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).
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).
42.36 | |
64.88 | |
83.64 | |
98.25 |
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
\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?
\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?
Line charging capacitor of finite value is present in all three lines | |
Line charging capacitor of finite value is present in line 2-3 only | |
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 | |
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 |
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.
\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
4r | |
1.414r | |
2r | |
1.723r |
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
\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:
\begin{bmatrix} -t_it_jY & t_j^2 Y\\ t_i^2 Y & -t_it_jY \end{bmatrix} | |
\begin{bmatrix} t_it_jY & -t_j^2 Y\\ -t_i^2 Y & t_it_jY \end{bmatrix} | |
\begin{bmatrix} t_i^2 Y & -t_it_jY\\ -t_it_jY & t_j^2 Y \end{bmatrix} | |
\begin{bmatrix} t_it_jY & -(t_i-t_j)^2Y\\ -(t_i-t_j)^2Y & t_it_jY \end{bmatrix} |
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:
1 | |
3.61 | |
5 | |
6.25 |
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.
=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.