Easy to build Rotary drum turbine cheap and easy

[oztules]

ok couldn' rectify and get hz as well so put 150w floodlight accross each phase for to give it some load figures

So these figures are for current in 1 leg  (.5r voltage drop), open voltage (no load any phase), voltage at the current indicated and voltage as indicated.... something like this:

rpm (6Xhz in this case) 20 poles so 20hz=120rpm.. yes?

 rpm     loaded V    no load V current in single leg

 90      13v         20v        .43A /phase leg

 120     37v         44v        .48A

 150     55v         62v        .54A

 180     68v         78v        .59A

 210     83v         91v        .63A

 240     96v        106v        .68A

 270    108v        119v        .72A

 300    124v        133v        .76A

 330    134v        147v        .81A

 360    150v        160v        .84A

 390    160v        173v        .88A

 420    172v        185v        .92A

 450    185v        200v        .95A

 480    198v        212v        missed reading

 510    210v        225v        missed reading

 534    218v        242v        missed reading

The clutch started slipping here, so had to give up. The missed readings were because when i ran the second time to verify the volts and measure the current, the clutch gave out and she wouldn't go above 450rpm in the second run. Should have measured it on the first run as well.

Due to this torque problem I ran a single phase test only at 80hz (480rpm) and the current across the resistor gave a reading of only .594A where before with three phases, running in star, the reading was .95A  roughly .6 times less current to drive the same single load. So the load in the legs of a star circuit will require some reading in my behalf.

Hope this helps quench your curiosity... what can you tell me about it, coz my guesses were obviously wrong.

So my stator loss conclusions may be all wrong. by the way it stayed cold through out this series of tests. A lot were done at 400w too.

.........oztules

[oztules]

After re-reading I'll clarify the loaded figures, the loaded voltage is with three phases evenly loaded with 1x150W floodlight each.

The current measured was in this config, but measured only 1 leg, the others must be the same i hope.

Unloaded was with no load of any kind.

Hope that is clear now.

Current was calculed by voltage drop accross 1R resistor in series with one of the flood lights. 1r was added to each of the other legs to even out the load.

.......oztules

[pepa]

hi Willib, i cleaned up seven fan motors today. use a small pully puller to remove bearings, pull wire from center of shaft. one side of shaft is larger than stator hole so put that end into the end of a short length of iron pipe. next screw several of the thin nuts on exposed shaft threads and screw down about one half inch. next place a socket just large enough to go over threads and set on nuts. next several good licks with a machinest hammer will drive the shaft out east with no damage done. i did find one that had ring clamps insted of a raised section. pepa.

[willib]

this is your no load volts vs RPM

i used ' create a graph '

http://nces.ed.gov/nceskids/graphing/classic/line.asp

to graph the data

[oztules]

Nice Willib,

Are you able to superimpose power and loaded volts on the one graph as well?

grovel grovel

.........oztules

[willib]

It would be real interesting to rewire the stator

 , by using less coils in series and at the same time paralleling the outputs through rectifiers , sort of jerry phased,  that would lower the output voltage and reduce the stator resistance at the same time

for example :

if you were to use 1/2 the coils in series you could lower the output voltage by 1/2, 1/3 the coils in series lower the output by 1/3 , drop the tranny , and rectify each third of each phase ect,,,

lots of potential there

pun intended

[willib]

there is one glaring mistake , if you were to break each phase into three parts the output Voltage would be cut by three..

so at 400 RPM your loaded voltage was 160V : it would then be 53V (at three times the current).and your 15 Ohm /phase stator cut in three sections of 5 Ohms each would yeald 1.66 Ohms if you paralleled them before rectification.

 a little minor surgery on the stator would yeald some nice results..

[oztules]

Thanks very much for that Willib

in comment 33 i think it was I said this:

"Due to this torque problem I ran a single phase test only at 80hz (480rpm) and the current across the resistor gave a reading of only .594A where before with three phases, running in star, the reading was .994A  roughly .6 times less current to drive the same single load. So the load in the legs of a star circuit will require some reading in my behalf."

The power curve seems too high. I'm guessing that the .6 deviation has something to do with star connections. Someone here will know how and why.... and it's not me (i thought I did, but I'm wrong)

Somehow higher current flows in the leg resistor, because it could not have reached 500W at less then 240v (lights were not fully bright), so can you do it again multiplying the current by .6. This will (I think) give a truer vale for power. How .994A can be flowing but the load only seeing .594A I don't know.

It may be a power factor thing, and the extra current is wattless current (no voltage in phase with it ) or some other explanation....

I'm not sure if that makes any sense.

And I'm after a transformer setup to relieve the stator of mismatch and heat. It also allows for the stator to be matched to the load better. Transformer losses appear to be minimul with this pole number.

Thanks for your help

.....oztules

[oztules]

It's ok Willib, I got the hang of it now. Here is the phase currrent corrected graph

Tomorrow I will test again and update this graph... fill in the holes.

........oztules

[willib]

i think your readings are correct ,

0.594A times root 3 is 1.02A .. close enough to your reading of 0.994A

thats where i went wrong with the graph , i didnt multiply the current per phase by root 3 , it was late , really no excuse , but it was late..

in the graph i was going for the whole alternator output , not just the power per phase .

so if you want to make another graph , take the current per phase , multiply by root 3 then multiply that by the voltage per phase , that should get you the total power output at that load level.

[oztules]

Yes thanks for your help here Willib,

So if I load all three phases (which I did) and measure the current in 1 leg (which I did), i will need to multiply that current reading by .6 Then use that number and multiply by the voltage and then multiply by 3 for total power out.... Yes?

Coz thats how I derived the graph, which more correctly mirrors what I found in practice (ie globes not fully lit @198v so must be less than 450W.... 150w globesX3)

If I just use 1 phase to test under load,(other phases open) I can measure the current in that phase, multiply by the voltage to get the power, (no root 3 thing here)and then multiply by 3 to get projected total 3ph. output. Is this true.

Thanks Willib

........oztules

[willib]

 the square root of three is used because all currents and voltages are not positive at the same moment in time.

that 0.6 number you came up with is so close to 1/(sqroot(3))= 0.577 its scary!

to answer your first question yes ... because if you multiply the current by 0.6 then later multiply by three ... what you are really doing is multiplying the current by 1/root3 then later multiplying that by three ...is exactly the same multiplying by root three in the first place because 3/root3 = root3 ..!

try it on your calculator...

[oztules]

Willib, in light of this, i did the tests again and came up with this lot:

Then with all the phases we get this:

and to get the total power out:

Something is still wrong here, power is slightly too high at the higher end, but I think it may be because the current resistor was warmer at the higher levels, and it's R went up slightly, skewing the current reading a tad. But It should represent this alts performance up to 500 od rpm.

Have I blundered anywhere?

........oztules

[willib]

i may be wrong , but this is what i got,..

using your previous data for the 150W bulb test

 90      13v         20v        .43A /phase leg

 120     37v         44v        .48A

 150     55v         62v        .54A

 180     68v         78v        .59A

 210     83v         91v        .63A

 240     96v        106v        .68A

 270    108v        119v        .72A

 300    124v        133v        .76A

 330    134v        147v        .81A

 360    150v        160v        .84A

 390    160v        173v        .88A

 420    172v        185v        .92A

 450    185v        200v        .95A

 480    198v        212v        missed reading

 510    210v        225v        missed reading

 534    218v        242v        missed reading

because this is the only data that i have to go by,

i get a respectable 304 Watts total projected output (at the load level of one single 150Watt bulb) using 185V and 0.95A at 450 RPM .It will be higher the more load that you put on it

try loading the alt down more maybe two 450 watt bulbs in parallel connected to one phase and the star point