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22-Mec-A5 Electrical and Electronics Engineering · May 2015

Question 3 of 8: Spoke-Type dc 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 Examination 07-Mec-A5 Electrical & Electronics Engineering, May 2015 — 3 hours, closed book, Casio or Sharp approved calculator only. Eight questions of equal value; any five constitute a complete paper. All eight are solved here, since the set is intended as a study resource.

Constants printed on the front page. $\pi = 3.14159$, $1\ \text{hp} = 746\ \text{W}$, and the permeability of free space $\mu_0 = 4\pi \times 10^{-7}\ \text{H}\,\text{m}^{-1}$.

Reference texts for this subject.

Check — printed constants, checked against the printed paper. All three constants in Note [8] of this examination paper are printed correctly and are used exactly as given: $\pi = 3.14159$, $1\ \text{hp} = 746\ \text{W}$ and $\mu_0 = 4\pi \times 10^{-7}\ \text{H}\,\text{m}^{-1}$. The flux-path lengths in the Question 4 specification table are likewise printed correctly as $3.77 \times 10^{-2}$ m and $7.54 \times 10^{-2}$ m, and the negative exponents are used as printed; with them the reluctances come out to exactly $1.5 \times 10^{6}$ and $3.0 \times 10^{6}$ A/Wb, which is how the question was designed.

Question 3: Spoke-Type dc 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 homopolar (acyclic) machine whose armature is a set of eight straight radial spokes spanning the annulus between two coaxial slip rings, spinning in a uniform axial field.

Given data
QuantitySymbolValue
Inner (brush ring) radius$R_1$0.05 m
Outer (brush ring) radius$R_2$0.20 m
Number of radial conductors$N$8
Flux density (vertical, uniform)$B$0.5 T
Rotational speed$n$3000 rev/min
Total brush-to-brush current$I$500 A
Conversion printed on the front page—1 hp = 746 W

Find. [a] the brush-to-brush emf at 3000 rev/min; [b] the electromagnetic torque at 500 A and the corresponding output power in horsepower.

R2 = 0.20 m R1 = 0.05 m dr r ROTATION, ω outer ring brush inner ring brush TOP VIEW (8 spokes) UNIFORM VERTICAL FIELD B = 0.5 T FRONT VIEW Total brush current I = 500 A, n = 3000 rpm
Figure 3 — Spoke-type dc machine. Eight radial conductors are connected between the same pair of ring brushes, so electrically they are eight branches in parallel. The shaded element is the elemental radial length $dr$ suggested by the hint.

Approach. Integrate the motional emf $\mathrm{d}e = B\,v\,\mathrm{d}r = B\,\omega r\,\mathrm{d}r$ along one spoke to get the brush voltage, then integrate the elemental Lorentz torque $\mathrm{d}T = r\,B\,i\,\mathrm{d}r$ over all eight spokes; the product $T\omega$ must reproduce $eI$, which is the free check on both answers.

  1. Convert the speed to an angular velocity. Every point on a spoke moves on a circle, so its linear speed depends on radius through $v = \omega r$. $$\omega = \frac{2\pi n}{60} = \frac{2\pi (3000)}{60} = 314.16\ \text{rad/s}$$
  2. Write the emf of a single elemental length. A conductor element $\mathrm{d}r$ at radius $r$ cuts the vertical field at speed $\omega r$, and $B$, $v$ and $\mathrm{d}\boldsymbol{r}$ are mutually perpendicular, so the elemental emf is simply the product of the three. $$\mathrm{d}e = B\,v\,\mathrm{d}r = B\,\omega\,r\,\mathrm{d}r$$
  3. Integrate along one spoke, from the inner brush ring to the outer one. The integrand is linear in $r$, so the integral is the difference of two squares. $$e = \int_{R_1}^{R_2} B\,\omega\,r\,\mathrm{d}r = \frac{B\,\omega\,(R_2^{2} - R_1^{2})}{2}$$
  4. Substitute and box the brush emf. $$e = \frac{(0.5)(314.16)\left(0.20^{2} - 0.05^{2}\right)}{2} = \frac{(0.5)(314.16)(0.0375)}{2}$$ $$\boxed{e = 2.945\ \text{V}}$$

The single most common error on this question is to multiply that result by the eight conductors. The spokes all run between the same inner ring and the same outer ring, so they are eight identical sources connected in parallel, not in series: the open-circuit brush voltage is the emf of one spoke, and what the eight spokes actually buy is eight times the current capability (and one-eighth of the armature resistance). This is the defining feature of a homopolar machine and the reason such machines are inherently low-voltage, high-current devices.

  1. Split the total current among the parallel spokes. By symmetry the 500 A divides equally. $$i = \frac{I}{N} = \frac{500}{8} = 62.5\ \text{A per spoke}$$
  2. Integrate the elemental torque on one spoke. The force on an element carrying $i$ in a field $B$ is $\mathrm{d}F = B\,i\,\mathrm{d}r$, acting tangentially, so its moment about the shaft is $r\,\mathrm{d}F$. $$T_{\text{spoke}} = \int_{R_1}^{R_2} r\,B\,i\,\mathrm{d}r = \frac{B\,i\,(R_2^{2} - R_1^{2})}{2} = \frac{(0.5)(62.5)(0.0375)}{2} = 0.5859\ \text{N}\cdot\text{m}$$
  3. Sum over the eight spokes and box the torque. Note that $N$ cancels against the current division, so the answer depends only on the total current. $$T = N\,T_{\text{spoke}} = \frac{B\,I\,(R_2^{2} - R_1^{2})}{2} = \frac{(0.5)(500)(0.0375)}{2}$$ $$\boxed{T = 4.688\ \text{N}\cdot\text{m}}$$
  4. Convert torque and speed to output power. $$P = T\,\omega = (4.688)(314.16) = 1472.6\ \text{W}$$ $$\boxed{P = \frac{1472.6}{746} = 1.974\ \text{hp}}$$
  5. Close the energy balance as an independent check. The electrical power crossing the brushes must equal the mechanical power on the shaft in a lossless converter. $$e\,I = (2.945)(500) = 1472.6\ \text{W} = T\,\omega \quad\checkmark$$
Final results — Question 3
PartQuantityResult
[a]Angular velocity $\omega$314.16 rad/s
[a]Brush-to-brush emf $e$2.945 V
[b]Current per spoke $i$62.5 A
[b]Torque per spoke0.586 N·m
[b]Total electromagnetic torque $T$4.688 N·m
[b]Output power $P = T\omega$1472.6 W
[b]Output power in horsepower1.974 hp
Check — engineering reality check. Slightly under 3 V at 3000 rev/min is characteristic of homopolar machines and is why they are almost never built for ordinary drive duty: delivering the 1.97 hp computed here requires 500 A through sliding carbon ring brushes, and the brush contact drop (typically 1–2 V per brush pair) would be comparable to the generated emf itself. Brush drop, spoke resistance, windage and bearing friction are all neglected above, exactly as the question intends; a real machine would deliver noticeably less than 1.97 hp at the shaft.