Improve small system efficiency

[salty]

Hello again all, I have been running a small micro hydro system but would like to explore ways to improve the efficiency. Mostly because my neighbor who shares a waterline from the lake with me gets twice the power I get . He is about 100 feet closer to the lake than me and has a very nice but pricey axial flux machine while mine is a DIY job. This is what I've done; the head is around 130 ft. static pressure about 58psi, the penstock is 1200ft of 2 inch poly and then another 100 ft. of 1 1/2, the turbine is a 20 blue spoon turgo. My last attempt to improve efficiency was to add a second jet but there was no noticeable change in power output, jets are 0.172 and 0.187 in.and these seem to be optimum size. The wheel is driving a 12V  SEV Marchal (don't ask me where I got that) automotive type alternator in which I have removed the electro magnet and replaced it with a Neo donut magnet .My output is about 5 A. with my machine only running off waterline.
  I'm looking for suggestions on improving the power output of this machine. The waterline is what it is and won't be altered. A couple of things I've been thinking about are rewinding the stator ( not sure weather to go fewer or more turns) or using fewer spoons on the turgo to try to get the revs up. What say ye electron gurus?

[XeonPony]

run 2 inch all the way, the 1 1/2 is your biggest issue, idealy 4inch would be best!

you convert water to torque then energy, torque is voloucity and volume of water, with out improving either the pressure or volume you simply do not have the energy pressent.

[joestue]

optimum power is .48 no load rpm range for the turbine.

if you disconnect the alternator, does it reflect that?
also, when you attach the alternator without any electrical loading, what is its rpm, relative to the maximum turbine rpm without a mechanical load of any kind?

[salty]

Sorry I don't have any way to measure rpm?When I connect the alternator to the batteries the RPM drops.I can't say by what percentage.

So would you say I am getting optimum power output for that waterline?

[hydrosun]

Check the voltage output loaded and unloaded. If it is close to half you have the correct windings. Car alternators are not made with efficiency in design. A purpose built alternator would  be more efficient but you have to decide how much you are willing to spend.
Chris

[salty]

I ran a test and got 21 V open circuit and 14 V when I put a 2.1A load onto it. I'm not sure how to interpret these results with the size of the windings or to correct it to get one half Volts under load.
I would like to make a purpose built alternator for this but need some direction. Have metal lathe and machining skills so that would be a good start.

[joestue]

do you have a meter that reads hz or rpm? that's the only way to know for certain.

you're getting about 30 watts out of the alternator.. thats probably how much mechanical power it takes just to spin the alternator.
however, the 21 volt open circuit says otherwise.
ignoring the electromagnetic reasons for voltage drop under load, then with a load the turbine is spinning 2/3rds as fast with a 30 watt load.
you could probably get a little more power out by increasing the load to get 12 volts out, but my guess is that the alternator in this case is about 70% efficient.

and you're within 20% of the turbine optimal point.
combine the two  together you should be able to get 50% more power out by finding the optimum rpm, by rewinding the alternator, and by selecting the right number of spoons/diameter of the turgo wheel.

[bob g]

you mention "removing the magnets in favor of a ring magnet"

does the alternator have a stainless steel shaft?  probably not seeing how one would not be strictly needed if the rotor was fitted with individual magnets, however

if you have a standard steel shaft, and are using a ring magnet, most of the flux path will follow the path of least reluctance!  guess where that is?

yup, it will route through the shaft, the shaft effectively becomes a magnetic shunt in the flux path and you don't get much across the airgap cutting stator windings to make power.

check that shaft, and see if a magnet is drawn to it, if so it is steel, if not or very little then it is stainless and all i have said is of no use to your problem.

i am not familiar with your alternator choice, but given the price of most of these converted units, it is unlikely the manufacture replaced the shaft with a stainless one when the only need would be one of corrosion abatement.


+1 on going the whole distance with 2inch pipe, you have a pretty long run and likely are loosing some power to pipe friction in my opinion.

fwiw
bob g

[salty]

Bob,
The flux pass of the magnet is axial, it magnetizes the poles of the rotor. I think it is effective as I can feel the alternator cogging when I turn it by hand.

I agree that the waterline is long and small but the only improvement I can make there is to bump the 100 ft of 1 1/2 to 2 inch.

[hiker]

perhaps your alt is stalling--have you tried a stock car alt with the field windings?
you could lower the voltage going to the field coil-- and get extra voltage and maybe a few more amps....ive played around with a few stock car alts with the field windings--and found out i get more output with a lower field coil voltage..
all tests done on a pedgen...your alt right now is made for max amps output--
your water supply is not enough to do this--so it stalls out..lower your field stringth or smaller mags--and you should see a incress in output...???
try it..................... ;D

[bob g]

i am not saying that all the flux will be shunted through the steel shaft, just a significant portion of it will be shunted.

the end effect is you have less flux through the air gap and lower alternator output than you might expect or have witnessed with the same machine using individual magnets out on the periphery of the rotor.

bob g

[dnix71]

Salty The flow in a pipe goes with the square of the diameter.  1 1/2 inches is 3/2. Square that and you have 2.25.

2 squared is 4. Going from 1 1/2 to 2 inches almost doubles the flow capacity of your pipe. That's an easy system boost to make.

[salty]

Thanks for all the good advice.Changing the alternator shaft to SS is simple enough. Upgrading the waterline doable. I would like to play around with the stator also but need some guidance.                                                                                           Joe, I took some frequency measurements with a fluke 117.
on O.C.  AC V= 17   f = 274 Hz   DC V = 24

2.1 A load   AC V= 12.5   f= 210  DC V = 15.1

AC volts and frequency were equal across the 3 phases.

I'm not sure but I'm thinking that if I pull the stator out and count the number of slots and divide by 3, I will come up with RPM?

[XeonPony]

F  = (# poles * RPM) / (2* 60)

RPM = (F / poles) * (2*60)

Poles = (F / rpm) * (2*60)

Average Altenator from a car is 16 poles, that gives us 1,575rpm loaded

[salty]

First, I am doing tests in my basement sink. We have good pressure here but it has to go through a lot of small pipes.

I put a rheostat in as load and adjusted it to an output of 12.5V dc. At this I was down to 174 Hz and 3A output. By the formulas given my OC RPM was 2055 and my loaded RPM was 1305 which is 0.63 no load RPM.

How can I modify the alternator to obtain optimal power RPM of 0.48 ?

[joestue]

2740 rpm sound about right?

to get from frequency to rpm, multiply Hz by 120, and divide by the number of poles. almost all alternators are 12 poles.

[joestue]

First, I am doing tests in my basement sink. We have good pressure here but it has to go through a lot of small pipes.

I put a rheostat in as load and adjusted it to an output of 12.5V dc. At this I was down to 174 Hz and 3A output. By the formulas given my OC RPM was 2055 and my loaded RPM was 1305 which is 0.63 no load RPM.

How can I modify the alternator to obtain optimal power RPM of 0.48 ?

short of rewinding it, you can reduce the number of spoons to cause the turbine to spin faster.

[salty]

It may be easier  to rewind as to modifying  the housing to the new pitch diameter of the runner. There doesn't seem to be a lot of info out there on rewinding stators

[Flux]

It looks as though you need a bit more flux to drop the on load speed. Just changing to a stainless shaft may make a difference.

I don't think you are far enough off the mark to be worth rewinding the stator. If you wind with more turns you will introduce more resistance and probably gain very little.

If you were very careful with the winding you may possibly be able to get in a few more turns of the same gauge wire but that will increase resistance. if you have to go to thinner wire it will have an even worse effect.

I am not sure what the iron loss is on a car alternator at 3000 rpm but it certainly won't be negligible so you will never compete with an air gap alternator on light loads.

If you can manage it, the ideal way would be to wind with 50% more turns and use a buck converter to get the perfect match. You can get the match with a wound field alternator but in addition to iron loss you then add field loss. You wont win doing that with so little input power.

Any improvement you get will be marginal unless you can increase water flow.

Flux

[BadFordRanger]

That 100' of 1 - 1/2" pipe is robbing you of nearly half your ability to produce electricity!
I fail to understand how you understand how to work on alternators as well as you seem to be able to, but fail to see the significance in the size of the pipe!
Yet on the flip side, I do see the significance in the size of the pipe but couldn't rebuild an alternator if my life depended on it! 
For the cost of replacing the 1 1-2" with 2" pipe, you will be close to doubling the juice you get from you system!

Godspeed

Ranger