1. TRANSFORMER TECHNOLOGY
Prof. Chandrabhan Sharma
2. TRANSFORMER TECHNOLOGY
WINDINGS
Material → * mainly Cu
* Al also used
Insulation: Most common - Paper
- paper board (Press Board)
Some cases - Wood (laminated varnish impregnated),
- Synthetic enamel.
Fluid insulant provides two (2) functions:
- inter-part insulation
- removal of heat
3. - Oil mainly used
- Air/SF6 for dry type. (H2 also used in the past)
- Oil dielectric breakdown → 40 kV
test using 2 kV/s voltage between 13mm spheres
spaced 2.5mm apart.
4. DRY TYPE INSULATING MATERIALS
Note : Most Air transformer are type C
Conductors are usually Cu or Al
Form : Wire, strip, foil, sheets
Choice of form f(current)
5. WINDING TYPE
•Helical (high ‘I’; low ‘V’)
Construction •Crossover (low ‘I’; high ‘V’)
•Mulitlayer HV (V ≥ 132 kV)
•Continuous Disc (Arc Furnaces)
Note:
Because of poor space factor, wire only found in coils of distribution
and instrumentation transformers.
- No space problem
- Less wire needed
- More efficient
6. TAPPINGS
Depending on MVA rating, tap position can be selected so as to
offset out-of-balance conditions.
Losses:
- I2R
- Eddy current
- Losses in clamps (leakage flux)
- Tank Losses (due to tank being magnetic)
7. TRANSFORMER CORES
Materials : Iron-silicon alloys used for power transformers.
Produced by cold rolling process which causes crystals to
align along the direction of rolling.
a - silectron
b - unisil
8. Insulating Materials:
Magnesia is added to the core surface during manufacture.
Then given a coat of acid phosphates prior to annealing.
This is sufficient for cores up to 30 MVA.
Additional insulation can be:
- Paper
- Mixtures of china, clay and flour
- Varnishes
9. FORM
(1)Cross-section:
Windings for core are cylindrical in shape → circular fill.
⇒ core section should be circular
(2)Pattern:
10.
11. Clamping (of Laminations) :
Disadvantages:
- Bolts are potential source of core faults.
- Bolt must be well insulated, else cooling slots have to be
provided.
- This causes increase in flux densities and iron losses.
- Complications of punching process.
- Eddy current losses.
- Difficult getting optimum cross-sectional areas. (Bolt heads,
insulation and stiffening plates).
12. LOSSES
Building factor ≅ actual losses
(B.F.) ideal losses
Design is to get B.F. → 1
Core joints cause B.F. >1
NOISE:
Source is the core, caused by longitudinal mangnetostrictive
vibrations of the iron.
Flatness and proper bending strain, when clamped, will reduce noise.
13. TRANSFORMER REPRESENTATION
Winding connections suppress 3rd harmonic voltages by providing path
for 3rd harmonic current to flow.
Equipment ckt: (Two winding)
xm >> Ƶp, Ƶs
xm, magnetizing impedance neglected
14. THREE WINDING
Only earthed star connection and delta connection allow I0 currents
to flow by providing a path.
** Unearthed star connection does not allow I0 to flow.
17. Let Ƶoc = Ƶm ( xm >> Ƶp )
∴ Ƶsc = Ƶp + Ƶs for 2 winding transformer
For 3 winding transformer
Ƶsc can be either
(i) Secondary (s) is grounded or
(ii) Tertiary (t) is grounded
For (i) → Ƶsc = Ƶps = Ƶp + Ƶs
or (ii) → Ƶsc = Ƶpt = Ƶp + Ƶt
18. From which we can write:
Ƶp=½[Ƶps + Ƶpt – Ƶst]
Ƶs=½[Ƶps + Ƶst – Ƶpt]
And
Ƶt=½[Ƶpt + Ƶst – Ƶps]
where Ƶst = Ƶs + Ƶt
Ƶst → read as impedance seen from the secondary with the
tertiary short-circuited.
19. AUTO TRANSFORMERS
Consider YY∆ transformer earthed through Ƶe.
np 1 np
Let ratios be : = and =1
ns n nt
Then the unit current into primary is:
20. For the secondary:
∴ VƵe = (1-n) Ƶe
Ƶe
∴ Ƶsn = (1-n) Ƶe
Referring to primary
⇒ Ƶsu = [(1-n)/n2 ]( Ƶe )
(where Vp is the base voltage)
Ƶtu = Ƶe/n
pu
22. For a Generator solidly earthed with a 3 winding auto-transformer
we have, as seen from the secondary:
n - 1 3Z
( Zg + Z p ) + 3 Ze Z t + e
1 - n n n
i.e. Z 0 = Zs + 3 Ze +
n n - 1 3Z
(Zg + Z p ) + 3 Ze + Z t + e
n n
23. (Zg + Z p ) (Zg + Z p ) (Zg + Z p ) 2
Zs +
Z t Zs n -1
+ 3Zs +
Z t + 3
2
+ 3Z t
Ze Ze Ze n n
Z0 =
(Zg + Z p ) Z t
+ +3
Ze Ze
but Z e >> Zg , Z p , Zs , Z t
(Zg + Z p ) n −1
2
⇒ Z 0 ≅ Zs + + Zt
n2 n
Or in terms of short circuit impedances
2
1 1 1 n −1
Z 0 = [Z ps + Zst − Z pt ] + 2 [2Zg + Z ps + Z pt − Zst ] − [ Z pt + Zst − Z ps ]
2 2n 2 n
Z ps n -1 1 - n Zg
OR Z0 = + Zst + Z pt 2 + 2
n n n n
24. POWER TRANSFORMERS
RATING:
Flux Density in core limited for 3 reasons
(i) Heating of core
(ii) Introduction of current or voltage harmonics
(iii) Excessive magnetizing current in one or more windings
⇒ Limit on number of volts/turn
Maximum voltage which may be applied is limited to 105%
of rating!!
25. ** The current carrying capacity of a winding is limited by
heating due to resistive losses.
Windings normally capable of carrying up to 105 % of rated current
Example:
For a winding rated ± 10 %
rating on –10 % of tap ≃ 110% plus 5 % over-rating.
∴ Current carrying capacity = 115 %
26. For the + 10% tap -:
Voltage rating/capability = 110 % V
Also current rating/capability = 110 % I
∴Capability of transformer = 121% rating
N.B.
This rating cannot be utilised because the untapped winding is
limited to 100 % i.e. rated MVA.
Present practice is to specify transformer ratio/turns ratio/voltage
ratio at no load.
Similarly tapping range is in terms of turns ratio at no load.
27. CAPACITIVE VOLTAGE TRANSFORMERS
As system voltage rise, insulation cost of wound transformers
become prohibitive.
→ C.V.T used in these cases where
V ≥ 132 kV
The capacitor divider can also be used as the coupler for PLC
(Power Line Carrier) signal; hence costs reduced.
Accuracies of better than 1 % for operating frequency range (for 50
Hz system can be 47≤ f ≤ 51)
28. To achieve this performance in a capacitor divider very large
capacitance would be required.
Problem simplified if source capacitance tuned to the mid- frequency
by a series reactor.
Further alleviation obtained if divider reduces voltage to an
intermediate value (10 ∼ 22 kV) and then a standard wire wound
transformer is used to step this down to 110 V.
30. Equivalent circuit referred to primary becomes:
Where:
Vp = system phase voltage
k = C1/(C1+C2)
C1 + C2 = effective source capacitance
Req = equivalent resistance referred to primary
31. kVp R eq
Output voltage = Vout = ........................(1)
1
R eq + j ωL1 −
ω(C1 + C 2 )
For resonance :
1 1
ω0 L1 = or ω L1 =
2
0
ω0 (C1 + C 2 ) (C1 + C 2 )
Sub into (1)
kVp R eq
⇒ Vout =
ω ω0
R eq + j ω0 L1
ω − ω
0
kVp
=
ω0 L1 ω ω0
1+ j
ω − ω
R eq 0
32. - Vout + Vin kVp − Vo
But error = =
Vin kVp
Vo
=1-
kVp
Vo 1
∴maximum error = f(D) ⇒ =
kVp 1 + jD
33. Vo 1 1
or = =
kVp 1 + jD 1 + D2
or 1 + D 2 ≅ 1 + error (as error gets small)
Where D governs the maximum acceptable error.
ω ω0
Let this frequency deviation = ω − ω
0
This has a maximum value (δ) at ω0
Req is determined by maximum load Wmax
(kVp ) 2
i.e. Wmax =
R eq
34. δ Wmax
⇒D=
ω (C + C ) (kV ) 2
2
or Wmax = Dω0 (C1 + C 2 ) k 2 Vp
0 1 2 p
Since D, ω0, Vp and δ are fixed
⇒ maximum burden ∝ C1 and k
Since, due to cost, necessary to keep C1 and k as small as possible.
35. Example:
Vphase = 275 kV ; k = 1/25
error = 0.5 % ; frequency range = 51.5 – 47.5
Soln:
For frequency range
51.5 + 47.5
⇒ ω0 = = 49.5
2
51.5 49.5
∴δ = − = 0.08
49.5 51.5
Wmax = 200 VA (given)
36. magnitude of error = 1 + D 2
= 1 + 0.005
∴ 1 +D2 ≑1.01
D2 = 0.01 ⇒ D = 0.1
≑ 500pf
38. C3 – stray capacitance of E.M. unit
R1- loss of tuning cct.
Req – includes loss in magnetizing inductance L2
This is a band-pass network with an upper and lower resonant
frequency.
(a) For the higher frequency, Ceq is low compared to L1 and L2
and high compared to C3.
For lower frequency reverse is true.
39. HIGH FREQUENCY c.c.t.
for p = j 2 πfu = jωu − Complex frequency
kVp 1 R eq
p
Vo =
'
R eq 1 + pC 2 R eq
R 1 + pL1 +
1 + pC 2 R eq
'
40. kVp
R eq
⇒ Vo =
p
R + (R + pL )(1 + pC ' R )
eq 1 1 2 eq
Looking at denominator →
' ' '
R eq + (R 1 + pL1 )(1 + pC 2 R eq ) = p 2 L1C 2 R eq + p[C 2 R 1R eq + L1 ] + R 1 + R eq
' 2 2
= L1C 2 R eq p 2 + p + ω h
T
2 R 1 + R eq
where ω h = '
and
L1C 2 R eq
2 R1 1
= + '
T L1 C 2 R eq
2
1 1 2
roots are : α1 = − + −ω h
T T
2
1 1 2
α 2 = − − −ω h
T T
41. 2
2 1
For oscillation : ω h >
T
R eq + 1
R
2 R1
2
1
⇒ > +
' 2 2
L1C 2 R eq L1 '
C 2 R eq
2
True for high R eq
(i) R eq = 0
⇒ transfer function is pL1 + R 1
Simple exponential transient
(ii) R eq = ∞
1
and T = ( L1 R 1 )
2 2
⇒ ωh = '
L1C 2
42. LOW FREQUENCY c.c.t
pL 2 R eq
kVp pL + R
2 eq
Vo =
p 1 pL 2 R eq
R1 + +
pC1 pL 2 + R eq
'
43. kVp pL 2 R eq
Vo =
1
pR 1 + ' (pL 2 + R eq ) + p 2 L 2 R eq
C1
Looking at the denominator it can be re-written as:
L2
p R 1R eq + '
p 2 + C1
+ R eq
(R 1L 2 + R eq L 2 )
L 2 (R 1 + R eq ) '
L 2 C1 (R 1 + R eq )
L2
R 1R eq + '
2 C1 R 11 1
Let = = + '
T L 2 (R 1 + R eq ) L 2 C1 (R 1 + R eq )
R 1R eq
where R 11 =
(R 1 + R eq )
44. 2 R eq
and ω L = '
L1C1 (R 1 + R eq )
Roots are :
2
1 1 2
α1 = − + −ω L
T T
2
1 1 2
α 2 = − − −ω L
T T
For oscillation:
2
1
( ωL ) 2
>
T
45. This yields:
1
2 2
R eq R 1 R eq 1 R 1R eq
> + +
4C1 (R 1 + R eq ) 2 L 2 C1 (R 1 + R eq )
' 2 2 ' 2
L1C1 (R 1 + R eq ) 4L 2 (R 1 + R eq ) 2 '
2 2
1 1 R 1 R eq 1
or R eq − R 11 > +
L 2 C1 (R 1 + R eq )
' 2 '2
4L 2 (R 1 + R eq )
2
2 4C1 (R 1 + R eq )
Again, oscillation occurs for large Req.
Compromise in design between Ceq i.e. (C1+C2) and accuracy of
response required.
Generally, a large as possible value of C1 and k is used.
46. SURGES ON TRANSFORMER WINDING
Initial Voltage Distribution
Assuming that the capacitance is uniformly
distributed along the winding
Define: l = length of winding
Cg = total capacitance to ground
Cs = total series capacitance
47. Cg
∇ ground capacitance/unit length =
l
and series capacitance/unit length = Cs l
Let: E = voltage to ground at any point in the winding
Ig = total current in the ground capacitance
Is = total currents in the series capacitance
Consider an elementary length of winding ∆x
Cg
∴ Capacitance/ground = ∆x
l
∴ Current through this capacitance :
C g ∆xωE
∆I g = ..................................(1)
l
48. dI s
But ∆I g = ∆x..................................(2)
dx
Combining (1) and (2)
dI s C g ωE
⇒ = ..................................(3)
dx l
l Cs
But the series capacitance of unit winding is
∆x
dE
and the voltage across the element = ∆x............................(4)
dx
dE
and current in the series capacitance = C = ............................(5)
dt
49. dE
Sub. into (3) ⇒ Is = l Cs ω ..................................(6)
dx
dI s d 2E
Differentiating equation (6) ⇒ = l Cs ω 2 ..................................(7)
dx dx
C g ωE d 2E
From (3) and (7) ⇒ = l Cs ω 2 ..................................(8)
l dx
d 2E 1 Cg
or 2
− 2 E = 0..................................(9)
dx l Cs
Equation (9) is independent of ω and has a solution in the form
E = A1e bx + A 2 e -bx .........................................(10)
1
1 Cg 2
where b =
l Cs
50. Solving for A1 and A2 using the boundary conditions:
x=0 ⇒ E=0 Grounded
neutral
And x=l ⇒ E=V
where V is the amplitude of the step function surge
Sub. Into (10) ⇒ A1 + A2 = 0
A1e bl + A 2 e -bl = V
V
or A1 = - A 2 = bl
e − e -bl
51. 1
Cg
2
Let bl = = α
Cs
V
A1 = - A 2 = ...............................(11)
2 sinh α
V
Sub. (11) into (10) ⇒ E = [e bx - e -bx ]
2 sinh α
∴ E=V
sinh α x( l)
sinh α
For isolated neutral the boundary conditions are :
x=l ; E=V
dE
and x=0 ; Is = 0 or =0
dx
52. ⇒ b(A1 - A 2 ) = 0
A1e bl + A 2 e -bl = V.............................(12)
From which we get :
V V
A1 = A 2 = = .........................(13)
2 cosh(bl) 2 cosh α
V
Sub. (13) into (10) ⇒ E = [e bx + e -bx ]
2 cosh α
=V
cosh α x ( l)
cosh α
53. (a) Grounded Neutral :
( )
x
dE α cosh α l
= V
dx l sinh α
dE αV
for x = l ⇒ = cothα
dx l
54. (b) Isolated Neutral :
( )
x
dE α sinh α l
= V
dx l cosh α
dE αV
for x = l ⇒ = tanhα
dx l
55. WINDING OSCILLATIONS
Cg
Let the capacitance/unit length = C1 =
l
Looking at a length ∆x ⇒ the capacitance to ground = C1.∆x
Let the interturn capacitance = C 2 /length
or = bC 2 [for length of turn = b]
Let L1 = self inductance/unit - length
= L1.∆x [for elementary unit]
And for one turn with length ' b' = bL1
56. ∴ within the limits, the induced voltage is
∂I ∂E ∂I Where I is the
∂E = L ∂x or = L .......................(1)
∂t ∂x ∂t current flowing
∂E
Also, the capacitance to ground for one turn = bC1 ............................(a)
∂t
Capacitance current to turn 2 from turn 1 is
∂ ∂E
= b C2 ∂x b ............................(2)
∂t 1,2
∂ ∂E
and from turn 3 is = - b C2 ∂x b ............................(3)
∂t 2,3
57. ∂E ∂E ∂ E b
2
It is known that : ∂x = ∂x + ∂x 2 . 2 ............................(4a)
1,2 2 2
∂E ∂E ∂ E b
2
and ∂x = ∂x − ∂x 2 . 2 ............................(4b)
2,3 2 2
∂ 2E
where 2 = space rate of change of voltage gradient at conductor 2
∂x 2
Sub. (4a) and (4b) into (2) and (3) gives the net charging current from adjacent turns
∂ ∂E ∂ 2 E b ∂ ∂E ∂ 2 E b
i.e. bC 2 + 2 . .b - bC 2 − 2 . .b
∂t ∂x 2 ∂x 2 2
∂t ∂x 2 ∂x 2 2
∂ 3E
= C2b 3
............................(5)
∂t ∂x 2
58. Subtract from this the capacitance current to ground (a), we have
the net rate of change of the total current per turn length
∂I 3 ∂ E
3
∂E
-b = C2b - C1b ............................(6)
∂x ∂t ∂x 2
∂t
[-ve because ' x' is measured from the neutral]
Differentiating (1) w.r.t. x
∂ 2I 1 ∂ 2E
⇒ = ............................(7)
∂x∂t L ∂x 2
Differentiating (6) w.r.t. ' t'
∂ 2I ∂ 4E ∂ 2E
⇒ = - C2b 2
+ C1 2 ............................(8)
∂x∂t ∂t ∂x
2 2
∂t
59. Equating (7) and (8)
1 ∂ 2E ∂ 4E ∂ 2E
⇒ = - C 2 b 2 2 + C1 2
2
L ∂x 2
∂t ∂x ∂t
∂ 2E ∂ 2E ∂ 4E
or − LC1 2 + LC2 b 2 2 = 0................(9)
2
∂x 2
∂t ∂t ∂x
Sub : E = Vγ e jωt e j γ x
From which ⇒ - γ 2 + LC1ω 2 + LC2 b 2 γ 2 ω 2 = 0
1/ 2
LC1ω 2
or γ=
1 - LC2 b 2 ω 2
γ
and ω=
(LC1 + LC2 b 2 γ 2 )1/ 2
60. From the denominator for ' γ'
⇒ if LC2 b 2 ω 2 > 1 then γ is imaginary
This corresponds to a critical frequency ω c such that
1
ωc =
b LC2
Therefore, the winding behaves like a filter blocking all frequencies > ωc
62. FIELD DRYING OF TRANSFORMER INSULATION
Residual water content < 0.5 % of wt. of paper insulation
Method #1 :- CIRCULATING HOT OIL (Slow process)
Requirements:
(a) Oil Filter → either vacuum drier type or blotter press.
(b) Heater suitable of raising oil temp to 85oC
63. Procedure:
- Fill transformer to top of core with oil.
- Circulate oil through filter with heater on.
- Reduce heat losses by:
- Close off radiators
- Blanket outside of transformer tank.
- Continue circulation until dryness is achieved.
N.B. The rate of drying can be increased by application of a vacuum to
the surface of the oil!
64. Method #2 :- Short – Circuited Windings: Vacuum (Most Rapid
Method)
Requirements:
- Power source to drive current thru transformer
coil/winding → heating source [winding temperature <
95oC
- Vacuum Pump
- Refrigerated condenser trap in vacuum line.(this serves
to collect/condensed extracted moisture)
- Blanket the outside of transformer tank so as to
minimise heat loss.
65. Procedure:-
- Transformer filled with oil above core and coil
allowing room for expansion.
- Vacuum pump connected to suitable valve at top
of tank.
- With one winding s/c, connect supply to other.
Circulate current up to FLC. [Winding temperature not
to exceed 95oC]
- For forced cooled transformer, oil pumps should
be operating → however, radiators should be off.
66. Procedure:-(cont’d)
- After desired temperatures reached, disconnect supply
and drain oil.
- Start vacuum pump and continue until water extraction
ceases.
- Repeat procedure if necessary to achieve desired
dryness.
N.B.
Water extraction will stop WHEN the V.P. of water in the
insulation equals the partial pressure of water vapour in the
tank at the prevailing temperature and pressure.
67. Method #3 :- Using High Vacuum (Specially suited for EHV T/F)
Requirements:
- Vacuum pump capable of vacuum of <7 Pa absolute.
- Cold trap to collect moisture.
- Possibly additional heat may be required.
- Transformer must be designed to withstand full
vacuum.
68. Procedure:-
- Drain oil while at same time fill with nitrogen (dry)
- Remove heat exchangers and external pipe connections
- Install cold trap as close to the transformer as possible
- Seal transformer and pressure test for leaks
N.B.
Water extraction from the facilities start when the
residual vapour pressure in tank is less than v.p. of
water insulation.
- Vacuum maintained in the order of 1mm Hg (130 pa)
- End point is when the rate of water condensate in the
cold trap is less than 3g/hr per 1000kg of insulation wt.
69. Method #4: -Using Hot Air (Hazardous)
Tank should first be blanketed in order to minimize heat required (loss)
Procedure:-
Volume of Hot Air Required
Area of Base (m2) 2.8 5.6 11.6 14.0
Volume of Air (m3/min) 28 56 114 140
_______________________________________________________________
i.e → Volume Required (m3/min) = 10 (Area of tank base m3)
___________________________________________________________
- Clean dry hot air is blown through air opening at
base of tank, over coils exiting at top.
- Temperature of inlet and outlet streams noted
(Temperature of inlet should be ≃ 100oC)
Note: Fire hazard as flash point for transformer oil is 145oC.
70. Variation of winding (R & p.f.) with drying time (t)
N.B.: As soon as the transformer is considered dry, it should
be immediately filled with oil to cover the core and
windings
71. Vacuum Treatment prior to oil filling
Principal function of vacuum treatment is to remove trapped
air/moisture from the insulation so that the insulation can achieve
full dielectric strength.
For V ≤ 345 kv → Vacuum used ≈ 5mmHg
V > 345 kv → Vacuum used ≃ 1mmHg
Vacuum should be held for 4 hours or more as per manufacturers
instruction.
72. PRE-ENERGIZATION TESTING:-
- Insulation resistance tests (megger)
- Power factor test → windings, bushings
- Ratio test
- Operation of attached auxiliary parts and sensors
- Dielectric strength, power factor of oil
- O2 content and total combustible gas content of
Nitrogen Gas Cushion
After energizing, the transformer should “soak” unloaded for 8 hours.
73. TESTING (BS 171)
Routine Test → All transformers are subjected to these:
- Voltage Ratio and polarity
- Winding resistance
- Impedance voltage, s/c impedance and load test
- Dielectric tests
- No-load losses
- On-load tap changer where installed.
74. TYPE TESTS
Test made on one transformer is representative of all.
- Temperature-rise test
- Lightning Impulse test
75. SPECIAL TESTS
As agreed between vendor and purchaser
- Dielectric test
- Zero sequence Impedance (3-phase t/f only)
- s/c test
- Acoustic sound level
- Magnetizing on no load
- Power taken by auxiliaries ( fans, pumps etc.)
