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I would like to share some of the tips which can help in ascertaining failure of a distribution/power transformer and also in repair of the same.  SPM/TRE Engineers in particular must have these on finger tips.

All most all the three phase distribution/power transformers we have in the system are delta/star connected with capacity ranging from 16 KVA to 12500 KVA.  We have to conduct many tests to establish failure of a transformer but we have to have alternate three phase/single phase supply which may not be available at the failure locations. At such instance a simple tong tester and insulation tester will be sufficient and serves the purpose in more than 90% of the cases

1) Tests to be conducted:
a) Continuity test:
Tests the continuity of the winding (Healthy transformer must have continuity in all the phases but transformer with shorted windings can also have).

Testing terminals:
HV           LV
RY           r n
YB           y n
BR           b n

b) IR values:
Tests insulation resistance between windings and winding to earth. A healthy transformer should have resistance >0 in Mega Ohms.

If   HV—E=0 (HV winding is earthed)
LV—E=0 (LV winding is earthed)
HV—LV=0 (HV and LV winding are shorted)

Any of the HV three terminals and any of the LV four terminals can be used for conducting the tests (If continuity in winding is available).

We should never test charge a DTR/PTR if the IR values are 0 in any of the above three parameters.

c) DC resistance test
A healthy transformer will have approximately same resistance in all the three phase in HV winding though its value changes with the capacity of the transformer.  LV winding resistance cannot be measured with a tong tester since it has less number of turns and higher size conductor when compared to the HV winding.  Continuity of the LV winding can only be established.

For example:
a) If the resistance of coils in one phase is 30 Ohms then for a healthy transformer resistance between RY, YB & BR terminals will be equal and 20 Ohms.

i.e. 1/R=1/30+1/60=3/60=1/20

i.e. R=20 Ohms

Diagram Delta

b)  If winding is opened/cut in one phase.
RY=60, YB=30, BR=30   (R winding is open)
YB=60, BR=30, RY=30   (Y winding is open)
BR=60, RY=30, YB=30   (B winding is open)

i.e. resistance measured is double across one winding in case the winding is open and other two winding are healthy.

c) In case two windings are open and one winding healthy.
RY=30, YB=infinity, BR=infinity (Y&B windings open)
YB=30, BR=infinity, RY=infinity (B&R windings open)
BR=30, RY=infinity, YB=infinity (R&Y windings open)

From the above it is understood that:

  • When resistance values are equal in all the HV phases the PTR/DTR may be OK.
  • If resistance across one winding is double when compared to other individual winding it is clear that winding is open in one phase and PTR/DTR can be declared failed without going for other tests.
  •  If one winding is showing resistance and others are showing open (infinity) it can be suspected that two phases are open and one phase is healthy.

Note:  But it should be kept in mind that even opening of a jumper wire leading to HV bush rod may give similar value and can result in minor failure instead of two limb failure.  So verification of jumper connection is to be done before declaration of a failure.

The failure in windings causes erosion of winding material which results in collection of gas in bucholz relay and transformer tank.

If the DTR/PTR is still found healthy in above tests it can be subjected for further test like 1. ratio tests, 2. short circuit test, 3. magnetic balance test/arranging LT phase test supply to take final decision for declaration of failure/healthiness.

Design aspects in repairing of DTR/PTR
LV & HV windings of a power transformer and LV winding of a failed distribution transformer will be generally reinsulated during repairs.  Conductor to the extent of damaged portion will be replaced for PTRs and damaged HV winding coils will be replaced for distribution transformers.

For rewinding of coils after re-insulation the winder generally takes into consideration the dia-metre and height of the damaged coil.  But the winder may endup with a coil with less/excess number of turns and slight variation in height and dia-metre.  Hence the AE/Winder should have the knowledge of minimum size of conductor, number of turns, number of conductors any capacity distribution/power transformer should have to deliver the rated output.

a) Number of turns in LV winding:
ET =K√Q
ET =Voltage per turn that can be allowed
Q=Rated KVA
K=Constant (0.32 to 0.35 for aluminum transformer and 0.37 to 0.45 for copper transformer)

Example:
For 100 KVA aluminum DTR
ET =0.32 √100=0.32 volts
LV turns =250V / 3.2V=78 turns (minimum)

Hence 100 KVA aluminum distribution transformer should have a minimum of 78 turns to have voltage/turn within limits.

Similarly,
For 25 KVA aluminum DTR
ET =0.32 √25=1.6

Hence LV turns=250V/1.6V=156 turns 

It means lesser capacity DTR/PTR will have more number of turns when compared to higher capacity transformer

Number of turns in HV winding can be easily arrived by formula.
Number of turns in HV winding = voltage ratio X Number of turns in LV winding.
                                                = 11000/250V X LV turns
= 44 X LV turns

b) Size of the conductor:
The size of the conductor depends on the current density of the material used for winding.  Current density that can be allowed is.

Aluminum       = 1.5Amps/sqmm
Copper           = 3.0Amps/sqmm

Hence cross sections of LV conductor can be arrived at follows:

For 100 KVA Alu. DTR LV cross section  =  Rated phase current/current density
=  133.3/1.5Amps=88.8 sqmm
Cross section for HV conductor             =  Rated phase current/current density
=  3.03/1.5Amps=2.02 sqmm

The above values are minimum size of cross section the conductor that can be utilized for replacement of damaged conductor.  However higher size can be utilized provided minimum clearances are maintained between windings and core.

c) Shape of conductor:
If the dia-metre of the conductor comes more than 3.5mm dia, rectangular shaped strip to be used since strip has better surface length and more space factor than that of a round conductor.

Hence for distribution transformer round conductor is used for HV windings and rectangular strip is used for LV windings.

Example:
For 100KVA HV winding wire, cross section of the conductor required is 2.02 sqmm
i.e. ∏r2=2.02

r=√2.02/∏
r=0.80
diameter=1.6

The dia-metre of the HV conductor being used for a 100 KVA is 1.6mm i.e. 16 SWG.

Selection of size of rectangular strip of equal cross section is more importance since we can have many sizes for the same cross section.

In general for all practical purposes for easy handling.

  • Width of the strip is to be more than double the depth of the strip.
  • Minimum depth of the strip is to be 2.25mm

From the above it is evident that more number of conductor in parallel can be utilized instead of higher sized single connector.  Utilization of multiple conductors results in more flexibility during winding and also decreases skin-effect since surface length of conductor is more with multiple conductors.

Example:
For 100KVA aluminum transformer LV conductor of 88 sqmm cross section can be had by having two conductors of size 44 sqmm each in parallel.  Depth of strip should be minimum a 2.25mm and width can be calculated to accommodate required number of the LV turns within the height of LV coils in single layer or double layer.

d) Diameter of the core:
Dia-metre of the core can be calculated by arriving at gross core area by using following formula:

Ag=Et/4.44 X f X Bm X 0.97 X 10-4
f=frequency
Ag=gross core area in sqcm
Bm=maximum flex density in tesla = 1.6
Et=phase voltage/turn
0.97=stacking factor (assumed)
Once Ag is calculated dia-metre of the core can be calculated by formula
Ag=K1 X ∏dX ¼
K1=0.92 for six step core
K1=0.925 for seven step core
K1=0.93 for eight step core
K1=0.935 for nine step core etc

By Er. K Sadasiva Reddy
SE/Operation/Mahaboobnagar

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