Comments on: Hydropower Energy calculations and GPM-PSI Unit conversions

By: frenchriverland

frenchriverland — Wed, 05 May 2010 09:38:41 +0000

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

By: Daimen

Daimen — Thu, 15 Apr 2010 16:14:32 +0000

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?

By: shankarji

shankarji — Tue, 01 Dec 2009 09:29:03 +0000

Its good information but i need detailed information as i am hydropower engineer i need scientific information