Question 6.6: Symmetrical Short Circuit of a Generator at No Load Derive t......

From Equations 6.89 through 6.94,v_{d }= -ri_{d} – \frac{dθ}{dt}λ_{q} – \frac{dλ_{d}}{dt}      (6.89)
v_{F} = r_{F}i_{F} + \frac{dλ_{F}}{dt}     (6.90)
v_{D} = r_{D}i_{D} +\frac{dλ_{D}}{dt} = 0     (6.91)
v_{q}= -ri_{q} + \frac{d\theta }{dt}\lambda _{d} – \frac{d\lambda _{q}}{dt}     (6.92)
v_{Q} = r_{Q}i_{Q} + \frac{dλ_{Q}}{dt} = 0   (6.93)
λ_{0}= L_{0}i_{0}      (6.94)

Zero sequence
v_{0} = -L_{0} \frac{di_{0}}{dt} = 0
Direct axis
v_{d} = -ω_{0}λ_{q} – \frac{dλ_{d}}{dt} = 0
v_{f} = \frac{dλ_{F}}{dt}
Quadrature axis
v_{q} = ω_{0}λ_{d}- \frac{dλ_{q}}{dt}=0
The flux linkages can be expressed in terms of currents by using Equations 6.95 and 6.96:
\left|\begin{matrix} \lambda _{d} \\ \lambda _{F} \\ \lambda _{D} \end{matrix} \right| = \left|\begin{matrix} L_{d} & kM_{F} & kM_{D} \\ kM_{F} & L_{F} & M_{R} \\ kM_{D} & M_{R} & L_{D} \end{matrix} \right| \left|\begin{matrix} i _{d} \\ i _{F} \\ i _{D} \end{matrix} \right|             (6.95)

\left|\begin{matrix} \lambda _{q} \\ \lambda _{Q} \end{matrix} \right| = \left|\begin{matrix} L_{q} & kM_{Q} \\ kM_{Q} & L_{Q} \end{matrix} \right| \left|\begin{matrix} i _{q} \\ i _{Q} \end{matrix} \right|            (6.96)
ω_{0}L_{q}i_{q} + L_{d}\frac{di_{d}}{dt}+ kM_{F} \frac{di_{F}}{dt}= 0
v_{F} = kM_{F}\frac{di_{d}}{dt} + L_{F} \frac{di_{F}}{dt}
-ω_{0}L_{d}i_{d} – ω_{0}kM_{F}i_{F} + L_{q}\frac{di_{q}}{dt}=0
These equations can be solved using Laplace transformation. The initial conditions must be considered. In Example 6.4, we demonstrated that at no load, prior to fault, the terminal voltage is equal to the generated voltage and this voltage is
\sqrt{2} E =ω_{0}M_{F}i_{F}e^{ jδ}
and this is a quadrature axis quantity. Also, v_{d} = 0. The effect of short circuit is, therefore, to reduce v_{q} to zero. This is equivalent to applying a step function of -v_{q}1 to the q axis. The transient currents can then be superimposed on the currents prior to fault. Except for current in the field coil, all these currents are zero. The solution for i_{F} will be superimposed on the existing current i_{F0}.
If we write
kM_{F} = L_{ad} = \frac{X_{ad}}{ ω}
L_{F} = \frac{(X_{f} \ + \ X_{ad})}{ω}
The expressions converted into reactances and using Laplace transform, dx/dt = sX(s) – x(0^-)
[where X(s) is Laplace transform of x(t)] reduce to
0 =(\frac{ 1}{ω}) (X_{ad} + X_{f})si_{F} +(\frac{ 1}{ω})X_{ad}si_{d}          (6.108)
0 = (\frac{ 1}{ω})(X_{d})si_{d} + (\frac{ 1}{ω})X_{ad}si_{f} + (X_{q})i_{q}       (6.109)
-v_{q} = (\frac{ 1}{ω})X_{q}si_{q} – X_{ad}i_{F} – X_{d}i_{d}       (6.110)
The field current from Equation 6.108 is
i_{F} = \frac{-i_{d}X_{ad}}{(X_{ad} \ + \ X_{f})}          (6.111)
The field current is eliminated from Equations 6.109 to 6.110 by substitution. The quadrature axis current is
i_{q} = \frac{1}{X_{aq} \ + \ X_{l}} [\frac{X_{aq}X_{f}}{X_{aq} \ + \ X_{f}}+ X_{l}] \frac{s}{ω}i_{d}
= – (\frac{X^{\prime}_{d}}{X_{q}})\frac{s}{ω}i_{d}
and
i_{d} = \frac{ω^2}{X^{\prime}_{d}}[\frac{1}{s^2 \ + \ ω^2}]v_{q}
Solving these equations gives
i_{d} = \frac{( \sqrt{3} |E|)}{X^{\prime}_{d}}(1 – \cos ωt)
i_{q} = – \frac{( \sqrt{3} |E|)}{X_{q}} \sin ωt
Note that k = \sqrt{3/2}. Apply
\overline{i}_{abc} = \overline{P} \ \overline{i}_{0dq}
with θ = ωt + π/2 + δ, the short-circuit current in phase a is
i_{a} = \sqrt{2} |E|[(\frac{1}{X^{\prime}_{d}}) \sin (ωt+δ)+ \frac{X_{q} \ + \ X^{\prime}_{d}}{2X^{\prime}_{d}X_{q}} \sin δ – \frac{X_{q} \ – \ X^{\prime}_{d}}{2X^{\prime}_{d}X_{q}} \sin (2ωt+δ)]        (6.112)
The first term is normal-frequency short-circuit current, the second is constant asymmetric current, and the third is double-frequency short-circuit current. The 120 Hz component imparts a nonsinusoidal characteristic to the short-circuit current waveform. It rapidly decays to zero and is ignored in the calculation of short-circuit currents.
When the damper winding circuit is considered, the short-circuit current can be expressed as
i_{a} = \sqrt{2} E [(\frac{1}{X_{d}}) \sin (ωt+δ)+(\frac{1}{X^{\prime}_{d}}-\frac{1}{X_{d}})e^{-t/T^{\prime}_{d}} \sin (ωt+δ) + (\frac{1}{X^{\prime \prime}_{d}}-\frac{1}{X_{d}^{\prime}})e^{-t/T^{\prime \prime}_{d}} \sin (ωt+δ)
– \frac{(X^{\prime \prime}_{d} \ + \ X^{\prime \prime}_{q})}{2X^{\prime \prime}_{d}X^{\prime \prime}_{q}}e^{-t/T_{a}} \sin δ – \frac{(X^{\prime \prime}_{d} \ – \ X^{\prime \prime}_{q})}{2X^{\prime \prime}_{d}X^{\prime \prime}_{q}}e^{-t/T_{a}} \sin (2ωt+δ)]      (6.113)

● The first term is final steady-state short-circuit current.
● The second term is normal-frequency decaying transient current.
● The third term is normal-frequency decaying subtransient current.
● The fourth term is asymmetric decaying dc current.
● The fifth term is double-frequency decaying current.