Can somebody tell me what part of the gravitational energy of water at Hoover Dam can be transformed in electrical power in the ideal case?
Here is a general link: http://www.wvic.com/hydro-works.htm Current electrical generators operating at design can run 95 - 98% efficient. That is using sophisticated water cooled windings etc. The general plant efficiency will vary as to how much is done to recover losses, etc. Hydro-turbines are generall high in efficiency as well. Also in the mid 90% range when properly operated. Collectively a general figure would be in the 80- 90% range I would think, depending on the internal equipment power draw vs the operating output.
MacM responded: "Here is a general link: http://www.wvic.com/hydro-works.htm Current electrical generators operating at design can run 95 - 98% efficient. That is using sophisticated water cooled windings etc. The general plant efficiency will vary as to how much is done to recover losses, etc. Hydro-turbines are generall high in efficiency as well. Also in the mid 90% range when properly operated. Collectively a general figure would be in the 80- 90% range I would think, depending on the internal equipment power draw vs the operating output." 1. I do not know how else one should formulate question, to be understood in right way... Question was not about efficiency of turbine or generator. They are assumed ideal, i. e. with 100% efficiency. 2. Question is about part of gravitation energy, accumulated in water in reservoir before dam, which can be obtained as a useful energy of electrical current produced by 100% efficient generator rotating due to 100% efficient hydro-turbine.
Hey, people, what is going on? So simple problem can not be solved in our Forum? There are a lot specialists and nobody has an opinion about how big part of potential (gravitational) energy of water in the reservoiar before dam can be usefully converted in energy of electric current due to ideal turbo-generator? Common, guys, make your guesses...
Only the water that is above the level of the generators can be converted to kinetic energy to power the generators. A zero gravitational potential energy 'line' would drawn at this level and the gravitational potential energy that could be converted into kinetic energy would depend on the mass of the water above this level, 100% conversion if 'ideal'.
So far 65 people visited this thread and only two have expressed their (similar) answer: 100% of potential of energy of water in upper reservoir (before dam) can be converted into energy of the electric current by the ideal (100% efficiency) turbo-generator. Let me put on the table a little help... If level of water in upper reservoir is H and level on the exit of turbine's blades is 0 then the specific energy of water (the energy of the unit mass of water) in this reservoir is E = dgH, where d is specific density of water (1 kg/m^3). Let us assume that answer we get is a right one. Then whole this energy can be transformed into energy of the generated electricity. Yes? But does not it mean that water should stop? What energy will force it to maintain the flow needed for a continuos work of turbine? So, we have to leave some part of E = dgH in the form of the kinetic energy of stream of water. How much this part should be? When you will get the right answer, you will be amazed. I guarantee it...
I'm not sure I understand your question. Gravitational potential energy is converted into kinetic energy when the water 'falls' from one level to another. If the turbo-generators are 100% efficent in converting this kinetic energy into electric current, then they are 100% efficient. That is not possible, but that was YOUR assertion, Yuriy. Are you now asking what per centage of the kinetic energy from the stream of water is actually used to power the generators? Then the turbo-generators would not be 'ideal' (100% efficient).
Dear 2inquisitive, As any other mechanism, turbine's efficiency is ration of useful output energy to the totally spend one. When stream of water turn the turbine it cames to turbine's blades with kinetic energy K1 and leaves them with kinetic energy K2. So the spent energy becomes equal to K1-K2. The turbo-generator is an ideal one if this whole K1-K2 becomes transformed into energy of the generated current. If part of this K1-K2 goes in heat, mechanical loses and so on turbo-generator is not an ideal and its efficiency is less than 100%. But we are talking about efficiency of conversion of the energy of water in the reservoir into electricity due to ideal turbo-generator. Some people think that this efficiency is 1oo%. I am trying to convince them to see deeper in the origins of this phenomenon. That is it...
My initial take on this is based on the fact that I am pretty sure that water must flow through the turbine at at least the same rate as when it enters the turbine. IOW, the velocity or decrease through the turbine can either stay the same or increase. If the water slows down temporarily the pressure acting on the turbine will slowly increase to the point where it at least matches the velocity that it was coming in at. By this logic, an ideally efficient turbine would keep the velocity increase of the water to zero as it travels through the turbine. As such, an ideally efficient turbine would convert all the potential kinetic energy of the water as it enters that turbine into electricity. Cheers,Please Register or Log in to view the hidden image!
I get it. A quizz! Would I be anywhere near the point in drawing a parallel with the 2nd Law of Thermodynamics? i.e. as long as there is an energy resevoir there can be no such thing as 100% energy transfer. Come Yuriy, tell us the answer
OK I thought about this a bit more and I was wrong. My point is that there are two places you can potentially extract energy from a flow of water, using a turbine, the velocity of that water as it reaches the turbine and the force of gravity acting upon that water, while it is in that turbine. I neglected to consider that while water has to leave a turbine at, at least the same rate as it enters(in volume/time), the speed that the water is travelling when it leaves the turbine may be less. For instance, if the outlet pipe is much wider than the input pipe, the same rate of water flow can be maintained with a much lesser velocity on the part of the water. Hypothetically, an infinitely large outlet pipe could allow a turbine to extract all the kinetic energy from the water, which combined with the gravitational energy would allow for an ideal turbine. Cheers,Please Register or Log in to view the hidden image!
The right answer is the following. If whole process takes volume of water Q = vS, where v is velocity of flow and S is cross-section of channel with a turbine, then we spent the power of water P = dgHvS, but the outcoming from channel stream has a kinetic power W = vS dv^2/2 (d everywhere is water's density). So, the used power for transformation into electricity will be the following: N = dgHvS - vS d v^2/2 = dSgH (v - v^3/2gH)............(*) As one can see, N as a function of v is equal 0 at v=0 and at v = Vtorr = (2gH)^1/2. Physical sense is very simple: there is no electric power if v = 0 (no water's flow) and at v = (2gH)^1/2, when water is in regime of a free falling (Torricelli outflow) and does not pass any energy to turbine. N has a maximum at v = Vmax = (2gH/3)^1/2...........................................(**) and this maximum is: Nmax = 2*(1/3)^3/2 * dgHSVtorr = 2*(1/3)^3/2 *Ptorr ....(***) It means that the maximal power of water's flow that can be transformed into electric power is 2/3^3/2 = 0.385 or 38.5% This number shocked me when I for the first time recognized the solution of this problem. More shock I got when I found out that ... hydroengineers do not know about it... Then I wrote two books about the new approach to such type problems. I called this approach "Qualitics" and now any of you can read this .... book on my site www.minescience.com You will find there a lot of absolutely amazing things...
In my textbook, they had the same approach on the problem of finding the highest power output attainable from a generater driven by wind. Only they did not use the same approach, when reasoning about what should be the highest attainable power from a water-turbine. Weird that.
AndersHermansson' Thanks a lot. unfortunately, Google does not translate from Swedish yet, but I will try to do that. Guys, If somebody knows an easy way to translate from Swedish to English, or Russian, or Franch on Internet please help me and teach me how ....
Yuriy, the textbook is rather superficial and for a superficial course, it might be a waste of time and money buying it and translating. The information must be out on the net in english somewhere. I'm sure the author isn't a lone carrier of that information =)
