Gday, just been trying to work out this problem. Have an idea of what it is all about, but I can never work out the correct answer. --- A person with a remote mountain cabin plans to install her own hydroelectric plant. A nearby stream is 3.00m wide and 0.500m deep. Water flows at 1.20m/s over the brink of a waterfall 5.0m high. The manufacturer promises only 25.0% efficiency in converting the potential energy of the water-Earth system into electric energy. Find the power she can generate. --- My take on it is that I need to work out the mass of water going down the waterfall at a given time, then find the weight of that...Then work out the kinetic energy of the water as it lands at the bottom, and then times by .25... Am I on the right track? If not where am I going wrong? Thansk
Sounds ok to me. Once you have the energy, you need to know how much potential energy is being converted per unit time - that's the power.
since the speed of the stream is 1.2 m/s, a "slab" of water 3 m wide, .5 m deep, and 1.2 m long falls off the edge every second. using the density of water as 1000 kg/m^3, the mass of the "slab" is 1800 kg. at the top of the waterfall, the "slab" has both kinetic energy + gravitational potential energy ... Please Register or Log in to view the hidden image! since this total energy is transferred every second, the waterfall's power is Please Register or Log in to view the hidden image! 25% efficiency would rate the power of the hydroelectric plant as approximately 22 kW.
In general for efficient control, water is first stored in the reservoir for Hydel plants, and then water is dropped from a height through appropriate size pipe lines, the control of flow is done by the proportional valve just before the water is allowed to hit the turbine buckets.. Thus the kinetic Energy of the stream at the brink will not be of any help, it is just the potential Energy (Head Height - Reservoir output to turbine inlet point) that will matter for conversion.
About 88kw of theoretical power is available. But a lousiest hydraulic turbine would have an efficiency of 90.% or above. Friction losses and generator losses would account for another 15%, not more. System should give a net efficiency of at least 60% or more, taking care of all losses. I think the manufacturer is conning her to buy a bigger system OR he is a LOUSY designer/manufacturer.
