Question

A \(30 \mathrm{KV}, 3\) phase, \(50 \mathrm{HZ}\) synchronous generator has a synchronous reactance of 9 \(\Omega\) phase and an armature resistance of \(2 \Omega\) phase. The generator is delivering 50 MVA at a 0.8 power factor lagging at the rated terminal voltage to an infinite bus.

a) Determine the excitation voltage \(E\) per phase and the power angle.

b) With excitation held constant at the value found in (a), the driving torque is reduced until the generator is delivering \(25 \mathrm{MW}\). Determine the armature current, the power factor and the load reactive power. (Use Newton-Raphson to find \(\delta\) ).

c) If the generator is operating at the excitation voltage of part (a), what is the steady-state maximum power the machine can delivered? Also find the armature current corresponding to this maximum power, power factor and the load reactive power.

d) With excitation held constant at the value found in (a), the generator is delivering only active power. Determine the armature current and the generator efficiency. (Use Newton-Raphson to find \(\delta\) ).

e) If the generator is delivering 40 MVA at a minimum armature current, determine the excitation voltage \(E\) and the power angle.

f) Construct \(\mathrm{V}\) - curve if the generator is delivering \(40 \mathrm{MW}\) with varying field excitation from 0.4 pf leading to 0.4 pf lagging. Assume the open-circuit characteristic in the operating region is given by \(E_{p h}=2000 I_{f}\)

g) The generator excitation voltage is kept constant at the value found in part (a). Plot active power-power angle curve for \(R=0,2,5\) and \(10 \Omega\) phase.

h) The generator excitation voltage is kept constant at the value found in part (a). Plot reactive power-power angle curve for \(R=0,2,5\) and \(10 \Omega\) phase.

Answer 1

bowee feccter coup = Ź U= so k u; 3-0 i 5 = sono gur R=2010 S pour > somun & pf zool Xs = SUL Uppht (R+Jx) Jq -Castos 3 o ono ss +(2-134X (Sox106 DL 53 X 30k 13.5 G = 24.739 | 13.5° (f) line = 24.739 XJS = 4284 KU power and le (6) bout as mw : Cf=(24.739 ko )) han P=Sxup P = BUL IL Coup = cos } -60 Ef LS = V+ Lo + (Rtjx) Jele et? - v gecece G² = ( ut could + Ja Pey Je pa t + [ut sin ottanu? (24.9594)> (30 x + * x2] +(Berf + 3x3)? o cus** 434 ² + 4 Jq2 + (15)²+ 81 742 85 Ty + Ja / 50 +2 1032 +1514152(Q4 731)20 C + 30 x 19X2 + 1SX9X 27en 53 Tq = 0.831KA

load Peactic powe= 3 XULX J x sing power facter colp= ² = 53x30kX 0.83) KX Singo OR => 37.395 MUAR 25 Imay= V. Ercas (8) UL. Casas - VIE for maximum pour = 925 Prna =) @of tabya) - (30k) s J85 30 KX 42.84K 585 85- Pomor 3 1 18.22MW (d) P.F = 1 P= JUL IL Ces y P TL 5 6x106 0.962K4 JULI 53xzok sox 106 Creneut effiua Pout part fler - sox108 + 3X (0.9625x2 234 = 019 = 90/ ,