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Transformer Riddle No.66 - Impedance of a transformer
In order to increase the impedance to tertiary winding in a single phase autotransformer tertiary winding is placed on outer limb of a 3 limbed single phase or a 4 limbed single phase configuration.

Regulating winding is also placed on outer limb to reduce transfer surges to tapping winding from reliability point of view. This arrangement requires a compensating winding on a main limb feeding the regulating winding on outer limb so that distribution of 50% flux to outer limb is ensured.

I shall be thankful if somebody can help me to decide

i) current rating of compensating winding

ii) calculate % impedance from HV to tertiary and LV to tertiary by classical formula.


Regards
Sreelatha
Author : Sreelatha - From: India
 
#1
Mon, May 9th, 2011 - 19:02
1-For the purpose of illustration it is possible to put some typical values on the booster arrangement of Figure below. This might be required to provide a phase shifting capability of 18° on a 400 kV transmission line having a current rating of 1200 A. The tangent of 18° is 0.325 so that the series winding must produce a maximum quadrature output of 0.325 x 400/√3 = 75 kV. With a rated current of 1200 A, the series unit will have a three-phase rating of 3 x 75 x 1200 =270 MVA. (Note that this unit has its output winding in open delta so that 75 kV and 1200 A are the respective phase voltage and current. The three phase unit thus has a rating of 3 times the product of these phase quantities, and no √3 factor is involved.) A typical current rating of tapchanger which might be considered for such a unit is 1600 A. If maximum use is to be made of tapchanger current capability, the open-delta/delta series transformer must have a voltage ratio such as to produce this line current from the delta winding, that is the delta winding must have a phase current of 1600/√3 = 924 A and hence a phase voltage of 97.4 kV. Connecting the regulating windings in star means that each phase must have a total all-taps-in voltage of 97.4/√3 =56.2 kV. This represents the maximum voltage across the range for the tapchanger and based on an 18 step, 19 position, tapchanger is equivalent to 3120 V per step. These values are just about on the limit of the capability of a commercial 1600 A tapchanger. The rating of the shunt transformer will be 3 x 56.2 x 1600 = 270 MVA, and the combined rating of the complete unit will be 540 MVA. At full output for the series winding the shunt winding will draw a current (270/√3 x 106)/400 000 =390 A from the 400 kV system.



2- As you know the leakage reactance of a transformer arises from the fact that all the flux produced by one winding does not link the other winding. As would be expected, then, the magnitude of this leakage flux is a function of the geometry and construction of the transformer. Figure below shows a part section of a core-type transformer taken axially through the centre of the wound limb and cutting the primary and secondary windings. The principal dimensions are marked in the figure, as follows:

l is axial length of windings (assumed the same for primary and secondary)
a is the radial spacing between windings
b is the radial depth of the winding next to the core
c is the radial depth of the outer winding



If mlt is then the mean length of turn of the winding indicated by the appropriate subscript, mltb for the inner winding, mltc for the outer winding and mlta for a hypothetical winding occupying the space between inner and outer windings, then the leakage reactance in per cent is given by the expression:

%X = KF(3a.mlta _ b.mltb _ c.mltc)/Φml

Where:
K is a constant of value dependent on the system of units used
F is equal to the ampere-turns of primary or secondary winding, that is m.m.f. per limb
Φm is the maximum value of the total flux in the core

The above equation assumes that both LV and HV windings are the same length, which is rarely the case in practice. It is also possible that a tapped winding may have an axial gap when some of the tappings are not in circuit. It is usual therefore to apply various correction factors to l to take account of these practical aspects. However, these corrections do not change the basic form of the equation. 
Author : Hamid - From: Iran
 
#2
Tue, May 10th, 2011 - 08:36
Dear Sir,

Thank you very much for your reply. You have given the impedance calculation for normal arrangement. But I wanted the calculation of impedance between windings placed on different limbs.

Kindly help me in this regard.

Thanks & Regards
Sreelatha 
Author : Sreelatha - From: India
 
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