A three phase 50 MVA, 11 kV generator is subjected to the various faults and the currents so obtained in each fault are: 2000 A for a three phase fault; 1800 A for a line-to-line fault and 2200 A for a line-to-ground fault. Find the sequence impedances of the generator. (AMIE Summer 2022, 10 marks)
Case 1
Consider the conditions w.r.t. the three phase fault
\(\begin{array}{l}{I_f} = {I_a} = {I_{a1}} = {E_{a1}}/{Z_1}\\2000 = 11000/(\sqrt 3 {Z_1})\\{Z_1} = 3.18\Omega \end{array}\)
Case 2
Consider the conditions w.r.t. the LL fault
\(\begin{array}{l}{I_{a1}} = [{E_{a1}}/({Z_1} + {Z_2})]\\{I_f} = {I_b} = - {I_c} = \sqrt 3 {I_{a1}}\\ = \sqrt 3 {E_{a1}}/({Z_1} + {Z_2})\,or\\({Z_1} + {Z_2}) = \sqrt 3 (11000/\sqrt 3 )/1800\\{Z_2} = 2.936\Omega \end{array}\)
Case 3
Consider the conditions w.r.t. a LG fault
\(\begin{array}{l}{I_{a1}} = \frac{{{E_{a1}}}}{{({Z_1} + {Z_2} + {Z_0})}}\\{I_f} = 3{I_{a1}} = 3\frac{{{E_{a1}}}}{{({Z_1} + {Z_2} + {Z_0})}}\\({Z_1} + {Z_2} + {Z_0}) = 3\frac{{{E_{a1}}}}{{{i_f}}}\\{Z_0} = 2.55\Omega \end{array}\)
If a 250 MVA, 11/400 kV three-phase power transformer has
leakage reactance of 0.05 per unit on the base of 250 MVA and the
primary voltage of 11 kV. then find the actual leakage reactance of the
transformer referred to the secondary side of 400 kV. (AMIE Summer 2022,
4 marks)
\(\begin{array}{l}{Z_p}(actual) = {Z_{pu}}.{Z_{base}}\\ = 0.05x\frac{{400x400}}{{250}} = 32\Omega \end{array}\)
A sending end bus transfer power P0 through a transmission line of p.u. impedance 0.1 If the steady-state stability margin is 40% and the bus voltages at both the ends is 1.0 p.u. Find the operating power angle and magnitude of P0.
(AMIE Summer 2021, 10 marks)
Steady-state stability margin
\( = \frac{{{P_{\max }} - {P_0}}}{{{P_{\max }}}}x100\)
\(40 = \left( {1 - \frac{{{P_0}}}{{{p_{\max }}}}} \right)x100\)
0.4Pmax - Pmax = -P0
P0 = 0.6Pmax = Pmaxsinδ
δ = sin-1(0.6) = 36.870
\({P_0} = \frac{{{V_1}{V_2}}}{{{X_{12}}}}\sin {\delta _0} = \frac{1}{{0.1}}\sin {\delta _0}\)
= 6 p.u.
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