surge arresters, etc.) disconnected and grounded as these items may influence the resistance reading.
etc. for more info
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