22-Elec-A6 Power Systems and Machines · December 2013
Question 2 of 6: Magnetic Circuit with Air Gap and Nonlinear Core
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: Professional Engineers Ontario — 07-Elec-A6 Power Systems and Machines, Fall 2013 (closed book, formula sheet supplied). Six questions of equal value; any five constitute a complete paper. All ac voltages/currents are rms; three-phase voltages are line-to-line unless noted. All six questions are solved below as a complete study resource.
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill); T. Wildi, Electrical Machines, Drives, and Power Systems, 6th ed. (Pearson); Fitzgerald, Kingsley & Umans, Electric Machinery, 6th ed.
Question 2: Magnetic Circuit with Air Gap and Nonlinear Core (20 marks)
Given. Symmetric double-window (E-I type) core (silicon steel, Figure 3 B-H curve). Two identical coils $N_1=N_2=500$ turns carry equal currents oriented so their fluxes add in the central leg C, which contains the air gap.
Given data
Air-gap flux $\Phi_C$
3.6 mWb
Cross-section $A$ (all legs)
30 cm² = 30×10−4 m²
Path A (left loop) $\ell_A$
90 cm
Path B (right loop) $\ell_B$
90 cm
Path C (centre) $\ell_C$
45 cm
Air gap $g$
0.4 mm
Turns per coil
500
μ0
4π×10−7 Wb/(A·t·m)
Find. The coil current $I$ (with $|I_1|=|I_2|=I$) needed to drive 3.6 mWb across the gap.
[Figure not reproduced: Figure 2 (redrawn): double-window core, coils on the outer legs, air gap in the central leg C. Both mmf sources drive flux downward through C. See the official exam paper.]
[Figure not reproduced: B-H curve for silicon steel. See the official exam paper or the cited reference text.]
Figure 3: silicon-steel B-H curve (reproduced from the exam). Reads used: B = 0.60 T → H ≈ 560 A·t/m; B = 1.20 T → H ≈ 1225 A·t/m.
Approach. The central leg carries the full 3.6 mWb; by symmetry paths A and B each carry half. Convert each flux to a flux density, read H from the curve, and write Ampere's law around one outer loop (outer leg + centre leg iron + air gap). The mmf of one coil equals the sum of the H·ℓ drops.
Flux densities. Central leg: $B_C=\dfrac{\Phi_C}{A}=\dfrac{3.6\times10^{-3}}{30\times10^{-4}}=1.20\ \text{T}$. By symmetry the outer legs each carry $\Phi_A=\Phi_B=\tfrac12\Phi_C=1.8\ \text{mWb}$, so $B_A=B_B=\dfrac{1.8\times10^{-3}}{30\times10^{-4}}=0.60\ \text{T}.$
Field intensities from the curve. From Figure 3, $B=1.20\ \text{T}\Rightarrow H_C\approx1225\ \text{A}\!\cdot\!\text{t/m}$ (just below the 1.25 T crossing near 1375 A·t/m) and $B=0.60\ \text{T}\Rightarrow H_A\approx560\ \text{A}\!\cdot\!\text{t/m}$ (on the steep segment between about 0.48 T at 500 and 0.90 T at 750 A·t/m).
Air-gap field. The gap is linear: $H_g=\dfrac{B_C}{\mu_0}=\dfrac{1.20}{4\pi\times10^{-7}}=9.549\times10^{5}\ \text{A}\!\cdot\!\text{t/m}$ (fringing neglected, so $B_g=B_C$).
Ampere's law around the left loop. The loop threads coil 1, the left leg (path A), the central leg (path C) and the gap:
$$N I=\mathcal{F}_A+\mathcal{F}_C+\mathcal{F}_g=504.0+551.3+382.0=1437\ \text{A}\!\cdot\!\text{t}$$
By symmetry the right loop gives the identical equation, consistent with $|I_1|=|I_2|$ and the fluxes adding in C.
Solve for the current.
$$\boxed{I=\frac{NI}{N}=\frac{1437}{500}=2.87\ \text{A}}$$
Check: $H_C$ and $H_A$ are graphical reads with roughly ±5% tolerance; the iron drops (551 and 504 A·t) and the gap drop (382 A·t) are comparable, so the answer is sensitive to the curve read to about ±0.1 A. The value 2.87 A uses $H_C=1225$, $H_A=560$ A·t/m.
