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The Power of Tides

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The Power of Tides

Energy for the Lower East Side

Alannah Bennie, James Davis, AndriyGoltsev, Bruno Pinto, Tracy Tran

- Introduction
- Motivation
- Methodology
- Results
- Impact

- We want to harness the power of tidal currentsinto energy that can be used as electricity
- How many turbines can we put into the East River around downtown Manhattan to power the Lower East Side?

- Tidal energy is a clean alternative that we can use efficiently without harm to the environment
- Highly reliable: Tides will always exist due to the gravitational forces exerted by the Moon, Sun,Â and the rotation of the Earth
- Predictable: The size and time of tides can be predicted very efficiently

- Tides â€“ the alternate rising and falling of the sea, usually twice in each lunar day at a particular place, due to the attraction of the moon and sun
- Currents generated by tides

- Flood tide â€“ tide propagates onshore
- High tide â€“ water level reaches highest point
- Ebb tide â€“ tide moves out to sea
- Low tide â€“ water level reaches lowest point
- Slack tide â€“ period of reversing wave (low current velocity)

- Not a river
- A tidal strait connecting the Atlantic Ocean to the Long Island Sound
- Semidiurnal tides
- Flow of the river
- What makes up velocity
- Direction

- Our region is bounded by:
(40.715 N,73.977 W)

(40.707 N,73.997 W)

(40.704 N,73.996 W)

(40.708 N,73.976 W)

- Video of Tidal Turbine
- Size of turbine
- Each turbine has a rotor diameter of 4 meters

- Type of turbine
- Modeled after turbines used by Verdant Power (2007)

- Efficiency
- We are looking at a turbine efficiency of around 40%

- Data from the National Oceanic and Atmospheric Administration (NOAA)
- Tidal velocity
- Daily
- 2007-2011

- Use polynomial interpolation to gather a velocity field
- Interpolation is a method of constructing new data points within the range of aÂ discrete setÂ of known data points.
- Polynomial interpolation is theÂ interpolation of a givenÂ data set by a polynomial

- Since we are working with four data points, we need to find a third degree polynomial of the form:
- Thus, given any set of coordinates in our region, (x,y), we can use this polynomial to determine the velocity at that point

P(x,y) = a0+ a1 x+ a2y+ a3x2+ a4xy+ a5y2+ a6x3+ a7x2 y+ a8 x y2+ a9y3

- Because we know the velocities at our four collection points, we will use polynomial interpolation to find a set of polynomials which go exactly through these points

- Begin by defining the matrix that will be used to create our interpolating polynomials
- The matrix is a 4 x 10 since there are 4 data points with coordinates and 10 terms in the polynomial that we are seeking

- Now create a system of equations, so that we can solve for the coefficients of our interpolating polynomials
- Here is the average tidal velocity at

- Finally, we have found our coefficients and therefore our interpolating polynomials
- Since we looked at the average tidal velocities (mps) per month over the course of 5 years, we have 12 separate polynomials (one for each month)
- We can use these polynomials to find the velocity at any location in any month

P1[x,y] = 0.000299263 + 0.00984117 x + 0.362105 x2 + 15.4992 x3 + 0.0120129 y + 0.200942 x y + 0.679683 x2 y + 0.366569 y2 - 14.844 x y2 + 5.37202 y3

P2[x,y] = 0.000289367 + 0.00957213 x + 0.35658 x2 + 15.4945 x3 + 0.0115257 y + 0.191041 x y + 0.67966 x2 y + 0.348566 y2 - 14.8392 x y2 + 5.37048 y3

P3[x,y] = 0.000288423 + 0.00954475 x + 0.355857 x2 + 15.4786 x3 + 0.0114819 y + 0.190195 x y + 0.678976 x2 y + 0.347027 y2 - 14.8239 x y2 + 5.36498 y3

P4[x,y] = 0.000281853 + 0.00937133 x + 0.352788 x2 +15.5226 x3 + 0.01115 y + 0.183318 x y + 0.681046 x2 y + 0.334526 y2 - 14.8659 x y2 + 5.38031 y3

P5[x,y] = 0.00029242 + 0.00965699 x + 0.358498 x2 + 15.5127 x3 + 0.011673 y + 0.193988 x y + 0.680411 x2 y + 0.353925 y2 - 14.8568 x y2 + 5.37678 y3

P6[x,y] = 0.000297207 + 0.00978715 x + 0.361172 x2 + 15.5152 x3 + 0.0119087 y + 0.198776 x y + 0.680429 x2 y + 0.362631 y2 - 14.8593 x y2 + 5.37758 y3

P7[x,y] = 0.00029066 + 0.00960398 x + 0.356925 x2 + 15.4654 x3 + 0.0115946 y + 0.192525 x y + 0.678349 x2 y + 0.351262 y2 - 14.8114 x y2 + 5.36038 y3

P8[x,y] = 0.00029513 + 0.00971291 x + 0.35798 x2 + 15.3541 x3 + 0.0118348 y + 0.197724 x y + 0.673347 x2 y + 0.360707 y2 - 14.7051 x y2 + 5.32176 y3

P9[x,y] = 0.000282421 + 0.00938159 x + 0.352511 x2 + 15.4761 x3 + 0.0111863 y + 0.184186 x y + 0.67898 x2 y + 0.336101 y2 - 14.8214 x y2 + 5.36418 y3

P10[x,y] = 0.000279165 + 0.00929402 x + 0.350803 x2 + 15.4833 x3 + 0.0110244 y + 0.180872 x y + 0.679356 x2 y + 0.330076 y2 - 14.8281 x y2 + 5.36669 y3

P11[x,y] = 0.000291449 + 0.00963447 x + 0.358401 x2 + 15.5473 x3 + 0.0116189 y + 0.19279 x y + 0.681959 x2 y + 0.351751 y2 - 14.8898 x y2 + 5.38879 y3

P12[x,y] = 0.000292155 + 0.00965048 x + 0.358429 x2 + 15.5188 x3 + 0.0116588 y + 0.193682 x y + 0.680686 x2 y + 0.35337 y2 - 14.8626 x y2 + 5.37891 y3

- Our polynomials appear similar which is due to the fact the tidal velocities have minimal seasonal change
- This was verified when we plotted our contour maps of the velocities and saw that they all looked the same

- Pros
- No error at the data points
- Easy to program
- Able to determine an interpolating polynomial just given a set of points

- Cons
- It is only an approximation
- Accuracy dependent on the number of points you interpolate
- Not the best technique for multivariate interpolation

- Each Turbine needs to be approx. 9.8 â€“ 24.4 meters (32-80 ft) apart (Verdant Power, 2007)
- 1 degree of latitude = 111047.863 meters (364330.26 ft)
- 1 degree of longitude = 84515.306 meters (277281.19 ft)

- We decided to place the turbines 12.2 meters (40 ft) apart

- Using Mathematica, given a min/max latitude and longitude we were able find all points that lie 40 feet apart from one another in a set area
- We then had to use basic mathematics to confine the points to our particular area

- Using the fact that the
line thru pt1 and pt2 y = -5.80563 x + 310.327

line thru pt2 and pt3y = 0.327377 x + 60.6708

line thru pt3 and pt4 y = -5.70626 x + 306.265

line thru pt4 and pt1 y = 0.292453 x + 62.0709

We used these lines to constrain the points to our study area

- In an optimal environment, the available power in water can be calculated from the following equation:
= turbine efficiency

= water density ( kg/m3 )

A = turbine swept area ( m2 )

V= water velocity ( m/s )

P = power (watts)

- Total number of homes in the Lower East Side: 1546
- On average, a household in America uses 10,000 kWh per year
- Total energy needed: 15,460,000 kWh per year

- Total number of turbines: 3794
- Total energy from turbines: 21893.9 kWh
- Total power output in a year: 1.91791 Ã— 108 kWh
- Total # of homes we could power in a year: 19179.1

- Limiting parameters
- Velocity
- Turbine efficiency

- The turbines cost $2,000-$2,500 per kilowatt installed
- Total Cost for 3794 turbines: 44 - 54 million dollars
- Who pays:
- In 2010, conEd gained a revenue of 25.8 Â¢ per kWh to residents and 20.4 Â¢ per kWh for commercial and industrial.
- The average yearly revenue for residencies alone would be approximately 49 million dollars. conEd would start profiting from the turbines in about a year after they are installed.

- Hardisty, Jack. "The Analysis of Tidal Stream Power." West Sussex, UK: John Wiley & Sons, Ltd, 2009. 109-111.
- NOAA. Tidal Current Predictions. 25 1 2011. <http://tidesandcurrents.noaa.gov/curr_pred.html>.
- Power, Verdant. The RITE Project.2007. 2011 <http://www.theriteproject.com/>.
- Yun Seng Lim, Siong Lee Koh. "Analytical assessments on the potential of harnessing tidal currents for electricity generation in Malaysia." Renewable Energy (2010): 1024-1032.