faroun wrote:Fish, i have not done what you recommended, the bulb read .8ohms and that did not calculate correctly according to the formula you left on the other post.
i decided to focus on the no load vs rpm for now.
Yes, if it is a 12V light bulb, and the resistance is 0.8 ohms, this would imply a current of 15A, and a power rating of 180W (216W @ 14.4V). That would be a pretty high-wattage 12Vdc bulb (similar to a pair of car head lamps). It may well be that the DC resistance of the tungsten filament increases with temperature, and/or it could be your VOM is not accurate below 10 ohms. In either case, you can easily determine the actual resistance by simultaneously measuring the voltage drop across the bulb and the current through the bulb:
E/I = R
Using the "charge controller" between your alternator output and your batteries obfuscates any data you are attempting to collect to the point where it is all but meaningless. Even at 300RPM your output frequency is only 60hz. At 21RPM your output is 5.5hz. In all cases your waveforms will be far from sinusoidal which means your VOM will give you very misleading results for both current and voltage. Both will be "pulsating" DC. If you insist on using simple VOMs for your measurements (rather than your scope), you will need to add a fairly large capacitor to the output of your rectifier and make certain that your load is 99.9% resistive (that is: NOT a charge controller connected to batteries). A light bulb would be a decent choice, just use a high enough wattage bulb to handle your largest possible load. (Obviously if your charge controller has a digital power display or data logger then you CAN use it to measure your output power, but I did not get the impression yours had either of those features).
If you want to seriously test your alternator output at various RPM, I would strongly urge you to use a more powerful motor to turn it. You can use step-pulleys to achieve a wide range of speeds as long as the motor itself is at least 2x as powerful as your alternator's anticipated output (4x would be better). Any power reading taken from the driving motor's current draw can be used ONLY as a rough estimate for your "power input" to your alternator. To get more than a rough estimate, you would need to know the driving motor's efficiency curve (Torque/RPM/Current) and subtract the drive-train losses; none of these will be linear. The only thing that will be certain is that you will not get MORE than ~40% of the power from your alternator that it takes to power the driving motor (most likely less than 20%). If you have access to a drill press/milling machine/Lathe, they would make a great source for your driving motor.
Note: Both of the following formula are flawed (one pessimistic, the other highly optimistic):
P = I^2 * R => (2.1A ^ 2) * 1.2ohms = 5.3W
P = E^2/R => (27V ^ 2) / 1.2ohms = 607W
Assuming your 22W input is accurate, I would guess the 5.3W is fairly close, this would imply an overall efficiency of 24%, and this would be exceptional for the configuration. Sadly, this tells us almost nothing about what we might expect with a 2kW input at 200 RPM.
If you want more detailed guidance on how to test your alternator under load with your scope (my first choice) or with your VOMs, please just let me know. I will be happy to draw up schematics and/or explain the set up in detail.
I ASSUME your RPM figures:
@123rpm=>14.1vdc
@212rpm=>27.0vdc
@300rpm=>39.0vdc
.
.
.
@104rpm=>27.0vdc
@148rpm=>39.0vdc
@190rpm=>50.0vdc
@285rpm=>76.0vd
Are all "NO LOAD" RPMs? If not, then knowing the current at the various RPMs sure would be helpful ;-)
As always, Great Work! I respect and understand how much time it takes to build and set up everything you have done; it is my desire to make sure you get the best data from your efforts to improve the chances of your success.
Fish
Fri Dec 31, 2010 9:58 pm