Power System Transients

Question 1
Consider a step voltage of magnitude 1 pu travelling along a lossless transmission line that terminates in a reactor. The voltage magnitude across the reactor at the instant travelling wave reaches the reactor is
A
-1pu
B
1pu
C
2pu
D
3pu
GATE EE 2010   Power Systems
Question 1 Explanation: 


If a step voltage of magnitude 'V' travels through transmission line which is terminated with a inductive load Z_L=L_1s then voltage transmitted (Induced) across the Z_L=L_1s
\begin{aligned} V'(t)&=2Ve^{-Z_st/L_1} \\ \text{At time }&t=0 \\ V'(t)&=2V \\ \text{If }V&=1 \\ V'&=2p.u. \end{aligned}
Question 2
The insulation strength of an EHV transmission line is mainly governed by
A
load power factor
B
switching over-voltages
C
harmonics
D
corona
GATE EE 2005   Power Systems
Question 2 Explanation: 
NOTE: At transmission line voltages upto around 230 kV, the insulation level is dictated by the requirement of protection against lighting. For voltages from 230 kV to 700 kV, both switching transients and lightning overlotages must be accounted for in deciding the insulation levels. In EHV (\gt 700 kV) switching surges cause overvoltages than lightning and are therefore mainly responsible for insulation level decision.
Question 3
Total instantaneous power supplied by a 3-phase ac supply to a balanced R-L load is
A
zero
B
constant
C
pulsating with zero average
D
pulsating with the non-zero average
GATE EE 2004   Power Systems
Question 3 Explanation: 
Impedance of the load
=Z_L=R+j\omega L=|Z|\angle \theta _L
where, \theta _L=\tan ^{-1}\left ( \frac{\omega L}{R} \right )
Voltages of 3-\phi supply
\begin{aligned} V_a&=V_m \sin \omega t \\ V_b&=V_m \sin (\omega t-120^{\circ}) \\ V_c &=V_m \sin (\omega t+120^{\circ}) \\ I_a&=\frac{V_a}{Z_L}=\frac{V_m \sin \omega t}{|Z|\angle \theta _L} \\ &= I_M \sin (\omega t-\theta _L)\\ \text{where, } I_m&=\frac{V_m}{|Z|}\\ \text{Similarly,}&\\ I_b&=I_m \sin (\omega t-120-\theta _L)\\ I_c&=I_m \sin (\omega t+120-\theta _L)\\ & \text{Instantaneous power}\\ &=P=V_aI_a+V_bI_b+V_cI_c\\ P&=V_mI_m[\sin \omega t\cdot \sin (\omega t-\theta _L)\\ &+\sin (\omega t-120^{\circ}) \cdot \sin (\omega t-120-\theta _L)\\ &+\sin (\omega t+120^{\circ})\cdot \sin (\omega t+120-\theta _L)]\\ &=\frac{V_mI_m}{2}[(\cos \theta _L -\cos(2\omega t-\theta _L) ) \\ &+(\cos \theta _L -\cos(2\omega t-240-\theta _L) )\\ &+(\cos \theta _L -\cos(2\omega t+240-\theta _L) )]\\ P&=\frac{3V_mI_m}{2}\cos \phi = \text{constant} \end{aligned}
Question 4
A surge of 20 kV magnitude travels along a lossless cable towards its junction with two identical lossless overhead transmission lines. The inductance and the capacitance of the cable are 0.4 mH and 0.5 \muF per km. The inductance and capacitance of the overhead transmission lines are 1.5 mH and 0.015 \muF per km. The magnitude of the voltage at the junction due to surge is
A
36.72 kV
B
18.36 kV
C
6.07 kV
D
33.93 kV
GATE EE 2003   Power Systems
Question 4 Explanation: 


where, z_1=z_2
Parameter of cable:
Inducatance=L_C=0.4 mH/km and
Capacitance =C_C=0.5\mu F/km
Surge impedance of the cable
=Z_C=\sqrt{\frac{L_C}{C_C}}=\sqrt{\frac{0.4 \times 10^{-3}}{0.5 \times 10^{-6}}}=28.28\Omega
Parameters of OH
Transmission lines
InductanceL_1=1.5 mH/km and
Capacitance =C_1==0.015\mu F/km
Surge impedance of the line =Z_1=Z_2=\sqrt{\frac{L_1}{C_1}}=\sqrt{\frac{1.5 \times 10^{-3}}{0.015 \times 10^{-6}}}=316.22\Omega
E= 20 kV
Junction voltage
\begin{aligned} E_j&=\frac{2E \times \frac{1}{Z_C}}{\frac{1}{Z_C}+\frac{1}{Z_1}+\frac{1}{Z_2}} \\ &= \frac{2 \times 20 \times \frac{1}{28.28}}{\frac{1}{28.28}+\frac{2}{316.22}}\\ &=33.93kV \end{aligned}
There are 4 questions to complete.
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