Figure 1: Single Line to Ground Fault.

Consider an unloaded 3-Φ generator with its neutral being grounded through an impedance Z_n. Let a single line to ground fault occurs on phase ‘a’ through a fault impedance Z_f as shown in figure (1). Under the fault condition the currents and voltage are given as

\[{{I}_{b}}={{I}_{c}}=0\]

\[{{V}_{a}}={{I}_{a}}{{Z}_{f}}…(1)\]

The symmetrical components of current are given as,

\[\left[ \begin{matrix}
{{I}_{a0}} \\
{{I}_{a1}} \\
{{I}_{a2}} \\
\end{matrix} \right]=\frac{1}{3}\left[ \begin{matrix}
1 & 1 & 1 \\
1 & a & {{a}^{2}} \\
1 & {{a}^{2}} & a \\
\end{matrix} \right]\left[ \begin{matrix}
{{I}_{a}} \\
{{I}_{b}} \\
{{I}_{c}} \\
\end{matrix} \right]\]

Substituting I_b = I_c = 0 in the above equation we get

\[\left[ \begin{matrix}
{{I}_{a0}} \\
{{I}_{a1}} \\
{{I}_{a2}} \\
\end{matrix} \right]=\frac{1}{3}\left[ \begin{matrix}
1 & 1 & 1 \\
1 & a & {{a}^{2}} \\
1 & {{a}^{2}} & a \\
\end{matrix} \right]\left[ \begin{matrix}
{{I}_{a}} \\
0 \\
0 \\
\end{matrix} \right]\]

\[=\frac{1}{3}\left[ \begin{matrix}
{{I}_{a}}+0+0 \\
{{I}_{a}}+0+0 \\
{{I}_{a}}+0+0 \\
\end{matrix} \right]=\frac{1}{3}\left[ \begin{matrix}
{{I}_{a}} \\
{{I}_{a}} \\
{{I}_{a}} \\
\end{matrix} \right]\]

\[{{I}_{a0}}={{I}_{a1}}={{I}_{a2}}=\frac{1}{3}{{I}_{a}}…(2)\]

For an unloaded 3-Φ generator we know that

\[{{V}_{a0}}=-{{I}_{a0}}{{Z}_{0}}\]

\[{{V}_{a1}}={{E}_{a}}-{{I}_{a1}}{{Z}_{1}}\]

\[{{V}_{a1}}=-{{I}_{a2}}{{Z}_{2}}\]

Also,

\[{{V}_{a}}={{V}_{a0}}+{{V}_{a1}}+{{V}_{a2}}\]

Substituting V_a0, V_a1 and V_a2 in the above equation, we get

\[{{V}_{a}}=-{{I}_{a0}}{{Z}_{0}}-{{I}_{a1}}{{Z}_{1}}-{{I}_{a2}}{{Z}_{2}}\]

\[{{V}_{a}}={{E}_{a}}-\left( {{I}_{a0}}{{Z}_{0}}-{{I}_{a1}}{{Z}_{1}}-{{I}_{a2}}{{Z}_{2}} \right)\]

\[{{V}_{a}}={{E}_{a}}-\left( \frac{{{I}_{a}}}{3}{{Z}_{0}}+\frac{{{I}_{a}}}{3}{{Z}_{1}}+\frac{{{I}_{a}}}{3}{{Z}_{2}} \right)\]

\[{{V}_{a}}={{E}_{a}}-\frac{{{I}_{a}}}{3}\left( {{Z}_{0}}+{{Z}_{1}}+{{Z}_{2}} \right)…(3)\]

Substituting equation (1), in equation (3), we get

\[{{I}_{a}}{{Z}_{f}}={{E}_{a}}-\frac{{{I}_{a}}}{3}\left( {{Z}_{0}}+{{Z}_{1}}+{{Z}_{2}} \right)\]

\[3{{I}_{a}}{{Z}_{f}}=3{{E}_{a}}-{{I}_{a}}\left( {{Z}_{0}}+{{Z}_{1}}+{{Z}_{2}} \right)\]

\[3{{E}_{a}}=3{{I}_{a}}{{Z}_{f}}+{{I}_{a}}\left( {{Z}_{0}}+{{Z}_{1}}+{{Z}_{2}} \right)\]

\[3{{E}_{a}}={{I}_{a}}\left( {{Z}_{0}}+{{Z}_{1}}+{{Z}_{2}}+3{{Z}_{f}} \right)\]

\[{{I}_{a}}=\frac{3{{E}_{a}}}{{{Z}_{0}}+{{Z}_{1}}+{{Z}_{2}}+3{{Z}_{f}}}\]

From equation (2), we have

\[{{I}_{a0}}={{I}_{a1}}={{I}_{a2}}=\frac{1}{3}{{I}_{a}}\]

\[=\frac{1}{3}\left( \frac{3{{E}_{a}}}{{{Z}_{0}}+{{Z}_{1}}+{{Z}_{2}}+3{{Z}_{f}}} \right)\]

\[=\frac{{{E}_{a}}}{{{Z}_{0}}+{{Z}_{1}}+{{Z}_{2}}+3{{Z}_{f}}}\]

The fault current is,

\[{{I}_{f}}={{I}_{a}}=\frac{3{{E}_{a}}}{{{Z}_{0}}+{{Z}_{1}}+{{Z}_{2}}+3{{Z}_{f}}}\]

Figure 2: Interconnection of sequence Network of Single Line to Ground Fault.

Hence, from the above equation it is observed that, inter connection of sequence networks to simulate L-G fault contains positive, negative and zero sequence networks.  As the three sequence networks carry the same current they are connected in series as shown in figure (2). If an alternator neutral is solidly grounded, Z_n = 0.

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or

The equation which describes the behavior of a synchronous machine during transient period is known as swing equation.

Figure 1.

Derivation of Swing Equation

Consider the synchronous machines showing the flow of mechanical and electrical power as shown in figure (1)

Let,

θ – Angular displacement in radians

ω – Angular velocity in radian/sec

α – Angular acceleration in radians/sec²

T_m – Turbine torque in N-m

T_e – Electromagnetic torque in N-m

J – Moment of inertia in kg-m².

When the synchronous machine is acting as a generator (see Figure 1), T_m is the input and T_e is the output and in case of generator, the accelerating torque is given by,

\[{{T}_{a}}={{T}_{m}}-{{T}_{e}}…(1)\]

Multiplying equation (1) with ‘ω’, we get,

\[\omega {{T}_{a}}=\omega {{T}_{m}}-\omega {{T}_{e}}\]

Since,

\[P=\omega T\]

\[{{P}_{a}}={{P}_{m}}-{{P}_{e}}…(2)\]

Where,

P_a = Accelerating power

P_m = Mechanical power

P_e = Electrical power.

We know that,

\[{{T}_{a}}=J\alpha \]

\[{{P}_{a}}=J\alpha \omega \]

\[=\left( J\omega  \right)\alpha \]

But,

\[M=J\omega \]

\[{{P}_{a}}=M\alpha …(3)\]

Where, M is the angular momentum.

Figure 2

The variation of angular position and velocity with respect to synchronously rotating axis is shown in figure (2). From figure (2),

\[\theta ={{\omega }_{s}}t+\delta …(4)\]

Where, δ = Load angle (or) Torque angle

Differentiating equation (4) with respect to t, we get,

\[\frac{d\theta }{dt}={{\omega }_{s}}+\frac{d\delta }{dt}\]

Again differentiating above equation with respect to t,

\[\frac{{{d}^{2}}\theta }{d{{t}^{2}}}=\frac{{{d}^{2}}\delta }{d{{t}^{2}}}\]

But,

\[\frac{{{d}^{2}}\theta }{d{{t}^{2}}}=\alpha \]

Thus

\[\alpha =\frac{{{d}^{2}}\delta }{d{{t}^{2}}}…(5)\]

From equation (3),

\[{{P}_{a}}=M\alpha \]

\[\alpha =\frac{{{P}_{a}}}{M}…(6)\]

Using equation (2), in equation (6), we get,

\[\alpha =\frac{{{P}_{m}}-{{P}_{e}}}{M}…(7)\]

Substituting equation (5) in equation (7), we get,

\[\frac{{{d}^{2}}\delta }{d{{t}^{2}}}=\frac{{{P}_{m}}-{{P}_{e}}}{M}\]

\[M\frac{{{d}^{2}}\delta }{d{{t}^{2}}}={{P}_{m}}-{{P}_{e}}…(8)\]

or

\[\frac{H}{\pi f}\frac{{{d}^{2}}\delta }{d{{t}^{2}}}={{P}_{m}}-{{P}_{e}}\]

Equation (8) is called the swing equation. It explains the rotor dynamics for a synchronous machine. It also describes the behavior of a synchronous machine during transient period.

Swing equation is a second order non-linear differential equation since the electrical power, P_e = P_max sin δ is based on the sine of angle δ.

The swing equation is given by,

\[M\frac{{{d}^{2}}\delta }{d{{t}^{2}}}={{P}_{m}}-{{P}_{e}}\]

Or

\[\frac{H}{\pi f}\frac{{{d}^{2}}\delta }{d{{t}^{2}}}={{P}_{m}}-{{P}_{e}}\]

Where,

M = Inertia constant in p.u

H = Inertia constant in NIJ/MVA

f = Frequency in Hz

δ = Power angle in electrical radians

P_m = Mechanical power input in p.u

P_e = Electrical power output in p.u.

Swing Curve

Swing Curve Swing curve is a graph drawn between the torque angle (δ) on Y-axis and time (t) on X-axis. It is generally drawn for a transient state to study the nature of variations in torque angle (δ) for sudden disturbances. The variation of ‘δ’ with respect to time ‘t’ is shown in figure 3,

Figure 3: Swing Curve.

Swing curve explains the behavior of a synchronous machine during transient and gives information regarding the stability of a power system for any disturbance. These curves are very useful in determining the effectiveness of relay protection on power system with respect to the fault clearance before one or more machines become unstable and lose synchronism.

Importance of Swing Equation

In power system, the swing equation has a great importance for the study of transient stability. The swing equation is used to determine the stability of a rotating synchronous machine within a power system. When swing equation is solved, the expression for ‘δ’ is obtained, which the function of time. The graph of this solution is known as ‘Swing Curve’ of the machine. By investigating the swing curves of overall machines connected to the system we can know whether the machines continue in synchronous or not after a disturbance. The following graph represents the variations in δ with respect to time t. From the above graph the condition $\frac{d\delta }{dt}=0$ indicates that the system is stable and if, $\frac{d\delta }{dt}>0$, then the system is unstable.

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In the gas turbine power plant, electrical energy generation takes place by using gas turbine as the prime mover.

Layout and Working  of Gas Turbine Power Plant

The layout of gas turbine influences the performance of the plant. The sharp bends in the pipelines may cause reduction in the power developed. Thus, the gas turbine plant must be designed carefully as per the standards. The air from atmosphere enters into low pressure compressor through air filters and then high pressure compressor through intercooler. This pressurised air is sent to heat exchanger, where it gets heated. Further, it is passed into combustion chamber and undergoes combustion process. The products of combustion enter high pressure turbine where they get expanded and then passes into the low pressure turbine. for further expansion. Therefore, the turbine power (rotation) is converted into electrical energy by means of a generator. Layout of a typical gas power plant is shown in below figure 1.

Figure 1: Layout of Gas Turbine Power Plant

Components of Gas Turbine Power Plant

Compressor

The function of compressor is to compress the atmospheric air to high pressure. The compressor used for gas turbine plant is of axial type. It produces high pressures upto 15 bars. It is run by coupling it with the shaft of turbine.

Following two types of compressors are used in gas turbines.

Centrifugal Compressor: In this compressor, air flow is in radial direction. The pressure ratio per stage is 4-5. The efficiency of the compressor is about 80

Axial Flow Compressor: In this type. air flow is parallel to axis of compressor. The pressure ratio is about 1 – 2. The efficiency of the compressor is 86

Combustion Chamber

The high pressure air coming out of the compressor enters into the combution chamber. The fuel is injected into the combustion chamber, where the combustion of air-fuel mixture takes place. It is located between the compressor and the turbine, The fuels used for combustion are coal gas, producer gas, oil petrol, etc.

Turbine

It is an essential component of the open cycle gas turbine plant. High temperature and high pressure gases are expanded in it. A jet of hot gases and air mixture is made to flow over rings of moving blades of the turbine wheel. The velocity of the jet decreases as its kinetic energy is absorbed by the rings of blades, imparting rotary motion to the shaft of the turbine. The mechanical power obtained from the turbine is then converted into electrical power by coupling a generator to the turbine shaft.

Regenerator

The temperature of the gases leaving the turbine is considerably higher than that of air delivered from the compressor. The heat of exhaust gases can be used to preheat the air delivered from the compressor, thus reducing the mass of fuel supplied in the combustion chamber. A counter flow heat exchanger, which traufers heat from hot gases to air is known as regenerator.

Intercooler

In gas turbine power plants, inter-cooler is used only when the pressure ratios are very large and the compression is accomplished. Cold water is used for cooling the compressed air. The purpose of intercooler is to reduce the temperature of working fluid between the stages, approximately to initial temperature. Cross-flow type of intercooler is mostly used for obtaining higher efficiency.

Advantages of Gas Turbine Power Plant

1. The capital cost required to generate the power is less.
2. Gas turbine power plant has high reliability
3. Gas turbines are more flexible during the operation.
4. These type of power plants can be started quickly by using various types of ihels.
5. The overall efficiency of gas turbine and conventional system power plant is same and is about 35

Applications of Gas Turbine Power Plants

The various applications of gas turbines are as follows,

1. Power Generation: Gas turbines are extensively employed for generation of electricity. Depending upon the utilization of power, the gas turbines are classified into three types,

Small Sized Gas Turbine: It generates power less than 2 MW. The turbine contains centrifùgal compressor. which is driven by radial inward flow turbine.

Medium Sized Gas Turbine: This turbine gen rates power in the range of 5 MW to 50 MW. It consists of axial-flow compressors and axial flow turbines.

Large Sized Gas Turbine: These turbines generates power more than 50 MW. They are operated at a pressure ratio of 35:1 and a very high temperature of about 13-15°C.

2. Aviation: The gas turbines are widely employed in the aircraft field to drive air compressors in turbo-jets and turbo-propeller engines. These engines are operated at a temperature range of about 800°C- 1000°C.

3. Supercharger: The gas turbines can also be used in superchargers to produce the compressed air. These superchargers are fitted in the aviation gasoline engines and also heavy duty diesel engines.

4. Marine Engines: The gas turbines can also be used in marine engines, since these turbines does not require water storage tanks or distillation plants.

5. Railway Engines: The main purpose of gas turbine is to drive the air compressor in the railway engines.

Other applications of gas turbines are to run turbo pumps, rotary compressors and can also be used in high speed racing cars and hovercrafts.

The post What is Gas Turbine Power Plant? Working, Diagram & Applications appeared first on ElectricalWorkbook.

Types of Electrical Loads

The categorization of loads are done in order to have different tariffs (band of power that a consumer can use and pay amount corresponding to that slab) can be collected from the consumers. By having different slab rates for the power consumer, the consumers who consume the least (e.g., household) are made to pay judiciously as compared to the consumer who consumes heavy power (e.g., industry). The loads may be resistive, capacitive and inductive in nature. The various types of loads are:

Domestic Load

They are single phase loads connected to a three-phase transformer. They are household loads. The household loads may have fans, air conditioners, refrigerators, lights, television, etc. They have their demand for 24 hours and may vary from time to time in a day and in a year. The load factor is low at 10

Commercial Load

Commercial loads are the ones like the shops and malls. Their main power consumption is for lighting and airconditioning. They need the power during a particular time duration in a day. There may be special occasions like festival times and discount sales times where they may work for more time. The types of commercial loads are canteens/pubs, cloths and shoes, surgeries, grocery and bakery, etc.

Industrial Loads

They may be small scale, large scale and medium scale industries. The industries may be petrochemical, pesticides, pharmaceuticals, pulp and paper, tannery, etc.

Agriculture Loads

They are generally pumps which runs during nights. They are the pumps installed to supply to water for the irrigation and agriculture.

Traction and Railways

These types of loads include rope cars, trolley busses, tram cars, railways, etc. They may be public transport and hence will be maximum during the morning and evening hours.

Importance of Electrical Load

Loads are important part of the power system. They are the need or demand as it is technically called as. The power producers promise the consumers who use the loads with certain standards of power supply, like voltage within certain band and frequency also within certain band. If this is violated in today’s scenario the power producing company may be taken action legally. Hence proper supply system should be ensured to the consumers. This can be done by an in-depth understanding of the loads and its various types, its characteristics, etc. Most of our household appliances are single phase, Examples are television, microwave oven, air conditioner, hair drier, iron box are single phase. They are operated at different times of the day. Their usage may differ for a month and an entire year. During a month there may be increasing or decreasing number of family members. During an year there may be different seasons like summers and winters. On the whole it is the duty of the substations nearer to the household consumers or small consumers for single phase power, to deliver the right amount of balanced power. In case of shortage of power like loss of a generator or imbalance in the power system these consumers can be given first preference for load shedding.

The industrial consumers are the important lot a consumers, whose productions cannot be stopped. If done so, may cause crores of economical loss for the industry and to the nation. They are essentially three-phase consumers, be it small scale industries or large scale industries. The substations connected to them should ensure “peak hour” supply of power to these industries. They are not an option for load shedding unless there is a major fault and grid separation.

Effects of Variable Electrical Loads on Power System

When a particular part of the power (city, colony, industries, etc.) systems is planned, the load of that part is calculated on certain assumptions. Based on the total load, the active and reactive power requirements, the generation, transmission and distribution is planned. Once the planning is done, the installation is done based on the demand taking some projected upper limit. After the city is planned and after some years the demand of the same area goes above the projected limit. These may be due to the sophistication or development of that particular area. In order to cater to these demands, the following measures have to be taken.

Increasing Generation

The existing generating capacity may be increased by installing a generator in an existing plant or planning and installing a new power plant. The increasing demand may also be satiated by spinning reserve. This increases the cost of production and the tariffs has to be revised.

Installing Higher Capacity Devices

If the generation and the demand increases the protective devices and the associated measuring devices, relays and others should be upgraded accordingly. This also increases the cost and has to be taken care in tariffs.

Power Factor Improvement

When the load is increased there are chances that the power factor of the system reduces. This means that there is more reactive power consumption. But the generator is deviced to deliver a certain amount of reactive power. Therefore special devices are installed at various points that are economical and improves the power.

Transformers, Cables and Others

As the capacity is increased the transformers, cables and protective devices should be upgraded. The appropriate capacity of transformers, cables and transmission lines should be installed. The effects of increasing the loads on the power system, incidentally increases the cost of production and transmission of power. This in turn puts burden on the consumer. If the capacity has to be increased, new transmission lines has to be errected. This posses getting right of way for the excess capacity.

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When in case of an overhead transmission line, all the three phases are supposed to be spaced symmetrically keeping the line constants, identical but due to some mechanical considerations it in not so, these are placed irregularly. It disturbs the line constants, as a result the voltage drop in the three lines will be different and unequal line voltage is achieved at the receiving end.

This difficulty is overcome by the use of transposition of conductors. The conductors of the transmission line are kept at a regular distance and (the system adopted is known as transposition of conductors as shown in Fig. 1.

Figure 1: Transposition of Transmission Line

In this case let the sequence by RYB, then after x/3 distance it is made by transposition as BRY and after again x/3 distance as YBR, by inter changing the line conductors, so the arrangement of changing the positions of overhead line conductors of the transmission line at regular distance in order to make line constants symmetrical is called transposition of conductors.

Procedure of Transposition of Transmission Line

As stated, the transposition of line conductors involves changing the positions of line conductors for one-third of the length of the line. Transposition of a transmission line over the length ‘x’ is shown in the figure 1. The phase conductor ‘R’ will take the position 1 for the first 1/3^rd of the length of the line, it will take position 2 and finally it occupies position 3 in the remaining 1/Y” of the length of the line. The phase conductors ‘Y’ and ‘B’ also changes their position in a similar manner as shown in the figure.

Need of Transposition of Transmission Line

The transmission line is transposed,

1. To maintain constant voltage drop across each conductor.
2. To maintain equal voltage at the receiving end of the transmission line for all phases.
3. To reduce disturbances to the nearby communication lines.

Advantages of Transposition of Transmission Line

The advantages of transposition of conductors are as follows.

1. Equal average inductance is appeared across each conductor/phase of the line.
2. Constant voltage drop is maintained across each conductor/phase of the line.
3. Equal voltages are appeared at the receiving end of the lines.
4. Electrostatically induced voltages are balanced along the balTel.
5. Electromagnetically induced e.m.f is reduced on wires.
6. The line constants are similar for all the three phases.
7. It also prevents the telecommunication or radio interferences from the neighbouring line

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Construction of Under Frequency Relay

This relay consists of four coils wound on the four poles of stator. These poles act as electromagnets when the coil is energized. The rotor has cup type structure as shown in Figure 1.

Figure 1: Under Frequency Relay.

By using the impedances that are dependent of frequency, the coils on the stator are connected in parallel across supply voltage through a voltage transformer.

Operating Principle and Working of Under Frequency Relay

Under frequency relay regularly monitors the frequency and works on the principle of Ferraris measurement. At normal frequency. the impedances are in balance condition. Thus, the rotor does not experience any torque. But when frequency decreases, these impedances vary hence imbalance condition occurs and torque is applied on the rotor. Thus, the relay operates.

The power system operates at normal frequency when it is in stable condition. During this condition. the mechanical power input of generators is exactly equal to the algebraic sum of all load demands and power losses. But if excessive loading of generator takes place i.e., if the generator is forced to give more output power than its rated capacity, then imbalance occurs in  the above power equation of generator and results in reduction of speed in generator. The relation between frequency and speed of generator is given by,

\[{{\text{N}}_{\text{S}}}=\text{ }\frac{120\text{ f}}{\text{P}}\]

Where,

f = System frequency

N_S = Synchronous speed

P = Number of poles.

From the above equation. it is clear that under speeding caused due to increment in load leads to reduction of frequency. The rate at which frequency decreases depends on overload percentage, variations in load and generator and the duration of overload on system.

Applications of Under Frequency Relay

Under frequency relay is used to maintain the stability of the system and it also avoids damage to generator caused due to overloading. They are used in several load stations i.e., substations and power generating units to trip specified feeders and generators respectively. The setting of these relays is based on maximum permissible overloading conditions.

The post What is an Under Frequency Relay? Working, Diagram & Construction appeared first on ElectricalWorkbook.

Abnormalities and Faults of Transformer

The faults occurring in power transformer are earth faults, phase to phase faults, inter turn faults, over heating because of overloading, core heating. The most common type of faults are; earth faults, inter turn faults.

Through fault: It is a fault that lies outside the protective zone of the transformer hut fed through the transformer.

Table 1 shows abnormal conditions; the protection necessary for power transformer.

Table 1

For other transformers the protections provided are as per Table 2.

Table 2

Information Needed for Protective Scheme Selection for Transformer

While selecting protection scheme for transformer the following in format ion is necessary.

1. kVA rating.
2. Whether it is indoor or outdoor.
3. Voltage ratio.
4. Connections of winding.
5. Percentage reactance.
6. Neutral point earthing, value of resistance.
7. Conservator is provided or not.
8. Value of system earthing resistance.
9. Details of connection of CTs and relays.
10. Fault level at transformer terminals.
11. Network diagram showing position of transformer.

The post What is Transformer Protection? Theory, Protection Relays & Types appeared first on ElectricalWorkbook.

Fig. 1: Earthing Switch

Installation of Earthing Switch

It is installed along with the isolator to discharge the voltage on circuit capacitance to earth and provides safety (o the operator doing maintenance/ repair/ inspection work.

Function of Earthing Switch

Earthing switch provides phase to ground discharge path to capacitive charges trapped in the circuit (particularly a high voltage circuit has inherent capacitance).

Even after opening the circuit by C.B. and isolating by isolator, the voltage remains trapped on the capacitors and there remains a chance of getting shock to the operator. Hence earthing switch is necessary.

Working of Earthing Switch

(1) During opening of circuit breaker

Fig. 2 shows the sequence of operations as

1. Opening C.B
2. Opening Isolator
3. Closing earth switch.

(2) During closing circuit breaker

Fig. 3 shows the sequence of operations as

1. Opening earth switch.
2. Closing Isolator
3. Closing C.B.

The post What is Earthing Switch? Working, Symbol, Installation & Function appeared first on ElectricalWorkbook.

Fig. 1: Reactance relay

The current in cylinder portion which is opposite operating coil pole reacts with the flux of that pole and produces torque ∝ I² tending to close the relay contacts. And current in the cylinder portion opposite to restraining coil pole reacts with the flux of that pole and produce a torque a VI tending to open the relay contacts

Torque Equation of MHO Relay

The general Torque equation is

\[\text{T}=~\text{ }{{\text{K}}_{\text{1}}}{{\text{I}}^{\text{2}}}\text{ – }{{\text{K}}_{\text{2}}}\text{V I cos}\left( \Phi \text{ }~\text{ }\!\!~\!\!\text{  – }~\text{ }\!\!~\!\!\text{  }\alpha \text{ }\!\!~\!\!\text{ } \right)\text{ – }~\text{ }\!\!~\!\!\text{  }{{\text{K}}_{\text{3}}}\]

where, T = Operating torque

V = Voltage to relay coil

I = Current to relay coil

Φ = Phase angle between V and I

α = Angle of maximum torque

K₃ = Spring constant

Angle of maximum torque ‘α’ = 90°

\[\text{cos}\left( \Phi ~-~\text{ }\alpha ~ \right)=\text{ cos}\left( \Phi ~-~\text{ }90~ \right)=\text{ sin }\Phi \]

When relay is on the verge of operation,

\[\text{T }=\text{ }~0\]

\[\text{0}=~\text{ }{{\text{K}}_{\text{1}}}{{\text{I}}^{\text{2}}}\text{ – }{{\text{K}}_{\text{2}}}\text{V I cos}\left( \Phi ~\text{ }-\text{ }\alpha  \right)\text{ -}~\text{  }{{\text{K}}_{\text{3}}}\].

\[{{\text{K}}_{\text{1}}}{{\text{I}}^{\text{2}}}=~\text{ }{{\text{K}}_{\text{2}}}\text{V I cos}\left( \Phi ~\text{ }-\text{ }\alpha  \right)\text{ -}~\text{  }{{\text{K}}_{\text{3}}}\]

Dividing both sides by K₂I²

\[\frac{{{\text{K}}_{\text{1}}}}{{{\text{K}}_{\text{2}}}}\text{ }=\text{ }\frac{\text{V sin }\Phi }{\text{I}}+\frac{{{\text{K}}_{\text{3}}}}{{{\text{K}}_{\text{2}}}{{\text{I}}^{2}}}\]

Neglecting K3,

\[\frac{\text{V}}{\text{I}}\sin \Phi =\frac{{{\text{K}}_{\text{1}}}}{{{\text{K}}_{\text{2}}}}=\text{X}\]

T sin Ø = = X where, X = reactance of potential circuit.

K₁ / K₂ = X so relay operates on reactance only.

So operating characteristic is a vector whose head lie on the characteristic such that it has a constant magnitude of X as shown in Fig. 2.

Fig. 2: Characteristics of Reactance Relay

The post What is Reactance Relay? Theory, Diagram, Torque Equation & Characteristics appeared first on ElectricalWorkbook.

Fig. 1: MHO Relay.

Torque Equation of MHO Relay

The general Torque equation is

\[\text{T }=\text{ }\!\!~\!\!\text{ }{{\text{K}}_{\text{1}}}\text{V I cos}\left( \Phi \text{ }\!\!~\!\!\text{  – }\!\!~\!\!\text{  }\alpha  \right)\text{ – }{{\text{K}}_{\text{2}}}{{\text{V}}^{\text{2}}}\text{ – }\!\!~\!\!\text{  }{{\text{K}}_{\text{3}}}\]

where, V = Voltage to relay coil

I = Currcnt to relay coil

Φ = p.f. angle

α = Angle of max. torque

K₃ = Constant of relay spring

At the time when relay is on the verge of operation

\[\text{T }=\text{ }~0\]

\[{{\text{K}}_{\text{2}}}{{\text{V}}^{\text{2}}}\text{ }=\text{ }~{{\text{K}}_{\text{1}}}\text{V I cos}\left( \Phi ~\text{ }-\text{ }~\alpha  \right)\text{ – }~{{\text{K}}_{\text{3}}}\]

Dividing both sides by K₂VI

\[\frac{\text{V}}{\text{I}}=\text{ Z}=\frac{{{\text{K}}_{\text{1}}}}{{{\text{K}}_{\text{2}}}}\text{ cos}\left( \Phi ~\text{ }-\text{ }~\alpha  \right)-\frac{{{\text{K}}_{\text{1}}}}{{{\text{K}}_{\text{2}}}\text{V I}}\]

Neglecting the effect of spring K₃ = 0

\[\frac{\text{V}}{\text{I}}=\frac{{{\text{K}}_{\text{1}}}}{{{\text{K}}_{\text{2}}}}\text{ cos}\left( \Phi ~\text{ }-\text{ }~\alpha \text{ }\!\!~\!\!\text{ } \right)\]

This is the equation of a circle of diameter K₁ / K₂.

Consider that mho relay is located at point A (Fig. 2). So relay operates only for fault occurring in line AB and it doesn’t operate for faults occurring in AC. Hence, from this operation it is clear that this relay is made inherently directional.

Fig. 2: Characteristics of MHO Relay

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