Q: Say you want to convert Psi & GPM to kW water power?
A: Good question – unfortunately there are a myriad of Unit conversions, like furlongs per fortnight. 
Follow my round about conversions, I like to focus on the most common formula to start my thinking.
(unfortunately in English Units)
So I stick with the following formula and convert units to fit it. Metric converts too…
I usually think in CFS & Ft head:
I usually start with this formula –> kW ideal = (CFS flow x Ft head)/11.8
But you have GPM & Psi for units
1 ft head = 0.433 Psi –> (1 ft/.433 psi) x Z Psi = ft head
There are 60 seconds to a minute –> 1 min = 60 sec
There are 7.48 Gal per Cubic Feet volume –> (1 CF /7.48 Gal) x (Gal /sec) = CFS
So say we have 78 Psi -> 78 psi /.433 psi/ft = 180.14 ft head
And now we’re given 1000 GPM = 1000 GPM x 1 Min/ 60 sec = 16.67 GPS
Now convert to CFS -> 16.67 gal/sec /7.48 Gal/CF = 2.23 CFS
Therefore our kW ideal is = (2.23 CFS x 180.14 ft)/11.8 = 34.02 kW Ideal –> Efficiency % not included
To factor efficiency in Multiply x ~72% for water to wire typical. (must use chosen Gen % x Turbine %).
So we get something like 34.02 kW x 72% = 24.49 kW Typical power as a more realistic measure of output potential.
I sure hope this helps,
Jess
DoradoVista, Inc.
PS: Mileage will vary, there are a lot of issues not modeled here. It is a start though.
April 24th, 2009 at 10:30 pm
This entry was from an email to “smallhydronewsletter@doradovista.com” this week. I regularly use this feedback as feed material for our blog and newsletter to get the discussions going among our community and as a means to answer questions that you may have that others have asked. You have been a very supportive group! Small & Micro
Feel free to comment and clarify that’s how we learn. One thing I ask, please try to keep it civil. I have set up the comments such that you will need to login to comment. If you want to use a picture avitar submit one by going to gravitar and getting yours. Just follow the instructions on the blog.
As I said please be polite and do not put multiple links or malicious links I would not like to have to edit out the comments for problems if you understand my drift
Have a great time on SMALLHYDRO.com – Welcome!
April 30th, 2009 at 11:32 am
What is the constant 11.8? The constant we use for the value of gravity is 9.81.
Water to wire efficiency varies depending on the turbine our double regulated Kaplan ranges from a low of 45% when the guide vanes are open 5% rising rapidly to 81.2% when the guidevanes are open 50% this falls to 80% when the guide vanes are open 80% and 76.9% when they are open 100%.
In Europe we use Cumecs (cu m/sec) for flow and m for head so
kW= cumecs x m x 9.81 x efficiency
flow = kW/(flow x cumecs x 9.81 x efficiency)
April 30th, 2009 at 11:35 am
In your formula you, as everyone else, never seem to include the eficency in the origional equation. Ultimately you do select 72%, and I realise that this may not always be the same. However, wouldn’t it make more sence to include it so no one forgets it. Q x H x E / 11.8. Afterall, they are all variables, and all necessary. Just my opinion. Harry
May 2nd, 2009 at 3:12 pm
Thanks Harry,
Good point!
The point you make about water power formulas needing an efficiency factor right from the start is a good one. In energy conversion processes just like other things in life, “There is no free Lunch!”
The energy losses in Hydro turbines can vary significantly for different penstock to hydro-turbine setups and at different operating flows. Energy efficiency varies typically from 100% loss to as little as 10-20% depending on a particular Hydropower systems design parameters and the current water flow rate. As you have noted losing 20% of a Hydro systems output corresponds directly to 20% loss of ROI or cash flow value for the project.
That 20% to 100% loss in turbine efficiency is a Big system modeling issue as you have correctly noted in your comment. In modeling hydro machinery to electrical power output there is also the loss created by converting mechanical to electrical energy and related losses in generators, switches, transformers and lines.
We’ll be getting to these modeling issues as we progress in our blog posts in the future. For now we are trying to steadily progress from simple to complex in our discussion.
As one of my advanced engineering calculus professors, Dr. Schmid-Schoenbein, once said, (when he was challenged by a student about not moving the spectrum analysis lecture fast enough,) “We must learn to climb hills before we climb mountains, we’ll be climbing mountains soon enough.” He was right, we really learned to climb – that was both a valuable and challenging class!
Lesson: There are no perfect loss-less energy conversions. If there were perfect conversions, we’d have perpetual motion!
Again, thanks for your feedback,
Jess
May 4th, 2009 at 7:52 pm
I have sent a small note trying to tell you what we are about and what we could do for both of our benifits.
We are starting an assembly plant in Kalama, Wa. right next to the Columbia and Kalama rivers. They are a little strick about sticking things in the rivers and it would be great to get your idea sold up in this neck of the woods.
Just an idea and hope be talking later.
Best Regards Ed
May 6th, 2009 at 12:11 pm
Ed,
Sounds like an interesting project and a promising Hybrid Renewable energy strategy to follow. Yes, there are rules on Small & Micro hydro. You’ll need to evaluate the site and then work toward getting enough data to begin the approval process. First establish site capability, then look into regulations & authority(s) that have jurisdiction.
Jess
December 1st, 2009 at 2:29 am
Its good information but i need detailed information as i am hydropower engineer i need scientific information
April 15th, 2010 at 9:14 am
I am a first year engineering student and am studying externally. It is my understanding that any given pump with a given impeller diameter and speed will raise a liquid to a specific height regardless of the weight of the liquid. Is this true? Apparently as the density of the liquid changes, so too will the pressure, but the height remains the same??
I assume this height to be the ‘head’ and am trying to calculate the amount of power needed by the pump in kW to supply H20 vertically through a 100mm inside diameter pipe to a height or ‘head’ of 25m(82ft). The volume flow rate at the ‘head’ must be 12L/s.
So, what I have done is convert 12L/s to 0.423CFS, and then following the above formula ‘ideal kW = (0.423CFSx82ft)/11.8 = 2.9395kW’.
Is this all that is needed to calculate the pump power?
May 5th, 2010 at 2:38 am
Hello:
I am new to this organization. I want to let you know that I have a website called http://www.frenchriverland.com. I have been designing small hydro systems and dams for the past 30 years. I have an extensive reference library on the subject. I have spent the last four years scanning reference books, trade catalouges and photos that I have taken of other peoples site. I have no formal training in web design and cold turkeyed the site from a copy of Microsoft front page. So pardon the amateurish page layout. If you need information on older turbines and generators it is the place to go. Having survived the hydro renaisance of the late 1970s, it is a pleasure to see so many younger folks interested in the subject.
Sincerely,
William K. Fay P.E.
West Ware, MA USA