Instrument Transformers Presentation2011-2

7/28/2019 Instrument Transformers Presentation2011-2

1/104

Instrument Transformers

For currents greater than 100A and

voltages higher than 500V, it is difficult to

construct ammeters and current coils of

wattmeters, energy meters and relays

carrying alternating currents greater than

100A.

Specially designed transformers knownas instrument transformers are usedfor this purpose.


2/104


3/104

Agenda

Current TransformersVoltage Transformers

Required Information for Specifying CTs & VTs

Take Home Rules for CTs & VTs

Applications


4/104

Why use Instrument Transformers?

Circuit Isolation

Reduce voltage and currentsto reasonable working levels.

Phasor combinations for

summing and measuringpower


5/104

CURRENT TRANSFORMERS

TYPES OF C.T. CONSTRUCTION

The most common type of C.T. construction is the

DOUGHNUT type. It is constructed of an iron toroid, which

forms the core of the transformer, and is wound with secondary

turns.

Secondary Winding Primary Conductor

Iron Core


6/104

Transformer ratio (TR)

Primary Current

(100 amps)

Secondary

Current

(5 amps)

Primary Current

Secondary CurrentTransformer Ratio =___________________

__

100

5___= 100:5 or 20:1


7/104

Polarity

Direction ofPrimary Current

Direction of

Secondary Current

H1

X1

Primary current into polarity =

Secondary current out ofpolarity

P1

IEEE

IEC

PrimaryPolarity

Marks

IEEE

IECS1

Secondary

Polarity

Marks

Remember:


8/104

Polarity

Direction of

Primary Current

Direction of

Secondary Current

H1

X1

P1

IEEE

IEC

Primary

Polarity

Marks

IEEE

IECS1

Secondary

Polarity

Marks

Primary current into non-polarity =

Secondary current out ofnon-

olarity

Remember:


9/104

Generator typical wiring


10/104

CT Rating Factor (RF) -- IEEE

Rated current x (RF) =Maximum continuous current carrying

capability:

Without exceeding temperature limits

Without loss of published accuracy

class

Typical rating factors for Metering CTs are:

1.0, 1.33, 1.5, 2.0, 3.0, 4.0


11/104

Short-Time Thermal Current Rating

One (1) second thermal ratingExpressed as value of RMS primary current

Main influencing factor:

CT primary & secondary wire size

Can be converted to thermal rating for any

time period (t) up to five (5) seconds:

(New RF at new Ambient/Stated RF at 30 degrees C)2=

85-New Ambient/New Ambient

Example: CT with rating factor of 4 at 30 degrees = rating factor of 2.95 at55 degrees

X2/42=85-55/55 X=2.95


12/104


The instrument transformers in a sub-

station are:

Current transformers:

Voltage transformer.

Capacitive voltage transformers (CVTs )

Voltage Transformers ( IVTS or PTs )


13/104

Importance of CTs &PTs

Many protection systems are required to

operate during the period of transient

disturbance in the output of the

measuring transformers that follows asystem fault.

The errors in transformer output may

abnormally delay the operation of theprotection, or cause unnecessary

operations.


14/104

Importance of CTs &PTs

Whenever the values of voltage orcurrent in a power circuit are too high topermit convenient direct connection of

measuring instruments or relays,coupling is made through transformers.

Such 'measuring transformers arerequired to produce a scaled downreplica of the input quantity to theaccuracy expected for the particularmeasurement


15/104

To provide insulation between primaryand secondary circuit for equipment and

personnel safety

To change the magnitude (but not thenature) of the quality (voltage or current)

being measured to a suitable level for

use with standard instruments (protectiverelays, metering equipment, etc).

Basic Function of Instrument

Transformers


16/104

Instrument Transformers (DO)

The quality of instrument transformers

willaffect directly the overall accuracyand performance of these metering

and monitoring systems.

Instrument transformer performance is

critical in protective relaying, since the

relays can only be as good as theinstrument transformers.


17/104

Basic Function of Instrument

Transformers

By using instrument transformers,electrical instruments have beenstandardized to operate on 110V and 5A

or 1A. They are essential parts of many

electrical systems, and are used for

Measuring (metering) andMonitoring (relaying) devices.


18/104

Normally both the above functions arecombined in one unit in such apparatusused in power systems. Hence the

general terminstrument transformers.

There are occasions where these areused exclusively for commercialmetering and in which case they arecalled Metering Transformers.


19/104

Types of Instrument

Transformers

Instrument transformers are of twotypes, depending upon whether it is used

to excite the current or voltage coil of the

measuring instrument

Current Transformers- CTs

Voltage Transformers VTs (also

referred to as Potential Transformers,

PTs).


20/104

Types of Instrument

Transformers (Contd)

Both of these types act as insulatorsbetween high-voltage primary and low-voltage secondary.

The ratio of primary to secondary voltageis in proportion to the turns of ratio and

will usually produce 110-120V at thesecondary terminals with rated primaryvoltage applied.


21/104

Accuracy of Instrument

Transformers

To be a useful part of a measurement system,

instrument transformers must change the

magnitude of the quantity being measured

without introducing any excessive unknownerrors.

The accuracy of an instrument transformer

must either be of a known value, so that errors

may be allowed for, or the accuracy must besufficiently high that errors introduced by the

instrument transformer may be ignored.


22/104

Factors Affecting Accuracy of


Design of the instrument transformer

Circuit conditions such as voltage,current and frequency

Burden connected to the secondarycircuit of the transformer


23/104

Burden of Instrument

Transformers

In instrument transformer operations, the

primary quantities are reduced by the

turns ratio to provide a secondary current

or voltage to energize protective relaysand other equipment.

The totality of the impedances of the

loads connected to current or voltagetransformers are referred to as burden.

B d f I


24/104

Burden of Instrument

Transformers (Do)

The burden consists of the impedances

of the following:

Secondary winding of the instrument

transformer

Interconnecting leads

Relay and/or other connected devices.


25/104

connections


26/104

Current Transformers - CTs

Current Transformers are used whenever

the magnitude of the operating current

has to be reduced to the value for which

instruments, meters and protectivedevices are designed. At the same time

current transformers isolate metering and

protective devices from the systemvoltage.


27/104

CTs

The essential requirement of a currenttransformers is to deliver on its secondarya quantity(current), which truly represents

the applied quantity on its primary. Thefailure of protective system to perform itsfunction correctly is due to incorrect

application of this transformers.


28/104

CTs

The requirements of a protective currenttransformer are quite different from thatof a metering C.T. The metering C.T. is

only required to perform its functionover the normal range of load current,while the protective C.T is required to

give satisfactory protection over a widerange of fault conditions.


29/104

CTs

Working ranges of CTs


30/104

CTs Characteristics

The curve above describe a characteristic,

indicating three regions namely:

(i) Ankle point

(ii) Linear or straight line region(iii)Knee point

The working range of a metering C.T., isfrom the Ankle point to the Knee point and

slightly beyond it.


31/104

CTs Characteristics

Thus the metering C.T., operates between 10%and 120% of the rated current and saturatesbeyond this in order to protect the metering

instruments.The working range of a protective C.T. extendsover the full range from the ankle point andbeyond. Generally the operating region of aprotective C.T. is beyond the knee point as it isrequired to operate at fault currents, which isseveral times the full load or rated current.


32/104

CTs Characteristics

The knee point voltage of a metering C.T. isgenerally around 60 to 120V and is keptlow so as to protect meters.

The knee point voltages of protective C.T.saregenerally quite high varying from 200V to1900V depending upon the requirements of therelay. The upper limit of 1900V is specified

because the secondary cables from a C.T. aregenerally rated to withstand 2KV for about 1 or3 minutes and 660 volts or 1100 voltscontinuously.


33/104

CTs Characteristics

The excitation voltages of metering and protective C.Ts is as follows:


34/104


35/104

Theory of Current Transformers

The current transformer operates likeany other transformer in that the voltage

ratio and the reciprocal of the current ratio

are proportional to the turns ratio i.e.Ep = Np

Es NsWhere: p and s denote primary and secondary.

E VoltageI Current

N Number of turns


36/104


The primary winding is connected inseries with the load and it is the latter

which determines the current induced in

the secondary winding. The secondary is connected to a burden,

which does not vary, and the primary

current is not influenced by the magnitudeof the secondary burden.


37/104


The current in the secondary isdetermined by the current in the primary

winding.


38/104

Phasor diagram of a C.T.


39/104


The secondary currentIs lags behind thesecondary induced voltage, Es by an

angle . This angle is determined by the

impedance of the external burden and theimpedance of the secondary winding.

The primary currentIp is the resultant of -

Is andIo the exciting current. The exciting

currentIo consists of two components

namelyIcthe core-loss component andIm

the magnetizing component.


40/104


The angle between Ip and (-Is) is thephase displacement error between theprimary and secondary currents. This

angle is expressed in minutes of arc.

The difference in lengths between Ip and

(-Is) is called the Ratio Error. When this

ratio error is expressed as a percentage of

the primary current Ip, it is calledPercentage Ratio Error.

Kn = Ip = N2 (the transformation ratio).

Is NI

Common definition of terms used


41/104

Common definition of terms usedwith Current Transformers

Composite error C Composite error C is the difference

between the ideal secondary current and

the actual secondary current under steady-state conditions. It includes amplitude

(Ratio) and phase errors and also the

effects of any possible harmonics in theexciting current


42/104

Common definitionThusEc= 1001o

t[(Kn is - ip)2dt]Ip T

Where

Ec is the composite errorT the time of one periodIp the rms value of the primary current in Ampsip the instantaneous value of the primary current in Amps.is the instantaneous value of the secondary current in

Amps.Kn the transformation ratio = Ip

Is


43/104

Common definition

At rated frequency and with rated burden connected, theamplitude error and phase error and composite error of a CT shall

not exceed the values given in the following table.

Marking: The accuracy class of a CT is written after the rated power. E.g.

10 VA 5P10, 15 VA 10P10, 30 VA 5P20


44/104


45/104


46/104


47/104


48/104


49/104

Common definition

Rated BurdenThis is the apparent resistance of the secondary circuit

expressed in ohms together with the power factor for

which the specified accuracy limits are valid.

Burden Zb = rb + jXb

Secondary winding impedance Zs = rs + jXs

Total secondary impedance Zt = Zb + Zs

= (rb + rs) + j(Xb+Xs)

Es =Is (Zb + Zs)


50/104

Common definition

Rated OutputThe rated output of a current transformer is theapparent power expressed in VA together with thepower factor, which the C.T. can deliver to thesecondary circuit at rated current and burden while stillmaintaining its accuracy in the specified class.

The rated output is equal to the product of the ratedsecondary current and the voltage drop in the

external secondary circuit due to this current.The standardised values of rated outputs are 2.5, 5, 7.5,10, 15, 30, 45, 60, 90, & 120 in VA.


51/104

Common definition

In BSS, the VA output is specified along

with the accuracy class. For example

30 5P 10 means a protection C.T. of

accuracy class s having a total error of5% with a VA of 30. The number 10 is

the ALF defined later However in IEC,

the VA is specified separately.


52/104

Common definition

Accuracy Limit Factor (ALF) The accuracy limit current is the highest

primary current at which a current

transformer still meets the specifiedrequirements as regards total error. Theaccuracy limit factor is the ratio of theaccuracy limit current to the rated

primary current. The standardized accuracy limit factors

are 5, 10, 15, 20 and 30.


53/104

Common definition

For example, as per IEC 5 p 20 means aC.T. for protection having maximumtotal error of 5% at 20 times the rated

current. Marking: Accuracy limit factor is written

after the accuracy class.

E.g. 10 VA 5P10, 15 VA 10P10, 30 VA5P20.


54/104

Common definition

Instrument Security Factor (ISF)The rated instrument security factor is thesmallest primary current at which an

instrumentation core exhibits a currenterror of 10%.The Instrument Security Factor ISF is the

ratio of the rated instrument safety currentto rated primary current.


55/104

Common definition

The instrument security factor defines thebehavior of a metering C.T. core underover-current conditions. The ISF is

specified to protect instruments connectedto the metering C.T. core from systemshort circuit currents. The ISF to bechosen should be as low as possible.It is expressed as a number n 5 or n10.


56/104

Common definition

Knee point voltage (Vk)This is the sinusoidal e.m.f of ratedfrequency applied to the secondary

terminals of the C.T., with all otherwindings being open circuited, which whenincreased by 10% causes the exciting

current to increase by 50% or more. Thisis illustrated below:

f


57/104

Common definition


58/104

Common definition

ExampleV1= 100 VV2= 110 V

Percentage Increase = 10%

Corresponding currents C1= 0.35AC2= 0.7A

Percentage increase = 50%V is the knee point voltage Vk.

The knee point voltage indicates the voltage above which

the C.T. enters into saturation and exciting current

increases rapidly with a very little increase in voltage


59/104

Common definition

The Vk is also limited by practicaldesign and manufacturing considerationas:

Vk = Rated output in VA x ALFSecondary rated current

d fi i i


60/104

Common definition

Rated Short Time Thermal Current (Ith)This is the rms value of the primary current,which the C.T. will withstand for one secondwithout suffering any internal damage or otherharmful effects with the secondary being short-circuited.

This rating is for a very short time and it is

usually assumed that the entire heat generatedis stored in the primary winding itself.

C d fi i i


61/104

Common definition

Rated short time thermal current isexpressed in KA. It is related to themaximum short circuit current at the

point of installation of the C.T., and alsoon the duration of the breaking time ofthe short circuit current

C d fi i i


62/104

Common definition

The following condition should be met with

Ith Iscx[t + 0.05 x 50] KA rms.f

WhereIth- Rated short time thermal current for 1 sec.Isc- Short circuit current at C.T. location in KA rmst - short circuit duration in sec.f - Rated system frequency.

For system frequency of 50 Hertz

Ith Isc[t + 0.05] KA rms.

C d fi iti


63/104

Common definition

Rated Dynamic Current (I dyn)It is the peak value of the primary current, whichthe transformer will withstand without

being damaged electrically or mechanically by theresulting electromagnetic forces, the secondarywinding being short- circuited.The maximum value of this current can be 2.5times the rated short time thermal current(Ith)

I dyn = 2.5 Ith

C d fi iti


64/104

Common definition

Rated Primary and Secondary CurrentThese are the values of the primary and secondarycurrent on which the performance of the currenttransformer is based.

Standard values of primary currents are:5, 10, 15, 20, 30, 60, 75, 50, 100, 150, 200, 300, 400,600, 800, 1000, 1500, 2000, 3000, 4000 and above.

Standard values of secondary currents as per BS 3938are 5A, 2A and 1A and as per IEC, 5A or 1A. Howeverthere are cases where occasionally ratings of 0.577A,0.866A or 2.87A have been used.


65/104


66/104

Tips on CT selection

The selection of the primary current of a C.T. shall always beadopted as closely as possible to the full load or rated current of theinstallation by rounding off to the next higher standard. However theC.T must be capable of continuously carrying the maximum expectedcurrent in service. It is advisable to consider a permitted overload of20% of the full load current while deciding the rated current.

Another factor to be considered is also the load growth and theincrease in capacity of an installation. It is for this reasonthat multi ratio primary currents are adopted like 800- 400 - 200 - 100 A.

The selection of the secondary current depends upon the secondarycurrent of the equipment already in service where interchangeability isa consideration.


67/104

The following are the advantages and disadvantages ofCTs with 5A and 1A secondary currents.

The number of turns required on the secondary side is less

for a 5A C.T. than for a 1A C.T. for a given primary current.

A thicker gauge wire is required for a 5A C.T than for a 1AC.T.

Both the above factors contribute to the cost reduction of a

5A C.T. when compared to a 1A C.T.

Since the number of turns is less for a 5A C.T, the voltage

induced on the secondary side during secondary saturation

or secondary open circuit is less when compared to a 1A

C.T.


68/104

The lead burden, however, becomes excessive for a 5A C.Tsince the same is proportional to the product of the square

of the current and resistance of the lead wire. The lead

burden in a 1A CT. will be very low.

In view of the reduced number of secondary turns in a 5AC.T., it is difficult to provide for turns compensation to

design and manufacture low current higher accuracy class

CTs. However in a 1A C.T. it is possible to achieve

the desired accuracy class because of the increased

number of turns and by providing compensating turns. The internal resistance of a 5A C.T. is comparatively less (

1 ohm) when compared to that of 1A C.T. (Generally 3 to

12 ohms)

E l


69/104

Examples

A case study of Estimation of Burden, Knee PointVoltage, Accuracy Class etc of a Protective CurrentTransformer

Requirement of a C.T. to protect a 15 MVA, 132/33 KVDelta/Star connected transformer.

Data available

% Impedance of Transformer = 10Fault level at 132KV side = 1400 MVA

E l


70/104

Examples

Transformer full load current per phase= 15 x 106_____

3 x 132 x 103= 65.61 A

Hence select primary current = 100 A

i.e. Ip = 100 A

Select secondary current Is as 5A. A 5A C.T secondary

has a winding resistant of less than 1.0 ohm. A typical

value may be chosen as 0.601 ohms.

E l


71/104

Examples

Assume(a)Distance from C.T to Relay control panel as 100 metres and

C.T. secondary leads of 10 sq mm. (RL = 0.1627 ohms for 100

metres)

(b)Connected relays are GEC CDG 11 over-current and earthfault relays with VA burden of 1.8 and 4 respectively.

E l


72/104

Examples

Relay burden =IS2RS + 2IS

2RL + VA of (OCR + EFR)

=(5) 20.601 + 2(5) 20.1627 + (1.8 + 4)

= 15.0 + 8.135 + 5.8= 28.935 VA

Hence select relay burden or output as 30 VA

Select Accuracyclass 5 P 20

Examples


73/104

Examples

Knee point voltage

Vk = VA x ALF__

Sec. current

= 30 x 20

5

= 6005

= 120 V

Examples


74/104

Examples

Fault current at C.T. installation = 1400 x 106

___3 x 132 x 103

= 6123.6 A

or 6.124 KA = Isc

IthIsc[t + 0.05] KA rms for 1 sec

Assume operating time of breakers, relays etc = 1 sec

Ith6.124[1.05]6.275 KA rms

Select Ithas 10 KA rms for 1 sec

Examples


75/104

Examples

Ith short time rating = 10 KA rms for 1 sec.

Idyn = 2.5 Ith =2.5 x 10 = 25 KA for 1 sec.

Hence complete specifications for this protection C T will be:

Voltage class: 132 KVPrimary current: 100 A

Highest System Voltage: 145 KV

Secondary current: 5 A

Accuracy class: 5 P 20

Vk: 120 V.Ith: 10 KA rms for 1 sec.

Idyn: 25 KA for 1 sec.

Polarity and Markings


76/104




77/104



78/104


79/104


80/104

Voltage Transformers


81/104


Types of VTs

Electromagnetic VT

Capacitive VT

Types of VTs for Protective


82/104

Relaying.

Voltage transformers have woundprimaries that are

Either connected directly to the power

systems (VTs)

Or across a selection of capacitor string

connected between phase and ground,

that is, coupling-capacitor voltagetransformers (CCVTs)

VTs For Relay Applications


83/104

VTs For Relay Applications

Voltage transformers, which step downsystem voltages to sufficiently low, safer,

measurable values, are required for

Indication of the voltage conditions. Energy meters and watt meters (kWh

and kW meters)

Protective relays Synchronizing

Points To Note About VTs


84/104

Points To Note About VTs

VTs are used at all power system voltages,andare usually connected to the bus.

Usually the CCTVs are connected to the line,

rather than to the bus, because the couplingcapacitor devices may also be used to couple

radio frequencies to the line for use in pilot

relaying

At about 115kV, the CCVT types becomesapplicable and generally more economical

than VTs at the higher voltages.

Points to Note About VTs (Do)


85/104

Points to Note About VTs (Do)

Either type of voltage transformer (VT orCCVT) provides excellent reproduction ofprimary voltage, both transient and steady-state, for protection functions.

Saturation is not a problem because powersystems should not be operated above normalvoltage, and faults result in a collapse orreduction in voltage.

VTs are normally installed with primary fuses,which are not necessary with CCTVs. Fusesare also used in the secondary.



86/104


HV and EHV Capacitor-coupled VT (CVT)

C1 & C2 are adjusted, so that a few kVs of voltage is obtains

across C2

Then, stepped down by T


87/104

Example: To calculate the capacitance requirements for aCVT to be used on a 132KV system.

Let (1) Total capacitance of capacitor be 20,000pF

(2) Burden requirement 100 VA

(3) Magnetic transformer designed for a standard primary

voltage of 10/3 KV


88/104

C1 = E2C2 E1

= 10/3_________

132/3 10/3

C1 = _10____ x C2 = 10_ x C2

132 10 122

or C2 = 122 C1

10

Also 1_ + 1 = 1C1 C2 C

Or C = C1C2_

C1 + C2


89/104

Substituting for C2 in the above eqn.

C = C1 x 122 C1

10____

C1 + 122 C1

10

= C12 x 122

10____10C1 + 122C1

10

= 122 C1

132

C1 = 132 C

122= 132 x 20000 = 21639.34pF

122

C2 = 122 x 21639.34 = 264000pF

10


90/104


91/104


92/104


93/104


94/104


95/104


96/104


97/104


98/104


99/104


100/104


101/104


102/104


103/104


104/104

Documents

Instrument Transformers Presentation2011-2