posted 3 years ago
From the experts here, a little help please  I ahve to opportunity to buy this property
Current situation
New England
brook with dam
flow at 22 cubic ft/sec
Flume to Powerhouse
16 foot overshot wheel
Produces 5kw max  runs at 4KW
Net metered and excess sold to group members
Here is the problem, the permit is for 32kw. I am willing to invest in making it more effecient and profitable HOWEVER.. I can't find a penstock & turbine combination that will get me the 32KW!
With 1920 ft of head and high flow of 22 ft3/sec I need to get the full 32 KW out of it.
Any thoughts?
Current situation
New England
brook with dam
flow at 22 cubic ft/sec
Flume to Powerhouse
16 foot overshot wheel
Produces 5kw max  runs at 4KW
Net metered and excess sold to group members
Here is the problem, the permit is for 32kw. I am willing to invest in making it more effecient and profitable HOWEVER.. I can't find a penstock & turbine combination that will get me the 32KW!
With 1920 ft of head and high flow of 22 ft3/sec I need to get the full 32 KW out of it.
Any thoughts?
posted 3 years ago
 1
You should be able to achieve the 32kw if you used multiple Vortex turbines but the groundworks would be rather expensive.
posted 3 years ago
A potion of the spreadsheet I'm building:
Amount of energy ( Power in Watts ) that can be generated from the flowing water Power(watts) = Fall Height(metres) x Gravity x Density(kg/m^3) x Flow Rate(m^3/second)
You have the biggest part.. Flow Rate
What is the Fall Height?
Gravity and Density are constants in this case.
Pass along the Fall Height and I'll show you how to figure the potential.
I have now reached the point of Penstocks and turbines to gain a further understanding.
What I'm learning so far, is the elusive attempt to gain the most of the potential possible.
My gut feel is Turgo or Francis styles.. with some sort of a draft tube..
But I'm still learning..
The Penstock seems to depend a lot on your available distance of use.
Curious as to what that is?
There's still more to this.. but for now, let's get a answer on the two questions and see where we go.
Amount of energy ( Power in Watts ) that can be generated from the flowing water Power(watts) = Fall Height(metres) x Gravity x Density(kg/m^3) x Flow Rate(m^3/second)
You have the biggest part.. Flow Rate
What is the Fall Height?
Gravity and Density are constants in this case.
Pass along the Fall Height and I'll show you how to figure the potential.
I have now reached the point of Penstocks and turbines to gain a further understanding.
What I'm learning so far, is the elusive attempt to gain the most of the potential possible.
My gut feel is Turgo or Francis styles.. with some sort of a draft tube..
But I'm still learning..
The Penstock seems to depend a lot on your available distance of use.
Curious as to what that is?
There's still more to this.. but for now, let's get a answer on the two questions and see where we go.
Liam McDuffie
Posts: 8
posted 3 years ago
Here are some answers and my thoughts.
The head or total fall is 20 ft top of dam to the mean water level of the brook
June measurement is 22 ft/sec2  that is the low period flow. The majority of the time the flow will be higher.
Tther info: the brook is the sole drainage for 24 square miles.
The state has it permitted at 32Kw
A crossflow turbine (JLA 59?) charts the flow and head for an output of 30+kw Some other charts rate the potential at only 16kw max. So, something is not right.
Distance from the dam where the 24 inch pipe is stubbed in, to the power house is 4045 feet. (Pipe is stubbed in under the flume)
There is a turbine bay poured into the power house. the right hand side of the power house has a 16ft steel overshot wheel in it right now  hooked up to a 5KW generator. Worst case is that I regen it for a 10 kw generator  there is enough torque there I'm sure. Calculated torque of the wheel is 7,253 ft lbs.
While the wheel is nice, I think a penstock and crossflow or other turbine will provide a huge increase in power output.
here is a pic, let me know if you want more info.
The head or total fall is 20 ft top of dam to the mean water level of the brook
June measurement is 22 ft/sec2  that is the low period flow. The majority of the time the flow will be higher.
Tther info: the brook is the sole drainage for 24 square miles.
The state has it permitted at 32Kw
A crossflow turbine (JLA 59?) charts the flow and head for an output of 30+kw Some other charts rate the potential at only 16kw max. So, something is not right.
Distance from the dam where the 24 inch pipe is stubbed in, to the power house is 4045 feet. (Pipe is stubbed in under the flume)
There is a turbine bay poured into the power house. the right hand side of the power house has a 16ft steel overshot wheel in it right now  hooked up to a 5KW generator. Worst case is that I regen it for a 10 kw generator  there is enough torque there I'm sure. Calculated torque of the wheel is 7,253 ft lbs.
While the wheel is nice, I think a penstock and crossflow or other turbine will provide a huge increase in power output.
here is a pic, let me know if you want more info.
Rj Howell
Posts: 15
Location: New Hampshire, USA
posted 3 years ago
What I would like you to do is calculate your Flow Rate by volume of available water, not ft/s2 from your pipe.
This is good info, but now that 'I see' what you have, there's a lot more there!
I have taken some liberties in calculating this, and this is what I've done.
I now know you have a 20ft head
I assume the generator is running from the 24" pipe (Penstock)
The pipe has 6.14 cubic feet, being there are 7.48 gallons to a cubic foot and 3.785 liters to a gallon
6.14 x 7.48 x 3.785 equals 173.83 liters a second or .0174 kg/m^3 (173.83 liters x 1000 = .017383 kg/m^3)
20 feet equals 6.1 meters
Here it comes:
Power in watts = Fall Height (meters) x Gravity (9.81) x Density (kg/m^3= 1000) x flow rate
6.1 x 9.81 x 1000 x .0174
10,412 possible watts of power (no consideration of flow losses, 4045ft will generate a bit)
Your Source is huge!
Multiple generators may be your solution.
I see a whole lot of water passing over what I think is a 40ft wide dam at (3" of depth?)
Would like to know the volume of your source.
There are better ways of doing, but I do like the simplicity of just determining the average depth over a 20ft stretch of outflow and timing how long a it takes a 2/3 full liter bottle to pass through.
Then again, in your case.. How wide is that dam and how deep is the over flow?
Oh yes, and is the 24" pipe flowing while your measuring??
This is good info, but now that 'I see' what you have, there's a lot more there!
I have taken some liberties in calculating this, and this is what I've done.
I now know you have a 20ft head
I assume the generator is running from the 24" pipe (Penstock)
The pipe has 6.14 cubic feet, being there are 7.48 gallons to a cubic foot and 3.785 liters to a gallon
6.14 x 7.48 x 3.785 equals 173.83 liters a second or .0174 kg/m^3 (173.83 liters x 1000 = .017383 kg/m^3)
20 feet equals 6.1 meters
Here it comes:
Power in watts = Fall Height (meters) x Gravity (9.81) x Density (kg/m^3= 1000) x flow rate
6.1 x 9.81 x 1000 x .0174
10,412 possible watts of power (no consideration of flow losses, 4045ft will generate a bit)
Your Source is huge!
Multiple generators may be your solution.
I see a whole lot of water passing over what I think is a 40ft wide dam at (3" of depth?)
Would like to know the volume of your source.
There are better ways of doing, but I do like the simplicity of just determining the average depth over a 20ft stretch of outflow and timing how long a it takes a 2/3 full liter bottle to pass through.
Then again, in your case.. How wide is that dam and how deep is the over flow?
Oh yes, and is the 24" pipe flowing while your measuring??
Liam McDuffie
Posts: 8
posted 3 years ago
The dam at the time pictured is during the spring thaw. Normal flow is about 34 inches over a 14ft or so? spillway opening. I have yet to measure the flow, however I will by Mid July at a point downstream where there is a flat bottomed section of the brook that can be easily measured for uniform depth and width and a good 30 ft section can be measured. For reference, here is a youtube video of the current water wheel operation. in early fall
Rj Howell
Posts: 15
Location: New Hampshire, USA
posted 3 years ago
Some quick numbers (and a few more liberties), 4 inches of water spilling over a 14ft spread is almost as much water as your currently producing from.
I think you see where that is leading..
You certainly have potential of much more than your doing.
I watched the video and understand a bit more of what is happening.
I didn't see the fore mentioned 24" pipe, sluiceway yes, so am I correct in assuming the 24' pipe is just the outlet from the dam entering the sluiceway.
I'll let someone better determine this, yet my belief would be, if the opening to the sluiceway is only 24" diameter, then no matter what the size of the sluiceway, it is still governed by the opening.
Being so, then the math still works.
Head height we need to work on.
This may be different than earlier stated.. Let's see!
Difference in height from pond surface to top of wheel.
Believe we could add half the height of wheel (14' wheel right?), with a little loss..
Very nice score!
I think you see where that is leading..
You certainly have potential of much more than your doing.
I watched the video and understand a bit more of what is happening.
I didn't see the fore mentioned 24" pipe, sluiceway yes, so am I correct in assuming the 24' pipe is just the outlet from the dam entering the sluiceway.
I'll let someone better determine this, yet my belief would be, if the opening to the sluiceway is only 24" diameter, then no matter what the size of the sluiceway, it is still governed by the opening.
Being so, then the math still works.
Head height we need to work on.
This may be different than earlier stated.. Let's see!
Difference in height from pond surface to top of wheel.
Believe we could add half the height of wheel (14' wheel right?), with a little loss..
Very nice score!
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