Hydro-Electric Dam Demo

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Energy from Water?

Big Ideas: 
  • The energy flow from water can be intercepted to produce useful power.

The Hydro-electric Dam article looks at the generation of power via a dam. This same phenomenon can be studied using a simple demo, consisting of an 18L water bottle and an impulse turbine which turns a small generator that powers four Light-Emitting Diodes (LEDs) in a model house. We can make measurements and apply some simple physics to measure the power output and compare it to an estimate of the power in the flowing water.

Figure 1: Set up. A water reservoir (jug) is set a few feet above a turbine, connected by a tube with a valve. h is the height from the top of the water level to the contact point with the turbine. When the valve is open, the water runs down the tube and into the turbine, producing power, which is then used to light up the LEDs.

We want to estimate the initial power output when the water jug is filled to the top and the valve is open, allowing the water to flow down into the turbine.

The power from the water flow is defined to be:

$ P_{flow} = \dfrac{mgh}{t} = \rho h Q g $

where $ \rho $ is the density of water (1000 kg/m3), h is the height of the water above the turbine in metres, Q is the flow rate of the water in m3/s, and g is the acceleration due to gravity (10 m/s2).

To measure the instantaneous flow rate when the reservoir is completely filled, fill the reservoir up to the top and record how long it takes to fill a small container (small compared to 18 L). Then, measure the volume of the water in the container. By dividing the volume of water by the time it took for that water to exit the reservoir and tube, you get the flow rate.

The electrical power generated is defined to be

$ P_e = \eta P_{flow} = VI $

Where $ \eta $ is the efficiency. We can use a voltmeter connected in parallel and an ammeter connected in series to determine the measured power output from the dam.

Hyrdo-Electric_Demo.mp4 (right-click and choose "Save Link As..." to download to your computer)


Height = 0.95 m

When the valve to the reservoir was first opened:

Flow Rate trial 1 = 3.3 L / 6.8 s = 0.48 L/s

Flow Rate trial 2 = 2.3 L / 4.9 s = 0.47 L/s

Average Flow Rate ≈ 0.5 L/s = 0.0005 m3/s

Voltage = 3 V

Current = 30 mA.


The power from the water flow:

<br />
 \begin{eqnarray}<br />
\textnormal{Power} &=& (1000 \textnormal{ kg} / \textnormal{m}^3)(0.95 \textnormal{ m})(0.0005  \textnormal{ m}^3 / \textnormal{s})(10 \textnormal{m/s}^2) \nonumber \\<br />
&=& 4.75 \textnormal{ kg}\cdot \textnormal{m}^2 \textnormal{/s}^3 \nonumber \\<br />
& \approx & 5 \textnormal{ W} \nonumber<br />
    \end{eqnarray}<br />

The electrical power generated:

$ P_e = VI = (30 \times 10^{-3} \textnormal{ A})(3 \textnormal{ V}) = 0.09 \textnormal{ W} $

The efficiency:

$  \eta = \dfrac{0.1 \textnormal{ W}}{5 \textnormal{ W}} \approx 2 \%  $

So when the container is full, the power output of the system is approximately 0.1 W and the efficiency is 2 %. This little turbine is not optimized and so we expect a low efficiency. A real turbine would have multiple stages to extract as much of the water's energy as possible, and be optimized to avoid energy loss by turbulence, splashing etc.

Issues to be aware of:

  • Variation in power. In this demonstration, the height of the water is not constant and so as the water level decreases, the power output also decreases. Here we only measure the initial power output, when the jug is completely filled and not the power while the jug is emptied. 
  • Power loss. Power is lost through the pipe connecting the water jug to the turbine.
  • Not all the kinetic energy is extracted from the water, as it obviously still has some when it comes out of the turbine.




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