can output be measured by pressure and flow?

[strider3700]

My house is fed by a 1" line from a spring up on the hill behind me.  The line runs down the hill buried all the way  and into my water system.   A float valve controls the input of water into my system.  

I don't know the head on that line because it's a ways up into the woods and without line of site it's really hard to say.  I'd guess something around 50 feet but I really am guessing.  anyways I'm thinking that since we really only care about pressure and flow  that I could put a pressure guage in line and measure the pressure when it is flowing.   Does anyone know the formula to estimate output based on pressure and flow?

Say I get 10 PSI and 20 gallons/minute    how do I work out the possible watts?

Since i have a glut of water in the winter and no sun,  but tons of sun in the summer but limited water  what I'm thinking of doing is putting a T before the float value and using the excess water in the winter to generate some power.  Then come summer I'll just turn off the power plant and use solar.     Excess water will be dumped into the drainage ditch beside my house  which runs constantly all winter anyways.  Really I should put the plant at the bottom of my driveway.  That will net me an extra 20 feet in head  and will only force me to run maybe 70 feet more of line and power cables coming back to the "shed"

Thoughts?

Jamie

[ghurd]

I think the PSI is 2.41 for every pound of static head.

That's a start

G-

[RP]

Each foot of elevation change is equal to 0.433 PSI of water pressure

[pyrocasto]

well strider, if you have 10 PSI that would be about 23ft of head.

Now you just use an equation:

Head(in feet) x Flow(in GPM) / 10 = Total watts

23 feet x 20 gpm / 10 = 46 watts

Though, that is without losses so you will only get maybe 3/4 that.

But say you get 30 watts usable, at 24 hours a day you've almost got 1hp a day.

Just remember if you put the plant down the hill, that the voltage will need to be high enough that your losses wont overcome your gain of potential power.

[ghurd]

Thanks.

Had the inverse.

G-

[Ungrounded Lightning Rod]

Just remember if you put the plant down the hill, that the voltage will need to be high enough that your losses wont overcome your gain of potential power.

You can run a partial vacuum in the tailstock and get the advantage of some of the drop after the turbine without actually moving the turbine down the hill and the power back up.  20 feet is pushing it a bit but might be doable.

You'll need to do a few things to get away with this:

 - Good seals on the turbine to keep it airtight or very close to it.

 - Discharge the water from the tailstock into a container that keeps the output pipe covered at all times, to prevent air from being sucked back into the tailstock.

 - Run your lines so that there are no high spots to trap air bubbles, and any air in the line will come back through the turbine to where you can get rid of it.

 - Provide some means to exhaust (bleed) trapped air.  This will require cutting off or restricting the tailstock flow to bring the pressure above atmospheric at the bleedout point.  (You can make it automatic by using a local standpipe to a height above the spring's height.

With enough flow and narrow enough pipes at a low slant, trapped air will tend to be sucked out the tailstock (though narrow pipes also cause friction, lowering the effective head).

Also note that running a vacuum means you'll depressurize the line for feeding water to your house - potentially risking backflow conatminating your water supply.  You can avoid this by running an additional line from the spring for the turbine - tapping the spring box at a slightly higher point to automatically cut off the turbine water if there isn't enough to supply the house.  Or you can make some arrangement of transfer valving (perhaps an automatic one using a valve like that used for a sink hand-sprayer), and depend on check valves and anti-siphon valves to defend your house plumbing from backflow contamination.

Note that depressurization of your house plumbing can also occur from running the turbine even if you DON'T run a below-atmospheric-pressure tailstock.  So you might consider that second headstock in any case.

[Nando]

Jamie:

Where are you located ?.

Static pressure is 0.433 PSI per foot of water column.

Can you convert to KG/M^2 if you need such value ?.

1. inch pipe may have a limitation of power.
   
10. PSI * 0.433 = 4.33 feet * 0.3048 = 1.3 meters
    
    

(20 gallon/min ) / 60 sec = 0.3333 gallon/sec * 3.78 liter / gallon = 1.26 liter / sec

Power available 1.3 meter * 1.26 liter/se * 5= 8.2 watts

To determine the available volume, once you know the static pressure, open a valve and set the dynamic pressure to 2/3 of the static pressure and measure the flow, that will be the optimum volume available due to pipe diameter and length.

Does this help ?.

Nando

[Nando]

Heck:

I made a big mistake

10 PSI / 0.433 = 23 feet * 0.3048 = 7 meters

(20 gallon/min ) / 60 sec = 0.3333 gallon / sec * 3.78 liter / gallon = 1.26 liter / sec

Power = 7 meters * 1.26 liter / sec * 5 = 44 watts

Sorry for the mistake

Nando

[strider3700]

I'm on vancouver island in BC.  

Well this is good news,  I'm just guessing that I have about this much pressure and flow. For possibly getting 40 watts it's well worth me spending a couple of bucks and getting a pressure guage to find the exact numbers.

  Now looking at that type of output I'm assuming I should be looking into building a pelton wheel type of device?  Has anyone built anything like that?  Can it be done relatively inexpensively?  I look online and see things like the waterbaby.  Nice,  but well outside of what I want to spend just for the generator.  Would a different type of device work better or be much easier to build and worth it?   This is a hobby not critical so I'm not needing the best of the best here.

[pyrocasto]

I would say rather something like the turgo or banki turbine. Banki's are good turbines really easy to get or make.

[strider3700]

Ok I've hooked up a guage.  unfortunately  it's a 100 psi guage.   Anyways when water is flowing straight out of the end of the hose it shows zero psi as I would have expected.  it's 2 inches from the end so not much pressure there.    When the end is blocked it goes to 50 PSI almost exactly   the bucket filled at  12L in about 28 seconds(+_2)    

so using the above formula

50. /.433 = 115.4 feet * 0.3048 = 35 meters ( this is believable,  it's pretty steep in places and a ways up there)
    
12. L/28 sec = .428 L/sec   1" only gives ya so much.  I should really look into doubling this hose with a couple hundred feet of 2" (not sure that the neighbours that share the spring will appreciate that though  )
    
    

so power = 35 * .428 * 5 = 74.9 watts  (where does the 5 come from?)

that works out to 54ish kwh/month  or 5% of my current usage or hopefully 10% of my usage once the final big hogs are upgraded.   at current rates that works out to $40 worth of electricity running year round or $20 in the six months of the year I would run it to prevent sucking the spring dry.  

Now of course there will be efficiency issues  would I get 50% of that?

If I add the 2"line I'm assuming that I'll get about 4X extra correct(my geometry isn't what it used to be but 2"pipe is more then 2X 1 inch pipe correct)?  so that would be about 2 L/second and I could then get about 350 watts? thats almost 1/4 of my usage.   1/2 after upgrades  thats so worth doing screw the neighbours

Anyone see any issues with this? or do I get almost 0 because the pressure shows as 0 when flowing?

[pyrocasto]

Well, are you taking the water away from the neighbors, or are the below you to where you can just let it run back into the spring?

I would say out of the 75 watts you'll get 50+.

[strider3700]

all of the houses are side by side

basically how it works is the hillside behind us is very very wet,  lots of little springs come out everywhere.  What we have is a large pond dug that we steer a bunch of water into and in the bottom of the pond we have our hoses.  The hoses then run a couple hundred feet down the hill to each of our houses.  

During the winter there is so much water the pond overflows and basically pours out into the rest of the hillside,  that then is mostly absorbed into the ground but pops out here and there before getting to the road below our houses.  at the road it seeps out of the hill everywhere and fills a ditch.  the ditch in front of my house is a foot across  and 6inches deep and runs fast non stop from late october until late may.  Like I said, lots and lots of water.  I intend to add the water from the system back into the ditch in front of my house directly.

during the summer the hillside gradually dries up and the pool slowly dries up depending on how much water we use.   Since I store 3000 gallons in my tank in the back I don't run out of water,  the little trickle that comes down in late august is nasty looking stuff but it is enough to keep my tank from going empty.  The neighbours however haven't had large tanks in their systems.  They have only been storing 200 gallons in the filtration systems and that runs dry during really dry periods.

The question is

at 24L/minute or 34500L/day will the pool be able to keep up?   I believe it will quite easily during the winter. I know it can't during the summer.

at 172800 L/day  I have my doubts that the pool won't be a sucking sound going into my generator

 

[pyrocasto]

You could put a overflow pipe and build the dam up over it higher so the output goes into your generator in the winter.

[Nando]

The 5 is the efficiency factor.

The true formula is Gross Watts = Volume (l / s) * Head (meters ) * 9.81 * 0.5 = ;

50 % efficiency because at low power it is generally lower than 60 %.

Regards

Nando

[ghurd]

WOW!

I never looked at it that way before!

G-

[ghurd]

(OK, they took out the post in Chinese)