Chapter B

Q. B.2

The two-pole stator-winding distribution of Fig. B.2 is found on an induction motor with an air-gap length of 0.381 mm, an average rotor radius of 6.35 cm, and an axial length of 20.3 cm. Each stator coil has 15 turns, and the coil phase connections are as shown in Fig. B. 10. Calculate the phase-a air-gap inductance L_{aa0} and the phase-a to phase-b mutual inductance L_{ab}.



Verified Solution

Note that the placement of the coils around the stator is such that the flux linkages of each of the two parallel paths are equal. In addition, the air-gap flux distribution is unchanged if, rather than dividing equally between the two legs, as actually occurs, one path were disconnected and all the current were to flow in the remaining path. Thus, the phase inductances can be found by calculating the inductances associated with only one of the parallel paths.

This result may appear to be somewhat puzzling because the two paths are connected in parallel, and thus it would appear that the parallel inductance should be one-half that of the single-path inductance. However, the inductances share a common magnetic circuit, and their combined inductance must reflect this fact. It should be pointed out, however, that the phase resistance is one-half that of each of the paths.

The winding factor has been calculated in Example B.1. Thus, from Eq. B.27,

L=\frac{4 \mu_{0} l r}{\pi g}\left(\frac{2k_{\mathrm{w}} N_{\mathrm{ph}}}{\text { poles }}\right)^{2}=\frac{16 \mu_{0} l r}{\pi g}\left(\frac{k_{\mathrm{w}} N_{\mathrm{ph}}}{\text { poles }}\right)^{2} \quad \quad \quad (B.27)

\begin{aligned}L_{\mathrm{aa} 0} & =\frac{16 \mu_{0} l r}{\pi g}\left(\frac{k_{\mathrm{w}} N_{\mathrm{ph}}}{\text { poles }}\right)^{2} \\& =\frac{16\left(4 \pi \times 10^{-7}\right) \times 0.203 \times 0.0635}{\pi\left(3.81 \times 10^{-4}\right)}\left(\frac{0.933 \times 30}{2}\right)^{2} \\& =42.4  \mathrm{mH}\end{aligned}

The winding axes are separated by α = 120°, and thus from Eq. B.28

\begin{aligned}L_{\mathrm{12}}&=\frac{4μ_0}{π}\left(\frac{2k_{\mathrm{w1}} N_{\mathrm{1}}}{\text{poles}}\right)\left(\frac{2k_{\mathrm{w2}} N_{\mathrm{2}}}{\text{poles}}\right) \frac{lr}{g} \cos α\\ \\&=\frac{16 \mu_{0}\left(k_{\mathrm{w1}} N_{\mathrm{1}}\right)\left(k_{\mathrm{w2}} N_{\mathrm{2}}\right) l r}{\pi g(\text { poles })^{2}} \cos \alpha \quad \quad \quad (B.28)\end{aligned}

L_{\mathrm{ab}}=\frac{16 \mu_{0}\left(k_{\mathrm{w}} N_{\mathrm{ph}}\right)^{2} l r}{\pi g(\text { poles })^{2}} \cos \alpha=-21.2  \mathrm{mH}