Unit 3 Lecture 4 Generation of High AC voltages & Resonant Transformers.pdf

Notes

1. HIGH
VOLTAGE
ENGINEERING
MANIKANDAN.P –CUFE
UNIT 3
GENERATION OF HIGH VOLTAGES AND CURRENTS

2. When test voltage requirements are less than about 300 kV, a single transformer can be used for test purposes. The impedance
of the transformer should be generally less than 5% and must be capable of giving the short circuit current for one minute or
more depending on the design.
GENERATION OF HIGH ALTERNATING VOLTAGES
2

3. Firsttransformerisatthegroundpotentialalongwithitstank.Thesecond
transformeriskeptoninsulatorsandmaintainedatapotentialof??????
2
Thehighvoltagewindingofthefirstunitisconnectedtothetankofthesecondunit.The
lowvoltagewindingofthisunitissuppliedfromtheexcitationwindingofthefirst
transformerwhichisinserieswiththehighvoltagewindingofthefirsttransformeratits
highvoltageend
Thehighvoltageconnectionfromthefirst
transformerwindingandtheexcitation
windingterminalaretakenthroughabushing
tothesecondtransformer.
3

4. In a similar manner, the third transformer is kept on insulators above the ground
at a potential of 2??????
2and is supplied likewise from the second transformer.
The number of stages in this type of arrangement are usually two to four, but very
often, three stages are adopted to facilitate a three-phase operation so that
3??????
2can be obtained between the lines
The rating of the primary or the low voltage winding is usually 230 or 400 V for
small units up to 100 kVA. For larger outputs the rating of the low voltage winding
may be 3.3kV, 6.6 kV or 11 kV.
4

5. The advantage of this scheme is that the natural cooling is sufficient and the transformers are light and
compact. Transportation and assembly is easy.
Also the construction is identical for isolating transformers and the high voltage cascade units. Three
phase connection in delta or star is possible for three units.
Testing transformers of ratings up to 10 MVA in cascade connection to give high voltages up to 2.25 MV
are available for both indoor and outdoor applications
�=�.??????
2
??????�
&#3627408451;&#3627408476;&#3627408484;&#3627408466;&#3627408479;&#3627408455;&#3627408479;&#3627408462;&#3627408475;&#3627408480;&#3627408467;&#3627408476;&#3627408479;&#3627408474;&#3627408466;&#3627408479;&#3627408480;&#3627408479;&#3627408462;&#3627408481;??????&#3627408475;&#3627408468;<1&#3627408448;????????????=1000&#3627408477;&#3627408441;
&#3627408451;&#3627408476;&#3627408484;&#3627408466;&#3627408479;&#3627408455;&#3627408479;&#3627408462;&#3627408475;&#3627408480;&#3627408467;&#3627408476;&#3627408479;&#3627408474;&#3627408466;&#3627408479;&#3627408480;&#3627408479;&#3627408462;&#3627408481;??????&#3627408475;&#3627408468;>1&#3627408448;????????????=1000−10&#3627408446;&#3627408477;&#3627408441;
&#3627408443;??????&#3627408451;&#3627408476;&#3627408484;&#3627408466;&#3627408479;&#3627408464;&#3627408462;&#3627408463;&#3627408473;&#3627408466;&#3627408480;250−300&#3627408477;&#3627408441;/&#3627408474;
&#3627408443;??????&#3627408451;&#3627408476;&#3627408484;&#3627408466;&#3627408479;&#3627408464;&#3627408462;&#3627408463;&#3627408473;&#3627408466;&#3627408480;50−80&#3627408477;&#3627408441;/&#3627408474;
5

6. 6

7. Resonant Transformers
ForAChighvoltagetestingdifferentkindsofequipmentareknownandavailable.
TheACtestinghereisbytheseriesresonanceprinciple.Thetuningoftheresonantcircuitwithitschangingtestcapacitances
howeverisnotachievedbyvaryingtheinductance,butvaryingthefrequency.
Thedescribedhighvoltagesourcesareavailableinavoltagerangeupto800kV,acurrentrangeupto200Aandafrequency
rangefrom30Hzto200Hz.
TheadvantageofvariablefrequencytestsystemscomparedtootherACtestsystems,suchastransformersandresonanttest
systemswithvariableinductance,istheiroptimumtestingpower/weightratio.Thisratioresultsinreducedweight,
dimensions,costandprice.
Singleunitresonanttest
systemsarebuiltforoutput
voltagesupto500kV
whilecascadedunitsfor
outputsupto3000kV
50/60Hzareavailable.
7

8. This kind of AC test equipment is especially suited for on site testing of capacitive test objects like
gas insulated bus bars, cables and GIS substations
The chief advantages of this principle are:
(a)it gives an output of pure sine wave (b) power requirements are less (5 to 10% of total kVA required)
(c) no high-power arcing and heavy current surges occur if the test object fails, as resonance ceases at the failure of the test
object
(d) cascading is also possible for very high voltages (e) simple and compact test arrangement
(f) no repeated flashovers occur in case of partial failures of the test object and insulation recovery.
8

9. A voltage regulator of either the auto-transformer type or the induction regulator type is connected to the
supply mains and the secondary winding of the exciter transformer is connected across the H.V.
reactor, L, and the capacitive load C.
In the parallel resonant mode the high voltage reactor is connected as an auto-transformer and the circuit is
connected as a parallel resonant circuit
The advantage of the parallel resonant circuit is that more stable output voltage can be obtained along with a
high rate of rise of test voltage, independent of the degree of tuning and the Q-factor
9

10. Therefore, the input voltage required for excitation is reduced by a factor 1/Q, and the output kVA required
is also reduced by a factor 1/Q. The secondary power factor of the circuit is unity.
&#3627408455;&#3627408440;&#3627408454;&#3627408455;&#3627408438;&#3627408450;&#3627408449;&#3627408439;&#3627408444;&#3627408455;&#3627408444;&#3627408450;&#3627408449;:??????&#3627408447;
??????+&#3627408447;=
1
??????&#3627408438;
The vertical scale B (flux density, Tesla) is proportional to the applied
voltage V
in
The horizontal scale H (magnetizing force) is proportional to current
Thesteepslope(highpermeability)isforacorewithoutanairgap
andthemoregradualslopeforthesamecorewithasmallairgap
Intersection with the non-gapped and gapped B/H loops for the
Square wave,
The margin betwn the (Bm) and the saturation value (Bsat) remains
the same with or without the air gap. Hence, if the applied voltage
were to be increased to the point of saturation, introducing an air
gap wouldn't help.
The Inductance of the reactor L
is varied by varying its air gap
10:1 and Q factor is around 50
10

11. A100kVA,400V/250kVtestingtransformerhas8%leakagereactanceand2%resistanceon100kVAbase.
Acablehastobetestedat500kVusingtheabovetransformerasaresonanttransformerat50Hz.Ifthe
chargingcurrentofthecableat500kVis0.4A,findtheseriesinductancerequired.Assume2%resistancefor
theinductortobeusedandtheconnectingleads.Neglectdielectriclossofthecable.Whatwillbetheinput
voltagetothetransformer?
&#3627408448;&#3627408462;&#3627408485;??????&#3627408474;&#3627408482;&#3627408474;&#3627408464;&#3627408482;&#3627408479;&#3627408479;&#3627408466;&#3627408475;&#3627408481;=
100∗10
3
250∗10
3
=0.4??????
??????
??????=&#3627408453;&#3627408466;&#3627408462;&#3627408464;&#3627408481;&#3627408462;&#3627408475;&#3627408464;&#3627408466;&#3627408476;&#3627408467;&#3627408481;ℎ&#3627408466;&#3627408464;&#3627408462;&#3627408463;&#3627408473;&#3627408466;
??????
??????=
??????
??????
&#3627408444;
=
500∗10
3
0.4
=1250&#3627408472;&#3627408476;ℎ&#3627408474;&#3627408480;
??????
??????=&#3627408447;&#3627408466;&#3627408462;&#3627408472;&#3627408462;&#3627408468;&#3627408466;&#3627408453;&#3627408466;&#3627408462;&#3627408464;&#3627408481;&#3627408462;&#3627408475;&#3627408464;&#3627408466;&#3627408476;&#3627408467;&#3627408481;ℎ&#3627408466;&#3627408481;&#3627408479;&#3627408462;&#3627408475;&#3627408480;&#3627408467;&#3627408476;&#3627408479;&#3627408474;&#3627408466;&#3627408479;
8%&#3627408447;&#3627408466;&#3627408462;&#3627408472;&#3627408462;&#3627408468;&#3627408466;&#3627408479;&#3627408466;&#3627408462;&#3627408464;&#3627408481;&#3627408462;&#3627408475;&#3627408464;&#3627408466;&#3627408476;&#3627408475;??????&#3627408481;&#3627408480;??????&#3627408474;&#3627408477;&#3627408466;&#3627408465;&#3627408462;&#3627408475;&#3627408464;&#3627408466;&#3627408483;&#3627408462;&#3627408473;&#3627408482;&#3627408466;=
??????
&#3627408444;
=
8
100
∗
250∗10
3
0.4
=50&#3627408472;&#3627408476;ℎ&#3627408474;&#3627408480;
11

12. ??????&#3627408481;&#3627408453;&#3627408466;&#3627408480;&#3627408476;&#3627408475;&#3627408462;&#3627408475;&#3627408464;&#3627408466;,&#3627408438;&#3627408453;=&#3627408444;&#3627408453;&#3627408454;&#3627408476;&#3627408462;&#3627408465;&#3627408465;??????&#3627408481;??????&#3627408476;&#3627408475;&#3627408462;&#3627408473;&#3627408479;&#3627408466;&#3627408462;&#3627408464;&#3627408481;&#3627408462;&#3627408475;&#3627408464;&#3627408466;&#3627408481;&#3627408476;&#3627408463;&#3627408466;&#3627408462;&#3627408465;&#3627408465;&#3627408466;&#3627408465;??????&#3627408480;1200&#3627408472;&#3627408476;ℎ&#3627408474;&#3627408480;
&#3627408444;&#3627408475;&#3627408465;&#3627408482;&#3627408464;&#3627408481;&#3627408462;&#3627408475;&#3627408464;&#3627408466;&#3627408476;&#3627408467;&#3627408462;&#3627408465;&#3627408465;??????&#3627408481;??????&#3627408476;&#3627408475;&#3627408462;&#3627408473;&#3627408479;&#3627408466;&#3627408462;&#3627408464;&#3627408481;&#3627408462;&#3627408475;&#3627408464;&#3627408466;&#3627408447;=
1200∗10
3
2∗50∗3.14
=3820&#3627408443;
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&#3627408476;ℎ&#3627408474;??????&#3627408464;&#3627408483;&#3627408462;&#3627408473;&#3627408482;&#3627408466;=
4
100
∗
250∗10
3
0.4
=25&#3627408472;&#3627408476;ℎ&#3627408474;&#3627408480;
&#3627408440;&#3627408485;&#3627408464;??????&#3627408481;&#3627408462;&#3627408481;??????&#3627408476;&#3627408475;&#3627408483;&#3627408476;&#3627408473;&#3627408481;&#3627408462;&#3627408468;&#3627408466;&#3627408440;
2&#3627408476;&#3627408475;&#3627408481;ℎ&#3627408466;&#3627408480;&#3627408466;&#3627408464;&#3627408476;&#3627408475;&#3627408465;&#3627408462;&#3627408479;&#3627408486;&#3627408480;??????&#3627408465;&#3627408466;=&#3627408444;∗&#3627408453;=0.4∗25∗10
3
=10&#3627408472;??????
&#3627408451;&#3627408479;??????&#3627408474;&#3627408462;&#3627408479;&#3627408486;&#3627408483;&#3627408476;&#3627408473;&#3627408481;&#3627408462;&#3627408468;&#3627408466;=
10∗10
3
∗400
250∗10
3
=16??????
12

13. A100kVA,400V/150kVtestingtransformerhas10%leakagereactanceand4%resistanceon100kVAbase.
Acablehastobetestedat400kVusingtheabovetransformerasaresonanttransformerat60Hz.Ifthe
chargingcurrentofthecableat500kVis0.5A,findtheseriesinductancerequired.Assume4%resistancefor
theinductortobeusedandtheconnectingleads.Neglectdielectriclossofthecable.Whatwillbetheinput
voltagetothetransformer?
Determinetheripplevoltageandregulationofa10stageCockcroftWaltontypedcvoltagemultipliercircuit
havingastagecapacitance=0.01microfaradandsupplyvoltage=100kVatafrequencyof400Hzandload
current=10mA
Avoltagedoublercircuithas&#3627408438;
1=&#3627408438;
2=0.01??????&#3627408441;anditissuppliedfromavoltagesourceof??????=
100&#3627408480;??????&#3627408475;314&#3627408481;&#3627408472;??????.Calculatetheoutputvoltageandtheripple
13