Last week I received a request for a working example of hydropower calculation procedures. Along with that request was another question asking what the 1/11.8 factor meant when working with feet vs. the 9.800 factor using meters and kW . The latter 9.800 factor was what that reader was familiar with. Please bear with me, there is a lot of interesting math ahead!
I will deal with both of these requests and a little bit more about hydropower efficiency computations in today’s SmallHydro.com blog post.
Part of the mystery hydro conversion factor issue stems from the different measurement units used (SI) or metric vs. (US) or British imperial units. These units of length, volume and time all play a role in getting the “right” answer when computing Small & Micro hydro power. Above all, memorize this relationship:
HydroPower = Efficiency x Pressure drop x Flow rate = η x (ρ · g · h) x Q
Another interesting note shown at the bottom of the chart is that when computing the sites turbine inlet hydrostatic pressure = (rho x g x h) the pressure coefficient value for (rho x g) is quite different for each measurement unit type. This coefficient varies with the inverse of the 4′th power of the length for a given unit, hence all the strange conversion factors, like 1/11.82 for (US- kW) and 9.807 for (SI-kW), etc.
Generally it is best to stick with a single unit standard when working these problems to avoid mistakes! Metric really makes the conversion factors simpler, but the US is still using the British imperial system long after others have moved on, so I include it for comparison. Of interest too is that this site is right on the edge of Small or Micro Hydropower qualification. It would make a nice renewable hydro power addition to a ranch, small village or community.
This handy chart shows an example of a 150 meter hydropower site with 0.095 cubic meters per second flow rate. The British (US) unit equivalents of 492.126 ft and 3.355 CFS are shown computed as well as incorporating both turbine and generator efficiency for accuracy.
Here is a PDF file of the same chart for printout: Hydropower-calculations-metric-si-british-us
Sincerely,
Jess
Hydropower Calculation Example SI & US units
May 18th, 2009 at 11:51 pm
dear sir
it is very useful for me cause i am in learning stage.
I want to get this in metric unit and detail design of other component like desander design parameter,surgeshaft, penstock, intake etc.
thanks
purushottam budhathoki
May 27th, 2009 at 4:33 pm
Dear Sir
Im interested in the cost of production of energy from run-of the river small hydro.
Ive got potential 65 sites within the Fiji Islands which will provide power to rural native population.
These sites have ranges of potential from as low as 10kW upto 3MW
Will be interested to know of the total cost of plants to construct locally and how best we can put these proposals together.
June 14th, 2009 at 1:55 pm
I´m from Mexico, and I’m interested to know of the total cost of plants, I’ve a location betwen 120 m to 160 m of head and a constant flow of 750 liters/seg.
June 14th, 2009 at 7:28 pm
Antonio,
That’s a bit more than 700 kW at 140 M head and 750 l/s flow. It sounds like a decent prospect if the water rights and site constraints are reasonable. What site flow variation data do you have?
July 4th, 2009 at 3:32 pm
Purushottam,
You’re welcome!
Jess
July 4th, 2009 at 3:35 pm
Pere,
We’ll need to get more info on the sites. Typically start with; 1. Head, 2. Flow, 3. Variation in flow(and head potentially), 4. Altitude (impacts reaction turbines), 5. Penstock length., 6. Water or stream name & location.
We look forward to that in an email us smallhydroblog@SmallHydro.com
Jess
August 10th, 2009 at 1:19 am
I am looking for that most basic formula – calculation of fluid volume per second.
I can measure the width of the tail race, and the depth of water flowing through it.. Any information on the applicable formula would be greatly appreciated.
Tom
August 15th, 2009 at 8:19 am
Tom,
It appears that you can measure the area of the channel, now you need to mesure its velocity of flow. By multiplying area and velocity together you will have the volumetric flow rate.
Note that the velocity measured at the top of the channel will not be be the same as the velocity at the sides and bottom of the channel due to frictional effects. As a broad estimate, assume that the average channel flow velocity is about 90% of that measured at the top.
I hope this helps.
Steve
February 11th, 2010 at 10:57 pm
how to calculate the Penstock diameter and thikness
September 14th, 2010 at 3:23 pm
I’d like to discuss possibility of usage of bigger, watermill-type turbine-wheel, especially at high-speed mountain rivers and creeks, waterfalls. Is there better or more efficient way to use transmission through gears with big wheel instead of Pelton turbine?
September 14th, 2010 at 3:47 pm
Another thing is electric grid installation. For instance, how to build a transformer for 110/220V output 50Hz? Any links to share? Thank you all
January 3rd, 2012 at 2:46 pm
Thank you very much for the post – most useful!