Question 1 |
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 1 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 2 |
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 2 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.
Question 3 |
A three-phase 50 Hz, 400 kV transmission line is 300 km long. The line inductance is 1 mH/km per phase, and the capacitance is 0.01 \mu F/km per phase. The line is under open circuit condition at the receiving end and energized with 400 kV at the sending end, the receiving end line voltage in kV (round off to two decimal places) will be ___________.
418.85 | |
256.25 | |
458.45 | |
369.28 |
Question 3 Explanation:
\begin{aligned} V_s&= 400kV\\ l&=300km\\ L_1&=1mH/km/phase\\ C_1&=0.01 \mu F/km/phase\\ v&=\frac{1}{\sqrt{L_1C_1}}\\ &=\frac{1}{\sqrt{1 \times 10^{-3} \times 0.01 \times 10^{-6}}}\\ &=3.16 \times 10^5 km/sec\\ \beta '&=\frac{2 \pi f l}{v}\\ &=\frac{2 \pi \times 50 \times 300}{3.16 \times 10^5}=0.29\\ A&=1-\frac{\beta ^2}{2}\\ &=1-\frac{(0.29)^2}{2}=0.955\\ V_R&=\frac{V_S}{A}=\frac{400}{0.955}=418.85kV \end{aligned}
Question 4 |
A three-phase load is connected to a three-phase balanced supply as shown in the figure. If V_{an}=100\angle 0^{\circ} V, V_{bn}=100\angle -120^{\circ}V and V_{cn}=100\angle -240^{\circ} V (angles are considered positive in the anti-clockwise direction), the value of R for zero current in the neutral wire is ___________\Omega (up to 2 decimal places).
5.77 | |
2.45 | |
4.75 | |
6.25 |
Question 4 Explanation:
From the given voltages,
\begin{aligned} I_R&=\frac{V_{RN}}{R}=\frac{100\angle 0^{\circ}}{R}\\ I_Y&=\frac{V_{YN}}{jX_L}=\frac{100\angle 120^{\circ}}{j10}\\ &= 10\angle -210^{\circ}\\ I_B&=\frac{V_{BN}}{-jX_C}=\frac{100\angle -240^{\circ}}{-j10}\\ &=10\angle -150^{\circ}\\ \text{For }\; I_N&=0\\ I_R+I_Y+I_B&=0\\ \frac{100}{R}+&10\angle -210^{\circ}+10\angle -150^{\circ}=0\\ R&=5.77\Omega \end{aligned}
\begin{aligned} I_R&=\frac{V_{RN}}{R}=\frac{100\angle 0^{\circ}}{R}\\ I_Y&=\frac{V_{YN}}{jX_L}=\frac{100\angle 120^{\circ}}{j10}\\ &= 10\angle -210^{\circ}\\ I_B&=\frac{V_{BN}}{-jX_C}=\frac{100\angle -240^{\circ}}{-j10}\\ &=10\angle -150^{\circ}\\ \text{For }\; I_N&=0\\ I_R+I_Y+I_B&=0\\ \frac{100}{R}+&10\angle -210^{\circ}+10\angle -150^{\circ}=0\\ R&=5.77\Omega \end{aligned}
Question 5 |
For the balanced Y-Y connected 3-Phase circuit shown in the figure below, the line-line voltage
is 208 V rms and the total power absorbed by the load is 432 W at a power factor of 0.6 leading.
The approximate value of the impedance Z is
The approximate value of the impedance Z is
33\angle -53.1^{\circ}\Omega | |
60\angle 53.1^{\circ}\Omega | |
60\angle -53.1^{\circ}\Omega | |
180\angle -53.1^{\circ}\Omega |
Question 5 Explanation:
For star connection,
\begin{aligned} P_{3-\phi }&=\left | \frac{V_L^2}{z} \right | \cos \phi \\ Z&=\frac{V_L^2}{P_{3-\phi }} \cos \phi \\ &=\frac{208^2 }{430} \times 0.6 \\ \cos \phi &=0.6 \text{ lead}\\ \phi &=53.13^{\circ}\\ \therefore \; Z&=60\angle -53.13^{\circ}\Omega \end{aligned}
\begin{aligned} P_{3-\phi }&=\left | \frac{V_L^2}{z} \right | \cos \phi \\ Z&=\frac{V_L^2}{P_{3-\phi }} \cos \phi \\ &=\frac{208^2 }{430} \times 0.6 \\ \cos \phi &=0.6 \text{ lead}\\ \phi &=53.13^{\circ}\\ \therefore \; Z&=60\angle -53.13^{\circ}\Omega \end{aligned}
Question 6 |
Consider an overhead transmission line with 3-phase, 50 Hz balanced system with conductors located at the vertices of an equilateral triangle of length D_{ab}=D_{bc}=D_{ca}=1m as shown in figure below. The resistance of the conductors are neglected. The geometric mean radius (GMR) of
each conductor is 0.01m. Neglecting the effect of ground, the magnitude of positive sequence reactance in \Omega/ km (rounded off to three decimal places) is ________
0.572 | |
5.721 | |
0.289 | |
2.892 |
Question 6 Explanation:
\begin{aligned} X_L&=2 \pi f L\\&=2 \pi f \times 2 \times 10^{-7} \ln \left ( \frac{D_m}{D_s} \right )\\ &=2 \pi \times 50 \times 2 \times 10^{-7} \ln \left ( \frac{1}{0.01} \right )\\ &=2.89 \times 10^{-4}\Omega /m\\ X_L&=0.289 \Omega /km \end{aligned}
Question 7 |
The nominal- \pi circuit of a transmission line is shown in the figure.
Impedance Z=100\angle 80^{\circ}\Omega and reactance X=3300\Omega. The magnitude of the characteristic impedance of the transmission line, in \Omega, is _______________. (Give the answer up to one decimal place.)
Impedance Z=100\angle 80^{\circ}\Omega and reactance X=3300\Omega. The magnitude of the characteristic impedance of the transmission line, in \Omega, is _______________. (Give the answer up to one decimal place.)
406.2 | |
201.5 | |
635.8 | |
52.4 |
Question 7 Explanation:
\begin{aligned} Z&=100\angle 80^{\circ}\Omega \\ X&=3300\Omega \\ \frac{y}{2}&=\frac{1}{X}=\frac{1}{3300} \end{aligned}
\begin{aligned} y&=\frac{2}{3300}=6.06 \times 10^{-4}\\ Z_C&=\sqrt{\frac{Z}{y}}=\sqrt{\frac{100}{6.06 \times 10^{-4}}}\\ &=406.2\Omega \end{aligned}
\begin{aligned} y&=\frac{2}{3300}=6.06 \times 10^{-4}\\ Z_C&=\sqrt{\frac{Z}{y}}=\sqrt{\frac{100}{6.06 \times 10^{-4}}}\\ &=406.2\Omega \end{aligned}
Question 8 |
A source is supplying a load through a 2-phase, 3-wire transmission system as shown in figure
below. The instantaneous voltage and current in phase-a are V_{an}=220sin(100\pi t)V and i_{a}=10sin (100\pi t)A, respectively. Similarly for phase-b the instantaneous voltage and current are V_{bn}=220cos (100\pi t)V and i_{b}=10cos( 100\pi tA, respectively
The total instantaneous power flowing form the source to the load is
The total instantaneous power flowing form the source to the load is
2200W | |
2200 sin^{2}(100\pi t)W | |
440W | |
2200 sin(100\pi t) cos(100 \pi t) W |
Question 8 Explanation:
\begin{aligned} V_{an}&=220 \sin (100 \pi t)V\\ i_a&=10 \sin (100 \pi t)A\\ V_{bn}&=220 \cos (100 \pi t)V\\ i_b&=10 \cos (100 \pi t)A\\ p&=V_{an}i_a+V_{bn}i_b\\ &=2200 W \end{aligned}
Question 9 |
At no load condition, a 3-phase, 50 Hz, lossless power transmission line has sending-end and
receiving-end voltages of 400 kV and 420 kV respectively. Assuming the velocity of traveling wave to be the velocity of light, the length of the line, in km, is ____________.
148.68 | |
224.45 | |
294.52 | |
326.86 |
Question 9 Explanation:
At no load,
\begin{aligned} V_s&=AV_R\\ 400 &=A420\\ A&=\frac{400}{420}=0.9524\\ A&=1+\frac{YZ}{2}\\ &=1+\frac{(r+j\omega L)()g+j\omega C)}{2}\\ &\text{For lossless line,}\\ r&=0,g=0\\ \text{then,}\\ A&=1-\frac{(\omega C)(\omega L)}{2}\\ \beta l&=\sqrt{\omega L\omega C}\\ A&=0.9524=1-\frac{\beta ^2l^2}{2}\\ \beta l&=0.3085\\ \beta &=\frac{0.3085}{l}\\ \frac{V}{f}&=\frac{2\pi}{\beta }\\ \frac{3 \times 10^5}{50}&=\frac{2 \pi}{\left ( \frac{0.3085}{l} \right )}\\ l&=294.52km \end{aligned}
\begin{aligned} V_s&=AV_R\\ 400 &=A420\\ A&=\frac{400}{420}=0.9524\\ A&=1+\frac{YZ}{2}\\ &=1+\frac{(r+j\omega L)()g+j\omega C)}{2}\\ &\text{For lossless line,}\\ r&=0,g=0\\ \text{then,}\\ A&=1-\frac{(\omega C)(\omega L)}{2}\\ \beta l&=\sqrt{\omega L\omega C}\\ A&=0.9524=1-\frac{\beta ^2l^2}{2}\\ \beta l&=0.3085\\ \beta &=\frac{0.3085}{l}\\ \frac{V}{f}&=\frac{2\pi}{\beta }\\ \frac{3 \times 10^5}{50}&=\frac{2 \pi}{\left ( \frac{0.3085}{l} \right )}\\ l&=294.52km \end{aligned}
Question 10 |
A single-phase transmission line has two conductors each of 10 mm radius. These are fixed at a center-to-center distance of 1 m in a horizontal plane. This is now converted to a three-phase transmission line by introducing a third conductor of the same radius. This conductor is fixed at an equal distance D from the two single-phase conductors. The three-phase line is fully transposed. The positive sequence inductance per phase of the three-phase system is to be 5% more than that of the inductance per conductor of the single-phase system. The distance D, in meters, is _______.
0.64 | |
1.25 | |
1.43 | |
2.36 |
Question 10 Explanation:
In first case,
\begin{aligned} L_1&=2 \times 10^{-7} \ln \frac{D}{r}\\ &=2 \times 10^{-7} \ln \left ( \frac{100}{0.7788} \right )\\ &=0.97\mu H/m\\ L_2&=1.05 \times 0.97=1.0185 \mu H/m\\ L_2&=2 \times 10^{-7} \ln \left ( \frac{\sqrt[3]{D^2 \times 100}}{0.7788} \right ) \end{aligned}
\begin{aligned} \ln \left ( \frac{\sqrt[3]{D^2 \times 100}}{0.7788} \right )&=\frac{1.0185 \times 10^{-6}}{2 \times 10^{-7}}\\ &=5.0925\\ e^{5.0925}&= \frac{\sqrt[3]{D^2 \times 100}}{0.7788} \\ D&=142.7cm=1.427m \end{aligned}
\begin{aligned} L_1&=2 \times 10^{-7} \ln \frac{D}{r}\\ &=2 \times 10^{-7} \ln \left ( \frac{100}{0.7788} \right )\\ &=0.97\mu H/m\\ L_2&=1.05 \times 0.97=1.0185 \mu H/m\\ L_2&=2 \times 10^{-7} \ln \left ( \frac{\sqrt[3]{D^2 \times 100}}{0.7788} \right ) \end{aligned}
\begin{aligned} \ln \left ( \frac{\sqrt[3]{D^2 \times 100}}{0.7788} \right )&=\frac{1.0185 \times 10^{-6}}{2 \times 10^{-7}}\\ &=5.0925\\ e^{5.0925}&= \frac{\sqrt[3]{D^2 \times 100}}{0.7788} \\ D&=142.7cm=1.427m \end{aligned}
There are 10 questions to complete.