Friday, 22 December 2017

HSV / NSV Rating of Electric Equipment

You might surprise to hear about HSV & NSV Rating of Electric Equipment. If you ever get a chance to visit a EHV Switchyard, you will notice that HSV / NSV rating is mentioned on every equipment connected to the Switchyard. I was also wondering to see this rating on every connected equipment in EHV Switchyard. 

HSV stands for Highest System Voltage and NSV for Nominal System Voltage. As you can guess, nominal system voltage rating is the expected continuous voltage of system in which the equipment is connected. For example, let us assume that a Circuit Breaker is connected in 400 kV Switchyard, the NSV will be 400 kV.

HSV is the highest continuous system voltage. Considering 400 kV Switchyard, it may happen under light load condition or because of some other reason that system voltage is maintain say 410 kV or 420 kV. Therefore it is expected that the connected equipment like CT, PT, and Breaker etc. should withstand this voltage without any damage. Therefore it is very important to study the system for highest continuous voltage which system / grid can achieve to assign or design HSV of equipment.

Highest continuous voltage of system should not be confused with switching surge or voltage surge due to lightening. HSV is based on highest continuous voltage of system. 

Thus HSV rating of equipment is an important factor for designing insulation requirement. Basically insulation requirement of equipment depends on switching voltage surge, lightening over voltage, highest power frequency withstand voltage and HSV. Insulation of equipment is so designed to withstand lightening over voltage for a time of the order of micro second, surge overvoltage for a time of the order of mili second and highest power frequency over voltage for 1 minute. But equipment insulation is designed based on HSV for continuous operation.

Wednesday, 20 December 2017

Instrument Safety Factor of Current Transformer

Instrument Safety Factor

Instrument Safety Factor (ISF) is defined as the ratio of CT saturation current to its rated current. Suppose the CT ratio is 2000/1 and the CT gets saturated if there is flow of 10 kA current through its primary, then Instrument Safety Factor is given as

Instrument Safety Factor, ISF = CT saturating Current / Rated current

                                                = 2000x5 / 2000 = 5

ISF is defined only for metering current transformer (CT). Metering CT is nothing but a CT used for metering purpose. 

Need of Instrument Safety Factor

Generally a CT have more than one core say 4 cores. Different cores are designed for different purpose like Core-1, Core-2 and Core-4 are meant for protection purpose whereas Core-3 is meant for metering as shown in figure below.

Instrument Safety Factor

Thus ISF will be defined for metering core i.e. core-3 of CT. Since meters are only designed for low value of current, therefore it is very important to protect them from high value of current. As meters are connected directly with the terminals of metering core of CT, it may happen so that during fault condition the secondary current of CT may be high which in turn will flow through the connected meter. This may lead to the damage of meter coil. Therefore some measure must be taken to protect meters from such event. This is the reason we define Instrument Safety Factor, ISF for metering CT. Now the question arises, how does defining ISF protects connected meters from over current?

Well, suppose there occurs some fault in the system. Assume that the fault current is 6 times the rated primary current of CT i.e. 6x2000 A. In this case, if the ISF value of CT is 5 then it is most likely to saturate and hence the secondary current of CT metering core will become zero. Thus there will not any flow of current through the connected meters during such fault. In this way, meters are protected from over current during fault. You may think of overload condition like everything is normal but the load current is say 4000 A. In this case CT secondary current will be 2 A which will flow through the meter but it should be noted that meters are designed for certain overloading. Based on meter overloading, Instrument Safety Factor of CT is chosen. Thus meters always remain protected.

Why secondary current of CT becomes zero during saturation?

Since during CT saturation, the magnetic flux in the core will become almost constant, this means that there will be no change in the flux and hence no induced emf. As there is no induced emf, hence there will not be any transformer action. This means that there will not be any CT secondary current.

If you see the name plate of metering core of a Current Transformer (CT), you will notice ISF value mentioned.

Compensating Device - Purpose

A compensating device is generally used in metering CT as shown in figure above. The basic purpose of this compensating device is to achieve the Instrument Safety Factor. This device is nothing but high resistance.

If you carefully observe the figure, you will notice that a secondary terminal S’ is connected to 3S1 through a compensating device. If we want to connect this CT terminal 3S1 and 3S4 then first of all S’ terminal is shorted with 3S4 and then 3S1 & 3S4 are connected to meters.

Thus compensating device forms a parallel path to the connected meter through high resistance. In case of fault, if CT do not saturate above its ISF (it is most likely to saturate above ISF), then excess current will be shunted through the compensating device. But under normal condition, the flow of current through the compensating device will be negligible.

Saturday, 16 December 2017

Why PT/VT Secondary Terminal should not be Shorted?

Before discussing "Why PT/VT Secondary Terminal should not be Shorted?", it is good to have a brief idea of Potential Transformers / Voltage Transformers. PTs/VTs are Instrument Transformer used for the purpose of protection and measurement. The construction of PT/VT is same as that of power transformer except for insulation level, cooling, sealing etc. PTs are designed for of specific voltage rating like 400 kV / 110 V. This means that when a PT primary is connected to 400 kV line, the secondary voltage will be 110 V. This secondary voltage is then connected to various measuring instruments like voltmeter, energy meter etc. and protection relays like distance relaydirection earth fault relay etc.

Thus we can say, PT steps down the primary line voltage to some lower voltage suitable for relays and meters. This means that PT design should be such that to have low voltage regulation to maintain its secondary voltage constant.

Why PT/VT Secondary Terminal should not be Shorted?

Let us now come to the point, why PT/VT Secondary Terminals should not be shorted? Unlike Current Transformer (CT), PTs are connected in line to ground as shown in figure below. Figure below depicts the connection of three PTs connected in three phases. Note that a neutral point is made by shorting a terminal of three PTs and then grounding the neutral point. 


You may like to read Difference between Current Transformer & Potential Transformer

Due to low voltage regulation, the secondary terminal voltage will remain constant and hence if we keep the PT terminals open, nothing is going to happen as the secondary voltage is low (110 / 1.732 = 63.5V). Mind that the same is not true for CT. CT secondary terminals should never be kept open. In normal condition, PT secondary is connected to some impedance offered by relay / measuring instrument. Therefore the current through the secondary circuit is low.

But when we short the secondary of PT, a high current will flow thorough the secondary circuit. This is because of low voltage regulation. PT will try to maintain its secondary voltage and for doing this it will try to flow high current through shorted terminals. This high current will lead to overheating and consequent damage to the PT.
To avoid damage due to short circuit of PT terminals, fuses are installed in PT junction box. In case of short circuit of secondary terminals, these fuses will blow out and thus will open the circuit. It shall be noted that fuse should be installed as near to the PT as possible to avoid heating of connecting cables.

Thursday, 9 November 2017

Core Balance Current Transformer

Core Balance Current Transformer or CBCT is a ring type current transformer through center of which a three core cable or three single core cables of three phase system passes. This type of current transformer is normally used for earth fault protection for low and medium voltage system. A typical Core balance Current Transformer is shown in figure below.


Secondary of CBCT is connected to Earth Fault Relay. During normal operating condition as the vector sum of three phase current i.e. (Īa + Īb + Īc =0) is zero therefore no residual current in the primary will be present. Here residual current means zero sequence current. Therefore there will not be any flux developed in the CBCT core and hence no current in the secondary circuit of CBCT.

Working Principle of CBCT:

Let Īa, Īb and Īc be the three line currents and Φa, Φb and Φc be corresponding components of magnetic flux in the core. Assuming that the CT is operating in the linear region (Read B-H Curve to get idea of linearity), magnetic flux because of individual phase current will be directly proportional to the phase current and hence we can write as below,

Φa = kIa

Φb = kIb

Φc = kIc

where k is constant of proportionality. Mind here that same constant of proportionality is used as all the three phase current are producing magnetic flux in the same core i.e. magnetic material.

Thus the resultant magnetic flux in the CBCT core,

Φr = k(Īa + Īb + Īc) …………………..(1)

But we know from theory of symmetrical components,

Īa + Īb + Īc = 3Ī0 = Īn

Where, Io is zero sequence current and In is neutral current. Hence we can write as

Φr = kĪn  …………………………(2)

Now let us consider two cases:

Case1: During normal condition

Īa + Īb + Īc = 0

Hence from equation (1),

Net resultant flux in the CBCT Core, Φr = 0 which means no secondary current and therefore the Earth Fault Relay won’t operate.

Case2: During earth fault, three phase current passing through the center of Core Balance Current Transformer will not be balanced rather a zero sequence current will flow. For example for single line ground fault,

If = 3Ia0 = In

Thus from equation (2),

Net magnetic flux in the CBCT core, Φr will have some finite value which in turn will induce current in the secondary circuit due to which earth fault relay will operate. Because of this reason, a Core Balance Current Transformer or CBCT is also called Zero Sequence Current Transformer.

Advantage of Core Balance Current Transformer:

The advantage of using CBCT for earth fault protectionis that only one CT core is used instead of three core as in conventional system where the secondary winding of three cores are connected residually. Thus the magnetizing current required for the production of a particular secondary current is reduced by one third which is a great advantage as the sensitivity of protection is increased.

Also, the number of secondary turn does not need to be related to the cable rated current because no secondary current flows under normal operating condition as the currents are balanced. This allows the number of secondary turns to be chosen to optimize the effective primary pick-up current.

Core Balance Current Transformer is normally mounted over a cable at a point close to the cable gland of the Switchgear. In case cables are already laid in a Switchgear, physically split core, which is also known as Slip-over type CT, are used.

Equal Area Criterion and Out of Step Condition

Equal Area Criterion is the method of studying the Transient Stability of system of two machines or single machine connected to infinite bus. The study of Transient Stability tells us whether or not synchronism is maintained i.e. whether or not load angle δ settles down to steady state stable value after the clearance of fault or disturbance. In this post we will focus on
  • Equal Area Criterion
  • Transient Stability Limit and Margin
  • How Transient Stability Margin depend on Breaker Opening time?
  • Meaning of Out-of-Step condition in power system
  • Meaning of Loss of Synchronism
  • Critical Clearing Angle

Concept of Equal Area Criterion:

As we know that under steady state operation of a generator, there exists a balance between the power generation and demand. In other words we can say that mechanical input to generator is balanced with the generated power due to which Generator rotates at a constant synchronous speed. Basically, the output electric power from the generator produces an electric torque that balances the mechanical torque applied to the generator rotor shaft. The generator rotor therefore runs at a constant speed with this balance of electric and mechanical torques. Mind that electromagnetic torque and mechanical torque opposes each other and because of this balance between the mechanical & electromagnetic torque, mechanical energy is converted to electrical energy.

Let us consider a two source system as shown in figure below. The equation of power transfer between the two sources assuming lossless system is given as

P = VsERSinδ / X

Where Vs = Sending end Voltage

VR = Receiving end voltage

X = Transmission Line Reactance + Source Reactance


Thus from the above equation, it is clear that power transfer between the two sources is inversely proportional to the Reactance. If the net reactance increases, power transfer will decrease and vice versa.

During fault condition like Single Line to Ground Fault, Double Line to Ground Fault, Three Phase Fault etc. the effective transmission reactance between the two sources increases depending upon the type of fault. Due to this increase in the effective transmission line Reactance the power transfer between the two sources will reduce.  Due to this reduction of power transmission, the electric torque that counters the mechanical torque is also decreased. Thus

Mechanical Torque > Electromagnetic Torque

If the mechanical power is not reduced during the period of the fault, the generator rotor will accelerate with a net input torque (equal to Mechanical Torque - Electromagnetic Torque).

Assume that the two source power system shown in above figure is initially operating at a balance point of δ0 and hence transferring electric power P0. After a fault, the power output is reduced to PF, the generator rotor therefore starts to accelerate, and δ starts to increase. Let’s say the fault is cleared when the angle difference between the sources reaches δC. At this point, fault reactance will not be considered as the system is has attained normal configuration and hence the power transfer will follow the Pre Fault Power Angle Curve as shown in figure below.

But at point δC power output PC is larger than the mechanical power input P0 which will result in deceleration of rotor as

Electromagnetic Torque > Mechanical Torque

However, because of the inertia of the rotor system, the angle does not start to go back to δ0 immediately. Rather, the angle continues to increase till δF such that energy lost during deceleration in Area 2 is equal to the energy gained during acceleration in Area 1. This is called Equal Area Criterion.

Let us consider two cease now for the sake of better understanding of out-of-step condition and application of Equal Area Criterion.

Case-1: If δF < δL or Area 1 < Area 2

Under this condition, the rotor of Generator will oscillate. But because of presence of damping, the amplitude of oscillation will continuously reduce and eventually δ will settle to balanced angle δ0.

Thus we can say that, in this situation the system is Transiently Stable. How much margin we have for transient stability?

If you carefully observe the above Power Angle Curve then you will definitely say that the margin is (δL- δF). Great!

But next question arises, who decides the value of δF?

You must say it is δC i.e. the point of fault clearing.  That is why δC is also known as Critical Clearing Angle.

How Transient Stability Margin depends on Breaker Opening Time?

Since the value of δC depends on the point of clearing fault. Therefore, Transient Stability margin i.e. (δL- δF) depends upon the fault clearing time. The less the fault clearing time, less will be the value of δC which means less value of δF. Hence Transient Stability margin will be more. Thus we can say that Transient Stability Margin is dependent on Breaker Opening time to clear the fault. Got it? I guess you get. Please write in comment box.

Case-2: If δF < δL or Area 1 > Area 2


If Area 2 is smaller than Area 1 then at the time the angle reaches δL, then further increase in angle δ will result in an electric power output that is smaller than the mechanical power input. Therefore, the rotor will accelerate again and δ will increase beyond recovery. This is a transiently unstable scenario, as shown in figure above. When an unstable condition exists in the power system, one equivalent generator rotates at a speed that is different from the other equivalent generator of the system. We refer to such an event as a loss of synchronism or an out-of-step condition of the power system.

All Day Efficiency of Distribution Transformer

All Day Efficiency is defined as the ratio of total energy output for 24 hrs i.e. for the whole day to the total energy input for the same day. All Day Efficiency is basically Energy Efficiency of Transformer calculated for a period of 24 hrs. Mathematically we can write,

All Day Efficiency = 1 – [Daily Losses in kWh / Daily Input in kWh]

What is the need of All Day Efficiency?

To understand the need of calculating All Day Efficiency, first we will discuss about the basic difference between Power Transformer and Distribution Transformer.

Power Transformers are used at sending and receiving ends of a long, high voltage transmission line to step up or step down the voltage level. These Transformers are manipulated to operate near their rated capacity. Therefore, Power Transformers are disconnected during light load periods. In view of this, a Power Transformer is designed to have maximum efficiency at or near rated capacity. Therefore, the choice of Power Transformer out of large number of competing Transformers shall be based on full load efficiency.

Now, Distribution Transformers are those which changes the voltage level to a level suitable for utilization purpose at the consumer’s premises. A Power Transformer doesn’t come in direct contact with the consumer’s terminal whereas a Distribution Transformer must have its secondary directly connected to consumer’s terminals. The load on a Distribution Transformer varies a wide range during 24 hours a day. For example, a Distribution Transformer may have practically no load or little load during considerable period of day but in evening time it may go up to the rated capacity. In this way, the primary of Distribution Transformer is always energized irrespective of load. Notice here the difference between the Power and Distribution Transformer. As the primary of Distribution Transformer is always energized, core loss will take place continuously. In view of this, Distribution Transformers are designed to have very low value of core loss.

But we know that,

kVA Load for maximum efficiency = Rated Load x n

where n = sq. root (Pc / Psc)

Pc = Core Loss

Psc = Full Load ohmic loss

Thus from the above equation, for the reduced core loss Pc, the maximum efficiency may occur at about one half of its rated kVA. Thus Distribution Transformer should not be judged by its full load efficiency which is usually much less than its maximum efficiency. However, the choice of Distribution Transformer out of large number of competing transformers shall be based on All Day Efficiency. This is the essence of the term All Day Efficiency.

Must Read,

Ammeter Shunt –Construction and Calculation

Ammeter Shunt is used for the measurement of heavy current using an Ammeter. Ammeter Shunt is basically a low resistance connected in parallel with the moving coil so that most of the current is bypassed by the Shunt and hence only a small current flows through the moving coil.

Why do we need Ammeter Shunt for Measuring High Current?

The basic movement of DC Ammeter is a PMMC Instrument. The coil winding of PMMC Instrument is small and light and is only designed to carry very low current since the construction of an accurate instrument with moving coil to carry current more than 100 mA is not feasible as bulk and weight of coil would be required. Thus if we simply connect the Ammeter for measuring higher current then flow of current through the moving coil will be much more than the for which it is designed and will burn.

Ammeter Shunt Material:

The basic requirement for Ammeter Shunt can be summarized as

1) The resistance of shunt should not vary with time

2) They should carry current without excessive rise in temperature

3) They should have low thermal electromotive force with copper

Well, Manganin is generally used for Shunt of DC Instruments as it gives low value of thermal emf with copper although it is liable to corrosion and difficult to solder. Constantan is used for AC circuit.

Calculation of Ammeter Shunt:

Let us assume that we want to measure a current of I while moving coil of Ammeter is only designed to carry a current of Im (full scale deflection current), therefore we need to use ammeter shunt. We will calculate the value of suitable shunt. Figure below shows the basic circuit of ammeter.

In the figure above, ammeter shunt resistance Rsh is connected in parallel with the moving coil (Resistance of moving coil is assumed to be Rm).

Therefore, IshxRsh = ImxRm

⇒ Rsh = (Im / Ish)xRm  …………………………….(1)

But Ish = I – Im

Therefore from equation (1), we can write as

Rsh = (Im xRm) /( I – Im)

Dividing Numerator and Denominator by Im,

 ⇒ Rsh = Rm/(I/Im – 1)

⇒ (I/Im – 1) = Rm/Rsh

I/Im = 1+Rm/Rsh

Here, the ratio of total current to be measured to the current in moving coil i.e. (I/Im) is called Multiplying Power of Ammeter Shunt.


Multiplying Power, m = I/Im

                                      = 1+Rm/Rsh

Resistance of Ammeter Shunt,

Rsh = Rm/(m-1)

Suggested Reads,

Voltmeter Multiplier - Construction and Calculation

A PMMC Instrument can be used as voltmeter by just connecting a series resistance with the moving coil. This series resistance is called Voltmeter Multiplier. This combination of moving coil and multiplier is connected across the point whose voltage is to be measured.

There are two main functions of voltmeter multiplier:

1) It limits the current through the PMMC moving coil to a value less than full scale deflection current and thus prevents moving coil from being damaged.

2) It minimizes the flow of current through the voltmeter and thus do not alter the circuit current whose voltage is to be measured. Ideally the resistance of voltmeter should be infinite.

Calculation of Voltmeter Multiplier:

The value of multiplier required to extend the voltage range is calculated as below.


Im = Ifs = Full scale deflection current of meter

Rm = Internal resistance of meter

Rs = Multiplier resistance

Vm = Voltage across the moving coil

V = Full range voltage of meter

From the simplified voltmeter circuit given below,


Vm = ImRm  ……………….(1)

and V = Im(Rm + Rs)  ………….(2)

Dividing equation (2) by (1) we get,

V/Vm = 1 + Rs/Rm


m = Multiplying Factor = 1 + Rs/Rm

The term V/Vm is called Multiplying Factor of Voltmeter. Multiplying Factor is basically the value by which the range of voltmeter can be extended. Let’s assume that the moving coil of voltmeter can sustain a voltage of 0.25 V and if we want to use it for measuring a voltage of 10 V then Multiplying Factor = 10/0.25 = 40. In this case the value of voltmeter multiplier which needs to connect in series with the moving coil can be calculated from equation (1) and (2). Let the value of internal resistance of meter Rm = 25 Ohm.

m = 1 + Rs/Rm

But m = 40 and Rm = 25

⇒ 40 = 1+Rs/25

⇒ 39 = Rs/25

⇒ Rs = 39x25 Ohm = 975 Ohm

Also, meter full scale deflection current = 0.25/25 = 0.01 A = 10 mA

Thus from the above example we observe that, to measure a voltage of 10 V (40 times greater than designed voltage) we need to connect a series resistance of 975 Ohm.

Material of Voltmeter Multiplier:

The essential requirements of voltmeter multiplier to fulfill are
  • Their resistance should not change with time.
  • Change in their resistance with temperature should be small
The resistance material used for multipliers is Manganin and constantan (copper nickel alloy).

Composition of Manganin, Cu (84%) + Mn (12%) + Ni (4%)

Working Principle of Moving Iron Instruments

Moving Iron Instruments are the most common type of ammeter and voltmeter used at power frequencies in laboratories. These instruments are very accurate, cheap and rugged as compared to other AC instruments.

Working Principle of Moving Iron Instruments:

In Moving Iron Instruments, a plate or van of soft iron or of high permeability steel forms the moving element of the system. The iron van is so situated that it can move in the magnetic field produced by a stationary coil. Figure below shows a simple moving iron instrument.


The stationary coil is excited by the current or voltage under measurement. When the coil is excited, it becomes an electromagnet and the iron van moves in direction of offering low reluctance path. Thus the force of attraction is always produced in a direction to increase the inductance of coil. Mind that as the van follows the low reluctance path, the net flux in air gap will increase which means increased flux linkage of coil and hence inductance of coil will increase. It shall also be noticed that, the inductance of coil is variable and depends on the position of iron van.

Torque Equation of Moving Iron Instruments:

Suppose that, at any instant of time current flowing in the coil is I. Thus the energy stored in the coil in the form of magnetic field = (1/2)LI2.

As soon as the current changes to (I+dI), deflection in the pointer becomes dƟ resulting into change in inductance of coil from L to (L+dL). Let this deflection in pointer is due to deflection torque Td.

Thus mechanical work done = Tdx dƟ ………………..(1)

Energy stored in Coil = (1/2)(L+dL)(I+dI)2

Change in stored energy of coil

= Final Stored Energy – Initial Stored Energy

= (1/2)(L+dL)(I+dI)2 - (1/2)LI2

= (1/2)[ (L+dL)(I+dI)2 – I2L]

= (1/2)[ (L+dL)(I2+2IdI+(dI)2 – I2L]

= (1/2)[ LI2+2LIdI+L(dI)2 + dLxI2+2IdIxdL+dLx(dI)2   – I2L]

Neglecting second order and higher terms of differential quantities i.e. L(dI)2, 2IdIxdL and dLx(dI)2   

= (1/2)[ 2LIdI+dLxI2]

= LIdI +(1/2)dLx I2  ……………………(2)

Again, just think, when there is a change of current from I to (I+dI), this change change of current must be accompanied by change in emf of coil. Thus we can write as

e = d(LI) / dt

   = IdL/dt + LdI/dt

But electrical energy supplied by the source = eIdt

                                                                        = (IdL + LdI) x I

                                                                        = I2dL + LIdI

According to law of conservation of energy, this electrical energy supplied by the source is converted into stored energy in the coil and mechanical work done for deflection of needle of Moving Iron Instruments.


I2dL + LIdI = Change in stored energy + Work done

⇒ I2dL + LIdI = LIdI +(1/2)dLx I2  + Tdx dƟ  ….[from (1) and (2)]

⇒ Tdx dƟ = (1/2)dLxI2 

Td = (1/2)I2(dL/dƟ) 

Thus deflecting torque in Moving iron Instruments is given as

Td = (1/2)I2(dL/dƟ) 

From the above torque equation, we observe that the deflecting torque is dependent on the rate of change of inductance with the angular position of iron van and square of rms current flowing through the coil.

In moving iron instruments, the controlling torque is provided by spring. Controlling torque due to spring is given as

Tc = KƟ

Where K = Spring constant

Ɵ = Deflection in the needle

In equilibrium state, deflecting and controlling torque shall be equal as below.

Deflecting Torque = Controlling Torque

⇒ Td = Tc

⇒ (1/2)I2(dL/dƟ) = KƟ

Ɵ = (1/2)(I2/K)(dL/dƟ)

From the above torque equation, we observe that the angular deflection of needle of moving iron instruments is square of rms current flowing through the coil. Therefore, the deflection of moving iron instruments is independent of direction of current.