- #31
voko
- 6,053
- 391
I have reviewed the discussion and it seems that we have completely neglected the question of efficiency, which has a major impact on the design, and even its feasibility.
1. Determination of the required EMF per coil. For any desired power rating ## P ## (of the entire generator), the power at the per-phase load is ##P_p = P / 3 ##. Let ## r_p ## be the resistance of one phase's internal wiring, ## R_p ## be the resistance of one phase's load, ## E_p ## the phase EMF, and ## I_p ## the phase current. Then, from Ohm's law, ## E_p = (r_p + R_p)I_p = r_pI_p + R_pI_p ##. Multiplying this with ## I_p ##, we have ## E_pI_p = r_pI_p^2 + R_pI_p^2 = r_pI_p^2 + P_p ##. The ## r_pI_p^2 = W_p ## term is the power wasted on heating the internal wiring, so we want to minimize that. Let's say we want ## W_p = wP_p ##, where ## w ## is the "waste factor": this is how much waste heat is produced for the rated power.
Since ## W_p = r_p I_p^2 = w P_p ##, we can determine the phase current from it: ## I_p = \sqrt { \frac {w P_p } {r_p} } ##, and the formula for the phase EMF then becomes: ## \displaystyle E_p = \frac {1 + w } {\sqrt {w}} \sqrt {P_p r_p} = \frac {1 + w } {\sqrt {3w}} \sqrt {P r_p}##.
Taking ## P = 2500 \ W ##, ## w = 0.03 ## (3% waste) and ## r_p = 1 \ Ω ##, we end up with ## E_p = 172 \ V ##. Since there are 6 coils per phase, the required EMF per coil ## E_c = 29 \ V ##. Not five volts as estimated originally!
2. Windings per coil. A more accurate formula for a coil's induced EMF is ## E_c = \frac {\pi N B_s S n f} {\sqrt {2}} ##, where ## N ## is the number of turns per coil, ## B_s ## is the average magnetic flux density in the air gap, ## S ## is the area of the coil's cross-section (and of the magnet at the same time), ## n ## is the number of magnets, ## f ## is the rotational frequency (Hz, revolutions per second). To find out the number of turns: ## \displaystyle N = \frac {\sqrt {2} E_c} {\pi B_s S n f} ##.
## B_s ## could be taken to be 0.5 T. Assuming 0.0025 mm2 square coils (and magnets, meaning 5 cm x 5 cm), ## N = 145 ##.
3. Wire gauge. Each coil has 145 turns, and each turn is 4*0.05 m, thus one coil has 29 m of wire, and one phase has 174 m. Thus the wire's resistance must be no greater than 1/174 = 0.0057 Ω/m = 5.7 mΩ/m. And here comes the catch: that corresponds to the 4 mm2 standard copper wire. That means the diameter of the conductor is 2.25 mm, and with insulation it is probably around 2.5 mm. If you arrange 145 turns of such wire in a single layer, you will end up with a coil over a foot long. That's clearly unacceptable. You could wind that in 7 layers, but then the thickness of the winding will be 35 mm, which is comparable to the size of the coil, which means the the formula we used to get EMF and the number of coils becomes invalid.
The reason we end up in this situation is because we set a very high standard for efficiency: just 3% losses. Let's see what we get if we make that 10%. Then ## E_p = 100 \ V, E_c = 16.7 \ V, \ N = 83 ##. This results in about 100 meters of wire per phase, thus requiring its resistance to be no greater than 10 mΩ/m, and that means the 2.5 mm2 standard copper wire. Its diameter is 1.8 mm; with insulation, 2 mm. Thus the length of a single-layer coil becomes 166 mm; a four-layer coil is then under 5 cm length, with the thickness of the winding at 16 mm, which will probably work.
1. Determination of the required EMF per coil. For any desired power rating ## P ## (of the entire generator), the power at the per-phase load is ##P_p = P / 3 ##. Let ## r_p ## be the resistance of one phase's internal wiring, ## R_p ## be the resistance of one phase's load, ## E_p ## the phase EMF, and ## I_p ## the phase current. Then, from Ohm's law, ## E_p = (r_p + R_p)I_p = r_pI_p + R_pI_p ##. Multiplying this with ## I_p ##, we have ## E_pI_p = r_pI_p^2 + R_pI_p^2 = r_pI_p^2 + P_p ##. The ## r_pI_p^2 = W_p ## term is the power wasted on heating the internal wiring, so we want to minimize that. Let's say we want ## W_p = wP_p ##, where ## w ## is the "waste factor": this is how much waste heat is produced for the rated power.
Since ## W_p = r_p I_p^2 = w P_p ##, we can determine the phase current from it: ## I_p = \sqrt { \frac {w P_p } {r_p} } ##, and the formula for the phase EMF then becomes: ## \displaystyle E_p = \frac {1 + w } {\sqrt {w}} \sqrt {P_p r_p} = \frac {1 + w } {\sqrt {3w}} \sqrt {P r_p}##.
Taking ## P = 2500 \ W ##, ## w = 0.03 ## (3% waste) and ## r_p = 1 \ Ω ##, we end up with ## E_p = 172 \ V ##. Since there are 6 coils per phase, the required EMF per coil ## E_c = 29 \ V ##. Not five volts as estimated originally!
2. Windings per coil. A more accurate formula for a coil's induced EMF is ## E_c = \frac {\pi N B_s S n f} {\sqrt {2}} ##, where ## N ## is the number of turns per coil, ## B_s ## is the average magnetic flux density in the air gap, ## S ## is the area of the coil's cross-section (and of the magnet at the same time), ## n ## is the number of magnets, ## f ## is the rotational frequency (Hz, revolutions per second). To find out the number of turns: ## \displaystyle N = \frac {\sqrt {2} E_c} {\pi B_s S n f} ##.
## B_s ## could be taken to be 0.5 T. Assuming 0.0025 mm2 square coils (and magnets, meaning 5 cm x 5 cm), ## N = 145 ##.
3. Wire gauge. Each coil has 145 turns, and each turn is 4*0.05 m, thus one coil has 29 m of wire, and one phase has 174 m. Thus the wire's resistance must be no greater than 1/174 = 0.0057 Ω/m = 5.7 mΩ/m. And here comes the catch: that corresponds to the 4 mm2 standard copper wire. That means the diameter of the conductor is 2.25 mm, and with insulation it is probably around 2.5 mm. If you arrange 145 turns of such wire in a single layer, you will end up with a coil over a foot long. That's clearly unacceptable. You could wind that in 7 layers, but then the thickness of the winding will be 35 mm, which is comparable to the size of the coil, which means the the formula we used to get EMF and the number of coils becomes invalid.
The reason we end up in this situation is because we set a very high standard for efficiency: just 3% losses. Let's see what we get if we make that 10%. Then ## E_p = 100 \ V, E_c = 16.7 \ V, \ N = 83 ##. This results in about 100 meters of wire per phase, thus requiring its resistance to be no greater than 10 mΩ/m, and that means the 2.5 mm2 standard copper wire. Its diameter is 1.8 mm; with insulation, 2 mm. Thus the length of a single-layer coil becomes 166 mm; a four-layer coil is then under 5 cm length, with the thickness of the winding at 16 mm, which will probably work.