1 As I understand it the 2/3 notion is simply a bulk efficiency taking account of hydraulic losses, generator/transformer/line efficiency. 2 The most logical/efficient means to harness hydro for heating is (ideally direct drive) heat pump. Ground source heat pumps are becoming popular for new-build houses even in Ireland. You can get 3.5 kW heat from the ground for every 1 kW pumping energy with a reasonable design implementation. We are just talking about a loop of pipe 1 m under ground and a heat exchanger. I would suggest for hydro people, who obviously have a lot of water flowing by, that they immerse the pipes in the water unless freezing pipes is a problem. It can also cool the house in summer and is ideal with underfloor heating!! 3 If you live near enough to the generator you may be able to gain heat from its inefficiency. i.e. if it is a 10 kW generator and 90% efficient it is giving off 1 kW in heat... Martin Leahy Millstream Energy -----Original Message----- From: ReadFamily [mailto:[EMAIL PROTECTED] Sent: 04 February 2005 14:13 To: [email protected] Subject: Re: [microhydro] Confused about potential energy. OK - so I am wrong about the potential energy - it is 1litre of water falling 10 metres gives up = 98 Joules (and not 980). A simple (and embarrassing) inability to multiply 9.8 x 10. So for the temperature difference in the penstock - for every 10m (33ft) of drop, the water temperature will rise 0.023 degrees Celsius (0.058 degrees Fahrenheit). That really is a small temperature rise. Using the most efficient generator it would take 3500 litres water falling 10m to heat 1 litre of water from room temperature to boiling point for a cup of coffee. The sooner we start capturing this energy, and the more of it we capture, the better off we will be when the oil reserves start drying up. Dominic Read ----- Original Message ----- From: Max Enfield To: [email protected] Sent: Friday, February 04, 2005 1:32 AM Subject: Re: [microhydro] Confused about potential energy. On 2 Feb I wrote: > In the second case, the turbine impedes the flow of water and so less energy is > lost as friction or needed for kinetic energy. However the water must flow for > the turbine to work and so some of the potential energy is still consumed as > friction or converted to kinetic energy. In the best case the turbine is able > to extract up to two thirds of the potential energy. As written, this is not correct and I humbly apologize to the group for possibly adding to the confusion rather than helping to alleviate it. Typically in micro-hydro applications the objective is to extract as much power as possible from the water stream. Under these conditions the "two thirds" property is correct. However if a lesser amount of power is required the flow rate can be reduced (with consequent reduction in frictional losses and kinetic energy) so the proportion of energy available rises above the two thirds value. The values for the proportion available to the turbine for different flow rates are: flow rate as %age of open flow proportion of potential energy available to the turbine 57.7% 67% (this provides maximum power) 40% 84% 20% 96% The situation resembles the discharge of a lead acid battery - the slower the rate of discharge, the greater is the total amount of energy that can be recovered. So in theory, if maximum energy is to be extracted from the loss of potential as a fixed amount of water falls through a fixed height then use as low a flow rate as possible. However, if there is adequate continuous flow then maximum power (thus maximum energy over a given interval of time) is obtained at 57.7% of open flow, even though one third is lost as friction and unrecoverable kinetic energy. In practise though it is not quite this simple. As the flow rate drops any turbine will be operating at such a low part flow value that the reduced efficiency of the turbine will hardly compensate for the increased proportion of available potential energy. Also in my experience, the flow rate through a penstock seldom exceeds 25% of its open flow value. So in considering the total picture of converting the power of flowing water to delivered electricity, penstock losses are usually of minor concern. Dominic Read replied with: > Following your best case that we could extract 2/3 of this energy for electrical power then each litre of water with a head > of 10m could produce 980 Joules. > > So a flow of 1 litre per second with a 10 meter head could theoretically produce nearly 1000 Watts. > > I wonder how close to that ideal any micro-hydro project has ever come? I haven't followed your calculations in detail, but at some point you are out by a factor of 10. The potential energy of 1 litre of water at 10m head is 98 joules. If this falls at a rate of 1 litre/sec the hydraulic power is 98 watts. The water-to-wire efficiency of micro-hydro projects varies greatly, however for systems delivering less than 1kW it usually lies between 35% and 60%. Frank Leslie replied with: > Is "two thirds limit to the power " related to the Betz Limit of 59.3% of > available power in the wind power domain? No, it is quite unrelated. The "two thirds" limit relates to the proportion of hydraulic power available to a turbine and says nothing about limits to which the turbine is able to convert the hydraulic power to mechanical (shaft) power. Over the past two or three years there have been some insteresting postings to this group regarding turbine (and also waterwheel) efficiency limits. I would think that the theoretical efficiency limit for the Gorlov helical water current turbine would come conceptually closest to the Betz limit for wind turbines. Regards, Max Enfield Planetary Power Does your company feature in the microhydro business directory at http://microhydropower.net/directory ? If not, please register free of charge and be exposed to the microhydro community world wide! NOTE: The advertisements in this email are added by Yahoogroups who provides us with free email group services. The microhydro-group does not endorse products or support the advertisements in any way. 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