For research purposes, when was this article published?
Published on: 20 Jun 2010 @ 12:00 Edit
Hello: My micro-hydro scheme locates at a small river. The flow of his river is 2 m3/s.And the head is 1.8 m. I would like to build a run of the water scheme without water storage. Is this feasible? Because of water species, I can’t utilize all the water in the river. My plan is using an intake make water going down to a small whirlpool turbine. My question is how many flow’s proportions should be used to generate electricity? 50% or 75? What is micro-hydro usually value?
Thanks Best regards
Hello: My micro-hydro scheme locates at a small river. The flow of his river is 2 m3/s.And the head is 1.8 m. I would like to build a run of the water scheme without water storage. Is this feasible? Because of water species, I can’t utilise all the water in the river. My plan is using an intake make water going down to a small whirlpool turbine. My question is how many flow’s proportions should be used to generate electricity? 50% or 75? What is micro-hydro usually value?
Can you give me your e-mail address so that I can send you attachment. Thanks
I am uncertain how to design the micro hydro or mini hydro system I need.. First of all what are the advantages and disadvantages of generating ac current versus dc current? Would I use a generator or an alternator to do this? Dayton makes a 120 volt generator/turbine combo. Other companies make a 24 dc generator/turbine combo… Which is better? If I understand correctly I can store the electricity in batteries with a dc system and convert it to ac current with an inverter, but would it produce 240 volts or only 120 volts? Am I correct in assuming that I cant store ac current? So what are the advantages of making dc current and converting it to ac current with an inverter?
Next thing is that I don’t really know if I want a stand alone system or if I want to be able to sell the power back to my local utility co. Can I do this with a dc system as well as an ac system? We are subject to frequent power outages here every time a tree branch falls on their power lines.. Earlier this month we were out of power for ten days due to heavy snow and trees falling all over the place…… How would utility company power outages affect my micro or mini hydro system? Would I have to install some kind of a switch so that I don’t send power back to the utility company when they are working on their transmission lines?
Finally I have a 1350 gallon water holding tank about 40 or 50 feet above my house… The point of diversion for ths tank is a creek about another 40 or 50 feet higher up the hill than that.. It will be a concrete enclosure placed in the stream with a 2 inch hole at the top of the enclosure for a pipe and a concrete over it so the that water will not enter the enclosure from the top. It will enter the enclosure from the bottom and filter up through about 12 inches of small rocks to reduce the sediment in the system. My question is how do I figure the head and flow.. Is the head measured from the highest point of diversion in the creek way up on the hill or is it measured from the 1350 gallon holding tank half way up the hill? The tank will have a two inch intake pipe and a two inch pipe overflo pipe located at the top of the tank going down to the turbine… Is that big enough or is it too big? Should I use a smaller pipe?.. I cant use a larger pipe as the tank has 2 inch holes at the top and bottom. There is another inch and a half hole to supply water to the house.
This is a lot of questions in a single post and we will not attempt to answer them all, but to comment on the difference between AC and DC supplies. You are correct that batteries can only store DC energy and not AC, so if you intend to use battery storage then you will need a DC generator, or an alternator with rectification. The use of DC or AC depends on your proposed system, and whether you want an off-grid or grid connected system and power consumption. For long cable runs an AC supply is better than a low voltage DC supply, due to voltage drops, cable sizing, etc. and finally inverters take a DC supply and convert it (hence the name) to single-phase or three-phase AC, either 120VAC, 240VAC or 400VAC. There are many inverter types to choose from.
The worked example needs to be multiplied by 1000 to come up with the 1,390 watts answer.
Interested in hydro and would like to request if you could provide information
How much it cost to generate 5kv electricity and does it produced round the clock and how long it sustained?
What protocol are followed for generating 5kv electric?
Really informative Article. Thanks for clearing my mind I am just recently entered in this field. Sir is there any ebook about the power generation of Electricity? If there is!! The let me know.
There will be lots, do a search in Google.
So useful information that clears the mind of some doupt…gravely thankful.
I can see how mini scale hydro electric generators would be connected directly to the grid. I have always been interested in cleaner forms of renewable energy. I definitely feel like we should look into using hydro electric power because it seems like an environmentally friendly alternative.
Please guide me to gain power from the discharge water/return water which is pump by 5hp pump 24 hrs
At present we have fabricated a bulcart wheel having 22 pelton buckets and erected and rpm is around 75RPM and with increase in pully size we are getting finaly 600-650 RPM .
Good article. I think you have an error in power calculation: In the equation: Power (P) = Flow Rate (Q) x Head (H) x Gravity (g). You miss the density of water =997.0475 kg/m3 (at 25ºC)
Correct equation: Power (P)[W] = Density Water (p) [kg/m3] x Flow Rate (Q)[m3/s] x Head (H)[m] x Gravity (g)[m/s2]
Best Regards. Domingos Goncalves
The energy released by a body of falling water is defined as the vertical distance of the fall times the weight or the downward force exerted. The force is a product of the water’s mass (m) and gravitational acceleration (g). The energy released therefore is the vertical distance (h), or the head, multiplied by the force.
Since power (P) is defined as energy per second, then the estimated power (W) = head (h) × flow (Q) × efficiency (n) × a multiplier constant
This multiplier constant consists of water density x gravitational acceleration of the water. As water density is defined as 1,000 kg/m3 this is equivalent to 1.0 kg/L. If the flow rate is given in litres/min, then the constant can be written as 1kg/L x 9.81 m/s2 or just 9.81
My question is this our company dischages wastage water for 8 hour daily by 5hp motor pump .I want to use discharge pressure of water for rotating turbine .How much electricity can generated please help me . I am working on project
Your email address will not be published. Required fields are marked *
What's the Answer * 9 × six =