Performance of Transmission Lines Questions and Answers

Answer: C

The most dangerous is a Direct stroke on line conductor.

When lightning strikes a phase conductor directly, the whole surge current is injected into the line and travels both ways as a steep, very high voltage wave. It can flash over insulators and reach terminal equipment such as transformers.

A stroke on the tower top or on the ground (earth) wire is largely drained to earth through the tower footing, which is exactly what the ground wire is installed to do. An indirect stroke, one that hits nearby ground and only induces a charge on the conductor, produces a much smaller surge.

Answer: A

Transmission efficiency increases as Voltage and Power factor both increase.

For a given power \(P\), the line current is \(I = \frac{P}{\sqrt{3}\,V\cos\phi}\), and the line loss is \(3I^{2}R\). Raising either the voltage \(V\) or the power factor \(\cos\phi\) reduces the current, so the loss falls as \(1/(V\cos\phi)^{2}\) and the efficiency rises.

Lowering both increases the current and the losses. The two mixed options pull in opposite directions, so neither gives a sure rise in efficiency.

Answer: D

The capacitance can be neglected when the line is shorter than 80 km.

Lines are classified by length: short lines up to about 80 km, medium lines from about 80 to 200 km (some texts say 250 km), and long lines beyond that. In a short line the charging current is small compared with the load current, so the line is represented by its resistance and inductance alone.

100, 160 and 200 km all fall in the medium range, where capacitance must be included, usually as a lumped nominal-T or nominal-pi model.

Answer: C

The key gives 100 - 200 ohms.

Surge impedance is \(Z_{0} = \sqrt{L/C}\), set by the line's inductance and capacitance per unit length, not by its length. Overhead lines have relatively high inductance and low capacitance, so their surge impedance is high, while underground cables, with much higher capacitance, are low, of the order of 40 to 60 ohms.

Measured values for overhead lines are commonly a few hundred ohms (300 to 500 ohms is typical); of the bands offered the textbook answer is 100 - 200 ohms. 10 - 20 and 50 - 60 ohms are cable-like values, and 1000 - 2000 ohms is far higher than any practical line.

Answer: D

For 500 km the voltage should be 150 KV - 220 KV.

Line loss falls as the square of the voltage, so longer distances need higher voltages. Still's empirical formula, \(V = 5.5\sqrt{\frac{L}{1.6} + \frac{P}{100}}\) kV with \(L\) in km and \(P\) in kW, gives about 97 kV for 500 km even with negligible power, and about 150 to 200 kV for loads of 50 to 100 MW.

33 - 66 KV is a sub-transmission level for tens of kilometres, and 66 - 100 KV is still below the formula's floor for this distance. 110 - 150 KV covers only light loads at 500 km, so the highest band is the one that suits bulk transmission over this distance.

Answer: A

The relay used on long transmission lines is the Mho's Relay.

Distance relays are matched to line length. The mho (admittance) relay has a circular characteristic passing through the origin, so it is inherently directional, occupies the least area on the R-X diagram, and is the least affected by power swings, which are severe on long lines.

The reactance relay suits short lines, where arc resistance is large compared with the line impedance and must not affect the reach. The impedance relay is used on medium lines. Long lines certainly need protection, so 'No Relay is used' is wrong.

Answer: D

The losses become about 1.57 W.

For the same power delivered at the same voltage, the line current is inversely proportional to the power factor, and the line loss \(I^{2}R\) is proportional to the square of the current, so the loss varies as \(1/\cos^{2}\phi\).

\[ \begin{aligned} W' &= \frac{W}{0.8^{2}} = \frac{W}{0.64} \\ &= 1.5625\,W \approx 1.57\,W \end{aligned} \]

1.25 W divides by 0.8 only once, forgetting that the loss depends on the square of the current, and 0.8 W multiplies instead of dividing.

Answer: D

The constants are 1, Z, 0 and 1.

A short transmission line is modelled as a series impedance \(Z\) only, with its capacitance neglected. So the sending-end current equals the receiving-end current, and the sending-end voltage is the receiving-end voltage plus the drop in \(Z\):

\[ \begin{aligned} V_{S} &= V_{R} + Z I_{R} \\ I_{S} &= 0 \cdot V_{R} + I_{R} \end{aligned} \]

Comparing with \(V_{S} = AV_{R} + BI_{R}\) and \(I_{S} = CV_{R} + DI_{R}\) gives \(A = 1\), \(B = Z\), \(C = 0\), \(D = 1\), and the check \(AD - BC = 1\) holds. The other options put Z in the place of A or C, which would make the units wrong.