Problems with Preexisting Hydro

[Jonesy]

Hi All,

I just recently found the board.  Surprising since I have been researching all our hydro troubles online for months.  Couldn't find what I need in old postings, though.

We moved into a rental powered by an old harris.  Owners really weren't informed on the setup as the previous renters of ten years had put it in.  After months of no output due to problems with the alternator and time delays with it's rebuild, we have been getting some output, but far below what I expected to see.

Site and System:

40 ft head--16.5 psi dynamic

approx 420' penstock, 4" PVC

Harris four nozzle (1/2 inch nozzles) 120gpm potential I believe, flow accomidates

Delco style alt, shop guy said it had a 24V stator.

285' 2/0 aluminum triplex, creek to batts*

trace C40 with 450W @ 12V diversion

12V battery array

*recently replaced old #8 copper thhn.  New 2/0 is actually 315' at present as it has not been trimmed and installed permanently yet.

Obviously 8 awg was way to small, but there were some clues that seemed to show that the alternator was underproducing.  We got 6.5 amps at the house with the #8, and now we get 7.5.  After seeing that, I pulled a battery from my truck, and lugged it down the hill with a 1 ohm resister in order to remove the transmission line from the equation.  Hooking the alternator to the battery with jumper cables and connecting the resister to the battery I saw 10 amps.  Head and flow should give me 250 watts assuming 30% alt efficiency, right?  Did I not present a big enough load perhaps?

I also drug the wire back up the hill and compared the charge from an automotive charger connected directly to the batteries, and to the transmission line, at the 2, 10, and 50 Amp settings.

Batt connection gave: 5.5, 9.5, and 21 amps, respectively

Trans lines gave:     3.8, 6.2, and 13 amps

This also seems to point to the alternator producing in the neighborhood of 10-11 amps.

At this point, I feel pretty certain we need a new alternator to see any noticable improvement.  However, I'd like to save a bundle and not buy a Harris PM upgrade. So, now, all the questions:

How do I find expected turbine RPM?

How much will turbine damage effect efficiency (bent out on the inside corners of some buckets--bent somewhat to position)?

If X amps at 24 volts was applied with our C40, the charge would only be X amps at battery voltage, NOT 2X, correct?

If the input voltage is higher that batts, I know the batteries regulate it down to their voltage, but does that mean transmission voltage is also lowered?  I.E. would we gain reduce voltage drop from an alt producing a higher voltage without an MPPT or other stepdown method?

Look forward to meeting some of you, and hopefully figuring this out.

Any other suggestions to boost effieciency would be welcome also.

Justin

[chainsaw]

Curious why you are using a 24 volt alternator to charge a 12 volt bank. I think the efficiently would be greatly improved not to mention a reduction in wire size and a big reduction in wire loss from 12v to 24v. Running 12 volts 285 feet is going to be tough.

Have you considered using a PMA so as to take advantage of higher voltage for the transmission distance and reducing voltage at the battery bank?

Are you using a pelton or a turgo runner?

[Jonesy]

I don't know what the original intent was with the design.  I basically walked into a poorly designed system.  I definately understand the implications of trying to send 12 volts this distance.  But that's what was/is here.  

The alternator produces around 15 V in operation.

For my part, I was against the 600 dollar wire upgrade, in favor of a PM alternator producing higher voltage and a method of step down.  Which, it looks like is nescisary anyway.  But, now 24 volt transmission should be sufficient, as opposed to the 120 V I figured would be needed to overcome VD in the original wire.

The turbine is the standard harris pelton.  I don't understand what pitch diameter is, but it's about 5 inches total diameter.

[dbcollen]

The reason for the 24 volt stator is the alternator is only spinning 1300~ rpm at Vmp, 38 feet dynamic head is not very much for a pelton runner, it should work though. with 4, 1/2 inch nozzles you would be using 118Gpm, and assuming an 85% efficient runner and a 40% efficient alternator, you have 273 watts available. A standard Delco-Remy 10si alternator is usually 25-40% effecient in use as a hydro alternator, with a custom made permanent mag alternator being 45-65% effecient. In automotive use an alternator can see 50,000+ rpm and are way more efficient at high rpm, as the field current gets cut to near nothing. most people don't realize how fast an automotive alternator is designed to spin. The alternator on my Ford has a 2.5 inch pitch diameter pulley on the alternator (4.9 inch circumference) and an 8 inch pitch diameter pulley on the crank (50.25 inch circumference) the alternator spins aprox 10x the rpm of the crankshaft, and that motor sees 7000+ rpm, which means that alternator will spin 70,000+ rpm. so running a 12 volt alt at 1300 rpm is just barely above cut in, a 24 volt stator at 1300 rpm would be equivilent to running the 12 volt at stator at 2600 rpm, and would thus require less field current.

sorry for the rambling, but I hope this helps.

Dustin

[Jonesy]

Dustin, all your numbers are the same I've found, and you confirmed my assumption as to the reason for the stator.  But, like I said we only see about 100 watts.  Could the minor turbine damage account for all that difference?

[dbcollen]

Pitch diameter is from the radius of the center of where the  jet of water strikes the cups times 2.

As to the low output, your alternator at that head is probably lucky to be 20% efficient, cutting the expected output to 150 watts or so. Try dialing the field current with the rheostat on the turbine housing, and I adjust my nozzles by unscrewing the nozzles and putting a piece of drillrod, or a drillbit through the nozzle hole and putting the nozzle back in and use the drillrod to see where the jet of water is aimed, it should hit the center of the cups, centered on the bifercation, and all the water should hit the cup, not go past on the outside. As I recall on the harris runner the water jet should be hitting either 2 or 3 cups ay a time. It sounds as if your runner may be worn out, if you can post a picture of your runner I can tell you if it is the problem, a picture from the nozzles view would be best.

Dustin

[Flux]

Can't add a lot to what Dustin has covered.

I just want to confirm that you are using a rheostat not an internal regulator ( if that alternator is normally fitted with one). You need to strike a compromise between wasted field current and turbine speed.

Also we know that you have a 24v stator but we don't know if you have a 12v or 24v field. I suspect that with your line drop you would get enough field with the 24v rotor and have to waste less in a rheo and mismatched field winding.

The 24v stator running on 12v will not be very efficient. If you had means of measuring turbine speed I would be tempted to suggest that you try to get the speed up enough to try it at 24v ( not going to work if you have a 12v field without an electronic field regulator).

Flux

[hydrosun]

I've got one hydro with 15 psi and converted from the Motorcraft 70 amp alternator to the harris PM and greatly increased the output. The maximum I could get from the alternator was 10 amps at 12 volts. Now I get 8 amps at 24 volts. The difficulty of comparison is that is the amperage reading of the built in amp meter, which is notoriously inaccurate. Harris uses an inexpensive meter to make it easy to adjust the reastat to get the maximum output but isn't intended to give an accurate measurement of actual output. I have people ask if they are losing amperage in the wire because the meter at the pelton doesn't read the same as the one at the battery.  So what kind of meter are you using to determine the amperage output?

 You say you are reading 15 volts at the pelton but I didn't see a voltage at the battery. A maximum of 3 volts drop in the wire would be about 20% loss. Increasing the output at the same voltage would increase that loss. A Motorcraft alternator with the proper field and stator would slightly increase the output but I'd reccomend going to the PM if you really need more output. If the alternative is solar panels or running a gas generator it would make sense to get all you can from the hydro.

 You don't mention if you have done the checks on the nozzles to see if there are any obstructions. Or if the pressure guage is accurate. Mine seem to change over time.

Chris

[dbcollen]

There is no way to increase the rpm without gaining more head, or getting a smaller diameter runner. He is running the hydro at 12 volts, so the 12v rotor is fine. Switching to the 24v stator is just like winding an axial for lower cut in. There will be more I^2r losses in the stator windings, but it should be far offset by the lowering of the field current at his low output. there is also the possibility of using an Mx-60 to find the max power point, but there is the problem of huge inductive surges with the wound field when the mx60 unloads the turbine. Those surges don't happen with a PM alt, the voltage just doubles as the rpm doubles. (pelton runners typically have their max power when the cups are at 50% of jet velocity)

Dustin

[Ungrounded Lightning Rod]

Unless you have some sort of power tap or REALLY bad insulation creating a high-resistance short along the way, the current at the battery will be the current at the alternator.  This seems unlikely.

Differences in ammeter readings indicate that one of the meters is out of calibration or otherwise inaccurate (or both of them are but differently).

At these voltages and with these insulations, wiring losses are from voltage drop, not current diversion.  Exactly the same number of electrons go into the wire as go out of it, and all but a vanishingly small number go out at the far end rather than finding some other way across.  They just don't push their way out quite as hard as they were pushed in, thanks to a "pressure drop" from "friction" on their way from one end of the wire to the other.

[Ungrounded Lightning Rod]

This seems unlikely.

"This" being the high-resistance short.  (It's unlikely because the heating would turn it into a low-resistance short in rather little time.)

[Jonesy]

Here are some shots of the nozzle alignment.  I sighted down the nozzles as in the third pic, but looks like I could improve a bit with the drill bit.

Here is the damage to the runner.  Some slight pitting on the inner surfaces of the buckets in addition to the dinged up edges.

Pitch diameter is 4 inch.  

The alternator does use a rheostadt (12 ohm, I think) to control the field.  Just the other night however, it apparently shorted out, and will now only supply field when turned all the way down.

Voltage reading at present is 13.2v at alt., 12.2v applied to batteries, and 11.5v on the field.

The hydro unit's ammeter does not work and has been bypassed.  Amperage at the alternator has been taken with a multimeter.  Meter is only rated for 10 amps--when I first used it I opened nozzles slowly to see when I would hit 10, but since it hit it just before reaching max flow, I decided to risk passing 10.  

Amps read at the house are as shown on the system monitor.

The nozzles are not clogged, and the pressure gauge is new, has read consistent, and is consistant with surveyed head.

I think that addresses everyones q's.

[dbcollen]

There is significant wear to the cups, but still servicable, aiming looks good, maybe slightly to the inside. you need to get a new rheostat, and hook up your ammeter inline, open all 4 nozzles, and dial the field for maximum amperage out of the turbine.  If you have a junkyard around you may try several different stators, they come in many ratings from 37 amps to 100+ all are wound differently, and you may find one that works better. I have rewoung many of them for my permanent magnet conversions I used to do. I build high voltage inductive alternators and a step down transformer array to drop the voltage at the batteries. I can build you one if you like.

I have a 1hp 6 pole that would be well suited to your setup.

Contact me at madriverwind at gotsky.com

Dustin

[dbcollen]

we spoke off board in an email and I thought I would post my reply here as well.

http://www.windbluepower.com/category_s/1.htm, this is the page he is looking at.

The voltage of an inductive system is 150- 600 volts, there are 3 transformers that set the transmission voltage. it would take me a few weeks to build and test one for you, as I would want to duplicate your head and flow here on my test bench and dial the system in for you. the cost would be around $700.

You can't use the c-40 at a higher voltage on the turbine with 12 volt batteries. The C-40 should be connected to the batteries and be set in diversion mode with a proper load hooked up to it. The wires from the turbine should be connected directly to the batteries and the charge control just takes excess power from the batteries to keep them from overcharging.

If you buy the rotor only  from that site, you can put it in your existing alternator and then use an MX-60 to track the max power point. but you first need to test the open circuit voltage of the alternator, to do that you would  open all 4 nozzles and check the voltage at the alternator to make sure the voltage is not over about 125 volts with the alternator running unloaded.  the MX 60 can take 150 VDC absolute maximum, and will stop working temporarily at 135 VDC. If your open circuit voltage was 120 vdc, your max power voltage would be around 60 VDC  which would greatly cut your transmission losses.

Dustin

[Jonesy]

Thanks Folks,

I think I've got what I need on this topic.  Am going to upgrade alternator and use an MX-60 for step down.  Will post results.

Thanks again.

[74VDC]

I'm stumped..and I'm probably going to ask more questions and present more problems than solutions...but:

My calculations suggest that you should be expecting much, much more power than you

indicate.

 (head in feet) x (flow in gals. per sec.) x (efficiency of turbine) / 11.8 = kw

420 (feet) x 2 (gals. per sec.) x .8 (pelton efficiency) / 11.8 = 56.9492 kw

Assuming a generator efficiency of around 30% that would equate to 17 kw.

Assuming additional losses like penstock frictional losses, poor nozzle aiming, the slightly damaged runner, parasitic losses (field current), I could see this number reduced by as much as half.

Still...I'm thinking around 9kw. Part of me says this cannot be right, but then 420 feet of head...WOW...and 120 gpm.

My numbers must be wrong...they probably are, but these meager outputs you describe gotta be way off. This is definitely NOT a electrical line loss issue if my power calculations are even close to being right.

I can give you exact calculations for line loss and all of the electrical side stuff. I need to know how to calculate the hydro side correctly if I'm wrong before anything else.

BTW...I never did understand why everyone wants to use auto alternators...give up way too much for the convenience of being easy to obtain etc.

 

[74VDC]

OOOPS!

Just reread the initial post and I see it was 40' head and a 420 foot long penstock....ok, now that certainly makes a difference.

Still I'm coming up with 5.42 kW.

So after generator losses I get 1.63 kW.

The 420 feet of penstock friction, runner damage, possible slight nozzle misalignment (I know this is critical)and other deducting factors would account for some loss for sure. Your measured current at generated voltage is barely breaking 100W.

The batteries may pull down the voltage depending on their state of charge, but while doing so, current will increase. I like to get things simple and unmistakable for testing things like this. I'd get several 5 0hm wire wound ceramic resistors  along with several other smaller higher value resistors and putting them in various connections of series/parrallel begin making measurements for optimal power. I would do this because I have all this stuff.

I live in Maryland and we have net metering, so I'd never consider charging batteries, but instead grid tying. You will lose so much power with batteries it isn't even funny. Far, far more than a grid tie inverter.

I still cannot imagine expecting only 240W with 120 gpm and 40' of head. I also have a hard time beleiving so much drop on your xmission line, particularily with DC, but 630' is a long, long way at 15v. I'll crunch the numbers later when I can look of the circular mil area of 2/0 and specific resistance of Al, although I think it's 17.7 and soft drawn Cu is around 10.4 I think.

I do not fully understand your set-up or what you're exact goals are.

You seem to be set on DC, so why then did you spend $600 on triplex...were you wanting the possibility of future 3 phase?

If you really think you have line loss issues parallel the old #8 AWG conductors alongside the 2/0's...just for temporary test purposes...I'm thinking you're gonna see little difference. If it turns out that it helps...get some al/cu connectors, Penatrox and leave it. It would not be code approved, but unless you grid-tie your not subject to the NEC, besides, there are numerous paralleling provisions in the NEC for instances such as these. You could have saved a bundle of cash by utilizing the #8's and paralleling to begin with. You're voltage is low, but so is your current.

  I'd find a good 30VDC PM genny like another poster suggested....like fast!