Here’s some questions from Ben, one of our Small & Micro Hydropower Newsletter readers. He asked these questions regarding turbine sizes, efficiency and penstock design.
Sincerely,
Jess
Hydro System Design Questions:
——– Original Message ——–
Subject: turbine size
From: Ben
Date: Thu, June 11, 2009 8:16 am To: jess.blog@smallhydro.com
Jess,
- Was wondering, does a smaller hydro propeller, ( not sure of terminology) add rpm’s thus giving more electric production?? If so, what is a general increase??
- Also, in the formula you sent me about converting to psi, how can you increase the effeciency rating of 72%, in general terms???
- Finally, on the penstock, how much do you typically decrease the intake from the main water line?? In other words, with a 4″ line is the penstock intake 4″ or does it decrease to 3″, and what if you decreased it to 2″ penstock intake?? Will that increase the flow pressure and what impact does it have on spinning the turbine?? I still don’t have specifics but working on it.
Thank you,
Ben
Hi Ben,
Looking at your questions I’ll try and answer them the best I can. Turbine deisgn and choice is complex and efficiency of hydro design is also. Mixing both in one response is a great simplification. With that said here are my Answers.
Jess
Hydro System Design Answers:
- Choosing Smaller or larger propeller or water wheels is not the issue. Fluid flow over the hydro generator propulsion device gives rise to rotational motion. Rotor specific speed is the issue you refer to and different designs have higher or lower specific speeds due to the physics involved with that wheels geometry. That is the rotational speed of a rotor of one design compared to others will be higher or lower depending on the actual design or geometry involved regardless of scale. Specific speed is a comparative “unitless” number much like a Mach number is used for supersonic flow. Specific speed often stated something like this “ηs= η/η0″. Efficiency at a given flow/pressure regime depends on this number a lot. Enough on that topic for now. What a smaller diameter wheel (using a pelton wheel for example) does is imply that for a given water jet velocity they must spin at a faster rate to produce their optimum power output. Max Pelton efficiency is at about 47% of the water jet velocity. This speed maximizes power output for given a pelton impulse design. So with all other factors being fixed the wheels pitch or diameter will increase rotational speed to keep the center line of water impact moving at the same 47% linear speed of the water jet. Power will be about the same, but generator frequency will change with rotational speed which may make a speed reducer/increaser a requirement.
- Increasing hydro system %efficiency is a complex art, starting with intake and Penstock friction, then turbine choice & performance curve optimization, combined with inlet valve and draft tube design if required, then any speed transmission or reducer friction, next will be generator/alternator efficiency, finally the transmission losses including wires, transformer/inverter systems. The 72% efficiency example I give is a very simple average of a model 85% turbine x 85% geneator/transmission output. Mileage for your site and turbine choices will vary. The more accurate you make your hydropower model the better the result will be. You will be modeling power production and loss until the first water flows through the system, then you get your course grade based on how close your “Real/Model” score approaches 100% of actual results.
- The last part of your request has more to do with jet design and penstock friction. Typically the penstock will be much larger diameter than the jet intake. The pipe friction means that there is a large dynamic head loss over long distance flows. The jet or turbine intake design then takes the pressure drop and accelerates water through the intake throat, a much smaller distance. These cross sectional intake areas are much smaller on higher head units like pelton and turgo units vs. lower head pressure ones like Francis, PAT and Kaplan/Popeller ones. That’s part of why the low head turbine prices rise so rapidly with flow rate, there’s a large opening to a much larger volume of machine for the same output power. Generally you should seek to keep penstock intakes and penstocks free of friction loss, then reducers at the turbine may be possible if required. For example a PAT I know of takes a >30inch diameter main and necks it down to 10inches right at the PAT input valve. The overall system efficiency is just a part of the equation, remember those big penstocks cost big bucks so make sure you balance the equation for optimum water power system ROI$.
I hope this helps clarify things,
Sincerely,
Jess
DoradoVista, Inc.
PS. I plan to post this set of question/answers on the Blog OK? Generic name & content only.
Ben replied,
Use the question, so far I am 2 for 2 in getting my questions in the blog. Batting a thousand here.
Thank you for the answers.
Ben
You’re welcome!
Jess
= Efficiency × Pressure difference × Volume rate of water flow 
= Water volume flow rate (m³/s) or (ft³/s)
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