greenheatman Posted July 6, 2012 Report Share Posted July 6, 2012 I find it incredible that the wave energy industry continues to use flawed mathematics to calculate the power in a wave. it is well known that the kinetic energy in a wave is equal to the potential energy in the same wave. What Prof MacKay does in his 'without the hot air' book on wave power (page 308 I think) is calculate the potential power, assumes that he is right, and simply doubles it to give his formula - which is wrong as it happens. The beauty of mathematics is that it is a pure science but you must prove your formula to confirm that you are correct. In wave power they simply have not bothered to do a mathematical proof. What I have done below in calculated first the potential energy and came up with pretty much the same answer as Prof MacKay except that he has a 4 in his denominator but this has to be pi. I then do the same with kinetic energy and come up with the same answer - adding them together gives the correct and proven formula for wave power. Assuming a perfect sinusoidal oceanic wave, one metre high between crest and trough with a period of 10 seconds in deep water, its velocity is calculated using the formula (gT/2Ï€) = 15.613 m/s The wavelength, Î» = vT =(gT^2/2Ï€) = 156.131 m The potential energy passing per unit time, per unit length, is Ppotential =m*gh/T where m* is the mass per unit length and h is the change in height at the centreline of the sinusoidal mass of seawater. The cross-section area of the waveâ€™s amplitude is calculated using the following formula Area = (2 x base x height)/pi, (2 (Î»/2) a)/ Ï€ because a = h/2 this can be rewritten as (Î»h/2Ï€). Ppotential =m*g h/T = (2 Î» h g h Ï) / 4 Ï€ T = (Î» h2 g Ï) / 2Ï€ T and because Î»/T is velocity, substituting we get, Ppotential = h^2 g Ï v / 2Ï€ = (1 x 1 x 9.81 x 1025 x 15.613)/ 2Ï€ = 24.986kW/m The potential and kinetic energies in waves are exactly equal. I now will calculate the kinetic energy of deep water dispersive waves independently to see if the answers are the same. P kinetic = Â½ m v^2 (kW/m) The simultaneous vertical descending and ascending velocity of all oceanic waves irrespective of their height is always âˆš2 (m/s) which I will call vd and va respectively. Substituting we get P kinetic = Â½ [(vd h)(va h)(g v Ï)]/2 Ï€ (kW/m) = Â½ [(âˆš2 h) (âˆš2 h) (g v Ï)]/2 Ï€ (kW/m) = Â½ 2h^2 g v Ï)/2 Ï€ (kW/m) Therefore, P kinetic = (h^2 g v Ï)/2Ï€ (kW/m) = 24.986 (kW/m) = P potential P potential + P kinetic =24.986 + 24.986 = 49.972 kW/m Therefore, P potential + P kinetic = (h^2 g v Ï)/ Ï€ (kW/m) so that for a dispersant pelagic ocean wave in deep water, the following formula is true; Ptotal = (h^2 g v Ï)/ Ï€ Questions & Answers Q1) The idea of converting wave shaft power into heat, storing it and converting it back into (generator) shaft power sounds very inefficient â€“ wouldnâ€™t it be better to convert the wave power into electricity directly? A1) I have shown that the near shore raw resource power is 474kW/m. Conventional â€˜real timeâ€™ wave energy to electricity converters have very poor resource to wire conversion efficiencies. The following link describes how an oscillating water column device, the Limpet 200 could only generate @21kW at best during optimum wave conditions. The effective length of the device was 10m so 2.1kW/m from a possible 474kW/m gives this failed wave energy converter a resource to wire efficiency of just 0.44%http://www.inference.phy.cam.ac.uk/withouthotair/c12/page_75.shtml The Pelamis Sea Snake developers claim that their device gleans a raw resource power of 72kW/m in deep water well off-shore. This still has be converted into intermittent pulses of electricity using hydraulics and, assuming a generous efficiency of 25%, only 18kW(e)/m is generated from a possible 1,333kW/m giving a resource to wire efficiency of 1.35%. http://www.icrepq.com/icrepq-08/380-leao.pdf Gentec WaTS on the other hand, in the same deep water, has a resource to wire efficiency of 48%. Q2) If you invention is as good as you say â€“ why wasnâ€™t it invented before? A2) Well, it has all been invented before! Every element of my invention is well proven science â€“ all I have done is â€˜rearrangedâ€™ these components to solve the â€˜intermittency problemâ€™ and generate over 50 times more secure thermally generated electricity as a result. Converting the power in waves in direct proportion into heat and storing it is pretty obvious when you consider that almost all of the worldâ€™s electricity is generated thermally. Providing thermal power stations with an alternative â€˜heat sourceâ€™ without burning anything or splitting atoms is all that is needed to generate electricity using well proven thermodynamic processes. Quote Link to comment Share on other sites More sharing options...

crofter Posted July 6, 2012 Report Share Posted July 6, 2012 Good to see you back greenheatman. Did you ever get the gentec venturi to work? I think your new idea is much better, if only because it will not need any moorings. http://www.shetnews.co.uk/news/newsbites/4367-vattenfall-looking-for-missing-wave-buoy.html Quote Link to comment Share on other sites More sharing options...

greenheatman Posted July 6, 2012 Author Report Share Posted July 6, 2012 No - I couldn't get any funding but I have walked away from Gentec venturi because, despite it being capable of generating 10 times more 24-7-365 electricity my wave machine will generate 50 times more 24-7-365 electricity per linear metre of wave front. I made the classic mistake of being up too close to Gentec venturi to realise that Neap tides make too big a 'hole' every other week. Quote Link to comment Share on other sites More sharing options...

crofter Posted July 6, 2012 Report Share Posted July 6, 2012 I made the classic mistake of being up too close to Gentec venturi to realise that Neap tides make too big a 'hole' every other week. Surely that will be an issue for all tidal generators? Even during neap tides there is some energy available, and it is not as if tidal will be a big part of the mix anytime soon - so what if there is "too much" during spring tides? How does the cost per unit compare between venturi and WaTS? Quote Link to comment Share on other sites More sharing options...

greenheatman Posted July 8, 2012 Author Report Share Posted July 8, 2012 Of course it is an issue with all 'real time' tidal stream generators because they all use the mean Neaps' velocity which approximates to a half of the means Springs' mean velocity. This means that only 1/8th of the tidal stream resource is actually used because velocity is cubed in the power equation (1/2 x 1/2 x1/2 = 1/8) These low tech machines struggle to convert more than 40% of this 1/8th into electricity so that an absolute maximum of 5% of the resource becomes low quality intermittent insecure power - pretty pathetic really. My Gentec venturi converted almost all of the power in each and every tide into heat and this amounts to around 50 times more thermal power put into storage. Covering it back into generator shaft power @ at a lowly 20% thermal efficiency gives you the ability to generate 10 times more electricity to meet fluctuating electrical demand 24-7-365 As to costs of production my price per MWh will be around 2 pence - real time TSGs are 'happy' with 6 pence per kWh! Quote Link to comment Share on other sites More sharing options...

Brian86 Posted July 11, 2012 Report Share Posted July 11, 2012 greenheatman can you maybe expand on what you post here, I'm guessing there was maybe another thread that explained more what you're doing. Are you using some maths thing to your advantage? Are you getting higher efficiency through some engineering work? Quote Link to comment Share on other sites More sharing options...

greenheatman Posted July 12, 2012 Author Report Share Posted July 12, 2012 Brian, to understand what I am proposing you have to understand the dodgy engineering practices by the self-promoting 'experts' in the field of marine renewables. It is easier to to put hard and fast numbers on tidal streams but the same principles apply to other renewables. Let's look at Bluemull Sound. The water courses through there about 2 days after a full or new moon at, say a nice round 5m/s on the zenith Spring tide. However, our academic 'experts' are intent on generating electricity in real time from this prime mover and are not interested in Spring tides and design their Tidal Stream Generators (TSGs) to work on the mean Neap tide velocity which, for convenience, we will say is half of the Mean Spring Tide (actually it is around 56%).Mathematically, they are simply discarding 7/8ths of the resource because velocity is cubed in the power equation (ie 1/2 x 2/2 x 1/2 = 1/8). This 'extra' power is discarded by feathering the blades of their TSGs The may be able to convert about 35% of this remaining 1/8 into predictable varying pulses of practically worthless electricity with a maximum capacity factor of ~22% so that ~4% of this huge resource is converted into dodgy electricity that requires fossil backup for 78% of the year! (please ignore claims by MCT of 66% because they have reduced the hours in the denominator by a third to attract inward investment!) Now some people will say that this is 'brilliant engineering' but I don't. Simply by converting ALL of the power under the power graph for each and every tide my calculations show that you can generate 10 times more thermal electricity 24-7-365 that does not require any back up whatsoever. Quote Link to comment Share on other sites More sharing options...

MuckleJoannie Posted July 12, 2012 Report Share Posted July 12, 2012 For new readers discussion on greenheatman's earlier invention, the Gentec Venturi, can be found in this thread http://www.shetlink.com/forum/viewtopic.php?p=45449&sid=0fd6144962121f1df0c33cf3739072b2 Quote Link to comment Share on other sites More sharing options...

Brian86 Posted July 12, 2012 Report Share Posted July 12, 2012 Ok, so if I understand this right, you're basically proposing a wave power thing that operates using spring tides which have twice the velocity of neap tides. Since this is such a well known fact, why do companies not do this already? One thing I would like ask is that, although the power equation gives a nice result for spring tides, would larger and faster waves not have the potential to generate much more turbulence which would make things a bit more complicated than just the power equation lets on? Perhaps this won't really apply so much in deep water, or even effect the wave machine that much, but it's still something I'd like to ask. Do you have a prototype for this? If not have you approached anyone for funding? Quote Link to comment Share on other sites More sharing options...

greenheatman Posted July 12, 2012 Author Report Share Posted July 12, 2012 Brian, no I am not proposing that at all! We're discussing tidal streams so that figures can be put on it. I pointed out that simplistic TSGs discard 7/8ths of the power and attempt to extract as much of the remaining 1/8th as possible - this is bad engineering because a) the output is intermittent and the output is a tiny ~4% of the resource. Maths can simply prove this to be true. If you have a 1MW TSG it can only generate a maximum of 1MW irrespective of how much shaft power is available during a Spring tide. To calculate the power in one tide the formula is P = (4 x h x b)/(3 x pi) so for a Spring tide maxxing at 8MW there will have been 26MWh of mechanical power available on that tide (for conversion to heat in my system). The electromechanical 1MW TSG may have generated a rather pitiful 3 to 4MWh electrical on the same tide throwing away >22MW(mech) in the process. Quote Link to comment Share on other sites More sharing options...

Brian86 Posted July 12, 2012 Report Share Posted July 12, 2012 Ok so I thiiink I might be following you now but really I'd like to see some sort of full report or something to get an idea of what you're really proposing. Oh well good luck with it, I'll follow the thread if you plan to update it. Quote Link to comment Share on other sites More sharing options...

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