Power System Transients

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}

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.

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}

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}