Question 1 |

Consider the two bus power system network with given loads as shown in the figure. All
the values shown in the figure are in per unit. The reactive power supplied by generator G_1 and G_2 are Q_{G1} and Q_{G2} respectively. The per unit values of Q_{G1}, Q_{G2}, and line reactive
power loss (Q_{loss}) respectively are

5. 00, 12.68, 2.68 | |

6.34, 10.00, 1.34 | |

6.34, 11.34, 2.68 | |

5.00, 11.34, 1.34 |

Question 1 Explanation:

At G_2 load demand is 20 pu, G_2 supplies only 15 pu. Remaining supplied by G_1 through transmission line.

\begin{aligned} P_S&=\left | \frac{V_SV_R}{X_L} \right | \sin \delta \\ 5&=\frac{1 \times 1}{0.1} \sin \delta \\ \delta &=30^{\circ}\\ Q_S&=\frac{V_S^2}{X_L}-\frac{V_SV_R}{X_L} \cos \delta \\ &=\frac{1^2}{0.1}-\frac{1 \times 1 }{0.1} \cos 30^{\circ}\\ &=1.34 p.u.\\ Q_R&=\left | \frac{V_SV_R}{X_L} \right | \cos \delta -\left | \frac{V_R^2}{X_L} \right | \\ 5&=\frac{1 \times 1}{0.1} \cos 30^{\circ}-\frac{1^2}{0.1} \\ &=-1.34p.u.\\ Q_{loss}&+Q_S-Q_R\\ &=1.34-(-1.34)=2.68p.u.\\ \text{At }G_1: \; Q_{G1}&=Q_{load}+Q_S\\ &=5+1.34=6.34p.u.\\ \text{At }G_2: \; Q_{G2}&=Q_{load}-(-Q_R)\\ &=10-(-1.34)=11.34p.u. \end{aligned}

Question 2 |

A load is supplied by a 230 V, 50 Hz source. The active power P and the reactive power Q
consumed by the load are such that 1 kW \leq P \leq 2kW and 1{kVAR} \leq Q \leq {kVAR}. A capacitor connected across the load for power factor correction generates 1 kVAR reactive power. The worst case power factor after power factor correction is

0.447 lag | |

0.707 lag | |

0.894 lag | |

1 |

Question 2 Explanation:

Load p.f.

\begin{aligned} \cos \phi &=\cos \left [ \tan ^{-1}\frac{Q}{P} \right ]\\ &=\cos \left [ \tan^{-1}\frac{1}{1} \right ]\\ &=0.707 \text{ lag}\\ Q_c&=P[\tan \phi _1 -\tan \phi _2]\\ 1&=(\tan \cos ^{-1} 0.707- \tan \phi _2)\\ \phi _2&=45^{\circ}\\ \Rightarrow \; \cos \phi _2&=0.707 \text{ lag} \end{aligned}

Question 3 |

The inductance and capacitance of a 400 kV, three-phase, 50 Hz lossless transmission line are 1.6 mH/km/phase and 10 nF/km/phase respectively. The sending end voltage is maintained at 400 kV. To maintain a voltage of 400 kV at the receiving end, when the line is delivering 300 MW load, the shunt compensation required is

capacitive | |

inductive | |

resistive | |

zero |

Question 3 Explanation:

\begin{aligned} Z_n&=\sqrt{\frac{L}{C}} \\ &= \sqrt{\frac{1.6 \times10^{-3}}{10 \times 10^{-9}}}\\ &= 400\Omega \\ \text{SIL} &=\frac{400 \times 400}{400}=400 MW \end{aligned}

In the second case SIL decreases means Z_n increasess.

Z_n increases with increase in inductance 'L'.

So, it is inductive.

Load \lt SIL means, line behaves capacitive to compensate it inductor to be placed.

In the second case SIL decreases means Z_n increasess.

Z_n increases with increase in inductance 'L'.

So, it is inductive.

Load \lt SIL means, line behaves capacitive to compensate it inductor to be placed.

Question 4 |

The complex power consumed by a constant-voltage load is given by (P_1+jQ_1),

where, 1 kW \leq P1 \leq 1.5 kW and

0.5 kVAR \leq Q_1 \leq 1 kVAR.

A compensating shunt capacitor is chosen such that |Q| \leq 0.25 kVAR, where Q is the net reactive power consumed by the capacitor load combination. The reactive power (in kVAR) supplied by the capacitor is ____.

where, 1 kW \leq P1 \leq 1.5 kW and

0.5 kVAR \leq Q_1 \leq 1 kVAR.

A compensating shunt capacitor is chosen such that |Q| \leq 0.25 kVAR, where Q is the net reactive power consumed by the capacitor load combination. The reactive power (in kVAR) supplied by the capacitor is ____.

0.25 | |

0.50 | |

0.75 | |

0.90 |

Question 4 Explanation:

\begin{aligned} \text{Given, }S_{load}&=P_1+jQ_1 \\ 1kW \leq P_1 &\leq1.5 kW \\ 0.5 \text{ kVAR} \leq Q_1 &\leq 1\text{ kVAR} \end{aligned}

Shunt capacitor will only supply reactive power and the total reactive power consumed by the load and the capacitor together will be less than 0.25 kVAR

i.e. Q \leq 0.25\text{ kVAR}

Therefore, reactive power supplied by the capacitor will be

Q _C=(1-0.25)=0.75 \text{ kVAR (lagging)}

Shunt capacitor will only supply reactive power and the total reactive power consumed by the load and the capacitor together will be less than 0.25 kVAR

i.e. Q \leq 0.25\text{ kVAR}

Therefore, reactive power supplied by the capacitor will be

Q _C=(1-0.25)=0.75 \text{ kVAR (lagging)}

Question 5 |

Shunt reactors are sometimes used in high voltage transmission systems to

limit the short circuit current through the line | |

compensate for the series reactance of the line under heavily loaded condition | |

limit over-voltages at the load side under lightly loaded condition. | |

compensate for the voltage drop in the line under heavily loaded condition. |

Question 5 Explanation:

Shunt reactor is basically an inductor having no-core or core is replaced by cement slab to avoid saturation problem. These are widely used to reduce the fault level or the short circuit current through the line.

There are 5 questions to complete.