Question 6.2: Calculate the time variation of the direct axis, quadrature ......

As the generator is operating without load, i_{abc} = i_{0dq} = 0. Therefore, from Equations 6.94 through 6.96
λ_{0}=L_{0} i_{0}    (6.94)

\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)

λ_{0} = 0 \ \ \ λ_{d} = kM_{F}i_{F} \ \ \ λ_{F} = L_{F}i_{F} \ \ \ λ_{q} = 0
From Equations 6.88 through 6.92,
v_{0}=ri_{0} – \frac{d\lambda _{0}}{dt}           (6.88)
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_{0} = 0
v_{d} = – \frac{ d λ_{d}}{dt} = -kM_F  \frac{di_{E}}{dt}
v_{F} = r_{F}i_{F} + L_{F}  \frac{di_{F}}{dt}
Therefore, as expected, the time variation of field current is
i_{F} = \frac{ 1}{r_{F}} \ (1 \ – \ e^{(-r_{F}/L_{F})t})
The direct axis and quadrature axis voltages are given by
v_{d} = – \frac{ kM_{F}}{L_{F}}e^{(-r_{F}/L_{F})t}
v_{q} = – \frac{ \omega kM_{F}}{r_{F}} \ v_{F} (1 \ – \ e^{(-r_{F}/L_{F})t})
The phase voltages can be calculated using Park’s transformation.