Tidal Turbines

The Power of TidesEnergy for the Lower East SideAlannah Bennie, James Davis, Andriy Goltsev, Bruno Pinto, Tracy TranOutlineIntroductionMotivationMethodologyResultsImpactProblemWe want to harness the power of tidal currents into energy that can be used as electricityHow many turbines can we put into the East River around downtown Manhattan to power the Lower East Side?Why Tidal EnergyTidal energy is a clean alternative that we can use efficiently without harm to the environmentHighly reliable: Tides will always exist due to the gravitational forces exerted by the Moon, Sun,and the rotation of the Earth Predictable: The size and time of tides can be predicted very efficientlyWhat are tides?Tides 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 sunCurrents generated by tides

TidesFlood tide tide propagates onshoreHigh tide water level reaches highest pointEbb tide tide moves out to seaLow tide water level reaches lowest pointSlack tide period of reversing wave (low current velocity)The East RiverNot a riverA tidal strait connecting the Atlantic Ocean to the Long Island SoundSemidiurnal tidesFlow of the riverWhat makes up velocityDirection

BackgroundOur 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)

Region and data collection9BackgroundVideo of Tidal TurbineSize of turbineEach turbine has a rotor diameter of 4 metersType of turbineModeled after turbines used by Verdant Power (2007)EfficiencyWe are looking at a turbine efficiency of around 40%

MethodologyData from the National Oceanic and Atmospheric Administration (NOAA)Tidal velocityDaily2007-2011

Monthly Mean Tidal Velocities

MethodologyUse polynomial interpolation to gather a velocity fieldInterpolation is a method of constructing new data points within the range of adiscrete setof known data points.Polynomial interpolation is theinterpolation of a givendata set by a polynomialPolynomial InterpolationSince 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+ a9y314Polynomial InterpolationBecause 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

Polynomial Interpolation 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

Polynomial InterpolationNow create a system of equations, so that we can solve for the coefficients of our interpolating polynomials

Here is the average tidal velocity at

Polynomial InterpolationFinally, we have found our coefficients and therefore our interpolating polynomialsSince 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 monthOur PolynomialsP1[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 y3P2[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 y3P3[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 y3P4[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 y3P5[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 y3P6[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 y3P7[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 y3P8[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 y3P9[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 y3P10[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 y3P11[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 y3P12[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 y3Polynomial InterpolationOur polynomials appear similar which is due to the fact the tidal velocities have minimal seasonal changeThis was verified when we plotted our contour maps of the velocities and saw that they all looked the same

Tidal Velocity Contour (mps)

Polynomial InterpolationProsNo error at the data pointsEasy to programAble to determine an interpolating polynomial just given a set of points

ConsIt is only an approximationAccuracy dependent on the number of points you interpolateNot the best technique for multivariate interpolationPlacement of the TurbinesEach Turbine needs to be approx. 9.8 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) apartPlacement of the TurbinesUsing 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 areaWe then had to use basic mathematics to confine the points to our particular areaPlacement of the TurbinesUsing the fact that the line thru pt1 and pt2 y = -5.80563 x + 310.327 line thru pt2 and pt3y = 0.327377 x + 60.6708line 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

MethodologyIn 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)

FactsTotal number of homes in the Lower East Side: 1546On average, a household in America uses 10,000 kWh per yearTotal energy needed: 15,460,000 kWh per yearResultsTotal number of turbines: 3794Total energy from turbines: 21893.9 kWhTotal power output in a year: 1.91791 108 kWhTotal # of homes we could power in a year: 19179.1DiscussionLimiting parametersVelocityTurbine efficiency

ComparisonEfficiency40%50%60%70%80%90%Power (kWh)21893.927367.432840.938314.343787.849261.3

ComparisonEfficiency40%50%60%70%80%90%# of Homes19179.123973.828768.633563.338358.143152.9

CostsThe turbines cost $2,000-$2,500 per kilowatt installed Total Cost for 3794 turbines: 44 - 54 million dollarsWho pays:In 2010, conEd gained a revenue of 25.8 per kWh to residents and 20.4 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.BibliographyHardisty, Jack. "The Analysis of Tidal Stream Power." West Sussex, UK: John Wiley & Sons, Ltd, 2009. 109-111.NOAA. Tidal Current Predictions. 25 1 2011. .Power, Verdant. The RITE Project.2007. 2011 .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.