This is an archival version of the original KnowledgePoint website.

Interactive features have been disabled and some pages and links have been removed.

Visit the new KnowledgePoint website at https://www.knowledgepoint.org.

 
0

Information regarding groundwater pumping

KnowledgePointAdmin
RedR CCDRR

A question from Vicky: We have a pumped groundwater to surface water system which currently extracts and discharge approximately 10,000 m3 per day through a 15" pipe. As such we are looking at investing in a small micro-hydro plant to see if it would be possible to capture any of this energy which is currently being lost. I do not know if this is a feasible idea, the efficiency rate for such a plant or even if this is something you could offer. Any information or response would be gratefully received.

Comments

We are currently in a position where we need to lower an aquifer table by approximately 27m, this is a 24 hour 7 days per week process. As such I was trying to find a way of them re capturing or producing energy form the pumped volume of water as a method of trying to off-set some of the initial pumping costs.

We have a series of 4 pumps which pump into a large chamber and therefore this has a 1m or so head. The water is then sent to a settlement lagoon where it is discharged into a surface water stream via a single pipe which flows full bore constantly.

Binaya gravatar imageBinaya ( 2015-07-14 09:48:25 )

2 Answers

0

The amount of energy you will regain is in proportion to the head difference between what you are pumping, and from what you are generating, so if you are pumping around 30 m head, and generating from 1 m head, then you will recover approximately 1/30 of the power usage.

0

It is important to know the available head between the average water level in "the large chamber" and the "settlement lagoon". You wrote the "water is sent then..." does is mean this is gravity flow? no pumping from the chamber to the settlement? If it is gravity flow then you can roughly estimate the available potential with the formula: P(kW) = Q (m3/s) x h (m) x 7; meaning you multiply the flow in the pipe (... m3/s) with the available difference in level (between chamber and settlement) and then multiply it with 7 which is a figure that considers all kind of losses (pipe losses, turbine efficiency, generator efficiency etc.). More detailed inform on using hydropower in existing infrastructure you can also find in the short flyer: https://www.skat.ch/activities/prartic... and more comprehensive information you find in the the following article: www.skat.ch/activities/prarticle.2012... If you need further support, please don't hesitate to write me an e-mail (hedi dot feibel at skat dot ch)