oil filled cts (72!5!420 kv)
TRANSCRIPT
1
EMEK ELEKTRIK ENDUSTRISI A.S
PO Box 648 Ulus 06044Ankara-Turkey
Tel : ++90 312 398 07 95 & 96Fax : ++ 90 312 398 09 03e-mail : [email protected] site : www.emek.com.tr
OIL FILLED CTís
2
DESIGN CARETERIADESIGN CARETERIA
In the following pages brief and simplified guidelines are given for the selection and design of current transformers for a particular duty.
The information and data presented in these notes is not sufficient to be used alone for selecting an instrument transformer for a particular duty. Instrument transformers are ìcustom made î and it is recommended that
each and every parameter should be carefully considered before aselection is made.
Instrument transformers are not low priced items. In general they make up of about 2 to 4 % of the cost of a conventional (open terminal) substation, excluding land. Therefore economics is as important as
technical details when selecting CTís.
In case instrument transformers are ìover designedî or ìover ratedîmoney will be simply wasted on technical characteristics that will never
be used in the field life of the instrument transformers.
We hope these simple guidelines will be of assistance to readers.
3
Dry types are cast epoxy types. These are most common up to 36 kV applications. In Western Europe, Australia, cast epoxy types are also commonly used up to 72.5 kV.
Current Transformers
Oil FilledSF6 gas filledDry Type
Type of Insulations
SF6 gas filled CTís are most commonly used for very high voltageapplications, usually above 500 kV. Below 500 kV, SF6 filled CTísare not economical.
Oil filled CTís are
the oldest, the
most reliable and
the most field
tested type of CTís.
4
Type of structure
Oil Filled CTís
Porcelain TypeInverted TypeTank Type
All three types listed above can perform the same job, that is they can reduce the current to small magnitudes which can be safely handled. However some types have some particular characteristics that have to be discussed and considered very carefully before a selection is made
5
Porcelain TypeInverted TypeTank Type
Type of structure
6
Comparison Between Types
In tropical climates, the steel tank will absorb sunlight and the oil will operate at high temperatures. Increased oil temperature will decrease oil life.
7
The tank is subject to corrosion. If the tank is only painted it will start to corrode in a few years in a tropical climate. If the tank is hot galvanized, scars caused during transportation or installation will cause corrosion
Comparison Between Types
8
Center of gravity is very high. Therefore transportation and erection are very difficult. Earthquake performance is very poor
Center ofGravity
Center ofGravity
Center ofGravity
The center of gravity is low, therefore earthquake performance is very good, Transportation and erection is easy.
Comparison Between Types
The center of gravity is low, therefore earthquake performance is very good, Transportation and erection is easy.
9
The hot oil rises to the top and at the same time direct sunlight heats the oil on the top, therefore the windings are always in the hottest part of the CT. Hot oil and hot paper loose their good insulation properties very quickly.
The warm oil rises to the top, therefore the windings operate in a cool region.
The warm oil rises to the top, therefore the windings operate in a cool region.
Comparison Between Types
10
A very small amount of oil loss will leave the windings without oil. Therefore internal flashovers are common if the CT losses some oil.
Small oil loss will not affect the insulation. Actually more than half of the oil has to be lost before the windings are without oil.
Comparison Between Types
Small oil loss will not affect the insulation. Actually more than half of the oil has to be lost before the windings are without oil.
11
F1F4
F3 F2
F2F1
During a short circuit the dynamic forces make it into a round shape. If the winding is already a round shape then these radial dynamic forces cancel each other and the resultant force is minimum
Comparison Between Types
12
HOW TO SELECT A CURRENT TRANSFORMER
At least the following parameters must be known when a CT is to be selected for a certain job.
i) Highest System Voltage ( Insulation Level )i) Highest System Voltage ( Insulation Level )
This parameter must be selected in accordance with a Standard. This value will also determine the power frequency and impulse test levels of the CT.
As an example, according to IEC, 170/325/750 kV insulation level means the CT will operate at 170 kV phase to phase voltage continuously, the one minute power frequency test voltage is 325 kV rms. and the basic impulse test voltage is 750 kV peak.
It is very important and advisable not to select insulation levels different from standards. For example if an insulation level of 170/325/900 kV is asked for, the manufacturers will have to custom design the insulation and the resulting CT will be unnecessarily very expensive.
13
Rated kV Power Frequency BIL
3.6 10 407.2 20 6012 28 75
17.5 38 9524 50 12536 70 17052 95 250
72.5 140 325123 230 550145 275 650170 325 750245 460 1050300 460 1050362 510 1175420 630 1425
HOW TO SELECT A CURRENT TRANSFORMER
IEC STANDARD
14
ii) Frequencyii) Frequency
Frequency will be usually 50 or 60 Hz. Current transformers and inductive voltage transformers can usually operate at both frequencies without major difference in their performance, however this is not a rule. Capacitive voltage transformers must be designed for a specific frequency.
iii) Ratioiii) Ratio
It is advisable to have as little number of interchangeable ratios as possible. However for practical and operational reasons multi ratio CTís are usually preferred.
There are three ways of changing the ratio of CTís :
a) by changing the number of turns of the primary windingb) by changing the number of turns of the secondary winding c) by changing the number of turns of both windings
It is advisable to change ratios by changing the number of turns of the primary winding. This way the CT will easily satisfy the accuracy, burden and composite error requirements on all ratios.
HOW TO SELECT A CURRENT TRANSFORMER
15
HOW TO SELECT A CURRENT TRANSFORMER
Primary SecondaryA Turn AT A Turn AT Connection
200 1 200 1 200 200 S1-S2400 1 400 1 400 400 S1-S3
Primary SecondaryA Turn AT Connection A Turn AT Connection
200 2 400 Parallel 1 400 400 S1-S2400 1 400 Series 1 400 400 S1-S2
P1 P2
S1 S2
200 200
1
RATIO : 200-400/1 A
S3
SECONDARY TAP
P1 C2 C2 P2
S1 S2
400
22
RATIO : 200-400/1 A
PRIMARY TAP
16
HOW TO SELECT A CURRENT TRANSFORMER
Primary SecondaryA Turn AT Connection A Turn AT Connection
200 4 800 Parallel 5 160 800 S1-S2300 4 1200 Parallel 5 240 1200 S1-S3400 2 800 Series 5 160 800 S1-S2800 2 1600 Series 5 320 1600 S1-S4
P1 C2 C2 P2
S1 S2 S3 S4
160 80 80
22PRIMARY WINDINGS
SECONDARY WINDINGS
RATIO : 200-300-400-800/5 A
17
iv) Rated Burden in VAiv) Rated Burden in VA
It is important to select the correct VA requirement. It is not true that a CT with 80 VA burden is better or stronger than a CT with 30 VA burden.
It is easy to calculate the required VA if we know :
a) the length and cross section of the conductor between the CT and the control room or panels
b) the VA requirement of the instrument and/or relay
the simple addition of these loads will give the VA. It is advisable to spare some VA for future addition of instruments into the circuit, however, this should not be overdone. Unnecessarily high VA will require big and bulky CTís that will be more expensive.
HOW TO SELECT A CURRENT TRANSFORMER
18
P1S1
S2
kWhI>A
P2 d
a² . 2dBc = --------
cs . 57
Bc = Burden in VA presented by copper leads(out and return conductor
a = amperd = Distance in meter, one waycs = cross section in mm²
HOW TO SELECT A CURRENT TRANSFORMER
19
Burden in VA2d 1A, cross section in mm2 5A, cross section in mm2m 1 2.5 4 6 2.5 4 6 10 5 0.18 0.07 1.78 1.12 0.74 0.44 6 0.21 0.09 2.14 1.34 0.89 0.54 7 0.25 0.10 2.50 1.56 1.04 0.63 8 0.29 0.11 2.86 1.79 1.19 0.71 9 0.32 0.13 3.21 2.01 1.34 0.80 10 0.36 0.14 0.09 0.06 3.57 2.24 1.49 0.89 20 0.71 0.29 0.18 0.12 7.10 4.50 3.00 1.80 30 1.07 0.43 0.27 0.18 10.70 6.70 4.50 2.70 40 1.43 0.57 0.36 0.24 14.30 8.90 6.00 3.60 50 1.78 0.72 0.45 0.30 17.80 11.20 7.40 4.40 60 2.14 0.86 0.54 0.36 13.40 8.90 5.40 70 2.50 1.00 0.63 0.42 15.60 10.40 6.30 80 2.86 1.14 0.71 0.48 17.90 11.90 7.10 90 3.21 1.29 0.80 0.54 21.10 13.40 8.00 100 3.57 1.43 0.89 0.60 22.40 14.90 8.90
HOW TO SELECT A CURRENT TRANSFORMER
20
Following are two simple examples of how to choose burdens.
Example 1 : 1 Ampere Secondary Current Transformer
- Distance from CT to control panel : 100 meters- Cross section of conductor between CT and panel : 4 mm²- Instruments connected to the CT : i) ammeter 2 VA
ii) wattmeter 5 VAiii) O/C relay 15 VA
Burden presented by the copper conductor = Bc
a² . 2d 1² . 2.100Bc = -------- = ------------- = 0.9 VA
cs . 57 4 . 57
Total Actual Burden = 0.9 + 2 + 5 + 15 = 22.9 VASpare 25% = 5.8 VABurden to Select = 22.9 + 5.8 = 28.7 ≅ 30 VA
HOW TO SELECT A CURRENT TRANSFORMER
21
Example 2 : 5 Ampere Secondary Current Transformer
- Distance from CT to control panel : 70 meters- Cross section of conductor between CT and panel : 10 mm²
Instruments connected to the CT : i) ammeter 2 VA ii) wattmeter 5 VAiii) O/C relay 15 VA
Burden presented by the copper conductor = Bc
a² . 2d 5² . 2.70Bc = -------- = ----------- = 6.15 VA
cs . 57 10 . 57
Total Actual Burden = 6.15 + 2 + 5 + 15 = 28.15 VASpare 25% = 7 VABurden to Select = 28.15 + 7 = 35.15 ≅ 40 VA
HOW TO SELECT A CURRENT TRANSFORMER
22
Another very important point to consider is that, many international standards require the accuracy to be satisfied between 100% and 25% of the rated burden. If the burden is selected too high and the actual burden is less than 25% of the selected burden, the CT accuracy class is not guaranteed.
HOW TO SELECT A CURRENT TRANSFORMER
%5In
0.5
0.5
%5In
In%50In %100In %120In%20In
%25 VA
%100 VA
Required60 VAClass is guaranteedbetween 15 - 60 VA
Actual 20 VAClass is not guaranteedbetween 5 - 20 VA
23
HOW TO SELECT A CURRENT TRANSFORMER
Calculation of Core Size (IEC) :
VkVk x 10000 x 10000 S = S = ------------------------------------------------
4.44 x B x N2 x f4.44 x B x N2 x f
S = cross section of Core (cm²)Vk = Knee point voltage (V)B = Flux Density (1.4 Tesla)N2 = Number of Secondary Turnf = Frequency
S1 S2
160 Turn5 A
P1 P2
1 Turn
800 A
S1 S2
160 Turn5 A
Core 1 : 30 VA, Class 0.5 Fs5, 800/1 AVk = n x VA/Is = 5 x 30 VA /1 A = 150 V
150 x 10000 150 x 10000 S = S = --------------------------------------------------= 30 cm = 30 cm ²²
4.44 x 1.4 x 160 x 504.44 x 1.4 x 160 x 50
Core 2 : 30 VA, Class 5P20, 800/1 AVk = n x VA/Is = 20 x 30 VA /1 A = 600 V
600 x 10000 600 x 10000 S = S = --------------------------------------------------= 120 cm = 120 cm ²²
4.44 x 1.4 x 160 x 504.44 x 1.4 x 160 x 50
24
HOW TO SELECT A CURRENT TRANSFORMER
Calculation of Core Size (BS) :
VkVk x 10000 x 10000 S = S = ------------------------------------------------
4.44 x B x N2 x f4.44 x B x N2 x f
S = cross section of Core (cm²)Vk = Knee point voltage (V)B = Flux Density (1.4 Tesla)N2 = Number of Secondary Turnf = Frequency
Calculation of Io
H x L H x L Io = Io = ----------------
N2 N2
Io = Magnetising Current (A)H = Magnetesing Force AT/mL = Length of Magnetic Part (m)N2 = Number of Secondary Turn
B Tesla
H AT/m
1.4
32
Calculation of Rct
l l RctRct = = δδ x x ----------------
AA
Rct = Secondary resistance (Ω)δ = Constant (0.0172) for copperl = Length of secondary winding A = cross section of secondary winding
25
v) Accuracy Classv) Accuracy Class
The accuracy should be selected for the job. If only indicating instruments like ammeters and wattmeters are to be connected, then Class 1.0 is sufficient for measuring cores. If energy meters are to be connected and also if revenue is to be calculated with these meters then Class 0.5 can be selected. Class 0.2 should be selected only when very large amounts of energy transfer and revenue is in question.
It must be remembered that as the CT becomes more accurate in class, its VA rating drops. As a rough example if a CT can provide Class 1.0 at 40 VA, it can provide Class 0.5 at 25 VA and perhaps only 10VA at Class 0.2.
For measuring cores, the instrument security factor ( FS ) has to be stated alongside the Accuracy Class. This figure is usually 5 or 10.
A typical expression (in IEC ) is 0.5FS5 which means the Accuracy Class is 0.5 and the Instrument Security Factor is 5.
HOW TO SELECT A CURRENT TRANSFORMER
26
Instrument Security Factor FS5 means, during a short circuit when there is a very high current flowing in the primary winding the measuring core will saturate around 5 times its rated value. If this core is a 1 Ampere core, the secondary current flowing through the instruments will not rise above 5 Amps. This is a precaution to prevent instruments being burn out during a short circuit.
At lower burdens than the rated burden, the saturation value increases approximately by n.
SnSn + + RctRct x Isx Is²² Examplen = Fs x n = Fs x ------------------------------------------------ Rct x Is² = 1
S + S + RctRct x Isx Is²² Sn = 50 VAS = 20 VA
Sn = Rated Burden (VA) Fs = 5S = Actual Burden (VA) n = 5 x 2.4 = 12Is = Rated secondary Current (A)Rct = Internal Resistance at 75° (Ω)
HOW TO SELECT A CURRENT TRANSFORMER
27
The IEC classes for protective current transformers are 5P and 10P
The main characteristics of these CTís :
ï Low accuracy (larger errors permitted than measuring transformers).ï High Saturation value
Saturation voltage is given by ALF (Accuracy limit Factor) ,indicates the over current as a multiple of the the rated primary current up to which the rated accuracy is maintained with the rated burden connected. It is given as a minimum value.
For protection cores, selection of accuracy class and saturation factor is more complicated. In order to make a good selection, the characteristics for the protection relays should be known. All reputable protection relay manufacturers publish charts, tables, graphs and formulae for CT accuracy class and burden calculations. If these documents are followed accurately, then optimum selections can be made.
HOW TO SELECT A CURRENT TRANSFORMER
28
HOW TO SELECT A CURRENT TRANSFORMER
Relation between Relation between VkVk and VAand VA
Ratio : 800/1 A, Vk > 600 V, 5P20, VA = ?20 x Vs = 600 V Vs = 30 V VA = 30 V x 1 A = 30 VA
Ratio : 800/1 A, 5P20, 30 VAVs = 30 VA/1 A = 30 V, Vk = 30 V x 20 = 600 V
Es
Np/Ns x Ip Is
IµIf
Rj (winding resistance)
Burden
Ie
Exciting impedance
Transformers diagram Transformers diagram convertedconverted to the secondary sideto the secondary side
29
vi ) Standardsvi ) Standards
When selecting a CT for a particular job, the standard should be firmly defined from the beginning. It is not advisable to use more than one standard for one particular job because the standards may have conflicting clauses. It is not good practice, for example, to select the insulation level from BS and then to select the accuracy class from IEC.
vii) Environmental Conditionsvii) Environmental Conditions
The altitude, climate and seismic conditions do affect the design of CTís. Therefore these must be considered when a CT is selected for a particular job.
HOW TO SELECT A CURRENT TRANSFORMER
30
HOW TO SELECT A CURRENT TRANSFORMER
1.28
Example :
145 KV Current Transformers
Normal Insulation Level :145 / 275 /650 KV
After Correction :145 / 275/832 KV
Inner insulation (hermetically sealed) does not effect from altitude. Basically this factor is valid for correction of lightning impulse level.
31
viii) Short Circuit Ratingviii) Short Circuit Rating
The short circuit rating is usually defined as follows :
Ith = 40 KA/1 sec rms. Idyn = 2.5 Ith peak
Ith is the thermal capacity of the CT. In this example the CT will withstand 40KA for 1 second without any burning of the windings, insulation or the oil (250 °C). Ith is expressed in rms. value.
Idyn is the dynamic capacity of the CT to withstand mechanical forces during the short circuit. These forces act to break apart the CT. Idyn is expressed in peak value. Generally Idyn is 2.5 times Ith.
HOW TO SELECT A CURRENT TRANSFORMER
32
The relation between 1 seconds and 3 seconds performance can be shown with the formula
I² x t = constant
If a CT is rated at 40 KA/1 seconds, its rating at 3 seconds can be calculated as below :
40² x 1 sec = constant = 1,600(I2)² x 3 sec = 1600I2 = 23 KA / 3 sec
Where I2 is the maximum current the CT can withstand for 3 seconds.
HOW TO SELECT A CURRENT TRANSFORMER
33
HOW TO SELECT A CURRENT TRANSFORMER
Calculation of Primary Cross Section :
In x 1.2 In x 1.2 IthIth (1 sec) (1 sec) IthIth x x √√3 (3 sec)3 (3 sec)cs1 = cs1 = -------------------------- cs2 = cs2 = -------------------------- or or ------------------------------------
1.8 1.8 180180 180180
cs1 = Cross section of primary (mm²) cs2 = Cross section of primary (mm²) 1.2 = Continuous current Ith = Short time current KA1.8 = Constant for copper 180 = Constant for copper
Example : In = 300 A, 1.2 Cont, 31.5 KA/1 sec
cs1 = 200 mm² cs2 =175 mm²
200 mm200 mm²² > 175 mm> 175 mm²² ?? select 200 mmselect 200 mm²²
34
ix) ix) CreepageCreepage
Creepage is the length of the surface path from the live part to the grounded part of a transformer. The creepage length is determined by the exterior shape of the bushing sheds. Most standards have recommendations for light, normal, heavy and very heavily polluted areas.
High creepage distance bushings cost more, therefore creepage distance has a major effect on the CT price. Unnecessary high values of creepage distances should be avoided.
The total creepage length is calculated by multiplying the creepage distance with the maximum system voltage. For example, in Turkey where 170 kV system is used , the total length of medium pollution creepage is
170 kV x 20 mm/kV = 3,400 mm.
HOW TO SELECT A CURRENT TRANSFORMER
35
HOW TO SELECT A CURRENT TRANSFORMER
kV Type mm/KV Total
72.5 Light 16 1,160 72.5 Medium 20 1,450 72.5 Heavy 25 1,813 72.5 Very Heavy 31 2,248
123 Light 16 1,968 123 Medium 20 2,460 123 Heavy 25 3,075 123 Very Heavy 31 3,813
145 Light 16 2,320 145 Medium 20 2,900 145 Heavy 25 3,625 145 Very Heavy 31 4,495
170 Light 16 2,720 170 Medium 20 3,400 170 Heavy 25 4,250 170 Very Heavy 31 5,270
245 Light 16 3,920 245 Medium 20 4,900 245 Heavy 25 6,125 245 Very Heavy 31 7,595
300 Light 16 4,800 300 Medium 20 6,000 300 Heavy 25 7,500 300 Very Heavy 31 9,300
362 Light 16 5,792 362 Medium 20 7,240 362 Heavy 25 9,050 362 Very Heavy 31 11,222
420 Light 16 6,720 420 Medium 20 8,400 420 Heavy 25 10,500 420 Very Heavy 31 13,020
36
EMEK DESIGN
General Descriptionï Primary and secondary cores are housed in a single porcelain bushing.ï Transformers are hermetically sealed.ï Well proven ìhair pinî technology is used.
Frequencyï 50 Hz, 60 Hz
Secondary Currentï 1 A, 2 A or 5 A, others on request.
Primary Currentï up to 4000 A
Thermal and Dynamic Ratingsï Short time thermal current ratings up to 40 KA/ 3 sec and dynamic current
ratings up to 100 KA peak.
Continuous Thermal Currentï 1.2 x In continuous (up to 2.0 x In on request).
Standardsï IEC, BS, ANSI, AS, other standards on request.
37
1. Aluminium Rain Cover
2. Stainless Steel Bellows
3. Oil Level Indicator
4. Primary Terminals
5. Aluminium Head
6. Primary Conductor
7. Steel Tube (filled with epoxy resin to eliminate effect of
dynamic forces during short circuit)
8. Paper Insulation (HV side)
9. Primary Windings
10. Porcelain Bushing
11. Paper Insulation (LV side)
12. Secondary Windings
13. Secondary Windings Support
14. Oil (Current transformer is completely filled by oil)
15. Base Plate
16. Lifting holes
17. Secondary Terminals
18. Oil Sampling Valve (Opposite side of the secondary
box)
19. Oil Filling Tap
1
23
5
4
6
7
8
10
12
11
13
14
15
17
16
18
9
19
EMEK DESIGN
38
Coresï Grain oriented silicon steel is used. ï The core shapes are toroidal without air gap.
Insulation of cores by paper
EMEK DESIGN
39
Secondary winding on the cores
Secondary Windingsï Enamelled electrolytic copper wire is used.
EMEK DESIGN
40
Max. 6 cores can be accommodated Paper Insulation on the cores
EMEK DESIGN
41
Primary WindingsElectrolytic copper suitable to withstand rated continuous current and short time thermal current is used.
Paper insulation on primary winding
EMEK DESIGN
42
Primary and secondary winding insulation is completed.
How to calculate thickness of paper :
145 KV Current TransformersPower Frequency Withstand Voltage : 275 KVDielectric characteristic of paper = 5.5 mm/KVTotal Thickness : 275 KV/5.5 mm = 50 mm
Safety Margin : 50 mm x 1.1 = 55 mm
Primary Insulation : 55 x 2/3 = 36.6 mmSecondary Insulation : 55 x 1/3 = 42.3 mm
InsulationOil impregnated paper, high dielectric strength, low dielectric losses.
Graded layers with well rounded edges enable a uniformly distributed field over the whole unit
EMEK DESIGN
43
Epoxy
Steel Tube
Forces during short time
EMEK DESIGN
44
Inserting porcelain bushing onto windings.
Installation of Base PlateBase of the transformers are fabricated from steel and galvanised
EMEK DESIGN
45
How to measure creepage distance :
WRONG CORRECT
Brown glazed porcelain (other colours on request).Extra high creepage distance is also possible on request
EMEK DESIGN
46
Installation of HeadThe housing head is fabricated from corrosion protected aluminium.
Every housing is subject to pressure andvacuum test before installation.
Spring
Bracelet
BraceletsAll bracelets used on transformers are madefrom aluminium alloy.
EMEK DESIGN
47
Primary Terminalsï Primary terminals are made from tin plated copper. ï Vertical and horizontal primary terminals are available.ïStud, nema pad or clamp type primary
terminals are available.
Dimensions depend on rated primary current (stud type).1-800 A = ∅ 30 & 80 mm801-1600 A = ∅ 40 & 125 mm1601-2500 A = ∅ 50 & 125 mm2501-3000 A = ∅ 60 & 130 mm3001-4000 A = ∅ 65 & 125 mm
EMEK DESIGN
48
Oil ImpregnationHeat and vacuum are applied to the fully insulated windings after assembly of thecurrent transformer for drying purposes (around 3-4 days), followed immediately by the oil filling under vacuum.
Heaters
Vacuum &Oil Filling pipes
EMEK DESIGN
49
Insulating OilThe insulating oil is mineral oil in conformity with IEC 296.
Before impregnation, moisture and gases are removed from the oil by special processes.
Drain plug for taking samples and oil refilling facilities are provided on request.
Floating type oil level indicator is standard supply.
Oil Filling Tap
Oil Level Indicator
EMEK DESIGN
50
Hermetic Seal Oil volume variations are compensated with stainless steel bellows placed in the head of the current transformers. With this facility there will be no contact between air and oil in the transformers.
The dielectric strength will remain unchanged for the life time of the CT.
EMEK DESIGN
51
Changing of RatioRatio change can be achieved by
primary series-parallel connection
Primary ratio changing terminal is located in the primary connection box which is on the head of transformer.
Changing of RatioRatio change can be achieved by secondary taps located in the secondary terminal box..
EMEK DESIGN
52
Secondary Terminal BoxHinged type boxes are used and are sealableon request. They are made from aluminiumor stainless steel.
Clamp type secondary terminals are used.
Other secondary terminals are available onrequest
Name PlateName plates are stainless steel and weatherproof.
EMEK DESIGN
53
Earthing M10 or M16 earthing terminals are usedand are located on the base of thetransformers. Clamp type terminals areavailable on request.
External Binding ElementsAll external bolts, nuts, washers etc. are either stainless steel or hot dip galvanisedsteel.
PaintingExternal surface of the transformers,
which are already corrosion protected, areadditionally painted grey . Other colours are available on request.
EMEK DESIGN
54
Optional Accessories
ïSteel support structure
ï Primary connectors
ï Marshalling box
EMEK DESIGN
55
Earthquake DesignStandard design is up to 0.6g.
Mechanical StrengthApplied load to the primary terminal inhorizontal and vertical is in conformity with IEC 44-1.
Ambient TemperatureStandard design is suitable for temperature range -50° C to +50° C.
Other ranges on request.
AltitudeStandard design is suitable up to 1000 mabove sea level. Higher values on request.
MaintenanceMaintenance free, simple recommendations : Clean the porcelain bushing at regularintervals depending on the degree of pollution.Ensure that there is no oil leakage.
Spare PartThere are no user serviceable parts, therefore no spare parts are required.
Special ToolsNo special tools are required for
maintenance
EMEK DESIGN
56
AssemblyBefore the assembly, following visual checks should be made.ï There are no oil leaks,ï The porcelain is not damaged,ï The aluminium protective cover on the bellows is not damaged,ï There is no damage on the primary terminals,ï The oil level indicator indicates sufficient oil level.
Storageï It is recommended to store the transformers packed prior to installation.
Packingï The transformers are packed in rugged wooden crates which are suitable for overseas transportation.ï The transformers are encapsulated in polyethylene covers to keep them from environmental effects such as rain-fall, dust and sand etc. (during transport and storage).ï Wooden crates can be lifted by either a fork lift or by a crane with slings.
EMEK DESIGN
57
TOP
EMEK
TOP
EMEK
TOP
EMEK
max 15°
CORRECT MAX INCLINATION WRONG
TRANSPORTATIONTransportation is only possible in the upright (vertical) position. The CT should never be
transported or stored in the horizontal position. The crates should not be put on top of each other. Attention should be paid to the red ìUPî arrows on the crate.
For transformers which have been transported and/or stored in horizontal position for any reason, factory level re-conditioning and testing is necessary.
The vehicle used for transportation must be of adequate size. The packing must be tightly tied to the chassis of the transportation vehicle.
If unpaved and rough roads are to be travelled, care should be given to prevent rigorous vibration. Prolonged and rough vibration may lead to oil disturbances.
EMEK DESIGN
58
NEVERNEVERUse the primary terminalsfor lifting the transformers
Lift the transformersby using the holes
provided on the base
EMEK DESIGN
59
CASCADED TYPE (245-420 KV)
CASCADED TYPE CURRENT TRANSFORMERS
In this part we will explain basic technical characteristics of cascaded type current transformers.
The CTís have both dielectric and magnetic cascade arrangements. The upper part and lower part have independent oil compartment therefore no oil connection is made between the units. Each unit has a magnetic circuit, high voltage ( high amps ) and low voltage ( low amps ) winding, insulation, oil and stainless steel bellows for oil volume compansation. In other words each unit is an independent CT itself such that when they are connected in series they share the line voltage and the turns ratio of the cascaded CT is the combination of turns ratio of each unit.
No connections are necessary between the two units. Each unit has its own stainless steel bellow for compansating the change in the oil volume. A simple electrical connection is made between the upper part and lower part.
Cascade design CTís are not new. They have been around since system levels of 400kV and above have been in use. They are feasible for levels of 245kV and above. For lower levels the single piece design has an economic advantages.
60
1600 A
150 A
150 A
1 A 1 A
U P
P E
R P
A R
TL
O W
E R
P A
R T
Upper part primary winding
Upper part secondary winding
Lower part primary winding
Lower part secondary winding
Cores
Cores
SINGLE LINE DIAGRAM OF 1600/1-1 A
CASCADECURRENT TRANSFORMER
U P
P E
R P
A R
TL
O W
E R
P A
R T
CASCADED TYPE (245-420 KV)
61
1. Aluminium Rain Cover2. Stainless Steel Bellows3. Oil Level Indicator4. Primary Terminals5. Aluminium Head6. Primary Winding7. Insulating Oil8. Paper Insulation 9. Porcelain Bushing10. Secondary Support11. Oil Sampling Valve 12. Stainless Steel Bellows13. Paper Insulation 14. Porcelain Bushing15. Secondary Cores16. Insulating Oil17. Lifting Holes18. Secondary Terminal Box19. Oil Sampling Valve
1
2
3
5
4
6
8
9
11
15
12
13
16
14
19 18
17
10
7
CASCADED TYPE (245-420 KV)
62
CASCADED TYPE (245-420 KV)
Cascade type current transformers are made up from two individual and independent CTís which are
connected in series.
The upper and lower part have independent oil compartments therefore no oil connection is made
between the units.
Each unit has its own stainless steel bellow for compensating the change in the oil volume.
Minimum ferrous surfaces are subjected to the environment to minimise corrosion. Ferrous surfaces
are hot dip galvanised and painted.
The height of the units is suitable for upright (vertical) transport on normal trucks and they will
pass under standard bridges. This way complications of horizontal transport are avoided.
63
Both units are assembled on site. Simple electrical connection is made between the upper part and the lower part. No special
tools are required for connection.
CASCADED TYPE (245-420 KV)
64
The line (and test) voltage is shared
equally by both units. The insulation
task of each unit is actually half of the
task of a single piece unit. Therefore
there is considerably less strain on the
insulation. Less insulation strain means
longer life.
V
0
V/2
275/√3/2 = 137/√3 KV
275/√3/2 = 137/√3 KV
CASCADED TYPE (245-420 KV)
65
Better Short Circuit Characteristics : Only the upper part is prone to thermal and dynamic effects resulting from a short circuit. The bottom unit is never effected with thermal and dynamic forces because the upper part saturates around 30 to 50 times ( according to core characteristics ) the rated current and anything above this level is not passed onwards to bottom unit.
Because of its short height, the upper part is not severely effected from a short circuit. The primary conductors carrying the short circuit current is far shorter than the ones in a single piece hairpin design. Shorter conductor means less heating and smaller dynamic forces.
40 KA / 1 sec
CASCADED TYPE (245-420 KV)
7.5 KA is the short timecurrent pass through bottompart which is negligibablecompare to 40 KA
Core saturate at 50 times of rated secondary 50 x 150 A = 7.5 KA