Homopolar (Disc-Rotor) DC Machine: 25-Nav-B6 Ocean Engineering and Offshore Structures, May 2013, Q3

Question 3: Homopolar (Disc-Rotor) DC 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.

Symbol	Quantity
$D$	effective outer diameter of the disc rotor (m)
$d$	effective inner diameter, at the inner ring brush (m)
$B$	uniform vertical flux density, perpendicular to the disc (T)
$\omega$	angular speed of the rotor (rad s$^{-1}$)
$I$	radial current fed through the disc between the brushes (A)
$1\ \text{hp}$	$746\ \text{W}$ (given on the front page)

The disc is horizontal, the field is vertical, and the current is radial: the three vectors are mutually perpendicular everywhere, so no direction cosines appear.

Find. [a] the brush-to-brush emf $e$; [b] the electromagnetic torque $T$ on the rotor and the output power expressed in horsepower.

Figure 3 — disc rotor with the elemental annulus of radius $r$ and radial length $\mathrm{d}r$ suggested by the hint. The field is vertical, the current radial, the velocity tangential.

Approach. Take the hint’s elemental annulus, write the motional emf $\mathrm{d}e = B v\,\mathrm{d}r$ and the Lorentz force $\mathrm{d}F = B I\, \mathrm{d}r$ on it, integrate each from the inner to the outer radius, then close with the electromechanical power balance $T\omega = eI$.

Set up the element. Work in radius, with $r_1 = d/2$ at the inner brush and $r_2 = D/2$ at the outer brush. An element at radius $r$ of radial length $\mathrm{d}r$ moves tangentially with speed $$v = \omega r .$$ Because $\vec v$ (tangential), $\vec B$ (vertical) and the element $\mathrm{d}\vec r$ (radial) are mutually orthogonal, all cross products reduce to simple products.
[a] Motional emf of the element. The motional-emf density is $(\vec v \times \vec B)$, and integrating it along the radial path between the brushes gives $$\mathrm{d}e = B\,v\,\mathrm{d}r = B\,\omega r\,\mathrm{d}r .$$ Every radial path from inner to outer brush is equivalent, so the disc behaves as a single conductor of that emf rather than as many parallel conductors of differing emf.
Integrate for the emf. $$e = \int_{d/2}^{D/2} B\omega r\,\mathrm{d}r = B\omega\left[\frac{r^{2}}{2}\right]_{d/2}^{D/2} = \frac{B\omega}{2}\left(\frac{D^{2}}{4}-\frac{d^{2}}{4}\right),$$ that is $$\boxed{\;e = \frac{B\,\omega\,(D^{2}-d^{2})}{8}\ \ \text{volts}\;}$$ Equivalently $e = B\omega(r_2^2-r_1^2)/2$, and note that it is a steady dc emf: the geometry never reverses, which is what distinguishes the homopolar machine from a conventional commutator machine.
[b] Force on the element. With the current $I$ flowing radially, the same element carries $I$ over a length $\mathrm{d}r$ in the field $B$, so $$\mathrm{d}F = B\,I\,\mathrm{d}r,$$ directed tangentially. Its moment about the shaft is $\mathrm{d}T = r\,\mathrm{d}F$.
Integrate for the torque. $$T = \int_{d/2}^{D/2} B I r \,\mathrm{d}r = \frac{BI}{2}\left(\frac{D^{2}}{4}-\frac{d^{2}}{4}\right),$$ giving $$\boxed{\;T = \frac{B\,I\,(D^{2}-d^{2})}{8}\ \ \text{N}\cdot\text{m}\;}$$
Output power and horsepower. The mechanical power developed is $P = T\omega$, and substituting the two boxed results shows the electromechanical balance closes exactly: $$P = T\omega = \frac{B I \omega (D^{2}-d^{2})}{8} = e\,I .$$ Dividing by the front-page conversion, $$\boxed{\;\text{hp} = \frac{P}{746} = \frac{B\,I\,\omega\,(D^{2}-d^{2})}{8\times 746}\;}$$

Check: numerical sanity check The paper gives no numbers, so the closed forms above are the answer. As a scale check, a machine with $D = 0.60$ m, $d = 0.20$ m, $B = 0.80$ T, $\omega = 200$ rad s$^{-1}$ and $I = 50$ A yields $e = 6.4$ V, $T = 1.6$ N·m and $P = 320$ W = 0.429 hp. The very low voltage at a very high current is characteristic of homopolar machines and is precisely why they are rare outside specialised high-current applications.

Quantity	Result
Element emf	$\mathrm{d}e = B\omega r\,\mathrm{d}r$
[a] Brush-to-brush emf	$e = B\omega(D^{2}-d^{2})/8$ V
Element force	$\mathrm{d}F = BI\,\mathrm{d}r$
[b] Torque	$T = BI(D^{2}-d^{2})/8$ N·m
[b] Developed power	$P = T\omega = eI = BI\omega(D^{2}-d^{2})/8$ W
[b] Output horsepower	$\text{hp} = BI\omega(D^{2}-d^{2})/5968$

Topic. The homopolar (Faraday disc) machine — motional emf and Lorentz torque in a continuous conductor.

Key relations. Motional emf $e=\int(\vec v\times\vec B)\cdot \mathrm{d}\vec\ell$ with $v=\omega r$; Lorentz force $\mathrm{d}\vec F = I\,\mathrm{d}\vec\ell\times\vec B$; torque $\mathrm{d}T=r\,\mathrm{d}F$; electromechanical power balance $T\omega = eI$ for a lossless conversion.

Why this works. Both integrals share the same integrand shape, $(\text{constant})\times r\,\mathrm{d}r$, because the velocity grows linearly with radius in the emf integral and the moment arm grows linearly with radius in the torque integral. That is why $e$ and $T$ come out with an identical $(D^2-d^2)/8$ geometry factor, and why the power balance $T\omega=eI$ closes identically rather than only approximately — a useful self-check on the algebra. Physically the disc is a single-turn machine whose “winding” is the metal itself, so it produces a smooth dc emf with no commutator, but a very small one: doubling the output voltage would require quadrupling the disc area.

Common pitfall. Working in diameters and forgetting the factor of four — the integration limits are radii $d/2$ and $D/2$, which is where the 8 in the denominator comes from; using $(D^2-d^2)/2$ is the single most common slip. A second pitfall is integrating $\mathrm{d}F$ to get a force and then multiplying by the mean radius, which is not the same as integrating $r\,\mathrm{d}F$ unless the current distribution happens to be symmetric about that radius.

Source. Chapman, Electric Machinery Fundamentals, 5th ed., §1.6–1.8 (induced voltage and induced force on a conductor in a magnetic field) and §7.1 (the linear dc machine as a prototype).

Question 3 of 8: Homopolar (Disc-Rotor) DC Machine

Question 3: Homopolar (Disc-Rotor) DC Machine (20 marks)