Question 11.12: The three-phase, 230-V, 60-Hz, 12-kW, four-pole induction mo......
The three-phase, 230-V, 60-Hz, 12-kW, four-pole induction motor of Example 6.7 and Example 11.11 is to be driven by a field-oriented speed-control system (similar to that of Fig. 11.21b) at a speed of 1740 r/min. Assuming the controller is programmed to set the rotor flux linkages λ_{DR} to the machine rated peak value, find the rms amplitude of the armature current, the electrical frequency, and the rms terminal voltage, if the electromagnetic power is 9.7 kW and the motor is operating at a speed of 1680 r/min.
We must first determine the parameters for this machine. From Eqs. 11.66 through Eq. 11.74
\begin{array}{c}L_{\mathrm{m}}=\frac{X_{\mathrm{m}0}}{\omega_{\mathrm{e0}}}=\frac{18.7}{120 \pi}=49.6 \mathrm{mH} \\ \\L_{\mathrm{S}}=L_{\mathrm{m}}+\frac{X_{10}}{\omega_{\mathrm{e0}}}=49.6 \mathrm{mH}+\frac{0.680}{120 \pi}=51.41 \mathrm{mH} \\ \\L_{\mathrm{R}}=L_{\mathrm{m}}+\frac{X_{20}}{\omega_{\mathrm{e0}}}=49.6 \mathrm{mH}+\frac{0.672}{120 \pi}=51.39 \mathrm{mH} \\ \\R_{\mathrm{a}}=R_{1}=0.095 \Omega \\ \\R_{\mathrm{aR}}=R_{2}=0.2 \Omega\end{array}
The rated rms line-to-neutral terminal voltage for this machine is 230/\sqrt{3}= 132.8 V and thus the peak rated flux for this machine is
\left(\lambda_{\text {rated }}\right)_{\text {peak }}=\frac{\sqrt{2}\left(V_{\mathrm{a}}\right)_{\text {rated }}}{\omega_{\mathrm{e}}}=\frac{\sqrt{2} \times 132.8}{120 \pi}=0.498 \mathrm{~Wb}
For the specified operating condition
\omega_{\mathrm{m}}=n\left(\frac{\pi}{30}\right)=1680\left(\frac{\pi}{30}\right)=176 \mathrm{rad} / \mathrm{sec}
and the mechanical torque is
T_{\text {mech }}=\frac{P_{\text {mech }}}{\omega_{\mathrm{m}}}=\frac{9.7 \times 10^{3}}{176}=55.1 \mathrm{~N} \cdot \mathrm{m}
From Eq. 11.77, with \lambda_{D R}=\lambda_{\text {rated }}=0.498 \mathrm{~Wb}
T_{\text {mech }}=\frac{3}{2}\left( \frac{\text{poles}}{2}\right) \left( \frac{L_{\mathrm{m}}}{L_{\mathrm{R}}}\right) λ_{DR}i_Q \quad \quad \quad (11.77)
\\ \begin{aligned}i_{\mathrm{Q}} & =\frac{2}{3}\left(\frac{2}{\text { poles }}\right)\left(\frac{L_{\mathrm{R}}}{L_{\mathrm{m}}}\right)\left(\frac{T_{\text {mech }}}{\lambda_{\mathrm{DR}}}\right)\\ \\& =\frac{2}{3}\left(\frac{2}{4}\right)\left(\frac{51.39 \times 10^{-3}}{49.6 \times 10^{-3}}\right)\left(\frac{55.1}{0.498}\right)=38.2 \mathrm{~A}\end{aligned}
From Eq. 11.79,
\lambda_{\mathrm{DR}}=L_{\mathrm{m}}i_{\mathrm{D}} \quad \quad \quad (11.79)
i_{\mathrm{D}}=\frac{\lambda_{\mathrm{DR}}}{L_{\mathrm{m}}}=\frac{0.498}{49.6 \times 10^{-3}}=10.0 \mathrm{~A}
The rms armature current is thus
I_{\mathrm{a}}=\sqrt{\frac{i_{\mathrm{D}}^{2}+i_{\mathrm{Q}}^{2}}{2}}=\sqrt{\frac{10.0^{2}+38.2^{2}}{2}}=27.9 \mathrm{~A}
The electrical frequency can be found from Eq. 11.81
\omega_{\mathrm{e}}=\omega_{\mathrm{me}}-R_{\mathrm{aR}}\left(\frac{i_{\mathrm{QR}}}{λ_{\mathrm{DR}}}\right)\quad \quad \quad (11.81)
\omega_{\mathrm{e}}=\omega_{\mathrm{me}}+\frac{R_{\mathrm{aR}}}{L_{\mathrm{R}}}\left(\frac{i_{\mathrm{Q}}}{i_{\mathrm{D}}}\right)
With \omega_{\mathrm{me}}=(\text{poles}/ 2) \omega_{\mathrm{m}}=2 \times 176=352 \mathrm{rad} / \mathrm{sec}
\omega_{\mathrm{e}}=352+\left(\frac{0.2}{51.39 \times 10^{-3}}\right)\left(\frac{38.2}{10.0}\right)=367 \mathrm{rad} / \mathrm{sec}
and f_{\mathrm{e}}=\omega_{\mathrm{e}} /(2 \pi)=58.4 \mathrm{~Hz}
Finally, from Eq. 11.89, the rms line-to-neutral terminal voltage
\begin{aligned}V_{\mathrm{a}} & =\sqrt{\frac{v_D^2+v_Q^2}{2}}=\sqrt{\frac{\left(R_{\mathrm{a}} i_{\mathrm{D}}-\omega_{\mathrm{e}}λ_Q\right)^2+(R_ai_Q+ω_e λ_D)^2}{2}}\\ \\ & =\sqrt{\frac{\left(R_{\mathrm{a}} i_{\mathrm{D}}-\omega_{\mathrm{e}}\left(L_{\mathrm{S}}-\frac{L_{\mathrm{m}}^{2}}{L_{\mathrm{R}}}\right) i_{\mathrm{Q}}\right)^{2}+\left(R_{\mathrm{a}} i_{\mathrm{Q}}+\omega_{\mathrm{e}} L_{\mathrm{S}} i_{\mathrm{D}}\right)^{2}}{2}}\quad \quad \quad (11.89)\end{aligned}
\begin{aligned}V_{\mathrm{a}} & =\sqrt{\frac{\left(R_{\mathrm{a}} i_{\mathrm{D}}-\omega_{\mathrm{e}}\left(L_{\mathrm{S}}-\frac{L_{\mathrm{m}}^{2}}{L_{\mathrm{R}}}\right) i_{\mathrm{Q}}\right)^{2}+\left(R_{\mathrm{a}} i_{\mathrm{Q}}+\omega_{\mathrm{e}} L_{\mathrm{S}} i_{\mathrm{D}}\right)^{2}}{2}}\\ \\& =140.6 \mathrm{~V} \text { line-to-neutral }=243.6 \mathrm{~V} \text { line-to-line }\end{aligned}