Way remote power part 2

Ok guys,
I posted a new one to bring it up in the pages a bit. Im looking for some math help at this point. I did a lot of reading and digging and still have no answer, so I hope someone out there has a close answer for me. I have a 4" smooth pipe, and its 230 feet long. The head in this pipe is 16 feet from the inlet to the discharge. The creek is still dry, so I cant calculate flow. Using basic formulas (that I cant find) is it possible to calculate volume and pressure off of the unknown? What I want to know is how much volume and the velocity if this pipe was running full. There are three 45 degree bends in the pipe to snake it arround the creek, but other than that it is straight and all down hill.

gomango wrote in news:b7ce0813-3154-49bb-8b67- 1aeb696edfea@d45g2000hsc.googlegroups.com:

Ok guys,
I posted a new one to bring it up in the pages a bit. Im looking for some math help at this point. I did a lot of reading and digging and still have no answer, so I hope someone out there has a close answer for me. I have a 4" smooth pipe, and its 230 feet long. The head in this pipe is 16 feet from the inlet to the discharge. The creek is still dry, so I cant calculate flow. Using basic formulas (that I cant find) is it possible to calculate volume and pressure off of the unknown? What I want to know is how much volume and the velocity if this pipe was running full. There are three 45 degree bends in the pipe to snake it arround the creek, but other than that it is straight and all down hill.

Here is a win32 application someone wrote that I use. I assume its accurate
http://www.energyalternatives.ca/Downloads/MicroHydroCalc.exe
you can plug some numbers in there and get an idea maybe
good luck!

On Sun, 19 Oct 2008 12:52:00 -0500, z wrote:

gomango wrote in news:b7ce0813-3154-49bb-8b67- 1aeb696edfea@d45g2000hsc.googlegroups.com:
Ok guys,
I posted a new one to bring it up in the pages a bit. Im looking for some math help at this point. I did a lot of reading and digging and still have no answer, so I hope someone out there has a close answer for me. I have a 4" smooth pipe, and its 230 feet long. The head in this pipe is 16 feet from the inlet to the discharge. The creek is still dry, so I cant calculate flow. Using basic formulas (that I cant find) is it possible to calculate volume and pressure off of the unknown? What I want to know is how much volume and the velocity if this pipe was running full. There are three 45 degree bends in the pipe to snake it arround the creek, but other than that it is straight and all down hill.
Here is a win32 application someone wrote that I use. I assume its accurate
http://www.energyalternatives.ca/Downloads/MicroHydroCalc.exe
you can plug some numbers in there and get an idea maybe
good luck!

You now have to register to download, so just use the http://www.energyalternatives.ca link and click downloads
John

Boy, that takes all the brain work out of it.
Hmmmm, Now this is telling me that I have 276.78 watts available. I think this program will also tell me enough to build the banki turbine as well. Great link guys. Thanks.
Im a bit discouraged about what I found. I think that there would have been a lot more energy from a 4" pipe than that. I think that a 6" pipe will suck the creek dry, so thats out of the question. I was hoping for at lease 1 HP.

On Mon, 20 Oct 2008 06:08:32 -0700 (PDT), gomango wrote:

Boy, that takes all the brain work out of it.
Hmmmm, Now this is telling me that I have 276.78 watts available. I think this program will also tell me enough to build the banki turbine as well. Great link guys. Thanks.
Im a bit discouraged about what I found. I think that there would have been a lot more energy from a 4" pipe than that. I think that a 6" pipe will suck the creek dry, so thats out of the question. I was hoping for at lease 1 HP.

Since the creek appears to run continuously during the wet season, you can multiply 276 by 24 to get 6624 watt hours per day - if you have sufficient battery storage you should be able to power necessities at a small cabin (small, high efficiency fridge; a few CFLs (better yet, LED bulbs), etc.
John

news@picaxe.us wrote in news:jp2pf4poqip0pf109tnpe8lts8ogoh0eah@4ax.com:

On Mon, 20 Oct 2008 06:08:32 -0700 (PDT), gomango gomangodave@gmail.com> wrote:
Boy, that takes all the brain work out of it.
Hmmmm, Now this is telling me that I have 276.78 watts available. I think this program will also tell me enough to build the banki turbine as well. Great link guys. Thanks.
Im a bit discouraged about what I found. I think that there would have been a lot more energy from a 4" pipe than that. I think that a 6" pipe will suck the creek dry, so thats out of the question. I was hoping for at lease 1 HP.
Since the creek appears to run continuously during the wet season, you can multiply 276 by 24 to get 6624 watt hours per day - if you have sufficient battery storage you should be able to power necessities at a small cabin (small, high efficiency fridge; a few CFLs (better yet, LED bulbs), etc.
John

And once you have your power infrastructure setup its fairly easy to add solar to augment it. I just keep an eye out for a good deal and add a panel when I can.

On Sun, 19 Oct 2008 07:43:01 -0700 (PDT), gomango wrote:

Ok guys,
I posted a new one to bring it up in the pages a bit. Im looking for some math help at this point. I did a lot of reading and digging and still have no answer, so I hope someone out there has a close answer for me. I have a 4" smooth pipe, and its 230 feet long. The head in this pipe is 16 feet from the inlet to the discharge. The creek is still dry, so I cant calculate flow. Using basic formulas (that I cant find) is it possible to calculate volume and pressure off of the unknown? What I want to know is how much volume and the velocity if this pipe was running full. There are three 45 degree bends in the pipe to snake it arround the creek, but other than that it is straight and all down hill.

The simple answer to your questions is "yes" - but in practice the bends in the pipe and the actual "pipe loss figure" for your pipe may change that. However here's a few figures -
There is a rule of thumb which says "the maximum power for any given pipe occurs when the loss of head in the pipe is approx one third of the static head". ( I still think its a half, but that's what the books say!) So in round terms this might occur when the 16 ft (approx 5m) of head is reduced to about 10ft (3 m). (You will see I prefer metric units - they are so much easier to deal with).
Now you don't say what SORT of pipe you have - steel and plastic pipes have different loss figures.
I would be good to find a pipe nomograph from the web. I have one in a book for steel pipe - but my experience (depite what the books say), is that I haven't found smooth steel pipe and plastic pipe to be all that different - though steel pipe changes over time as it rusts.
My nomograph tells me that at somewhere between about 9 - 10 litres per sec, you will get about 2 ft of head loss per 100 ft of pipe (yes its an American book - ft, inches, pounds etc!).
So putting this information together, you might expect the maximum power at somewhere around 3 m of (net) head and 10 litres per second of flow. This is around 300 watts of power at the jet.
If you are lucky/ do everthing as carefuly as possible, this may translate to around 150 watts output from you generator.
If you can do more than 150 watts, its a bonus; but don't count on it.
Incidentally, a head of only 3m might be a bit hard to utilise. A Kaplan turbine (which I have built here for my holiday home) would probably improve things - though I have seen a 150mm turgo wheel working happily at this head - but it had a specially wound pm alternator for direct drive at 24 volts, and 4 one-inch jets. If you decide to use a belt drive to gear up the speed, use a toothed belt. You could easily lose a quarter of your power in a vee belt. You would definitely have lower loss with direct drive.
I did send a reply to your email earlier, but not sure if you got it.
Hope this helps
Eric Sears.

On Oct 20, 10:34�am, z wrote:

n...@picaxe.us wrote innews:jp2pf4poqip0pf109tnpe8lts8ogoh0eah@4ax.com:

On Mon, 20 Oct 2008 06:08:32 -0700 (PDT), gomango gomangod...@gmail.com> wrote:
Boy, that takes all the brain work out of it.
Hmmmm, Now this is telling me that I have 276.78 watts available. �I think this program will also tell me enough to build the banki turbine as well. �Great link guys. Thanks.
Im a bit discouraged about what I found. �I think that there would have been a lot more energy from a 4" pipe than that. �I think that a 6" pipe will suck the creek dry, so thats out of the question. �I was hoping for at lease 1 HP.
Since the creek appears to run continuously during the wet season, you can multiply 276 by 24 to get 6624 watt hours per day - if you have sufficient battery storage you should be able to power necessities at a small cabin (small, high efficiency fridge; a few CFLs (better yet, LED bulbs), etc.
John
And once you have your power infrastructure setup its fairly easy to add solar to augment it. �I just keep an eye out for a good deal and add a panel when I can.

Solar is already a reality, but it is poor in the winter, thus the hydro venture.
Quick question.... This alternator has a couple windings on it. It has the main windings, and the spark power source. Then it has two coils that are devoted to producing 12 v at 8 amps for battery charging. If I opt not to use the extra windings, should I place a dummy load on those coils to keep them from overloading, or would that place an extra strain on the alternator thus reducing the output. Should I just totally eliminate them and remove the windings? What do you guys think?

"gomango" wrote in message

Quick question.... This alternator has a couple windings on it. It has the main windings, and the spark power source. Then it has two coils that are devoted to producing 12 v at 8 amps for battery charging. If I opt not to use the extra windings, should I place a dummy load on those coils to keep them from overloading, or would that place an extra strain on the alternator thus reducing the output. Should I just totally eliminate them and remove the windings? What do you guys think?

I think you should just leave the redundant coils disconnected. You may find a need for them in the future, particularly those 12 volt coils. If the extra coils have no current through them, they can not overload and will not place any strain on the alternator nor take any power from your prime mover.
Vaughn

In article , "Vaughn Simon" wrote:

I think you should just leave the redundant coils disconnected. You may find a need for them in the future, particularly those 12 volt coils. If the extra coils have no current through them, they can not overload and will not place any strain on the alternator nor take any power from your prime mover.
Vaughn

Vaughn is right here, BUT you should make sure that you isolate the windings with Good High Insolation Tape, as they will have a significant voltage on them when the alternator is spinning, and you don't want them to arc, or short.
-- Bruce in alaska add <path> after <fast> to reply

Bruce in alaska wrote:

... make sure that you isolate the windings with Good High Insolation Tape, as they will have a significant ...

"High Insolation"? Only for a solar power setup.

... make sure that you isolate the windings with Good High Insolation Tape, as they will have a significant ...
"High Insolation"? �Only for a solar power setup.

I think silicone will do this job just fine.
Ok, update.....
I spent a good 4 hours studying the design math and theroy of the banki turbine, and I am now at point where I need to figure out optimal dia for the end disks. Im going to construct a 12" X 3.3" for now, just to test things out. Once I get a running turbine, I will refine the math, and go for a smaller turbine tuned for the actual measured flow. Who knows. the 12" might do just fine. I would rather have a 8" turbine that turned much faster than a slow-high torque turbine. Im still debating this. I did the layout of the endplate, and cut the buckets for the 12". Now I need to do some fancy cuting to get the end disks slotted for the buckets. I think that the nozel will be a bit difficult since the turbine will be narower than the pipe feeding it, but I think I can manage. The plan is to use a 3" X 3" square tube to fabricate the nozel from. If any of you out there have any experiance building a banki turbine, I would love to share my thinking with you and swap notes.

"Eric Sears" wrote in message

On Sun, 19 Oct 2008 07:43:01 -0700 (PDT), gomango gomangodave@gmail.com> wrote:
Ok guys,
I posted a new one to bring it up in the pages a bit. Im looking for some math help at this point. I did a lot of reading and digging and still have no answer, so I hope someone out there has a close answer for me. I have a 4" smooth pipe, and its 230 feet long. The head in this pipe is 16 feet from the inlet to the discharge. The creek is still dry, so I cant calculate flow. Using basic formulas (that I cant find) is it possible to calculate volume and pressure off of the unknown? What I want to know is how much volume and the velocity if this pipe was running full. There are three 45 degree bends in the pipe to snake it arround the creek, but other than that it is straight and all down hill.
The simple answer to your questions is "yes" - but in practice the bends in the pipe and the actual "pipe loss figure" for your pipe may change that. However here's a few figures -
There is a rule of thumb which says "the maximum power for any given pipe occurs when the loss of head in the pipe is approx one third of the static head". ( I still think its a half, but that's what the books say!) So in round terms this might occur when the 16 ft (approx 5m) of head is reduced to about 10ft (3 m). (You will see I prefer metric units - they are so much easier to deal with).
Now you don't say what SORT of pipe you have - steel and plastic pipes have different loss figures.
I would be good to find a pipe nomograph from the web. I have one in a book for steel pipe - but my experience (depite what the books say), is that I haven't found smooth steel pipe and plastic pipe to be all that different - though steel pipe changes over time as it rusts.
My nomograph tells me that at somewhere between about 9 - 10 litres per sec, you will get about 2 ft of head loss per 100 ft of pipe (yes its an American book - ft, inches, pounds etc!).
So putting this information together, you might expect the maximum power at somewhere around 3 m of (net) head and 10 litres per second of flow. This is around 300 watts of power at the jet.
If you are lucky/ do everthing as carefuly as possible, this may translate to around 150 watts output from you generator.
If you can do more than 150 watts, its a bonus; but don't count on it.
Incidentally, a head of only 3m might be a bit hard to utilise. A Kaplan turbine (which I have built here for my holiday home) would probably improve things - though I have seen a 150mm turgo wheel working happily at this head - but it had a specially wound pm alternator for direct drive at 24 volts, and 4 one-inch jets. If you decide to use a belt drive to gear up the speed, use a toothed belt. You could easily lose a quarter of your power in a vee belt. You would definitely have lower loss with direct drive.
I did send a reply to your email earlier, but not sure if you got it.
Hope this helps
Eric Sears.

Odd values there.
From my pocket ref:
Using "the Hazen-Williams equation for flow in pipes". "Feet of Head Loss values are subject to the following conditions: a) Pipes carrying clear water at approximately 60 F (15.6 C). b) Pipes are flowing full. c) Velocities of water are generally less than 10 feet per second."
For "New Clean" Steel or plastic pipe:
"Head Loss/100 Feet Pipe Due To Friction" 4" Diameter Pipe: at 30 GPM = 0.1' 40 GPM = 0.1' 50 GPM = 0.2' 60 GPM = 0.3' 70 GPM = 0.3' 80 GPM = 0.4' 90 GPM = 0.5' 100 GPM = 0.7'
150 GPM = 1.4' 200 GPM = 2.4' 250 GPM = 3.6' 300 GPM = 5.1' 400 GPM = 8.6' 500 GPM = 13.0' 600 GPM = 18.3' 700 GPM = 24.3' 800 GPM = 31.1' 900 GPM = 38.6' 1000 GPM = 47.0' 1200 GPM = 65.8' 1500 GPM =99.4'
Luck; Ken

On Mon, 27 Oct 2008 03:35:11 -0500, "Ken Maltby" wrote:

Odd values there.
From my pocket ref:
Using "the Hazen-Williams equation for flow in pipes". "Feet of Head Loss values are subject to the following conditions: a) Pipes carrying clear water at approximately 60 F (15.6 C). b) Pipes are flowing full. c) Velocities of water are generally less than 10 feet per second."
For "New Clean" Steel or plastic pipe:
"Head Loss/100 Feet Pipe Due To Friction" 4" Diameter Pipe: at 30 GPM = 0.1' 40 GPM = 0.1' 50 GPM = 0.2' 60 GPM = 0.3' 70 GPM = 0.3' 80 GPM = 0.4' 90 GPM = 0.5' 100 GPM = 0.7'
150 GPM = 1.4' 200 GPM = 2.4' 250 GPM = 3.6' 300 GPM = 5.1' 400 GPM = 8.6' 500 GPM = 13.0' 600 GPM = 18.3' 700 GPM = 24.3' 800 GPM = 31.1' 900 GPM = 38.6' 1000 GPM = 47.0' 1200 GPM = 65.8' 1500 GPM =99.4'
Luck; Ken
Ken - thanks for the numbers. I have just been reading some info

about the Hazen-Williams equation and downloading some nomographs.
My figures are in about the same ballpark as your pipe-loss numbers above - though I agree, I used "mixed" odd units. However, 10 litres per sec is about 10 x 60 / 4.5 gallons per min = 133 gallons/min. From your table, this incurs a loss of somewhere approaching 1.4 ft/ 100 ft ( I guessed about 2 - but I tend to err on the high side, because few pipelines are ideal - eg bends etc). Interpolating from your table, 2ft/100ft is closer to 170gallons per minute, or about 12 litres per sec.
But its good to see some other tables.
Thanks
Eric Sears.