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Energy Transformations: Chemical to Electrical, Rotational and Wind Energy Efficiency Analysis

This document analyzes the transformation of energy through various forms, specifically focusing on the efficiencies of converting chemical energy to electrical, electrical to rotational, and subsequently to wind energy. It calculates specific energy measurements, including joules produced and efficiencies at each transformation step, revealing the challenges in energy conversion efficiency. Using mathematical formulas, we discuss moment of inertia, angular velocity, and energy conservation principles. The overall efficiency from chemical to kinetic energy illustrates the effectiveness of these transformations.

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Energy Transformations: Chemical to Electrical, Rotational and Wind Energy Efficiency Analysis

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  1. Energy Transformations Arbaaz Khan Quinn Calvert Rishi Savaliya RishantDwivedi

  2. Chemical → Electrical • Chemical= m x SE • m= Mass of Chemical (Kg) • SE= Specific Energy (MJ/Kg) • .045 x 500,000=22500J • 22500J = 22.5KJ • Electrical = IV/t • I= Current (Amps) • V= Voltage (Volts) • t= time (seconds) • (.0038 x 8.52)/.539=.06J Efficiency = .06/22500 = .00267%

  3. Electrical → Rotational • Electrical = .06J • Rotational = ½Iω2 = ½ .0000245(25) = .007656 J • I = Moment of Inertia (kg*m) • Moment of Inertia = mr2 = 0000245 (Kg*m) • m = mass (Kg) • r = radius (meters) • ω = Angular Velocity (revs/s) • Angular Velocity = θ/t = 25 (revs/s) • θ = angular displacement (rads/revs) • t = time (seconds) Efficiency = .00001964/.069 = .0285 %

  4. Transformation Video:Electrical → Rotational

  5. Rotational → Wind • Rotational Energy = .007656 Joules • Wind Energy == ½(1.172)(3.1293)3()(.035)2 = .69 J • = Air Density (kg*m3) = 1.172 kg*m3 • = Velocity (m/s) = 3.1293 m/s • = Radius (m) = .035m Efficiency = .069/.007656 = 11%

  6. Wind → Kinetic1 • m = mass • v = velocity Efficiency = .00001964/.069 = .0285 %

  7. Potential → Kinetic2 • m = mass • g = acceleration of gravity • h = height • m=mass • v=velocity=x/t • X=Distance • t=time Efficiency = .018/(.557+.00001964) = 3.23%

  8. Overall Efficiency • Efficiency = Wout/Win = .018/(22500+.557) = .00008% • Ein= 22500J • Eout= .018J

  9. References The Engineering Toolbox. (n.d.). Retrieved February 13, 2011, from Engineering Toolbox website: http://www.engineeringtoolbox.com/ Battery energy -- What battery provides more? (n.d.). Retrieved February 17, 2011, from AllAboutBatteries.com website: http://www.allaboutbatteries.com/Battery-Energy.html Lithium. (n.d.). Retrieved February 17, 2011, from Duracell website: http://www1.duracell.com/procell/chemistries/lithium.asp

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