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Which of the following is NOT a disadvantage of EHV lines?
1. Corona problems increase in the voltage gradient.
2. Generation and distribution of power is in AC only.
3. Insulation coordination based upon switching impulse level.
4. Over voltage problem due to switching surge

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Correct Answer - Option 2 : Generation and distribution of power is in AC only.

Advantages of EHV Transmission System:

Electrical energy is generated at a voltage of about 11 kV using alternators.

This voltage is then stepped up to 132, 220, or 400 kV for transmission purposes and later it stepped down easily with a transformer for distribution.

For transmission of electric power high voltage is preferred because of the following advantages,

1) Reduction in the current Power in transmission:

P = √3 VI cos ϕ

Where, V is transmission voltage, I load current and cos ϕ is power factor

Hence, I = \(\dfrac{P}{\sqrt3 Vcos\phi}\)

From the above expression it can be seen that for the constant power and power factor, the load current is inversely proportional to the transmission voltage.

With the increase in transmission voltage, load current gets reduced.

As the current gets reduced, the size of the conductor required also reduces for transmitting the same amount of power.

2) As the current decreases the copper loss also decreases.

3) Reduction in volume of conductor material required.

The current (I) flowing through the system is given as,

\(I = \frac{P}{{\sqrt 3 \;V\;cos\phi }} \)

Resistance per conductor, \(R = \frac{{\rho l}}{a} \)

Total Power loss (W) is given as, 

\(W = 3{I^2}R = 3{\left( {\frac{P}{{\sqrt 3 \;V\;cos\phi }}} \right)^2} \times \frac{{\rho l}}{a} = \left( {\frac{{{P^2}\rho l}}{{\;{V^2}{{\cos }^2}\;\phi \;a}}} \right) \)

 \(W\propto \frac{1}{{{V^2}}} \)

Where, 

l = length of the line in meters

R = resistance per conductor in ohms

ρ = resistivity of conductor material

a = area of X-section of conductor

4)  Increases transmission efficiency

Input power = P + Total losses

= P + \( \left( {\frac{{{P^2}\rho l}}{{\;{V^2}{{\cos }^2}\;\phi \;a}}} \right) \)

Assuming J to be the current density of the conductor, then,

a = I/J

Now efficiency = Output Power/Input Power

After solving it we get,

Transmission Efficiency ≈ \([1-\frac{\sqrt3 J\rho l}{Vcos\phi}]\)

Hence, from the above expression, it is clear that Current density increases as voltage increases with charging current.

5) The line can be easily tapped and or extended as the power need not be converted.

6) Can easily use in an interconnected system.

Limitations of EHV:

  • The increased cost of insulating the conductors.
  • There is an increase in surge impedance loading as it is directly proportional to the transmission voltage.
  • The increased cost of transformers, switchgear, and other terminal apparatus
  • The surface-voltage gradient on conductors becomes larger as the voltage increases.

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