Timeline for Magnetic induction with varying rpm: Dealing with excess energy
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 6, 2019 at 1:28 | comment | added | Russell McMahon♦ | @BendingBender That 20V x I_load is for the example I gave where the alternator "tops out" at some voltage. This is usually the case. You can also clamp excess current so that the alternator saturates and this limiots voltage and power dissipated. You can .... MANY MANY things but as long as we have no clude as to what you are doing we can't offer much good advice. | |
| Jul 5, 2019 at 16:11 | vote | accept | BendingBender | ||
| Jul 5, 2019 at 16:11 | vote | accept | BendingBender | ||
| Jul 5, 2019 at 16:11 | |||||
| Jul 5, 2019 at 16:10 | comment | added | BendingBender | "P_load = I_load x 5V. P_alt = 20 V x I_load " Thank you Russell, I think this helped me. :) | |
| Jul 5, 2019 at 14:43 | comment | added | Russell McMahon♦ | @BendingBender Assumption that can be changed is " ... if the load resistance is constant ..." -> As noted above, if you use a voltage regulator then the load will "see" a fixed voltage and load current is constant after Vdesired is reached. So IF voltage increases linearly with voltage (and it would be exceptionally unusual to have Vmax = 20 x Vwanted) then power will be 20 x (linear with V) and not square related. As we still no ~~= nowt about what you are doing, allowed complexity, cost, ... ANYTHING ... we know not if a small smps would be OK but, if so, then Pmax is a few times Pwanted. | |
| Jul 5, 2019 at 14:06 | comment | added | BendingBender | For the induced voltage I used: V_ind = -NxAxΔB/Δt N: turns of inductor A: cross-sectional area of inductor B: magnetic flux t: time of magnetic flux increase/decrease Therefore if the rpm (revolutions per minute of the shaft) increase the voltage increases linear. But Power is voltage*current. If the load resistance is constant, the current increases linear as well. Therefore 2xrpm = 2x voltage = 2x current = 4x power. With 20000 rpm it would be 20x20=400 times the power generated. Means 4 W and not 200 mW, or am I wrong? | |
| Jul 5, 2019 at 13:59 | comment | added | BendingBender | For the induced voltage I used: V_ind = -NAΔB/Δt | |
| Jul 5, 2019 at 13:46 | history | edited | Russell McMahon♦ | CC BY-SA 4.0 |
added 862 characters in body
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| Jul 5, 2019 at 13:38 | history | answered | Russell McMahon♦ | CC BY-SA 4.0 |