Compensation Techniques and Voltage profile Control

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
GATE EE 2018   Power Systems
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
GATE EE 2017-SET-1   Power Systems
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
GATE EE 2016-SET-2   Power Systems
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.
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 ____.
GATE EE 2014-SET-3   Power Systems
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)}
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.
GATE EE 2014-SET-2   Power Systems
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.
Question 6
For the system below, S_{D1} \; and \; S_{D2} are complex power demands at bus 1 and bus 2 respectively. If |V_2|=1 pu, the VAR rating of the capacitor (Q_{G2}) connected at bus 2 is
0.2 pu
0.268 pu
0.312 pu
0.4 pu
GATE EE 2012   Power Systems
Question 6 Explanation: 

\begin{aligned} &\text{Real power,} \\ P_r&= \frac{|V_S||V_R|}{|X} \sin \delta \\ 1 &=\frac{1.0 \times 1.0}{0.5}\sin \delta \\ \delta &= \sin^{-1} (0.5)=30^{\circ}\\ &\text{Reactive power,} \\ Q_r&= \frac{|V_S||V_R|}{|X} \cos \delta -\frac{|V|^2}{|X|}\\ &= \frac{1.0 \times 1.0}{0.5}\cos 30-\frac{1^2}{0.5} \\ &= \frac{\left ( \frac{\sqrt{3}}{2} \right )}{\frac{1}{2}}-2\\ &= 1.732-2=-0.268\\ \text{But, } & Q_r+Q_C=0\\ Q_C&=-Q_r=0.268\;p.u. \end{aligned}
Question 7
For enhancing the power transmission in along EHV transmission line, the most preferred method is to connect a
Series inductive compensator in the line
Shunt inductive compensator at the receiving end
Series capacitive compensator in the line
Shunt capacitive compensator at the sending end
GATE EE 2011   Power Systems
Question 7 Explanation: 
Series capacitance compensation reduces reactance of the line.
Power flow in line \propto \frac{1}{\text{reactance of the line}}
As reactance of line decreases, power flow in the line {C_a} increases.
Question 8
Match the items in List-I (To) with the items in the List-II (Use) and select the correct answer using the codes given below the lists.
a - 2, b - 3, c - 4, d - 1
a - 2 , b - 4,  c - 3,  d - 1
a - 4 , b - 3,  c - 1,  d - 2
a - 4 , b - 1, c - 3, d - 2
GATE EE 2009   Power Systems
Question 8 Explanation: 
Shunt capacitor are used to provide part of the reactive VAR's required by the load to keep the voltage within desirable limits and to improve factor.
Series reactor reduce current ripple.
Shunt reactors are used across capacitive loads or ligjtly loaded lines to absorb some of the leading VARs to control the voltage across the load to within certain desirable limit.
Series capacitor compensation reduces the series impedance of the line. Power flow in line \propto \frac{1}{X_L}, power flow in line increases, as {X_L} decreases.
Question 9
A loss less transmission line having Surge Impedance Loading (SIL) of 2280 MW is provided with a uniformly distributed series capacitive compensation of 30%. Then, SIL of the compensated transmission line will be
1835 MW
2280 MW
2725 MW
3257 MW
GATE EE 2008   Power Systems
Question 9 Explanation: 
Surge impedance loading= SIL =2280 MW
\begin{aligned} Z_c&= \text{Surge impedance}\\ \text{SIL}_{2}&=\frac{\text{SIL}_{1}}{\sqrt{1-\gamma _{se}}}\\ \text{SIL}_{2}&=\frac{2280}{\sqrt{1-0.3}}\\ &=2725.12 \text{ MW} \end{aligned}
Question 10
A 230 V (Phase), 50 Hz, three-phase, 4-wire, system has a phase sequence ABC. A unity power-factor load of 4 kW is connected between phase A and neutral N. It is desired to achieve zero neutral current through the use of a pure inductor and a pure capacitor in the other two phases. The value of inductor and capacitor is
72.95 mH in phase C and 139.02 \muF in Phase B
72.95 mH in Phase B and 139.02 \muF in Phase C
42.12 mH in Phase C and 240.79 \muF in Phase B
42.12 mH in Phase B and 240.79 \muF in Phase C
GATE EE 2007   Power Systems
Question 10 Explanation: 

Taking V_{AN} as the reference
\begin{aligned} V_{AN}&=230\angle 0^{\circ} V\\ V_{BN}&=230\angle -120^{\circ}V \\ V_{CN}&=230\angle 120^{\circ}V \\ I_A&= \frac{P}{V_{AN} \cos \phi }\\ &= \frac{4 \times 10^3}{230 \times 1}\\ &=17.4\angle 0^{\circ} \end{aligned}
If an pure inductor is present in phase B, then I_B lags V_{BN} by 90^{\circ}.
If a pure capacitor is present in phase C, then I_C leads V_{CN} by 90^{\circ}.
If current through neutral is to be zero.
\begin{aligned} I_B \sin 30^{\circ} &=I_C \sin 30^{\circ} \\ \Rightarrow \; I_B &=I_C=I \\ I_A&=I_B \cos 30^{\circ} +I_C \cos 30^{\circ} \\ &=2I \cos 30^{\circ}=\sqrt{3}I \\ |I_B|&=|I_C|=|I|=\frac{|I_A|}{\sqrt{3}} \\ &= \frac{17.04}{\sqrt{3}}=10.05\\ |I_B|&=\frac{|V_{BN}|}{\omega L} \\ 10.05 &=\frac{230}{2 \pi \times 50 \times L} \\ L&=72.847 mH \\ |I_C| &=|V_{CN}| \cdot \omega C\\ 10.05&=230 \times 2 \times \pi \times 50 \times C \\ C&=139.087 \mu F \end{aligned}
There are 10 questions to complete.
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