Advanced Motor Analyzer

\[ n_s = \frac{120 \, f}{p} \quad (\text{RPM}) \] \[ \omega_s = \frac{2 \pi \, n_s}{60} \quad (\text{rad/s}) \]

Where \(f\) is the operating frequency (Hz) and \(p\) is the number of poles.

\[ \text{star: } V_{ph} = \frac{V_{LL}}{\sqrt{3}}, \quad I_{line} = I_1 \qquad \text{delta: } V_{ph} = V_{LL}, \quad I_{line} = \sqrt{3}\, I_1 \]

The equivalent circuit is per phase. For star-connected parameters each phase sees the line-to-neutral voltage — Chapman, Electric Machinery Fundamentals, Example 6-5, solves its 460 V Y-connected motor with \(V_\phi = 460/\sqrt{3} = 265.6\) V. Torque and power are three times the per-phase values in both cases; the reported currents are line currents. A single-phase motor uses the supply voltage directly.

\[ V_{eff} = V_{ph} \times \min\!\left(\frac{f_{op}}{f_{line}},\; 1\right) \] \[ X_{eff} = X_{rated} \times \frac{f_{op}}{f_{line}} \]

Up to rated (base) frequency the voltage scales with frequency to keep the flux constant. Above base frequency the drive cannot exceed rated voltage, so \(V\) is held at \(V_{ph}\) while the reactances keep rising: the flux falls as \(f_{line}/f_{op}\) and the pull-out torque roughly as \((f_{line}/f_{op})^2\) — the constant-power (field-weakening) region (Chapman, Electric Machinery Fundamentals, §6-9). Resistances \(R_1\) and \(R_2\) remain unchanged. In DOL mode, \(f_{op} = f_{line}\) so no scaling occurs.

\[ V_{th} = V_{eff} \times \frac{X_m}{\sqrt{R_1^2 + (X_1 + X_m)^2}} \]

The stator impedance and magnetizing branch are combined into a single Thévenin source as seen from the rotor terminals:

\[ |Z|^2 = R_1^2 + (X_1 + X_m)^2 \] \[ R_{th} = \frac{(-X_1 X_m) R_1 + (R_1 X_m)(X_1 + X_m)}{|Z|^2} \] \[ X_{th} = \frac{(R_1 X_m) R_1 - (-X_1 X_m)(X_1 + X_m)}{|Z|^2} \]

\[ I_2 = \frac{V_{th}}{\sqrt{\left(R_{th} + \dfrac{R_2}{s}\right)^2 + (X_{th} + X_2)^2}} \] \[ T = \frac{3 \, I_2^2 \left(\dfrac{R_2}{s}\right)}{\omega_s} \quad (\text{N} \cdot \text{m}) \] \[ I_1 = \frac{V_{ph}}{\left| R_1 + jX_1 + jX_m \parallel \left(\dfrac{R_2}{s} + jX_2\right) \right|} \quad (\text{stator phase current}) \]

Where \(s\) is the slip. The factor of 3 accounts for all three phases. \(I_2\) (rotor current referred to the stator) gives the torque only; the current curve and the starting currents report the stator current \(I_1\) (as line current: \(\sqrt{3}\,I_1\) for delta), which includes the magnetising current and is what the supply delivers (Chapman, Electric Machinery Fundamentals, Example 6-5). \(I_2\) under-reads it, most at light load where the magnetising current dominates.

The pulsating main-winding field is resolved into forward and backward rotating fields. Each sees half of the magnetising reactance and rotor; the backward field runs at slip \(2 - s\). One stator current flows through the stator impedance and both half-circuits in series (Chapman, Electric Machinery Fundamentals, 4th ed., §9-3; Fitzgerald & Kingsley, Electric Machinery, Appendix C):

\[ Z_F = \frac{(R_2/s + jX_2)\, jX_M}{R_2/s + j(X_2 + X_M)}, \qquad Z_B = \frac{(R_2/(2-s) + jX_2)\, jX_M}{R_2/(2-s) + j(X_2 + X_M)} \] \[ I_1 = \frac{V}{\left| R_1 + jX_1 + 0.5 Z_F + 0.5 Z_B \right|} \] \[ P_{AG,F} = I_1^2 \cdot 0.5 R_F, \qquad P_{AG,B} = I_1^2 \cdot 0.5 R_B \] \[ \tau_{ind} = \frac{P_{AG,F} - P_{AG,B}}{\omega_s}, \qquad P_{conv} = (1 - s)(P_{AG,F} - P_{AG,B}) \]

\(R_1, X_1\) are the main-winding values and \(R_2, X_2, X_M\) the full (not halved) rotor and magnetising values. At \(s = 1\), \(Z_F = Z_B\) and the torque is zero: the main winding alone cannot start the motor. Chapman's Example 9-1 (110 V, 6-pole split-phase motor at \(s = 0.05\): \(I_1 = 4.45\) A, \(\tau_{ind} = 2.11\) N·m) is reproduced exactly.

\[ I_1 = \frac{V}{\left| R_1 + jX_1 + jX_m \parallel \left(\dfrac{R_2}{s} + jX_2\right) \right|} \quad (\text{three-phase stator current}) \] \[ P_{scl} = 3 \, I_1^2 \, R_1 \quad (\text{three-phase}), \qquad P_{scl} = I_1^2 \, R_1 \quad (\text{single-phase}) \] \[ P_{ag} = T \, \omega_s \quad (\text{three-phase air gap power}) \] \[ P_{in} = P_{ag} + P_{scl} + P_{core} \quad (\text{single-phase: } P_{AG,F} + P_{AG,B} + P_{scl} + P_{core}) \] \[ P_{conv} = P_{ag,net} (1 - s) \quad (\text{converted mechanical power}) \] \[ P_{out} = P_{conv} - P_{fw} \quad (\text{core loss is already counted in } P_{in}) \]

The stator copper loss uses the stator current \(I_1\) of the full equivalent circuit, which includes the magnetising current (Chapman, Electric Machinery Fundamentals, Example 6-5: \(I_1 = 18.88\) A, \(P_{SCL} = 685\) W at \(s = 0.022\)). Core-loss convention: the equivalent circuit has no core-loss resistor, so \(P_{core}\) is added to the input power rather than subtracted from the converted power as Chapman does. Both conventions conserve energy, but efficiencies differ slightly from textbook answers that deduct the core loss at the shaft.

\[ \eta = \frac{P_{out}}{P_{in}} \times 100\% \]

The 101-point sweep locates the peaks; the pull-out torque and peak efficiency are then found between the neighbouring sweep points by golden-section search, so the reported maxima are the continuous maxima rather than the best grid point (which can read 0.1–0.2 percentage points low for efficiency).

pytest calcs/components/c0013_motor_analyser/ -v
pytest calcs/test_c0013* -v