To find the reactive part of the zero-sequence current we have to find the zero-sequence current.

$I_{R0}=I_{B0}=I_{Y0}=\frac{1}{3}\left(I_R+I_Y+I_B\right)$

I_R = 3 + j5 A

I_Y = 2 + j2 A

I_B = -2 – j1A

$I_{R0}=I_{Y0}=I_{B0}=\frac{1}{3}\left(I_R+I_Y+I_B\right)$

$=\frac{1}{3}[\left(3+j5\right)+\left(2+j2\right)+\left(-2-j1\right)$

$=\frac{1}{3}\left[3+j6\right]$

= 1 + j2 A

$X_{pu}\alpha \frac{MVA}{{\left(kV\right)}^2}$ 

$\frac{{\left(X_{pu}\right)}_{new}}{{\left(X_{pu}\right)}_{old}}=\frac{MV{A}_{new}}{MV{A}_{old}}\times {\left(\frac{k{V}_{old}}{k{V}_{New}}\right)}^2$ 

Given that, MVA_new = 3 MVA_old­

KV_New = 4 kV_old

$\Rightarrow \frac{{\left(X_{pu}\right)}_{New}}{{\left(X_{pu}\right)}_{old}}=\frac{3}{1}\times {\left(\frac{1}{4}\right)}^2$ 

Given information

Z_pu = 5 pu

Base MVA = 10 MVA

Base kV = 10 kV

Formula used:

$Z_{actual}=Z_{pu}\times \frac{{\left(k{V}_{base}\right)}^2}{MV{A}_{base}}$

Calculation:

${\mathrm{Z}}_{\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{u}\mathrm{a}\mathrm{l}}=5\times \frac{{\left(10\right)}^2}{10}=50\mathrm{\Omega }$

The given load is balanced star connected load. In balanced system only positive sequence component is present, both negative and zero sequence components are absent.

Zero sequence component, I_ao = 0 A

Given that, I_a = 50A

Positive sequence component,

The formula to find new per unit value for new base values is given below.

$X_{pu\left(New\right)}=X_{pu\left(old\right)}\times \frac{MV{A}_{\left(new\right)}}{MV{A}_{\left(old\right)}}\times {\left[\frac{K{V}_{\left(old\right)}}{K{V}_{\left(new\right)}}\right]}^2$

X_pu(old) = 0.15

MVA_(new) = 50

MVA_(old) = 10

KV_(old) = 13.2

KV_(new) = 13.8

For generator:

$X_{G1\left(new\right)}=0.15\times \frac{50}{10}\times {\left(\frac{13.2}{13.8}\right)}^2=0.686pu$

$X_{G2\left(new\right)}=0.15\times \frac{50}{15}\times {\left(\frac{13.2}{13.8}\right)}^2=0.457pu$

For transmission line:

$X_L=10\times \frac{10}{{\left(13.8\right)}^2}=2.625pu$

For motors:

$X_{M_1\left(new\right)}=0.2\times \frac{50}{8}\times {\left(\frac{12.5}{13.8}\right)}^2=1.025pu$

Zero sequence equivalent reactance is given by

Equivalent zero sequence impedance across bus 1

$X_{0\left(eq\right)}=j0.15||j0.45=\frac{j0.15\times j0.45}{j0.6}=j0.1125$

Given that, X_s = 0.2 pu

X_m = 0.05 pu

X_1eq = X_s - X_m = 0.2 – 0.05 = 0.15 pu

X_2eq = X_s - X_m = 0.2 – 0.05 = 0.15 pu

X_0eq = X_s + 2X _m = 0.2 + 2 (0.05) = 0.3 pu

X_s = 0.85 pu

X_T1 = X_T2 = 0.157

X_L1 = X_L2 = 0.35 pu

E = 1.5 pu

V = 1.0 pu

P₀ = 1.0 pu

Equivalent circuit of above data can be drawn as

X_T = j0.85 + j0.157 + j(0.35||0.35) + j0.17

X_T = 1.339 pu

As $P_0=\frac{EV}{X}\sin \delta$

P₀ = 1pu

$P=\frac{EV}{X}\sin \delta$

For the minimum value of E, sin δ = 1.

$E_{min}=\frac{P_{0}X}{V}=\frac{1\times 1.339}{1}=1.339pu$

Concept:

The relation between the line currents in terms of the symmetrical components of currents is given below.

$\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& a^2& a\\ 1& a& a^2\end{array}\right]\left[\begin{array}{c}{I}_{a0}\\ I_{a1}\\ I_{a2}\end{array}\right]$

$\left[\begin{array}{c}{I}_{a0}\\ I_{a1}\\ I_{a2}\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& a& a^2\\ 1& a^2& a\end{array}\right]\left[\begin{array}{c}{I}_a\\ I_b\\ I_c\end{array}\right]$

I_a0 = Zero Sequence Component of Current

I_a1 = Positive Sequence Component of Current

I_a2 = Negative Sequence Component of Current

a = 1∠120°, which represents the rotation of 120° in clockwise direction.

a² = 1∠-120° or 1∠240° in anticlockwise direction or in clockwise direction, respectively.

Calculation:

I_a = 100∠0°, I_b = 141.4∠225°, I_c = 100∠90°

Zero sequence component of current,

$I_{a0}=\frac{1}{3}\left(I_a+I_b+I_c\right)=\frac{1}{3}\left(100\mathrm{\angle }{0}^{\circ }+141.4\mathrm{\angle }{225}^{\circ }+100\mathrm{\angle }{90}^{\circ }\right)$

$I_{a0}=\frac{1}{3}\left(100\mathrm{\angle }{0}^{\circ }+141\left(\mathrm{c}\mathrm{o}\mathrm{s}{225}^{\circ }+\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{n}{225}^{\circ }\right)+100(\mathrm{c}\mathrm{o}\mathrm{s}{90}^{\circ }+\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{n}{90}^{\circ }\right))$

$\Rightarrow I_{a0}=\frac{1}{3}\left(0.02+i0.02\right)=0.007\mathrm{\angle }{45}^{\circ }$

Positive sequence component of current,

$I_{a1}=\frac{1}{3}\left(I_a+a{I}_b+a^2{I}_c\right)=\frac{1}{3}\left(100\mathrm{\angle }{0}^{\circ }+1\mathrm{\angle }{120}^{\circ }\left(141.4\mathrm{\angle }{225}^{\circ }\right)+1\mathrm{\angle }-{120}^{\circ }\left(100\mathrm{\angle }{90}^{\circ }\right)\right)$

⇒ I_a1 = 111∠15°

Negative sequence component of current,

$I_{a2}=\frac{1}{3}\left(I_a+a^2\ast I_b+a\ast I_c\right)=\frac{1}{3}\left(100\mathrm{\angle }{0}^{\circ }+1\mathrm{\angle }-{120}^{\circ }\left(141.4\mathrm{\angle }{225}^{\circ }\right)+1\mathrm{\angle }{120}^{\circ }\left(100\mathrm{\angle }{90}^{\circ }\right)\right)$

Given that %X₁ = j0.3, %X₂ = j0.1, %X₀ = j0.05

d = 5 m

r = 0.5 cm = 0.005 m

Conductors are arranged in the equilateral spacing of 5 m side and 1 cm diameter.

L/phase $=2\times {10}^{-7}ln\left(\frac{d}{r^{\prime }}\right)$

$=2\times {10}^{-7}ln\left(\frac{5}{0.005\times 0.7788}\right)=1.4315\times {10}^{-6}H/m$

Now, X_L = 2πfL/phase × 20 km

X_L = 2 × π × 50 × 1.4315 × 10^-6 × 20 = 9 Ω

Base Impedance $=\frac{{\left(KV\right)}^2}{MVA}$

$=\frac{{\left(30\right)}^2}{10}=90\mathrm{\Omega }$

X₂(Ω) = j0.1 × X_base

= j0.1 × 30 = j9 Ω

Now, negative sequence reactance up to fault location

X_{2 equivalent} = X₂ + X_L = 9 + 9 = 18 Ω

X_{2 eq} = 18 Ω