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Author Topic: Calculations - Is this about right  (Read 1350 times)
Lurk
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« on: March 04, 2012, 09:56:31 AM »

I took a look around a friends garden, within which is the site of a dis-used water wheel  Grin
I measured the following and came out with an approximate potential power- I would appreciate it if someone could take a look over the numbers and advise where Ive gone wrong / or otherwise.

The existing feeder is a stone flag lined water channel  = 1200mm wide x 35mm water depth -  running at an average of 7.167 seconds to travel 4 meters = 0.018 cu mtrs / sec
The remains of the wheel suggest an overshot design, with the wheel 6ft in diameter x full channel width.
I have assumed a head of 1.8 mtrs  (wheel diameter), and efficiency of 80% (wheel, inverter etc)

Which based on the flow rate recorded would generate 255 watts/hr - does this seam about right for what looks like a big wheel ?

The site has the advantage of a >10 meter  fall, and I was thinking it would be more cost effective to install one or more 1kw turbine and pipework rather than re-instate the wheel - I have yet to measure the fall it may be more.

When I work out how to get the photo's off my HTC phone I will post if interested.
Lurk
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Lurk
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« Reply #1 on: March 04, 2012, 10:28:14 AM »

water wheel remains:-


* IMAG0083a.jpg (79.54 KB, 560x420 - viewed 240 times.)
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Lurk
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« Reply #2 on: March 04, 2012, 10:28:58 AM »

Water feed channel


* IMAG0086a.jpg (69.39 KB, 420x560 - viewed 242 times.)
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dhaslam
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« Reply #3 on: March 04, 2012, 11:33:42 AM »

Are you sure there is 10 metres fall?  If that is the case a turbine would work very well but you could  could still have the wheel in  place anyway and just redirect water when it is needed but not attach a generator.     These old water wheels  are always of historic interest.   Some mills had quite a lot of water storage  capacity in their ponds so they could increase  the power when needed.
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Lurk
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« Reply #4 on: March 04, 2012, 11:42:04 AM »

There is approx 5 mtr  head above the wheel site, and at least 6 mtrs below - I would like to have both the wheel on a small generator and 1 kw turbine at the lowest part of the site with multi nozzel feed to take advantage of the huge variable flow conditions. Wish it was my garden though !
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guydewdney
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« Reply #5 on: March 04, 2012, 11:52:01 AM »

That is quite a small wheel. Its also a composite wheel - the buckets would be wooden, and quite a low efficiency design.

To give you a comparison, my wheel is 16 feet in diameter, and 6 feet across - and I generate a max of 6 or 7kw in reality.

your mates wheel is less than half the size - 2.666 times less diameter and 4 feet across - ie 0.66 the width

so taking the size as a direct ratio - it should be about 1.7kw max.
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Pic of wheel on day 1
7.2kW Waterwheel and 9.8kW PV
CeeBee
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« Reply #6 on: March 04, 2012, 11:57:40 AM »

I measured the following and came out with an approximate potential power- I would appreciate it if someone could take a look over the numbers and advise where Ive gone wrong / or otherwise.

The existing feeder is a stone flag lined water channel  = 1200mm wide x 35mm water depth -  running at an average of 7.167 seconds to travel 4 meters = 0.018 cu mtrs / sec
The remains of the wheel suggest an overshot design, with the wheel 6ft in diameter x full channel width.
I have assumed a head of 1.8 mtrs  (wheel diameter), and efficiency of 80% (wheel, inverter etc)

Which based on the flow rate recorded would generate 255 watts/hr - does this seam about right for what looks like a big wheel ?

OK - just trying to repeat the calculation, for fun:

Volume of water in 4 meters of channel 1200mm wide x 35mm water depth = 4 * 1.2 * 0.035 = 0.168 cu mtrs.
This is the amount flowing in 7.167 seconds, so divide by that to get 0.023 cu mtrs / sec (you said 0.018, but perhaps this is where you applied the 80%?).

Mass of water (assuming 1 cu mtr = 1000 kg) 23 kg/sec.

Gravitational potential energy (m * g * h) in that, for a fall of 1.8 meters: 23 * 9.8 * 1.8 = 406 joules / sec = 406 watts.

I've not allowed any 80% there, but still I'm getting a slightly larger number than you - albeit in the same ballpark. Don't say e.g. "255 watts/hr". It's just 255 watts or whatever (no hours involved). One could say "255 watt hours per hour" but that would be silly.

Once we agree on the figure (I'm just as likely to have messed up), then at least we know that if you could use 10 meters head instead of 1.8, then the output would be 5.5 times more.

Just out of interest: thought I'd see what the amount of kinetic energy in the moving water is (that's not saying that you'd necessarily extract ths, after all the water is still moving afterwards...). This time, it's 0.5 * m * v * v.
Velocity is 4 / 7.167 = 0.558 m/sec, so 0.5 * 23 * 0.558 * 0.558 =  3.6 watts. So insignificant in comparison to potential energy.
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Lurk
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« Reply #7 on: March 04, 2012, 11:59:14 AM »

Out of interest, if a new wheel was designed and installed with a larger (for ease say double) 3.6mtr diameter, and the feed supply provided by raised constructed channel,  would the power output potential double as the 'head' had effectively doubled ?
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Lurk
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« Reply #8 on: March 04, 2012, 12:08:14 PM »

CEEBEE

Thats a different set of calculations - I think, but for the sake of it I worked on the following which I found on the EAs Hydro Site pdf download (I think)
0.036 (A Cross Sectional Area)
4 (L Distance over which time measured)
0.9 (C coefficient for river channel)
7.167 (Time)
0.1296 = ALC/T
7.167

= 0.018 Cu/mtr / sec
Potential Power = Head x Flow x G constant(9.81)x Efficiency @ 80%
1.8 x18.084x9.81x0.8 = 255 w


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Lurk
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« Reply #9 on: March 04, 2012, 12:14:18 PM »

Its the C. Coefficient of river bed used in calculating flow - that stands out as the main difference (according to the blurb its to allow for variation in flow through differing river channel beds - not really applicable - but that is only 10%..... so I'm a bit stumped as to why your calc's are giving a higher figure ? Huh
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Lurk
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« Reply #10 on: March 04, 2012, 12:23:29 PM »

Ok sorted the base figure is  423 watts - not allowing for % efficiency or stream flow coefficients.

Thanks CeeBee etc
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CeeBee
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« Reply #11 on: March 04, 2012, 12:26:21 PM »

Its the C. Coefficient of river bed used in calculating flow - that stands out as the main difference (according to the blurb its to allow for variation in flow through differing river channel beds - not really applicable - but that is only 10%..... so I'm a bit stumped as to why your calc's are giving a higher figure ? Huh

We're both bascically doing the same calc. If I take my 406 watts, then allow for 0.9 factor, and also your assumed 0.8 which I hadn't included so far, then I get 406*0.9*0.8 = 292 watts. Still a bit higher? You've got 0.036 for your 'A' (cross-sectional area). Isn't it 0.035 * 1.2 = 0.042? That accounts for most of remaining discrepancy - I hope the remaining bit is just due to me dropping decimal places here and there.

I see your post while I was writing - I think we're now close enough.
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Tombo
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« Reply #12 on: March 04, 2012, 05:13:37 PM »

From the photos it looks like a garden feature to me. I imagine the wheel has been taken from its original site and installed for aesthetics rather than to be used.
Historically they would have put the biggest wheel in they could to get the most power, so the fact there is more unused head available backs up the theory.
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guydewdney
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« Reply #13 on: March 04, 2012, 05:30:58 PM »

I agree - location is all wrong. Its also in backwards.
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Pic of wheel on day 1
7.2kW Waterwheel and 9.8kW PV
RobNute
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« Reply #14 on: March 04, 2012, 07:40:32 PM »

Looks like a good site, can't quite see how that wheel ever worked where it is though but anyway if you have enough head for 6 ft and the full width you should get quite a fast turning wheel which will help with the gearing side of things, my guess would be around 1kw of power max - sorry if thats the wrong spelling  -KwH, KWh etc but you know what I mean. Go for the wheel, it will drive you mad , you do want to be mad don't you, I quite like it...
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