Ferranti Effect in AC Transmission Lines

 

 Ferranti Effect 

• Under the Ferranti effect, no-load and light load conditions make the receiving end voltage greater than the sending end voltage. If the receiving voltage exceeds the limit, it damages the connected load, which is why reducing the Ferranti effect is essential. 

• The influence of inductance and capacitance on the receiving end voltage of AC transmission lines at light load conditions is the root cause of the Ferranti effect. 

• Shunt compensation and series compensation can be employed in transmission lines to reduce the Ferranti effect.

Electrical transmission lines are subjected to various effects, including the Ferranti effect

Electrical transmission lines are subjected to various effects such as the skin effect, proximity effect, corona discharge, and the Ferranti effect. Specific conditions and circumstances lead to these effects in the transmission line. 

The influence of inductance and capacitance on the receiving end voltage of AC transmission lines at light load conditions is the root cause of the Ferranti effect. Under this effect, no-load and light load conditions make the receiving end voltage greater than the sending end voltage. If the receiving voltage exceeds the limit, it damages the connected load. This is why it is so essential to learn how to reduce the Ferranti effect, which is what we will be discussing in this article. 

The Ferranti Effect in Transmission Lines

Transmission lines can be classified as short, medium, and long. Among these classifications, the long transmission line is composed of the highest amount of capacitance and inductance distributed along the length. 

Consider a nominal pi model of a long transmission line. When the long transmission line is at no load or lightly loaded, the distributed capacitance of the transmission line draws more current. The capacitor charging current through the distributed inductance of the transmission line creates a voltage drop across it (which is in-phase with the sending end voltage). The resultant receiving end voltage becomes greater than the sending end voltage. This phenomenon of overvoltage in a transmission line’s receiving end at light load conditions is called the Ferranti effect. Under the Ferranti effect, the reactive power generated is more than the reactive power absorbed, and this causes the voltage to rise in the receiving end. 

The Disadvantages of the Ferranti Effect

The Ferranti effect is an undesirable effect in electrical AC power systems. All power systems follow the specifications for receiving end voltage with some tolerance level. The loads connected to the system are usually rated for this voltage and safely operate under normally loaded conditions. 

Potential Damage Caused by the Ferranti Effect

However, under light load conditions, the Ferranti effect introduces temporary overvoltage at the receiving end. These overvoltages are capable of limiting the performance of transmission lines and damaging the loads and equipment connected to the receiving end. The damage of voltage-sensitive process controls, controllers, and automated systems leads to a loss of utility and temporary shutdowns. The impact of monetary losses associated with the Ferranti effect can also be extremely damaging to a project’s budget.

How to Reduce the Ferranti Effect

High voltage at the receiving end of a transmission line is hazardous to equipment and personnel. So, how do we reduce the Ferranti effect? Here are a few ways to reduce this effect.

Protection Systems and the Ferranti Effect

Usually, switchgear and protection systems in transmission lines are designed for sending end voltage. When a rise in voltage is experienced in the transmission line due to the Ferranti effect, circuit breakers and other protection devices operate and break the circuit for safety. However, bringing the transmission line switchgears back to a normal state requires maintenance. Reactive compensation in a transmission line is essential, as the reactive power generated is greater than that which is absorbed. 

Passive Compensation

Shunt reactors and series capacitors can reduce the voltage rise if placed at suitable locations in the transmission line. The transmission line inductance can be compensated by connecting series capacitors and the capacitance of the line can be controlled by placing shunt reactors. The series capacitors are placed along the transmission line length, reducing the effective reactance (inductive reactance and capacitive reactance) of the transmission line. The compensation of the transmission line inductance by inserting capacitors in series results in low voltage at the receiving end compared to the sending end voltage.

Shunt reactors are positioned at the ends of the lines and at the junctions where two or more lines meet. Shunt reactors can also be connected across the tertiary winding of the power transformers in electrical transmission systems. The shunt reactors are constructed in the same way as power transformers, with one difference—non-magnetic gaps between the packets of reactor core steel. In 3-phase systems, 3-limbed and 5-limbed core reactors are used alternatively. The neutral of these reactors can be left unearthed, directly earthed, or earthed through the earthing reactor. 

Active Compensation

The Ferranti effect can be mitigated by using FACTS devices for reactive power compensation. Thyristor-controlled reactors and thyristor switched capacitors can be connected to the transmission line, and the proper switching of these devices can help control the Ferranti effect on transmission lines. Compensators such as STATCOM, dynamic voltage restorers, and unified power flow controllers (UPFC) can be introduced into electrical transmission systems for reactive power compensation, which aids in the reduction of the Ferranti effect.

There are passive and active solutions available when it comes to the question of how to reduce the Ferranti effect on transmission lines. Cadence’s software can assist power system engineers in choosing the most effective method of compensation for the given transmission system. 

Surge Impedance Loading

 Surge Impedance Loading

Capacitance and reactance are the main parameters of the transmission line. It is distributed uniformly along the line. These parameters are also called distributed parameters. When the voltage drops occur in transmission line due to inductance, it is compensated by the capacitance of the transmission line.

The transmission line generates capacitive reactive volt-amperes in its shunt capacitance and absorbing reactive volt-amperes in its series inductance.The load at which the inductive and capacitive reactive volt-amperes are equal and opposite, such load is called surge impedance load.

It is also called natural load of the transmission line because power is not dissipated in transmission. In surge impedance loading, the voltage and current are in the same phase at all the point of the line. When the surge impedance of the line has terminated the power delivered by it is called surge impedance loading.

Shunt capacitance charges the transmission line when the circuit breaker at the sending end of the line is close. As shown below

Let V = phase voltage at the receiving end
L = series inductance per phase
XL = series inductance reactance per phase
XC = shunt capacitance reactance per phase
Zo = surge impedance loading per phase

Capacitive volt-amperes (VAr) generated in the line

The series inductance of the line consumes the electrical energy when the sending and receiving end terminals are closed.

Inductive reactive volt-amperes (VAr) absorbed by the line

Under natural load, the reactive power becomes terminated, and the load becomes purely resistive.

And it is calculated by the formula given below

Surge impedance loading is also defined as the power load in which the total reactive power of the lines becomes zero. The reactive power generated by the shunt capacitance is consumed by the series inductance of the line.

If Po is its natural load of the lines, (SIL)1∅ of the line per phase

Since the load is purely resistive,

Thus, per phase power transmitted under surge impedance loading is (VP2)/ZO watts, Where Vp is the phase voltage.

If kVL is the receiving end voltage in kV, then

Surge impedance loading depends on the voltage of the transmission line. Practically surge impedance loading always less than the maximum loading capacity of the line.

If the load is less than the SIL, reactive volt-amperes are generated, and the voltage at the receiving end is greater than the sending end voltage. On the other hand, if the SIL is greater than the load, the voltage at receiving end is smaller because the line absorbs reactive power.

If the shunt conductance and resistance are neglected and SIL is equal to the load than the voltage at both the ends will be equal.

Conclusion

Surge impedance load is the ideal load because the current and voltage are uniform along the line. The wave of current and voltage is also in phase because the reactive power consumed are equal to the reactive power generated by the transmission line.

Basic Definitions in Magnetic circuit

 

Magnetic Circuit

The closed path followed by magnetic lines of forces is called the magnetic circuit. In the magnetic circuit, magnetic flux or magnetic lines of force starts from a point and ends at the same point after completing its path.

Magnetic flux

The magnetic lines of force passing through a magnetic circuit is known as magnetic flux. It is denoted by a symbol ϕ and given by a formula ϕ = BA, where B is the magnetic flux density and A is the area of the cross-section in m2. The unit of magnetic flux is weber.

Magneto-motive force

Magneto-motive force or MMF is the cause for producing the magnetic flux. The MMF in a magnetic circuit depends on the number of turns(N) and the amount of current(I) flowing through it.

It is given by a formula, MMF = NI and its unit is ampere turns.

Magnetic flux density

It is the amount of magnetic flux per unit area at right angles to the flux. The unit of magnetic flux density is weber/m2 and denoted by B. The formula is given by,

  

Magnetic field intensity

Magnetizing force or Magnetic field intensity or magnetic field strength is the MMF required to magnetize a unit length of the magnetic flux path. The unit of magnetic field intensity is AT/m and is denoted by H.

  

Reluctance

It is the opposition that the magnetic circuit offers for the flow of magnetic flux. We can also define the reluctance as the ratio of magneto-motive force to the magnetic flux. It is denoted by S and its unit is ampere-turns per weber.

  

Permeance

Permeance is the reciprocal of reluctance. The ease with which the flux can pass through the material is known as permeance. Weber/AT is the unit of permeance.

  

Permeability

It is the measure of the resistance of a material against the formation of a magnetic field. In simple words, the permeability of material means its conductivity for magnetic flux. The reciprocal of magnetic permeability is magnetic reluctivity. Greater permeability, greater is its conductivity.

Magnetic permeability is represented by a greek letter μ. It is given by a formula,

  

Relative permeability

It is the ratio of flux density of a magnetic material to the flux density produced in air by the same magnetizing force.

The formula for relative permeability is,

  

where μr – relative permeability of the magnetic material.

μ0 – absolute permeability of air or vacuum.

μ – absolute permeability of the magnetic material.

Analogy between Magnetic circuit and Electric Circuit

Magnetic Circuit	Electric Circuit
A closed path for a magnetic flux forms a magnetic circuit.	A closed path for an electric current form an electric circuit.
Magnetic flux does not flow in a magnetic circuit.	Electric current always flows in an electric circuit.
MMF is the cause for producing flux.	EMF is the cause for producing current.
Weber is the unit of flux.	Ampere is the unit of current.
Reluctance opposes the flow of flux.	Resistance opposes the flow of current.
Flux density,	Current density,
Magnetic field intensity,	Electric field intensity,
Magnetic flux lines flow from the North pole to the South pole.	Electric current flows from the positive to negative terminal.

Leakage Flux And Fringing

Leakage flux is defined as the magnetic flux which does not follow the particularly intended path in a magnetic circuit. 

Taking an example of solenoid you can explain the leakage flux and the fringing both.

When a current is passed through a solenoid, magnetic flux is produced by it.

Most of the flux is set up in the core of the solenoid and passes through the particular path that is through the air gap and is utilised in the magnetic circuit. This flux is known as Useful flux φu.

As practically it is not possible that all the flux in the circuit follows a particularly intended path and sets up in the magnetic core and thus some of the flux also sets up around the coil or surrounds the core of the coil, and is not utilised for any work in the magnetic circuit. This type of flux which is not used for any work is called Leakage Flux and is denoted by φl.

Therefore, the total flux Φ produced by the solenoid in the magnetic circuit is the sum of the leakage flux and the useful flux and is given by the equation shown below:

Leakage coefficient

The ratio of the total flux produced to the useful flux set up in the air gap of the magnetic circuit is called a leakage coefficient or leakage factor. It is denoted by (λ).

Fringing

The useful flux when sets up in the air gap, it tends to bulge outward at (b and b’) as shown in above figure, because of this bulging, the effective area of the air gap increases and the flux density of the air gap decreases. This effect is known as Fringing.

Fringing is directly proportional to the length of the air gap that means if the length increases the fringing effect will also be more and vice versa.

 Superposition Theorem 

    Superposition Theorem state that in any linear bilateral network having more than one source, the response in any one of the element is equal to algebraic sum of the response caused by individual source while rest of the sources are replaced by their internal resistances.

1.We consider one independent source at a time while all other independent sources are turned off. This implies that we replace voltage source by short circuit and current source by open circuit. This way we obtain a simpler and more manageable circuit.

2.Turn off all independent sources except one source. Find the output (voltage or current) due to that active source.

3.Repeat  the above step for each of the other independent sources.

4.Find the total contribution by adding algebraically all the contributions due to the independent sources.

Consider the following linear circuit with two sources: one current source and one voltage source. The two sources are the inputs to the function. For this problem we happen to want to find two outputs, currents $i_1$i, start subscript, 1, end subscript and $i_2$i, start subscript, 2, end subscript.

Let’s analyze this circuit using superposition.

First, we suppress the current source and analyze the circuit with just the voltage source acting alone. To suppress the current source, we replace it with an open circuit.

With just the voltage source, the two output currents are:

$i_{1V} = 0 \qquad i_{2V} = \dfrac{\text{Vs}}{\text R2}$i, start subscript, 1, V, end subscript, equals, 0, i, start subscript, 2, V, end subscript, equals, start fraction, start text, V, s, end text, divided by, start text, R, end text, 2, end fraction

Where $i_{1V}$i, start subscript, 1, V, end subscript and $i_{2V}$i, start subscript, 2, V, end subscript are the currents in $\text R1$start text, R, end text, 1 and $\text R2$start text, R, end text, 2 caused by the voltage source.

Next, we restore the current source and suppress the voltage source, to calculate the contribution of the current source acting alone.

With just the current source, the two output currents are:

$i_{1I} = \text{Is} \qquad i_{2I} = 0$

Where $i_{1I}$i, start subscript, 1, I, end subscript and $i_{2I}$i, start subscript, 2, I, end subscript are the currents in $\text R1$start text, R, end text, 1 and $\text R2$start text, R, end text, 2 caused by the current source. 

[zero current in R2?]

We complete the analysis by adding the contributions from each source:

$i_1 = i_{1V} + i_{1I} = 0 + \text{Is}= \text{Is}$i, start subscript, 1, end subscript, equals, i, start subscript, 1, V, end subscript, plus, i, start subscript, 1, I, end subscript, equals, 0, plus, start text, I, s, end text, equals, start text, I, s, end text

$i_2 = i_{2V} + i_{2I} = \dfrac{\text{Vs}}{\text R2} + 0 = \dfrac{\text{Vs}}{\text R2}$