22-Mec-A5 Electrical and Electronics Engineering · December 2013

Question 3 of 8: Homopolar Disc Machine — emf, Torque and Output Power

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. PEO / Engineers Canada National Examinations, December 2013 — 07-Mec-A5, Electrical & Electronics Engineering (Mechanical Engineering discipline). Three hours, closed book, one approved Casio or Sharp calculator. Eight questions; any five constitute a complete paper and each question is of equal value, so each counted question carries 20 marks. All eight questions are solved below, because this set is a study resource rather than an exam script.

Reference texts. A. S. Sedra and K. C. Smith, Microelectronic Circuits, 8th ed. (BJT current mirrors, op-amp frequency response); M. M. Mano and M. D. Ciletti, Digital Design, 6th ed. (universal-gate synthesis, DeMorgan reduction); S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (dc machines, induction-motor testing and slip); C. K. Alexander and M. N. O. Sadiku, Fundamentals of Electric Circuits, 7th ed. (phasor analysis, power factor, first-order transients); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (magnetic circuits and induced emf).

Check: front-page constant. Note [8] of the examination paper prints the permeability of free space as μ₀ = 4π × 10⁷ H·m⁻¹. The exponent has plainly lost its minus sign in printing; the physical constant is 4π × 10⁻⁷ H·m⁻¹, and that is the value used in Question 8. The other two constants, π = 3.14159 and 1 hp = 746 W, are used exactly as printed.

Question 3: Homopolar Disc Machine — emf, Torque and Output Power (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 conducting disc of outer diameter $D$ and inner diameter $d$ (so the active radii run from $r=d/2$ to $r=D/2$), spinning at $\omega$ rad/s in its own horizontal plane, immersed in a uniform vertical flux density $B$. A radial current $I_2$ is injected through ring brushes at the inner and outer radii. The field is everywhere perpendicular to the disc and to the radial current path, so all cross products take their maximum value.

Find. [a] the magnitude of the generated emf $e$ between the brushes; [b] the electromagnetic torque on the rotor and the output power expressed in horsepower.

[Figure not reproduced: Figure 3 redrawn. Left: plan view of the disc showing the elemental annulus at radius $r$ of radial width $dr$ suggested by the hint. Right: elevation showing the uniform vertical field through the horizontal rotor. See the official exam paper.]

Approach. Take the elemental annulus offered by the hint, apply the motional-emf law $d\mathcal{E}=(\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell}$ to it, and integrate radially. Then apply the Lorentz force law $d\mathbf{F}=I\,d\boldsymbol{\ell}\times\mathbf{B}$ to the same element to get the torque, and confirm the two results are consistent by checking that $T\omega$ equals $e I_2$.

Write the velocity of the element. A point of the disc at radius $r$ moves tangentially at $$v=\omega r$$ This velocity, the vertical field $\mathbf{B}$ and the radial direction are mutually perpendicular, which is what makes the machine work at all: the induced electric field $\mathbf{v}\times\mathbf{B}$ points radially, i.e. along the path between the two brushes.
Integrate the motional emf radially. The elemental contribution across the annulus of width $dr$ is $d\mathcal{E}=vB\,dr=B\omega r\,dr$, so $$e=\int_{d/2}^{D/2} B\omega r\,dr = B\omega\left[\frac{r^{2}}{2}\right]_{d/2}^{D/2}=\frac{B\omega}{2}\left(\frac{D^{2}}{4}-\frac{d^{2}}{4}\right)$$ $$\boxed{\;e=\frac{B\,\omega\left(D^{2}-d^{2}\right)}{8}\;}$$ Note that the answer depends on the difference of the squares of the diameters, not on the difference of the diameters — doubling the outer diameter roughly quadruples the emf.
Find the force on the same element. The injected current $I_2$ flows radially, so every element of the current path carries the full $I_2$ over its radial length $dr$, and the force is tangential: $$dF=I_2 B\,dr$$ This force acts at moment arm $r$ about the shaft, so the elemental torque is $dT=r\,dF=r I_2 B\,dr$.
Integrate the torque. The integral is identical in form to the emf integral: $$T=\int_{d/2}^{D/2} I_2 B r\,dr=\frac{I_2 B}{2}\left(\frac{D^{2}}{4}-\frac{d^{2}}{4}\right)$$ $$\boxed{\;T=\frac{I_2\,B\left(D^{2}-d^{2}\right)}{8}\;}$$
Form the output power and convert to horsepower. Mechanical output is torque times angular speed: $$P=T\omega=\frac{I_2 B\omega\left(D^{2}-d^{2}\right)}{8}=e\,I_2$$ which is the electromechanical energy-conversion identity and the strongest available check on parts [a] and [b] together: the mechanical power out must equal the electrical power associated with the induced emf. Using the front-page conversion 1 hp = 746 W, $$\boxed{\;P_{\text{hp}}=\frac{e I_2}{746}=\frac{I_2\,B\,\omega\left(D^{2}-d^{2}\right)}{5968}\;}$$
Substitute representative numbers to make the result concrete. The question is symbolic, but a worked case fixes the magnitudes. For $B=0.80$ T, $\omega=300$ rad/s, $D=0.50$ m, $d=0.10$ m and $I_2=200$ A: $$e=\frac{0.80\times300\times(0.50^{2}-0.10^{2})}{8}=\frac{0.80\times300\times0.240}{8}=7.20\ \text{V}$$ $$T=\frac{200\times0.80\times0.240}{8}=4.80\ \text{N}\cdot\text{m},\qquad P=4.80\times300=1440\ \text{W}=1.930\ \text{hp}$$ and indeed $e I_2 = 7.20\times200=1440$ W, matching $T\omega$ exactly.

The characteristic signature of the homopolar machine is visible in these numbers: a very large current at a very small voltage. Seven volts from a half-metre disc spinning at nearly 3000 rpm in a strong field is a poor voltage, and it is why the topology never displaced the commutated dc machine for general use. Where it does earn its keep is in applications that genuinely want tens of kiloamperes at a few volts — homopolar pulse generators for electromagnetic launchers, and low-voltage electrolytic and welding supplies — because there is no commutator to switch that current.

Final results — Question 3
Quantity	Symbolic result	Worked case
[a] Generated emf	$e=B\omega(D^{2}-d^{2})/8$	7.20 V
[b] Electromagnetic torque	$T=I_2B(D^{2}-d^{2})/8$	4.80 N·m
[b] Output power	$P=T\omega=eI_2$	1440 W
[b] Output in horsepower	$P_{\text{hp}}=I_2B\omega(D^{2}-d^{2})/5968$	1.930 hp

Topic. The homopolar (Faraday disc) machine: motional emf and Lorentz torque on a rotating conductor.

Key relations. Motional emf $d\mathcal{E}=(\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell}$ with $v=\omega r$; Lorentz force $d\mathbf{F}=I\,d\boldsymbol{\ell}\times\mathbf{B}$; torque $dT=r\,dF$; power balance $T\omega=eI$; 1 hp = 746 W.

Why this works. Both integrals reduce to $\int r\,dr$ because the tangential velocity and the moment arm each grow linearly with radius while the field is uniform. That shared structure is not a coincidence — it is the reason the electrical and mechanical powers come out equal, which is a statement of energy conservation in a lossless electromechanical converter. Setting up the problem with the annulus that the hint suggests turns what looks like a three-dimensional field problem into a one-line integral, and any solution that starts by writing $e=BLv$ with a single guessed radius will be wrong by whatever factor that guess implies.

Common pitfall. Working in diameters where radii belong. The limits of integration are $d/2$ and $D/2$, and forgetting the factor of two in each limit produces an answer four times too large; the denominator 8 in the final expressions is exactly the residue of those two halvings. A second pitfall is treating the machine as generating $B\omega D^{2}/8$ and ignoring the inner diameter, which is only valid if the shaft is vanishingly small.

Source. Hayt & Buck, Engineering Electromagnetics, 9th ed., Ch. 9 (motional emf); Chapman, Electric Machinery Fundamentals, 5th ed., Ch. 1 (the linear machine and force/emf duality).

← Question 2 Next: Question 4 →