How much hydro-electricity would a cup of water generate?

Question

I would like to know the minimum amount of water needed to create hydo-electricity. EG one cup of water linked to the equipment needed to convert to electricty. (I am not worried about the cost of the conversion equipment being prohibitive.) I am keen to know the minimum amount. Or is it more of a time factor. Would the rate of pouring one cup of water create more electricity if poured faster or slower.
Many thanks for reading this.

Answer 1

NONE, producing hydroelectricity is not as easy as knowing the ammount of water.

Let us work from the generator and backwards.

First of all, for the generator to prduce electricity, there has to be movement. For synchcronous machines (Typical for generators), a pice in the center of the machine called the rotor, will "rotate", towards an axis. In the rotor a magnetic field is created artificially. This field needs to be ainming always in the same radial direction (wether it be from the center to the outside or backwards). The voltage at the genarator terminals will be proprcional to the intensity of this field. In the genarators installed in hydrocentral, this field is created by an electromagnet (current circuling trough a winding), because there are no permanet magnets big enough for a generator that size. However for samller genarators the magnetic field can be created by putting permanent magnets in the rotor. If you want to drag electricity out of a cup of water you can use small permanent magnets to build a small rotor.

It is important to know, that to genarate a good AC sine wave the rotor needs to be move at a constant speed.

The energy needed to move the rotor to an specific speed (will be determinated later) will be a function of several variables.

First, the rotor needs to be accelerated from zero to the terminal speed, the energy needed to perform this will be proporctional to the rotor inercia momentum (this involved the mass and geometry of the rotor). Also, there will be some energy loses due to friction betwenn the rotor and its supports.

so to accelerate the rotor we have E=(I*w^2)/2

where I is the inercia momentum of the rotor, w it's final angular speed of the rotor

the friction on a rotative system is a function of speed so the enrgy loses will be

E=k*w where k is a constant depending on the materials use and w is the rotor instantaneus speed

the rest of the energy needed to move the rotor at it's terminal speed will be determinated by the total load conected to its terminals (current requirementes).

Surrounding the rotor there's another part of the genarator called the stator. This part does not move. It consint on an iron core winded by conductors (similar to a transformer). The moving magnetic field of the rotor will induce a electromotive force in the windings of the stator (this will be seen as a voltage in the generator terminals).

When conecting a load there will be a current flow trough the windings of the stator. This current flow will create another induced magnetic field that will opose the magnetic field coming from the rotor. The two magnetic field will interact and there will be a counter-force that will try to slow down the rotor.

(The induced electromotive forces and magnetig fields are determined by the faraday-lenz and amper laws)

It's also important to know that not all of the induced voltages and currents are used by the conected load, but there's an amount of power loses inside the iron core, and the conductors of the winding)

So there will be an electric eficciency intrinsic on the machine. That means that not all the energy that enters the rotor can be used as electrical energy.

So by now we've covered the electrical loses, the loses due to friction and the energy required to accelerate the machine from zero to its terminal speed. All these is lost energy because it cannot be used as electrical energy.

Know there are other factors to consider.

Water cannot move the rotor directly. The rotor needs to be dry at all times, otherewise there would be electrical problems.

That means we need to move a different structure and conect it to the rotor by some kind of mechanical transmision.This structure is called the turbine.

The turbine is a mechanical device that moves whenever a fluid passes trough it, using a mechanism of paddles.

Here we need to consider two more loses of energy. First in the transmision mechanism conecting the rotor and the turbine there will be mechanical loses. Second, a fluids dynamics law, establicsh that for a fluid passing trough a turbine not all the kinetikal energy can be used to move the turbine. That means that only a fraction of the water kinetical energy can be used to move the turbine in an hydraulic turbine.

Finally we nned to consider that for maximium efficiency, the water flow needs to be conduced towards the turbine trough some kind of channel or pipe. Because all fluids behave in a wave that makes them "stick" to the solid surface trough wich they are flowing, there will be a fluidic resistance asociated with the channel used to carry the water towards the turbine. That means that not all the pottencial energy can br converted in kinetical energy.

So now we have all the main factors that produce energy loses when generating hydroelectricity.
Electrical loses (inside the iron core of the stator)
Electrical loses (in the conductors of the winding)
Mechanical loses(Due to friction between the genarator movable parts)
Mechanical loses (In the mechanical transmision)
Fluid loses (Not all kinetical energy of the fluid can be converted to move the turbine)
Fluid loses (fluidical resistance in the conducting channel)
And finally the ammount of energy needed to carry the generator from it's initial off state (zero state), to it's final operational speed (steady state)

Modern hydroelectricity centrals, design by high-tech engineering, where everything is design to reduce the mentionated loses to it's minimium have an efficiency of about 40%. That means that in a hydroelectric central, only 40% of the water potencial energy can be used as electrical energy.

So building a "micro generator" that works with a cup of water is rather impossible. First of all, the cost would be incredible prohibitive. Second, calculating that a cup of water can be put vertically to a 2.5meter height (I'm assuming a cup of water having 250 cubic centmeters of water, and the water flow having a 1 square centmeter of area), then the potencial energy of that will be about 6.125 Joules. (water density equal 1gr/cc so 250 cc of water would have a 250gr mass, then potencial energy equeal mass*gravity aceleration*height equal 0.25Kg*9.8m/s*2.5m)
then assuming that you could be able to build a "micro hydroelectrical central" as efficient as high-tech centrals then you could produce 40/100*6.125 that is less than 1.5 joules of energy.

That means you would have lithere more than a watt of power for about a second.

But for all the technical impediments of the task I don't thing that a cup of water could generate enough electricity to light the samllest light bulb. It's just unpractical and maybe impossible.

Answer 2

The amount of generated electricity is directly proportional to the height at which it is dropped. When Ontario Hydro built the first turbine plant above the famous Niagara Falls, the Chief Engineer argued in favour of an aquaduct from Lake Erie across the Niagara Peninsula to Lake Ontario. The resulting drop would have been significantly higher than at the Falls, (about 9 times) He was outvoted by politicians because of the "eyesore" such an aquaduct would have presented. Oh, by the way, the engineer's name was Reginald Fessenden, a futurist before there was a word for it who designed the world's first AM radio and was its first voice, way ahead of Marconi's scratchy letter "s"

Answer 3

We must make further asumptions
Let´s suppose that the mass is 0.250kg and it falls from 100m(very high) .Its energy is mgh = 0.250*9.8*100=245 joules. It realeses it's energy in 1 second the power would be 25watt.
You should think that large water tutbine have a power of more than 100Mw =100*10^6 watt and the fall is more often less than 100m.
Obviosly if poured faster it would give you more
power but THT SAME ENERGY
( Think of the flash of a Photocam.It has a very large power but as E = power *time and the times are less then 1/100 s the energie needed is small and you can use small batteries.

Answer 4

The maximum energy that you could get depends on the height that the water falls through with this equation. So if you can get the water to fall from an aircraft at 5 miles you can get a lot of electricity.

Energy=mass x gravity x height.

The rate at which you pour the water only affects the power delivery. The total energy remains the same.

ie. You can get a lot of Watts for a short time, or not much Watts for a long time.

Answer 5

Hydro electric is generated by water turning turbines. One cup of water would have no effect on a conventional turbine other than to wet it. You'd need a very small turbine which in turn would produce a very small amount of electricity, not even enough to illuminate a flashlight.

Answer 6

It is directly proportional to the velocity.

Answer 7

the energy produced will be proportional to its potential energy.