Download “MINIMUM ENERGY” AS IT RELATES TO IG STAGE CONNECTIONS

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Opto-isolator wikipedia , lookup

Voltage optimisation wikipedia , lookup

Distributed generation wikipedia , lookup

Life-cycle greenhouse-gas emissions of energy sources wikipedia , lookup

Transcript
“MINIMUM ENERGY” AS IT RELATES TO IG STAGE CONNECTIONS
Reference:
Proposed minimum IG energy- P. Riffon
The “Minimum Energy” concept being discussed in the “WG Revision to Impulse Tests” (that
will meet at 3:15 p.m. on Tuesday, March 18th ), relates the kVA rating of a transformer to a
certain “minimum available” IG energy requirement for impulse testing. THE ENERGY BEING
REFERRED TO IN THIS DISCUSSION IS THE ENERGY AVAILABLE AT THE
CHARGING VOLTAGE LEVEL REQUIRED TO OBTAIN THE TEST BIL.
To illustrate how this impacts IG stage requirements consider the following example:
Based on an IG voltage efficiency of 80%, a 95 kV BIL test would require a total charging
voltage of [95kV/0.8] or 118.75 kV. This would require a 100 kV per stage IG to have two series
stages charged to 59.375 kV per stage, or one 200 kV per stage IG charged to the 118.75 kV
level. For both cases the total charging capability of the IG connection being used would be 200
kV. To obtain an available energy of 10 kJ at 118.75 kV an IG with a rated or “nameplate”
energy of at least 28.4 kJ would be required.
If the 200 kV per stage IG had a rated energy of 10 kJ per stage, 3 stages in parallel
would be required to meet the “available’ energy requirement.
If the 100 kV per stage IG had a rated energy of 5 kJ per stage a total of 6 stages would
be required, 2 series stages each of 3 stages in parallel.
If the 100 kV per stage IG had a rated energy of 2.5 kJ per stage a total of 12 stages
would be required, 2 series stages each of 6 stages in parallel.
To determine the number of IG stages required to achieve a given minimum energy, first
determine the number of series stages that will be required to obtain the total charging voltage
for the BIL, from
Series_Stages = BIL/( * Vrated)
Then, knowing the number of series stages required, obtain the total stages from:
Total_Stages = Emin/Erated * (Vrated * Series_Stages *  /BIL)^2
Where:
Total_Stages
Emin
Erated
Vrated
Series_Stages

BIL
= Required number of IG stages, rounding up to whole stage numbers.
= Available Energy
= Rated or nameplate energy of an IG stage
= Rated voltage of an IG stage
= Number of series IG stages required to obtain total charging voltage
= factor for voltage efficiency (i.e. 80% is 0.8)
= Basic Insulation Level
1
The following conditions for the use of alternative methods are also included in the above
referenced “Minimum Energy” proposal
The formula included in the “Minimum Energy” proposal yields the energy level that would be
required to obtain a 40 s wavetail for various transformer ratings. For cases where the
calculated value is less than the value obtained from the table, the test equipment would be
required to have at least the calculated value available for the tests. If the calculated value is
higher than the “table value”, than the test equipment requirement is the “table value” for the
tests.
For the special case of transformers with a BIL level of 75 kV or lower the “Minimum Energy”
proposal allows the output capacitance of the IG to be limited to a minimum value of 4
microfarad. The output capacitance being the “net” capacitance of the series/parallel stage
combination being used.
The use of alternative methods of extending the wavetail i.e loading resistors etc, will only be
allowed for cases where the minimum available energy, as mentioned above, is available from
the test equipment and that energy does not yield the required minimum wavetail.
Tests with a wavetail less than 40 s will only be allowed when, at least the minimum energy is
used and the addition of loading resistors to the non impulsed terminals does not produce the
minimum tail time.
Addendum
Note, for those of you more used to thinking in terms of IG capacitance rather than IG energy the
following may be of interest.
Since Emin is the energy stored by the minimum IG capacitance at the required charging voltage
(that is, the BIL/), if we substitute [Cmin/2*(BIL/)^2] for “Emin” in the minimum energy
equation and then simplify, the result is:
Cmin = 2 * [2 *  * f * (t2)^2 * VA/(z * U^2)]
where Cmin is the equivalent series capacitance of the IG connection required to produce the
minimum tail on the terminal being tested, in farads All other parameters are as defined in the
Minimum Energy Proposal
A.Molden 3/24/03
2