Harnessing Gravity for Power. Introduction. Module 3, Lesson 6

Transcription

1 Module 3, Lesson 6 Harnessing Gravity for Power Objective: Why is this beautiful valley in north-eastern BC about to be flooded? The energy flow in the water cycle can be intercepted to produce useful power. Introduction Hydro-electricity can be generated in numerous ways: dams, rivers, tides and waves. Here we describe the simplest and oldest method, the dam. However, the analysis can easily be applied to run-of-the-river power generation. For dams, the energy transformations are as follows: 1. Solar radiation evaporates water from the ocean 2. Rain falls on mountains, and runs into lakes. 3. The runoff is interrupted by a reservoir and dam. 4. Water is extracted at the base of the dam to turn generators. Peace River Dam, Hudson's Hope BC

2 Consider a reservoir (which may be a natural lake) of surface area A, in which water is extracted at a height h below the water level, and used to turn electricity generating turbines. The total rate (in m3/s) at which water is extracted is Q. In principle one can calculate a maximum value of Q from h and the size of the hole using Bernoulli s principle. However, in real hydro-electric dams Q is largely determined by the amount of rainfall landing on the catchment area A of the lake. One cannot extract more water out of a reservoir than is falling into it as rain. Example: the W.A.C. Bennett Dam Let s look at some numbers for the big W.A.C. Bennett Dam in north-eastern British Columbia1. The catchment area is about km2, and the mean rainfall in this area is about 600 mm per year2. Therefore the maximum possible Q (ignoring all evaporation) is given by: Q = ( km2)(106 m2/km2)(0.6 m/y)/( s/y) = 1300 m3/s The published height of the dam is 186 m. We will take this to be the height difference between the water level and the turbines, h. Consider a body of water which starts at the surface of the

3 reservoir and eventually moves through the turbines. Its potential energy per unit volume at the surface is ρgh (in J/m3) compared to the level of the turbines, where ρ is the density of the water (1000 kg/m3). The rate at which the water moves through the turbines is Q, and so the rate at which the available potential energy passes the turbines is ρghq. Now consider that useful electrical energy is generated with an efficiency η and so we can write the power generated P as follows. % % P = ηρghq Assuming for now that the efficiency of the generators η is 1; we calculate the maximum available power to be 2.4 GW. The maximum power rating of the dam and its 10 turbines is given as 2.73 GW. Plainly this maximum power cannot be sustained as it is more than the number we have come up using all the rainfall and also assuming 100% efficiency and no evaporation. However, the annual average power generated is given to be GWh per year. If we divide this number by (24)(365) h/y, we obtain a mean delivered power of 1.5 GW. This is of the same order as, but comfortably less than, our ideal maximum value of 2.4 GW. Greenhouse Gas Emissions While GHG emissions from hydro-electric projects are small to other means of electricity generation, they are not negligible. They arise from the initial construction and the subsequent decay of biomass in the flooded valley (if that is the way the project is constructed). Take for example the proposed Site-C dam which would be just downstream of the Bennett Dam discussed above. This project will have a rated power of about 1 GW, and will cause hectares to be flooded. A study predicts that the equivalent of about tonnes of CO 2 will be emitted each year until the second decade after completion. About half will come from construction (and the associated levelling of forest) and half will arise from biomass decay in the artificial lake. This number should be compared with about tonnes per GWy of electricity produced by coal-fired plants.

4 "Site C", Hudson's Hope BC For more information see these links BC Hydro. The Peace/Williston Fish and Wildlife Compensation Program (PWFWCP) (online). [12 May 2010]. 2. Natural Resources Canada. The Atlas of Canada (online). english/maps/environment/climate/precipitation/precip [12 May 2010]. 3. The Vancouver Sun. Site C Dam No Green Power Project, Critics Say (online). babb-1a20e58664ab&k=49198 [9 April 2010]. 4. MacKay DJC. Sustainable Energy - Without the Hot Air (online). UIT Cambridge. [12 May 2010]. Physics and Astronomy Outreach Program at the University of British Columbia (Chris Waltham )

5 Energy from Water? 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.

6 The power from the water flow is defined to be: where 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 Where 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) Measurements: Height = 0.95 m

7 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 = m3/s Voltage across LEDs = 3 V Current through LEDs = 30 ma. Calculations: The power from the water flow: The electrical power generated: The efficiency: 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.

8 Question: As we have seen, energy can be extracted from moving water. When it rains, water that lands on roofs is collected in gutters and carried to the ground via downspouts. In a typical heavy rain in your city, how much power could be harnessed from the rain flowing out of your downspouts? How much power could you extract averaged over the course of the entire year? Answers: A) A typical heavy rain storm in Vancouver produces 30 mm of rain in 6 hours, or 5 mm an hour. The roof area of my two car garage is 49m2(7 m by 7 m). % volume flow rate of rain, Q = 50m*0.005m/hour= 0.25m3 /hour=250 L/hour Each hour, 250 L of rain falls on the roof, flows into the gutter and is carried down the downspout towards the ground. What we know: ρ = 1000 kg/m3 g = 10 m/s2 h = 3 m Q = 250 L/h = 7 x 10-5 m3/s If the water simply runs off the roof and into the downspout, the power available during the rainstorm is: % P = ηρghq =(1000 kg/m )(10 m/s )(3m)(7 x 10 m /s) = 2 W This is enough to light a single small LED. B) According to the Weather Network, the annual precipitation at Vancouver International Airport is 1100 mm. volume flow rate of rain (50 m )(1.1m/year)(1 year/ 3*107 s) = 1.8 x 10 m /s % P = ηρghq = (1000 kg/m )(10 m/s )(3m)(2 x 10 m /s) = 0.06 W When averaged over an entire year, only 0.06 W is available which is of no use to anyone. Physics and Astronomy Outreach Program at the University of British Columbia (Brittany Tymos )

9 Wave Power Will wave power save us? Waves of any sort move energy, but not matter, from one place to another. Energy required to disturb the surface of water: Before we ask how much power we can extract from waves, we have to estimate how much energy there is in a wave. Consider a moving disturbance in the surface of a body of water that looks something like this: The energy required to create this disturbance is just the work done in moving the water that was in the trough of the wave up to the crest of the wave. Thus a body of water height h, width w and length L has been moved vertically up a distance h (centre-of-mass to centre-of-mass). If the density of water is ρ then the mass m involved in the move is ρwhl. As the work done W to raise this mass a height h is mgh, Power in a wave-train In the open ocean such disturbances usually occur repetitively, with a frequency f and a spacing λ, i.e. the speed of the waves, v = f λ. If the disturbances are packed together, λ = 2w.

10 We won t worry about the unnatural shape of the waves for now. Rectangles are easier to deal with than real wave shapes. If we have some kind of energy absorber at the end of this wave train capable of using wave power to generate, say, electricity, the rate at which waver energy crosses this absorber is Consider a typical ocean wave train: h = 1 m, λ = 10 m, f = 0.1 Hz (period = 10 s), i.e. v = 1 m/s. Assume the energy absorber is L = 10 m long, typical for such an installation: Although a crude approximation, this result is very close to that obtained with a much more sophisticated water-wave model. It shows that a significant amounts of power are potentially available in water waves. However, one must remember that 50 kw from this 10-metre size device would only service the energy needs of a few single-family dwelling. And we haven t said anything yet about how this power can be extracted [1]. For real current wave data, suitable for use in student projects, see the U.S. NOAA National Data Buoy Centre. Zoom in on a buoy and click to see current conditions2. 1. Wikipedia. Wave Power (online). [27 May 2010]. 2. National Oceanic and Atmospheric Administration. National Data Buoy Centre (online). [27 May 2010]. Problem Sets: Tidal Power Problem Set & Solutions Physics and Astronomy Outreach Program at the University of British Columbia (Chris Waltham )