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Electrical Power Systems

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92 <strong>Electrical</strong> <strong>Power</strong> <strong>Systems</strong><br />

Similar equivalent circuits can be obtained for reactances along the quadrature axis. These<br />

reactances, x² q, x¢ q, and x q may be considered for cases when the circuit resistance results in a<br />

power factor above zero and the armature reaction is not totally on the direct axis.<br />

The armature current following sudden short circuit to the armature of an initially unloaded<br />

machine can be expressed as:<br />

Where<br />

i a (t) = 2<br />

E<br />

o<br />

L<br />

NMHG<br />

I<br />

<br />

HG<br />

- t t¢¢ - t t¢<br />

I<br />

d d<br />

1 1 1 1 1<br />

e<br />

e<br />

wt<br />

Xd¢¢ Xd<br />

Xd Xd<br />

X d<br />

- +<br />

¢ KJ ¢ - + sin( + d)<br />

...(4.35)<br />

KJ P<br />

t² d = direct axis short-circuit subtransient time constant<br />

t¢ d = direct axis short-circuit transient time constant.<br />

4.8 DC COMPONENTS O STATOR CURRENTS<br />

The expression for the armature current as given by eqn. (4.35), the unidirectional transient<br />

component has not been considered. As we know from the consideration of the simple R-L<br />

circuit, there will be a dc offset depending on when the voltage is applied. Similarly, in a<br />

synchronous machine, the dc offset component depends on the instantaneous value of the stator<br />

voltage at the time of the short circuit. The time constants associated with the decay of the dc<br />

component of the stator current is defined as the armature short circuit time constant, t a .<br />

Most of the decay of the dc component current occurs during the subtransient period and<br />

because of this reason, the average value of the direct axis and quadrature axis subtransient<br />

reactances is used for calculating t a . It’s approximate value is given by,<br />

( Xd¢¢ + Xq¢¢<br />

)<br />

ta =<br />

2 Ra<br />

Typical value of ta is around 0.045 to 0.18 second.<br />

The dc component current for phase ‘a’ is given by<br />

idc = 2 Eo<br />

Xd¢¢<br />

sin. d e<br />

-t/<br />

ta<br />

Combination of eqns. (4.35) and (4.37) will give an asymmetrical wave form, i.e.,<br />

i asy (t) = 2<br />

E<br />

o<br />

L<br />

NMHG<br />

I<br />

<br />

HG<br />

1 1 1 1 1<br />

X X e<br />

X X e<br />

t<br />

t<br />

d¢¢ d<br />

d d X<br />

-<br />

+<br />

¢ KJ ¢ -<br />

+ KJ<br />

I<br />

O<br />

QP<br />

O<br />

Q<br />

Eo<br />

sin( wt + d) + 2 sin d.<br />

e<br />

X ¢¢<br />

...(4.36)<br />

...(4.37)<br />

- / td¢¢ - / td¢ -t/<br />

ta<br />

d<br />

d<br />

The maximum possible initial magnitude of the dc component current is<br />

max Eo<br />

idc = 2<br />

Xd<br />

¢¢<br />

Therefore, the maximum rms current at the beginning of the short circuit is<br />

...(4.38)<br />

...(4.39)

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