Magnetic Circuit of a Two-Pole Machine: 22-Elec-A6 Power Systems and Machines, May 2013, Q5

Question 5: Magnetic Circuit of a Two-Pole Machine (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A two-pole machine whose flux path runs pole–gap–rotor–gap–pole and returns through the stator yoke (splitting into $\Phi/2$ each side).

Given data
Rotor radius / axial length	$r = 120\ \text{mm},\ \ell = 200\ \text{mm}$
Pole arc	$40^\circ = 0.6981\ \text{rad}$
Turns (two coils in series)	$N = 2\times 360 = 720$
Air-gap length	$g = 1.5\ \text{mm}$
Yoke outer radius / thickness	$r_o = 210\ \text{mm},\ t = 25\ \text{mm}$
Target air-gap flux density	$B_g = 0.8\ \text{T}$

Find. Air-gap reluctance, pole flux, coil current (iron ignored), yoke flux density, and coil current including yoke mmf.

Figure 5.1 — equivalent magnetic circuit: series mmf $NI$, two air-gap reluctances, and the yoke reluctance; the rotor and pole iron are treated as ideal.

Approach. The pole-face area sets the air-gap reluctance and the flux; Ampère’s law around the loop ($NI = \sum H\ell$) gives the coil current, first with only the two gaps and then adding the yoke drop from the B–H curve.

Air-gap reluctance (a). The pole-face (air-gap) area is the arc length times the axial length: $A_g = (r\,\theta)\,\ell = (0.120)(0.6981)(0.200) = 0.01676\ \text{m}^2$. Then $$\mathcal{R}_g = \frac{g}{\mu_0 A_g} = \frac{1.5\times 10^{-3}}{(4\pi\times 10^{-7})(0.01676)}.$$ $$\boxed{\mathcal{R}_g = 7.12\times 10^{4}\ \text{A}\cdot\text{t/Wb}}$$
Pole flux (b). $\Phi = B_g A_g = (0.8)(0.01676)$. $$\boxed{\Phi = 0.01340\ \text{Wb}}$$
Coil current, iron ignored (c). With the iron ideal, all mmf is dropped across the two series air gaps. Using $H_g = B_g/\mu_0 = 6.366\times 10^{5}\ \text{A/m}$, $$NI = 2\,H_g\,g = 2(6.366\times 10^5)(1.5\times 10^{-3}) = 1910\ \text{A}\cdot\text{t}\;\Rightarrow\; I = \frac{1910}{720}.$$ $$\boxed{I = 2.65\ \text{A}}$$
Yoke flux density (d). At the yoke the pole flux splits, so each side carries $\Phi/2$. The yoke cross-section is thickness × axial length, $A_y = (0.025)(0.200) = 0.005\ \text{m}^2$: $$B_y = \frac{\Phi/2}{A_y} = \frac{0.006702}{0.005}.$$ $$\boxed{B_y = 1.34\ \text{T}}$$
Coil current including yoke mmf (e). From the magnetization curve at $B_y = 1.34\ \text{T}$, $H_y \approx 400\ \text{A/m}$. The yoke mean radius is $r_m = r_o - t/2 = 197.5\ \text{mm}$, and each $\Phi/2$ branch traverses a half-circumference $\ell_y = \pi r_m = 0.6205\ \text{m}$. Adding the yoke drop to the gap mmf, $$NI = 2H_g g + H_y \ell_y = 1910 + (400)(0.6205) = 2158\ \text{A}\cdot\text{t}\;\Rightarrow\; I = \frac{2158}{720}.$$ $$\boxed{I = 3.00\ \text{A}}$$

Check: $H_y \approx 400\ \text{A/m}$ is read graphically from the Figure 1(b) magnetization curve at $B = 1.34\ \text{T}$ (on the saturating solid curve, just below the $B = 1.35\ \text{T}$, $H = 400\ \text{A/m}$ point). A read anywhere in the 380–420 A/m band changes the coil current by only about ±0.01 A, so the result rounds robustly to 3.0 A.

The comparison of parts (c) and (e) is the pedagogical point: the two air gaps dominate the magnetic circuit (1910 of 2158 ampere-turns, about 89 %), while the highly permeable iron yoke — even near saturation at 1.34 T — needs only about 248 ampere-turns. Ignoring the iron underestimates the required current by only ~12 %.

Question 5 — results
Quantity	Value
(a) Air-gap reluctance	$7.12\times10^4$ A·t/Wb
(b) Pole flux	0.0134 Wb
(c) Coil current (iron ignored)	2.65 A
(d) Yoke flux density	1.34 T
(e) Coil current (with yoke)	3.00 A

Topic. Magnetic-circuit (reluctance) analysis of a rotating machine, including saturating iron.

Key relations. $\mathcal{R} = \ell/(\mu A)$; $\Phi = BA$; $H = B/\mu_0$ (air); Ampère’s law $NI = \sum H_k \ell_k$; series/parallel flux splitting.

Why this works. The magnetic circuit is the exact analogue of a resistive circuit: mmf ($NI$) is the source, flux is the “current,” and reluctance is the “resistance.” Because the two air gaps sit in series with the flux, their ampere-turns add; because the yoke splits the return flux into two equal parallel halves, its cross-section carries only $\Phi/2$ and thus a lower flux density than the pole.

Common pitfall. Using the full pole flux (rather than $\Phi/2$) for the yoke density, or using the straight-through diameter instead of the semicircular mean path for the yoke length. Both errors overstate the yoke mmf.

Source. Chapman, Electric Machinery Fundamentals, 5th ed., Ch. 1 (magnetic circuits and machine magnetics).

Question 5 of 6: Magnetic Circuit of a Two-Pole Machine

Question 5: Magnetic Circuit of a Two-Pole Machine (20 marks)