Gareth Lewis
Posts: 9
posted 3 years ago
Hi, im new to this forum and was wandering if anyone could help me with my school project ?
I have been given a task to find out the best way to generate electricity with hydro electricity
Pelton wheel, Archimedes screw or turbine ?
We have been given a value of 16gallons per minuit in flow and a drop of 10m(straight down or gradually)
Can anyone help
I have been given a task to find out the best way to generate electricity with hydro electricity
Pelton wheel, Archimedes screw or turbine ?
We have been given a value of 16gallons per minuit in flow and a drop of 10m(straight down or gradually)
Can anyone help
Marcos Buenijo
pollinator
Posts: 583
Location: Southwest U.S.
12
posted 3 years ago
Try an overshot water wheel (also see "backshot" water wheel which is fundamentally the same).
Are you tasked with actually generating electricity? If so, then you'll have to consider the most efficient means there as well. The most practical solution is probably a small permanent magnet alternator. My calculations show that a perfect system would achieve 100 watts electrical (probably why the instructor selected the parameters). The water wheel should be slow moving for optimal efficiency, and the alternator should be fairly slow (for an alternator). I recommend a chain drive between the two. Alternators designed for small wind turbines often generate power at very low speeds, and these might be a good choice.
Are you tasked with actually generating electricity? If so, then you'll have to consider the most efficient means there as well. The most practical solution is probably a small permanent magnet alternator. My calculations show that a perfect system would achieve 100 watts electrical (probably why the instructor selected the parameters). The water wheel should be slow moving for optimal efficiency, and the alternator should be fairly slow (for an alternator). I recommend a chain drive between the two. Alternators designed for small wind turbines often generate power at very low speeds, and these might be a good choice.
Peter Mckinlay
Posts: 182
1
posted 3 years ago
Efficiency of the Archimedes screw is not known to me other than it is less than the Pelton, Francis and Kaplan turbine.
Water having a vertical fall of ten meters produces a 1 bar flow pressure.
16 gallons per minute is 67.2 litres per minute or 1.12 litres per second.
A Pelton wheel turbine is 82% efficient. For each 1 litre per second of 9 bar pressure water flow it produces 720 watts.
80 watts each one litre water flow per second at 1 bar pressure.
Pelton wheel outcome = 89.6 watts.
Archimedes screw will produce less, yet I not know by how much.
Having the water flow by slope not a vertical fall will reduce the wattage generated whichever form of energy converter is used.
Good luck with your project.
Gareth Lewis wrote:Hi, im new to this forum and was wandering if anyone could help me with my school project ?
I have been given a task to find out the best way to generate electricity with hydro electricity
Pelton wheel, Archimedes screw or turbine ?
We have been given a value of 16gallons per minuit in flow and a drop of 10m(straight down or gradually)
Can anyone help
Efficiency of the Archimedes screw is not known to me other than it is less than the Pelton, Francis and Kaplan turbine.
Water having a vertical fall of ten meters produces a 1 bar flow pressure.
16 gallons per minute is 67.2 litres per minute or 1.12 litres per second.
A Pelton wheel turbine is 82% efficient. For each 1 litre per second of 9 bar pressure water flow it produces 720 watts.
80 watts each one litre water flow per second at 1 bar pressure.
Pelton wheel outcome = 89.6 watts.
Archimedes screw will produce less, yet I not know by how much.
Having the water flow by slope not a vertical fall will reduce the wattage generated whichever form of energy converter is used.
Good luck with your project.
Gareth Lewis
Posts: 9
posted 3 years ago
Hi and thank you both for the help, the calculations help a lot, but im not sure if that is 86.9 watts per minute does that mean I can times that by 60 to get the watts per hour or in this case Kw/h ??
Im not sure if we are allowed to generate our own electricity here but im sure we'll do something as a class although it might not be as big as 5kwh lol
Im not sure if we are allowed to generate our own electricity here but im sure we'll do something as a class although it might not be as big as 5kwh lol
Peter Mckinlay
Posts: 182
1
Marcos Buenijo
pollinator
Posts: 583
Location: Southwest U.S.
12
posted 3 years ago
No, 86.9 watts sustained for one hour would equate to 86.9 watt hours, or 0.0869 kilowatt hour.
BTW, here are some definitions, figures, and calculations that I believe will be helpful:
1 horsepower = 550 foot pounds/second = 746 watts
1 gallon of water weighs 8.35 pounds
1 meter is about 3.3 feet
So, if you have 16 gallons of water falling a vertical distance of 10 meters, then you have about 134 pounds of water falling a vertical distance of about 33 feet, and this equates to about (134)(33) = 4420 foot pounds. Since this takes place over one minute (i.e. 16 gpm), then it's about 74 foot pounds per second (i.e. 4420/60). Well, this is (74/550) = 0.135 horsepower, or (0.135)(746) = 100 watts. If you constructed a perfect system with zero losses, then it could in theory generate electricity at a rate of 100 watts. Of course, in reality, the system will see much less. The goal it seems is to devise a system that gets closest to 100 watts electrical output.
NOTE: The most practical device is probably a Pelton wheel as a water wheel would be too large.
Gareth Lewis wrote:Hi and thank you both for the help, the calculations help a lot, but im not sure if that is 86.9 watts per minute does that mean I can times that by 60 to get the watts per hour or in this case Kw/h ??
No, 86.9 watts sustained for one hour would equate to 86.9 watt hours, or 0.0869 kilowatt hour.
BTW, here are some definitions, figures, and calculations that I believe will be helpful:
1 horsepower = 550 foot pounds/second = 746 watts
1 gallon of water weighs 8.35 pounds
1 meter is about 3.3 feet
So, if you have 16 gallons of water falling a vertical distance of 10 meters, then you have about 134 pounds of water falling a vertical distance of about 33 feet, and this equates to about (134)(33) = 4420 foot pounds. Since this takes place over one minute (i.e. 16 gpm), then it's about 74 foot pounds per second (i.e. 4420/60). Well, this is (74/550) = 0.135 horsepower, or (0.135)(746) = 100 watts. If you constructed a perfect system with zero losses, then it could in theory generate electricity at a rate of 100 watts. Of course, in reality, the system will see much less. The goal it seems is to devise a system that gets closest to 100 watts electrical output.
NOTE: The most practical device is probably a Pelton wheel as a water wheel would be too large.
Peter Mckinlay
Posts: 182
1
posted 3 years ago
Hello Gareth Lewis,
Pelton wheel technology is improved upon when following Kaplan copy of the steam train. The funnel increases the draw of exit and in doing so increases the water flow past the turbine. This achieves greater wattage output than by open exit.
That which has been posted relies upon Pelton receiving water not gas. Pelton suited to both.
Water flow temperature is an ideal heat stripper when the water pass through a Radiator.
Turbine drive closely follows that commonly found in electricity generation using a steam turbine.
Many gas have high pressure at air temperature.
The variance between Water heat and Air heat provides the available work pressure.
1 litre per second at 9 bar pressure provides 720 watts.
Carbon Dioxide has most benefits in such application.
The attached is self cooling (CO2R744) though the expander though the expander may be used in Radiator type cooling if wanted.
Pelton wheel technology is improved upon when following Kaplan copy of the steam train. The funnel increases the draw of exit and in doing so increases the water flow past the turbine. This achieves greater wattage output than by open exit.
That which has been posted relies upon Pelton receiving water not gas. Pelton suited to both.
Water flow temperature is an ideal heat stripper when the water pass through a Radiator.
Turbine drive closely follows that commonly found in electricity generation using a steam turbine.
Many gas have high pressure at air temperature.
The variance between Water heat and Air heat provides the available work pressure.
1 litre per second at 9 bar pressure provides 720 watts.
Carbon Dioxide has most benefits in such application.
The attached is self cooling (CO2R744) though the expander though the expander may be used in Radiator type cooling if wanted.
Gareth Lewis
Posts: 9
posted 3 years ago
Water having a vertical fall of ten meters produces a 1 bar flow pressure.
16 gallons per minute is 67.2 litres per minute or 1.12 litres per second.
A Pelton wheel turbine is 82% efficient. For each 1 litre per second of 9 bar pressure water flow it produces 720 watts.
80 watts each one litre water flow per second at 1 bar pressure.
Pelton wheel outcome = 89.6
Mr. Peter McKinlay
Ok Im getting a little confused with all this now, with the above you say at the end that you have 80 watts each one litre water flow per second at 1 bar pressure, and you get this from 720 watts divide by the 9 bar.
Then because you have 1.12 litres per second you times that by the 80 watts to get the final figure of 89.6 watts
But surely this is still 89.6 watts per second and not per hour
16 gallons per minute is 67.2 litres per minute or 1.12 litres per second.
A Pelton wheel turbine is 82% efficient. For each 1 litre per second of 9 bar pressure water flow it produces 720 watts.
80 watts each one litre water flow per second at 1 bar pressure.
Pelton wheel outcome = 89.6
Mr. Peter McKinlay
Ok Im getting a little confused with all this now, with the above you say at the end that you have 80 watts each one litre water flow per second at 1 bar pressure, and you get this from 720 watts divide by the 9 bar.
Then because you have 1.12 litres per second you times that by the 80 watts to get the final figure of 89.6 watts
But surely this is still 89.6 watts per second and not per hour
Peter Mckinlay
Posts: 182
1
posted 3 years ago
Hello Gareth,
For a standard hydro generator setup using river flow, it is correct,
Water having a vertical fall of ten meters produces a 1 bar flow pressure.
16 gallons per minute is 67.2 litres per minute or 1.12 litres per second.
A Pelton wheel turbine is 82% efficient. For each 1 litre per second of 9 bar pressure water flow it produces 720 watts.
80 watts each one litre water flow per second at 1 bar pressure.
Pelton wheel outcome = 89.6 per second minute or hour).
The additional information on Pelton is correct however not commonly exploited in private use.
A river water flow is very easy to exploit, its temperature slightly more complicated.
 1
Hello Gareth,
For a standard hydro generator setup using river flow, it is correct,
Water having a vertical fall of ten meters produces a 1 bar flow pressure.
16 gallons per minute is 67.2 litres per minute or 1.12 litres per second.
A Pelton wheel turbine is 82% efficient. For each 1 litre per second of 9 bar pressure water flow it produces 720 watts.
80 watts each one litre water flow per second at 1 bar pressure.
Pelton wheel outcome = 89.6 per second minute or hour).
The additional information on Pelton is correct however not commonly exploited in private use.
A river water flow is very easy to exploit, its temperature slightly more complicated.
Gareth Lewis
Posts: 9
Peter Mckinlay
Posts: 182
1
posted 3 years ago
Hello Gareth Lewis,
You be correct , that is the wattage one could expect if the Pelton is driven by the litres and force per second carried by stream
flow that you mention.
The alternate wattage mentioned only applies if the stream flow is used for other purpose than direct water strike to the Pelton wheel,
or if the Pelton wheel housing is modified such as the Kaplan turbine.
Hello Gareth Lewis,
You be correct , that is the wattage one could expect if the Pelton is driven by the litres and force per second carried by stream
flow that you mention.
The alternate wattage mentioned only applies if the stream flow is used for other purpose than direct water strike to the Pelton wheel,
or if the Pelton wheel housing is modified such as the Kaplan turbine.
Gareth Lewis
Posts: 9
posted 3 years ago
Ok so our task here is to find out how much electricity can be generated over a day with 16gpm/1.12lps with a 10m head, a crude version for me would be a big tank/lake with a vertical pipe dropping the water by 10m onto the pelton(or whatever is best) wheel through a nozzle.
Im taking for granted that the 10m head pipe would have to be a certain size to accomodate 1.12lps, and could I start with a slightly bigger pipe and reduce the pipes diameter maybe on 5m to create more pressure ?
Im taking for granted that the 10m head pipe would have to be a certain size to accomodate 1.12lps, and could I start with a slightly bigger pipe and reduce the pipes diameter maybe on 5m to create more pressure ?
Marcos Buenijo
pollinator
Posts: 583
Location: Southwest U.S.
12
posted 3 years ago
Well, if you're not actually tasked with generating the electricity, then the project is a lot simpler. I went through the basic calculation that shows 100 watts is the ideal power output. So, a perfect system would see (100 watts)(24 hours) = 2400 watt hours, or 2.4 KWh over a 24 hour period. If you are tasked to consider the losses involved, then that requires specific knowledge of the components used. Perhaps your instructor is not requiring this? If so, then your task just got even simpler. If the instructor wants to know the losses involved, then you would determine the losses and subtract from the total.
Gareth Lewis wrote:Ok so our task here is to find out how much electricity can be generated over a day with 16gpm/1.12lps with a 10m head, a crude version for me would be a big tank/lake with a vertical pipe dropping the water by 10m onto the pelton(or whatever is best) wheel through a nozzle.
Im taking for granted that the 10m head pipe would have to be a certain size to accomodate 1.12lps, and could I start with a slightly bigger pipe and reduce the pipes diameter maybe on 5m to create more pressure ?
Well, if you're not actually tasked with generating the electricity, then the project is a lot simpler. I went through the basic calculation that shows 100 watts is the ideal power output. So, a perfect system would see (100 watts)(24 hours) = 2400 watt hours, or 2.4 KWh over a 24 hour period. If you are tasked to consider the losses involved, then that requires specific knowledge of the components used. Perhaps your instructor is not requiring this? If so, then your task just got even simpler. If the instructor wants to know the losses involved, then you would determine the losses and subtract from the total.
Marcos Buenijo
pollinator
Posts: 583
Location: Southwest U.S.
12
posted 3 years ago
I just wanted to clear up something. The power achieved by this particular system does not depend on how long water flow is maintained. Rather, it depends only on the rate of flow. In other words, the system will generate electricity at a constant rate (i.e. power) as long as the flow rate of 16 gpm is maintained. For example, let's consider a horse on a large treadmill working at a rate of one horsepower (also equal to 746 watts). Over a course of one hour, the horse will have generated work equal to 746 watt hours (or 0.746 kilowatt hour). Over a course of two hours, the horse will have generated (746 watts)(2 hours) = 1492 watt hours of work. Over a course of one second, the horse will generate (746 watts)(1/3600 hour) = 0.207 watt hour of work.
Power = work/time (power is the rate at which work is done). Therefore, to find the work done simply multiple the power times the time: (power)(time)= (work/time)(time)= work.
NOTE: work=energy
Gareth Lewis wrote:So in conclusion, if I have a constant flow of 16 gpm or 1.12 litres per second for a whole hour or even longer I will only produce 89.6watts?
...But surely this is still 89.6 watts per second and not per hour....im not sure if that is 86.9 watts per minute does that mean I can times that by 60 to get the watts per hour or in this case Kw/h ?
I just wanted to clear up something. The power achieved by this particular system does not depend on how long water flow is maintained. Rather, it depends only on the rate of flow. In other words, the system will generate electricity at a constant rate (i.e. power) as long as the flow rate of 16 gpm is maintained. For example, let's consider a horse on a large treadmill working at a rate of one horsepower (also equal to 746 watts). Over a course of one hour, the horse will have generated work equal to 746 watt hours (or 0.746 kilowatt hour). Over a course of two hours, the horse will have generated (746 watts)(2 hours) = 1492 watt hours of work. Over a course of one second, the horse will generate (746 watts)(1/3600 hour) = 0.207 watt hour of work.
Power = work/time (power is the rate at which work is done). Therefore, to find the work done simply multiple the power times the time: (power)(time)= (work/time)(time)= work.
NOTE: work=energy
Gareth Lewis
Posts: 9
Marcos Buenijo
pollinator
Posts: 583
Location: Southwest U.S.
12
posted 3 years ago
Convert the time to hours. If you keep all time units in hours and all power units in kilowatts, the you can take the product of the two to get energy in KWh.
Gareth Lewis wrote:Thank you Morcos, I understand your horsey theory, where as I dont understand how, when you calculate something in seconds (89.6) or minutes it does not multiply by the said number to achieve Kwh
Convert the time to hours. If you keep all time units in hours and all power units in kilowatts, the you can take the product of the two to get energy in KWh.
Gareth Lewis
Posts: 9
posted 3 years ago
Sorry  Marcos not Morcos
I understand that although I dont understand how Mr McKinlay can work out the ecuation in litres per seconds and get a figure in Watts (per second) and it does not multiply to per minute/hour although I do realise it would be very high in Kwh on that ecuation
Is there a better system, more efficient that would work with this amount of flow or head ?
I understand that although I dont understand how Mr McKinlay can work out the ecuation in litres per seconds and get a figure in Watts (per second) and it does not multiply to per minute/hour although I do realise it would be very high in Kwh on that ecuation
Is there a better system, more efficient that would work with this amount of flow or head ?
Marcos Buenijo
pollinator
Posts: 583
Location: Southwest U.S.
12
posted 3 years ago
It seems Mr. McKinlay is using a thumb rule. Note that the "watt" unit has 'per second' built into it (watt = joule/second).
If you're not having to actually build a device, then just go with an overshot (or undershot) slow moving water wheel used to drive a low speed permanent magnet alternator with a chain drive. At this low head and low flow rate the simple water wheel is more efficient. Pelton wheels tend to be most efficient at fairly high pressure and low flow rates.
Gareth Lewis wrote:I understand that although I dont understand how Mr McKinlay can work out the ecuation in litres per seconds and get a figure in Watts (per second) and it does not multiply to per minute/hour although I do realise it would be very high in Kwh on that ecuation
Is there a better system, more efficient that would work with this amount of flow or head ?
It seems Mr. McKinlay is using a thumb rule. Note that the "watt" unit has 'per second' built into it (watt = joule/second).
If you're not having to actually build a device, then just go with an overshot (or undershot) slow moving water wheel used to drive a low speed permanent magnet alternator with a chain drive. At this low head and low flow rate the simple water wheel is more efficient. Pelton wheels tend to be most efficient at fairly high pressure and low flow rates.
Peter Mckinlay
Posts: 182
1
posted 3 years ago
Hello Gareth Lewis,
Generating 89.6 watts per second means over a 24 hour period 7,741,440 watts in total are produced, never at any time is a Kilowatt being produced by the generator only 89.6 watts.. 89.6 x 60 = minute 5376 watts x 60 = hour 322,560 watts x 24 = 7,741,440 a day in total.
Yes there are better power generating means available using flowing water, however the efficiency remains the same but the energy delivery to the turbine multiplies by the thousands. This means exploits the heat difference between the water and the air. Where the water is hotter than the air the water converts to heater and air to cooler.
From the scale attached use the air/ground heat as top mark on the scale and the cold water as the bottom mark on the scale. Note the pressure difference between these two marks. That is the force per litre per second the turbine receives. It requires one litre of water per second at 9 bar pressure to produce 720 watts. Wattage increases by bar pressure or litres flow per second.
The two charts one is for heat 40*C up to +32.5*C the other for heat 40*C up to +100*C.
This type of electricity production actually outputs far more wattage if doing its own cooling rather than relying upon an outside cooling source such as water.
Generating 89.6 watts per second means over a 24 hour period 7,741,440 watts in total are produced, never at any time is a Kilowatt being produced by the generator only 89.6 watts.. 89.6 x 60 = minute 5376 watts x 60 = hour 322,560 watts x 24 = 7,741,440 a day in total.
Yes there are better power generating means available using flowing water, however the efficiency remains the same but the energy delivery to the turbine multiplies by the thousands. This means exploits the heat difference between the water and the air. Where the water is hotter than the air the water converts to heater and air to cooler.
From the scale attached use the air/ground heat as top mark on the scale and the cold water as the bottom mark on the scale. Note the pressure difference between these two marks. That is the force per litre per second the turbine receives. It requires one litre of water per second at 9 bar pressure to produce 720 watts. Wattage increases by bar pressure or litres flow per second.
The two charts one is for heat 40*C up to +32.5*C the other for heat 40*C up to +100*C.
This type of electricity production actually outputs far more wattage if doing its own cooling rather than relying upon an outside cooling source such as water.
CO2Critical.png
Carbon_dioxide_pressuretemperature_phase_diagram_svg.png
Marcos Buenijo
pollinator
Posts: 583
Location: Southwest U.S.
12
posted 3 years ago
Peter, respectfully, I think your discussion is beyond the scope of Gareth's project. Also, in the interest of clarity, I wish to intervene about your use of units. A "watt per second" is not a proper way to express power. Expressing it this way can encourage sloppy thinking on the topic. The watt includes a time dimension (watt = joule / second), so a "watt per second" is not merely redundant  it is incorrect, and conceptually misleading.
89.6 watts = 89.6 joules/second. So, over a 24 hour period there is (89.6 joules/second)(24 hours)(3600 seconds/hour) = 7,741,440 joules of work produced.
A more familiar way to express the work done is to use kilowatt hours: (89.6 watts)(24 hours)(1 kilowatt/1000 watts) = 2.15 kilowatt hours
89.6 watts = 89.6 joules/second. So, over a 24 hour period there is (89.6 joules/second)(24 hours)(3600 seconds/hour) = 7,741,440 joules of work produced.
A more familiar way to express the work done is to use kilowatt hours: (89.6 watts)(24 hours)(1 kilowatt/1000 watts) = 2.15 kilowatt hours
Gareth Lewis
Posts: 9
Gareth Lewis
Posts: 9
Marcos Buenijo
pollinator
Posts: 583
Location: Southwest U.S.
12
posted 3 years ago
I assume you mean to generate AC electricity at a rate of 5 kilowatts (5000 joules per second).
You can easily determine the required flow rate based on ideal considerations (meaning no losses). However, you have to know the specifications of the components used in the system to determine the actual required flow rate. All you have to do is consider the underlying definitions:
One hp is 746 watts. Therefore, 5 KW, or 5000 watts, is (5000/746) = 6.70 hp. Since one hp is 550 foot pounds per second, then 6.70 hp = (6.70)(550) = 3685 foot pounds per second. 10 meters is (10)(3.3) = 33 feet. Therefore, you need (3685/33) = 112 pounds of water per second. Since one gallon of water weighs 8.35 pounds, and there are 60 seconds in a minute, then you need (112/8.35)(60) = 805 gallons per minute. If the head is 20 meters, then you halve this flow rate. Now, when you consider the losses involved, then you will have to inflate these numbers considerably. For example, let's assume the alternator is 80% efficient, 10% pressure drop in the piping, and the pelton wheel is 80% efficient. The losses amount to (0.80)(0.90)(0.80) = 0.58. Therefore, the actual flow rate for the 10 m head would be (805/0.58 ) = 1388 gpm.
I highly recommend you use this kind of thinking in lieu of thumb rules. Thumb rules can be useful, but only in the proper context. If someone is using the same components on a regular basis, then they make sense. If not, then thinking in terms of underlying principles and definitions is more useful.
Gareth Lewis wrote:So on another note, if I wanted to generate 5kw/h of AC current with 10 or 20m head, is there an equation to calculate how many litres/gallons per minute I need
I assume you mean to generate AC electricity at a rate of 5 kilowatts (5000 joules per second).
You can easily determine the required flow rate based on ideal considerations (meaning no losses). However, you have to know the specifications of the components used in the system to determine the actual required flow rate. All you have to do is consider the underlying definitions:
One hp is 746 watts. Therefore, 5 KW, or 5000 watts, is (5000/746) = 6.70 hp. Since one hp is 550 foot pounds per second, then 6.70 hp = (6.70)(550) = 3685 foot pounds per second. 10 meters is (10)(3.3) = 33 feet. Therefore, you need (3685/33) = 112 pounds of water per second. Since one gallon of water weighs 8.35 pounds, and there are 60 seconds in a minute, then you need (112/8.35)(60) = 805 gallons per minute. If the head is 20 meters, then you halve this flow rate. Now, when you consider the losses involved, then you will have to inflate these numbers considerably. For example, let's assume the alternator is 80% efficient, 10% pressure drop in the piping, and the pelton wheel is 80% efficient. The losses amount to (0.80)(0.90)(0.80) = 0.58. Therefore, the actual flow rate for the 10 m head would be (805/0.58 ) = 1388 gpm.
I highly recommend you use this kind of thinking in lieu of thumb rules. Thumb rules can be useful, but only in the proper context. If someone is using the same components on a regular basis, then they make sense. If not, then thinking in terms of underlying principles and definitions is more useful.
Peter Mckinlay
Posts: 182
1
posted 3 years ago
Pelton, Francis and Turgo produce 720 watts for each 1 litre of 9 bar pressured water per second. 10 meter vertical fall is 1 bar pressure 89.6 watts, 20 meter vertical fall is 2 bar 178 watts.
56 litres water per second having vertical fall of 10 meters, or 28 litres waters per second having a vertical fall of 20 meters =5KW per second. (every minute every hour, every day). It not accumulate unless stored in a capacitor where a higher voltage may be drawn off for a short period of time.
Figures are by the book turbine maker, and California University. 720 watts for each 1 litre of 9 bar water per second applies to any 82% efficient hydro turbine  see Pelton, Turgo and Francis.
Kaplan turbine is 91% efficient and Mitchell is 60% efficient.
Gareth Lewis wrote:So on another note, if I wanted to generate 5kw/h of AC current with 10 or 20m head, is there an equation to calculate how many litres/gallons per minute I need
Pelton, Francis and Turgo produce 720 watts for each 1 litre of 9 bar pressured water per second. 10 meter vertical fall is 1 bar pressure 89.6 watts, 20 meter vertical fall is 2 bar 178 watts.
56 litres water per second having vertical fall of 10 meters, or 28 litres waters per second having a vertical fall of 20 meters =5KW per second. (every minute every hour, every day). It not accumulate unless stored in a capacitor where a higher voltage may be drawn off for a short period of time.
Figures are by the book turbine maker, and California University. 720 watts for each 1 litre of 9 bar water per second applies to any 82% efficient hydro turbine  see Pelton, Turgo and Francis.
Kaplan turbine is 91% efficient and Mitchell is 60% efficient.
steve lorentzen
Posts: 1
posted 3 years ago
 1
Im not sure why everyone is so mathematical about hydro electricity, if you have common sense you can just see what needs to be done and soforth, Ive built many hydro units most of them just .25 scale [money], and found the hardest part of it is keeping the power steady as the generator demands more hp . that's when I started thinking about old cassette decks and how they kept the steady power so the tape wouldn't sound funky, so I added 2 cement wheels , one was a hair smaller then the water wheel, And the one that runs the generator was half to .75 the size of water wheel, they were on a band type gearing so it would be no jerking , And yes its a bit of an eye sore but it works and once in motion there vary hard to stop . Ive had many ideas that have worked vary well. but I don't own a home so im limited on things I can do, So most projects end up as a past thought or a drawing on a piece of paper, anyway just wanted to throw my 2 cents in and see how much guff I get.
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