RELAY OPERATION PRINCIPLES

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2.1 Detection of fault :- In general, as faults (short-circuit) occur, current increase in magnitude, and

voltages go down. Besides these magnitude changes of the ac quantities, other changes may occur in one or more of the following parameters:

Phase angles of current and voltage phasors. Harmonic components. Active and reactive power. Frequency of the power system. Relay operating principles may be based upon detecting these changes, and

identifying the changes with the possibility that a fault may exist inside its assigned zone of protection . We will divide relays into categories based upon which of these input quantities a particular relay responds to :-

Level detection :- this is the simplest of all relay operating principles. Fault current magnitudes are almost always greater than the normal load current that exist in a power system.

As an example, consider the motor connection to a 4kv power system. The full load current for the motor is I=245 ampers.

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Allowing for an emergency overload capability of 25%, a current of I =1.25×245=306 amperes or lower should correspond to normal operation . Then any current above a set level(346 amperes ) may be taken to mean that a fault, or some other abnormal condition exists inside the zone of protection of the motor. The relay should be designed to operate and trip the circuit breaker for all currents above the setting value.The level above which the relay operate is known as the pickup setting of the relay . For all currents above the pickup , the relay operates, and for currents smaller than pickup value, the relay takes no action. The operating characteristic of an overcurrent relay can be presented as a plot of the operating time of the relay versus the current in the relay. It is best to normalize the current as a ratio of the actual current to the pickup setting(Ip).

In=𝐼

𝐼𝑃,the operating time for In less than 1 is infinite, which for values>1 ,

the relay operates. The actual time for operation will depend upon the design of the relay.

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The characteristics show the ideal and practice relay level detector relay also known as over current relay ( OV relay )

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b) Magnitude comparison

This operation principle is based upon the comparison of one or more operating quantities with each other. For example , a current balance relay may compare the current in one circuit with the current in another circuit ,which should have equal or proportional magnitudes under normal operating conditions . The relay will operate when the current division in the two circuit varies by a given tolerance .

The figure below shows two identical parallel lines which are connected to the same bus at either end . One could use a magnitude comparison relay which compares the magnitude of the two line current 𝐼𝐴 > 𝐼𝐵 + ϵ

Then with the line B is not open , would declare a fault on line A and trip it ,where ϵ is a suitable tolerance .

Similar logic would be used to trip line B .

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c) Differential comparison

It is one of the most sensitive and effective methods of providing protection against the faults . Consider the generator winding shown in the figure , then as the winding is electrically continuous , current entering one end I1`must equal the current leaving the other end I2` .

One could use a magnitude comparison relay described below to test for a fault on the protected winding .

When a fault occurs between tow ends the two current are no longer equal alternatively , one could from an algebraic sum of the two current entering the protected winding ,(I1-I2) and use a level detector relay to detect the presence of a fault. In either case , the protection is termed a differential protection . In general , the differential protection principle is capable of detecting very small magnitudes of a fault current .

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Thy only drawback of this technique it requires current from both ends of zone of protection which restricts its application to power apparatus ( Transformer , Generators , Motors , buses , capacitors , reactors …) .

d) Phase angle comparison

This type of relay compares the relative phase angle between to Ac quantities . Phase angle comparison is commonly used to determine direction of a current ( w.r.) to a reference quantity . For instance , the normal power flow in given direction will result in the phase angle between the voltage and the current varying around its power factor angle say approximately ±30˚ . When power

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flow in opposite direction ,this angle will become (180˚±30˚)

similarly , for a fault in forward or revers direction the phase angle of the current ( w.r.t ) the voltage will be –Φ and ( 180˚-Φ ) respectively ( Φ is impedance angle of the fault circuit ) as shown in figure .

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This difference in phase relationships created by a fault is given by making relay which response to phase angle differences between two inputs ( fault voltage and fault current ) .

E ) Distance Measurement :-As discussed above , most positive and reliable type of protection compares the current entering the circuit with the current leaving the it on transmission lines and feeders , the length ,voltage, and configuration of the line may make this principle uneconomical. Instead of comparing the local line current with the far end line current ,the relay compares the local current with the local voltage .

This in effect , is a measurement of the impedance of the line as seen from the relay terminal .

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An impedance relay relies on the fact that the length of the line (its distance ) for a given conductor diameter spacing determines its impedance .

F ) pilot relaying :-Certain relaying principles are based upon information obtained by the relay from a remote location , the information could be in the form of the contact status ( open or closed ). The information is sent over a communication channel using power line carrier , microwave , or telephone circuits .

G ) Harmonic content :-Current and voltage in a power system usually have a sinusoidal waveform of the fundamental power system frequency . There are however , deviations from a pure sinusoid , such as the third harmonic voltage and currents

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produced by generators , that are present during normal system operation , other harmonic occur during abnormal system conditions , such as the odd harmonics associated with transformer saturation or transient components . These abnormal conditions can be detected by sensing the harmonic content through filters in electromechanical or solid – state relays , or by calculation in digital relays. After the detection of these harmonics , a decision can be made and control action is required .H ) Frequency sensing :-Normal power system operation is at 50 or 60 Hz . Any deviation from these values indicates that a problem exists . Frequency can be measured by filter circuit , by counting of

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zero crossing wave forms in a unit of time , or by special sampling and digital computer technique frequency – sensing relays may be used to bring system frequency back to normal by some corrective actions .

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Relay designThe various input quantities described above , upon which fault detection is based ,may be used either single or in any combination to calculated power ,power factor , directionality , impedance ….. Some relays are also designed to respond to mechanical devices such as fluid detectors , pressure or temperature sensors . Relay may be constructed from electro mechanical element such as solenoids , induction discs , solid – state elements , digital computers using analog – to – digital converter and microprocessors . However , the construction of a relay does not inherently change the protection of a relay dose not inherently change the protection concept , although there are advantages and disadvantages associated with each type .

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Some relay design aspects are :

Relays are devices requiring low level inputs ( voltages , currents , or contacts ).

Derive their inputs from transducers such as current or voltage transformers and switch contacts .

They are fault detecting devices only and require an associated interrupting device ( a circuit breaker ) to clear the fault .

Separating the fault detection function from the interruption function gave the relay designer an ability to design a protection system that match the needs of the

power system .

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Fuses The fuse is the oldest and simplest of all the

protective device . The main characteristics are :-

1. It is a level detector .

2. It is both the sensor and the interrupting device .

3. It is installed in series with the equipment beingprotected .

4. Operating by melting a fusible element in response to thecurrent flow . The melting time is inversely proportionalto the magnitude of the current flowing in the fuse .

5. It is one – shot device since the fusible link is destroyed inthe process of interrupting the currents.

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6. Fuse may only be able to interrupt current up to their maximum short–circuit rating before it reaches its maximum value .

7. Its application is restricted for radial feeders such as distribution lines or auxiliary systems of power plants .

The two major dis advantages are :-

a) The single –shot feature requires that a blown fuse be replaced before service can be restored ,this means a delay ,and the need to have the correct spare fuses . To avoid this, it is possible a multi-shot feature by installing a number of fuses in parallel with a control unit to transfer to another one.

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b) In a three-phase circuit ,a single phase-to- ground fault will cause one fuse to blow, de energizing only one phase, permitting the connected equipment-such as motors to stay connected to the remaining phases , with excessive heating and vibration because of the unbalanced voltage supply .

2.3 Electromechanical Relays :-

The early replay designs utilized actuating forces that were produced be electromagnetic interaction between currents and fluxes. Some relays were also based upon the forces created by expansion of metals caused by temperature rise due to a flow of current. Two main types of electromechanical replays and in use , the plunger-type relays are usually driven by a single actuating quantity , while the induction –type relays may be activated by single or multiple inputs .

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a) plunger –Type Relay Consider a round moving a stationary electromagnet as shown in the figure. With no current in the coil, the plunger is held partially out side the coil by the force Fs produced by a spring. Let X be the position of the plunger tip inside the upper opening of the coil. When the coil is energized by a current i , and saturation phenomena are neglected ,the energy W(λ, i) and the co-energy w1(i,x),stored in the magnetic field are given by :

𝑤 𝜆 , 𝑖 = 𝑤1 𝑖, 𝑥 =1

2𝑙𝑖2 ______(2.1)

Where λ is the flax linkage of the coil and L is the inductance of the coil which can be given for this magnetic circuit as

𝐿 =𝛍 ̻𝜋 𝑑2𝑁2

4(𝑋+𝑔𝑑

4𝑎)

, N :number of turns in the coil

a: height of the pole –face

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The force which tries to pull the plunger inside the coil given by

𝑓𝑚 =𝛅

𝛅𝑥w` 𝑖, 𝑥 = 𝑘

𝑖2

𝑥+𝑔𝑑

4𝑎

2 ……………(2.2)

Where k is a constant depending upon constants of the magnetic circuit shown on the figure .The plunger moves when 𝑓𝑚 exceeds 𝑓𝑠 .if the current is sinusoidal with an rms value of I ,the average force is proportional to I² . The valve of the current (𝐼𝑝) at which the plunger just begins to move is known as the pickup setting of the replay is given by :

𝑓𝑚 = 𝑓𝑠 = 𝑘𝐼𝑝²

( 𝑥+𝑔𝑑

4𝑎)²

then

𝐼𝑝=[𝑓𝑠

𝑘] x +

𝑔𝑑

4𝑎………… ……(2.3)

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Where x is the displacement of the plunger when no current is flowing in the coil (stating position ). The plunger travels some distance from x to x1 before it closes its contacts and hits a stop . The energizing current must drop below a value 𝐼𝑑, known as the dropout current before the plunger can return to tis original position x . The dropout current is given by :

𝐼𝑑=[𝑓𝑠

𝑘] 𝑥1 +

𝑔𝑑

4𝑎………… ……(2.4)

𝐼𝑑<𝐼𝑝 𝑎𝑠 x < 𝑥1

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The operating time of the relay depends upon the mass of the plunger .For a normalized current of magnitude 𝐼𝑛 (actual current divided by the pickup current ), the accelerating force on the plunger is :

F=𝐹𝑚 − 𝐹𝑆 = 𝑘𝐼𝑛 𝐼𝑃

2

𝑋+𝑔𝑑

4𝑎

2 − 𝐹𝑠 , substituting for 𝐼𝑝 from equation

(2.3) then ,

F= 𝐹𝑠[(𝑥0+𝑔𝑑/4𝑎)²

(𝑥+𝑔𝑑

4𝑎)²

𝐼𝑛² − 1] …………(2.5)

The equation of motion for the plunger is ,M 𝑋= -F …………….(2.6)Where m is the mass of the plunger.Equation (2.6)can be integrated twice to provide the operating time of the relay (the time it takes the plunger to travel from 𝑥 𝑡𝑜 𝑥1). The integrals in eqn (2.6) are elliptic integrals and must be evaluated numerically for given displacements .

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The general shape of the replay characteristic list operating time plotted as a function of the current through the coil is as shown in this figure. Operating time calculation is given in the next example.Ex: consider a plunger-type relay with a pickup current of 5 amperes (rms).the pole face has a height of 1.5 cm ,while the spring holds the plunger 1cm out of the coil when the current is below the pickup value. The air gap g=0.2cm , and gd/4a=0.05 , let the spring force be a constant ,with a value of 0.001 Newton , and let the mass of the plunger be 0.005kg . Let the travel of the plunger be 3mm before it hits a stop and closes its contacts . Calculate :a)The dropout current to its pickup current b)The accelerating force on the plunger for 𝐼𝑛=2, and x=0.8 cmc)The operating time of the relay for closes its contacts.

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Sol: use the sam construction of plunger-type relay was given in section (2.3)then :

a)From eqns (2.3)and(2.4),we have for dimension in cm:-

𝐼𝑑𝐼𝑝=

(𝑥1 +𝑔𝑑4𝑎

)

(𝑥0+𝑔𝑑/4𝑎)=(0.7 + 0.05)

(1 + 0.05)= 0.714

b)From eqn (2.5),the accelerating force F is :

F=𝐹𝑠𝑥0+

𝑔𝑑

4𝑎

2

𝑥+𝑔𝑑

4𝑎

2 𝐼𝑛2− 1 = 0.001

1.05 2

𝑥+0.005 2 2 2 − 1

F=0.001 4.41

0.7225− 1 = 5.1 × 10−3N.

c)Using eqns (2.5)and (2.6) them

0.005 𝑥 =-0.001[(1.05)²

(𝑥+0.005)²𝐼𝑛² -1 ]= -F

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The operating time can be calculated using a constant force equal to the average taken over its travel from 𝑥0𝑡𝑜 𝑥1 then from the above eqn use x= 𝑥0 =1cm and x= 𝑥1=0.7 cm and calculate F in both cases.

For x= 1cm

𝐹1 = 0.001( 𝐼𝑛² - 1 ) newton

For x= 0.7cm

𝐹2 = 0.001(1.96 𝐼𝑛² - 1 ) newton

Fav = 0.001(1.48 𝐼𝑛² - 1 )newton

Using this expression for the force ,then the approximate equation of motion for the plunger is :

0.005 𝑥 = −0.001[1.48 𝐼𝑛² - 1 ] 𝑥 = 0.2[1.48 𝐼𝑛² − 1 ]

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𝑥0𝑥1 𝑑𝑥 = 0.2[1.48 𝐼𝑛² - 1 ]

0

𝑡𝑑𝑡

t²=10(𝑥0−𝑥1)

1.48 𝐼𝑛² − 1

Or t =10(𝑥0−𝑥1)

1.48 𝐼𝑛² − 1=

0.3

1.48 𝐼𝑛² − 1=

0.3

0.48= 0.79 sec. approximately

This relation shows the inverse –time behavior of the relay for larger values of 𝐼𝑛 and can be drown for 𝐼𝑛 ≥ 1 .

Most plunger relays also have several taps available on the winding of the coil to adjust the pickup current over a wide range (tap setting 1,2,….10 amperes ). Also the pickup can be controlled by adjusting the plunger within the coil. Plunger –type relays will operate on dc as well as on ac current .

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b) Induction – type relays These relays are based upon the principle of operation of the single phase ac motor. As such ,they cannot be used for dc currents. There are two types of these fairly standard : one with an induction disc, and the other with an induction cup. In both cases, the moving element (disc or cup) is equivalent to the rotor of the induction motor. The figure below shows the principle of construction of an induction disc relay. The moving element acts as a carrier of rotor currents, while the magnetic circuit is completed through stationary magnetic elements. Induction type relays require two sources of alternating magnetic flux in which the moving element may turn. The two fluxes must have a phase difference between them, otherwise no operating torque is produced. Let us assume that the two currents in the coils of the relay, 𝑖1 𝑎𝑛𝑑 𝑖2 are sinusoidal :𝑖1 (t)=𝐼𝑚1cos wt , and 𝑖2 (t)=𝐼𝑚2cos (wt+θ)λ1 𝑡 = 𝐿𝑚 𝐼𝑚1cos wt , and λ2 𝑡 ² = 𝐿𝑚 𝐼𝑚2cos (wt+θ) where Lm is mutual inductance

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Each of these flux linkages in turn induces a voltage in the rotor, and since the rotor is a metallic structure with low self-inductance, a rotor current in phase with the inducted voltages flows in the rotor. Assuming the equivalent rotor resistance to be 𝑅1, the induced rotor current are given by :

𝑖𝑟1 𝑡 =1

𝑅𝑟

𝑑λ1

𝑑𝑡= -

ω𝐿𝑚𝐿𝑚1

𝑅𝑟sin ωt ………….(2.7)

𝑖𝑟2 𝑡 =1

𝑅𝑟

𝑑λ2

𝑑𝑡= -

ω𝐿𝑚𝐿𝑚2

𝑅𝑟sin (ωt+θ) ………….(2.8)

Each of the rotor current interacts with the flux produced by the other coil, producing a force. The two forces are in opposite direction w.r.t each other, and the net farce, or the corresponding net torque T is given by :

T [λ1 𝑖𝑟2-λ2𝑖𝑟1], substituting for λ,i and simplify to get

T= k 𝐼𝑚1𝐼𝑚2

[cos ωt+θ)-cos(ωt+θ)sin ωt] ……….(2.9)

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Or using trigonometric identity, the net torque is

T = k 𝐼𝑚1𝐼𝑚2

sinθ ………………(2.10)

Not that the net torque is constant in this case and dose not change with time. If the phase angle between the two coil currents is zero, there is no torque produced. By an appropriate choice of the source of the two coil currents, this relay could be made to take on the characteristic of :

A level detector (relay)

A directional relay

A ratio relay

For example, by using the same current to flow through the two coil, one could make a level detector. The phase shift between currents can be produce by placing in parallel with one of the coil a shunt with an impedance angel that is different from that of the coil, then the relay will produce a torque .

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For 𝐼𝑚1= 𝐼𝑚2

= 𝐼 𝑡ℎ𝑒𝑛 𝑒𝑞𝑛 2.10 reduced to

T = 𝑘1I² …………(2.11)Where 𝑘1=k sin θ

When the torque produced by the current (the pickup current of the relay) just exceeds the spring torque Ts, the disc begins to turn. After turning and angel φ ,the relay closes its contacts .

E 𝒙𝟎:- consider an induction disc relay, designed to perform as an overcurrent relay. The spring torque Ts is 0.001 Nm, and the pickup current of the relay is 10 amperes. The constant 𝑘1is 10−5 , the moment of inertia of the disc is 10−4kg-m²

a) Give an expression for the accelerating torque.

b) Develop the inverse relation of the operating time w.r.t the normalized current for angel of rotation (0__2˚)

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c) Calculated the operating time of the relay to close its contacts over rotation period of (0≤φ≤2˚).

Sal:-

a) From eqn (2.11), the magnetic torque is

𝑇𝑚 = 𝑘1 𝐼𝑛𝐼𝑝2= 10−5 10𝐼𝑛

2 = 10−3𝐼𝑛²

The accelerating torque on the disc is the difference between the magnetic torque and the spring torque :

T=𝑇𝑚 − 𝑇𝑠 = 10−3(𝐼𝑛² -1)

b) The equation of motion of the disc is

J θ=T where J :moment of inertia

10−4 θ = 10−3 (𝐼𝑛² -1)

(θ is angel of rotation of the disc o≤θ≤φ)

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For φ=2˚=0.035 radian, in tegreting the equation of motion twice to get θ=5(𝐼𝑛² − 1)𝑡² and the operating time of the relay is

t=0.035

5(𝐼𝑛²−1)sec

c) The operation time after relay closes its contact (𝐼𝑛 ≥ 1)take 𝐼𝑛= 1.01 then

t = 0.035

5[ 0.01 2−1]= 0.59 sec

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2.4 solid –state relays:-The expansion and growing complexity of modern power systems have brought a need for protective relays with higher level of performance and more sophisticated character is tics .this has been made possible by the development of solid-state or static relays where all of the functions and characteristics available with electromechanical relays has can be performed by solid-state disadvantages as follow:Advantages:-1-Use low power compare and devices.2-Springs and driving torques from the input quantities are not presented.3-High reliability than other electromechanical relays.4-Performance and economic advantages.5-More flexible and reduced size devices.6-More accurate in setting process.

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7-Their characteristics can be shaped by adjusting logic elements as the fixed characteristics of induction relays.

8-Solid-state relays are not affected by vibration or dust and require less mounting space.

Disadvantages:-

1- Limited capability to high temperature and humidity, over voltages and over currents .

2-It requires independent power supplies .

3-Solid-state relays are designed, assembled and tested as a system which puts the overall responsibility for proper operation on the manufacturer.

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Types:- Solid-state relay circuits may be divided into tow categories:a) Analog circuits that are either fault- sensing or measuring

circuits.b) Digital logic circuits for operation on logical variables.These circuits may arrange to provide desired relay characteristics such as : a) Solid – state instantaneous over current relays .b) Solid – state distance relays .We shall consider these as a circuit configuration and its principle of operation and characteristics as this are in use .a) Solid – state instantaneous overcurrent relays :Consider the circuit shown below . The input current I is passed through the resistive shunt R , full – wave rectified by the bridge rectifier , filtered to remove the ripple by the R-C filter , and applied to a high – gain summing amplifier A .

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The other input of the summing amplifier is supplied with an adjustable reference voltage er . when the input an the positive input of the summing amplifier exceeds the reference setting, the amplifier output goes high and this step change is delayed by a time-delay circuit in order to provide immunity against spurious transient signals in the input circuit. wave forms at various points in this circuit are shown in figure far an assumed input fault current of a magnitude above the pickup setting er of the relay. By making the time-delay circuit adjustable, and by making the mount of delay depend upon the magnitude of the input current, a time-delay overcurrent relay character is tic can be obtained b)Solid –state distance relays :-The type of relays are designed to be used as impedance, mho relays for the application of transmission lines protection. It measures the voltage and current at the location of the relay at on end of T.L. then the impedance, reactance or the distance to fault location is measured. The impendence characteristics is represented as R-X circle and the measured impedance is projected on this diagram to detect the fault and producing the trip or block signals. An analog circuit may be designed to measure the angle between the input current and voltage and then these is supplied to a logic circuit to get the tripping signal for the internal fault .

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2.5 computer relays :-With the advent of rugged, high performance micro process it is obvious that a digital computer can perform the same function as other techniques for detecting and decision. Since the usual relay input consist of power system voltages and currents, it is necessary to obtain a digital representation of these parameters. This is done by sampling the analog signals, and using an appropriate

Computer algorithm to create suitable digital representation of the signals. This is done by a digital filter algorithm. The functional blocks shown represent a possible configuration far a digital relay(computer relay) the current and voltage signals from the power system are processed by signal conditioners consisting of analog circuits such as transducers, surge suppression circuits, anti-aliasing fitters before being sampled and converted to digital form by the analog to digital converter. The sampling clock provides pulses at sampling frequencies (8-32)time the power system frequency. The relaying algorithm processes the sample data to produce a digital output. Computer relays will play a very important role in the protection ,control, and monitoring of power system it has many advantages such as :

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1. Its ability to diagnose a complex cases without great effort, cost, and complexity as for analog relay.

2. It provides a communication capability that allows it to warn system operation when it is not functioning properly.

3. Permits remote diagnostics, and possible correction, and provides local and remote readout of its settings and operation .

4. Another dimension was added to the reliability of the protection system.

5. The ability to adapt itself in real time to changing system conditions, and setting, and other characteristics.

6. Data sharing abilities of microprocessors with the central control computer or other protection terminals.

One of important disadvantages is the need to the interface unit for mixing of digital and analog relays within a common overall protection system.Indicating light are used for targeting and trace the tripping sequence and other information in the system.

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Numerical relays :-The can viewed as natural development of digital relays as a result of advances in technology, they use a specialized digital signal processor (DSP) as the computational hardware together with the associated software tools.

The input analogue signals are converted into digital representation and processed using mathematical algorithm. Processing is carried out using a specialized microprocessor that is optimized for signal processing application (DSP). Single item of hardware is used to provide a rang of functions.

Advantages :-

1-several setting groups 2-wider range of parameter

3- remote communication built in 4-internal fault diagnosis

5-power system measurements available , 6-distance to fault locator 7- disturbance recorder 8- CB monitoring (state, condition) 9- backup protection function in – built

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