Performance of Transmission Lines, Line Parameters and Corona


Question 1
A 50 \mathrm{~Hz}, 275 \mathrm{kV} line of length 400 \mathrm{~km} has the following parameters:

Resistance, R=0.035 \Omega / \mathrm{km};
Inductance, \mathrm{L}=1 \mathrm{mH} / \mathrm{km};
Capacitance, \mathrm{C}=0.01 \mu \mathrm{F} / \mathrm{km};

The line is represented by the nominal -\pi model. With the magnitudes of the sending end and the receiving end voltages of the line (denoted by V_{S} and V_{R}, respectively) maintained at 275 \mathrm{kV}, the phase angle difference (\theta) between V_{S} and V_{R} required for maximum possible active power to be delivered to the receiving end, in degree is ____ (Round off to 2 decimal places).
A
42.36
B
64.88
C
83.64
D
98.25
GATE EE 2023   Power Systems
Question 1 Explanation: 
We have,
\mathrm{P}_{\mathrm{R}}=\frac{\mathrm{V}_{\mathrm{S}} \mathrm{V}_{\mathrm{R}}}{\mathrm{B}} \cos (\beta-\delta)-\frac{\mathrm{AV}_{\mathrm{R}}^{2}}{\mathrm{~B}} \cos (\beta-\alpha)

At max. power,
\delta=\beta

where, \beta= angle of T-parameter of B.

\pi-Model :

[T]=\begin{bmatrix} 1+\frac{yz}{2} &z \\ y\left ( 1+ \frac{yz}{4}\right )& 1+ \frac{yz}{2}\\ \end{bmatrix}=\begin{bmatrix} A & B\\ C&D \end{bmatrix}
\therefore \quad B=z
=(0.035 \times 400)+j(2 \pi \times 50 \times 10^{-3} \times 400) = 126.44\angle 83.643^{\circ}\Omega
Question 2
The bus admittance \left(Y_{\text {bus }}\right) matrix of a 3-bus power system is given below.

\begin{bmatrix} -j15 &j10 &j5 \\ j10& -j13.5 &j4 \\ j5& j4 & -j8 \end{bmatrix}

Considering that there is no shunt inductor connected to any of the buses, which of the following can NOT be true?
A
Line charging capacitor of finite value is present in all three lines
B
Line charging capacitor of finite value is present in line 2-3 only
C
Line charging capacitor of finite value is present in line 2-3 only and shunt capacitor of finite value is present in bus 1 only
D
Line charging capacitor of finite value is present in line 2-3 only and shunt capacitor of finite value is present in bus 3 only
GATE EE 2023   Power Systems
Question 2 Explanation: 
From Y_{\text {Bus }} matrix
\begin{aligned} & y_{10}=-j 15+j 10+j 5=0 \\ & y_{20}=j 10-j 13.5+j 4=j 0.5 \\ & y_{30}=j 5+j 4-j 8=j 1 \end{aligned}

Power system network :

Hence, option (A) and (C) will not be correct.


Question 3
The geometric mean radius of a conductor, having four equal strands with each strand of radius 'r', as shown in the figure below, is

A
4r
B
1.414r
C
2r
D
1.723r
GATE EE 2022   Power Systems
Question 3 Explanation: 
Redraw the configuration:

\therefore \; GMR=(r' \times 2r\times 2r\times 2\sqrt{2}r)^{1/4}
Where, r'=0.7788r
Hence, GMR=1.723r
Question 4
Two buses, i and j, are connected with a transmission line of admittance Y, at the two ends of which there are ideal transformers with turns ratios as shown. Bus admittance matrix for the system is:
A
\begin{bmatrix} -t_it_jY & t_j^2 Y\\ t_i^2 Y & -t_it_jY \end{bmatrix}
B
\begin{bmatrix} t_it_jY & -t_j^2 Y\\ -t_i^2 Y & t_it_jY \end{bmatrix}
C
\begin{bmatrix} t_i^2 Y & -t_it_jY\\ -t_it_jY & t_j^2 Y \end{bmatrix}
D
\begin{bmatrix} t_it_jY & -(t_i-t_j)^2Y\\ -(t_i-t_j)^2Y & t_it_jY \end{bmatrix}
GATE EE 2020   Power Systems
Question 4 Explanation: 

\begin{aligned} I&=Y(t_{i}V_{i}-V_{j}t_{j}) \\ I_{i}&=t_{i}I \\ &=t_{i}^{2}YV_{i}-t_{i}t_{j}YV_{j} \\ I_{j}&=-t_{j}I \\ &=-I_{i}t_{j}YV_{i}+t_{i}^{2}YV_{j} \\ \begin{bmatrix} I_{i}\\ I_{j} \end{bmatrix}&=\begin{bmatrix} t_{i}^{2}Y &-t_{i}t_{j}Y \\ -t_{i}t_{j}Y &t_{j}^{2}Y \end{bmatrix}\begin{bmatrix} V_{i}\\ v_{j} \end{bmatrix} \end{aligned}
Question 5
A lossless transmission line with 0.2 pu reactance per phase uniformly distributed along the length of the line, connecting a generator bus to a load bus, is protected up to 80% of its length by a distance relay placed at the generator bus. The generator terminal voltage is 1 pu. There is no generation at the load bus. The threshold pu current for operation of the distance relay for a solid three phase-to-ground fault on the transmission line is closest to:
A
1
B
3.61
C
5
D
6.25
GATE EE 2020   Power Systems
Question 5 Explanation: 
I_{f}=\frac{1}{Z_{Th}}=\frac{1}{0.2}
=5 pu for 100% of line
Relay is operated for 80%
Z_{f}=0.8\, Z_{t}\Rightarrow 0.8\times 0.2=0.16\, p.u.
For 80% of line,
I_{f}=\frac{1}{0.16}=6.25\: p.u.


There are 5 questions to complete.