Question 11.2: Derive angular speed ωψrT of the rotor-flux-linkage referenc...
Derive angular speed w_{ψrT} of the rotor-flux-linkage reference frame Equation (11.51) using Equation (11.42).
u_{rT}=R_{r}i_{rT}+(\omega _{\Psi _{r}T}-P\Omega _{r})\Psi _{r\Psi }=0 (11.42)
\omega _{\Psi _{\Psi r}T}=\frac{R_{r}}{L_{r}}\frac{L_{m}}{\Psi _{r\Psi }} i_{sT}+P\Omega _{r}=\frac{L_{m}}{\tau _{r}\Psi _{r\Psi }} i_{sT}+P\Omega _{r} (11.51)
The equation for angular speed from Equation (11.42) is u_{rT} = R_{r}i_{rT} + (ω_{ψ_{r}T} – pΩ_{r} ) ψ_{rψ} = 0. Solving it for the angular velocity yields the following.
ω_{ψ_{r}T}ψ_{rψ} = pΩ_{r}ψ_{rψ} – R_{r}i_{rT}Rotor current written in terms of its components is
i_{r} = i_{rψ} + ji_{rT} = (ψ_{r} – L_{m}i_{s}) / L_{r} = (ψ_{rψ} + jψ_{rT} – L_{m} ( i_{sψ} + ji_{sT} )) / L_{r} ⇒ i_{rT} = (ψ_{rT} – L_{m}i_{sT} )/ L_{r}Because ψrT = 0 in the rotor-flux-linkage reference frame, irT =- Lm /Lr. Substitution results in this equation for angular velocity.
ω_{ψrT} = \frac {pΩ_{r}ψ_{rψ} +R_{r}i_{sT}L_{m} / L_{r} }{Ψ_{rΨ}} =pΩ_{r} + \frac {R_{r}L_{m}i_{sT} } {L_{r}ψ_{rψ} } = pΩ_{r} + \frac {R_{r}L_{m} } { L_{r}ψ_{rψ} } i_{sT} = pΩ_{r} + \frac {1 L_{m} }{τ_{r}ψ_{rψ} } i_{sT}The above is identical to Equation(11.51).