Steady State Analysis of Transmission Analysis MCQ Quiz - Objective Question with Answer for Steady State Analysis of Transmission Analysis - Download Free PDF

Concept:

Voltage regulation of Transmission line:​

• When a transmission line is carrying a current, there is a voltage drop in the line due to the resistance and inductance of the transmission line.
• Finally, the receiving end voltage ( VR ) of the line is generally less than the sending end voltage (VS)
• The difference in voltage at the receiving end of the transmission line between the conditions of no-load and the full load is called voltage regulation
• Voltage regulation is expressed as a percentage of the receiving end voltage.

 $V.R=\frac{V_S-V_R}{V_R}\times 100=\frac{\left|V_R^{NL}\right|-\left|{\mathrm{V}}_{\mathrm{R}}^{\mathrm{F}\mathrm{L}}\right|}{\left|{\mathrm{V}}_{\mathrm{R}}^{\mathrm{F}\mathrm{L}}\right|}\times 100$

Note: At no load, $|V_S|=|V_R|$

Ferranti Effect:

• The effect in which the receiving end voltage is higher than sending end voltage in the long transmission line.
• It mainly occurs because of a light load or open circuit at the receiving end.
• It is due to the charging current of the line.
• When an alternating voltage is applied, the current that flows into the capacitor is called the charging current.
• The charging current increases in the line when the receiving end voltage of the line is larger than the sending end.
   

Method to Minimize Ferrant Effect:

• This voltage can be controlled by placing the shunt reactors at the receiving end of the lines.
• A shunt reactor is an inductive current element connected between line and neutral to compensate for the capacitive current from transmission lines.
• When this effect occurs in long transmission lines, shunt reactors compensate the capacitive VAr of the lines and therefore the voltage is regulated within the prescribed limits.

Ferranti on long overhead lines can't be experienced when

• The line heavily loaded.
• The power factor is unity.

Given: Impedance = j0.8 Ω, Current = 500 A, Voltage = 300 V

We know that

V_S = V_R + IZ_S

Here V_S = Sending voltage, V_R = Receiving end voltage, Z_S = Line impedance

V_S = V_R + IZ_S = 300 + 500 × 0.8j = 300 + 400j = 500∠53.13°

If we want to transmit the same number of watts, the current will be $\frac{1}{n}$ when the voltage is increased by n.

For the same amount of loss, the new diameter of the wire will be 1/n.

The amount of copper will be $\frac{1}{n^2}$

Concept:

The maximum power transfer capacity of the transmission line is given by

$P=\frac{\left|V_s\right|\left|V_r\right|}{X}\sin \delta$

Where,

|Vs| is sending end voltage of transmission line

|Vr| is receiving end voltage of transmission line

X is the series reactance of transmission line

δ is load angle

Explanations:

For maximum power transfer, sin δ should be equal to 1, so δ = 90°

Transmission lines are classified based on three criteria.

a) Length of transmission line

b) Operating voltage

c) Effect of capacitance

The table below summaries the classification of transmission lines.

The correct answer is option 2):((a) and (b) only)

Concept:

• A feeder is a conductor having constant current density. The size of the feeder is designed based on current-carrying capacity. For V ≤ 220 kV, the selection of conductor is done based on the current-carrying capacity.
• The main criteria for the design of a feeder are its current carrying capacity which accounts for thermal limits rather than voltage drops.
• A distributor has variable loading along its length due to the service conditions of tapping by the individual consumers. The voltage variation at the consumer end must be kept under ±5%. So, the main criterion for the design of a distribution feeder is voltage regulation or voltage drop.
• Hence c) Voltage variation at the consumer's terminal is not considered while designing the distributor is wrong

Types of Transmission Line

The transmission line is divided into three types on the basis of transmission length:

When the physical size of the element is small compared to the wavelength of the electromagnetic wave propagation, it is considered as a lumped element.

When the physical size of the element is very large compared to the wavelength of the electromagnetic wave propagation, it is considered as a distributed element.

In the transmission line, the capacitance is connected as an admittance.

Since admittance is directly proportional to the length of the line, hence capacitor is neglected in the case of the short transmission line.

In the case of a medium transmission line, the effect of capacitance is seen, but its effect is very small. Hence, the capacitor is considered as a lumped element in the medium transmission line.

In the case of the long transmission line, the effect of capacitance is very significant. Hence, the capacitor is considered as a distributed element in the long transmission line.

Under no-load condition or light load condition, medium and long transmission lines may operate at leading power factor due to the capacitance effect.

So that receiving end voltage becomes greater than sending end voltage.

In this case, shunt reactors are needed to bring down receiving end voltage at light loads.

The leading power factor can be changed to a lagging power factor by using a shunt reactor. By using a shunt reactor, it will compensate for the effect of capacitance and changes the power factor.

Note:

• The shunt capacitor is used to improve the power factor.
•  A series reactor smoothens the wave shape.
• A Series capacitor reduces the net reactance in a line.
• Shunt inductor reduces the Ferranti effect by limiting overvoltages at the load side under lightly loaded conditions.

For Distortion less Transmission line:

r = Resistance per unit length

l = Inductance per unit length

g = Shunt conductance per unit length

c = Capacitance per unit length

$\frac{r}{l}=\frac{g}{c}$

or rc = lg

Additional Information

Phase constant

$\beta =\omega \sqrt{\mathrm{l}\mathrm{c}}$

Attenuation constant

$\alpha =\sqrt{RC}$