Power Systems Test 7

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:

\({{\rm{Z}}_{{\rm{actual}}}} = 5 \times \frac{{{{\left( {10} \right)}^2}}}{{10}} = 50{\rm{\Omega }}\)

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,

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

Concept:

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

\(\left[ {\begin{array}{*{20}{c}} {{I_a}}\\ {{I_b}}\\ {{I_c}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&1&1\\ 1&{{a^2}}&a\\ 1&a&{{a^2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{I_{a0}}}\\ {{I_{a1}}}\\ {{I_{a2}}} \end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}} {{I_{a0}}}\\ {{I_{a1}}}\\ {{I_{a2}}} \end{array}} \right] = \frac{1}{3}\left[ {\begin{array}{*{20}{c}} 1&1&1\\ 1&a&{{a^2}}\\ 1&{{a^2}}&a \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{I_a}}\\ {{I_b}}\\ {{I_c}} \end{array}} \right]\)

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,

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

\({I_{a0}} = \frac{1}{3}\left( {100\angle 0^\circ + 141\left( {{\rm{cos}}225^\circ + {\rm{i\;sin}}225^\circ } \right) + 100({\rm{cos}}90^\circ + {\rm{i\;sin}}90^\circ } \right))\)

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

Positive sequence component of current,

\({I_{a1}} = \frac{1}{3}\left( {{I_a} + a{I_b} + {a^2}{I_c}} \right) = \frac{1}{3}\left( {100\angle 0^\circ + 1\angle 120^\circ \left( {141.4\angle 225^\circ } \right) + 1\angle - 120^\circ \left( {100\angle 90^\circ } \right)} \right)\)

⇒ I_a1 = 111∠15°

Negative sequence component of current,

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