Thursday, 8 March 2018

Auto Transfer System – Basic Concept

Auto Transfer System (ATS) or Bus Transfer System (BTS) is a scheme adopted in industries to increases the power supply reliability to their auxiliary loads. Auxiliary loads are essential loads for normal plant operation and production. A complete / partial loss of power supply to these loads may result into plant outage and hence production loss. Let us discuss this system in detail.

Let us consider the applicability of ATS for a power generation utility. Figure below shows a simple power supply arrangement for power generation utility.


Following points should be noted form the above figure:

1)   Two sources are connected to the motor BUS, one is through UAT CB and second one is through SUT CB. UAT means Unit Auxiliary Transformer while SUT means Start-up Transformer. Primary of UAT is connected to Generator output and secondary to the motor bus. UAT CB remains normally close and thus feeds the motor loads connected to the bus.

2)   Primary of Start-up Transformer (SUT) is connected to Grid while its secondary to the motor bus. SUT is basically meant to import auxiliary power from the Grid in case of fault in the UAT or Generator tripping and GCB (Generator Circuit Breaker) open.

3)  Power evacuation to Grid is through the Generator, GCB and Generator Transformer (GT). Auxiliary loads can also be supplied through the GT-UAT combination in case of Generator tripping and GCB open.

Keeping all the above points in mind, when there is a fault in UAT, the power supply to motor bus will be lost and subsequently motor loads will trip on under-voltage protection (normally set at 20% of nominal voltage). This will result into plant outage and loss of production. To avoid such outage, Auto Transfer System (ATS) is used. ATS senses the fault in old source and restores the power supply through alternate source so that the motor loads are not tripped. For our example, when ATS relay senses fault in UAT, it gives tripping command to UAT CB and closing command to SUT CB. Thus power supply to motor bus is restored. The operation of ATS relay must be fast enough to ensure that motor loads are not tripped on under-voltage. So the question arises how fast ATS should be? How do we decide the time window within which the auto transfer cycle should be completed?

The time window for auto transfer to operate is dependent on motor bus spin down characteristics. Let us first understand the bus spin down characteristics to answer the questions.

Motor Bus Spin down Characteristics:

The bus voltage will instantaneously become zero if all the loads connected to the bus are resistive in nature. Isn’t it?  Yes, but since the loads connected to the bus are motor loads i.e. inductive loads therefore on loss of supply to the bus, the voltage of the bus will not instantaneously decay to zero. It will take some definite time to decay down. The voltage decay time of the motor bus will depend on the time constant (L/R ratio). This time constant can easily be calculated by the equivalent open circuit parameter of connected loads. The more the inductance of the system, the more will be the time constant and hence the more will be time for voltage to decay down to zero.


Let us consider the same thing in different but practical way. Suppose the motor loads are operating normally and suddenly power supply to the bus is lost. What will happen? The motor will slowly decelerate. The rate of deceleration will be inversely proportional to the inertia of the connected load to the motor. The more is the inertia of load to the motor; the lower will be rate of deceleration. This means that a heavy motor will take more time to stall. During the period of deceleration, the motors will behave like an induction generator due to trapped magnetic flux in its air gap. This will result into back feeding of voltage at to bus till the trapped residual magnetic flux is dissipated. The frequency and phase of residual bus voltage will also decay with the deceleration of motor and eventually will become zero. Thus there are three important parameters from auto transfer point of view. These are

1)    Magnitude of the residual voltage

2)    Decay time of this residual voltage

3)    The phase angle of the residual voltage with respect to healthy bus

From the above discussion of Spin down Characteristics, it is quite clear that the motor bus voltage should not decay to such an extent that under-voltage relay is operated and motor is tripped. Again for ATS to restore power supply through alternate source, the alternate source and residual motor bus voltage should be in synchronism. This means that the condition for synchronism should be satisfied. Thus the whole cycle of auto transfer must be completed within a specified time decided by the decay time and mode of auto transfer. Thus an ATS relay must take bus voltage, old source voltage, new source voltage and breaker status as an input as shown in figure below.


Bus auto transfer can either be sequential or parallel. In parallel mode, new source breaker is first closed and then old faulty source breaker is opened. Whereas in sequential mode, first old faulty source breaker is opened and then new healthy source breaker is closed. But the parallel mode has some disadvantage. The main disadvantage is availability of dual source fault current and the high reactive currents which may result due to angular difference between the two sources. Sequential transfer mode eliminates this problem. Fast, in-phase and slow transfer are the available sequential transfer modes.

Modes of Auto Transfer:

Based on the system condition like voltage, frequency and phase angle of motor bus, auto transfer can take place in three different modes: Fast Transfer, In-Phase Transfer and Slow Transfer.

Fast Transfer: Fast transfer is a supervised open circuit transfer scheme. Supervised open circuit transfer means that the alternate source breaker is closed with synchronism check with motor bus and open circuit means alternate source breaker is closed when the faulty source breaker is confirmed to be open by the ATS relay.  This scheme minimizes the interruption of power to the bus, while avoiding a potentially unsafe parallel operation of both the sources. In the fast transfer mode, the decision to supply transfer is done on the basis of comparing the phase angle and voltage magnitude of the bus with that of the new source. The time window for fast transfer will be governed by spin down characteristics of motor bus of the utility. Thus for one utility it may be 200 ms while for some other utility it may be 250 ms. Similarly time frame for in-phase transfer will vary from industry to industry.

In-Phase Transfer: In-Phase Transfer takes place when the first opportunity to supply transfer is missed. Careful observation of spin down characteristic of motor bus reveals the zone for fast transfer. Once the time window for fast transfer is elapsed, ATS Relay need to wait till the phase angle of residual bus voltage again comes within synchronism limit (normally this limit is 20 degree). Zone AB in the spin down characteristics shows the region where in-phase transfer can take place. Thus the time window for in-phase transfer is generally more. This mode is also an open circuit transfer scheme. The main advantage of the in-phase transfer mode is the ability to safely transfer the bus to the new source, even if the fast transfer is blocked, without necessarily having to trip the motors.

Slow Transfer: The slow transfer or residual voltage transfer method is the slowest transfer of the bus to the new source. Basically slow transfer is a dead bus transfer scheme where the new source is connected to bus when the bus is dead. During this time all the motor loads are tripped on U/V protection. Tripping of motor loads is necessary to avoid simultaneous re-acceleration on supply restoration.

Various microprocessor based high speed programmable ATS Relays are available. The above three modes may be enabled in the relay with appropriate setting like time window for fast and in-phase transfer, residual voltage magnitude, phase difference between the healthy source and residual bus voltage etc. This relay decided for the suitable transfer mode and direction of transfer based on its setting and breaker configuration. Breaker configuration means, if UAT CB is open and SUT CB is close, ATS will detect transfer direction from SUT to UAT and UAT CB will be closed on fault in SUT. Breaker auxiliary contacts are wired up to the relay.

Wednesday, 28 February 2018

RVDT – Construction and Working Principle

Rotary Variable Differential Transformer or RVDT is an inductive transducer which converts angular displace to an electrical signal. Unlike LVDT, the input of this transducer is differential value of rotary variable i.e. angular rotation (dƟ) to generate voltage output.


Like every transformer, RVDT has two types of winding i.e. Primary winding and Secondary winding. The primary and secondary winding are wound on a former. There are two secondary winding having equal number of turns. These winding are placed on either side of the primary winding identically. A cam shaped magnetic core made of soft iron is connected to a shaft. This magnetic core can be thus be rotated in between the winding. Carefully observe the figure below to understand the construction and working principle.


The construction of LVDT and RVDT is almost same. The only difference in their construction is that in RVDT, the core is cam shaped and may be rotated between the windings by means of a shaft. You should read LVDT – Construction and Working Principle to understand the constructional detail.

Working Principle:

The reluctance seen by the primary mmf changes with the rotation of cam shaft. This results in change in the magnetic flux with rotation of the cam shaft. Due to this change in magnetic flux with rotation of cam, the flux linkage of secondary winding also changes. Therefore, as per the transformer action, an emf is induced in secondary winding. The magnitude of induced emf will depend on the rate of change of rotation. The more the rate of change of rotation, the more will be the rate of change of flux w.r.t. and hence more emf will be induced.

As can be seen from the figure, the two secondary winding are connected in series but in phase opposition. This is done to get a single output voltage from the transducer. If Es1, Es2 and E0 be the emf induced in the two secondary winding S1 & S2 and output voltage respectively then

E0 = Es1 – Es2

Under normal condition of RVDT, the flux linkage of both the secondary winding are same due to their symmetrical placing with respect to primary and core. Therefore, the induced emf Es1 and Es2 are equal and hence output voltage E0 of the transducer in such condition is zero. Therefore, normal position of RVDT is called NULL position.

Clockwise rotation of cam causes an increasing voltage Es2 in one of the one secondary winding while counter clockwise rotation leads to increase in voltage Es1 of another secondary winding. Thus the direction as well as magnitude of angular rotation can be ascertained from the magnitude and phase of transducer output voltage. Phase of transducer output voltage means whether (Es1 – Es2) is positive or negative. In case of anti-clockwise rotation, the value of Es1 will be more than that of Es2 and hence (Es1 – Es2) will be positive. In this case we say that output voltage E0 is in phase with the primary voltage. With the same logic, when cam is rotated in clockwise direction, the output voltage will be negative i.e. out of phase with primary voltage.

Tuesday, 27 February 2018

LVDT – Construction and Working Principle

What is LVDT?

Linear Variable Differential Transformer, LVDT is the most used inductive transducer for translating linear motion into electrical signal. This transducer converts a mechanical displacement proportionally into electrical signal.


LVDT is a transformer consisting of one primary winding P and two secondary winding S1 & S2 mounted on a cylindrical former. The two secondary winding have equal number of turns and placed identically on either side of the primary winding as shown in figure below.

A movable soft iron core is placed inside the former. Actually the movable core is made of nickel iron with hydrogen annealed. Hydrogen annealing is done to eliminate harmonics, residual voltage of core and thus provides high sensitivity. The movable core also is laminated in order to reduce eddy current loss. The assembly of laminated core is placed in a cylindrical steel housing and end lids are provided for electromagnetic and electrostatic shielding. The displacement to be measured is attached to this movable soft iron core.

LVDT- Working Principle:

Since the primary winding of Linear Variable Differential Transformer (LVDT) is supplied with AC supply, it produces an alternating magnetic flux in the core which in turn link with the secondary winding S1 and S2 to produce emf due to transformer action. The electrical equivalent circuit of LVDT is shown below.

LVDT NUL position

Let us assume that the emf produced in secondary winding S1 is Es1 and that in S2 is Es2. The magnitude of Es1 and Es2 will depend upon the magnitude of rate of change of flux (dØ / dt) as per the Faraday’s Law. The lower the value of ‘dt’, the more will be the emf induced. But lower value of ‘dt’ means that core is moving faster. Thus we can say that the faster the movement of core, the greater will be the magnitude of emf induced in secondary windings.
To get a single output voltage from the Linear Variable Differential Transformer (LVDT), both the secondary winding are connected in series but in phase opposition as shown in figure below.
LVDT secondary connection

Due to this connection, the net output voltage E0 of the LVDT is given as below.
E0 = Es1 – Es2

Since the secondary windings of LVDT are identical and placed symmetrically on either side of core, therefore under normal position the flux linkage of both the secondary winding S1 & S2 will be same. This means Es1 = Es2 and hence net output voltage E0 of LVDT = 0. This position of soft iron core is called NULL position. Thus NULL position of Linear Variable Differential Transformer is the normal position of movable core where the net output voltage is zero.

Now, as the core can either be moved toward right or left to the null position. Let us now consider such movement of core under two cases.

Case-1: Core is moved left to the NULL position
LVDT core moving left

When core of LVDT is moved to the left of the NULL position ‘O’ as shown in figure above, the flux linkage of secondary winding S1 will become more than that of winding S2. This means the emf induced in winding S1 will be more than S2. Hence Es1 > Es2 and net output voltage E0 = (Es1 – Es2) = Positive. This means that the output voltage E0 will be in phase with the primary voltage.

Case-2: Core is moved right to the NULL position

LVDT core moving right

When the core of LVDT is moved toward right of NULL position ‘A’, you can guess what will happen? Obviously the emf induced in secondary winding S2 will be more than that of S1. This means Es2 > Es1 and hence net output voltage E0 = (Es1 – Es2) = negative. This means that the output voltage of LVDT will be in phase opposition (180 degree out of phase) with the primary voltage.

From the above two cases, we can have the following conclusions:

1) The direction of movement of a physical quantity can be identified by the output voltage of LVDT. If the output voltage E0 is positive, this means the physical quantity is moving toward left.

2) If the output voltage E0 is negative, this will mean that the physical quantity is moving in the right direction from the NULL position.

3) The amount / magnitude of displacement is proportional to the magnitude of output voltage. The more the output voltage, the more will be displacement. But here is a clue. You can’t take core out of the former; otherwise the output voltage will become zero.

4) In fact corresponding to both the cases i.e. whether core is moving left or right to the NULL position, the output voltage will increase lineally up to a displacement of around 5 mm from the NULL position. After 5mm, output voltage E0 becomes non-linear. The graph of variation of E0 with displacement is shown below.

LVDT output with displacement

Carefully observe the above graph. It may be noted from the graph that even at NULL position (i.e. when there is no displacement) there is some output voltage of LVDT. This small output is due to the residual magnetism in the iron core.


LVDT is used in those applications where displacement ranging from fraction of a mm to few cm. As a primary transducer, it converts the mechanical displacement into electrical signal.

Acting as a secondary transducer, it is sued for measurement of force, pressure, weight etc.

Saturday, 24 February 2018

Understanding of Phantom Loading

Phantom Loading is a loading condition in which an energy meter is connected to factious load for testing of energy meter with high current rating. Such loading is favorable to avoid wastage of energy during the test of measurement instrument.

Testing of energy meters are carried out to verify the actual registration as well as the adjustment done to bring the meter error within acceptable limit. An energy meter is subjected to various kind of test like Creep Test, Starting Test etc.

Pressure coil (PC) is connected across the source and Current coil in series with the load for the test purpose. In another words, we can say that meter is directly connected to the load. This is known as direct loading. Direct loading of meter for testing is only adopted where the current rating of meter is considerably low. This method of directly connecting the meter to the circuit leads to wastage of appreciable amount of energy when the current rating of energy meter is high. Therefore we must devise a new way to test energy meters having high current rating. Phantom Loading is that new way.

In Phantom Loading, the pressure coil is connected to the normal supply voltage and the current coil (CC) circuit is connected to a low voltage supply (phantom voltage). As the impedance of CC is low, therefore it is possible to circulate rated current through the CC with low voltage supply. The arrangement is shown below. In the figure value of voltages are just taken for example. These values will vary with rating of meter.

Phantom loading

Under phantom loading condition, the total power required for this test will be sum of power loss in PC and CC. Mathematically, power required to carry out test is given as below.

= V2 / Rpc + VphIcc

Where V = Normal rated voltage of meter

 Rpc = Resistance of pressure coil

Icc = Rated current of energy meter

Vph = phantom voltage connected with current coil

Since Rpc is very high and Vph is very low (when compared with rated voltage of meter), therefore the total power required for the test under phantom loading will be very less. Let us now consider an example to better understand the facts discussed above.

Example: A 250 V, 5A DC Energy Meter is tested at its marked rating. The resistance of PC and CC are 8800 Ω and 0.1 Ω respectively. Calculate the power consumed when testing the meter with

1) Direct loading arrangement

2) Phantom loading with current circuit excited by a 6 V battery.


Case-1: Direct Loading Arrangement

A schematic diagram for direct loading arrangement is shown below.

Phantom loading-energy-meter

Power consumed in PC circuit = (250)2 / 8800

                                                 = 7.1 Watt

Power consumed in CC circuit = 250x5

                                                 = 1250 Watt

Total power consumed during test = 1257.1 Watt

(Don’t think that we are only calculating loss in CC rather we are calculating loss in whole CC circuit which includes load too. This is the reason we used VI for power calculation instead of using I2R as value of load resistance is not known.)

Case-2: Phantom Loading

The schematic diagram of phantom loading is given below.

Phantom loading

Power consumed in PC circuit = (250)2 / 8800

                                                 = 7.1 Watt

Power consumed in CC circuit = 6x5

                                                 = 30 Watt

Total power consumed during test = 37.1 Watt

Carefully observe the results of case-1 & 2. You will notice that loss in PC circuit is not affected. But loss in CC circuit is now only around 2.5 % of loss under direct loading condition. This percentage reduction in loss in CC circuit is equal to the percentage of phantom voltage w.r.t. rated voltage i.e. (6 / 250) x 100 = 2.4 %. This means the lower the phantom voltage required to circulate the rated current in CC circuit, lower will be the power required for testing.

Friday, 23 February 2018

Plant Load Factor and its Calculation

Plant Load Factor (PLF) is the ratio of average power generated by the plant to the maximum power that could have been generated for a given time period. Thus mathematically it can be written as,

PLF = Pavg / Pmax…………….(1)

As it is the ratio of same quantity, hence it is a unit less quantity.

Note that maximum power Pmax is dependent on load. During peak load period a power plant is supposed to operate at its maximum capacity. Plant Load Factor may either be calculated on daily basis, weekly, monthly or early basis. It is clear from the mathematical representation of PLF that value of Plant Load Factor is less than one since Pavg < Pmax. But when a plant is operated above its rating for a certain period of time, then obviously for that period average power supplied by the Plant i.e. Pavg will be more than the PmaxTherefore for such case value of PLF become more than unity. But it is usually not the practice to operate a Generator beyond its rating. Thus, normally PLF is less than 1.

PLF can also be defined in terms of energy supplied by the plant for a given period of time. Let us suppose that we are only interested in calculating PLF for a period of T. Then plant load factor PLF from (1),

PLF = (PavgxT) / (PmaxxT) 

        = Average Energy Supplied / Energy Supplied at maximum demand ….....(2)

Thus in terms of energy, Plant Load Factor is the ratio of average energy supplied for a given time period to the energy that could have been supplied at maximum loading condition for the same time period.

Plant Load Factor is one of the performance parameter of a power plant. It is a degree of plant capacity utilization for a period of time. More the PLF, more will be the revenue of the plant. Alternatively, higher the PLF, lesser will be cost of per unit (kWh) energy generated.

Calculation of Plant Load Factor:

We already discussed the mathematical representation of PLF. It is calculated for a certain time period as per the definition. Therefore it may either be calculated from Load Curve or Load Duration Curve. If we have the load curve, then Plant Load Factor can be calculated using (1). But for calculation from load duration curve, it is calculated by using energy interpretation i.e. (2). Let us consider a load curve as shown below and try to calculate the Plant Load Factor.

Plant load factor calculationAs per the definition,

Plant Load Factor, PLF =   Pavg / Pmax

From the load curve, Pmax = 200 MW but we need to calculate the average power Pavg.
Average power Pavg can be calculated as below.

Pavg = Units (kWh) Generated per day / 24 hrs

But units (kWh) generated per day is the area under the curve i.e. area of ABCFILM.

Area under the curve = Area of (rectangle ABLM + Trapezoid BCFI)

                                  = [(100x16) + (1/2)(12+4)x100]

                                  = 1600 + 800

                                  = 2400 MWh = 2400x103 kWh

                                  = 2400x103 Units

                                  = 2.4 Million Units (MUs)

Thus the average power Pavg = 2400 MWh / 24 h

                                                 = 100 MW

PLF = 100 / 200 = 0.5

Second Method:

Let us assume that the plant is operated through the day at its maximum loading i.e. 200 MW. Then the total energy that could have been generated will be equal to the 24x200 MWh = 4800 MWh (This is simple multiplication as the power output is assumed constant and time period is known).

But the actual energy generated by the plant in 24 hours is equal to the area under the curve ABCFILM. Thus by using energy interpretation of plant load factor i.e. (2),

PLF = area under the curve ABCFILM / 4800

        = 2400 / 4800 = 0.5

Thus the second method is not different from the first method but actually both are same.